__Practice Aptitude Questions__:

**With Solutions SSC CGL, CHSL Exams Set-28**
Dear Readers, Important Aptitude Questions for SSC CGL,CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

**1).**If x – y = (x + y) / 7 = xy / 4, the numerical value of xy is

a)
4 / 3

b)
3 / 4

c)
1 / 4

d)
1 / 3

**2).**The area of a field in the shape of trapezium measures 1400 m

^{2}. The perpendicular distance between its parallel side is 24 m. If the ratio of the parallel sides is 5 : 3, the length of the longer parallel side is

a)
75 m

b)
45 m

c)
120 m

d)
60 m

**3).**The area of the greatest circle, which can be inscribed in a square whose perimeter is 120 cm, is

a)
(22 / 7) × (15)

^{2}cm^{2}.
b)
(22 / 7) × (7 / 2)

^{2}cm^{2}
c)
(22 / 7) × (15 / 2)

^{2}cm^{2}
d)
(22 / 7) × (9 / 2)

^{2}cm^{2}**4).**If (x + y – z)

^{2}+ (y + z – x)

^{2}+ (z + x – y)

^{2}= 0, then the value of x + y – z is

a)
√3

b)
3√3

c)
3

d)
0

**5).**Starting from his house one day, a student walks at a speed of 2(1 / 2) km / h and reaches his school 6 min late. Next day at the same time, he increases his speed by 1 km/h and reaches the school 6 min early. How far is the school from his house?

a)
2 km

b)
1(1 / 2) km

c)
1 km

d)
1(3 / 4)km

**6).**The average of a marks in Mathematics for 5 students was found to be 50. Later, it was discovered that in the case of 1 student, the marks 48 were misread as 84. The correct average is

a)
40.2

b)
40.8

c)
42.8

d)
48.2

**7).**D and E are the points in the sides AB and AC respectively in ∆ ABC, AD = 8 cm, DB = 12 cm, AE = 6 cm and EC = 9cm, then BC is equal to

a)
2 / 5 DE

b)
5 / 2 DE

c)
3 / 2 DE

d)
2 / 3 DE

**8).**A vertical stick 15 m long casts a shadow 12 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is

a)
60 m

b)
62 m

c)
62.5 m

d)
63 m

**9).**A man gets Rs. 13 more by selling an article at a profit of 12(1 / 2)%, then selling it at a loss of 12(1 / 2)%. The cost price of the article is

a)
Rs. 25.50

b)
Rs. 38

c)
Rs. 52

d)
Rs. 65

**10).**A walks at a uniform rate of 4km an hour and 4h after his start, B bicycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A?

a)
16.7 km

b)
18.6 km

c)
21.5 km

d)
26.7 km

__Answers__**:**

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

__Solution:__**1).**x – y = (x + y) / 7 = xy / 4 = k

x – y = k; x + y = 7k; xy = 4k

(x + y)

^{2}– (x – y)^{2}= 49k^{2}– k^{2}
k = 1 / 3

xy = 4k = 4 × (1 / 3) = 4 / 3

**Answer: a)**

**2).**Let the parallel sides be 5x m and 3x m.

Area of trapezium = 1 / 2 (Sum of parallel
sides) × Distance between them

1440 = 1 / 2(5x + 3x) × 24

12 × 8x = 1440

x = 1440 / (12 × 8) = 15

The longer parallel side = 5x = 5 × 15 = 75 m

**Answer: a)**

**3).**Side of the square = 120 / 4 = 30 cm

Clearly, diameter of the greatest circle =
side of the square = 30 cm

Radius = 30 / 2 = 15 cm

Required area = π × (Radius)

^{2}
= (22 / 7) × (15)

^{2}cm^{2}**Answer: a)**

**4).**(x + y – z)

^{2}+ (y + z – x)

^{2}+ (z + x – y)

^{2}= 0

(x + y – z) = 0

[∵ (a – b)

^{2 }+ (b – c)^{2}+ (c – a)^{2}= 0 provides a – b = 0]**Answer: d)**

**5).**Let the required distance be x km.

Difference of time = 6 + 6 = 12 min

= 12 / 60 = 1 / 5 h

According to the question,

[x / (5 / 2)] – [x / (7 / 2) = 1 / 5

(2x / 5) – (2x / 7) = 1 / 5

(14x – 10x) / 35 = 1 / 5

4x / 35 = 1 / 5

x = 35 / (4 × 5) = 7 / 4 = 1(3 / 4)km

**Answer: d)**

**6).**Total marks obtained by 5 students = 50 × 5 = 250

Now, In this total marks, 84 is included
Instead of 48,

Correct total marks = 250 – 84 + 48 = 214

Correct average = 214 / 5 = 42.8

**Answer: c)**

**7).**In ∆ ADE and ∆ABC

AD / AB = 8 / 20 = 2 / 5

AE / EC = 6 / 15 =2 / 5

So, AD / AB = AE / AC

And ∠A = ∠A (common)

∆ ADE ~ ∆ ABC

DE / BC = AD / AB ⇒ DE / BC = 2 / 5

BC = 5 / 2 DE

**Answer: b)**

**8).**Let AB be a vertical stick and AC be its shadow

Also, let PQ be a tower having shadow PR.

As, ∆ABC ~ ∆ PQR

AB / PQ = AC / PR

15 / x = 12 / 50

x = (15 × 50) / 12 = 62.5m

Hence, height of the tower is 62.5 m.

**Answer: c)**

**9).**Let the CP of article be Rs.x

According to the question,

[100 + (25 / 2)] % of x - [100 - (25 / 2)] %
of x = 13

⇒ (x / 100) [100 + (25 / 2)] - [100 - (25 /
2)] = 13

(x / 100) × 25 = 13

x = 13 × 4 = Rs. 52

**Answer: c)**

**10).**Distance covered by A in 4 h = 4 × 4 = 16km

Relative speed of B with respect to A = 10 – 4
= 6 km/h

Time taken to catch A = 16 / 6 = 8 / 3h

Required distance = (8 / 3) × 10 = 80 / 3 =
26.7 km

**Answer: d)**