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1)
If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319?
Sol) log 0.317=0.3332 and log 0.318=0.3364, then
log 0.319=log0.318+(log0.318-log0.317) = 0.3396
2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?
Sol) x= 2 kg Packs
y= 1 kg packs
x + y = 150 .......... Eqn 1
2x + y = 264 .......... Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36
ANS : Number of 2 kg Packs = 114.
3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00
Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM
4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ?
Sol) Since it is moving from east to west longitide we need to add both
ie,40+40=80
multiply the ans by 4
=>80*4=320min
convert this min to hours ie, 5hrs 33min
It takes 8hrs totally . So 8-5hr 30 min=2hr 30min
So the ans is 10am+2hr 30 min
=>ans is 12:30 it will reach
5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation. Very urgent.
Sol) see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple....
6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes.
(n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure
7) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week
Ans:4 good, 1 fair n 2 bad days
Sol) Go to river catch fish
4*9=36
7*1=7
2*5=10
36+7+10=53...
take what is given 53
good days means --- 9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days.
fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10) if u assume 2 then there is no chance for bad days.
bad days means -------5 fishes so remaining 10---10/5=2days.
Ans: 4 good, 1 fair, 2bad. ==== total 7 days.
x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------eq3
from eq2 and eq3
2*y+4*z=10-----eq4
since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4
for first y=1,z=2 then from eq1 x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied
so finally there are two solution of this question
(x,y,z)=(4,1,2) and (3,3,1)...
8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?
Sol) Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35
9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?
suppose total income is 100
so amount x is getting is 80
y is 70
z =60
total=210
but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r getting 30% discount
10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?
Sol) incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
ans:12/19
11) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?
ans: -1
bcoz g(2)=g(1)-g(0)=1+1=2
g(3)=1
g(4)=-1
g(5)=-2
g(6)=-1
12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages?
Sol) A can copy 50 pages in 10 hrs.
A can copy 5 pages in 1hr.(50/10)
now A & B can copy 70 pages in 10hrs.
thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x--> no. of pages B can copy in 10 hrs.]
so, B can copy 9 pages in 1hr.
therefore, to copy 26 pages B will need almost 3hrs.
since in 3hrs B can copy 27 pages.
13) what's the answer for that :
A, B and C are 8 bit no's. They are as follows:
A -> 1 1 0 0 0 1 0 1
B -> 0 0 1 1 0 0 1 1
C -> 0 0 1 1 1 0 1 0 ( - =minus, u=union)
Find ((A - C) u B) =?
To find A-C, We will find 2's compliment of C and them add it with A,
That will give us (A-C)
2's compliment of C=1's compliment of C+1
=11000101+1=11000110
A-C=11000101+11000110
=10001001
Now (A-C) U B is .OR. logic operation on (A-C) and B
10001001 .OR . 00110011
The answer is = 10111011,
Whose decimal equivalent is 187.
14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........
Sol) it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000
so starting add of 2x3 will be 3012.
15) In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5).
Sol) initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 -1) * 4 = 212 -----( -1 is to deduct the 1 element ) 3000 + 212 = 3212
16) Which of the following is power of 3 a) 2345 b) 9875 c) 6504 d) 9833
17) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ?
Sol) M=sqrt(100N)
N is increased by 1%
therefore new value of N=N + (N/100)
=101N/100
M=sqrt(100 * (101N/100) )
Hence, we get M=sqrt(101 * N)
18)
1)SCOOTER --------- AUTOMOBILE--- A. PART OF
2.OXYGEN----------- WATER ------- B. A Type of
3.SHOP STAFF------- FITTERS------ C. NOT A TYPE OF
4. BUG -------------REPTILE------ D. A SUPERSET OF
1)B 2)A 3)D 4)C
19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand
this is the step by step solution:
a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min.
after this speed of return increase by 50% so 50%of 18 mph=9mph
Total speed of returnig=18+9=27
Then in return it take 27/27=1 hour
then total time in joureny=1+1:30+00:30 =3 hour
so it will come at 8+3 hour=11 a.m.
So Ans==11 a.m
20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ?
Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338.
21) The temperature at Mumbai is given by the function: -t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM?
22) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low temperature and highest temp. is 100 degrees. Then what is the low temp?
Sol) Let highest temp be x
so low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2
total temp=x+x/6+x/2=100
therefore, x=60
Lowest temp is 40
23) In Madras, temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. Ans. At 9pm 7.5 more
Sol) In equestion first put t=9,
we will get 34.5...........................(1)
now put t=4,
we will get 27..............................(2)
so ans=34.5-27
=7.5
24) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product?
a) 780 b) 1040 c) 1590 d) 1720
Sol) x*53-x*35=540=> x=30 therefore, 53*30=1590 Ans
25) How many positive integer solutions does the equation 2x+3y = 100 have?
a) 50 b) 33 c) 16 d) 35
Sol) There is a simple way to answer this kind of Q's given 2x+3y=100, take l.c.m of 'x' coeff and 'y' coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula--- constant / (l.cm of x coeff and y coeff)
26) The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders.
a) 65 b) 55 c) 50 d) 45
Sol)
Let a = fixed cost and k = variable cost and n = number of boarders
total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k
total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k
solving above 2 eqns, 3000-1750 = 25k i.e. 1250 = 25k i.e. k = 50
therefore, substituting this value of k in either of above 2 eqns we get
a = 500 (a = 3000-50*50 = 500 or a = 1750 - 25*50 = 500)
so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500
so cost per head = 5500/100 = 55
27) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens?
a) 37.5% b) 62.5% c) 50% d) None of these
Sol)
Let, 5 pens + 7 pencils + 4 erasers = x rupees
so 10 pens + 14 pencils + 8 erasers = 2*x rupees
also mentioned, 6 pens + 14 pencils + 8 erarsers = 1.5*x rupees
so (10-6) = 4 pens = (2-1.5)x rupees
so 4 pens = 0.5x rupees => 8 pens = x rupees
so 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid byamal)
i.e 5/8 = 500/8% = 62.5% is the answer
28) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race?
Sol)
x + x+6 = rs 68
2x + 6 = 68
2x = 68-6
2x = 62
x=31
x is the amt lost in I race
x+ 6 = 31+6=37 is lost in second race
then my friend lost 37 + 4 = 41 Rs
29) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms. i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10 iii) Same question with n=9
to me the chances are
when n=3
(i) nC1= 3C1 =3 for 10 boxes .. 10*3=30
(ii) 10C1=10 for 10 boxes ....10*10=100
(iii)9C1=9 for 10 boxes .....10*9=90
30) (1-1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
leaving the first numerater and last denominater, all the numerater and denominater will cancelled out one another. Ans. 5/(n+5)
31) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part?
Sol) the clock normally has 12 hr
three parts x,y,z
x+y+z=12
x=y+z
2x=12
x=6
so the largest part is 6 hrs
32) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels
Sol) 4/5 full tank= 12 mile
1 full tank= 12/(4/5)
1/3 full tank= 12/(4/5)*(1/3)= 5 miles
33) wind blows 160 miles in 330min.for 80 miles how much time required
Sol) 160 miles= 330 min
1 mile = 330/160
80 miles=(330*80)/160=165 min.
34) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was travelling at a speed of 35 mph. find the speed limit
Sol)
(x+10)=(x+35)/2
solving the eqn we get x=15
35) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3. what is the answer he actually has to get.
Sol) Assume 1
1* 3 = 3
1*1/3=1/3
so he has to got 1/3
this is the exact answer
36) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot.
37) Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp.
ans is 40 c.
Sol) let x be the highest temp. then,
x+x/2+x/6=100.
therefore, x=60 which is the highest temp
and 100-x=40 which is the lowest temp.
38) car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up.
Sol) let feul consumed in coming up is x. thus equation is: x+1.25x=4.5ans:2gallons
39) A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person
Sol) Two people work together in 24 mins.
So, their one day work is
(1/A)+(1+B)=(1/24)
One man can complete the work in 40mins
one man's one day work (1/B)= (1/40)
Now,
(1/A)=(1/24)-(1/40)
(1/A)=(1/60)
So, A can complete the work in 60 mins.
40) In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability? That a randomly chosen employee is a male or female?
Sol) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is 0.264
41) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin
Sol) the minimum number of wieghtings needed is just 5.as shown below
(1) 80->30-30
(2) 15-15
(3) 7-7
(4) 3-3
(5) 1-1
42) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples?
2x+3y+4z=15
3x+2y+z=10 adding
5x+5y+5z=25
x+y+z=5 that is for 1 orange, 1 bannana and 1 apple requires 5Rs.
so for 3 orange, 3 bannana and 3 apple requires 15Rs.
i.e. 3x+3y+3z=15
43) In 8*8 chess board what is the total number of squares refers
Sol) odele discovered that there are 204 squares on the board We found that you would add the different squares - 1 + 4 + 9 + 16+ 25 + 36 + 49 + 64.
Also in 3*3 tic tac toe board what is the total no of squares
Ans 14 ie 9+4(bigger ones)+1 (biggest one)
If you ger 100*100 board just use the formula
the formula for the sum of the first n perfect squares is
n x (n + 1) x (2n + 1)
______________________
6
if in this formula if you put n=8 you get your answer 204
44) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish.
Sol) Faster one can do 1/2 of work in one hourslower one can do 1/3 of work in one hourboth they do (1/2+1/3=5/6) th work in one hour.so work will b finished in 6/5=1.2 hour i e 1 hour 12 min.
45) If Rs20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive?
Here is the answer Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42) ==> (42 * 20 )/70 ==> 12 Rs
46) An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she cann t type.?
47) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?
the answer is 138 matches
Sol) since one player gets a bye in each round,he will reach the finals of the tournament without playing a match.
therefore 137 matches should be played to detemine the second finalist from the remaining 138 players(excluding the 1st player)
therefore to determine the winner 138 matches shd be played.
48) One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate?
Sol) Vol. of rect. plate= 8*11*2=176
area of rod=(22/7)*(8/2)*(8/2)=(352/7)
vol. of rod=area*length=vol. of plate
so length of rod= vol of plate/area=176/(352/7)=3.5
49) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank?
