1. 4, 6, 10, 14, 22, 26, 34, 38, 46, _ ? what is next term in the
series.
Sol:
Divide each number by 2. Then we get 2, 3, 5, 7, 11, 13, ......, 23.
This is a prime number series. So next number will be 2 x 29 = 58
2. y, _?, q, m, i
1. w
2. u
3. t
4. l
Sol:
Difference of 3 is in between two alphabets
i + 3 = m (j,k,l)
m + 3 = q (n,o,p)
q + 3 = u (r,s,t)
u is the answer.
3. What is the next number in the series 3,7,13,19....
Sol:
Prime numbers from 3 on wards are 3, 5, 7, 11, 13, 17, 19, 23, 29 . . .
Write alternate primes numbers starting from 3.
3, 7, 13, 19, 29
Answer is 29
4. Data Sufficiency Question:
Is w a Whole number?
Statement 1: 3w is an Odd number.
Statemet 2: 2w is an Even number
Sol:
Statement 2 is enough to solve this
3w is Odd means w may be Odd or Fraction like 5/3 we can not guess what
w is from the first statement.
2w is Even.
So must and should w either Odd r Even
i.e which is whole number.
No Fraction will give Even output.
5. Joe's age, Joe's sister's age and Joe’s fathers age sums up to a
century. When son is as old as his
father, Joe's sister will be twice as old as now. When Joe is as old as
his father then his father is twice as old as when his sister was as old as her
father.Age of her father ?
Sol:
Joe + sister + father = 100
After x years let us consider Joe's age is equal to his father
Joe + x = father
Therefore,
sister + x = 2 x sister
⇒ sister = x
Joe + sister = father
Therefore,
2 × father = 100
Hence, Father = 50
6. The sum of series represented as
1/(1×5)+1/(5×9)+1/(9×13)+−−−−+1/(221×225) is
a) 28/221
b) 56/221
c) 56/225
d) None of these
Sol:
11×5 + 15×9 + 19×13 + - - - - 1221×225
= 14×[(5–1)1×5 + (9–5)5×9 + (13–9)9×13 + - - - - (225–221)221×225]
= 14×[(1–15)+(15–19)+(19–113)+...(1221–1225)]
= 14×(1–1225)
= 14×224225
= 56225
7. What are the next three terms in the series 3, 6, 7, 12, 13, 18, 19,
24, _ _ _?
Sol:
This is a mixed series. 6, 12, 18, 24, . . . . form a series. Adding 1
to 6, 12, 18, forms another series. So next three terms are 25, 30, 31.
8. What is the next number in the series. a, b, d, h, _?
Sol:
a = 1
b = 2
d = 4
h = 8
This is a 2n series starting with n = 0, 1, 2, ...
24 = 16 which is p.
Ans = p
9. Find the letter that comes in the place of " - "
b, _, d, d, e, d, f, g, d.
Sol:
[b c] d [d e] d [f g] d [h i] d....
The series follow above manner. Answer will be c.
10. The number of zeros at the end of the product of all prime numbers
between 1 and 1111 is?
Sol:
Prime numbers between 1 & 1111 are 2,3,5,7,11,...
There is no other prime no. ending with 5 as unit digit, except one '5'
2 x 5 = 10 gives only one zero in the product of all prime numbers
So, number of zeros at the end of the product = 1
11. A train goes from stations A to B. One day there is a technical
problem at the very beginning of the journey & hence the train travels at
3/5 of it's original speed and so it arrives 2 hours late. Had the problem
occurred after 50 miles had been covered, the train would have arrived 40 min
earlier(i.e., only 120-40 = 80 min late). What is the distance between the 2
stations?
Sol:
For 1 mile the train is late by 40 / 50 min or 4/5 minutes. Or it
is late by 1 minute for every 5/4 miles. For 120 minutes late it has to travel
120 x 5/4 = 150 miles.
12. Due to some defect in our elevator, I was climbing down the
staircase. I’d climbed down just 7 steps when I saw a man on the ground
floor. Continuing to walk down, I greeted the man and I was surprised to see
that when I was yet to get down 4 steps to reach the ground floor, the man had
already finished climbing the staircase. He perhaps climbed up 2 steps for
every 1 of mine. How many steps did the staircase have?
Sol:
Let us consider x be the number of steps
7 + x + 4 = 2x
As old man takes 2 steps for every one steps he takes and he has to
complete 4 steps,
So x = 11 and total steps = 2x = 22
39. A card board of size 34 × 14 has to be attached to a wooden
box and a total of 35 pins are to be used on the each side of the card box.
find the total number of pins used.
Sol:
Total 35 pins are there and 4 sides of card board.
So 35 x 4 = 140
Now in the rectangle 4 vertices have 4 pins which is common to the
sides.
So 140 – 4 = 136.
13. In the Garbar Jhala, Ahmadabad a shopkeeper first raises the price
of Jewellery by x% then he decreases the new price by x%. After one such up
down cycle, the price of a Jewellery decreased by Rs. 21025. After a second
updown cycle the jewellery was sold for Rs. 484416. What was the original price
of the jewellery.
Sol:
Let the original price be "p":
I cycle:
Up by x% means new price is p + px100
Down by x% on current price means new price is (p + px100)
– (p + px100) ×x100
Price after one up down cycle is (p – 21025)
Thus, (p + px100) – (p + px100) ×x100
= $(p – 21025) = p' - - - - - (1)
II cycle:
Up by x% means new price is p′ + p′x100
Down by x% on current price means new price is
(p′ + p′x100) – (p′ + p′x100)
×x100
Price after second up down cycle is 484416.
Thus,(p′ + p′x100) – (p′ + p′x100)
×x100 = 484416 - - - - - - - (2)
Putting value of p' = p – 20125 in equation (2) and dividing (1) &
(2) to eliminate x.
We get a quadratic equation in p:
p2–526466p–(21025)2=0
The equation has real roots in the form 525625, 841.
