__Quantitative Aptitude Practice Word Problems – Set 26__**1. Local trains leave from a station at an interval of 15 minutes at a speed of 20 km/h. A man moving from opposite side meets the trains at an interval of 10 minutes. Find the speed of the man.**

(a) 10 km/h

(b) 8 km/h

(c) 9 km/h

(d) 7.8 km/h

**2. The number of laptops sold by a company increases by 10% every year. In 2015, the number of laptops sold was 24200. The price of laptop increases by 20% every year. The price of a laptop in 2014 was Rs. 30000. What was the average price of a laptop sold by this company considering years 2014, 2015 and 2016?**

(a) Rs. 35992.15

(b) Rs. 36819.34

(c) Rs. 37122.78

(d) Rs. 36713.96

(e) Cannot be determined

**3. Two containers are taken such that capacity of first is one tenth of second. In both of them, 3 litres of milk and 7 litres of water is added. Now, from second container, 4 litres of the mixture is taken out and added to first container. The first container gets full after this process. With a mixture of milk and water, the remaining portion of second container is to be filled so that the final ratio of milk and water in second container becomes 2:3. What should be the ratio of milk and water in the added mixture?**

(a) 1:3

(b) 91:133

(c) 271:399

(d) 27:80

(e) 9:31

**4. 4 iron-men, 6 winter soldiers and 8 X-men defeat an entire army of frost giants in 18 hours. A iron-man does double the work the winter soldier does and an X-man does half the work a winter soldier does. How many X-men alone can defeat an army of frost giants in 9 hours?**

(a) 24

(b) 72

(c) 144

(d) 108

(e) None of these

**5. Geeta invests Rs. 45,000 to start a business. Four months later Shreya joins her by investing Rs. 72,000 and another two months later Saina joins them both by investing Rs. 87,500. At the end of one year the business earns a profit of Rs. 21,880. What is Saina’s share in the profit?**

(a) Rs. 7,608

(b) Rs. 7,200

(c) Rs. 7,680

(d) Rs. 7,000

(e) None of these

**6. The age of my father is equal to sum of my mother’s age and my age. When I was born, the ratio of ages of my mother and father was 5 : 6. After 8 years, the ratio of my age and my mother’s age will be 3 : 8. What was my mother’s age when I was 2 years old?**

(a) 22 years

(b) 26 years

(c) 24 years

(d) 32 years

(e) 30 years

**7. George invested a sum of money of Rs. 13000 in a scheme that works on the principle of compound interest. The rate of interest is 20% per annum. After 2 years, it was notified that the scheme will now work on the principle of simple interest. After further 2 years, George withdrew all the money from investment. Due to pre mature withdrawal, a penalty of Rs. 1000 was charged. How much money will George get?**

(a) Rs. 25208

(b) Rs. 26208

(c) Rs. 24208

(d) Rs. 23708

(e) Rs. 23208

**8. Gary has two weighing machines, and he uses them in the ratio 2 : 3. The first machine is defective, and shows a weight 100 grams more than actual. Gary sold 1 kg of rice to each of 15 customers. If Gary sells at cost price, how much profit would he make?**

(a) 6.57%

(b) 5.38%

(c) 4.17%

(d) 3.18%

(e) 3.97%

**9. The number of ways in which 6 different marbles can be put in two boxes of different sizes so that no box remains empty is**

(a) 62

(b) 64

(c) 36

(d) 60

(e) None of these

**10. Inside a circle of radius of 20 cm, largest possible square is drawn such that vertices of square lie on the circle. Now, the size of the circle is reduced so that its area becomes 0.36 times of the original. However, the centre of circle is kept fixed. What will be the area (in square cm) that lies inside the square but outside the circle? (Take π = 3.14)**

(a) 800

(b) 347.84

(c) 352.16

(d) 452.16

(e) 1252.16

**Solutions:**

**1. A)**Let the speed of the man be x km/h.

We know,

Distance = Speed × Time ⇒ Time = Distance/Speed

Here, the distance
between two trains is actually the distance travelled by one train before the 2

^{nd}train starts.
∵ the 2

^{nd}train starts 15 minutes after the 1^{st}train,
In 15 minutes the 1

^{st}train goes = 20 × (15/60) = 5 km
The man moving from
opposite side meets the trains at an interval of 10 minutes i.e. 1/6 hours.

