# Cisco aptitude questions

Cisco aptitude questions

1     What is the probability of getting at least one six in a single throw of three unbiased dice?

(1) 1 / 6
(2) 125 / 216
(3) 1 / 36
(4) 81 / 216
(5) 91 / 216

Correct choice is (5) and Correct Answer is 91 / 216
Every die has got six sides. Each of the sides is numbered from 1 to 6.
When a single unbiased die is thrown you can have six possible outcomes.
When two dice are thrown simultaneously, the total number of outcomes will be 6 * 6 = 36
Similarly, when three dice are thrown simultaneously, the total number of outcomes will be 6*6*6=216.
We need to find out the number of cases in which at least one of the facing sides shows 6.
At least one means – either one dice or two dice or all three.

Case 1: Let us take the easiest case first – all three dice showing ‘6’ – There is only one such possibility.

Case 2: The number of cases in which two of the dice show 6 and one of them is a different number.
For eg an event like 6 6 5 will be one of the outcomes for case 2.

As two of the dice show ‘6’ , it can happen in only one way. The third die shows a different number, a number other than 6, and it can be any of the 5 other numbers. Therefore, there will 5 possible options i.e. (1, 6, 6), (2, 6, 6), (3, 6, 6), (4, 6, 6), (5, 6, 6).

However, each of these possibilities can have three different arrangements depending upon where the third different digit appears. For example take (1, 6, 6) case – it will have three options (1, 6, 6), (6, 1, 6), (6, 6, 1).

Therefore, the total number of events in which 2 of the dice will show ‘6’ and one will show a different number = 5*3 = 15 cases.

Case 3: When only one of the die shows ‘6’ and the other two show numbers other than ‘6’.

The die showing ‘6’ has only one option. The other two dice can have any of the ‘5’ options. Therefore, the total number of possibilities = 1*5*5 = 25.

However, the die showing ‘6’ can either be the first die or the second die or the third die.

Therefore, there are a total of 25 * 3 = 75 possibilities.

Total possible outcomes where at least one die shows ‘6’ = Case 1 + Case 2 + Case 3 = 1 + 15 +75 = 91.

Therefore, the required probability = 91/216

Alternate approach:

Find the number of cases in which none of the digits show a ‘6’.

i.e. all three dice show a number other than ‘6’, 5 * 5 *5 = 125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a ‘6’ = Total possible outcomes when three dice are thrown – Number of outcomes in which none of them show ‘6’.

= 216 – 125 = 91.

The required probability = 91/216

2     When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 11?

(1) 5 / 6
(2) 11 / 12
(3) 1 / 6
(4) 1 / 12
Solution:

Instead of finding the probability of this event directly, we will find the probability of the non-occurrence of this event and subtract it from 1 to get the required probability.

Combination whose sum of 12 is (6,6)

Combinations whose sum of 11 is (5,6), (6,5).

Therefore, there are totally 3 occurrences out of 36 occurrences that satisfy the given condition.

Probability whose sum of two numbers is greater than or equal to 11 = 3 / 36 = 1 / 12.

Hence probability whose sum of two numbers is lesser than 11 = 1 – 1 / 12 = 11 / 12.

3     The average monthly salary of 12 workers and 3 managers in a factory was Rs. 600. When one of the manager whose salary was Rs. 720, was replaced with a new manager, then the average salary of the team went down to 580. What is the salary of the new manager?
Rs 570
Rs 420
Rs 690
Rs 640
Rs 610
Correct Answer – Rs 420. Choice (2)

The total salary amount = 15 * 600 = 9000

The salary of the exiting manager = 720.

Therefore, the salary of 12 workers and the remaining 2 managers = 9000 – 720 = 8280

When a new manager joins, the new average salary drops to Rs.580 for the total team of 15 of them.

The total salary for the 15 people i.e., 12 workers, 2 old managers and 1 new manager = 580 *15 = 8700

Therefore, the salary of the new manager is 9000 – 8700 = 300 less than that of the old manager who left the company, which is equal to 720 – 300 = 420.

An alternative method of doing the problem is as follows:

The average salary dropped by Rs.20 for 15 of them. Therefore, the overall salary has dropped by 15*20 = 300.

Therefore, the new manager’s salary should be Rs.300 less than that of the old manager = 720 – 300 = 420.

4     If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?

(1) 24
(2) 31
(3) 32
(4) 30
Solution:

The 5 letter word can be rearranged in 5! Ways = 120 without any of the letters repeating.

Then the 25th word will start will CA _ _ _. The remaining 3 letters can be rearranged in 3! Ways = 6. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _. The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24 + 6 + 2 = 32.

5     The time in a clock is 20 minute past 2. Find the angle between the hands of the clock.
(1) 60 degrees
(2) 120 degrees
(3) 45 degrees
(4) 50 degrees
Correct choice (4). Correct Answer – 50 degrees
Solution:

Time is 2:20. Position of the hands: Hour hand at 2 (nearly).
Minute hand at 4
Angle between 2 and 4 is 60 degrees [(360/12) * (4-2)]
Angle made by the hour hand in 20 minutes is 10 degrees, since it turns through ½ degrees in a minute.
Therefore, angle between the hands is 60 degrees – 10 degrees = 50 degrees
6    .A 12% stock yielding 10% is quoted at:

A.Rs. 83.33
B.Rs. 110
C.Rs. 112
D.Rs. 120
To earn Rs. 10, money invested = Rs. 100.
To earn Rs. 12, money invested = Rs.100x 12= Rs. 120.
10 Market value of Rs. 100 stock = Rs. 120

7 A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

A.5.15 kmph
B.4.14 kmph
C.4.25 kmph
D4.4 kmph

8 100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:
9    What was the day of the week on 28th May, 2006?
A.Thursday
B.Friday
C.Saturday
D.Sunday
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2)  6 odd days
Jan. Feb. March April May
(31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day)  1 odd day.Total number of odd days = (0 + 0 + 6 + 1) = 7  0 odd day.Given day is Sunday.

10 The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

A.279
B.283
C.308
D.318
Other number =11 x 7700 = 308.275

11     How many digits will be there to the right of the decimal point in the product of 95.75 and .02554 ?

A.5
B.6
C.7
D.None of these
Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.

12  One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
A.81 min.
B.108 min.
C.144 min.
D.192 min.
Let the slower pipe alone fill the tank in x minutes.

13    A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:
A 0%
B.5%
C.7.5%
D.10%
C.P. = Rs. 3000.

S.P. = Rs.

3600 x 100

= Rs. 3000.
100 + (10 x 2)
Gain = 0%.

14     Find out the wrong number in the series.

6, 12, 48, 100, 384, 768, 3072

A.768
B.384
C.100
D.48
E.12