14. Three football teams are there. Given below is the group table. Fill
in the x's
P - Played
W - Won
L - Lost
D - Draw
F - Goals For
A - Goals Against
P W L D F A
A 2 2 x x x 1
B 2 x x 1 2 4
C 2 x x x 3 7
Sol:
P W L D F A
A 2 2 0 0 7 1
B 2 0 1 1 2 4
C 2 0 1 1 3 7
Total goals for = Total goals against
1 + 4 + 7 = 3 + 2 + x
x = 7
A has played two and won 2 therefore lost = 0, draw = 0
B has played 2 but one is draw as A has 0 draw, it should be against C
i.e C draw = 1
C played 2 draw = 1 therefore lost = 1 because A has won both matches
played against them.
So we can conclude that each team has played a match with every other
team.
15. A dog takes 4 leaps for every 5 leaps of hare but 3 leaps of dog is
equal to 4 leaps of hare compare speed?
Sol:
Dog and hare speeds according to the number of leaps = 4 : 5
But their leap lengths are in the ratio = 4 : 3 (3 x D = 4 x H )
Multiplying number of leaps and leap lengths we get their speeds as = 4
x 4 : 5 x 3 = 16 : 15
Answer = 16 : 15
16. A bird keeper has got P pigeons, M mynas and S sparrows. The
keeper goes for lunch leaving his assistant to watch the birds. Suppose p
= 10, m = 5, s = 8 when the bird keeper comes back, the assistant informs the x
birds have escaped. The bird keeper exclaims: "Oh no! All my sparrows are
gone."
How many birds flew away?
When the bird keeper comes back, the assistant told him that x birds
have escaped. The keeper realized that atleast 2 sparrows have escaped.
What is minimum no of birds that can escape?
Sol:
This question can be solved using the pigeonhole principle.
I guess the answer for the first question is 23 (10 + 5 + 8 ).
Since if all the birds are escaped, then only he can be sure that all
sparrows are gone.
And for the second one, answer is 17 (10 p + 5 m + 2 s ).
If 17 birds escaped then best case such that least number of sparrows
escaped will be like 10 pigeon, 5 myna and 2 sparrows escaped.
17. 3,4,7,10,13,16,19,22, . . . Find 10th term in series
Sol:
3
3×1 = 3 + 1 = 4
3×2 = 6 + 1 = 7
3×3 = 9 + 1 = 10
3×4 = 12 + 1 = 13
3×5 = 15 + 1 = 16
3×6 = 18 + 1 = 19
3×7 = 21 + 1 = 22
3×8 = 24 + 1 = 25
3×9 = 27 + 1 = 28
10th term = 28
18. a,d,i,p,? what is next term
a) q
b) r
c) s
d) t
Sol:
a = 1×1
d = 2×2
i = 3×3
p = 4×4
Next will be
5×5 = 25 = Y
19. Marbles are to be distributed. Ann gets 1,Mary gets 2, Rose gets 3
and Lisa gets 4. John Brown gets as much as his sister.Tim Smith gets 2 times
as much as his sister. Neil Johnson gets 3 times as much as his sister. Sam
Paul gets 4 times as much as his sister.Find the surnames of Ann,Mary,Rose and
Lisa ?
Sol:
Ann's brother is Neil John (1×3=3).
Mary 's brother Sam paul ( 2×4=8).
Rose's brother John Brown (3×1=3).
Lisa's Brother is Tim Smith (4 \times 2 = 8$).
1 + 2 + 3 + 4 + 3 + 8 + 3 + 8 = 32
20. A shop has 4 shelf, 3 wardrobes, 2 chairs and 7 tables for sell.
You have to buy
a. 1 shelf
b. 1 wardrobe
c. either 1 chair or 1 table
How many selection can be made?
Sol:
The way to answer this question
4C1×3C1×2C1+4C1×3C1×7C1 = 108
1. X Z Y + X Y Z = Y Z X.
Find the three digits
Sol:
2nd column, Z + Y = Z shows a carry so, Z + Y + 1 = 10 + Z ⇒ Y = 9
1st column, X + X + 1 = 9 ⇒ X = 4
so, Z = 5
459 + 495 = 954
X = 4, Y = 9, Z = 5
2. In a 5 digit number, 3 pairs of sum is 11 each.last digit is 3 times
first one,3rd digit is 3 less than 2nd, 4th digit is 4 more than the second
one. Find the number.
Sol:
1st Digit ⇒ a
2nd Digit ⇒ b
3rd Digit ⇒ (b – 3)
4th digit ⇒ (b + 4)
5th Digit ⇒ 3a
So the number is : (a)(b)(b – 3)(b + 4)(3a)
Now, Let's analyze 1st and the 5th digit :
Possible combinations -
1 - 3
2 - 6
3 - 9
(Since 4 will yield 12 which is obviously more than 2 digits)
Now Let's analyze 2nd,3rd and 4th Digits :
Possible Values of 2nd Digit i.e 'b' is :
5,4,3
As, (b – 3) > 0 i.e 3rd Digit and (b + 4) 1 + 3 + 7 = 11
Similarly, 24186 for 4 – 1 – 8 and 6 + 4 + 1 = 11
3rd Combination 5 – 2 – 9 will get no possible match.
Hence, 2 solutions : 13073 and 24186
If Repetitions not allowed then Ans should be 24186
3. GOOD is coded as 164 then BAD as 21. If UGLY coded as 260 then JUMP?
Sol:
G O O D = 7 + 15 + 15 + 4 = 41
41 x 4 = 164
Similarly
B A D = 2 + 1 + 4 = 7
7 x 3
U G L Y = 21 + 7 + 12 + 25 = 65
65 x 4
Similarly,
J U M P = 10 + 21 + 13 + 16 = 60
60 x 4 = 240
4. Supposing a clock takes 7 seconds to strike 7. How long will it take
to strike 10?
Sol:
7 strike of a clock have 6 intervals
While 10 strikes have 9 intervals.
Required time = (76×9) seconds =10 1/2 seconds.
Because time is only moving ahead ! so when we say between 1 to 2 hours,
that means we assume only 1 hours not 2 hours.
