TCS Aptitude Question

                                                                                           TCS Aptitude Question

                                                                                                SET-01

1.An exam was conducted and the following was analyzed. 4 men were able to check some exam papers in 8 days working 5 hours regularly.

 What is the total number of hours taken by 2 men in 20 days to check double the number of exam papers?

Ans:- Answer: 8 hours

 Assuming that 1 unit of work is done in 1 hour

Let’s calculate the total number of working hours:

=> 4 * 8 * 5 = 160 units

Now the work is doubled:

=> 160 * 2 = 320 units

Let ‘x’ be the number of hours taken by 2 men to complete the work in 20 days.

Therefore,

=> 2 * 20 * x = 320

=> x = 8 hours (Answer).

2. The numbers from 101 to 150 are written as, 101102103104105…146147148149150. What will be the remainder when this total number is divided by 3?

Ans:- Answer: 2

Solution:-

The divisibility rule for 3 is that the sum of all digits of a number should be divisible by 3. Let’s calculate the sum of the digits:

There are 50 1’s (unit place) = 50

There are 10 1’s (tens place) = 10

There are 10 2’s (tens place) = 20

There are 10 3’s (tens place) = 30

There are 10 4’s (tens place) = 40

There is one 5 (tens place) = 5

For each number 1 to 9, there are 5 sets of sum 45(1+2+…+9) = 225

=> So sum of all digits = 380

=> 380 / 3 = 2 (Answer)

3. If the alphabets are written in the sequence of a, bb, ccc, dddd, eeeee, ffffff, …. What will be the 120th letter?

Ans:- Answer: O

Solution:-

it can be seen that the letter are in AP sequence, So applying the formula we get,

n(n+1)/2 <= 120

We find that n = 15 fits the equation

The 15th letter in the English alphabet = O

So 15th term contains O.

4. There is a tank whose 1/7 th part is filled with fuel. If 22 liters of fuel is poured into the tank, the indicator rises to 1/5 th mark of the tank. So what is the total capacity of the tank?

Ans:- Answer: 385

Solution:-

Let the total capacity of the tank be ‘x’ liters.

According to the question,

=> x/7 + 22 = x/5

=> x/5 – x/7 = 22

=> x = 385 litres(Answer)

5. How many prime numbers lie between 3 and 100 (excluding the values) that satisfies the condition:

4x + 1

5y – 1

Ans:- Answer: 2

Solution:-

There are 23 prime numbers between 3 and 100 (excluding the values) of which all are odd.

For 5y – 1 to be odd, 5y must be even

For 5y to be even y should be even.

Taking y as 2 we get 5y – 1 as 9.

Now looking at all those prime numbers ending at 9 = 9, 19, 29, 39, 59, 79, 89

Out of these, the numbers satisfying both the equations(integer is assumed) are 9, 29 and 89.

We cannot consider 9 as it violates the constraint of number should be greater than 3.

Therefore answer is 2 (29 and 89)

6. In the given figure, find the ratio of area of the square to area of the triangle:

a) 3:2

b) 2:3

c) 2:1

d) 1:2

Ans:- Answer: c) 2:1

Solution:-

Let the side of the square be ‘2’ units

Area of the square = (side)^2 = 2^2 = 4 unit

Side of triangle, using Pythagoras Theorem = \sqrt{5} unit

Height of triangle = side of square (Using Pythgorean theorem)

Area of triangle = 1/2 * base * height

here, base = height = side

Area of triangle = 1/2 * (side)^2

=> 1/2 *2 * 2

=> 2 units

Therefore the ratio = 4:2

=> 2:1 (Answer)

7. There is a fairy island where lives a Knight, a Knave, and a Spy. You go there and meet three people suppose A, B, and C, one of whom is a knight, one a knave, and one a spy.

 It is known that the knight always tells the truth, the knave always lies, and the spy can either lie or tell the truth.

A says: “C is a knave.”

B says: “A is a knight.”

C says: “I am the spy.”

So who is the knight, who the knave, and who the spy?

Ans:- Answer: A = Knight, B = Spy, C = Knave

Solution:-

Let us say A is the Knight, then he speaks the truth and C is Knave who lied and finally B is Knave, who speaks the truth regarding A. So this condition holds.

Let us say B is the knight. then it contradicts the answer since a knight always speaks the truth and there cannot be two knights.

Same goes with C.

8. Find the number of perfect squares in the given series 2013, 2020, 2027, ……………., 2300? (Hint 44^2=1936)

a) 2

b) 1

c) 3

d) None of the above

Ans:- Answer: b) 1

Solution:-

We can see that the series is in the form of AP with common difference of 7.

So the series is in the form of 2013 + 7d

The hint is actually a shortcut:

44^2 = 1936

45^2 = 2025

46^2 = 2116

47^2 = 2209

48^2 = 2304

Therefore among these numbers, we need to find which of them are in the form of 2013 + 7d

Only one number 2209 can be written in the form 2013 + 7*28.

9. In the series of 7^1+7^2+7^3+7^4…….+7^204+7^205, how many numbers are there with the unit place as 3?

Ans:- Answer: 51

Solution: 

According to the cyclicity of 7, the unit digit follows the pattern of 7, 9, 3, 1 and this repeats. So in every 4 numbers, 

we get one 3 in the unit place. Dividing 205 by 4 we get 51 which is the answer to the following question.

10. Find the number of divisors of 1728(including 1 and the number itself)

Ans:- Answer: 28

Solution:-

There is a direct formula for this:

Number = p^a.q^b.r^c...

where p, q and r are prime numbers. Simply we need to prime factorize the Number.

Then, (a+1).(b+1).(c+1) is the number of divisors.

For 1728 = 2^6.3^3

Therfore, (6+1).(3+1) = 28

                                                                                                        SET-02

1. There is a set of 30 numbers. The average of first 10 numbers is equal to the average of last 20 numbers. What is the sum of last 20 numbers?

a) Twice the sum of the first ten numbers

b) Sum of first 10 numbers.

c) Twice the sum of the last ten numbers

d) Cannot be determined.

Ans:- 

a) Twice the sum of the first ten numbers

Solution:

Let the sum of the first 10 numbers is equal to ‘x’

Let the sum of the last 20 numbers is equal to ‘y’

According to the question:

x/10 = y/20

Therefore, y = 2x

2. There is a town called Metron, where wheels of the front and rear of vehicles are of different size. The measurement unit followed in the town is the metre.

 The circumference of the front wheel of the car is 133 metres and that of rear wheels is 190 metres.

So what is the distance travelled by the cart in metres when the front wheel has done nine more revolutions than the rear wheel?

a. 1330

b. 572

c. 399

d. 3990

Answer: d) 3990

Solution:

At first, we calculate the LCM of 133 and 190 which is 1330. So, the front wheels take 10 rounds to cover 1330 metres and the rear wheels take 7 rounds to cover the same. 

So to take 9 extra revolutions the vehicle would have travelled 1330 * 3 = 3990 metres.