3 ft 7.5 inches
50) A man has to get air-mail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive?
ans:10min:::assume he started at 1:00,so at 1:30 he met cart. He returned home at 2:00.so it took him 1 hour for the total jorney.by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.so the actual time of plane is at 1:40.The cart travelled a time of 10 min before it met him.so the plane is 10 min early.
51) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked?
52) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years.
Sol) THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4
ans: 1.5 mt 4 (x+(3/5)x)=88x/5=2x=5/4 after 2 years x=(3/5)*(5/4)*2 =1.5
53) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code?
sol) Assume the digit xx xx xx (six digits)
First Two digit xx * 3=111
xx=111/3=37
( first two digits of 1 is not divisible by 3 so we can use 111)
Second Two digit xx*6=222
xx=222/6=37
( first two digits of 2 is not divisible by 6 so we can use 222)
Thrid Two digit xx*9=333
xx=333/9=37
( first two digits of 3 is not divisible by 9 so we can use 333)
54) There are 4 balls and 4 boxes of colours yellow, pink, red and green. Red ball is in a box whose colour is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball?
ans is green...
Sol) Yellow box can have either of pink/yellow balls.
if we put a yellow ball in "yellow" box then it wud imply that "yellow" is also the colour of the box which has the red ball(becoz acordin 2 d question,d box of the red ball n the ball in the yellow box have same colour)
thus this possibility is ruled out...
therefore the ball in yellow box must be pink,hence the colour of box containin red ball is also pink....
=>the box colour left out is "green",,,which is alloted to the only box left,,,the one which has yellow ball..
55) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color.
Ans:he should pick 6 ball totally.
Sol) Suppose he picks 5 balls of all different colours then when he picks up the sixth one, it must match any on of the previously drawn ball colour.
thus he must pick 6 balls
56) What is the number of zeros at the end of the product of the numbers from 1 to 100
Sol) For every 5 in unit palce one zero is added
so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place.
Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 11 zeros are added
for 25,50,75 3 extra zeros are added
so total no of zeros are 10+11+3=24
57) 10 Digit number has its first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4's..till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?.(6marks)
ans:2100010006
2---shows that two 1's in the ans
1---shows that one 2 in ans
0---shows no 3 in the ans
0---shows no 4 in the ans
0---shows no 5 in the ans
1---shows one 6 in the ans
0---shows no 7 in the ans
0---shows no 8 in the ans
0---shows no 9 in the ans
6---shows six 0's in the ans
58) There are two numbers in the ratio 8:9. if the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9. Find the larger number?
sol) 8x:9x initialy
8x+ 12 : 9x - 19 = 5x:9x
8x+12 = 5x
-> x = 4
9x = 36 not sure about the answer ..
59) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time
A-B = 3
B-c = 5
a-c = 8
so sum of diff = 8+3+5 = 16 kgs
60) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark?
ans is 11
coz each had atleast 1 shot done so 10 + 1 = 11
n 9 + 2 = 11
so d ans is 11
61) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time?
sol) A shows 11:57, B shows 12:03, C shows 12:02, and D shows 11:06 therefore, max time is for B
62) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall.
Sol) The falling height is proportional to the squere of the time.
Now, the falling height is 64cm at 2sec
so, the proportional constant is=64/(2*2)=16;
so, at 6sec the object fall maximum (16*6*6)cm=576cm;
Now, the object may be situated at any where.
if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec.
63) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings
Ans) first 50 ings.- run= 50*50=2500
51st ings.- avg 51. so total run =51*51=2601.
so run scored in that ings=2601-2500=101 runs.
64) Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over?
a) 3rd b) 6th c) 9th d) 7th
Ans is d) 7th day
Sol) In d 1st day Anand does 1/7th of total work
similarly,
Bithu does 1/8th work in d 2nd day
hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168
remaining work = (168-73)/168 = 95/168
again after 6 days of work, remaining work is = (95-73)/168 = 22/168
and hence Anand completes the work on 7th day.(hope u understood.)
65) A man, a women and a child can do a piece of work in 6 days,man can do it in 14 days, women can do it 16 days, and in how many days child can do the same work?
The child does it in 24 days
66)
A: 1 1 0 1 1 0 1 1
B: 0 1 1 1 1 0 1 0
C: 0 1 1 0 1 1 0 1
Find ( (A-B) u C )==?
Hint : 109
A-B is {A} - {A n B}
A: 1 1 0 1 1 0 1 1
B: 0 1 1 1 1 0 1 0
by binary sub. a-b = 01100001 (1-0=1, 1-1=0,0-0=0, n for the 1st 3 digits 110-011=011)
now (a-b)uc= 01100001
or 01101101
gives 1101101... convert to decimal equals 109
_________________________________________________________________________Sol) log 0.317=0.3332 and log 0.318=0.3364, then
log 0.319=log0.318+(log0.318-log0.317) = 0.3396
2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?
Sol) x= 2 kg Packs
y= 1 kg packs
x + y = 150 .......... Eqn 1
2x + y = 264 .......... Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36
ANS : Number of 2 kg Packs = 114.
3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00
Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM
4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ?
Sol) Since it is moving from east to west longitide we need to add both
ie,40+40=80
multiply the ans by 4
=>80*4=320min
convert this min to hours ie, 5hrs 33min
It takes 8hrs totally . So 8-5hr 30 min=2hr 30min
So the ans is 10am+2hr 30 min
=>ans is 12:30 it will reach
5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation. Very urgent.
Sol) see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple....
6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes.
(n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure
7) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week
Ans:4 good, 1 fair n 2 bad days
Sol) Go to river catch fish
4*9=36
7*1=7
2*5=10
36+7+10=53...
take what is given 53
good days means --- 9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days.
fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10) if u assume 2 then there is no chance for bad days.
bad days means -------5 fishes so remaining 10---10/5=2days.
Ans: 4 good, 1 fair, 2bad. ==== total 7 days.
x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------eq3
from eq2 and eq3
2*y+4*z=10-----eq4
since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4
for first y=1,z=2 then from eq1 x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied
so finally there are two solution of this question
(x,y,z)=(4,1,2) and (3,3,1)...
8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?
Sol) Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35
9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?
suppose total income is 100
so amount x is getting is 80
y is 70
z =60
total=210
but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r getting 30% discount
10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?
Sol) incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
ans:12/19
11) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?
ans: -1
bcoz g(2)=g(1)-g(0)=1+1=2
g(3)=1
g(4)=-1
g(5)=-2
g(6)=-1
12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages?
Sol) A can copy 50 pages in 10 hrs.
A can copy 5 pages in 1hr.(50/10)
now A & B can copy 70 pages in 10hrs.
thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x--> no. of pages B can copy in 10 hrs.]
so, B can copy 9 pages in 1hr.
therefore, to copy 26 pages B will need almost 3hrs.
since in 3hrs B can copy 27 pages.
13) what's the answer for that :
A, B and C are 8 bit no's. They are as follows:
A -> 1 1 0 0 0 1 0 1
B -> 0 0 1 1 0 0 1 1
C -> 0 0 1 1 1 0 1 0 ( - =minus, u=union)
Find ((A - C) u B) =?
To find A-C, We will find 2's compliment of C and them add it with A,
That will give us (A-C)
2's compliment of C=1's compliment of C+1
=11000101+1=11000110
A-C=11000101+11000110
=10001001
Now (A-C) U B is .OR. logic operation on (A-C) and B
10001001 .OR . 00110011
The answer is = 10111011,
Whose decimal equivalent is 187.
14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........
Sol) it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000
so starting add of 2x3 will be 3012.
15) In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5).
Sol) initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 -1) * 4 = 212 -----( -1 is to deduct the 1 element ) 3000 + 212 = 3212
16) Which of the following is power of 3 a) 2345 b) 9875 c) 6504 d) 9833
17) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ?
Sol) M=sqrt(100N)
N is increased by 1%
therefore new value of N=N + (N/100)
=101N/100
M=sqrt(100 * (101N/100) )
Hence, we get M=sqrt(101 * N)
18)
1)SCOOTER --------- AUTOMOBILE--- A. PART OF
2.OXYGEN----------- WATER ------- B. A Type of
3.SHOP STAFF------- FITTERS------ C. NOT A TYPE OF
4. BUG -------------REPTILE------ D. A SUPERSET OF
1)B 2)A 3)D 4)C
19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand
this is the step by step solution:
a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min.
after this speed of return increase by 50% so 50%of 18 mph=9mph
Total speed of returnig=18+9=27
Then in return it take 27/27=1 hour
then total time in joureny=1+1:30+00:30 =3 hour
so it will come at 8+3 hour=11 a.m.
So Ans==11 a.m
20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ?
Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338.
21) The temperature at Mumbai is given by the function: -t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM?
22) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low temperature and highest temp. is 100 degrees. Then what is the low temp?
Sol) Let highest temp be x
so low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2
total temp=x+x/6+x/2=100
therefore, x=60
Lowest temp is 40
23) In Madras, temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. Ans. At 9pm 7.5 more
Sol) In equestion first put t=9,
we will get 34.5...........................(1)
now put t=4,
we will get 27..............................(2)
so ans=34.5-27
=7.5
24) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product?
a) 780 b) 1040 c) 1590 d) 1720
Sol) x*53-x*35=540=> x=30 therefore, 53*30=1590 Ans
25) How many positive integer solutions does the equation 2x+3y = 100 have?
a) 50 b) 33 c) 16 d) 35
Sol) There is a simple way to answer this kind of Q's given 2x+3y=100, take l.c.m of 'x' coeff and 'y' coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula--- constant / (l.cm of x coeff and y coeff)
26) The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders.
a) 65 b) 55 c) 50 d) 45
Sol)
Let a = fixed cost and k = variable cost and n = number of boarders
total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k
total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k
solving above 2 eqns, 3000-1750 = 25k i.e. 1250 = 25k i.e. k = 50
therefore, substituting this value of k in either of above 2 eqns we get
a = 500 (a = 3000-50*50 = 500 or a = 1750 - 25*50 = 500)
so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500
so cost per head = 5500/100 = 55
27) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens?
a) 37.5% b) 62.5% c) 50% d) None of these
Sol)
Let, 5 pens + 7 pencils + 4 erasers = x rupees
so 10 pens + 14 pencils + 8 erasers = 2*x rupees
also mentioned, 6 pens + 14 pencils + 8 erarsers = 1.5*x rupees
so (10-6) = 4 pens = (2-1.5)x rupees
so 4 pens = 0.5x rupees => 8 pens = x rupees
so 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid byamal)
i.e 5/8 = 500/8% = 62.5% is the answer
28) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race?
