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Directions (Q. 1-5): What value should come in place of question mark (?) in the following questions?
(5/3) of? – (2/3) of (12/17) of 3162 = 42 % of 850
Correct
Answer c
(5x/3) – (2/3) * (12/17) * 3162 = (42/100) * 850
(5x/3) – 1488 = 357
(5x/3) = 1845
X = (1845*3)/5 = 1107
Incorrect
Answer c
(5x/3) – (2/3) * (12/17) * 3162 = (42/100) * 850
(5x/3) – 1488 = 357
(5x/3) = 1845
X = (1845*3)/5 = 1107
Question 4 of 20
4. Question
Directions (Q. 1-5): What value should come in place of question mark (?) in the following questions?
25 % of x – 1664 ÷ 13 + 222 ÷ 6 = 67
Correct
Answer d
25 % of x – 1664 ÷ 13 + 222 ÷ 6 = 67
(25/100)*x – 128 + 37 =67
(25/100)*x = 67 + 128 -37
(25/100)*x = 158
X = (158*100)/25
X = 632
Incorrect
Answer d
25 % of x – 1664 ÷ 13 + 222 ÷ 6 = 67
(25/100)*x – 128 + 37 =67
(25/100)*x = 67 + 128 -37
(25/100)*x = 158
X = (158*100)/25
X = 632
Question 5 of 20
5. Question
Directions (Q. 1-5): What value should come in place of question mark (?) in the following questions?
√x × 13 – 26 % of 2650 – 12 × 28 = 15
Correct
Answer d
√x × 13 – 26 % of 2650 – 12 × 28 = 15
√x × 13 – (26/100)*2650 – 336 = 15
√x × 13 – 689 – 336 = 15
√x × 13 = 15 +689 + 336
√x × 13 = 1040
√x = 1040/13 = 80
X = 80^{2} = 6400
Incorrect
Answer d
√x × 13 – 26 % of 2650 – 12 × 28 = 15
√x × 13 – (26/100)*2650 – 336 = 15
√x × 13 – 689 – 336 = 15
√x × 13 = 15 +689 + 336
√x × 13 = 1040
√x = 1040/13 = 80
X = 80^{2} = 6400
Question 6 of 20
6. Question
Directions (Q. 6 – 10): What value should come in place of (?) in the following series?
809, 811, 820, 848, 913, ?
Correct
Answer d
The difference is, 1^{3}+1, 2^{3}+1, 3^{3}+1, 4^{3}+1, 5^{3}+1,..
The answer is, 1039
Incorrect
Answer d
The difference is, 1^{3}+1, 2^{3}+1, 3^{3}+1, 4^{3}+1, 5^{3}+1,..
The answer is, 1039
Question 7 of 20
7. Question
Directions (Q. 6 – 10): What value should come in place of (?) in the following series?
87, 87, 83, 99, 63, ?
Correct
Answer b
The pattern is, +0^{2}, -2^{2}, +4^{2}, -6^{2}, +8^{2},..
The answer is, 127
Incorrect
Answer b
The pattern is, +0^{2}, -2^{2}, +4^{2}, -6^{2}, +8^{2},..
The answer is, 127
Question 8 of 20
8. Question
Directions (Q. 6 – 10): What value should come in place of (?) in the following series?
294624, 24552, 2232, 248, ?
Correct
Answer c
The pattern is, ÷ 12, ÷ 11, ÷ 9, ÷ 5,..
The answer is, 49.6
Incorrect
Answer c
The pattern is, ÷ 12, ÷ 11, ÷ 9, ÷ 5,..
The answer is, 49.6
Question 9 of 20
9. Question
Directions (Q. 6 – 10): What value should come in place of (?) in the following series?
856, 213, 52.25, ? , 2.015625
Correct
Answer c
The pattern is, ÷ 4 -1
The answer is, 12.0625
Incorrect
Answer c
The pattern is, ÷ 4 -1
The answer is, 12.0625
Question 10 of 20
10. Question
Directions (Q. 6 – 10): What value should come in place of (?) in the following series?
727, 513, 390, 328, ?
Correct
Answer a
The pattern is, -(6^{3} – 2), -(5^{3} – 2), -(4^{3} – 2), -(3^{3} – 2),..
The answer is, 303
Incorrect
Answer a
The pattern is, -(6^{3} – 2), -(5^{3} – 2), -(4^{3} – 2), -(3^{3} – 2),..