3. Let a number ‘x’ when divided by 406 leaves a remainder 115. What will be the number when the number is divided by 29?

Answer: 28

Solution:

According to the question, the number is equal to 406x + 115.

Since 406 is divisible completely by 29, therefore any multiple of 406 that is 406x when divided by 29 leaves remainder 0. Now 115, when divided by 29, leaves remainder 28.

4. A sequence of an alpha-numeric is to be formed. The sequence consisting of two alphabets followed by two numbers is to be formed with no repetitions. In how many ways can it be formed?

a. 65000

b. 64320

c. 58500

d. 67600

Answer: c) 58500

Solution:

The first can be filled in 26 ways.

The second place can be filled in 25 ways.

The third place can be filled in 10 ways.

The last digit can be filled in 9 ways

5. According to a particular code language, A=0, B=1, C=2, …, Y=24, Z=25 then can will ONE+ONE (in the form of alphabets only) be coded?

a) DABI

b) CIDA

c) BDAI

d) ABDI

Answer: c) BDAI

Solution:

This is a 26 base question. Just like there is the Decimal system consisting of 10 digits from 0 to 9, the Base 26 system consist of 26 alphabets where A = 0, B = 1, Z = 25 and so on.

Let’s calculate, O N E + O N E

For E(4),

=> E + E

=> 4 + 4

=> 8

=> I

For N(13),

=> 13 + 13

=> 26

On converting 26 to Base 26 we get 1 0. Keeping 0(A) and taking 1 as carry

For O(14),

=> O + O + 1

=> 29

Dividing 29 by 26 we get 1(B) 3(D)

So answer is BDAI

6. In the sequence of problemsolvingproblemsolvingproblemsolving… what is the 2015th alphabet?

a) p

b) g

c) r

d) n

Answer: d) n

Solution:

‘problemsolving’ consist of 14 letters. On dividing 2015 by 14 we get 13. So the 13th letter is n and hence the answer.

7. What is the remainder when the number 101102103104105106107…148149150 is divided by 9?

Answer: 2

Solution:

The divisibility rule for 9 is that the sum of all digits of a number should be divisible by 9. Let’s calculate the sum of the digits:

There are 50 1’s (unit place) = 50

There are 10 1’s (tens place) = 10

There are 10 2’s (tens place) = 20

There are 10 3’s (tens place) = 30

There are 10 4’s (tens place) = 40

There is one 5 (tens place) = 5

For each number 1 to 9, there are 5 sets of sum 45(1+2+…+9) = 225

=> So sum of all digits = 380

=> 380 / 9 = 2 (Answer)

8. 30 litres of 78% of a concentrated acid solution is to be prepared. How many litres of 90% concentrated acid needs to be mixed with 75% solution of concentrated acid to get the result?

a) 10

b) 6

c) 3

d) 4

Answer: b) 6

Solution:

Let’s apply the weighted-average formula.

Let there be n1 litre of 90% acid solution and n2 litre of 75% solution

Therefore,

=> 78 = ((90 * n1) + (75 * n2))/(n1 + n2)

We get,

=> n1/n2 = 1 / 4

So 30 litres needed to be divided in the ratio of 1:4, which gives us 6 litre as the answer.

9. Ram will be 32 years old in eight years from now. In 4 years, Ram’s fathers age will be twic as Ram’s age and two years ago, his mother’s age will be twice as his age. What will be the present age of Ram’s father and mother?

Answer: Father's age = 52, Mother's age = 46

Solution:

Ram will be 32 years old in next 8 years. So his present age is 32 – 8 = 24 years old

After 4 years Ram will be 28 years old. So his father will be 28 * 2 = 56 years old.

Therefore, fathers present age is 56 – 4 = 52 years old

Two years ago Ram was 22 years old. So his mothers age the was 22 * 2 = 44 years old

Therefore mothers present age is 44 + 2 = 46 years old.

10. In a class, the number of boys is equal to the number of girls. What was the total number of students if twice the number of boys as girls remain when 12 girls entered out?

Answer: 48

Solution:

Let ‘b’ be the number of boys and ‘g’ be the number of girls. According to the question:

=> b / (g – 12) = 2 / 1

Since b = g;

we get g = 24.

So the total number of students = 24 + 24 = 48

                                                                                                          SET-03

1. If f(x) = ax^4 – bx^2 + x + 5 and given f(-3) = 2, then f(3) = ? (a^b = a raised to power b)

a) 3

b) 8

c) 1

d) -2

Answer: b) 8

Solution:

We can directly solve:

=> f(-3) = a(-3)^4 – b(-3)^2 + (-3) + 5 = 2

=> 81a – 9b + 2 = 2

=> 81a – 9b = 0

Now solving f(3),

=> f(3) = 81a – 9b + 8

=> f(3) = 0 + 8 = 8(Answer)

2. There is a chocolate factory which distributes chocolates to a class. It supplies chocolates to a class of 50 students for 30 days, 

keeping in mind that all students get an equal number of chocolates. For the first 10 days, only 20 students were present. 

How many students be accommodated into the group so that all the chocolates get consumed?

a) 70

b) 55

c) 60

d) 45

Answer: d) 45

Solution:

Let each student get 1 chocolate each, so the total number of chocolates = 50 * 30 = 1500 chocolates.

For first 10 days 20 students were present, so total chocolates consumed = 20 * 10 = 200 chocolates.

Chocolates left = 1300. These are to be distributed for the next 20 days. Therefore in each day 1300 / 20 chocolates were to be consumed which = 65 chocolates per day.

So the required answer = 65 – 20 = 45 chocolates.

3. Given, log(0.318) = 0.3364 and log(0.317) = 0.3332, find log(0.319)?

a) 0.3396

b) 0.3394

c) 0.3393

d) 0..390

Answer: a) 0.3396

Solution:

=> log(0.319) = log(0.318) + (log(0.318) – log(0.317))

= 0.3364 + (0.3364 – 0.3332)

= 0.3364 + 0.0032

= 0.3396 (Answer)

4. There are a set of 20 students out of which 18 are boys and 2 are girls. They are to be seated in a circular manner so that the two girls are always separated by a boy.

 In how many ways can the students be arranged?

a) 12

b) 18!x2

c) 17×2!

d) 17!

Answer: b) 18!x2

Solution:

There are in all 20 places out of which if one girl sits in one position then the other girl may sit either to her left or right skipping one place,

which is to be filled by a boy. So total number of ways the boys can sit = 18! ways and girls may alternate there sits so the total answer would be = 18! * 2 ways.

5. Ram appears for an exam. In paper A he scores 18 out of 70. In paper B he scores 14 out of 30. So in which paper did he perform better?

a) Paper A

b) Paper B

Answer: b) Paper B

Solution:

We just need to calculate the percentage he scored in each paper.

In paper A: (18/70) * 100 = 25.7%

In paper B: (14/30) * 100 = 46.6% (Answer)

6. A flight takes off at 2 a.m. from a place at 18N 10E and landed at 36N 70W, 10 hours later. What is the local time of the destination?

a) 6:00 a.m.

b) 6:40 a.m.

c) 7:40 a.m.

d) 7:00 a.m.

e) 8:00 a.m.

