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Quadratic Equation
Directions (1 – 5): 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,
a) If x> y
b) If x ≥ y
c) If x = y or relationship can’t be determined.
d) If x< y
e) If x ≤ y
1)
I) x2 + 12x + 20 = 0
II) y2 + 13y + 22 = 0
Â
2)
I) 5x2 + 15x + 10 = 0
II) y2 – y – 2 = 0
Â
3)
I) x2 – 27x + 182 = 0
II) y2 – 23y + 132 = 0
Â
4)
I) 2x2 – 18x + 36 = 0
II) y2 – 12y + 32 = 0
5)
I) x2 + 23x + 120 = 0
II) y2 + 13y + 42 = 0
Application Sums
6) If the difference between the simple interest and Compound interest on a certain sum for 2 years at 8 % is Rs. 160. Find the sum?
a) Rs. 25000
b) Rs. 28000
c) Rs. 32000
d) Rs. 36000
e) None of these
7) 6 years ago, the ratio of ages of A and B is 2: 3. After 6 years, the ratio of ages of A and B is 3: 4. Find the sum of the present ages of A and B?
a) 64 years
b) 72 years
c) 68 years
d) 66 years
e) None of these
8) Pipe A opened to fill the tank and after 4 hours is closed and then Pipe B opened and fills the remaining tank in 10 hours. If Pipe B alone fill 25% of the tank in 5 hours, in how many hours pipe A and B together fill the tank completely?
a) 8(5/7) hours
b) 4(5/7) hours
c) 5(5/7) hours
d) 6(5/7) hours
e) None of these
9) A bag contains x+4 pink, 6 green and 8 brown colour balls; if two balls are taken random and the probability of getting both are green colour balls is 5/92, then find the difference between the no. of pink colour balls and the no. of brown colour balls
a) 1
b) 2
c) 3
d) 4
e) 0
10) Average ages of 5 persons A, B, C, D and E is 38 years. Ratio of the ages of A to C is 3:4 and after 5 years the ratio of the ages of B to D becomes 2: 3. If the difference between the ages of A and B is 2 years, then what is the present age of E?
a) 10 years
b) 39 years
c) 40 years
d) 48 years
e) Cannot be determine
Answers :
Directions (1-5) :
1) Answer: C
I) x2 + 12x + 20 = 0
(x + 2) (x + 10) = 0
x= -2, -10
II) y2 + 13y + 22 = 0
(y + 2) (y + 11) = 0
y = -2, -11
Relationship between x and y cannot be established.
2) Answer: E
I) 5x2 + 15x + 10 = 0
5x2 + 10x + 5x + 10 = 0
5x(x + 2) + 5(x + 2) = 0
(5x + 5)(x + 2) = 0
x = -1, -2
II) y2 – y – 2 = 0
(y + 1) (y – 2) = 0
y = -1, 2
Hence x ≤ y
3) Answer: A
I) x2 – 27x + 182 = 0
(x – 14) (x – 13) = 0
x = 14, 13
II) y2 – 23y + 132 = 0
(y – 11) (y – 12) = 0
y = 11, 12
Hence x > y
4) Answer: C
I) 2x2 – 18x + 36 = 0
2x2 – 12x – 6x + 36 = 0
2x(x – 6) – 6(x – 6) = 0
(2x – 6)(x – 6) = 0
x = 3, 6
II) y2 – 12y + 32 = 0
(y – 4) (y – 8) = 0
y = 4, 8
Hence Relationship between x and y cannot be established.
5) Answer: D
I) x2 + 23x + 120 = 0
(x + 15) (x + 8) = 0
x = -15, -8
II) y2 + 13y + 42 = 0
(y + 6) (y + 7) = 0
y = -6, -7
Hence x < y
Directions (6-10) :
6) Answer: A
Diff = Sum*(r/100)2
160 = Sum*(8/100)2
160*100*100/64 = Sum
Sum = Rs. 25000
7) Answer: B
6 years ago, the ratio of ages of A and B = 2: 3 (2x, 3x)
Present ages of A and B = 2x + 6, 3x + 6
After 6 years, the ratio of ages of A and B = 3: 4
According to the question,
(2x + 12)/(3x + 12) = (3/4)
8x + 48 = 9x + 36
x = 12
Sum of the present ages of A and B = 2x + 3x + 12 = 72 years
8) Answer: C
B = 5 * 4/1 = 20 hours
4/x + 10/20 = 1
4/x = ½
1/x = 1/8
A + B = 1/20 + 1/8
= 7/40
Time = 40/7 hours = 5(5/7) hours
9) Answer: B
Given,
6c2/(x+18)c2=5/92
X2+35x-246=0
Simplify the above equation we get x=6
Required difference = 10-8=2 balls
10) Answer: E
A + B + C + D + E = 190 years
A/C = 3/4
B + 5/D + 5 = 2/3
2D + 10 = 3B + 15
2D – 3B = 5
We cannot find the answer.
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