### Practice Aptitude Questions for SSC CGL,CHSL,CPO Exam

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 m2. 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 cm2.
b)    (22 / 7) × (7 / 2)2 cm2
c)    (22 / 7) × (15 / 2)2 cm2
d)    (22 / 7) × (9 / 2)2 cm2

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 = 49k2 – k2
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 cm2
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)