5. An escalator is descending at constant speed. A walks down and takes
50 steps to reach the bottom. B runs down and takes 90 steps in the same time
as A takes 10 steps. How many steps are visible when the escalator is not
operating?
Sol:
Lets suppose that A walks down 1 step / min and
escalator moves n steps/ min
It is given that A takes 50 steps to reach the bottom
In the same time escalator would have covered 50n steps
So total steps on escalator is 50 + 50n.
Again it is given that B takes 90 steps to reach the bottom and
time
taken by him for this is equal to time taken by A to cover 10 steps
i.e
10 minutes. So in this 10 min escalator would have covered 10n steps.
So total steps on escalatro is 90 + 10n
Again equating 50 + 50n = 90 + 10n we get n = 1
Hence total number of steps on escalator is 100.
6. Albert and Fernandes have two leg swimming race. Both start from
opposite ends of the pool. On the first leg, the boys pass each other at 18 m from
the deep end of the pool. During the second leg they pass at 10 m from the
shallow end of the pool. Both go at constant speed but one of them is faster.
Each boy rests for 4 seconds at the end of the first leg. What is the length of
the pool?
Sol:
The solution is :Let the length of swimming pool be : D
let their speed be x and y. So according to question the fast swimmer
(let x) would start from shallow end.
Thus
Let they first meet after time: t1
x×t1=D–18 (1)
y×t1=18 (2)
(2) / (1)we get
yx=18(D–18) --- (3)
Let t2 be the time after which they meet 2nd time (the 4 sec delay is
cancelled as both wait for 4 sec)
So
x×t2=2D–10 ---- (4)
(as x travelled one length complete to deep end + length from deep end
to 10 m before shallow end)
4y×t2=D+10 ----- (5)
(as y travelled one length complete to shallow end + 10 m from shallow
end)
(5) / (4)we get
yx=(D+10)(2D–10) ----- (6)
from (3) and (6)
18(D–18)=(D+10)(2D–10)
solving we get
D x (D – 44) = 0
Since D cannot be zero
So D = 44 m answer.
7. 16, 36, 100, 324, _ ?
Find the next term.
Sol:
This sequence can be written as a sequence of squares of numbers
as...
42,62,102,182
The differences between the successive numbers are in geometric
progression
which is of
2,4,8,?
21,22,23,24
The next number = (18+16)2 = 1156
8. How many ways can one arrange the word EDUCATION such that relative
positions of vowels and consonants remains same?
Sol:
The word EDUCATION is a 9 letter word with none of letters
repeating
The vowels occupy 3,5,7th & 8th position in the word & remaining
five positions are occupied by consonants
As the relative position of the vowels & consonants in any
arrangement should remain the same as in the word EDUCATION
The four vowels can be arranged in 3rd,5th,7th & 8th position in 4!
ways.
similarly the five consonants can be arranged in 1st ,2nd ,4th,
6th & 9th position in 5! ways
Hence the total number of ways = 5!×4!=120×24=2880
9. There are 8 digits and 5 alphabets.In how many ways can you form an
alphanumeric word using 3 digits and 2 alphabets?
Sol:
Select 3 digits from 8 digits i. e. 8C3 ways
And also select 2 alphabets from 5 alphabets i.e., 5C2 ways
Now to form a alphanumeric word of 5 characters we have to arrange the 5
selected digits.
So the answer is .8C3 × 5C2 × 5! = 43200
10. In an Octagon the number of possible diagonals are?
Sol:
Formula : Number of diagonals for n sided regular polygon =nC2–n
For Octagon n = 8
Number of diagonals =8C2 – 8 = 20
11. What is the next number of the following sequence 7, 14, 55, 110, _
?
Sol:
In that sequence first number is 7
7 + 7 =14
14 + 41 = 55
55 + 55 = 110
110 + 011 =121
Next number in that sequence = 121
12. How many numbers are divisible by 4 between 1 to 100
Sol:
Sequence of numbers that are divisible by 4 between 1 to 100 are as
follows
4,8,12,16, - - - - - - - - , 96
The series forms an Arithmetic Progression with
First number = a = 4
Common difference,d = 4
Last number = l = 96
Number of terms = n
Formula for last number in A.P. l = [a+(n – 1)×d]
96 = 4 + (n –1) ×4
n = 24
13. 5 cars are to be parked in 5 parking slots. there are 3 red cars, 1
blue car and 1 green car. How many ways the car can be parked?
Sol:
Total ways to park the cars having same color = 5!
But according to question ,there are 3 red cars,so no. of ways for
parking
3 red cars= 3!
and both blue & green in 1 ways
so, 5!1!×3!×1! = 20 ways
Hence correct answer is 20 ways.
14. 12 persons can complete the work in 18 days. after working for 6
days, 4 more persons added to complete the work fast. in how many more days
they will complete the work?
Sol:
Total work 12 x 18 =216 units
After 6 days, work finished 6 x 12 =72 units
Remaining work 216 - 72=144 units
Remaining days= 144(12+4)
Answer is 9 days
15. A set of football matches is to be organized in a
"round-robin" fashion, i.e., every participating team plays a match
against every other team once and only once. If 21 matches are totally played,
how many teams participated?
Sol:
Consider number of teams be n
nth has to with (n –1) matches
(n – 1)th team has to play (n – 2) matches,since every
participating team plays a match against every other team once and only
once.
Sequence folilows as
(n – 1), (n – 2), (n – 3) - - - - - - -,1
Formula for summation(x) for n terms = n(n+1)2
But we have (n –1) terms so formula becomes n(n–1)2
Equating formula to 21
n2 – n – 42=0
Factors = 7,–6
Number of teams =7
16. Next term in series 3, 32, 405, _
Sol:
First term 3×12=3
Second term 4×23=32
Third term 5×34=405
Fourth term 6×45=6144
17. A cube is divided into 729 identical cubelets. Each cut is made
parallel to some surface of the cube . But before doing that the cube is
colored with green color on one set of adjacent faces ,red on the other set of
adjacent faces, blue on the third set. So, how many cubelets are there
which are painted with exactly one color?
