# TCS Digital Quantitative Aptitude Questions & Answers - TCS

TCS Digital Quantitative Aptitude MCQs : This section focuses on "Quantitative Aptitude" of TCS Exam. These Quantitative Aptitude MCQs are asked in previous TCS Digital Exams and will help you to prepare for upcoming TCS Digital exam.

1. Find the greatest number that will divide 148 246 and 623 leaving remainders 4 6 and 11 respectively?

A. 20

B. 12

C. 6

D. 48

View Answer

Ans : B

Explanation: Remainder in case of 148 = 4

Number = 148 - 4 = 144

Remainder in case of 246 = 6

Number = 246 - 6 = 240

Remainder in case of 623 = 11

Number = 623 - 11 = 612

Now, we have find the H.C.F. of 144. 240 and 612

H.C.F. of 144, 240 and 612 = 2*2*3 = 12

Therefore, the required number is 12 that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.

2. Determine the metal required to make a 21 m long pipe if its inner and outer diameter is 12 m and 10 m respectively

A. 234 m3

B. 647 m3

C. 726 m3

D. 843 m3

View Answer

Ans : C

Explanation: volume of a hollow cylinder = pie * h * (R^2 - r^2)

= (22/7) * 21 * (36-25)m3

=66 x 11 m3

=726 m3

3. Age of Umesh will be 4 times the age of Reena in 6 years from today. If ages of Umesh and Mahesh are 7 times and 6 times the age of Reena respectively, what is present age of Umesh?

A. 64 years

B. 30 years

C. 48 years

D. 42 years

View Answer

Ans : D

Explanation: At present

Age of Umesh = x

Age of Reena is Y age of Mahesh is m

after 6 years

X+6=4(y+6)

X+6=4y+24

But it is given that age of Mahesh is equal to 7 times age of Reena

So, 7y+6=4y+24

3y+6=24

3y=18

Y=6

Age of Mahesh is equal to 7 times the age of Reena

X=7 x 6 = 42

4. The average of first 10 even numbers is?

A. 18

B. 22

C. 9

D. 11

View Answer

Ans : D

Explanation: Sum of 10 even numbers = 10 * 11 = 110

Average = 110/10 = 11

5. In a clock the long hand is of 8cm and the short hand is of 7cm. if the clock runs for 4 days find out the total distance covered by both the hands?

A. 1824π cm

B. 1724π cm

C. 1648π cm

D. 2028π cm

View Answer

Ans : C

Explanation: 4 days = 96 hours

long hand makes 1 round (i.e 2*pi*r) in 1 hour, so 96 rounds in 96 hours

short hand makes 1 round in 12 hours, so 8 rounds in 96 hours

Hence total distance = (96*2*pi*8) + (8*2*pi*7) = 1648*pi

6. If 43 times of two digit numbers is 34 times of two digit no .and sum of ( number and reverse of the number) is 14 then what is the number?

A. 68

B. 51

C. 53

D. 55

View Answer

Ans : A

Explanation: 43 time the no. And 34 time the reverse no.

43(10x+ y)=34(10y+ x)

430x+ 43y= 340y+34x

430x+43y-340y-34x=0

To solve this equation become

4x-3y=0 equation=2

Sum of no. And reverse no. Is 14

X+ Y= 14 equation=1

To solve both equation value of x and y

X= 6 and y= 8

number will be 68

7. Three partners X, Y, Z start the business. Y's capital has been four times Z's capital also twice X's capital has been equal to three time Y's capital. In case total profit has been Rs. 16500 at end of the year. Find out Y's share in it.

A. 4000

B. 5000

C. 6000

D. 9000

View Answer

Ans : C

Explanation: Rs. 6000

Y=4Z ---(i) and

2X=3Y =>Y= 2X/3 ---(ii)

To get the ratio of X, Y, Z we can represent Y and Z in terms of X.

From (i) & (ii), 2X/3 = 4Z => Z=X/6

So X : Y : Z = X : 2X/3 : X/6 = 6 : 4 : 1

So Y's share in Rs. 16500 = (4/11)*16500 =Rs. 6000

8. How many positive integer numbers not more than 4300 can be formed with the digits 0, 1, 2, 3, 4 if repetitions are allowed?

A. 625

B. 560

C. 565

D. 575

View Answer

Ans : D

Explanation: Ans is 575

One digit no = 4 (0 is not a positive integer)

Two digit no=4 x 5 = 20

Three digit no=4 x 5 x 5 = 100

Four digit no=3 x 5 x 5 x 5 = 375(the possibility for 1,2,3 will come in the first position)

Four digit no=1 x 3 x 5 x 5 (the possibility of 4 is fixed in the first position and then 0,1,2 is comes in second position)

And the last digit is 4300 we include this number also

Ans is 4 + 20 + 100 + 375 + 75 + 1 = 575

9. 100 students appeared for two different examinations 60 passed the first,50 the second and 30 both the examinations.Find the probability that a student selected at random failed in both the examination?

A. 0.8

B. 0.2

C. 0.6

D. 0.7

View Answer

Ans : B

Explanation: Total number of Student =100

Number of student passed first exam is 60, so probability is 60/100 = 0.6

Number of student passed second exam is 50/100, so probability is = 0.5

Number of student passed third exam is 30/100, so probability is = 0.3

From probability addition rule,

P(A ∪ B) = P(A) + P(B) - P(A∩B)

P(A ∪ B) = 0.6 + 0.5 - 0.3

P(A ∪ B) = 0.8

Thus, probability of student passing at least in one subject is 0.8.

Probability of student failing in both subject is given by,

{P(A U B)}= 1 - P(A ∪ B)

{P(A U B)} = 1 -0.8

{P(A U B)} = 0.2

Thus, Probability of student failing in both subjects are 0.2.

10. A farmer bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

A. 4

B. 5

C. 6

D. 8

View Answer

Ans : B

Explanation: here Gain%=20

now formula is gain%=(gain*100)/cp

i.e 20=(gain*100)/1

i.e 20/100=gain

gain=0.2

so 6 tofees will be sold at 1.2 rs to gain 20%

0.20 for each toffee

i.e 5 toffess will be sold at 1 rs

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