Answer: b) 6:40 a.m 

Solution:

Let’ calculate the difference in the number of latitudes = 70 + 10 = 80 degrees towards east.

We know 1 degree = 4 min, so 80 degrees = 80 * 4 = 320 mins

320 mins = 5 hr 20 minutes

Now the plane landed 10 hours later so time of landing = 12 hrs according to the starting place

So time at destination = 12 hrs – 5 hrs 20 min = 6 hr 40 mins(Answer)

7. An athletic run at 9 km/hr along a railway track. The track is 240m long and ahead of a train 120m long running at 45km/hr, in the same direction. how much time will the train take to completely cross the athlete?

a) 3.6 sec

b) 18 sec

c) 72 sec

d) 36 sec

Answer: d) 36 sec 

Solution:

Let’s try to find the relative speed = 45 – 9 = 36km/hr

= 36 * 5/18 = 10m/s

Now the total distance needed to be covered by the train to completely cross the athlete = 240 + 120 = 360m

So time = dist/speed = 360/10 = 36 seconds

8. A takes 3 days to complete a work while B takes 2 days. Both of them finish a work and earn Rs. 150. What is A’s share of money?

a) Rs. 70

b) Rs. 30

c) Rs. 60

d) Rs. 75

Answer: c) 60

Solution:

A completes 1/3rd of work in one-day and B completes 1/2 of work in one day. So the ratio of there work is:

A:B = 2:3

So A’s share = (2/5)*150 = 60 rupees(Answer)

9. Ram and Shyam salaries are in the ratio of 2:3. If both of there salaries are increased by Rs 4000 each, the new ratio becomes 40:57. What is Shyam’s present salary?

a) Rs. 17, 000

b) Rs. 20, 000

c) Rs. 25, 500

d) None of these

Answer: d) None of these

Solution:

Let Rams and Shyams salary be ‘x’. The ratio of there salary according to the question is 2x:3y

According to the question,

(2x+4000):(3x+4000) = 40:57

On solving we get 3x = 34000

Therefore, the present salary of Shyam is Rs. 34000

10. At what rate per cent per annum will the SI on a sum of money be 2/5 of the amount in 10 years?

a) 6%

b) 5 2/3 %

c) 4%

d) 6 2/3 %

Answer: c) 4%

Solution:

Let the sum of money be Rs ‘x’. So SI = 2x/5

So, rate = (SI*100)/(P*Time)

=> (2x*100)/(5*x*10)

=> 4 % (Answer)

                                                                                                        SET-04

1. A call centre agent has a list of 305 phone numbers of people in alphabetical order of names, but Anuj does not have any of the names. 

He needs to quickly contact Danish Mank to convey a message to him. If each call takes 2 minutes to complete, and every call is answered,

 what is the minimum amount of time during which he can guarantee to deliver the message to Danish? 

a) 206 minutes 

b) 610 minutes 

c) 18 minutes 

d) 34 minutes 

 

Answer: c) 18 minutes

Solution: 

We need to search for a particular name in a phone book. So we need to apply a method in which we can easily search a number in a minimum count.

 So we divide the list into two equal halves, i.e., 305/2 = 152.5 or let’s take 152.

 Now we can decide whether to check for Danish in the upper or lower half of 152. This is decided by the starting letter of the name in a page. Proceeding in the similar manner we get, 

152/2 = 76 

76/2 = 38 

38/2 = 19 

19/2 = 9 

9/2 = 4 

4/2 = 2 

2/2 = 0 

So we get 0 at the 9th time, hence this is the minimum number of the count to find Danish. So total time taken = 9 * 2 = 18 minutes.

2. There is an office consisting of 38 people. 10 of them like to play golf, 15 like to play football and 20 neither play golf nor football. How many like both golf and football? 

     a) 10 

     b) 7 

     c) 15 

     d) 18 

Answer: b) 7

Solution: 

Let the number of people liking golf = ‘A’ 

Let the number of people liking football = ‘B’ 

Let the number of people liking either golf or football = A U B = 38 – 20 = 18 

People liking both golf and football = 

A \cap B$   

= A + B – AUB = 10 + 15 -18 = 7

3. If a dice is rolled 2 times, what is the probability of getting a pair of numbers with sum equal to 3 or 4? 

     a) 6/36 

     b) 5/36 

     c) 1/9 

     d) 1/12 

Answer: b) 5/36

Solution: 

Total probability = 36 

We can get a sum of 3 or 4 in this many ways: 

=> (2, 1), (1, 2), (1, 3), (3, 1), (2, 2) = 5 

So probability = 5 / 36

4. A shopkeeper charges 12 rupees for a bunch of cakes. Anuj bargained to the shopkeeper and got two extra ones, and that made them cost one rupee for dozen less than first asking price.

 How many cakes did Anuj receive in 12 rupees? 

    a) 10 

    b) 14 

    c) 18 

    d) 16 

Answer: d) 16

Solution: 

Let the number of cakes = ‘x’ or ‘x/12’ dozen 

So, x/12 cost Anuj 12 rupees, or 1 dozen cost him = 144/x rupees 

Now, he gets two extra = 144/(x+2) in 1 rupees less, 

=> 144/x – 144/(x+2) = 1 

=> On putting 16, the equation is satisfied, hence the answer.

5. Ram alone can do 1/4th of the work in 2 days. Shyam alone can do 2/3th of the work in 4 days. So what part of the work must be done by Anil in 2 days, 

for them to complete the work together in 3 days? 

   a) 1/8 

   b) 1/20 

   c) 1/16 

   d) 1/12 

   Answer: d) 1/12 

Solution: 

Ram alone can complete the work in 2*4 = 8 days. 

Shyam alone can complete the work in 4*(3/2) = 6 days. 

Taking the lcm of 8, 6, 3 = 24 

Capacity of Ram = 24/8 = 3 

Capacity of Shyam = 24/6 = 4 

Capacity of Anil = 8 – (4+3) = 1 

Now in 2 days Anil can do 2 unit of work = 2/24 = 1/12 part of the work

6. Mr Mehta chooses a number and keeps on doubling the number followed by subtracting one from it. If he chooses 3 as the initial number and he repeats the operation 30 times then what is the final result? 

   a) (2^30) – 1 

   b) (2^30) – 2 

   c) (2^31) – 1 

   d) None of these 

Answer: d) None of these

Solution: 

According to the question, 

3 * 2 – 1 = 5 = 

2^2+1$   

5 * 2 – 1 = 9 = 

2^3+1$   

9 * 2 – 1 = 17 = 

2^4+1$   

Proceeding in the similar fashion, on 30 times we get 

2^{31}+1$  

7. Ram alone can paint a wall in 7 days and his friend Roy alone paints the same wall in 9 days. In how many days they can paint the wall working together? (Round off your answer) 

   a) 3 

   b) 5 

   c) 4 

   d) 7 

Answer: c) 4

Solution: 