Sol:
Total cubes created are 729
So a plane of big cube has 9 x 9 cubes
Out of that (n – 2) x (n – 2) = 7 x 7 = 49 are painted only one side
and a cube has six sides = 6 x 49 = 294
18. Find the radius of the circle inscribed in a triangle ABC.
Triangle ABC is a right-angled isosceles triangle with the hypotenuse as
62√
Sol:
Since hypotenuse is 62√ cm.
Sides are 6 cm each as it is an isosceles triangle.
Now, if we have an inscribed circle the property is the point where the
circle touches the sides are exactly 2/3 rd of the length of sides, i.e, 23×6 =
4 cm.
Now, if you drop 2 radii on the sides of triangle then they act as
perpendiculars on sides. So, it forms a small square of (6 – 4) = 2 cm each
side.
Thus, radius of the circle is 2 cm.
19. How many boys are there in the class if the number of boys in the
class is 8 more than the number of girls in the class, which is five times the
difference between the number of girls and boys in the class.
Sol:
Let number of boys = b
Number of girls = g
then
given
b = 8+g = 5(b – g) [ b – g = 8 from given equation]
b = 5 x 8
b = 40
20. If dolly works hard then she can get A grade
1. If dolly does not work hard then she can get A grade
2. If dolly gets an A grade then she must have worked hard
3. If dolly does not gets an A grade then she must not have worked hard
4. Dolly wishes to get A grad
Sol:
Option 3 is correct
as it is contrapositive of the given statement.
1. The hour hand lies between 3 and 4. Tthe difference between
hour and minute hand is 50 degree.What are the two possible timings?
Sol:
The angle between the hour hand and minute hand at a given time H:MM is given by
θ = 30×H – 211×MM
The time after H hours, hour hand and minute hand are at
MM = | 211×((30×H)±Î¸) |
given H = 3, MM = 50
Substituting the above values in the formula
θ = 8011, 28011
2. Jack and Jill went up and down a hill. They started from the bottom and Jack met Jill again 20 miles from the top while returning. Jack completed the race 1 min a head of Jill. If the hill is 440 miles high and their speed while down journey is 1.5 times the up journey. How long it took for the Jack to complete the race ?
Sol:
Assume that height of the hill is 440 miles.
Let speed of Jack when going up = x miles/minute
and speed of Jill when going up = y miles/minute
Then speed of Jack when going down = 1.5x miles/minute
and speed of Jill wen going up = 1.5y miles/minute
Case 1 :
Jack met jill 20 miles from the top. So Jill travelled 440 – 20 = 420 miles.
Time taken for Jack to travel 440 miles up and 20 miles down = Time taken for Jill to travel 420 miles up
440x+201.5x=420y
681.5x=420y
68y = 63x
y = 63x68 ---(1)
Case 2 : Time taken for Jack to travel 440 miles up and 440 miles down = Time taken for Jill to travel 440 miles up and 440 miles down – 1
440x+4401.5x=440y+4401.5y – 1
440×53(1y−1x)=1-----(2)
Substitute (2) in (1) we get
x = 440×5×53×63
t = 440×53(1x)
t = 12.6min
3. Data Sufficiency question:
A, B, C, D have to stand in a queue in descending order of their heights. Who stands first?
I. D was not the last, A was not the first.
II. The first is not C and B was not the tallest.
Sol:
D because A is not first neither C and B is not the tallest person. The only person will be first is D.
So option (C). We can answer this question using both the statements together.
4. One of the longest sides of the triangle is 20 m. The other side is 10 m. Area of the triangle is 80 m2. What is the another side of the triangle?
Sol:
If a,b,c are the three sides of the triangle.
Then formula for Area = (s(s–a)×(s–b)×(s–c))−−−−−−−−−−−−−−−−−−−−√
Where s = (a+b+c)2=12×(30+c)
[Assume a = 20 ,b = 10]
Now,
Check the options.
5. Data Sufficiency Question:
a and b are two positive numbers. How many of them are odd?
I. Multiplication of b with an odd number gives an even number.
II.a2 – b is even.
Sol:
From the 1st statement b is even, as when multiplied by odd it gives even
a2 – b = even
⇒ a is even
Here none of a and b are odd
6. Mr. T has a wrong weighing pan. One arm is lengthier than other. 1 kilogram on left balances 8 melons on right, 1 kilogram on right balances 2 melons on left. If all melons are equal in weight, what is the weight of a single melon.
Sol:
Let additional weight on left arm be x.
Weight of melon be m
x + 1 = 8 x m - - - - - - (1)
x + 2 x m = 1 - - - - - - (2)
Solving 1 & 2 we get.
Weight of a single Melon = 200 gm.
7. a, b, b, c, c, c, d, d, d, d, . . . . . . Find the 288th letter of this series.
Sol:
Observe that each letter appeared once, twice, thrice .... They form an arithmetic progression. 1+2+3......
We know that sum of first n natural numbers = n(n+1)2
So n(n+1)2 ≤ 288
For n = 23, we get 276. So for n = 24, the given series crosses 288.
Ans is X
8. If ABC =C3 and CAB = D3, Then find D3÷B3
Sol:
ABC = C3
So, look for a number, that has a 3 digit cube, and the last digit of the cube is same as the number itself: 53 = 125
So, CAB = 512 = 83
D = 8 and B = 2
83÷23
Answer = 64.
9. There are three trucks A, B, C. A loads 10 kg/min. B loads 13 1/3 kg/min. C unloads 5 kg/min. If three simultaneously works then what is the time taken to load 2.4 tones?
Sol:
Work done in 1 min =10 + 403 – 5= 553 kg/min
For 1 kg = 3/55 min
For 2.4 tonnes = 3/55 x 2.4 x 1000 = 130 mins = 2hrs 10min
10. If A = x3y2 and B=xy3, then find the HCF of A, B
Sol:
A=x3×y2
B = x×y3
To find the HCF of the above numbers, take minimum power of x and y in both the numbers.
