**A is able to do a piece of work in 15 days and B can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?**- 8/15
- 7/15
- 11/15
- 2/11
- Other than those given as options

Answer : Option A.

Total work done by A + B in 1 day = 1/15 + 1/(20 ) = 7/60

Work done in 4 days = 7/60 × 4 = 7/15

Therefore, fraction of work left = 1 - 7/15 = 8/15**Cost price of each of the articles A and B is Rs. 'X'. Article A was sold at a profit of 10% and article B was sold at a profit of 30%. If the overall profit earned after selling both the articles is Rs. 136/-, what is the value of 'X'?**- Rs.340/-
- Rs.300/-
- Rs.360/-
- Rs.380/-
- Rs.320/-

Answer : Option A.

Given : 0.1x + 0.3x = 136 => x = 340

**Population of a village increased by 5% from 2007 to 2008 and by 25% from 2005 to 2009. If the population of the village was 480 in 2007, what was its population in 2009?**- 640
- 610
- 630
- 620
- 650

Answer : Option C.

Population in 2007 = 480

In 2008 = 1.05 × 480 = 504

In 2009 = 1.25 × 504 = 630**24% of 150 × 3/4= ?**- 64
- 40
- 72
- 24
- 48

Answer : Option E.

24% of 150 × 3/4= x = 36 × 4/3 = 48**3/17 of 20% of 510 + 7 = x**^{2}- 5
- 10
- 21
- 15
- 27

Answer : Option A.

3/17 of 20% of 510 + 7 = x^{2}

3/17 × 20/100 × 510 + 7 = 18 + 7 = 25 => x = 5Nagarro-Software Preparation Links- Sample Aptitude Questions of Nagarro-Software
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**In the month of March, Hiten spent 45% of his monthly salary on paying bill and rent. Out of the remaining salary, he invested 60% in PPF and the remaining he deposited in bank. He deposited Rs. 15,400 in bank. If in April, he got an increment of 10%, what was his salary in April?**- Rs. 84,000
- Rs. 77,000
- Rs. 1,10,000
- Rs. 59,000
- Rs. 68,000

Answer : Option B.

Let the initial salary of Hiten be ‘S’. Then:

(0.55 × 0.4) S = 15400 => S = 70000

After increment, Hiten’s salary = 1.1 × 70000 = 77000**A person covers a certain distance by travelling at a uniform speed of 120 km/h for 90 minutes. At what speed will he have to travel in order to cover the same distance in 1 hour 20 minutes? (in km/h)**- 135
- 125
- 140
- 130
- 145

Answer : Option A.

Distance = 120 × 90/60 = 180 km Speed required to cover 180 km in 1 hr 20 mins ( 4/3 hrs) = 180 × ¾ = 135 km/hr**In Jar A, 120 litres milk was mixed with 24 litre water. 12 litre of this mixture was taken out and 3 litre water was added. If 27 litre of newly formed mixture is taken out, what will be the resultant quantity of water in the jar? (In litre)**- 20
- 10
- 15
- 30
- 25

Answer : Option A.

Ratio of milk : water in Jar A = 120 : 24 = 5 : 1

12 lts of this mixture is taken out => milk = 5/6 × 12 = 10 lts and water = 2 lts taken out

3 lts of water added = 24 – 2 + 3 = 25 lts => new ratio of milk : water

= 110 : 25 = 22 : 5

Now 27 lts of this mixture is again taken out =>

water taken out = 5/27 × 27 = 5 lts

water left = 25 – 5 = 20 lts**A boat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. What is the speed of the stream? (in km/hr)**- 10
- 6
- 5
- 4
- 15

Answer : Option C.

Speed of boat = 15 k/h

Let the speed of stream be ‘S’

Given : 30/(15+S) + 30/(15-S) = 9/2

Solving we get, S = 5 km/h**Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11:10. What is Sagar's present age?**- 16 years
- 18 years
- 20 years
- 22 years
- 25 years

Answer : Option A.

Let the ages of Kunal and Sagar be K and S respectively.

Given : (K-6)/(S-6) = 6/5 => 5K – 6S = -6 ……(i)

And: (K+4)/(S+4) = 11/10 => 10K – 11 S = 4 …..(ii)

Solving (i) & (ii) we get: S = 16 years.