Sol)
x + x+6 = rs 68
2x + 6 = 68
2x = 68-6
2x = 62
x=31
x is the amt lost in I race
x+ 6 = 31+6=37 is lost in second race
then my friend lost 37 + 4 = 41 Rs
29) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms. i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10 iii) Same question with n=9
to me the chances are
when n=3
(i) nC1= 3C1 =3 for 10 boxes .. 10*3=30
(ii) 10C1=10 for 10 boxes ....10*10=100
(iii)9C1=9 for 10 boxes .....10*9=90
30) (1-1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
leaving the first numerater and last denominater, all the numerater and denominater will cancelled out one another. Ans. 5/(n+5)
31) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part?
Sol) the clock normally has 12 hr
three parts x,y,z
x+y+z=12
x=y+z
2x=12
x=6
so the largest part is 6 hrs
32) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels
Sol) 4/5 full tank= 12 mile
1 full tank= 12/(4/5)
1/3 full tank= 12/(4/5)*(1/3)= 5 miles
33) wind blows 160 miles in 330min.for 80 miles how much time required
Sol) 160 miles= 330 min
1 mile = 330/160
80 miles=(330*80)/160=165 min.
34) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was travelling at a speed of 35 mph. find the speed limit
Sol)
(x+10)=(x+35)/2
solving the eqn we get x=15
35) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3. what is the answer he actually has to get.
Sol) Assume 1
1* 3 = 3
1*1/3=1/3
so he has to got 1/3
this is the exact answer
36) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot.
37) Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp.
ans is 40 c.
Sol) let x be the highest temp. then,
x+x/2+x/6=100.
therefore, x=60 which is the highest temp
and 100-x=40 which is the lowest temp.
38) car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up.
Sol) let feul consumed in coming up is x. thus equation is: x+1.25x=4.5ans:2gallons
39) A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person
Sol) Two people work together in 24 mins.
So, their one day work is
(1/A)+(1+B)=(1/24)
One man can complete the work in 40mins
one man's one day work (1/B)= (1/40)
Now,
(1/A)=(1/24)-(1/40)
(1/A)=(1/60)
So, A can complete the work in 60 mins.
40) In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability? That a randomly chosen employee is a male or female?
Sol) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is 0.264
41) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin
Sol) the minimum number of wieghtings needed is just 5.as shown below
(1) 80->30-30
(2) 15-15
(3) 7-7
(4) 3-3
(5) 1-1
42) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples?
2x+3y+4z=15
3x+2y+z=10 adding
5x+5y+5z=25
x+y+z=5 that is for 1 orange, 1 bannana and 1 apple requires 5Rs.
so for 3 orange, 3 bannana and 3 apple requires 15Rs.
i.e. 3x+3y+3z=15
43) In 8*8 chess board what is the total number of squares refers
Sol) odele discovered that there are 204 squares on the board We found that you would add the different squares - 1 + 4 + 9 + 16+ 25 + 36 + 49 + 64.
Also in 3*3 tic tac toe board what is the total no of squares
Ans 14 ie 9+4(bigger ones)+1 (biggest one)
If you ger 100*100 board just use the formula
the formula for the sum of the first n perfect squares is
n x (n + 1) x (2n + 1)
______________________
6
if in this formula if you put n=8 you get your answer 204
44) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish.
Sol) Faster one can do 1/2 of work in one hourslower one can do 1/3 of work in one hourboth they do (1/2+1/3=5/6) th work in one hour.so work will b finished in 6/5=1.2 hour i e 1 hour 12 min.
45) If Rs20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive?
Here is the answer Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42) ==> (42 * 20 )/70 ==> 12 Rs
46) An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she cann t type.?
47) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?
the answer is 138 matches
Sol) since one player gets a bye in each round,he will reach the finals of the tournament without playing a match.
therefore 137 matches should be played to detemine the second finalist from the remaining 138 players(excluding the 1st player)
therefore to determine the winner 138 matches shd be played.
48) One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate?
Sol) Vol. of rect. plate= 8*11*2=176
area of rod=(22/7)*(8/2)*(8/2)=(352/7)
vol. of rod=area*length=vol. of plate
so length of rod= vol of plate/area=176/(352/7)=3.5
49) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank?
3 ft 7.5 inches
50) A man has to get air-mail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive?
ans:10min:::assume he started at 1:00,so at 1:30 he met cart. He returned home at 2:00.so it took him 1 hour for the total jorney.by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.so the actual time of plane is at 1:40.The cart travelled a time of 10 min before it met him.so the plane is 10 min early.
51) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked?
52) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years.
Sol) THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4
ans: 1.5 mt 4 (x+(3/5)x)=88x/5=2x=5/4 after 2 years x=(3/5)*(5/4)*2 =1.5
53) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code?
sol) Assume the digit xx xx xx (six digits)
First Two digit xx * 3=111
xx=111/3=37
( first two digits of 1 is not divisible by 3 so we can use 111)
Second Two digit xx*6=222
xx=222/6=37
( first two digits of 2 is not divisible by 6 so we can use 222)
Thrid Two digit xx*9=333
xx=333/9=37
( first two digits of 3 is not divisible by 9 so we can use 333)
54) There are 4 balls and 4 boxes of colours yellow, pink, red and green. Red ball is in a box whose colour is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball?
ans is green...
Sol) Yellow box can have either of pink/yellow balls.
if we put a yellow ball in "yellow" box then it wud imply that "yellow" is also the colour of the box which has the red ball(becoz acordin 2 d question,d box of the red ball n the ball in the yellow box have same colour)
thus this possibility is ruled out...
therefore the ball in yellow box must be pink,hence the colour of box containin red ball is also pink....
=>the box colour left out is "green",,,which is alloted to the only box left,,,the one which has yellow ball..
55) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color.
Ans:he should pick 6 ball totally.
Sol) Suppose he picks 5 balls of all different colours then when he picks up the sixth one, it must match any on of the previously drawn ball colour.
thus he must pick 6 balls
56) What is the number of zeros at the end of the product of the numbers from 1 to 100
Sol) For every 5 in unit palce one zero is added
so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place.
Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 11 zeros are added
for 25,50,75 3 extra zeros are added
so total no of zeros are 10+11+3=24
57) 10 Digit number has its first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4's..till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?.(6marks)
ans:2100010006
2---shows that two 1's in the ans
1---shows that one 2 in ans
0---shows no 3 in the ans
0---shows no 4 in the ans
0---shows no 5 in the ans
1---shows one 6 in the ans
0---shows no 7 in the ans
0---shows no 8 in the ans
0---shows no 9 in the ans
6---shows six 0's in the ans
58) There are two numbers in the ratio 8:9. if the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9. Find the larger number?
sol) 8x:9x initialy
8x+ 12 : 9x - 19 = 5x:9x
8x+12 = 5x
-> x = 4
9x = 36 not sure about the answer ..
59) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time
A-B = 3
B-c = 5
a-c = 8
so sum of diff = 8+3+5 = 16 kgs
60) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark?
ans is 11
coz each had atleast 1 shot done so 10 + 1 = 11
n 9 + 2 = 11
so d ans is 11
61) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time?
sol) A shows 11:57, B shows 12:03, C shows 12:02, and D shows 11:06 therefore, max time is for B
62) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall.
Sol) The falling height is proportional to the squere of the time.
Now, the falling height is 64cm at 2sec
so, the proportional constant is=64/(2*2)=16;
so, at 6sec the object fall maximum (16*6*6)cm=576cm;
Now, the object may be situated at any where.
if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec.
63) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings
Ans) first 50 ings.- run= 50*50=2500
51st ings.- avg 51. so total run =51*51=2601.
so run scored in that ings=2601-2500=101 runs.
64) Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over?
a) 3rd b) 6th c) 9th d) 7th
Ans is d) 7th day
Sol) In d 1st day Anand does 1/7th of total work
similarly,
Bithu does 1/8th work in d 2nd day
hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168
remaining work = (168-73)/168 = 95/168
again after 6 days of work, remaining work is = (95-73)/168 = 22/168
and hence Anand completes the work on 7th day.(hope u understood.)
65) A man, a women and a child can do a piece of work in 6 days,man can do it in 14 days, women can do it 16 days, and in how many days child can do the same work?
The child does it in 24 days
66)
A: 1 1 0 1 1 0 1 1
B: 0 1 1 1 1 0 1 0
C: 0 1 1 0 1 1 0 1
Find ( (A-B) u C )==?
Hint : 109
A-B is {A} - {A n B}
A: 1 1 0 1 1 0 1 1
B: 0 1 1 1 1 0 1 0
by binary sub. a-b = 01100001 (1-0=1, 1-1=0,0-0=0, n for the 1st 3 digits 110-011=011)
now (a-b)uc= 01100001
or 01101101
gives 1101101... convert to decimal equals 109
1.0 4 men can check exam papers in 8 days working
5 hours regularly. What is the total hours when 2 men will check the double of
the papers in 20 days?
Sol.
Let a man can do 1 unit of work in 1 hour.
Total units of work = 4 x 8 x 5 = 160 units.
Now work = 2 x 160 = 320 units.
Now 2 men work for 20 days. Let in x hours they have to work per day.
Now total work = 2×x×20 = 40 x
40x = 320 So x = 320/40 = 8 hours.
2.0 X = 101102103104105106107......146147148149150 (From numbers 101-150). Find out the remainder when this number is divided by 9.
Sol.
Let a man can do 1 unit of work in 1 hour.
Total units of work = 4 x 8 x 5 = 160 units.
