# Quantitative Aptitude Practice Test for NIACL

QUANTITATIVE APTITUDE PRACTICE TEST
1.A triangular based right prism having perimeter 35cm. The lateral surface area of the prism is 280 sq.cm. What is the hieight of the prism?
a)8cm
b)10cm
c)11cm
d)6.5cm
e)18cm

2.A train 300m long runs at a speed of 72 kmph. To pass a 200m platform it will take what time?
a)25s
b)20s
c)18s
d)15s
e)12s

3.A solid ornament box of dimensions (44cm × 27cm × 24cm) made off gold is melted to make balls of radius 3cm each. Find the number of balls:
a)200
b)252
c)250
d)240
e)244

4.A rectangular sized room having dimensions (12m × 11m × 10m). What is the Volume and total surface area of the room respectively?
a)1220 cu.m , 540 sq.m
b)1088 cu.m , 650 sq.m
c)1320 cu.m , 724 sq.m
d)1440 cu.m , 680 sq.m
e)1460 cu.m , 760 sq.m

5.Tanmoy's age is three times as his daughter's age. 10 years later, the sum of their ages will be 72 years. What is the ratio of the present age of Tanmoy to that of his daughter?
a)5:4
b)1:5
c)2:7
d)2:5
e)3:1

6.A rhombus having area of 78 sq.m and perimeter is 52m. What is the altitude of the rhombus?
a)7m
b)6m
c)5m
d)10m
e)11m

7.A train 200m long crosses a lamp post in 10 seconds. Another train travelling at the same speed crosses a 300m long platform in 20 seconds. Find the time taken by the second train to cross the first train, if the first train is stationary?
a)12 seconds
b)15 seconds
c)22 seconds
d)28 seconds
e)9 seconds

8.An article is marked 40% over its cost price. Two successive discounts of 15% and 10% are allowed on the marked price of the article. What is the loss/profit percentage?
a)7.1% profit
b)7.1% loss
c)9% profit
d)9% loss
e)11% loss

9.Amar, Akbar and Anthony entered into a partnership with the capital in the ratio of 4:7:5. If the end of one year the ratio of profits is 8:14:10. What is the ratio of their periods of investment?
a)2:2:2
b)1:3:4
c)3:5:6
d)1:7:8
e)3:3:1

10.X, Y and Z can complete a piece of work in 8, 12 and 24 days respectively. X and Z started working and Y joined them after one day. If Z left 2 days before completion of the work, in how many days was the work finished?
a)11/2 days
b)19/5 days
c)14/3 days
d)15/4 days
e)6/5 days

1.a)
2.a)
3.b)
4.c)
5.e)
6.b)
7.b)
8.a)
9.a)
10.c

Solution
1.Lateral surface area of the prism = Perimeter × Height
now, 35 × h = 280
or. h = 280/35 = 8cm

2.Sum of the length of train and platform = (300 + 200)m = 500m
Speed of the train = 72 kmph = 72 × 5/18 = 20 m/sec
Time taken by the train to cross the platform = 500/20 = 25sec

3.Number of gold balls = Volume of the box/Volume of the gold balls
= (44 × 27 × 24)/(4/3 × 22/7 × 3 × 3 × 3)
= 252

4.Volume = l × b × h = 12 × 11 × 10 = 1320 cu.m
Total surface area = 2(lb + bh + hl) = 2(12×11 + 11×10 + 10×12) = 724 sq.m

5.Let, the present age of Tanmoy's daughter = x years
Then, Tanmoy's age = 3x
After 10 years, Tanmoy's age = (3x + 10) years and Tanmoy's daughter age = (x + 10) years
Now, 3x + 10 + x + 10 = 72
or, x = 13
Tanmoy's age = 3 × 13 = 39
Required ratio = 39 : 13 = 3 : 1

6.We have perimeter of the rhombus = 52m and area of the rhombus = 78 sq.m
Now, side of the rhombus = (Perimeter/4) = 52/4 = 13m
Therefore, altitude of the rhombus = Area of rhombus/side = 78/13 = 6m

7.Speed of the 1st train = 200/10 = 20 m/s
Now, speed of the 2nd train = speed of the 1st train
Therefore, 20 = (x + 300)/20 [where x = speed of the 2nd train]
or, x = 100
Now, total length = 100 + 200 = 300m
Now, s = 20 m/s
Therefore, time = 300/20 = 15 seconds

8.Let, CP = Rs 100
Marked price = Rs 140
Now, 15% of 140 = Rs.21 and 10% of (140 - 21) = Rs.11.9
Therefore, SP = 140 - (21 + 11.9) = 107.1
Therefore, profit% = [(107.1 - 100)/100] × 100 = 7.1%

9.Ratio of periods of investment = (Ratio of profits)/(Ratio of capitals)
= (8/4):(14/7):(10/5)
= 2:2:2

10.One day's work of X = 1/8
Y's one day's work = 1/12
Z's one day's work = 1/24
Let, the number of days taken be s.
X worked for s days.
Y for (s-1) days
Z for (s-2) days
Now, (s/8) + (s-1)/12 + (s-2)/24 = 1 (total work)
or, s = 28/6 = 14/3 days 