This can be solved by applying a simple formula = ab/(a+b) 

or, (9*7)/(9+7) 

or, 63/16 = 3.9375 = 4 (answer)

8. Two vertical walls of the length of 6 meters and 11 meters are at a distance of 12 meters apart. Find the top distance of both walls? 

   a) 15 meters 

   b) 13 meters 

   c) 12 meters 

   d) 10 meters 

Answer: b) 13 meters

Solution: 

Let’s consider this figure, 

 We need to find the distance of AB, 

We know AC = 12 m and BC = 11-6 = 5 m 

So applying pythagoras theorem we get, 

AB = 

$\sqrt(12^2 + 5^2)$   

= 13 metres

9. For f(m, n) =45*m + 36*n, where m and n are integers (either positive or negative). What is the minimum positive value for f(m, n) for all values of m, n (this may be achieved for various values of m and n)? 

   a) 18 

   b) 12 

   c) 9 

   d) 16 

Answer: c) 9

Solution: 

To get the minimum value of f(m, n), put m = 1 and n = -1, we get 

f(, n) = 9

10. A white cube(with six faces) is to be painted blue on two different faces. In how many different ways can this be achieved 

(two paintings are considered same if on a suitable rotation of the cube one painting can be carried to the other)? 

   a) 30 ways 

   b) 18 ways 

   c) 4 ways 

   d) 2 ways 

Answer: d) 2

Solution: 

This can be achiededv in the following different ways;: 

First, painting on opposite faces can be achieved in 1 way. 

Second, painting on adjacent faces can be achieved in 1 way. 

Therefore in 2 ways.

                                                                                                             SET-05

1. Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?

a) 980

b) 618

c) 955

d) 797

Answer: d) 797

Solution:

Let the number added to 5678 be x to give a remainder 35 and quotient k when divided by 460.

So, 5678 + x = 460k + 35

or, 5643 + x = 460k

So 5643 + x must be divisible by 460

Ananysing from the options, we get on adding 797 to 5643, the number 6440 is divisible by 460.

2. Rahaman went to a stationery shop and bought 18 pencils for Rs.100. He paid 1 rupee more for each grey pencil than for each black pencil. What is the price of a grey pencil and how many grey pencils did he buy?

a) Rs.5, 10

b) Rs.6, 10

c) Rs.5, 8

d) Rs.6, 8

Answer: b) Rs. 6, 10

Solution:

The best way is to analyse from the mentions.

Let’s take option b in which 10 pencils are bought at Rs.6 each. So total cost of grey pencils = 6 * 10 = Rs.60. So Rahaman is left with 40 rupees. He buys 8 black pencils at Rs 5 each which is 1 rupee less than what he had spent in buying the grey ones. Thus satisfying the conditions.

3. Four people each roll a four die once. Find the probability that at least two people will roll the same number?

a. 13/18

b. 5/18

c. None of the above

d. 1295/1296

Answer: a) 13/18 ways

Solution:

Total possible outcomes = $6^4$ = 1296

Number of ways in which no two people get same number = 6*5*4*3 = 360 ways

The probability of no two people getting the same number = 360 / 1296 ways = 5/18 ways

So the probability of at least two people getting the same number = 1 – 5/18 = 13/18 ways

4. Ram said Shyam “If you give me half your money I will have Rs.75.” Shyam said, “If you give me one-third of your money, I will have Rs.75 How much money did Shyam have4

c) 48

d) 60

Answer: d) 60

Solution: 

Let Ram and Shyam be denoted by ‘R’ and ‘S’ respectively

According to the question,

Eqn 1. R + S/2 = 75

Eqn 2. R/3 + S = 75

Therefore, solving both the equations we get, R = 45 and S = 60.

5. Ram goes to the market to buy apples. If he can bargain and reduce the price per apple by Rs.2, he can buy 30 apples instead of 20 apples with the money he has. How much money does he have?

a. Rs.100

b. Rs.50

c. Rs.120

d. Rs.150

Answer: c) 120

Solution: 

Let the price per orange be Rs. x.

So total money Ram has in buying at original price = 20x.

On reducing the price by 2 rupees each the total money must be (x-2)*30

According to the question,

20x = (x-2)*30

On solving this we get x = 6 or the total money = Rs. 120

6. How many positive integers less than 500 can be formed using the numbers 1, 2, 3, and 5 for digits, each digit being used only once.

a) 68

b) 34

c) 66

d) 52

Answer: b) 34

Solution: 

Single digit numbers can be formed in 4 ways.

2 digit number can be formed in 4 * 3 = 12 ways

3 digit number less than 500 can be formed in 3 * 3 * 2 = 18 ways.

Total number of ways = 18 + 12 + 4 = 34 ways

7. A boy entered a shop and bought x number of books for y rupees. When he was about to leave the bookkeeper said, “if you buy 10 more books, you can have all the books for 2 rupees and you will also save 80 cents a dozen”. So what are x and y?

a) (5, 1)

b) (10, 1)

c) (15, 1)

d) Cannot be determined.

Answer: a) (5, 1)

Solution:

x number of books cost him y rupees.

So, 1 book will cost him y/x rupees.

12 books will cost him rupees 12 y/x.

The shopkeeper says,

x + 10 books cost him 12 rupees

1 book will cost him 12/(x+10) rupees

12 books will cost him 24/(x+10) rupees

We know that 80 cents = 4/5 of a dollar,

So, 12y/x – 24/(10+x) = 4/5

Analysing the given choices, we get (5, 1) satisfies the equation.

8. The perimeter of an equilateral triangle is equal to a regular hexagon. Find out the ratio of their areas?

a. 3:2

b. 1:6

c. 2:3

d. 6:1

Answer: c) 2:3

Solution:

Let the side of the equilateral triangle be a unit and that of the regular hexagon be b unit.

So perimeter of the triangle = 3a and perimeter of the hexagon is 6b unit.

or, 3a = 6b

or a/b = 2/1

The area of the equilateral tr

iangle = $\sqrt(3)a^2/4$

The area of the regular hexagon = $3 * \sqrt(3)a^2/2$

or, $\sqrt(3)a^2/4$ : $3 * \sqr  t(3)a^2/2$

Solving this and subsituting a/b we get the answer as 2 : 3

9. In the given series: 70, 54, 45, 41……. What will be the next number?

a) 40

b) 36

c) 35

d) 38

Answer: a) 40

Solution:

The series goes like:

70 – 54 = 16 (4^2)

54 – 45 = 9 (3^2)

45 – 41 = 4 (2^2)

41 – 40 = 1 (1^1)

10. A series of story books were published at an interval of seven years. When the seventh book was published the total sum of the publication year was 13524. In which year was the first book published?

a) 1910

b) 1911

c) 2002

d) 1932

Answer: b) 1911

Solution:

We get the series of publications as n, n+7, n+14, n+21, n+28, n+35, n+42.

Sum of publications = 13524 = 7/2[2n + (7-1)*7] (Using the sum of AP formula)

We get, n = 1911 (answer)

                                                                                                            SET-06

1. Crusoe hatched from a mysterious egg discovered by Angus was growing at a fast pace that Angus had to move it from home to the lake.