Now work = 2 x 160 = 320 units.
Now 2 men work for 20 days. Let in x hours they have to work per day.
Now total work = 2×x×20 = 40 x
40x = 320 So x = 320/40 = 8 hours.
2.0 X = 101102103104105106107......146147148149150 (From numbers 101-150). Find out the remainder when this number is divided by 9.
Sol:
The divisibility rule for 9 is sum of the digits is to be divisible by
9. So
We calculate separately, sum of the digits in hundreds place, tenths
place, and units place.
Sum of the digits in hundreds place: 1 x 50 = 50
Sum of the digits in tenths place : 0 x 9 + 1 x 10 + 2 x 10 + 3 x 10 + 4
x 10 + 5 x 1 = 105
Sum of the digits in units place : (1 + 2 + 3 + ...+ 9) x 5 = 225
So total = 380
So remainder = 380 / 9 = 2
3.0 A number is 101102103104...150. As 101 102 103
103.... 150. What is reminder when divided by 3?
Sol.
Divisibility rule for 3 also same as 9. so from the above discussion sum of the
digits = 380 and remainder = 380/3 = 2.
4. In 4 years, Raj's father age twice as raj, Two years ago, Raj's mother's age twice as raj. If Raj is 32yrs old in eight yrs from now, what is the age of Raj's mother and father?
4. In 4 years, Raj's father age twice as raj, Two years ago, Raj's mother's age twice as raj. If Raj is 32yrs old in eight yrs from now, what is the age of Raj's mother and father?
Sol. Raj
present age = 32 - 8 = 24.
After 4 years Raj's age is 28. and Raj's fathers age is 28 x 2 = 56, and his present age is 52.
Two years ago, Raj's age is 22. and his mother's age is 22 x 2 = 44. His mother's present age = 46
5. 7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?
After 4 years Raj's age is 28. and Raj's fathers age is 28 x 2 = 56, and his present age is 52.
Two years ago, Raj's age is 22. and his mother's age is 22 x 2 = 44. His mother's present age = 46
5. 7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?
Sol. Units
digits of first 4 terms are 7, 9, 3, 1. and this pattern repeats. So for every
4 terms we get one term with 3 in its unit digit. So there are total of 205/4 =
51 sets and each set contains one terms with 3 in its unit digit.
Ans is 51.
6. In paper A, one student got 18 out of 70 and in paper B he got 14 out of 30. In which paper he did fare well?
Ans is 51.
6. In paper A, one student got 18 out of 70 and in paper B he got 14 out of 30. In which paper he did fare well?
Sol. Find the
percentages. Paper A = 18/70 x 100 = 25.7
Paper B = 14/30 x 100 = 46.6
Paper B = 14/30 x 100 = 46.6
7.
Find the total no of divisors of 1728 (including 1 and 1728)
Sol. Direct
formula from our lesson on factors. Click Here.
The number of factors or divisors of a number N=ap.bq.cr... = (p+1).(q+1).(r+1)... where a, b, c ... prime numbers.
1728 = 26×33
So total number of divisors = (6 + 1).(3 + 1) = 28
8. The sum of two numbers is 45. Sum of their quotient and reciprocal is 2.05, Find the product of the numbers.
The number of factors or divisors of a number N=ap.bq.cr... = (p+1).(q+1).(r+1)... where a, b, c ... prime numbers.
1728 = 26×33
So total number of divisors = (6 + 1).(3 + 1) = 28
8. The sum of two numbers is 45. Sum of their quotient and reciprocal is 2.05, Find the product of the numbers.
Sol: Let a, b be
the numbers.
a + b = 45
ab+ba = 2.05
⇒a2+b2ab = 2.05
⇒(a+b)2−2abab=2.05
⇒(a+b)2 = 2.05ab + 2ab = 4.05ab
⇒ ab = 4524.05 = 500
9. A number is divided by 406 leaves remainder 115 , What will be the reminder when it will be divided by 29?
a + b = 45
ab+ba = 2.05
⇒a2+b2ab = 2.05
⇒(a+b)2−2abab=2.05
⇒(a+b)2 = 2.05ab + 2ab = 4.05ab
⇒ ab = 4524.05 = 500
9. A number is divided by 406 leaves remainder 115 , What will be the reminder when it will be divided by 29?
Sol. Let the
number be N.
So N = 406x + 115.
Now divide this number by 29. As 406 is exactly divisible by 29, we have to divide 115 by 29 and find the remainder. So remainder = 28
10. (p/q-q/p)=21/10. Then find 4p/q+4q/p?
So N = 406x + 115.
Now divide this number by 29. As 406 is exactly divisible by 29, we have to divide 115 by 29 and find the remainder. So remainder = 28
10. (p/q-q/p)=21/10. Then find 4p/q+4q/p?
sol.
Let p/q = a, then (a - 1/a) = 21/10
⇒a2−1=a.2110
⇒10a2−21a−10 = 0
Roots of the equation = −b±b2−4ac−−−−−−−√2a
a = 21±441+400−−−−−−−−√20
a = 21±2920 = 5/2 or -2/5
For a = 5/2, 4p/q+4q/p = 58/5
For a = -2/5, 4p/q+4q/p = -58/5
11. Two vertical ladders length of 6 m and 11 m are kept vertically at a distance of 12 m. Find the top distance of both ladders?
Let p/q = a, then (a - 1/a) = 21/10
⇒a2−1=a.2110
⇒10a2−21a−10 = 0
Roots of the equation = −b±b2−4ac−−−−−−−√2a
a = 21±441+400−−−−−−−−√20
a = 21±2920 = 5/2 or -2/5
For a = 5/2, 4p/q+4q/p = 58/5
For a = -2/5, 4p/q+4q/p = -58/5
11. Two vertical ladders length of 6 m and 11 m are kept vertically at a distance of 12 m. Find the top distance of both ladders?
Sol:
So distance between the top
points = AD = 122+52−−−−−−−√=13
So in paper B he
did well.
13.
Here is 15 dots. If you select 3 dots randomly, what is the probability that 3 dots make a triangle?
a. 440/455
b. 434/455
c. 449/455
d. 438/455
13.
Here is 15 dots. If you select 3 dots randomly, what is the probability that 3 dots make a triangle?
a. 440/455
b. 434/455
c. 449/455
d. 438/455
Sol.
There
seems to be a mistake with this question. Total ways of selecting 3 dots out of
15 is 15C3 = 455 If 3 dots are collinear
then triangle may not be formed. Now look at the above diagram. If we
select any 3 dots from the red lines they may not form a triangle. They are 5 x
5C3 = 50. If we select the
three letters from blue lines, they may not form a triangle. They are in total
5 ways. Also there are 6 others lines which don't form a triangle. Total
= 50 + 5 + 6 = 61. So we can form a triangle in 455 - 61 =
394. So answer could be 394/455.
14. In a series of numbers , the next number is formed by adding 1 to the sum of the previous numbers, and the 10th number is 1280. Then what is the first number in the series? (series will be like this x , x+1, (x+(x+1))+1,....... )
a. 1
b. 4
c. 5
d. None of these
Answer: Option B
Sol.
The given series is x, x + 1, 2x + 2, 4x + 4 ......
If you observe the pattern here, the coefficient of x + 1 is in the powers of 2. So 4th term has a power of 2, 5th term has a power of 3... 10th term has a power of 8. So tenth term would be 28(x + 1)
= 256(x+1).
Given 256(x+1) = 1280
x = 4.
15. The number of multiples of 10 which are less than 1000, which can be written as a sum of four consecutive integers is
a. 50
b.100
c. 150
d. 216
The given series is x, x + 1, 2x + 2, 4x + 4 ......
If you observe the pattern here, the coefficient of x + 1 is in the powers of 2. So 4th term has a power of 2, 5th term has a power of 3... 10th term has a power of 8. So tenth term would be 28(x + 1)
= 256(x+1).
Given 256(x+1) = 1280
x = 4.
15. The number of multiples of 10 which are less than 1000, which can be written as a sum of four consecutive integers is
a. 50
b.100
c. 150
d. 216
Answer: Option A
Sol:
We can write 10 = 1 + 2 + 3 + 4. So we have to find how many multiples of 10 can be written in this manner.
Let the first of the four numbers be n. So
n + (n+1) + (n+2) + (n+3) = 10k
4n + 6 = 10k
2n + 3 = 5k
n = 5k−32 = 2k – 1 + k–12
So n is intezer for k = an odd number. So for k = 1, 3, 5, .... 99 we can write a number as a sum of four consecutive intezers.
So there are 50 numbers.
16. Mr. Bean chooses a number and he keeps on doubling the number followed by subtracting one from it, if he chooses 3 as initial number and he repeats the operation for 30 times then what is the final result?
a. (2^30) – 1
b. (2^30) – 2
c. (2^31) – 1
d. (2^31) – 2
Ans: No option
Sol:
We can write 10 = 1 + 2 + 3 + 4. So we have to find how many multiples of 10 can be written in this manner.
Let the first of the four numbers be n. So
n + (n+1) + (n+2) + (n+3) = 10k
4n + 6 = 10k
2n + 3 = 5k
n = 5k−32 = 2k – 1 + k–12
So n is intezer for k = an odd number. So for k = 1, 3, 5, .... 99 we can write a number as a sum of four consecutive intezers.
So there are 50 numbers.