HCF = Common terms from both A & B and minimum powers = x×y2
11. HERE = COMES – SHE, (Assume s = 8)
Find value of R + H + O
Sol:
HERE = COMES – SHE
HERE
+ SHE
------------
COMES
------------
E + E = S = 8 => E = 4
3 digit no. + 4 digit no. = 5 digit no. ⇒ C = 1 ,O = 0, H = 9 etc
So 9454 + 894 = 10348
10348
– 894
--------
9454
-------
R + H + O = 5 + 9 + 0 = 14
12. A person is 80 years old in 490 and only 70 years old in 500 in which year is he born?
a) 400
b) 550
c) 570
d) 440
Sol:
He must have born in BC 570
Hence in BC 500 he will be 70 years
And in BC 490 he will be 80 years
13. Lucia is a wonderful grandmother and her age is between 50 and 70. Each of her sons have as many sons as they have brothers. Their combined ages give Lucia's present age.what is the age?
Sol:
The question basically states that if Lucia were to have say 10 sons, then each son would have 9 sons (Lucia's grandsons – since each son has 9 brothers). So the total in this case would be 9×10 grandsons + 10 sons = 100.
Let us assume Lucia has got x sons. Now each son has (x - 1) sons. So total = x + (x - 1) x. For x = 8 we get 64 which is in between 50 and 60. ( 7 x 8 grandsons + 8 sons = 64 )
14. A family X went for a vacation. Unfortunately it rained for 13 days when they were there. But whenever it rained in the mornings, they had clear afternoons and vice versa. In all they enjoyed 11 mornings and 12 afternoons. How many days did they stay there totally?
Sol:
Clearly 11 mornings and 12 afternoons = 23 half days
since 13 days raining means 13 half days.
so 23 – 13 =10 half days ( not affected by rain )
so 10 half days = 5 full days
Total no. of days = 13 + 5 = 18 days.
15. Find the unit digit of product of the prime number up to 50 .
Sol:
Prime number up to 50 are
2,3,5,7,11,...,43,47
Product = 2×3×5×7×11×−−−×43×47
There's a term 2×5=10
So unit digit of product = 0
16. HOW + MUCH = POWER Then P + O + W + E + R =
Sol:
HOW
+ MUCH
-------------
POWER
--------------
Here p = 1 and M = 9 because after adding carry bit it gives result 10. Hence O = 0,here three digits 0,1,9 have been used.
Now, put all remaining value in 3rd column and check which value is suitable for H,U and W and we get H = 7,U = 8 and W = 5 and 1 carry which will be added in 4th column.
Now in first column we have W + H = R means 5 + 7 = 2 and 1 carry will add in 2nd column
in 2nd column, 0 + C = E,0 + 3 + 1 = 4 so C = 3,E = 4
Therefore,
9837
+ 705
---------
10542
---------
so P + O + W + E + R = 1 + 0 + 5 + 4 + 2 = 12
17. Complete the series..
2 2 12 12 30 30 ?
Sol:
Answer is 56.
It follows the series as:
1 x 2 = 2
2 x 1 = 2
3 x 4 = 12
4 x 3 = 12
5 x 6 = 30
6 x 5 = 30
7 x 8 = 56
This is the required number for the series.
Sol:
The angle between the hour hand and minute hand at a given time H:MM is given by
θ = 30×H – 211×MM
The time after H hours, hour hand and minute hand are at
MM = | 211×((30×H)±Î¸) |
given H = 3, MM = 50
Substituting the above values in the formula
θ = 8011, 28011
2. Jack and Jill went up and down a hill. They started from the bottom and Jack met Jill again 20 miles from the top while returning. Jack completed the race 1 min a head of Jill. If the hill is 440 miles high and their speed while down journey is 1.5 times the up journey. How long it took for the Jack to complete the race ?
Sol:
Assume that height of the hill is 440 miles.
Let speed of Jack when going up = x miles/minute
and speed of Jill when going up = y miles/minute
Then speed of Jack when going down = 1.5x miles/minute
and speed of Jill wen going up = 1.5y miles/minute
Case 1 :
Jack met jill 20 miles from the top. So Jill travelled 440 – 20 = 420 miles.
Time taken for Jack to travel 440 miles up and 20 miles down = Time taken for Jill to travel 420 miles up
440x+201.5x=420y
681.5x=420y
68y = 63x
y = 63x68 ---(1)
Case 2 : Time taken for Jack to travel 440 miles up and 440 miles down = Time taken for Jill to travel 440 miles up and 440 miles down – 1
440x+4401.5x=440y+4401.5y – 1
440×53(1y−1x)=1-----(2)
Substitute (2) in (1) we get
x = 440×5×53×63
t = 440×53(1x)
t = 12.6min
3. Data Sufficiency question:
A, B, C, D have to stand in a queue in descending order of their heights. Who stands first?
I. D was not the last, A was not the first.
II. The first is not C and B was not the tallest.
Sol:
D because A is not first neither C and B is not the tallest person. The only person will be first is D.
So option (C). We can answer this question using both the statements together.
4. One of the longest sides of the triangle is 20 m. The other side is 10 m. Area of the triangle is 80 m2. What is the another side of the triangle?
Sol:
If a,b,c are the three sides of the triangle.
Then formula for Area = (s(s–a)×(s–b)×(s–c))−−−−−−−−−−−−−−−−−−−−√
Where s = (a+b+c)2=12×(30+c)
[Assume a = 20 ,b = 10]
Now,
Check the options.
5. Data Sufficiency Question:
a and b are two positive numbers. How many of them are odd?
I. Multiplication of b with an odd number gives an even number.
II.a2 – b is even.
Sol:
From the 1st statement b is even, as when multiplied by odd it gives even
a2 – b = even
⇒ a is even
Here none of a and b are odd
6. Mr. T has a wrong weighing pan. One arm is lengthier than other. 1 kilogram on left balances 8 melons on right, 1 kilogram on right balances 2 melons on left. If all melons are equal in weight, what is the weight of a single melon.
Sol:
Let additional weight on left arm be x.