Given the weights of Crusoe in its first weeks of birth as 5, 15, 30, 135, 405, 1215, 3645. Find the odd weight out.

a) 3645

b) 135

c) 15

d) 30

Answer: d) 30

Solution:

Looking at the series closely we find that the 3rd number is oddly placed.

The series is in the form:

5 * 3 = 15

15 * 3 = 45

45 * 3 = 135

135 * 3 = 405 and so on

2. Assume that f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1). For all natural numbers (Integers>0)m and n. What is the value of f(17)?

a) 5436

b) 4831

c) 5508

d) 4832

Answer: d) 4832

Solution:

We need to use f(1) to calculate the value of f(17)

f(17) can be written as f(1+16)

f(16) can be written as f(8+8)

f(8) can be written as f(4+4)

f(4) can be written as f(2+2)

f(2) can be written as f(1+1)

f(1) = 0, so f(2) = f(1+1) = f(1)+f(1)+4(9*1*1-1) = 32.

or, f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204.

or, f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980

or, f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260

or, f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832

3. A sum of Rs.3000 is distributed among P, Q, and R. P gets 2/3 of what Q and R got together and R gets 1/3 of what P and Q got together, R’s share is?

a) 750

b) 850

c) 800

d) 700

Answer: a) 750

Solution:

According to the question,

case 1: P = 2(Q + R)/3

or, (Q+R)/P = 3/2

case 2: Also, R = (P+Q)/3

or, (P+Q)/R = 3/1

Simply using componendo-dividendo, we get,

for case 1, (P+Q+R)/P = 3+2/2 = 5/2 = 20/8

for case 2, (P+Q+R)/R = 3+1/1 = 4/1 = 20/5

On solving we get, P = 8, Q = 7, R = 5

or R’s share = 5/(8+7+5) * 3000 = 750

4. In the given series 11, 23, 47, 83, 131, … What is the next number?

a) 145

b) 178

c) 191

d) 176

Answer: c) 191

Solution:

The given series follows the order of multiple of 12

23 – 11 = 12

47 – 23 = 24

83 – 47 = 36

131 – 83 = 48

x – 131 = 60

or x = 191

5. If a number is divided by 357 the remainder is 5, what will be the remainder if the number is divided by 17?

a) 9

b) 3

c) 7

d) 5

Answer: d) 5

 

Solution:

Let the number be N when divided by 357 leaves remainder 5 and quotient q.

So, N = 357k + 5 = 17 * 21 * k + 5

So, 357 is exactly divisible by 17 so remainder is 5

6. A pole of height 36m is on one edge of a road broke at a certain height. It fell in such a way that the top of the pole touches the other edge of the road. 

If the breadth of the road is 12m, then what is the height at which the pole broke?

a) 12

b) 16

c) 24

d) 18

Answer: b) 16

Solution:

Let the point at which the pole broke be ‘x’ from the ground, so the length of the broken piece be (36-x).

So applying Pythagoras theorem we get, $(36-x)^2 = x^2 + 12^2$

=> $1296 + x^2 - 72x = x^2 + 144$

=> 72x = 1296 – 144

=> x = 16

7. There is a hall consisting of 23 people. They are shaking hands together. So how many hands shakes possible if they are in a pair of cyclic sequence?

a) 23

b) 22

c) 253

d) 250

Answer: c) 253

Solution:

Since there are 23 people, number of handshakes possible = 23C2 = 253 handshakes.

8. In a basement, there are some bicycles and cars. On Tuesday there are 182 wheels in the basement. How many bicycles are there?

a) 20

b) 19

c) 18

d) 16

Answer: b) 19

Solution:

This is a very ambiguous question and must be calculated using the options.

If there are 20 bicycles, there must be 20*2 = 40 wheels

Remaining wheels = 182-40 = 142 wheels = 142/4 is not an integer so there cannot be 20 bicycles.

Similarly checking for 19 bicycles = 19*2 = 38 wheels

Remaining wheels = 182 – 38 = 144 = 144/4 = 36 cars hence this is the answer.

9. There is a rectangular ground 17 × 8 m surrounded by a 1.5 m width path. The depth of the path is 12 cm. Sand is filled and find the quantity of sand required.

a) 5.5

b) 10.08

c) 6.05

d) 7.05

Answer: b) 10.08

Solution:

Area of the inner rectangle = 17 * 8 = 136 meter-square

Area of the outer rectangle = (17 + 2*1.5) * (8 * 2*1.5) = 220 meter-square

So area of the remaining path = 220 – 136 = 84 meter-square

So sand required to fill the path = 84 * (12/100) = 10.08 meter-square

10. The numbers 272738 and 232342, when divided by n, a two digit number, leave a remainder of 13 and 17 respectively. Find the sum of the digits of n?

a) 5

b) 4

c) 7

d) 8

Answer: c) 7

Solution:

So according to the question, (272738 – 13) and (232342 – 17) are exactly divisible by n.

So if we find the HCF of these two numbers, we get n,

The HCF of 272725 and 232325 is 25

So the sum of the digits = 7.

                                                                                                    SET-07

1. The sticks of the same length are used to form a triangle as shown below. If 87 such sticks are used then how many triangles can be formed?

a) 42

b) 43

c) 44

d) 45

Answer: b) 43

Solution:

As we can see the first triangle can be formed using 3 sticks. So we have 87 – 3 = 84 sticks left.

So every next triangle can be formed using 2 sticks.

So we have 84/2 = 42 triangles and 43 triangles in all.

2. Find the next number in the series of 3, 12, 7, 26, 15, ?

a) 54

b) 55

c) 64

d) 74

Answer: a) 54

Solution:

3 * 2 + 1 = 7

12 * 2 + 2 = 26

7 * 2 + 1 = 15

26 * 2 + 2 = 54

3. There is a toy gun that made 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?

a) 1/16

b) 1/32

c) 1/48

d) 1/2

Answer: b) 1/32

Solution:

The probability of making the same sound every time = 1/2,

So, 1/2^5 = 1/32 (answer)

4. In how many possible ways you can write 3240 as a product of 3 positive integers?

a) 320

b) 420

c) 350

d) 450

Answer: d) 450 ways

Solution:

First let’s prime factorize the number 3240 = $2^3 * 3^4 * 5^1$

Let the three positive numbers be x, y and z

We have to distribute three 2’s to x, y and z ways in (3+3-1)C(3-1) = 5C2 ways = 10 ways

We have to distribute four 3’s to x, y, z in (3+4-1)C(3-1) = 6C2 ways

We have to distribute one 5 to x, y, z in 3 ways.

The total number of ways = 10×15×3=450 ways.

5. The marked price of a shirt was 40% less than the suggested retail price. Ram purchased the coat for half of the marked price at the 15th-anniversary sale.