The answer is, 303
Question 11 of 20
11. Question
The ratio between cost price of three articles A, B and C is 2:3:4 while the profit percentage earned in these articles is in the ratio of 5:4:3 and profit earned on article C is Rs 300. If the cost price of article A is Rs 1000, find the selling price of article A?
Correct
Answer c
Cost price article C = 1000*4/2= Rs 2000
Profit percentage earned on article C= 300*100/2000= 15%
Percentage profit earned in article A= 15*5/3= 25%
Profit earned in article A= 1000*25/100= Rs 250
Selling price of article A= 1000+250= Rs 1250
Incorrect
Answer c
Cost price article C = 1000*4/2= Rs 2000
Profit percentage earned on article C= 300*100/2000= 15%
Percentage profit earned in article A= 15*5/3= 25%
Profit earned in article A= 1000*25/100= Rs 250
Selling price of article A= 1000+250= Rs 1250
Question 12 of 20
12. Question
A and B invest in the ratio of 25:17. If they divide 40% of the profit equally and rest in the proportion of their investments, then find B’s profit if A’s profit is Rs 1950.
Correct
Answer b
Let the total profit be x
Then
A’s profit= 0.2x+0.6x*25/42
1950= 0.2x (1+75/42)
X= 1950*42/0.2*117= Rs 3500
B’s profit= 3500-1950= Rs 1550
Incorrect
Answer b
Let the total profit be x
Then
A’s profit= 0.2x+0.6x*25/42
1950= 0.2x (1+75/42)
X= 1950*42/0.2*117= Rs 3500
B’s profit= 3500-1950= Rs 1550
Question 13 of 20
13. Question
A shopkeeper sells a transistor at 15% above its cost price. If he had bought it 5% more than what he paid for it and sold it for Rs 6 more, he would have gained 10%. The cost price of the transistor is
Correct
Answer c
Let the cost of the transistor be x
SP = 115/100*x
New CP = x* 105/100= 105x/100
New SP= 115x/100+ 6=105x/100*110/100
=>115x/100-231x/200= -6
=>x=1200
Incorrect
Answer c
Let the cost of the transistor be x
SP = 115/100*x
New CP = x* 105/100= 105x/100
New SP= 115x/100+ 6=105x/100*110/100
=>115x/100-231x/200= -6
=>x=1200
Question 14 of 20
14. Question
The present age of Ravi’s father is four times Ravi’s present age. Five years back, Ravi’s father was seven times as old as Ravi was at that time. What is the present age of Ravi’s father?
Correct
Answer c
Let the present age of Ravi be x
Present age of Ravi’s father= 4x
Now 5 yr ago
Ravi’s father’s age= 7 * Ravi’s age
= 4x-5=7(x-5)
= x= 10
Ravi’s Father’s age= 4*10= 40 years.
Incorrect
Answer c
Let the present age of Ravi be x
Present age of Ravi’s father= 4x
Now 5 yr ago
Ravi’s father’s age= 7 * Ravi’s age
= 4x-5=7(x-5)
= x= 10
Ravi’s Father’s age= 4*10= 40 years.
Question 15 of 20
15. Question
The length and breadth of a rectangular piece of land are in the ratio of 5: 3. The owner spent Rs 3000 for surrounding it from all the sides at the rate of Rs 7.50 per meter. The difference between length and breadth is:
Correct
Answer a
Perimeter of the field = 3000/7.50= 400m
So,
2(5x + 3x) = 400 => x = 25
So, length = 125 m & breadth = 75 m
Difference between length & breadth
= (125-75)m = 50m
Incorrect
Answer a
Perimeter of the field = 3000/7.50= 400m
So,
2(5x + 3x) = 400 => x = 25
So, length = 125 m & breadth = 75 m
Difference between length & breadth
= (125-75)m = 50m
Question 16 of 20
16. Question
Directions (Q. 16 – 20): Study the following pie-charts and table to answer these questions.
What is the difference between the Masters male student and Graduate male student from Course ‘P’?