Weight of melon be m
x + 1 = 8 x m - - - - - - (1)
x + 2 x m = 1 - - - - - - (2)
Solving 1 & 2 we get.
Weight of a single Melon = 200 gm.
7. a, b, b, c, c, c, d, d, d, d, . . . . . . Find the 288th letter of this series.
Sol:
Observe that each letter appeared once, twice, thrice .... They form an arithmetic progression. 1+2+3......
We know that sum of first n natural numbers = n(n+1)2
So n(n+1)2 ≤ 288
For n = 23, we get 276. So for n = 24, the given series crosses 288.
Ans is X
8. If ABC =C3 and CAB = D3, Then find D3÷B3
Sol:
ABC = C3
So, look for a number, that has a 3 digit cube, and the last digit of the cube is same as the number itself: 53 = 125
So, CAB = 512 = 83
D = 8 and B = 2
83÷23
Answer = 64.
9. There are three trucks A, B, C. A loads 10 kg/min. B loads 13 1/3 kg/min. C unloads 5 kg/min. If three simultaneously works then what is the time taken to load 2.4 tones?
Sol:
Work done in 1 min =10 + 403 – 5= 553 kg/min
For 1 kg = 3/55 min
For 2.4 tonnes = 3/55 x 2.4 x 1000 = 130 mins = 2hrs 10min
10. If A = x3y2 and B=xy3, then find the HCF of A, B
Sol:
A=x3×y2
B = x×y3
To find the HCF of the above numbers, take minimum power of x and y in both the numbers.
HCF = Common terms from both A & B and minimum powers = x×y2
11. HERE = COMES – SHE, (Assume s = 8)
Find value of R + H + O
Sol:
HERE = COMES – SHE
HERE
+ SHE
------------
COMES
------------
E + E = S = 8 => E = 4
3 digit no. + 4 digit no. = 5 digit no. ⇒ C = 1 ,O = 0, H = 9 etc
So 9454 + 894 = 10348
10348
– 894
--------
9454
-------
R + H + O = 5 + 9 + 0 = 14
12. A person is 80 years old in 490 and only 70 years old in 500 in which year is he born?
a) 400
b) 550
c) 570
d) 440
Sol:
He must have born in BC 570
Hence in BC 500 he will be 70 years
And in BC 490 he will be 80 years
13. Lucia is a wonderful grandmother and her age is between 50 and 70. Each of her sons have as many sons as they have brothers. Their combined ages give Lucia's present age.what is the age?
Sol:
The question basically states that if Lucia were to have say 10 sons, then each son would have 9 sons (Lucia's grandsons – since each son has 9 brothers). So the total in this case would be 9×10 grandsons + 10 sons = 100.
Let us assume Lucia has got x sons. Now each son has (x - 1) sons. So total = x + (x - 1) x. For x = 8 we get 64 which is in between 50 and 60. ( 7 x 8 grandsons + 8 sons = 64 )
14. A family X went for a vacation. Unfortunately it rained for 13 days when they were there. But whenever it rained in the mornings, they had clear afternoons and vice versa. In all they enjoyed 11 mornings and 12 afternoons. How many days did they stay there totally?
Sol:
Clearly 11 mornings and 12 afternoons = 23 half days
since 13 days raining means 13 half days.
so 23 – 13 =10 half days ( not affected by rain )
so 10 half days = 5 full days
Total no. of days = 13 + 5 = 18 days.
15. Find the unit digit of product of the prime number up to 50 .
Sol:
Prime number up to 50 are
2,3,5,7,11,...,43,47
Product = 2×3×5×7×11×−−−×43×47
There's a term 2×5=10
So unit digit of product = 0
16. HOW + MUCH = POWER Then P + O + W + E + R =
Sol:
HOW
+ MUCH
-------------
POWER
--------------
Here p = 1 and M = 9 because after adding carry bit it gives result 10. Hence O = 0,here three digits 0,1,9 have been used.
Now, put all remaining value in 3rd column and check which value is suitable for H,U and W and we get H = 7,U = 8 and W = 5 and 1 carry which will be added in 4th column.
Now in first column we have W + H = R means 5 + 7 = 2 and 1 carry will add in 2nd column
in 2nd column, 0 + C = E,0 + 3 + 1 = 4 so C = 3,E = 4
Therefore,
9837
+ 705
---------
10542
---------
so P + O + W + E + R = 1 + 0 + 5 + 4 + 2 = 12
17. Complete the series..
2 2 12 12 30 30 ?
Sol:
Answer is 56.
It follows the series as:
1 x 2 = 2
2 x 1 = 2
3 x 4 = 12
4 x 3 = 12
5 x 6 = 30
6 x 5 = 30
7 x 8 = 56
This is the required number for the series.
1. A Lorry starts from Banglore to
Mysore At 6.00 a.m, 7.00 a.m, 8.00 a.m.....10 p.m. Similarly another Lorry on
another side starts from Mysore to Banglore at 6.00 a.m, 7.00 a.m, 8.00
a.m.....10.00 p.m. A Lorry takes 9 hours to travel from Banglore to Mysore and
vice versa.
(I) A Lorry which has started At 6.00 a.m will cross how many Lorries.
(II) A Lorry which has started At 6.00 p.m will cross how many Lorries.
Sol:
I. The Lorry reaches Mysore by 3 PM so it meets all the Lorries which starts after 6 a.m and before 3 p.m. So 9 lorries. Also the Lorry which starts at night 10 p.m on the previous day at Mysore reaches Bangalore in morning 7 a.m. So it also meets that Lorry. So the Lorry which starts at 6:00 am will cross 10 Lorries.
II. The lorry which has started at 6 p.m reaches destination by 3 a.m. Lorries which start at the opposite destination at 10 am reaches its destination at 7 pm. So all the lorries which starts at 10 am to 10 pm meets this lorry . So in total 13.