 What per cent less than the suggested retail price did Ram pay?

a) 70%

b) 20%

c) 60%

d) 30%

Answer: a) 70%

Solution:

Let the retail price of the shirt be Rs. 100

So according to the question, the market price will be = 100*0.6 = 60

Purchased price of Ram = 60/2 = 30

which is 70% less than retail price.

6. HCF of 2472, 1284 and a 3rd number, is 12. If their LCM is 8*9*5*103*107, then what is the number?

a) 2^2*3^2*7^1

b) 2^2*3^2*5^1

c) 2^2*3^2*8103

d) None of the above.

Answer: b) 2^2×3^2×5^1

Solution:

2472 = $2^3*3*103$

1284 = $2^2*3*107$

HCF = $2^2*3$

LCM = $2^3*3^2*5*103*107$

HCF of the number is the highest number which divides all the numbers. So N should be a multiple of 22×3

LCM is the largest number that is divided by the given numbers. As LCM contains 32×5 these two are from N.

So N = [Tex]$2^2×3^2×5^1$[Tex]

7. An old man takes 30 minutes and a young man takes 20 minutes to walk from apartment to office. If one day the old man started at 10.00 AM 

and the young man at 10:05 AM from the apartment to office, when will they meet?

a) 10:00

b) 10:15

c) 10.30

d) 10:45

Answer: b) 10:15

Solution:

Let the distance of the apartment from the office be 12 km

So the speed of the old man = 12 / (1/2) hr = 24 km/hr

The young man speed = 12 / (1/3) hr = 36 km/hr

Since the old man started 5 minutes earlier, he covers 24 × (5/60) = 2 km in 5 minutes.

Now the time taken to the young man to meets him = 2/(36-24) * 60 = 10 minutes

So the time of meet = 10:05 + 10 = 10 hr 15 min or 10:15

8. In the range of 112 to 375, how many 2’s are there?

a) 312

b) 156

c) 159

d) 160

Answer: b) 156

Solution:

The total number of 2’s in the units place = (122, 132, 142 … 192), (201, 212, 222, … 292), (302, 312, … 372) = 8 + 10 + 8 = 26 2’s

The total number of 2’s in tenth’s place = (120, 121, 122, …, 129) + (220, 221, …, 229) + (320, 321, …, 329) = 30

The total number of 2’s in hundred’s place = (200, 201, … 299) = 100.

So the total number of 2’s between 112 and 375 = 26 + 30 + 100 = 156

9. Ram walks 36 km partly at a speed of 4 km/hr and partly at 3 km/hr. If he had walked at a speed of 3km/hr when he had walked at 4 and 4 km/hr 

when he had walked at 3 he would have walked only 34 km. The time (in hours) spent by Ram in walking was

a) 10

b) 5

c) 12

d) 8

Answer: a) 10

Solution:

Let Ram walk ‘x’ hrs at 4 km/hr, and ‘y’ hrs at 3 km/hr.

Given,

4x + 3y = 36

3x + 4y = 34

Solving these two equations we get x + y = 10

10. What will be the 55th word in the arrangement of the letters of the word PERFECT?

a) CEPFRET

b) CEPFERT

c) CEPERFT

d) CEPRFET

Answer: b) CEPFERT

Solution:

Let’s arrange the word PERFECT in dictionary order = CEEFPRT

Here,

CEE(4!)=24

CEF(4!)=24

CEPF(3!)=6

So the 55th word is CEPFERT.

                                                                                                        SET-08

1. Identify the missing number in the series: 2, 5, __, 19, 37, 75?

a) 16

b) 12

c) 10

d) 9

Answer: d) 9

Solution:

2 * 2 + 1 = 5

5 * 2 – 1 = 9

9 * 2 + 1 = 19

19 * 2 – 1 = 37 and so on

2.A rectangle is divided into four rectangles with area 70, 36, 20, and x. What is the value of ‘x’?

a) 350/7

b) 350/11

c) 350/9

d) 350/13

Answer: c) 350/9

Solution:

Since the areas of the rectangles are in proportion we can say,

=> 70/x = 36/20

=> x = 350/9

3. If VXUPLVH is written as SURMISE, what is SHDVD written as?

a) PEASA

b) PBASA

c) PEBSB

d) None of the above

Answer: a) PEASA

Solution:

It is a question of coding-decoding where,

V is written as S (V – 3 = S)

X is written as U (X – 3 = U)

and so on.

Similarly, SHDVD will be written as PEASA

4. Aman owes Bipul Rs 50. He agrees to pay Bipul over a number of the consecutive day starting on Monday, paying a single note of Rs 10 or Rs 20 on each day. 

In how many different ways can Aman repay Bipul. (Two ways are said to be different if at least one day, a note of a different denomination is given)

a) 5

b) 6

c) 7

d) 8

Answer: d) 8

Solution:

Aman can pay Bipul in all 10 rupees note in 5 days = 5 * 10 = 50 rupees = 1 way

Aman can pay Bipul in 3 ten rupee note and 1 twenty rupee note = 4!/(3! * 1!) = 4 ways

Aman can pay Bipul in 1 ten rupee note and 2 twenty rupee note = 3!/(1! * 2!) = 3 ways

So in all Aman can pay Bipul in 8 ways.

5. Salim bought a certain number of oranges at a rate of 27 oranges for rupees 2 times M, where M is an integer. 

He divided these oranges into two equal halves, one part of which he sold at the rate of 13 oranges for Rs M and the other at the rate of 14 oranges for Rs M.

 He spent and received an integral no of rupees, but bought the least number of oranges. How many did he buy?

a) 980

b) 9828

c) 1880

d) 102660

Answer: b) 9828

Solution:

Let Salim buy 2x number of oranges.

So he buys 27 oranges at a price of 2M.

He buys 1 orange at a price of 2M/27

or, x oranges cost him Rs. 2Mx/27

Now he sells x oranges at the rate of 13 oranges for Rs. M

So he sells 1 orange at Rs. M/13

and x oranges at Rs Mx/13

Same goes for 14 oranges which are Mx/14,

According to the question, 2Mx/27, Mx/13, Mx/14 are integers

So, x oranges must be divisible by 27, 13 and 14

The lcm of 27, 13 and 14 = 4914 or 2x = 9828

6. In a football match, 16 teams participate and are divided into 4 groups. Every team from each group will play with each other once. 

The top 2 winning teams will move to the next round and so on the top two teams will play the final match. So how many minimum matches will be played in that tournament?

a) 40

b) 14

c) 43

d) 50

Answer: c) 43

Solution:

Total matches to be played = 4C2 = 6 matches.

So total number of matches played in the first round = 6 * 4 = 24 matches

Now top two teams from each group progress to the next round. These 8 teams are to be divided into 2 groups.

Total matches played in the second round = 6 × 2 = 12 matches

Now 4 teams progress to the next round. Total matches played in the third round = 6 * 1 = 6matches

From this round, 2 teams progress to the next round. And final will be played between them.