Correct
Answer b
Masters Male Student from Course ‘P’
= 24*(16/100)*(7/12) lakh
= 2,24,000
Graduate male student from Course ‘P’
= 32*(15/100)*(7/16) lakh
= 210000
Difference = 224000- 210000 = 14000
Incorrect
Answer b
Masters Male Student from Course ‘P’
= 24*(16/100)*(7/12) lakh
= 2,24,000
Graduate male student from Course ‘P’
= 32*(15/100)*(7/16) lakh
= 210000
Difference = 224000- 210000 = 14000
Question 17 of 20
17. Question
Directions (Q. 16 – 20): Study the following pie-charts and table to answer these questions.
What is the ratio of the Masters female student of Course ‘T’ to Graduate female student of Course ‘S’?
Correct
Answer d
Masters female student of course T
= 24*(20/100)*(7/16) lakh
Graduate female student of course S
= 32*(12/100)*(7/12) lakh
Required ratio = [24*(20/100)*(7/16)] : [32*(12/100)*(7/12)]
= 210 : 224
= 15 : 16
Incorrect
Answer d
Masters female student of course T
= 24*(20/100)*(7/16) lakh
Graduate female student of course S
= 32*(12/100)*(7/12) lakh
Required ratio = [24*(20/100)*(7/16)] : [32*(12/100)*(7/12)]
= 210 : 224
= 15 : 16
Question 18 of 20
18. Question
Directions (Q. 16 – 20): Study the following pie-charts and table to answer these questions.
The Masters female student of Course ‘R’ is what per cent of the Graduate female student of that Course?
Correct
Answer c
Masters female student of course R
= 24*(15/100)*(4/9) = 1.6 lakh
Graduate female student of course R
= 32*(18/100)*(5/9) = 3.2 lakh
Required % =(1.6/3.2)*100 = 50 %
Incorrect
Answer c
Masters female student of course R
= 24*(15/100)*(4/9) = 1.6 lakh
Graduate female student of course R
= 32*(18/100)*(5/9) = 3.2 lakh
Required % =(1.6/3.2)*100 = 50 %
Question 19 of 20
19. Question
Directions (Q. 16 – 20): Study the following pie-charts and table to answer these questions.
The Graduate male student of Course ‘R’ is what per cent of the Graduate male student of course ‘S’?
(5/3) of? – (2/3) of (12/17) of 3162 = 42 % of 850
783
857
1107
685
None of these
25 % of x – 1664 ÷ 13 + 222 ÷ 6 = 67
968
745
529
632
None of these
√x × 13 – 26 % of 2650 – 12 × 28 = 15
80
3600
60
6400
None of these
Directions (Q. 6 – 10): What value should come in place of (?) in the following series?
809, 811, 820, 848, 913, ?
1185
934
1257
1039
None of these
87, 87, 83, 99, 63, ?
98
127
143
156
None of these
294624, 24552, 2232, 248, ?
72
64.6
49.6
98
None of these
856, 213, 52.25, ? , 2.015625
32.50
24.075
12.0625
38.0225
None of these
727, 513, 390, 328, ?
303
286
212
311
None of these
The ratio between cost price of three articles A, B and C is 2:3:4 while the profit percentage earned in these articles is in the ratio of 5:4:3 and profit earned on article C is Rs 300. If the cost price of article A is Rs 1000, find the selling price of article A?
2500
2180
1250
600
1500
A and B invest in the ratio of 25:17. If they divide 40% of the profit equally and rest in the proportion of their investments, then find B’s profit if A’s profit is Rs 1950.
1650
1550
4500
5600
1420
A shopkeeper sells a transistor at 15% above its cost price. If he had bought it 5% more than what he paid for it and sold it for Rs 6 more, he would have gained 10%. The cost price of the transistor is
800
1000
1200
1400
None of these
The present age of Ravi’s father is four times Ravi’s present age. Five years back, Ravi’s father was seven times as old as Ravi was at that time. What is the present age of Ravi’s father?
84
70
40
35
None of these
The length and breadth of a rectangular piece of land are in the ratio of 5: 3. The owner spent Rs 3000 for surrounding it from all the sides at the rate of Rs 7.50 per meter. The difference between length and breadth is:
50m
100m
200m
150m
None of these
Directions (Q. 16 – 20): Study the following pie-charts and table to answer these questions.
What is the difference between the Masters male student and Graduate male student from Course ‘P’?
24000
14000
28000
36000
None of these
What is the ratio of the Masters female student of Course ‘T’ to Graduate female student of Course ‘S’?