16. Mr. Bean chooses a number and he keeps on doubling the number followed by subtracting one from it, if he chooses 3 as initial number and he repeats the operation for 30 times then what is the final result?
a. (2^30) – 1
b. (2^30) – 2
c. (2^31) – 1
d. (2^31) – 2
Ans: No option
Sol:
Step 1: (3 x 2) - 1 = 5 ( 2^2 + 1)
Step 2: (5 x 2) - 1 = 9 (2^3 + 1)
Step 3: (9 x 2) - 1 = 17 (2^4 + 1)
Step 4: (17 x 2) - 1 = 33 (2^5 + 1)
So After 30 steps we have 2^31 + 1
17. Tony alone can paint a wall in 7 days and his friend Roy alone can paint the same wall in 9 days. In how many days they can paint the wall working together? Round off the answer to the nearest integer.
a. 3
b. 4
c. 5
d. 7
Answer: Option B
Step 1: (3 x 2) - 1 = 5 ( 2^2 + 1)
Step 2: (5 x 2) - 1 = 9 (2^3 + 1)
Step 3: (9 x 2) - 1 = 17 (2^4 + 1)
Step 4: (17 x 2) - 1 = 33 (2^5 + 1)
So After 30 steps we have 2^31 + 1
17. Tony alone can paint a wall in 7 days and his friend Roy alone can paint the same wall in 9 days. In how many days they can paint the wall working together? Round off the answer to the nearest integer.
a. 3
b. 4
c. 5
d. 7
Answer: Option B
Sol. use formula (xy / x+y)
So nearest value for 3.93 = 4
________________________________________________________________________________So nearest value for 3.93 = 4
1. The figure shown can be folded into
the shape of a cube. In the resulting cube, which of the lettered faces
is opposite the face marked x?
a. c
b. a
c. d
d. b
Ans: C
Explanation: If you fold the above picture at the dotted lines, X and C are opposite to each other.
2. In how many ways a team of 11 must be selected from 5 men and 11 women such that the team must comprise of not more than 3 men?
a. 1565
b. 1243
c. 2256
d. 2456
Ans: C
Explanation;
The team may consist of 0 men + 11 women, 1 men + 10 women, 2 men + 9 women, or 3 men + 8 women.
So Number of ways are = 11C11+5C1×11C10+5C2×11C9+5C3×11C8 = 2256
3. Given that 0 < a < b < c < d, which of the following the largest ?
a.(c+d) / (a+b)
b.(a+d) / (b+c)
c.(b+c) / (a+d)
d.(b+d) / (a+c)
Sol: A
Explanation: Take a = 1, b = 2, c = 3, d = 4. option A is clearly true.
4. Eesha bought 18 sharpeners for Rs.100. She paid 1 rupee more for each white sharpener than for each brown sharpener. What is the price of a white sharpener and how many white sharpener did she buy ?
a. Rs.5, 10
b. Rs.6, 10
c. Rs.5, 8
d. Rs.6, 8
Sol: B
Explanation: Just check the options. If she bought 10 white sharpeners at Rs.6 per piece, She has spent Rs.60 already. And with the remaining Rs.40, she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener.
5.
b. a
c. d
d. b
Ans: C
Explanation: If you fold the above picture at the dotted lines, X and C are opposite to each other.
2. In how many ways a team of 11 must be selected from 5 men and 11 women such that the team must comprise of not more than 3 men?
a. 1565
b. 1243
c. 2256
d. 2456
Ans: C
Explanation;
The team may consist of 0 men + 11 women, 1 men + 10 women, 2 men + 9 women, or 3 men + 8 women.
So Number of ways are = 11C11+5C1×11C10+5C2×11C9+5C3×11C8 = 2256
3. Given that 0 < a < b < c < d, which of the following the largest ?
a.(c+d) / (a+b)
b.(a+d) / (b+c)
c.(b+c) / (a+d)
d.(b+d) / (a+c)
Sol: A
Explanation: Take a = 1, b = 2, c = 3, d = 4. option A is clearly true.
4. Eesha bought 18 sharpeners for Rs.100. She paid 1 rupee more for each white sharpener than for each brown sharpener. What is the price of a white sharpener and how many white sharpener did she buy ?
a. Rs.5, 10
b. Rs.6, 10
c. Rs.5, 8
d. Rs.6, 8
Sol: B
Explanation: Just check the options. If she bought 10 white sharpeners at Rs.6 per piece, She has spent Rs.60 already. And with the remaining Rs.40, she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener.
5.
The fourteen digits of a credit card are to be written in the boxes shown above. If the sum of every three consecutive digits is 18, then the value of x is :
a. 3
b. cannot be determined from the given information.
c. 2
d. 1
Sol : A
Explanation:
Let us assume right most two squares are a , b
Then Sum of all the squares = 18 x 4 + a + b .......... (1)
Also Sum of the squares before 7 = 18
Sum of the squares between 7, x = 18
and sum of the squares between x , 8 = 18
So Sum of the 14 squares = 18 + 7 + 18 + x + 18 + 8 + a + b (2)
Equating 1 and 2 we get x = 3
6. Four people each roll a four die once. Find the probability that at least two people will roll the same number ?
a. 5/18
b. 13/18
c. None of the given choices
d. 1295/1296
Sol: B
Explanation:
The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 x 5 x 4 x 3
Now total possibilities of rolling a dice = 64
The probability that a no one gets the same number = 6×5×4×364=518
So the probability that at least two people gets same number = 1−518=1318
7. Jake can dig a well in 16 days. Paul can dig the same well in 24 days. Jake, Paul and Hari together dig the well in 8 days. Hari alone can dig the well in
a. 96 days
b. 48 days
c. 32 days
d. 24 days
Sol:
Explanation: Simple one. Let the total work to be done is 48 meters. Now Jake can dig 3 mts, Paul can dig 2 mts a day. Now all of them combined dug in 8 days so per day they dug 48/8 = 6 mts. So Of these 8 mts, Hari capacity is 1 mt.
So he takes 48 /1 = 48 days to complete the digging job.
Updated :
8. Eesha bought 18 sharpeners for Rs.100. She paid 1 rupee more for each white sharpener than for each brown sharpener. What is the price of a white sharpener and how many white sharpener did she buy ?
a. Rs.5, 10
b. Rs.6, 10
c. Rs.5, 8
d. Rs.6, 8
Ans:
Explanation: This question can be solved easily by going through options.
A. White sharpener total cost: Rs.5 x 10 = Rs.50. Brown sharpeners cost = Rs.4 x 8 = 32. Total cost is only Rs.82. Wrong option.
B. White sharpener total cost: Rs.6 x 10 = Rs.60. Brown sharpeners cost = Rs.5 x 8 = 40. Total cost is Rs.100. Correct option.
9. The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is 109. If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order. Find the number.
a. 773
b. 683
c. 944
d. 863
Ans: D
Explanation: Check options. Sum of the squares should be equal to 109. Only Options B and D satisfying. When we subtract 495, only 863 becomes 368.
10. Mark told John "If you give me half your money I will have Rs.75. John said, "if you give me one third of your money, I will have Rs.75/- How much money did John have ?
a. 45
b. 60
c. 48
d. 37.5
Ans: B
Explanation: Let the money with Mark and John are M and J respectively.
Now
M + J/2 = 75
M/3 + J = 75
Solving we get M = 45, and J = 60.
11. Eesha has a wheat business. She purchases wheat from a local wholesaler of a particular cost per pound. The price of the wheat of her stores is $3 per kg. Her faulty spring balance reads 0.9 kg for a KG. Also in the festival season, she gives a 10% discount on the wheat. She found that she made neither a profit nor a loss in the festival season. At what price did Eesha purchase the wheat from the wholesaler ?
a. 3
b. 2.5
c. 2.43
d. 2.7
Ans: C
Explanation: Faulty spring balance reads 0.9 kg for a kg" means that she sells 1 kg for the price of 0.9 kgs, so she looses 10% of the price because of the faulty spring balance. She looses another 10% because of the discount.
So, she actually sells 1 kg for $3×0.9×0.9=$2.43 and since at that price she made neither a profit nor a loss, then Eesha purchase the wheat from the wholesaler for $2.43.
12. Raj goes to market to buy oranges. If he can bargain and reduce the price per orange by Rs.2, he can buy 30 oranges instead of 20 oranges with the money he has. How much money does he have ?
a. Rs.100
b. Rs.50
c. Rs.150
d. Rs.120
Ans: D
Explanation: Let the money with Raj is M. So M20−M30=2. Check options. Option D satisfies.
13. A city in the US has a basketball league with three basketball teams, the Aziecs, the Braves and the Celtics. A sports writer notices that the tallest player of the Aziecs is shorter than the shortest player of the Braves. The shortest of the Celtics is shorter than the shortest of the Aziecs, while the tallest of the Braves is shorter than the tallest of the Celtics. The tallest of the Braves is taller than the tallest of the Aziecs.
Which of the following can be judged with certainty ?
X) Paul, a Brave is taller than David, an Aziec
Y) David, a Celtic, is shorter than Edward, an Aziec
a. Both X and Y
b. X only
c. Y only
d. Neither X nor Y
Ans: B
Sol: We solve this problem by taking numbers. Let the shortest of Braves is 4 feet. Then tallest of Aziecs is less than 4. So let it be 3 feet.
A -> 2 - 3
B -> 4 - 6
C -> 1 - 7
From the above we can safely conclude X is correct. but Y cannot be determined.
14. There are 3 classes having 20, 24 and 30 students respectively having average marks in
an examination as 20,25 and 30 respectively. The three classes are represented by A, B and C and you have the following information about the three classes.
a. In class A highest score is 22 and lowest score is 18
b. In class B highest score is 31 and lowest score is 23
c. In class C highest score is 33 and lowest score is 26.
If five students are transferred from A to B, what can be said about the average score of A; and what will happen to the average score of C in a transfer of 5 students from B to C ?
a. definite decrease in both cases
b. can't be determined in both cases
c. definite increase in both cases
d. will remain constant in both cases
Ans: B
Explanation:
Class A average is 20. And their range is 18 to 22
Class B average is 25. And their range is 23 to 31
Class A average is 30. And their range is 26 to 33
If 5 students transferred from A to B, A's average cannot be determined but B's average comes down as the highest score of A is less than lowest score of B.
If 5 students transferred from B to C, C's average cannot be determined the B's range fo marks and C's range of marks are overlapping.
15. The value of a scooter depreciates in such a way that its value of the end of each year is 3/4 of its value of the beginning of the same year. If the initial value of the scooter is Rs.40,000, what is the value at the end of 3 years ?
a. Rs.13435
b. Rs.23125
c. Rs.19000
d. Rs.16875
Ans: D
Explanation: 40,000(34)3=16875
16. Rajiv can do a piece of work in 10 days , Venky in 12 days and Ravi in 15 days. They all start the work together, but Rajiv leaves after 2 days and Venky leaves 3 days before the work is completed. In how many days is the work completed ?
a. 5
b. 6
c. 9
d. 7
Ans: D
Explanation: Let the work be 60 units. If venky worked for 3 days, and the remaining days of work be x days, total days to complete the work be x + 3 days.