2. GOOD is coded as 164 then BAD coded as 21.if ugly coded as 260 then JUMP?
Sol:
Coding = Sum of position of alphabets x Number of letters in the given word
GOOD = (7 + 15 + 15 + 4 ) x 4 = 164
BAD = (2 + 1 + 4) x 3 = 21
UGLY = (21 + 7 + 12 + 25) x 4 = 260
So, JUMP = (10 + 21 + 13 + 16) x 4 = 240
3. If Ever + Since = Darwin then D + a + r + w + i + n is ?
Sol: Tough one as it has 10 variables in total. 4 digit number + 5 digit number = 6 digit number. So left most digit in the answer be 1. and S = 9, a = 0. Now we have to use trial and error method.
(I) A Lorry which has started At 6.00 a.m will cross how many Lorries.
(II) A Lorry which has started At 6.00 p.m will cross how many Lorries.
Sol:
I. The Lorry reaches Mysore by 3 PM so it meets all the Lorries which starts after 6 a.m and before 3 p.m. So 9 lorries. Also the Lorry which starts at night 10 p.m on the previous day at Mysore reaches Bangalore in morning 7 a.m. So it also meets that Lorry. So the Lorry which starts at 6:00 am will cross 10 Lorries.
II. The lorry which has started at 6 p.m reaches destination by 3 a.m. Lorries which start at the opposite destination at 10 am reaches its destination at 7 pm. So all the lorries which starts at 10 am to 10 pm meets this lorry . So in total 13.
2. GOOD is coded as 164 then BAD coded as 21.if ugly coded as 260 then JUMP?
Sol:
Coding = Sum of position of alphabets x Number of letters in the given word
GOOD = (7 + 15 + 15 + 4 ) x 4 = 164
BAD = (2 + 1 + 4) x 3 = 21
UGLY = (21 + 7 + 12 + 25) x 4 = 260
So, JUMP = (10 + 21 + 13 + 16) x 4 = 240
3. If Ever + Since = Darwin then D + a + r + w + i + n is ?
Sol: Tough one as it has 10 variables in total. 4 digit number + 5 digit number = 6 digit number. So left most digit in the answer be 1. and S = 9, a = 0. Now we have to use trial and error method.
Here E appeared 3 times, I, R, N two times each. Now E + I or E + I + 1 is a two digit number with carry over. What could be the value of E and I here. 8 and 7 are possible. But from the second column, 8 + C = 7 or 17 not possible. Similarly with 7 and 6. If E = 5, then the remaining value can be filled like above.
5653 + 97825 = 103478
Answer is 23
4. There are 16 hockey teams. find :
(1) Number of matches played when each team plays with each other twice.
(2) Number of matches played when each team plays each other once.
(3) Number of matches when knockout of 16 team is to be played
Sol:
1. Number of ways that each team played once with other team = 16C2. To play with each team twice = 16 x 15 = 240
2. 16C2 = 120
3. Total 4 rounds will be played. Total number of matches required = 8 + 4 + 2 + 1 = 15
5. 15 tennis players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?
A. 190
B. 200
C. 210
D. 220
E. 225
Sol:
Formula: 15C2 x 2. So 15 x (15 - 1) = 15 x 14 = 210
6. 1, 11, 21, 1211, 111221, 312211, . . . . . what is the next term in the series?
Sol:
We can understand it by writing in words
One
One time 1 that is = 11
Then two times 1 that is = 21
Then one time 2 and one time 1 that is = 1211
Then one time one, one time two and two time 1 that is = 111221
And last term is three time 1, two time 2, and one time 1 that is = 312211
So our next term will be one time 3 one time 1 two time 2 and two time 1
13112221 and so on
7. How many five digit numbers are there such that two left most digits are even and remaining are odd.
Sol:
N = 4 x 5 x 5 x 5 x 5 = 2375
Where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}
8. 13_46_8_180_210_75 = 64 . Use + and – in the empty places to make the equation holds good. Take m = number of + and n = number of – . Find m – n?
Sol:
13 – 46 – 8 – 180 + 210 + 75 = 64
m = 3
n = 4
m – n = – 1
10. If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that atleast 1 blue can and 1 red can remains in the refrigerator.
Sol:
Possible ways of keeping atleast 1 blue and 1 red ball are drawing cans like (6,2) (5,3) (4,4)
(6,2) ⇒7C6×5C2 ⇒ 710 = 70
(5,3) ⇒7C5×5C3 ⇒ 21 x 10 = 210
(4,4) ⇒7C4×5C4 ⇒ 35 x 5 = 175
70 + 210 + 175 = 455
11. Find the 8th term in series?
2, 2, 12, 12, 30, 30, - - - - -
Sol:
11 + 1 = 2
22 – 2 = 2
32 + 3 = 12
42 – 4 = 12
52 + 5 = 30
62 – 6 = 30
So 7th term = (72 + 7) = 56 and 8th term = ({82} – 8) = 56
Answer is 56
12. Find the next three terms of the series;
1, 4, 9, 18, 35 - - - - -
Sol:
21 – 1 = 1
22 + 0 = 4
23 + 1 = 9
24 + 2 = 18
25 + 3 = 35
So 26 + 4 = 68, 27 + 5 = 133, 28 +6 = 262
Answer is 68, 133, 262
13. Rahul took a part in cycling game where 1/5 ahead of him and 5/6 behind him then total number of participants =
Sol:
Let x be the total number of participants including Rahul.
Excluding rahul = (x – 1)
15(x–1)+56(x–1) = x
31x – 31 = 30x
Total number of participants x = 31
14. Data sufficiency question:
What are the speeds two trains travels with 80 yards and 85 yards long respectively? (Assume that former is faster than later)
a) they take 75 seconds to pass each other in opposite direction.
b) they take 37.5 seconds to pass each other in same direction
Sol:
Let the speeds be x and y
When moves in same direction the relative speed,
x – y = (85–80)37.5 = 0.13 - - - - - (I)
When moves in opposite direction the relative speed, x + y = 165/75 = 2.2 - - - - (II)
Now, equation I + equation II gives, 2x = 0.13 + 2.2 = 2.33 ⇒ x = 1.165
From equation l, x – y = 0.13 ⇒ y = 1.165 – 0.13 = 1.035
Therefore the speeds are 1.165 yards/sec and 1.035 yards/sec.
