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__SSC CHSL- Practice Aptitude Questions (Mixed Problems) With Solutions Set-45__
Dear Readers, Important Aptitude Questions for SSC CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1). Pipe A alone can fill a tank in 16 hours.
Pipe B can fill it in 12 hours. If both the pipes are opened and after 4 hours
Pipe A is closed, then the Pipe B will fill the tank in

a)
3 hours

b)
7 hours

c)
5 hours

d)
4 hours

2). Shyam goes to his office by a car at a
speed of 45 kmph and reaches there 12 minutes earlier. If he goes at a speed of
30 kmph he reaches there 10 minutes late. What is the distance of his office?

a)
33 km

b)
44 km

c)
36 km

d)
30 km

3). Radhani Express covers 20 km in 10
minutes. If the speed is increased by 20%, the time taken by Rajdhani to cover
1200 km will be

a)
8 hrs 25 minutes

b)
8 hrs 40 minutes

c)
8 hrs 15 minutes

d)
8 hrs 20 minutes

4). Water is flowing at the rate of 10 kmph
through a pipe of diameter 21 can into a
rectangular tank, which is 100m long and 22m wide. The time Taken (in hours)
for 14cm rise in the level of water in the tank is ?

a)
7 (11/19)

b)
8(5/19)

c)
8(8/9)

d)
7(1/9)

5). The average temperature on Mon, Tue and
Wed was 42

^{o}C, and the average temperature on Tue, Wed and Tue was 39^{o}C. If the temperature on Monday was 39^{o}C, then the temperature on Thursday was
a)
42

^{o}C
b)
40

^{o}C
c)
30

^{o}C
d)
36

^{o}C
6).If A : B = 7 : 10, B : C = 8 : 9 and C : D
= 4 : 5, then find the ratio of A : D.

a)
36 : 59

b)
19 : 20

c)
8 : 9

d)
112 : 225

7). The cost of a piece of diamond varies with
the square of its weight. A diamond valued at Rs.20,736 is cut into 3 pieces
whose weights are in the ratio of 1 : 2 : 3. Find the loss in value due to
cutting.

a)
Rs.12672

b)
Rs.528

c)
Rs.5808

d)
Rs.11616

8). Find the value of ∠A + ∠B + ∠C + ∠D + ∠E

a)
180

^{o}
b)
90

^{o}
c)
360

^{o}
d)
720

^{o}
9). Solve

(1/6) + (1/12) + (1/20) + (1/30) + (1/42) + (1/56)
= ?

a)
1

b)
3/8

c)
½

d)
1/8

10). If A = 7-4√3, Then find the value of √A+
(1/√A)

a)
2√3

b)
√3

c)
√2

d)
4

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

**Solutions :**

**1).**The part of tank filled by A in our hour 1/16

The part of tank
filled by pipe B in one hour = 1/12

The part of tank
filled by (A+B) in 1 hour = (1/16) + (1/12)
= (3+4)/48 = 7/48

The part of tank
filled by (A + B) in 4 hours = (7/48) × 4 = 7/12

Remaining part of
the tank = 1 – 7/12 = 5/12

Time taken by
pipe B to fill 5/12 part = 5/12 ×12 =
5hrs

**Answer: c)**

**2).**Required distance = { Multiplication of the Speeds / Difference of speeds } × Sum of time

= (45×30)/(45-30)
× (22/60)km

**Answer: a)**

**3).**Given distance = 20km

Time = 10 minutes
= 10 / 60 hrs = 1/6 hrs

Speed = Distance
/ time = 20×6 = 120 kmph

After 20%
increased in speed

The new speed
will be (120×120)/100 = 144 kmph

Distance = 1200
km

Time = Distance /
Speed = 1200 / 144

= 8 hrs 20
minutes

**Answer: d)**

4). Volume of
water = { (22×100×14)/100 }m

^{3}
= (22 × 100 × 100
× 100 × 14) cm

^{3}
Diameter = 21 cm

Radius = 21/2 cm

Volume of water
in 1 hr = 22/7 × 21/2 × 21/2 × 1000 × 100 cm

^{3}
Required time = {
(22×100×100×100×14×7×2×2)/(22×21×21×1000×100) }

= 80/9 hrs

= 8 (8/9) hrs

**Answer: c)**

**5).**Total temperature on Mon, Tue, Wed = 42 × 3 = 126

^{o}C

Temperature on
Mon = 39

^{o}C
Temperature on (Tue
+ Wed) = 126 – 39 = 87

^{o}C
Total temperature
on Tue, Wed and Thu = 39 × 3 = 117

^{o}C
Thu = 117 – 87 =
37

^{o}C**Answer: c)**

**6).**

A : B = 7 : 10

B : C = 8 : 9

C : D = 4 : 5

-------------

**-------**
A= 7×8×4 = 224

B= 10×8×4 = 320

C= 10×9×4 = 360

D= 10×9×5 = 450

Required ratio =
112 : 160 : 180 : 225

Hence ratio of A
: D = 112 : 225

**Answer: d)**

**7).**Suppose the weight of 3 pieces are x, 2x and 3x

Then, weight of
diamond = x + 2x + 3x = 6x

Value = (6x)

^{2}= 20736
Or, x

^{2}= 20736/36
x

^{2 }= 576 = 24
Total value of 3
pieces = x

^{2}+ (2x)^{2}+ (3x)^{2}= x^{2}+ 4x^{2}+ 9x^{2}= 14x^{2}
Loss

^{ }in value = 36x^{2}-14x^{2}= 22x^{2}= 22 × (24)^{2}= Rs.12672**Answer: a)**

**8).**

Let us call the
internal angles M, N, O, P and Q respectively ie. ∠OMN = m, ∠MNP = n and so on.

Now, sum of
internal angles = m + n + o + p + q = (2×5-4) 90

^{o}- 6× 90^{o }– 3 × 180^{o}…. (i)
Let us now
consider ∆AMN

Here ∠AMN = 180

^{o}– m, ∠AMN – 180^{o}= n and third angle = ∠ A
∠A + ∠AMN + ∠ANM =
180

^{o}
Or, ∠A + 180

^{o}– m + 180^{o}- n = 180^{o}
Or, ∠A = m + n –
180

^{o}
Similarly, ∠B = n
+ p – 180

^{o }and so on.
∠A + ∠B + ∠C + ∠D
+ ∠E = (m+n-180

^{o}) + (n+p-180^{o}) + (p+q-180^{o}) + (q+o-180^{o}) + (o + m – 180^{o})
= 2(m + n + o + p
+ q) – 5 × 180

^{o}
= 2 × 3 × 180

^{o}– 5 × 180^{o}------------ using (i)
= 180

^{o}**Answer: a)**

**9).**

Given expression –
1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56

= ½ - 1/3 + 1/3 –
¼ + ¼ - 1/5 +1/5 – 1/6 + 1/6 – 1/7 +1/7 – 1/8

= ½ - 1/8

=(4-1) / 8 = 3/8

**Answer: b)**

**10).**

A= 7 - 4√3

Or, A = 7 - 2√12

= (2)

^{2}+ (√3)^{2}– 2.2. √3
So. √A will be 2
- √3

Now, given
expression √A + 1/√A = 2 - √3 + [1/(2-√3)]
× [ (2+√3)(2+√3)

= 2 - √3 + (2+√3)/(4-3) { let a=2, b=-√3 }

= 2 - √3 + 2 + √3
= 4

**Answer: d)**

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