∵ the man is moving in a direction opposite to that of trains,

Relative velocity = (20 +
x) km/hr

So, we can write,

(x + 20) × (1/6) = 5

⇒ x + 20 = 30

⇒ x = 10

∴The speed of the man is 10 km/h.

**2. B)**The price of a laptop in 2014 was Rs. 30000. The price of a laptop in 2014 was Rs. 30000.

⇒ Price of laptop in 2015 = Rs. 30000 × (1 + 20/100) = Rs. 36000

⇒ Price of laptop in 2016 = Rs. 36000 × (1 + 20/100) = Rs. 43200

In 2015, the number of
laptops sold was 24200. The number of laptops sold by a company increases by
10% every year.

⇒ Number of laptops sold in 2014 = 24200/(1 + 10/100) = 22000

⇒ Number of laptops sold in 2016 = 24200 × (1 + 10/100) = 26620

∴ Average price of laptop in 2014, 2015 and 2016 = [(30000 ×
22000) + (36000 × 24200) + (43200 × 26620)]/(22000 + 24200 + 26620) =
2681184000/72820 = Rs. 36819.34

**3. C)**In containers, 3 litres of milk and 7 litres of water is added. Now, from second container, 4 litres of the mixture is taken out and added to first container. The first container gets full after this process.

⇒ Capacity of first container = (3 + 7 + 4) litres = 14 litres

⇒ Capacity of second container = 10 × 14 litres = 140 litres

Now, in second container,
amount of mixture = (3 + 7 - 4) litres = 6 litres

Empty part of second
container = (140 – 6) litres = 134 litres

Amount of milk is second
container = (3/10) × 6 = 1.8 litres

Amount of water is second
container = (7/10) × 6 = 4.2 litres

With a mixture of milk
and water, the remaining portion of second container is to be filled so that
the final ratio of milk and water in second container becomes 2:3.

Suppose the amounts of
milk and water in added mixture is M:(134-M).

⇒ (1.8 + M)/(4.2 + 134 – M) = 2/3

⇒ 5.4 + 3M = 276.4 – 2M

⇒ 5M = 271

⇒ M = 54.2

∴ Ratio of milk and water in added mixture = 54.2/(134-54.2) =
54.2/79.8 = 271/399

**4. B)**Consider a winter soldier alone can do 1 piece of work (defeat an army of frost giants) in n hours.

Thus, work done by a
winter soldier in 1 hour = 1/n

∴ Based on given data,

Work done by 1 iron-man
in 1 hour = 2/n,

Work done by 1 X-man in 1
hour = 1/2n

⇒ Work done by 4 iron-men, 6 winter soldiers and 8 X-men in 1
hour = (8/n), (6/n) and (4/n) respectively.

⇒ Work done in 18 hours = 18 × (8/n + 6/n + 4/n)

⇒ 18(8/n + 6/n + 4/n) = 1 ⇒ n = 324 ……… (1)

⇒ Work done by 1 X-man in 1 hour = 1/648

Let the number of X-men
required to complete the work in 9 hours = m

∴ Work done by m X-men in 1 hour = 1/ 9

⇒ (1/648) × m = 1/9

∴ m = 72

Thus, a total of 72 X-men
are required to do the job in just 9 hours.

**5. D)**Total profit at the end of one year, P = Rs.21,880

Time period for which
Geeta invested money, t

_{1}= 1 year = 12 months
Time period for which
Shreya invested money, t

_{2}= 12 – 4 = 8 months
Time period for which
Saina invested money, t

_{3}= 12 – (4+2) = 6 months
Amount invested by Geeta,
I

_{1 }= Rs. 45,000
Amount invested by
Shreya, I

_{2 }= Rs. 72,000
Amount invested by Saina,
I

_{3 }= Rs. 87,500
Share of profit of Saina
= (I

_{3 }× t_{3 }× P)/(I_{1 }× t_{1 }+ I_{2 }× t_{2 }+ I_{3 }× t_{3})
⇒ Share of profit of Saina = (87500 × 6 × 21880)/(45000 × 12 +
72000 × 8 + 875000 × 6) = Rs.7000

∴ Share of profit of Saina = Rs.7,000

**6. A)**Let my age be M years and my mother’s age be N years.

⇒ Age of my father = (M + N) years

When I was born, the
ratio of ages of my mother and father was 5 : 6.