7 : 5
5 : 7
16 : 15
15 : 16
None of these
The Masters female student of Course ‘R’ is what per cent of the Graduate female student of that Course?
40
62.5
50
52.5
None of these
The Graduate male student of Course ‘R’ is what per cent of the Graduate male student of course ‘S’?
180%
120%
110%
190%
None of these
What is the ratio of the Masters male student of Course ‘T’ to the Graduate female student of that Course?
28 : 35
35 : 28
32 : 45
45 : 32
None of these
Answers:
Directions (Q. 1-5):
Answer a
(32/16)*(15/2) + 216 + 279 = x
15 + 216 + 279 = x
X = 510
Answer d
8^{4} ÷ 64^{2} × 512^{5} = 8^{x}
8^{4} ÷ 8^{4} × 8^{15 }= 8^{x}
8^{4-4+15} = 8^{x}
8^{15} = 8^{x}
X = 15
Answer c
(5x/3) – (2/3) * (12/17) * 3162 = (42/100) * 850
(5x/3) – 1488 = 357
(5x/3) = 1845
X = (1845*3)/5 = 1107
Answer d
25 % of x – 1664 ÷ 13 + 222 ÷ 6 = 67
(25/100)*x – 128 + 37 =67
(25/100)*x = 67 + 128 -37
(25/100)*x = 158
X = (158*100)/25
X = 632
Answer d
√x × 13 – 26 % of 2650 – 12 × 28 = 15
√x × 13 – (26/100)*2650 – 336 = 15
√x × 13 – 689 – 336 = 15
√x × 13 = 15 +689 + 336
√x × 13 = 1040
√x = 1040/13 = 80
X = 80^{2} = 6400
Directions (Q. 6 – 10):
Answer d
The difference is, 1^{3}+1, 2^{3}+1, 3^{3}+1, 4^{3}+1, 5^{3}+1,..
The answer is, 1039
Answer b
The pattern is, +0^{2}, -2^{2}, +4^{2}, -6^{2}, +8^{2},..
The answer is, 127
Answer c
The pattern is, ÷ 12, ÷ 11, ÷ 9, ÷ 5,..
The answer is, 49.6
Answer c
The pattern is, ÷ 4 -1
The answer is, 12.0625
Answer a
The pattern is, -(6^{3} – 2), -(5^{3} – 2), -(4^{3} – 2), -(3^{3} – 2),..
The answer is, 303
Answer c
Cost price article C = 1000*4/2= Rs 2000
Profit percentage earned on article C= 300*100/2000= 15%
Percentage profit earned in article A= 15*5/3= 25%
Profit earned in article A= 1000*25/100= Rs 250
Selling price of article A= 1000+250= Rs 1250
Answer b
Let the total profit be x
Then
A’s profit= 0.2x+0.6x*25/42
1950= 0.2x (1+75/42)
X= 1950*42/0.2*117= Rs 3500
B’s profit= 3500-1950= Rs 1550
Answer c
Let the cost of the transistor be x
SP = 115/100*x
New CP = x* 105/100= 105x/100
New SP= 115x/100+ 6=105x/100*110/100
=>115x/100-231x/200= -6
=>x=1200
Answer c
Let the present age of Ravi be x
Present age of Ravi’s father= 4x
Now 5 yr ago
Ravi’s father’s age= 7 * Ravi’s age
= 4x-5=7(x-5)
= x= 10
Ravi’s Father’s age= 4*10= 40 years.
Answer a
Perimeter of the field = 3000/7.50= 400m
So,
2(5x + 3x) = 400 => x = 25
So, length = 125 m & breadth = 75 m
Difference between length & breadth
= (125-75)m = 50m
Directions (Q. 16 – 20):
Answer b
Masters Male Student from Course ‘P’
= 24*(16/100)*(7/12) lakh
= 2,24,000
Graduate male student from Course ‘P’
= 32*(15/100)*(7/16) lakh
= 210000
Difference = 224000- 210000 = 14000
Answer d
Masters female student of course T
= 24*(20/100)*(7/16) lakh
Graduate female student of course S
= 32*(12/100)*(7/12) lakh
Required ratio = [24*(20/100)*(7/16)] : [32*(12/100)*(7/12)]
= 210 : 224
= 15 : 16
Answer c
Masters female student of course R
= 24*(15/100)*(4/9) = 1.6 lakh
Graduate female student of course R