## Sample Aptitude Questions Of Cisco

**1.What is the equation of the line parallel to the line 2x + 3y = 12?**

A.2.4x + 1.8y = 16

**B.3.4x + 5.1y = 8**

C.4.2x + 3.3y = 18

D.13.2x + 4.8y = 24

**Answer & Explanation**

**Option: B**

Slope of the given line is (-2/3). So, slope of the line parallel to it should be the same.

As only in option 2, (-3.4/5.1) = -2/3 while the other equations do not have same slope.

Thus option 2 is the answer

**2.What is the difference between cost price and the marked price of an article, sold at a loss of 11% for Rs. 2136 with a discount of 20%?**

A.Rs. 178

B.Rs. 267

C.Rs. 240

**D.Rs. 270**

**Answer & Explanation**

**Option: D**

M.P. = 2136 /0.8 = Rs 2670 & C.P= 2136/0.89 = Rs 2400.

So required difference = 2670 – 2400 = Rs 270.

**3.A gives B a start of 200m and still wins a km race by 200m. B gives a start of 100m and still wins by 100m in another km race with C. How much start should A give C in a km race in order to beat him by 250m?**

**A.270**

B.480

C.520

D.230

**Answer & Explanation**

**Option: A**

Ratio of distance travelled by A and B ⇒ A : B = 1000 : 600, while that of B and C ⇒ B : C = 1000 : 800

Therefore, A : C = 1000 : 480.

Since A beats him by 250m therefore C should be 750m away from the starting point.

So, A should give C a start of (750-480) = 270m in a km race in order to beat him

**4.The simple interest accumulated in 4 years, with interest, R% p.a, on a particular sum is half of its principal. Find R.**

A.20%

B.6%

**C.12.5%**

D.8.33%

**Answer & Explanation**

**Option: C**

S.I. = P*R*T/100 ⇒ P/2 = P*R*4/100 ⇒ R = 12.5%

**5.If two dices are rolled simultaneously, then what is the probability of getting the sum of the numbers a prime number?**

A.14/36

B.5/17

C.15/29

**D.5/12**

**Answer & Explanation**

**Option: D**

Prime no.s = 2, 3, 5, 7 & 11. The no of ways to get :

2 as the sum is 1 i.e. 1 +1

3 as the sum are 2 i.e. 1 + 2, 2 + 1

5 as the sum are 4 i.e. 1 + 4, 4 + 1, 2 + 3, 3 + 2

7 as the sum are 6 i.e. 1 + 6, 6 + 1, 2 + 5, 5 + 2, 3 + 4, 4 + 3

11 as the sum are 2 i.e. 5 + 6, 6 + 5

So total no of ways are 1 + 2 + 4 + 6 + 2 = 15

So the probability is 15 / 36 = 5 / 12 (where 36 is total number of possible cases in case of two dices)

**6.A multiplex owner increases the price of a ticket by 25%, because of which the number of viewers decrease by 25%. What is the percentage change in the revenue?**

A.6.25% increase

B.9.375% increase

**C.6.25% decrease**

D.9.375% decrease

**Answer & Explanation**

**Option: C**

Using the formula a + b + ab/100, we get 25 – 25 – (625/100) i.e. -6.25.

So there is decrease of 6.25% in revenue.

**7.A and B start a business by investing 5 lks and 4 lks. B remains in the business for a complete year. How long did A remain in the business, if they received equal profit at the end of the year?**

**A.9.6 months**

B.2.4 months

C.8.4 months

D.6.9 months

**Answer & Explanation**

**Option: A**

Let N be the time period of A. Since the profits are equal, we get 4 x 12 = 5 x N,

On solving, we get N = 0.8 year or 9.6 months

**8.What is the remainder when addition of all the 2-digit multiples of 9 is divided by 11?**

A.0

B.1

**C.2**

D.3

**Answer & Explanation**

**Option: C**

Two digit multiples of 9 are 18, 27, ……, 99 which are 10 in number.

By A.P. formula we can find the addition of 2 digit multiples of 9.

Sum = [10/2(18+99)] = 585. So 585/11, we get remainder as 2.

**9.60% of the students learn German and 50% of the students learn French. If student in the class learns at least one subject out of the given two, then what percent of the students do not learn both the subjects?**

A.10

B.12

C.80

**D.90**

**Answer & Explanation**

**Option: D**

8. Total percentage of both german and english = 50+60 = 110%. So 10% learn both

Therefore, 100-10 = 90% do not learn both.

**10.In the inequality, x2 – 14x + 50 < 5, find the range of x.**

A.5 > x > 9

**B.5 < x < 9**

C.-5 < x < 9

D.-5 > x > -9

**Answer & Explanation**

**Option: B**

x2 – 14x + 50 < 5 ⇒ (x-9)(x-5)<0

Now, here 2 cases arises

Case 1: (x-9)<0 & (x-5)>0

So we get range to be 5 < x < 9

Case 2: (x-9)>0 & (x-5)<0

Which gives x > 9 and x < 5 which cannot occur simultaneously, hence rejected.