I was born M years back.

⇒ (N - M)/(M + N - M) = 5/6

⇒ 6N - 6M = 5N

⇒ N = 6M

After 8 years, the ratio
of my age and my mother’s age will be 3 : 8.

⇒ (M + 8)/(N + 8) = 3/8

⇒ 8M + 64 = 3N + 24

Put N = 6M

⇒ 8M + 40 = 18M

⇒ M = 40/10 = 4

⇒ N = 6M = 6 × 4 = 24

Mother’s age = 24 years

∴ I am 4 years old. When I was 2 years old, it would be 2
years back from now, my mother’s age was 22 years.

**7. A)**Money invested by George is Rs. 13000 for 2 years at 20% per annum in compound interest.

We know, in case of
compound interest,

A = P × (1 + R/100)^T

Amount George will have
after 2 years = 13000 × (1 + (20/100))

^{2 }= Rs. 18720
After 2 years, it was
notified that the scheme will now work on the principle of simple interest.
After further 2 years, George withdrew all the money from investment.

⇒ For next 2 years, George has invested Rs. 18720 at 20% per
annum in simple interest.

We know, in case of
simple interest,

A = P + (PTR/100)

Amount George will have
after next 2 years = 18720 + [(18720 × 20 × 2)/100] = Rs. 26208

Due to pre mature
withdrawal, a penalty of Rs. 1000 was charged.

Amount of money that
George will get = Rs. 26208 – Rs. 1000 = Rs. 25208

**8. C)**In 2 out of 5 cases, Gary will use defective machine. In these cases, Gary will give 900 grams of rice but will take price for 1 kg.

In ramining 3 out of 5
cases, Gary will give 1 kg of rice and will get price of 1 kg.

In total, for 5 kg,

Amount of rice given = 2
× 900 grams + 3 × 1 kg = 4 kg and 800 grams = 4.8 kg = 4800 grams

Price taken for amount of
rice = 5 kg

Profit will be price of
200 grams rice and cost price will be price of 4800 grams of rice.

We know, profit
percentage = (Profit/Cost Price) × 100 = (200/4800) × 100 = 4.17%

**9. A)**If we put 1 marble in the first box and the rest in the other, then the number of ways it can be done

=

^{6}C_{1}×^{5}C_{5 }= 6 × 1 = 6
If we put 2 marble in the
first box and the rest in the other, then the number of ways it can be done

=

^{6}C_{2}×^{4}C_{4 }= 15 × 1 = 15
If we put 3 marble in the
first box and the rest in the other, then the number of ways it can be done

=

^{6}C_{3}×^{3}C_{3 }= 20 × 1 = 20
If we put 4 marble in the
first box and the rest in the other, then the number of ways it can be done

=

^{6}C_{4}×^{2}C_{2 }= 15 × 1 = 15
If we put 5 marble in the
first box and the rest in the other, then the number of ways it can be done

=

^{6}C_{5}×^{1}C_{1 }= 6 × 1 = 6
The number of ways that 6
different marbles can be put in boxes of different sizes so that no box remain
empty is = 6 + 15 + 20 + 15 + 6 = 62

**10. B)**If the largest possible square is drawn inside a circle, length of diameter will be equal to length of diagonal of square.

Length of diameter = 2 ×
20 cm = 40 cm

Diagonal of square = 40
cm

Side length of square =
40 cm/√ 2 = 20√ 2 cm = 28.28 cm

The size of the circle is
reduced so that its area becomes 0.36 times of the original.

We know that area of a
circle is directly proportional to square of radius.

⇒ If area becomes 0.36 times, radius will become √ 0.36 times, i.e., 0.6 times.

⇒ Radius of reduced circle = 0.6 × 20 cm = 12 cm

⇒ Diameter of reduced circle = 2 × 12 cm = 24 cm

We observe that diameter
of reduced circle is less than side length of square.

⇒ Reduced circle will lie completely inside square.

∴ Area that lies inside the square but outside the circle =
Area of square – Area of reduced circle = (20√ 2 × 20√ 2) – (3.14 × 12 x 12) =
347.84 square cm