Now Capacities of Rajiv is 60/10 = 6, Venky is 5, Ravi is 4.
(6 + 5 + 4) 2 + (5 + 4) (x - 3) + 5 x 3 = 60.
Solving we get x = 4. So total days to complete the work is 7 days.
17. A man has a job, which requires him to work 8 straight days and rest on the ninth day. If he started work on Monday, find the day of the week on which he gets his 12th rest day.
a. Thursday
b. Wednesday
c. Tuesday
d. Friday
Ans: B
Explanation:
He works for 8 days and takes rest on the 9th day. So On the 12th rest day, there are 9 x 12 = 108 days passed. Number of odd days = (108 - 1) / 7 = 107 / 7 = 2. So the 12th rest day is wednesday.
18. On a 26 question test, five points were deducted for each wrong answer and eight points were added for each correct answer. If all the questions were answered, how many were correct, if the score was zero ?
a. 10
b. 12
c. 11
d. 13
Ans: A
Explanation:
Take options and check. If 10 are correct, his score is 10 x 8 = 80. But 16 are wrong. So total negative marking is 16 x 5 = 80. So final score is zero.
1. 2ab5 is a four digit number
divisible by 25. If a number formed from the two digits ab is a multiple of 13,
then ab is
a. 52
b. 45
c.10
d.25
Sol: For a number to be divisible by 25, last two digits of that number should be divisible by 25. So b must be either 2 or 7
it is given that ab must be divisible by 13 and in the options only 52 is divisible by 13.
2. The average temperature of Tuesday Wednesday and Thursday was 37 C. The average temperature of Wednesday and Thursday and Friday was 38 C. if the temperature on Friday was 39 C.
Find the temperature on Tuesday.
a. 37.33
b. 38.33
c. 36
d. None of the above
Sol:
(Tues + Wed + Thurs)/3=37
Tues + Wed + Thurs=111...(1)
(Wed + Thurs + Fri)/3=38
(Wed + Thurs + Fri) =114...(2)
Given friday is 39.
Then, (2) - (1) Fri - Tues = 3
So 39 - Tues = 3
Tuesday =36
3. There are 5 boxes in a cargo. The weight of the 1st box is 200 KG, the weight of the 2nd box is 20% higher than the third box, whose weight is 25% higher than the 1st box weight. The 4th box which weighs 350 KG is 30% lighter than the 5th box. Find the difference in average weight of the 4 heaviest boxes and the four lightest boxes.
Sol: weight of 1st box=200
weight of 3rd box=(125/100)*200=250
weight of 2nd box=(120/100)*250=300
weight of 4th box =350
weight of 5th box=(10/7)*350=500
average of 4 highest weighted boxes=(500+350+300+250)/4=350
average of 4 lightest boxes=(350+300+250+200)/4=275
therefore difference=350-275=75
4. The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved, while the length is doubled. Then the total area of the 4 walls of the room will be decreased by
a. 30%
b. 18.75%
c. 15%
d. 13.6%
Sol: Given l:b:h=3:2:1
let h=10, b = 20, and l = 30
area = 2(l+b)h
area= 2*(3x+2x)*x = 2(30+20)10=1000
Now after those adjustments in the measurements,
l=60, b=10, h=5
area= 2(l+b)h = 2(60+10)5=700
Percentage decrease= 1000−7001000×1000=30%
5. A circle circumscribes three unit circles that touch each other. What is the area of the larger circle? Note that p is the ratio of the circumference to the diameter of a circle ( 3.14159265).
Sol:
a. 52
b. 45
c.10
d.25
Sol: For a number to be divisible by 25, last two digits of that number should be divisible by 25. So b must be either 2 or 7
it is given that ab must be divisible by 13 and in the options only 52 is divisible by 13.
2. The average temperature of Tuesday Wednesday and Thursday was 37 C. The average temperature of Wednesday and Thursday and Friday was 38 C. if the temperature on Friday was 39 C.
Find the temperature on Tuesday.
a. 37.33
b. 38.33
c. 36
d. None of the above
Sol:
(Tues + Wed + Thurs)/3=37
Tues + Wed + Thurs=111...(1)
(Wed + Thurs + Fri)/3=38
(Wed + Thurs + Fri) =114...(2)
Given friday is 39.
Then, (2) - (1) Fri - Tues = 3
So 39 - Tues = 3
Tuesday =36
3. There are 5 boxes in a cargo. The weight of the 1st box is 200 KG, the weight of the 2nd box is 20% higher than the third box, whose weight is 25% higher than the 1st box weight. The 4th box which weighs 350 KG is 30% lighter than the 5th box. Find the difference in average weight of the 4 heaviest boxes and the four lightest boxes.
Sol: weight of 1st box=200
weight of 3rd box=(125/100)*200=250
weight of 2nd box=(120/100)*250=300
weight of 4th box =350
weight of 5th box=(10/7)*350=500
average of 4 highest weighted boxes=(500+350+300+250)/4=350
average of 4 lightest boxes=(350+300+250+200)/4=275
therefore difference=350-275=75
4. The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved, while the length is doubled. Then the total area of the 4 walls of the room will be decreased by
a. 30%
b. 18.75%
c. 15%
d. 13.6%
Sol: Given l:b:h=3:2:1
let h=10, b = 20, and l = 30
area = 2(l+b)h
area= 2*(3x+2x)*x = 2(30+20)10=1000
Now after those adjustments in the measurements,
l=60, b=10, h=5
area= 2(l+b)h = 2(60+10)5=700
Percentage decrease= 1000−7001000×1000=30%
5. A circle circumscribes three unit circles that touch each other. What is the area of the larger circle? Note that p is the ratio of the circumference to the diameter of a circle ( 3.14159265).
Sol:
By joining centers of 3 unit circles we
will get an equilateral triangle of length 2 unit. We have to find the length
of the orange line.
And center of the equilateral triangle will be the center of the big circle.
So radius of the big circle will be = (1 + Circum radius of the equilateral triagle)
Formula for Circul radius of the equilateral triangle = 23×(3√2a) here 3√2a is the height of the triangle. a is the side of the triangle
Circum radius of equilateral triangle = 23×3√2×2=23√
Area of big circle will be =Ï€r2=3.14×(1+23√)2 = 3.14×(1+43√+43)
=3.14×(1+43√+43) = 3.14×(73+43√)
=3.14×(7+43√3)
6. Rajesh calculated his average over the last 24 tests and found it to be 76. He finds out that the marks for three tests have been inverted by mistake. The correct marks for these tests are 87, 79 and 98. What is the approximate percentage difference between his actual average and his incorrect average?
Sol: No Change
Incorrect value is: 78, 97, 89
correct values are: 87, 79, 98
difference between correct and incorrect value is= 9 + 9 -18=0
7. Joke is faster than Paul, Joke and Paul each walk 24 KM. The sum of their speed is 7 Km per hour. And the sum of times taken by them is 14 hours. Then, Joke speed is
a. 3 KM/Hr
b. 4 KM/Hr
c. 5 KM/Hr
d.7 KM/Hr
Sol:
Speed=DistanceTime
let the speed of joke x then speed of paul will be 7-x
24x+247−x=14
Try to plugin the values from the options. If Joke speed is 4 the paul is 3.
8. The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one the two sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged.
SoL:
We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2×4! ways.
Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4!
9. In a certain city, 60% of the registered voters are congress supporters and the rest are BJP supporters. In an assembly election, if 75% of the registered congress supporters and 20% of the registered BJP supporters are expected to vote for candidate A, what percent of the registered voters are expected to vote for candidate A?
Sol: let the people in the city be 100
Congress supporters = 60% of 100 = 60
40% are BJP=40% of 100 = 40
out of 60,75% voted for congress=75%(60)=45
out of 40%,20% voted for congress=20%(40)=8
Total=45 + 8 = 53
Total percent= 53%
10. Anusha, Banu and Esha run a running race of 100 meters. Anusha is the fastest followed by Banu and then Esha. Anusha, Banu and Esha maintain constant speeds during the entire race. When Anusha reached the goal post, Banu was 10m behind. When Banu reached the goal post Esha was 10m behind. How far was behind Anusha when the latter reached the goal post.
option
a) 70
b) 81
c) 90
d) 80
Sol:
By that time Anusha covered 100m, Bhanu covered 90m. So ratio of their speeds = 10 : 9
By that time Bhanu reached 100m, Esha covered 90m. So ratio of their speeds = 10 : 9
Ratio of the speed of all the three = 100 : 90 : 81
By that time Anusha covered 100m, Esha Covers only 81.
11. Seven different objects must be divided among three persons. In how many ways this can be done if at least one of them gets exactly one object.
Sol: Division of m+n+p objects into three groups is given by (m+n+p)!m!×n!×p!
But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5
So The number of ways are (7)!1!×3!×3!×12! + (7)!1!×2!×4! + (7)!1!×1!×5!×12! = 70 + 105 + 21 = 196
12. George while driving along the highway saw road markers which are at equal distances from each other. He crosses the markers every 20 seconds. If he increases his speed by x meters per second, he crosses the markers at every 15 seconds. But if he increases his speed by y meters per second, he crosses the marker at every 10th second. If y-x = 40 meters per second, then what is the distance between two markers.
Sol: Let speed be =z m/s then Distance= 20z m
(z+x)15=20z; (z+y)10=20z
Also given that y - x = 40
solving we get 20z=1200
13. How many different 9 digit numbers can be formed from the number 223355888 by re-arranging its digits so that the odd digits occupy even position?
