### SSC CHSL- Practice Aptitude Questions (Mixed problems) With Solutions Set-45

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 42oC, and the average temperature on Tue, Wed and Tue was 39oC. If the temperature on Monday was 39oC, then the temperature on Thursday was
a)    42oC
b)    40oC
c)    30oC
d)    36oC

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)    180o
b)    90o
c)    360o
d)    720o

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 }m3
= (22 × 100 × 100 × 100 × 14) cm3
Diameter = 21 cm
Radius = 21/2 cm
Volume of water in 1 hr = 22/7 × 21/2 × 21/2 × 1000 × 100 cm3
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 = 126oC
Temperature on Mon = 39oC
Temperature on (Tue + Wed)  = 126 – 39 = 87oC
Total temperature on Tue, Wed and Thu = 39 × 3 = 117oC
Thu = 117 – 87 = 37oC
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, x2 = 20736/36
x2 = 576 = 24
Total value of 3 pieces  = x2 + (2x)2 + (3x)2 = x2 + 4x2 + 9x2 = 14x2
Loss in value = 36x2-14x2 = 22x2 = 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) 90o - 6× 90o – 3 × 180o …. (i)
Let us now consider ∆AMN
Here ∠AMN = 180o – m, ∠AMN – 180o = n and third angle = ∠ A
∠A + ∠AMN + ∠ANM = 180o
Or, ∠A + 180o – m + 180o - n = 180o
Or, ∠A = m + n – 180o
Similarly, ∠B = n + p – 180o and so on.
∠A + ∠B + ∠C + ∠D + ∠E = (m+n-180o) + (n+p-180o) + (p+q-180o) + (q+o-180o) + (o + m – 180o)
= 2(m + n + o + p + q) – 5 × 180o
= 2 × 3 × 180o – 5 × 180o   ------------ using (i)
= 180o
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)