Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Mains 2021 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.
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Quadratic equation
Directions (01-05): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give answer as,
1) I) x(5/3) ÷ 8 = 4.5 ÷ x(1/3)
II)Â y * 221 = 17 * (y2+ 42)
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
2) I)Â (x + 2)! = (x + 1)!
II)Â 2 * y! = (y + 1)!
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
3) I)2x2 – 20x + 32 = 0
II)3y2– 3y – 18 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
4) I)Â x2+ 5x – 84 = 0
II)Â y2+ 26y + 168 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
5) I)x2Â – 28x + 195 = 0
II) y2– y – 240 = 0
A.x > y
B.x ≥ y
C.x = y or relationship can’t be determined.
D.x < y
E.x ≤ y
Mixture and allegation
6) Vessel A contains 480 liters mixture of oil and water in the ratio of 5:3.If 5x liters of oil and 4x liters of water taken out and poured into vessel B, now the ratio of the oil and water in vessel A is 2:1. If the total quantity of vessel C contains (7x + 60) liters mixture of milk and water in the ratio of 3:2, then find the quantity of water in vessel C?
A.40 liters
B.60 liters
C.80 liters
D.50 liters
E.90 liters
Mensuration
7) Ratio of the curved surface area of the cone to volume of cone is 56:25. If radius of the cone to cylinder is 1:2 and the height of the cylinder is half of the height of cone, total surface area of the cone is 704 cm2. What is the difference between the volume of the cone and cylinder?
A.320 cm2
B.234 cm2
C.216 cm2
D.278 cm2
E.Cannot be determined
SI AND CI
8) Meena invests Rs.x in a compound interest scheme at the rate of 20% per annum for 2 years, Nila invests Rs.y in the same compound interest scheme at the rate of 20% per annum for 2 years. The total interest received by Meena and Nila is Rs.4092. If Nirmal invests Rs.2x in a simple interest scheme at the rate of 10% per annum for 2 years and Rani invests Rs.2y in a simple interest scheme at the rate of 15% per annum for 3 years and total interest received by Rani and Nirmal is Rs.5970, find the value of x?
A.Rs.4500
B.Rs.4800
C.Rs.5100
D.Rs.5400
E.Rs.5700
Probability
9) When 2 students are selected from a group of x boys and 7 girls, the probability of same gender is 31/66. What is the probability that when 3 people are selected at random, all 3 are boys? (x should be whole number)
A.1/22
B.3/22
C.5/21
D.4/21
E.2/11
Profit and loss
10) The shopkeeper sold the TV at the discount of 20% and earned the profit of 20%. The cost price of the table is two-fifth of the cost price of the TV. If the marked price of the TV is Rs.33000 and the marked price of the table is 35% above its cost price and sold it at the discount of 20%. Find the selling price of table?
A.Rs.9504
B.Rs.9498
C.Rs.9512
D.Rs.9478
E.Rs.9518
Answers :
1) Answer: E
x(5/3) ÷ 8 = 4.5 ÷ x(1/3)
x(5/3 + 1/3)Â = 4.5 * 8
x6/3Â = 36
x = 6, -6
y * 221 = 17 * (y2Â + 42)
y2 – 13y + 42 = 0
y2 – 6y – 7y + 42 = 0
y(y – 6) – 7(y – 6) = 0
y = 6, 7
x ≤ y
2) Answer: D
(x + 2)! = (x + 1)!
(x + 2)(x + 1)! = (x + 1)!
x+ 2 = 1
x= 1 – 2
x = -1
2 * y! = (y + 1)!
2 * y! = (y + 1)y!
y = 2 – 1
y = 1
x < y
3) Answer: C
2x2 – 20x + 32 = 0
2x2 – 16x – 4x + 32 = 0
2x(x – 8) – 4(x – 8) = 0
(2x – 4) (x – 8) = 0
x = 2, 8
3y2 – 3y – 18 = 0
3y2 – 9y + 6y – 18 = 0
3y(y – 3) + 6(y – 3) = 0
(3y + 6)(y – 3) = 0
y = 3, -2
Relationship between x and y cannot be established.
4) Answer: B
x2Â + 5x – 84 = 0
x2 + 12x – 7x – 84 = 0
x(x + 12) – 7(x + 12) = 0
(x – 7)(x + 12) = 0
x = 7, -12
y2Â + 26y + 168 = 0
y2Â + 14y + 12y + 168 = 0
y(y + 14) + 12(y + 14) = 0
(y + 12)(y + 14) = 0
y = -12, -14
x ≥ y
5) Answer: C
x2– 28x + 195 = 0
x2–13x – 15x + 195 = 0
x(x – 13) – 15(x – 13) = 0
(x – 13)(x – 15) = 0
x = 13, 15
y2 – y – 240 = 0
y2 – 16y + 15y – 240 = 0
y(y – 16) + 15(y – 16) = 0
(y + 15)(y – 16) = 0
y = -15, 16
Relationship between x and y cannot be established.
6) Answer: C
Oil in vessel A = 5/8 * 480 = 300 liters
Water in vessel A = 3/8 * 480 = 180 liters
(300 – 5x)/(180 – 4x) = 2/1
360 – 8x = 300 – 5x
3x = 60
x = 20
Quantity of vessel C = 7 * 20 + 60 = 200 liters
Water in C = 2/5 * 200 = 80 liters
7) Answer: E
TSA of cone = 22/7 * r * (l + r)
704 = 22/7 * r * (l + r)
r * (l + r) = 224
(1/3 * 22/7 * r * r * h)/(22/7 * r * l) = 25/56
rh/3l = 25/56
56 * r * h = 3 * l * 25
We cannot find the answer.
8) Answer: B
CI = P * (1 + R/100)n – P
SI = P * N * R/100
4092 = x * (1 + 20/100)2 – x + y * (1 + 20/100)2 – y
4092 = 0.44x + 0.44y
x + y = 9300—-(1)
5970 = (2x * 10 * 2/100) + (2y * 15 * 3/100)
597000 = 40x + 90y
4x + 9y = 59700—-(2)
5y = 22500
y = 4500
x = 9300 – 4500 = Rs.4800
9) Answer: A
(xC2/(x + 7)C2) + (7C2/(x + 7)C2)= 31/66
(x * (x – 1) + 7 * 6)/((x + 7) * (x + 6)) = 31/66
(x2 – x + 42) * 66 = 31 * (x2 + 6x + 7x + 42)
66x2 – 66x + 2772 = 31x2 + 403x + 1302
35x2 – 469x + 1470 = 0
5x2 – 67x + 210 = 0
5x2–25x – 42x + 210 = 0
5x(x – 5) – 42(x – 5) = 0
(5x – 42)(x – 5) = 0
x = 42/5, 5
Required probability = 5C3/12C3
= 60/1320
= 1/22
10) Answer: A
33000 * 80/100 = CP of TV * 120/100
CP of TV = 22000
CP of Table = 2/5 * 22000 = Rs.8800
SP of table = 8800 * 135/100 * 80/100
= 9504