Sol: Odd places are 4 and these are occupied by 3355. So this can be done in 4!/ (2! 2!) = 6
There are 5 even numbers which have to be placed at 5 odd places. So 5!/(2!3!) = 10 ways
so total number of ways of arranging all these numbers are 10 * 6 = 60 ways
14. In a vessel, there are 10 litres of alcohol. An operation is defined as taking out five litres of what is present in the vessel and adding 10 litres of pure water to it. What is the ratio of alcohol to water after two operations?
a) 1 : 5
b) 2 : 3
c) 1 : 6
d) 3 : 2
Sol: Final concentration = Initial concentration (1−replacement quantityfinal volume)
Final concentration = =1×(1−1015)=13
Final concentration = 13×(1−1020)=16
So ratio of alcohol : water = 1 : 5
And center of the equilateral triangle will be the center of the big circle.
So radius of the big circle will be = (1 + Circum radius of the equilateral triagle)
Formula for Circul radius of the equilateral triangle = 23×(3√2a) here 3√2a is the height of the triangle. a is the side of the triangle
Circum radius of equilateral triangle = 23×3√2×2=23√
Area of big circle will be =Ï€r2=3.14×(1+23√)2 = 3.14×(1+43√+43)
=3.14×(1+43√+43) = 3.14×(73+43√)
=3.14×(7+43√3)
6. Rajesh calculated his average over the last 24 tests and found it to be 76. He finds out that the marks for three tests have been inverted by mistake. The correct marks for these tests are 87, 79 and 98. What is the approximate percentage difference between his actual average and his incorrect average?
Sol: No Change
Incorrect value is: 78, 97, 89
correct values are: 87, 79, 98
difference between correct and incorrect value is= 9 + 9 -18=0
7. Joke is faster than Paul, Joke and Paul each walk 24 KM. The sum of their speed is 7 Km per hour. And the sum of times taken by them is 14 hours. Then, Joke speed is
a. 3 KM/Hr
b. 4 KM/Hr
c. 5 KM/Hr
d.7 KM/Hr
Sol:
Speed=DistanceTime
let the speed of joke x then speed of paul will be 7-x
24x+247−x=14
Try to plugin the values from the options. If Joke speed is 4 the paul is 3.
8. The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one the two sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged.
SoL:
We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2×4! ways.
Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4!
9. In a certain city, 60% of the registered voters are congress supporters and the rest are BJP supporters. In an assembly election, if 75% of the registered congress supporters and 20% of the registered BJP supporters are expected to vote for candidate A, what percent of the registered voters are expected to vote for candidate A?
Sol: let the people in the city be 100
Congress supporters = 60% of 100 = 60
40% are BJP=40% of 100 = 40
out of 60,75% voted for congress=75%(60)=45
out of 40%,20% voted for congress=20%(40)=8
Total=45 + 8 = 53
Total percent= 53%
10. Anusha, Banu and Esha run a running race of 100 meters. Anusha is the fastest followed by Banu and then Esha. Anusha, Banu and Esha maintain constant speeds during the entire race. When Anusha reached the goal post, Banu was 10m behind. When Banu reached the goal post Esha was 10m behind. How far was behind Anusha when the latter reached the goal post.
option
a) 70
b) 81
c) 90
d) 80
Sol:
By that time Anusha covered 100m, Bhanu covered 90m. So ratio of their speeds = 10 : 9
By that time Bhanu reached 100m, Esha covered 90m. So ratio of their speeds = 10 : 9
Ratio of the speed of all the three = 100 : 90 : 81
By that time Anusha covered 100m, Esha Covers only 81.
11. Seven different objects must be divided among three persons. In how many ways this can be done if at least one of them gets exactly one object.
Sol: Division of m+n+p objects into three groups is given by (m+n+p)!m!×n!×p!
But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5
So The number of ways are (7)!1!×3!×3!×12! + (7)!1!×2!×4! + (7)!1!×1!×5!×12! = 70 + 105 + 21 = 196
12. George while driving along the highway saw road markers which are at equal distances from each other. He crosses the markers every 20 seconds. If he increases his speed by x meters per second, he crosses the markers at every 15 seconds. But if he increases his speed by y meters per second, he crosses the marker at every 10th second. If y-x = 40 meters per second, then what is the distance between two markers.
Sol: Let speed be =z m/s then Distance= 20z m
(z+x)15=20z; (z+y)10=20z
Also given that y - x = 40
solving we get 20z=1200
13. How many different 9 digit numbers can be formed from the number 223355888 by re-arranging its digits so that the odd digits occupy even position?
Sol: Odd places are 4 and these are occupied by 3355. So this can be done in 4!/ (2! 2!) = 6
There are 5 even numbers which have to be placed at 5 odd places. So 5!/(2!3!) = 10 ways
so total number of ways of arranging all these numbers are 10 * 6 = 60 ways
14. In a vessel, there are 10 litres of alcohol. An operation is defined as taking out five litres of what is present in the vessel and adding 10 litres of pure water to it. What is the ratio of alcohol to water after two operations?
a) 1 : 5
b) 2 : 3
c) 1 : 6
d) 3 : 2
Sol: Final concentration = Initial concentration (1−replacement quantityfinal volume)
Final concentration = =1×(1−1015)=13
Final concentration = 13×(1−1020)=16
So ratio of alcohol : water = 1 : 5
1. A manufacturer of chocolates makes 6
different flavors of chocolates. The chocolates are sold in boxes of 10. How
many “different” boxes of chocolates can be made?
Sol:
If n similar articles are to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to take any number of articles then the total ways are n+r−1Cr−1
In this case x1+x2+x3......x6=10
in such a case the formula for non negative integral solutions is n+r−1Cr−1
Here n =6 and r=10. So total ways are 10+6−1C6−1 = 3003
2. In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.
a. 1/3
b. 1/2
c. 5/9
d. 17/36
Sol: Their sum can be 3,4,6,8,9,12
For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 - n) ways.
Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.
So probability is (20/36)=(5/9)
3. B alone can do piece of work in 10 days. A alone can do it in 15 days. If the total wages for the work is Rs 5000, how much should B be paid if they work together for the entire duration of the work?
a. 2000
b. 4000
c. 5000
d. 3000
Sol:
Time taken by A and B is in the ratio of = 3:2
Ratio of the Work = 2 : 3 (since, time and work are inversely proportional)
Total money is divided in the ratio of 2 : 3 and B gets Rs.3000
4. On a 26 question test, 5 points were deducted for each wrong answer and 8 points were added for right answers. If all the questions were answered how many were correct if the score was zero.
a. 10
b. 11
c. 13
d. 12
Sol:
Let x ques were correct. Therefore, (26- x) were wrong
8x−5(26−x)=0
Solving we get x=10
5. Arun makes a popular brand of ice cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain same, but the length and width will be decreased by some percentage. The new width will be,
a. 5.5
b. 4.5
c. 7.5
d. 6.5
Sol:
Volume =l×b×h = 6×5×2 = 60 cm3
Now volume is reduced by 19%.
Therefore, new volume = (100−19)100×60=48.6
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume: (x100×6)(x100×5)2=48.6
Solving we get x =90
thus length and width is reduced by 10%
New width = 5-(10% of 5)=4.5
6. If all the numbers between 11 and 100 are written on a piece of paper. How many times will the number 4 be used?
Sol: We have to consider the number of 4's in two digit numbers. _ _
If we fix 4 in the 10th place, unit place be filled with 10 ways. If we fix 4 in units place, 10th place be filled with 9 ways (0 is not allowed)
So total 19 ways.
Alternatively:
There are total 9 4's in 14, 24, 34...,94
& total 10 4's in 40,41,42....49
thus, 9+10=19.
7. If twenty four men and sixteen women work on a day, the total wages to be paid is 11,600. If twelve men and thirty seven women work on a day, the total wages to be paid remains the same. What is the wages paid to a man for a day’s work?
Sol: Let man daily wages and woman daily wages be M and W respectively
24M+16W=11600
12M+37W=11600
solving the above equations gives M=350 and W=200
8. The cost price of a cow and a horse is Rs 3 lakhs. The cow is sold at 20% profit and the horse is sold at 10% loss. Overall gain is Rs 4200. What is the cost price of the cow?
Sol:
Profit = 4200
Profit =SP - CP
4200=SP - 300000 therefore SP=304200
x+y = 300000
1.2x + 0.9y = 304200
Solving for x = 114000 = CP of cow.
9. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4......
In the above sequence what is the number of the position 2888 of the sequence.
a) 1
b) 4
c) 3
d) 2
Sol: First if we count 1223334444. they are 10
In the next term they are 20
Next they are 30 and so on
So Using n(n+1)2×10≤2888
Sol:
If n similar articles are to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to take any number of articles then the total ways are n+r−1Cr−1
In this case x1+x2+x3......x6=10
in such a case the formula for non negative integral solutions is n+r−1Cr−1
Here n =6 and r=10. So total ways are 10+6−1C6−1 = 3003
2. In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.
a. 1/3
b. 1/2
c. 5/9
d. 17/36
Sol: Their sum can be 3,4,6,8,9,12
For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 - n) ways.
Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.
So probability is (20/36)=(5/9)
3. B alone can do piece of work in 10 days. A alone can do it in 15 days. If the total wages for the work is Rs 5000, how much should B be paid if they work together for the entire duration of the work?
a. 2000
b. 4000
c. 5000
d. 3000
Sol:
Time taken by A and B is in the ratio of = 3:2
Ratio of the Work = 2 : 3 (since, time and work are inversely proportional)
Total money is divided in the ratio of 2 : 3 and B gets Rs.3000
4. On a 26 question test, 5 points were deducted for each wrong answer and 8 points were added for right answers. If all the questions were answered how many were correct if the score was zero.
a. 10
b. 11
c. 13
d. 12
Sol:
Let x ques were correct. Therefore, (26- x) were wrong
8x−5(26−x)=0
Solving we get x=10
5. Arun makes a popular brand of ice cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain same, but the length and width will be decreased by some percentage. The new width will be,
a. 5.5
b. 4.5
c. 7.5
d. 6.5
Sol:
Volume =l×b×h = 6×5×2 = 60 cm3
Now volume is reduced by 19%.
