Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Prelims 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
Simplification
Directions (01-06): What value should come at the place of question mark in the following questions?
1) ((598 + (729)1/2)/81 x (324/5) x (39/14) % of 1260 =? + 17500
A.40
B.50
C.60
D.70
E.None of these
2) 21161/2Â Â – 23Â % of 2700 + 180 = ?
A.11
B.10
C.12
D.9
E.None of these
3) 32/8 x 22÷ 2?/16 = 128
A.1
B.5
C.4
D.2
E.None of these
4) 70 % of 620 ÷ (2/9) +? = 452
A.79
B.77
C.72
D.75
E.None of these
5) 7291/2 ÷ 7291/3 x (8/3) % of 67800 – ? = 6251/2 ÷ 5 x 155
A.4649
B.5259
C.4640
D.4330
E.5950
6) (17281/3)2Â % x 6900/39 x 130 + ?2Â = 36369
A.60
B.49
C.59
D.57
E.None of these
Time and work
7) A is 1.2 times as efficient as that of B and C is 200% more efficient than B. If A and B take 15/11 days for completing 1/4th of the work then find the time taken by C to complete half of the work.
A.5 days
B.4 days
C.6 days
D.3 days
E.None of these
Mensuration
8) The base radius of right circular cylinder is 75% of height of cylinder which is 24 cm. If radius is increased by 33(1/3)% and height is reduced by 25%, find % change in volume of cylinder.
A.25%
B.33.33%
C.30.75%
D.28.5%
E.None of these
Boat and stream
9) The speed of boat downstream is 150% of speed of boat upstream. If the time required to cover 18 km downstream and coming back requires 12 hours, find speed of stream.
A.0.36
B.0.56
C.0.625
D.0.32
E.0.48
Average
10) The average age of A, his wife and their daughter is 20 years. His wife is 6 year younger than A and his wife was 21 years old when his daughter was born. What is the average age of A and his daughter?
A.17.5 Years
B.19.5 Years
C.18 Years
D.21 Years
E.None of these
Answers :
1) Answer: B
(598 + (729)1/2)/81 x (324/5) x (39/14) % of 1260 =? + 17500
(598 + 27)/81 x (324/5) x (39/14) x 1/100 x 1260=? + 17500
(625/81) x (324/5) x (39/14) x 1/100 x 1260 =? + 17500
17550 – 17500 =?
50 =?
2) Answer: B
21161/2 – 8/100 x 2700 + 180 =?
46 – 8 x 27 + 180 =?
46 – 216 + 180 =?
10 =?
3) Answer: A
32/8 x 22÷ 2? /16 = 128
22 x 22÷ 2?= 8
22 + 2 – ? = 23
2 + 2 – ? = 3
? = 1
4) Answer: C
70 % of 620 ÷ (2/9) +? = 452
70/100 x 620 x 9/2 +? = 2025
1953 +? = 2025
? = 2025 – 1953 = 72
5) Answer: A
7291/2 ÷ 7291/3 x (8/3) % of 67800 – ? = 6251/2 ÷ 5 x 155
27 ÷ 9 x 8/3 x 1/100 x 67800 – ? = 25/5 x 155
5424 – ? = 775
5424 – 775 = 4649 = ?
6) Answer: D
(17281/3)2Â % x 6900/39 x 130 + ?2Â = 36369
122/100 x 6900/39 x 130 + ?2Â = 36369
144 x 69/39 x 130 + ?2Â = 36369
33120 + ?2Â Â = 36369
?2 = 36369 – 33120 = 3249
? = 32491/2Â Â = 57
7) Answer: E
Let the efficiency of A and B be 6x and 5x units per day respectively.
So, efficiency of C = 5x + 5x * 200/100 = 15x units per day
Total work = (15/11) * 4 * 11x = 60x
Required time = 30x/15x = 2 days
8) Answer: B
Radius = (75/100) × 24 = 18 cm
Volume of right circular cylinder = πr2h = π × 182 × 24 = 7776π
If radius increased by 33.33% -> (133.33/100) × 18 = 24
If height decreased by 25% -> (75/100) × 24 =18
New volume = π × 242 ×18 = 10368π
Therefore % change in volume = [(10368Ï€ – 7776Ï€)/ 7776Ï€]×100
= [2592π/7776π]×100 = 33.33%
9) Answer: C
Sd = (150/100)Su
Sd =1.5 Su
S=D/T i.e T = D/S
Td +Tu = T
D/Sd + D/Su = 12
18/1.5Su + 18/Su = 12
(18 +27)/1.5Su = 12
45/12 = 1.5Su
Su = 2.5Â km/hr
Sd = 3.75
Speed of stream = (3.75 – 2.5)/2 = 0.625 km/hr
10) Answer: A
Present Age = 20 × 3 = 60
W = A – 6; W = D + 21
Now, A + W + D = 60
A + (A – 6) + (W – 21) = 60
A + (A – 6) + (A – 6 – 21) = 60
A + (A – 6) + (A – 27) = 60
3AÂ – 33 = 60
3A = 93
A=31
Age of A = 31; age of wife = A – 6 = 31 – 6 = 25; age of daughter= W – 21 = 4
Average age of A and his daughter = (31+4)/2 = 17.5 Years
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