15. Reversing the digits of father's age we get son's age. One year ago father was twice in age of that of his son? find their current ages?
Sol:
Let father's age = 10x + y
Son's age = 10y + x (As, it is got by reversing digits of fathers age)
At that point
(10x + y) – 1 = 2{(10y + x) – 1}
⇒ x = (19y – 1)/8
Let y = 3 then x = 7.
For any other y value, x value combined with y value doesn't give a realistic age (like father's age 120 etc)
So, this has to be solution.Hence father's age = 73.
Son's age = 37.
1. What is the 8th term in the series
1,4, 9, 25, 35, 63, . . .
Sol:
1, 4, 9, 18, 35, 68, . . .
The pattern is
1 = 21 – 1
4 = 22 – 0
9 = 23 + 1
18 = 24 + 2
35 = 25 + 3
68 = 26 + 4
So 8th term is 28 + 6 = 262
2. USA + USSR = PEACE ; P + E + A + C + E = ?
Sol:
3 Digit number + 4 digit number = 5 digit number. So P is 1 and U is 9, E is 0.
Now S repeated three times, A repeated 2 times. Just give values for S. We can easily get the following table.
Sol:
1, 4, 9, 18, 35, 68, . . .
The pattern is
1 = 21 – 1
4 = 22 – 0
9 = 23 + 1
18 = 24 + 2
35 = 25 + 3
68 = 26 + 4
So 8th term is 28 + 6 = 262
2. USA + USSR = PEACE ; P + E + A + C + E = ?
Sol:
3 Digit number + 4 digit number = 5 digit number. So P is 1 and U is 9, E is 0.
Now S repeated three times, A repeated 2 times. Just give values for S. We can easily get the following table.
USA = 932
USSR = 9338
PEACE = 10270
P + E + A + C + E = 1 + 0 + 2 + 7 + 0 = 10
3. In a cycle race there are 5 persons named as J, K, L, M, N participated for 5 positions so that in how many number of ways can M make always before N?
Sol:
Say M came first. The remaining 4 positions can be filled in 4! = 24 ways.
Now M came in second. N can finish the race in 3rd, 4th or 5th position. So total ways are 3 x 3! = 18.
M came in third. N can finish the race in 2 positions. 2 x 3! = 12.
M came in second. N can finish in only one way. 1 x 3! = 6
Total ways are 24 + 18 + 12 + 6 = 60.
Shortcut:
Total ways of finishing the race = 5! = 120. Of which, M comes before N in half of the races, N comes before M in half of the races. So 120 / 2 = 60.
4. If POINT + ZERO = ENERGY, then E + N + E + R + G + Y = ?
Sol:
4 digit number + 5 digit number = 6 digit number. So E = 1, P = 9, N = 0
Observe R + 0 = G. But R = G not possible. 1 + R = G possible. So R and G are consecutive. G > R.
1 + I = R, So I and R are consecutive. R > I. i.e., G > R > I. and G, R, I are consecutive. Now O + T should give carry over and O + Z also give carry over. So O is bigger number. Now take values for G, R, I as 8, 7, 6 or 7, 6, 5 etc. and do trial and error.
POINT = 98504, ZERO = 3168 and ENERGY = 101672.
So E + N + E + R + G + Y = 1 + 0 + 1 + 6 + 7 +2 = 17
5. There are 1000 junior and 800 senior students in a class. And there are 60 sibling pairs where each pair has 1 junior and 1 senior.1 student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?
Sol:
Junior student = 1000
Senior student = 800
60 sibling pair = 2 x 60 = 120 student
Probability that 1 student chosen from senior = 800
Probability that 1 student chosen from junior = 1000
Therefore,1 student chosen from senior and 1 student chosen from junior
n(s) = 800 x 1000 = 800000
Two selected student are from a sibling pair
n(E) = 120C2 = 7140
Therefore
P(E) = n(E)/n(S) = 7140⁄800000
6. SEND + MORE = MONEY. Then what is the value of M + O + N + E + Y ?
Sol:
Observe the diagram. M = 1. S + 1 = a two digit number. So S = 1 and O cannot be 1 but 0. Also E and N are consecutive. Do trial and error.
So E + N + E + R + G + Y = 1 + 0 + 1 + 6 + 7 +2 = 17
5. There are 1000 junior and 800 senior students in a class. And there are 60 sibling pairs where each pair has 1 junior and 1 senior.1 student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?
Sol:
Junior student = 1000
Senior student = 800
60 sibling pair = 2 x 60 = 120 student
Probability that 1 student chosen from senior = 800
Probability that 1 student chosen from junior = 1000
Therefore,1 student chosen from senior and 1 student chosen from junior
n(s) = 800 x 1000 = 800000
Two selected student are from a sibling pair
n(E) = 120C2 = 7140
Therefore
P(E) = n(E)/n(S) = 7140⁄800000
6. SEND + MORE = MONEY. Then what is the value of M + O + N + E + Y ?
Sol:
Observe the diagram. M = 1. S + 1 = a two digit number. So S = 1 and O cannot be 1 but 0. Also E and N are consecutive. Do trial and error.
SEND = 9567, MORE = 1085, MONEY = 10652
SO M + O + N + E + Y = 1 + 0 + 6 + 5 + 2 = 14
7. A person went to shop and asked for change for 1.15 paise. But he said that he could not only give change for one rupee but also for 50p, 25p, 10p and 5p. What were the coins he had ?
Sol:
50 p : 1 coin, 25 p : 2 coins, 10 p: 1 coin, 5 p : 1 coin, Total: 1.15 p
8. 1, 1, 2, 3, 6, 7, 10, 11, ?
Sol:
The given pattern is (Prime number - consecutive numbers starting with 1).
1 = 2 – 1
1 = 3 – 2
2 = 5 – 3
3 = 7 – 4
6 = 11 – 5
7 = 13 – 6
10 = 17 – 7
11 = 19 – 8
14 = 23 – 9
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