Total matches = 24 + 12 + 6 + 1 = 43

7. There are 12 letters and exactly 12 envelopes. There is one letter to be inserted randomly into each envelope. What is the probability that exactly 1 letter is inserted in an improper envelope?

a) 1

b) 0

c) 10!

d) None of these

Answer: b) 0

Solution:

This is a question of very common sense in which,

12 letters are to be inserted in 12 envelopes, 1 in each, so if one letter is inserted into a wrong envelope there has to be another letter 

which is to be inserted into another wrong envelope. So the probability of this is 0.

8. A hollow space on the earth surface is to be filled. The total cost of filling is Rs. 20000. The cost of filling per cubic-meter is Rs 225. 

How many times is a size of 3 cubic-meter soil required to fill the hollow space?

a) 29.62

b) 30.32

c) 88.88

d) 43.64

Answer: a) 29.62

Solution:

The total cost of filling = 20, 000

Cost of filling 1 cubic meter = Rs. 225

So cubic meters to be filled = 20, 000/225 = 88.89 meter-cube

Now we need to find the three times of 88.89 to be filled = 88.89/3 = 29.63

So the closest match is 29.62

9. A 7-digit number is to be formed with all different digits. If the digits at the extreme right and extreme left are fixed to 5 and 6 respectively, find how many such numbers can be formed?

a) 120

b) 30240

c) 6720

d) None of these

Answer: c) 6720

Solution:

If the digits at extreme left and right are fixed as 5 and 6, then the number of digits left = 8

So the in-between 5 places can be filled in 8 * 7 * 6 * 5 * 4 ways

= 6720 ways

10. There are five tires in a sedan (four road tires and one spare) which is to be used equally in a journey to travel 40, 000 km. The number of km of use of each tyre was

a) 32000

b) 8000

c) 4000

d) 10000

Answer: a) 32000

Solution:

The total km travelled by the sedan = 40, 000 km

Since every tire capacity’s = 40, 000/5 = 8000 km each

So total distance covered by each tire = 8000*4 = 32000 km each will be travelled by each tire after being worn out after every 8000 km.

                                                                                                              SET-09

1. When a + b is divided by 12 the remainder is 8, and when a – b is divided by 12 the remainder is 6. If a > b, what is the remainder when ab divided by 6?

a) 3

b) 1

c) 5

d) 4

Answer: b) 1

Solution:

According to the question,

a + b = 12k + 8

=> $(a+b)^2 = 144k^2 + 64 + 192k$

a – b = 12l + 6

=> $(a-b)^2 = 144l^2 + 36 - 144l$

Subtracting both the equations we get,

ab = $36(k^2 - l^2) + 48k - 36l + 7$

Now all the terms of ab is divisible by 6, except 7. So the remainder left is 1.

2. There is a set of 26 questions. For each wrong answer, five marks were deducted and eight points were added for each correct answer. 

Assuming that all of the questions were answered, and the score was 0, how many questions were answered correctly?

a) 12

b) 10

c) 11

d) 13

Answer: b) 10

Solution:

This can be easily solved using hit and trial method.

Let’s consider the first option. If 12 questions in all are answered correctly, then the total score = 12 * 8 = 96 marks.

If 12 questions are answered correctly, then 14 questions were wrongly answered. So total deductions = 14 * 5 = 70 marks.

So total score = 96 – 70 = 26 which is not correct.

Let’s consider the second option. If 10 questions in all are answered correctly, then the total score = 10 * 8 = 80 marks.

If 10 questions are answered correctly, then 16 questions were wrongly answered. So total deductions = 16 * 5 = 80 marks.

So total score = 80 – 80 = 0

Hence 10 is the correct option.

3. One day, Ramesh started 30 minutes late from home and driving at 25% slower than the usual speed, reached the market 50 minutes late.

 How much time in minutes does Ramesh usually take to reach the market from home?

a) 20

b) 40

c) 60

d) 80

Answer: a) 60

Solution:

Let the usual speed of Ramesh be ‘s’

Let the distance between home and market be ‘d’

So usual time took = d/s

Time took on that particular day = d/(3s/4)

So according to the question,

d/s(4/3 – 1) = 20

or, d/s = 60

4. Three containers A, B and C are having mixtures of milk and water in the ratio of 1:5, 3:5, 5:7 respectively. If the capacities of the containers are in the ratio 5:4:5, 

find the ratio of milk to water, if all the three containers are mixed together.

a) 54:115

b) 53:113

c) 53:115

d) 54:113

Answer: c) 53:115

Solution:

Using the weighted average formula we can calculate the weight of milk,

=> [5*(1/6) + 4*(3/8) + 5*(5/12)]/(5+4+5) = 53/168

So weight of water = 168 – 53 = 115

So the ratio of milk to water = 53:115

5. Aman participates in an orange race. In the race, 20 oranges are placed in a line of intervals of 4 meters with the first orange 24 meters from the starting point.

 Aman is required to bring the oranges back to the starting place one at a time. How far would he run in bringing back all the oranges?

a) 1440

b) 2440

c) 1240

d) 2480

Answer: d) 2480 

Solution:

Since every orange is placed at a difference of 4 meters and the first potato is placed at 24 meters from the starting position. Every orange is placed at 24m, 28m, 32m, 36m, ….20 terms.

Now to bring ever orange one at a time, Aman needs to cover the double of the distance = 48, 56, 64, …20 terms.

So putting the values in the sum of AP formula, a = 48, d= 8, n = 20.

Total distance travelled = 20/2 [2 * 48 + (20-1)*8]

= 2480 meters

6. There are two decks of cards each deck containing 20 cards, with numbers from 1 to 20 written on them. A card is drawn at random from each deck, getting the numbers x and y 

What is the probability that log x + log y is a positive integer. (Logs are taken to the base 10.)

a) 7/400

b) 29/100

c) 3/200

d) 1/80

Answer: a) 7/400

Solution:

We know that log x + log y = log xy

for log xy to be positive, we have the following choices:

(1, 10), (10, 1), (10, 10), (5, 20), (20, 5), (2, 5), (5, 2)

So the probability = 7/400

7. There is a conical tent which can accommodate 10 persons. Each person requires 6 sq.meter space to sit and 30 cubic meters of air to breathe. What will be the height of the cone?

a) 72 m

b) 15 m

c) 37.5 m

d) 155 m

Answer: b) 15 m

Solution:

All the persons are to sit on the ground forming the base of the cone.

Total base covered = pi * $r^2$ = 6*10 = 60 sq-meter.

The total volume of the tent will be equal to the total air to breathe by the 10 people = 30*10 = 300 cubic meter

So, 1/3(pi * $r^2$ * h) = 300

=> h = 15 meters.

8. Find the greatest power of 143 which can divide 125! exactly.

a) 11

b) 8

c) 9

d) 7

Answer: c) 9

Solution:

We can write 143 = 11 × 13.

So the highest power of 13 should be considered in 125!, which is 9 (13 * 9 = 117)

The highest power of 11 in 125! is 12 (11 * 11 = 121 and remaining 1).

That means, 125! = 11^12×13^9×…

So only nine 13’s are available. So we can form only nine 143’s in 125!. So maximum power of 143 is 9.

