Cisco Aptitude Previous Year Questions

by Ajinkya Kulkarni | Updated on 19 January 2024

Cisco Aptitude Previous Year Questions

1)The marked price of a watch has a profit component of x% over the cost price of Rs.X . if the marked price were to be Rs.416. what is it’s cost price?

A. 416

B. 260

C. 160

D. 360

Answer : C

Solution:

2) If the compound interest of a principle of Rs 4,000,000 over 3 years is Rs 2083500. What is the interest rate?

A. 15

B. 12

C. 14

D. 10

Answer: A

Solution:

3) A ten-digit number N is divisible by both 72 and 88. if N = x123y4566, how many distant values are possible for N?

A. 4

B. 5

C. 10

D. 6

Answer : B

solution:

N divisible by 72 and 88 means need to be divisible by 9,11,8

For 8 last 3 digit should be divisible by 8

For 9 sum should be divisible x123y4566 = 27+x+y

X+ y can be 9 or 18

For 11

Alternative sum difference should be 0, 11

X+y+ 13 , 18

X+y can be 5

4) A is thrice as good a workman as B, A takes 60 days less than B for a job. How many days will it take to complete the same job A&B together?

A. 20

B. 22.5

C. 25

D. 27.5

Answer: B

Solution:

The efficient ratio of A to B is 3:1. Therefore the time ratio is 1:3.

The difference in the ratio is 2, that corresponds to 60 days.

Hence A takes 30 days and B takes 90 days.

If they work together, they take 30*90/30+90 = 90/4 = 22.5 days.

5) For 2 events A & B, P(A U B) is equals to (P() refers to the probability)

A. P(A) + P(B) – P(A U B)

B. P(A) * P(B) – P(A

B)

C. P(A) - P(B) + P(A U B)

D. P(A) + P(B) – P(A

Answer: D

Solution: The formula for P(A U B) = P(A) + P(B) – P(A B)

6) The probability that a given positive integer lying between 1 and 100(both inclusive) is NOT divisible by 2, 3 or 5 is ?

A. 0.26

B. 0.49

C. 0.32

D. 0.22

Answer: A

Solution:
Number of integers in the set which are divisible by 2 or 3 or 5
= n(2∪3∪5)=n(2)+n(3)+n(5)−n(2∩3)−n(3∩5)−n(5∩2)+n(2∩3∩5)
=50+33+20−16−10−6+3=74
∴ The number of integers between 1 and 100 which are not divisible by 2 or 3 or 5
=n(¯2∩¯3∩¯5).
=100−n(2∪3∪5)
=100−74=26

∴ Required probability=26/100=0.26

7) Six people participated in a checker tournament. Each participant played exactly three games with each of the other participants. How many games were played in all?

A. 45

B. 60

C. 75

D. 30

Answer: A

Solution:

The first player plays 3 games with each of the other 5 players.

Therefore, the total number of games he played 5×3=15

The second player plays 3 games with 4 others,

So, the total number of games he played 3×4=12

So, the total number of games he played 3×3=9

Similarly, fourth players played 3×2=6 games

And fifth player played 3×1=3 games

Thus, the total number of games played = 15+12+9+6+3 = 45

Cisco Aptitude Previous Year Questions

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Cisco Aptitude Previous Year Questions

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