# Aptitude Test Paper

__Aptitude Questions__**Paper-1**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.

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.

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

9. 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.

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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}

16. 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

17. 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

18. 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(x1)+56(x1) = x

31x 31 = 30x

Total number of participants x = 31

19. 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 = (8580)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.

20. 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.

21. 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

? = 30×H 211×MM

The time after H hours, hour hand and minute hand are at

MM = | 211×((30×H)±?) |

given H = 3, MM = 50

Substituting the above values in the formula

? = 8011, 28011

22. 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 = 63×68 —(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

23. 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.

24. 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(sa)×(sb)×(sc))?????????????????????

Where s = (a+b+c)2=12×(30+c)

[Assume a = 20 ,b = 10]

Now,

Check the options.

25. 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

26. 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.

27. 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

28. 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

29. 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

30. 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 )

31. 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.

32. 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

33. 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.

34. 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 escalator is 90 + 10n

Again equating 50 + 50n = 90 + 10n we get n = 1

Hence total number of steps on escalator is 100.

35. 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

**Paper-2**

- (68-a)(68-b)(68-c)(68-d)(68-e) = 725 ? Find a+b+c+d= ?

Answer : 15 ways

- 23 people are there, they are shaking hands together, How many hand shakes possible, if they are in pair of cyclic sequence ?

Answer : 22

- A bus starts at 6:00 pm. from starting point at d speed if 18m/s. reached its destination. Waited for 40 minutes. And again returned back at the speed of 28m/s. Find the time to reach the destination ?

Answer : 232

4.10 men and 10 women are there, they dance with each other, is there possibility that 2 men are dancing with same women and vice versa.

Answer : Never

5.There are different things like p,q,r,s,t,u,v. We can take p and q together. If r and s are taken together then t must has to be taken. u and v can be taken together.v can be taken with p or s. every thing can be taken together except.

Answer : T

- There are 11 boys in a family. Youngest child is a boy. Probability is 1 that of all are boys out of?

Answer : 1024

- How many parallelograms are formed by a set of 4 parallel lines intersecting on other set of 7 parallel lines ?

Answer : 126

- Remainder of 2^47/47 ?

Answer : 2

- In school there are some bicycles and 4wheeler wagons. One Tuesday there are 58 wheels in the campus. How many bicycles are there?

Answer : 7

- If A works total in 10 days and B works total in 15 Days. If A+B of cost is Rs. 5000 then what is the share of B ?

Answer : 2000

- Key words in question (Fibonacci series, infinite series, in the middle of the question one number series is there. I got the series 3 12 7 26 15 b?

Answer : 54

- There are two pipes A and B. If A filled 10 liters in hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160. If B filled in (1/16) th of a tank in 3 hours, how much time will it take to fill completely?

Answer : 7 hours

- There is a toy train that can make 10 musical sounds. It makes 2 musical sounds after being defective. What is the probability that same musical sound would be produced 5 times consecutively?

Answer : 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32

- Six friends go to pizza corner there are 2 types of pizzas. And six different flavors are there they have to select 2 flavors from 6 flavors. In how many ways we can select?

Answer : 6C2

- 3, 15, x, 51, 53,159,161. Find X

Answer : 17

- A hollow space on earth surface is to be filled. Total cost of filling is Rs20000. The cost of filling per mt3 is Rs 225 .how many times a size of 3 mt3 soil is required to fill the hollow space?

Answer : 20000/225=88.88

88.88/3=29.62 .So 30 times of 3 mt3 is required to fill the space completely.

- Given a collection of points P in the plane, a 1-set is a point in P that can be separated from the rest by a line, .i.e the point lies on one side of the line while the others lie on the other side.The number of 1-sets of P is denoted by n1(P). The minimum value of n1(P) over all configurations P of 5 points in the plane in general position(.i.e no three points in P lie on a line) is:

Answer : 5

- The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8.A certain street in nigiet contains 1000 (in base 8) buildings numbered 1 to 1000.How many 3s are used in numbering these buildings?

Answer : 192

20) Here 10 programers, type 10 lines with in 10 minutes then 60 lines can type within 60 minutes. How many programmers are needed?

Answer :10

- . 1*1!+2*2!+3*3!…………….2012*2012!= ?

Answer : 2013!-1

- After the typist writes 12 letters and addresses 12 envelopes, she inserts the letters randomly into the envelopes (1 letter per envelope). What is the probability that exactly 1 letter is inserted in an improper envelope?

Answer : 0

- Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 20, 20 and 30, the number of points equidistant from all the 3 lines is…

Answer : 0

- For the FIFA world cup, Paul the octopus has been predicting the winner of each match with amazing success. It is rumored that in a match between 2 teams A and B, Paul picks A with the same probability as As chances of winning. Lets assume such rumors to be true and that in a match between Ghana and Bolivia, Ghana the stronger team has a probability of 2/3 of winning the game. What is the probability that Paul will correctly pick the winner of the Ghana-Bolivia game?

Answer : 5/9

- 36 people {a1, a2, , a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, , {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is

Answer : 18

- A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is:

Answer : 8/15

- Hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely3. A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race?

Answer : 37.80

- If VXUPLVH is written as SURMISE, what is SHDVD ?

Answer : PEASA

- What will be the 55th word in the arrangements of the letters of the word PERFECT ?

Answer : CEPFERT

- If DDMUQZM is coded as CENTRAL then RBDJK can be coded as…..

Answer : QCEIL

- A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?

Answer : 37 1/2