9. A car starts at 6:00 pm. from the starting point at a speed of 18 m/s, reached its destination. There it waited for 40 minutes and returned back at the speed of 28 m/s.

 Find the time taken to reach the destination.

a) 9:44 pm

b) 8:32 pm

c) 7:30 pm

d) 9:30 pm

Answer: a) 9:44 pm

Solution:

Let the distance covered be D m

The time to cover the starting distance = D/18 secs.

The time taken for the reverse journey = D/28 secs.

According to the quesiton,

D/18 – D/28 = (40 × 60)

On solving this we get,

D = 2400 × 252/5 = 120960 m

No the total time taken = (D/18) + (D/28) + 2400 = 13440 seconds

= 3 hours and 44 minutes

Therefore, the bus reaches back at 9:44 PM

10. The value of a house depreciates each year, by 3/4 of its initial value at the beginning of the year. If the initial value of the scooter is Rs. 40, 000. What will be the value at the end of 3 yrs?

a) Rs. 19000

b) Rs. 16875

c) Rs. 17525

d) Rs. 18000

Answer:  b) 16875

Solution:

This is the question of succession depreciation.

the starting amount = Rs. 40000

This reduces by 3/4 th of its initial value every year = (40, 000) * (3/4)^3 = 16875

                                                                                                              SET-10

1. All even numbers from 2 to 98 inclusive the both, are to be multiplied together. What is the unit digit of the product?

a) 2

b) 0

c) 6

d) 4

Answer: b) 0

Solution:

Let us look at the sequence of the multiplications,

2 * 4 * 6 * 8 *10* 12 * 14 * 16 * 18 *…* 98

If we look closely we will find that the units place of every number forms a sequence of 2, 4, 6 ,8 and 0 multiplying to a number whose units place is always 0 and in all we get 0. So the unit digit of final number = 0.

2. 10 programmers are able to type 10 lines in 10 minutes. How many programmer are required to type 60 lines in 60 minutes?

a) 10

b) 16

c) 60

d) None of the above

Answer: a) 10

Solution:

3. This is a simple question of logical reasoning. If 10 programmers can type 10 lines of code in 10 minutes then to type 60 lines of code, in 60 minutes, 

the same 10 coders will be required, since the lines of code and time are in proportion

Anil works for 8 straight days and rest on the 9th day. If he starts his work on Monday, then on which day he gets his 12th rest day?

a) Thursday

b) Tuesday

c) Wednesday

d) Friday

Answer: c) Wednesday

Solution:

Anil works for 8 days and rests on 9th day. In total 9 days are to be processed 12 times = 12 * 9 = 108.

If we calculate according to the week, we get 108 / 7 = remaining 3 days. So if Anil starts working on Monday, he will rest on third day of the week which is Wednesday.

4. Overfishing is a serious environmental issue. The scientists were able to determine that if the net of a trawler has a mesh size of ‘x’ cm (a square mesh), 

then the percentage of fish entering the net is caught in the net is expressed in form of the quadratic equation, 100 – 0.04x^2- 0.24x. For example, 

if the mesh size is zero, 100% of the fish that enter the net will be caught. A trawler with a net with a square mesh, that was suspect of using an illegal size net, 

dropped its net to the ocean floor near the Lakshadweep and the coast guard, arrested the crew. It was later looked at the size of the fish caught and estimated that for the net used by the trawler, 

at least 97.8% of the fish entering the net would be caught. What is the maximum value of x for the net used by the trawler?

a) 7

b) 4.5

c) 6

d) 5

Answer: d) 5

Solution:

According to the question,

for few values of x, the total fish caught is 97.8%. So

=> 100 – 0.04x^2- 0.24x = 97.8

=> 0.04x^2 + 0.24x = 2.2

=> 4x^2+ 24x = 220

=> x^2+ 6x – 55 = 0

Solving, we get x = 5 and -11

So, the value of x = 5 has to be positive and hence the answer.

5. The rejection rate for Audi production was 4 per cent, for Mercedes it was 8 per cent and for the 2 cars combined it was 7 per cent. What was the ratio of Audi production?

a) 4/1

b) 2/1

c) 3/1

d) 7/1

Answer: c) 3/1

Solution:

Using the simple weighted average formula we get,

(4x + 8y)/(x+y) = 7

or, 4x + 8y = 7x + 7y

or, a/b = 3/1

6. A team of 11 is needed to be formed who are to be selected from 5 men and 11 women, with the restriction of selecting not more than 3 men. In how many ways can the selection be done?

a) 1121

b) 1565

c) 1243

d) 2256

Answer: d) 2256

Solution:

Selecting 0 men and 11 women = 5C0 * 11C11 = 1

Selecting 1 men and 10 women = 5C1 * 11C10 = 55

Selecting 2 men and 9 women = 5C2 * 11C9 = 10 * 55 = 550

Selecting 3 men and 8 women = 5C3 * 11C8 = 10 * 165 = 1650

So total number of ways = 1650 + 550 + 55 + 1 = 2256 way

7. There are two bags containing white and black marbles. In the first bag there are 8 white marbles and 6 black marbles and in the second bag, 

there are 4 white marbles and 7 black marbles. One marble is drawn at random from any of these two bags. Find the probability of this marble being black.

a) 7/54

b) 7/154

c) 41/77

d) 22/77

Answer: c) 41/77

Solution:

Probability of drawing a black ball from the first bag is = 6C1 / 14C1

Probability of drawing a black ball from the second bag is = 7C1 / 11C1

Total probability = 1/2 * (6C1/14C1) * (7C1/11C1) = 41/77

8. There is a city where all 100% votes are registered. Among this 60% votes for Congress and 40% votes for BJP. Ram, gets 75% of congress votes and 8% of BJP votes. How many votes did Ram get?

a) 48.2 %

b) 56.6 %

c) 42.8 %

d) 64.4 %

Answer: a) 48.2 %

Solution:

Let the total number if votes = 100. So Ram gets,

75% of 60 = 60 * 0.75 = 45 votes

8% of 40 = 40 * 0.08 = 3.2 votes

Thus total number of votes that Ram gets = 48.2 %

9. John is faster than Peter. John and Peter each walk 24 km. Sum of the speeds of John and Peter is 7 km/h. Sum of time taken by them is 14 hours. Find John’s speed.

a) 4 km/h

b) 5 km/h

c) 3 km/h

d) 7 km/h

Answer: a) 4 km/h

Solution:

We know that John’s speed is greater than Peter’s speed and the sum of there speed is 7.

So the combinations are = (6, 1), (5, 2), (4, 3)

Now checking from the options if John’s speed is equal to 4, then Peter’s speed is 3,

or, the time taken by them = 24/4 + 24/3 = 14 hours.

10. If f(x) = 2x + 2 what is the value of f(f(3))?

a) 8

b) 64

c) 16

d) 18

Answer: d) 18

Solution;

f(f(3)) = 2(f(3)) + 2

=> 2(2(3) + 2) + 2

=> 16 + 2 = 18