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
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
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
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
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
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.
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 B)
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 B)
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
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
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
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
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