Therefore, new volume = (100−19)100×60=48.6
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume: (x100×6)(x100×5)2=48.6
Solving we get x =90
thus length and width is reduced by 10%
New width = 5-(10% of 5)=4.5
6. If all the numbers between 11 and 100 are written on a piece of paper. How many times will the number 4 be used?
Sol: We have to consider the number of 4's in two digit numbers. _ _
If we fix 4 in the 10th place, unit place be filled with 10 ways. If we fix 4 in units place, 10th place be filled with 9 ways (0 is not allowed)
So total 19 ways.
Alternatively:
There are total 9 4's in 14, 24, 34...,94
& total 10 4's in 40,41,42....49
thus, 9+10=19.
7. If twenty four men and sixteen women work on a day, the total wages to be paid is 11,600. If twelve men and thirty seven women work on a day, the total wages to be paid remains the same. What is the wages paid to a man for a day’s work?
Sol: Let man daily wages and woman daily wages be M and W respectively
24M+16W=11600
12M+37W=11600
solving the above equations gives M=350 and W=200
8. The cost price of a cow and a horse is Rs 3 lakhs. The cow is sold at 20% profit and the horse is sold at 10% loss. Overall gain is Rs 4200. What is the cost price of the cow?
Sol:
Profit = 4200
Profit =SP - CP
4200=SP - 300000 therefore SP=304200
x+y = 300000
1.2x + 0.9y = 304200
Solving for x = 114000 = CP of cow.
9. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4......
In the above sequence what is the number of the position 2888 of the sequence.
a) 1
b) 4
c) 3
d) 2
Sol: First if we count 1223334444. they are 10
In the next term they are 20
Next they are 30 and so on
So Using n(n+1)2×10≤2888
For n = 23 we get LHS as 2760. Remaining terms 128.
Now in the 24th term, we have 24 1's, and next 48 terms are 2's.
So next 72 terms are 3's.
The 2888 term will be “3”.
10. How many 4-digit numbers contain no.2?
Sol: Total number of four digit numbers =9000 (i.e 1000 to 9999 )
We try to find the number of numbers not having digit 2 in them.
Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here '0' cannot be taken
Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832
Total number of numbers having digit 2 in it = 9000-5832 =3168
10. How many 4-digit numbers contain no.2?
Sol: Total number of four digit numbers =9000 (i.e 1000 to 9999 )
We try to find the number of numbers not having digit 2 in them.
Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here '0' cannot be taken
Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832
Total number of numbers having digit 2 in it = 9000-5832 =3168
1. The value of diamond varies directly as the square of its
weight. If a diamond falls and breaks into two pieces with weights in the ratio
2:3. what is the loss percentage in the value?
Sol: Let weight be “x”
the cost of diamond in the original state is proportional to x2
when it is fallen it breaks into two pieces 2y and the 3y
x = 5y
Original value of diamond = (5y)2 = 25y2
Value of diamond after breakage = (2y)2+(3y)2=13y2
so the percentage loss will be = 25y2−13y225y2×100=48%
Sol: Let weight be “x”
the cost of diamond in the original state is proportional to x2
when it is fallen it breaks into two pieces 2y and the 3y
x = 5y
Original value of diamond = (5y)2 = 25y2
Value of diamond after breakage = (2y)2+(3y)2=13y2
so the percentage loss will be = 25y2−13y225y2×100=48%
2. Five college students met at a party and exchanged gossips. Uma
said, “Only one of us is lying”. David said, “Exactly two of us are lying”.
Thara said, “Exactly 3 of us are lying”. Querishi said, “Exactly 4 of us are
lying”. Chitra said “All of us are lying”. Which one was telling the
truth?
a)David
b)Querishi
c)Chitra
d)Thara
Sol: As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth.
Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. In the same way all the other statements should be checked. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly.
3. Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that.
a) 2676
b) 2
c) 445
d) 86
SOL: This is a number series problem nothing to do with the data given.
1x 1+1=2
2 x 2+2=6
6 x 3+3=21
21 x 4+4=88 and not 86
88 x 5+5 = 445
445*6+6 = 2676
4. The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order?
a) AOTDSP
b) AOTPDS
c) AOTDPS
d) AOSTPD
SOL:
In alphabetical order : A D O P S T
A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120. So the word starts with A.
A D _ _ _ _ : empty places can be filled in 4!=24
A O _ _ _ _ : the places filled with 4! ways = 24. If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.
A O D _ _ _ : 3!= 6
A O P _ _ _ : 3!=6
Till this 36 words are obtained, we need the 42nd word.
AOS _ _ _ : 3!= 6
Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.
So given word is AOSTPD
4. A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?
SOL: This is a case of without replacement. We have to multiply two probabilities. 1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.
6C114C1×8C113C1=2491
a)David
b)Querishi
c)Chitra
d)Thara
Sol: As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth.
Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. In the same way all the other statements should be checked. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly.
3. Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that.
a) 2676
b) 2
c) 445
d) 86
SOL: This is a number series problem nothing to do with the data given.
1x 1+1=2
2 x 2+2=6
6 x 3+3=21
21 x 4+4=88 and not 86
88 x 5+5 = 445
445*6+6 = 2676
4. The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order?
a) AOTDSP
b) AOTPDS
c) AOTDPS
d) AOSTPD
SOL:
In alphabetical order : A D O P S T
A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120. So the word starts with A.
A D _ _ _ _ : empty places can be filled in 4!=24
A O _ _ _ _ : the places filled with 4! ways = 24. If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.
A O D _ _ _ : 3!= 6
A O P _ _ _ : 3!=6
Till this 36 words are obtained, we need the 42nd word.
AOS _ _ _ : 3!= 6
Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.
So given word is AOSTPD
4. A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?
SOL: This is a case of without replacement. We have to multiply two probabilities. 1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.
6C114C1×8C113C1=2491
5. There are 6 red balls,8 blue balls and 7 green balls in a bag.
If 5 are drawn with replacement, what is the probability at least three are
red?
Sol: At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.
case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21
Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
= 312/16807
6. Total number of 4 digit number do not having the digit 3 or 6.
Sol:
consider 4 digits _ _ _ _
1st blank can be filled in 7C1 ways (0,3,6 are neglected as the first digit should not be 0)
2st blank can be filled in 8C1 ways (0 considered along with 1,2,4,5,7,8,9)
3st blank can be filled in 8C1 ways
4st blank can be filled in 8C1 ways
Therefore total 4 digit number without 3 or 6 is 7 x 8 x 8 x 8=3584
7. Find the missing in the series: 70, 54, 45, 41,____.
Sol: 40
70-54 = 16 = 42
54-45 = 9 = 32
45-41 = 4 = 22
41-40 = 1 = 12
Sol: At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.
case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21
Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
= 312/16807
6. Total number of 4 digit number do not having the digit 3 or 6.
Sol:
consider 4 digits _ _ _ _
1st blank can be filled in 7C1 ways (0,3,6 are neglected as the first digit should not be 0)
2st blank can be filled in 8C1 ways (0 considered along with 1,2,4,5,7,8,9)
3st blank can be filled in 8C1 ways
4st blank can be filled in 8C1 ways
Therefore total 4 digit number without 3 or 6 is 7 x 8 x 8 x 8=3584
7. Find the missing in the series: 70, 54, 45, 41,____.
Sol: 40
70-54 = 16 = 42
54-45 = 9 = 32
45-41 = 4 = 22
41-40 = 1 = 12
8. A school has 120, 192 and 144 students enrolled for its
science, arts and commerce courses. All students have to be seated in rooms for
an exam such that each room has students of only the same course and also all
rooms have equal number of students. What is the least number of rooms needed?
Sol: We have to find the maximum number which divides all the given numbers so that number of roots get minimized. HCF of 120,192 & 144 is 24. Each room have 24 students of the same course.
Then rooms needed 12024+19224+14424 = 5 +8 + 6 = 19
9. A farmer has a rose garden. Every day he picks either 7,6,24 or 23 roses. When he plucks these number of flowers the next day 37,36,9 or 18 new flowers bloom. On Monday he counts 189 roses. If he continues on his plan each day, after some days what can be the number of roses left behind? (Hint : Consider number of roses remaining every day)
a)7
b)4
c)30
d)37
Sol:
let us consider the case of 23. when he picks up 23 roses the next day there will be 18 new, so in this case., 5 flowers will be less every day. So when he counts 189, the next day 184, 179,174,169,................
finally the no. of roses left behind will be 4.
10. What is the 32nd word of "WAITING" in a dictionary?
Sol: Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W
Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways
so can't be arranged starting with A alone as it is asking for 32nd word so it is out of range
AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways. Out of range as it has to be within 32 words.
AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24
AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways
so,24+12 =36th word so out of range. So we should not consider all the words start with AGN
now AGNI_ _ _can be arranged in 3! ways =6 ways
so 24+6=30 within range
Now only two word left so, arrange in alphabetical order.
AGNTIIW - 31st word
AGNTIWI - 32nd word
Sol: We have to find the maximum number which divides all the given numbers so that number of roots get minimized. HCF of 120,192 & 144 is 24. Each room have 24 students of the same course.
Then rooms needed 12024+19224+14424 = 5 +8 + 6 = 19
9. A farmer has a rose garden. Every day he picks either 7,6,24 or 23 roses. When he plucks these number of flowers the next day 37,36,9 or 18 new flowers bloom. On Monday he counts 189 roses. If he continues on his plan each day, after some days what can be the number of roses left behind? (Hint : Consider number of roses remaining every day)
a)7
b)4
c)30
d)37
Sol:
let us consider the case of 23. when he picks up 23 roses the next day there will be 18 new, so in this case., 5 flowers will be less every day. So when he counts 189, the next day 184, 179,174,169,................
finally the no. of roses left behind will be 4.
10. What is the 32nd word of "WAITING" in a dictionary?
Sol: Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W
Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways
so can't be arranged starting with A alone as it is asking for 32nd word so it is out of range
AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways. Out of range as it has to be within 32 words.
AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24
AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways
so,24+12 =36th word so out of range. So we should not consider all the words start with AGN
now AGNI_ _ _can be arranged in 3! ways =6 ways
so 24+6=30 within range
Now only two word left so, arrange in alphabetical order.
AGNTIIW - 31st word
AGNTIWI - 32nd word
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