“20-20” Quantitative Aptitude | Crack SBI PO Prelims 2018 Day-122
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Directions (Q. 1 – 5): What approximate value should come in the place of question mark (?) in the following questions?
52 % of 2599 + (6/11) of 20350 – 14 % of 2599 =?
Correct
Answer c
(52/100)*2600 + (6/11)*20350 – (14/100)*2600 = x
X = 1352 + 11100 – 364
X = 12088 = 12090
Incorrect
Answer c
(52/100)*2600 + (6/11)*20350 – (14/100)*2600 = x
X = 1352 + 11100 – 364
X = 12088 = 12090
Question 4 of 20
4. Question
Directions (Q. 1 – 5): What approximate value should come in the place of question mark (?) in the following questions?
4502.98 ÷ 19.11 × 49.97 =? ÷ 25.089 × 50.01
Correct
Answer e
(4503/19)*50 = (x/25)*50
237 = x/25
X = 237*25
X = 5925
Incorrect
Answer e
(4503/19)*50 = (x/25)*50
237 = x/25
X = 237*25
X = 5925
Question 5 of 20
5. Question
Directions (Q. 1 – 5): What approximate value should come in the place of question mark (?) in the following questions?
39.85 % of 1540 +43.899 % of 1050 = 20 % of?
Correct
Answer b
40 % of 1540 + 44 % of 1050 = (20/100)*x
(40/100)*1540 + (44/100)*1050 = x/5
616 + 462 = (x/5)
1078*5 = x
X = 5390
Incorrect
Answer b
40 % of 1540 + 44 % of 1050 = (20/100)*x
(40/100)*1540 + (44/100)*1050 = x/5
616 + 462 = (x/5)
1078*5 = x
X = 5390
Question 6 of 20
6. Question
Directions (Q. 6 – 10): What value should come in the place of question mark (?) in the following number series?
56, 55, 102, 279, 1052,?
Correct
Answer e
The pattern is, *1 – 1^{3}, *2 – 2^{3}, *3 – 3^{3}, *4 – 4^{3}, *5 – 5^{3},..
The answer is, 5135
Incorrect
Answer e
The pattern is, *1 – 1^{3}, *2 – 2^{3}, *3 – 3^{3}, *4 – 4^{3}, *5 – 5^{3},..
The answer is, 5135
Question 7 of 20
7. Question
Directions (Q. 6 – 10): What value should come in the place of question mark (?) in the following number series?
72, 144, 576, 3456,? , 276480
Correct
Answer c
The pattern is, *2, *4, *6, *8, *10
The answer is, 27648
Incorrect
Answer c
The pattern is, *2, *4, *6, *8, *10
The answer is, 27648
Question 8 of 20
8. Question
Directions (Q. 6 – 10): What value should come in the place of question mark (?) in the following number series?
2400, 1295, 624, ? , 80, 15
Correct
Answer a
The pattern is, 7^{4}– 1, 6^{4}– 1, 5^{4}– 1, 4^{4} – 1, 3^{4} – 1, 2^{4} – 1….
The answer is, 255
Incorrect
Answer a
The pattern is, 7^{4}– 1, 6^{4}– 1, 5^{4}– 1, 4^{4} – 1, 3^{4} – 1, 2^{4} – 1….
The answer is, 255
Question 9 of 20
9. Question
Directions (Q. 6 – 10): What value should come in the place of question mark (?) in the following number series?
72000, 14400, 1440, 96, ?
Correct
Answer d
The pattern is, ÷ 5, ÷ 10, ÷ 15, ÷ 20,..
The answer is, 4.8
Incorrect
Answer d
The pattern is, ÷ 5, ÷ 10, ÷ 15, ÷ 20,..
The answer is, 4.8
Question 10 of 20
10. Question
Directions (Q. 6 – 10): What value should come in the place of question mark (?) in the following number series?
74, 75, 148, 447, ? , 8925
Correct
Answer b
The pattern is, *1 + 1, *2 – 2, *3 + 3, *4 – 4, *5 + 5,..
The answer is, 1784
Incorrect
Answer b
The pattern is, *1 + 1, *2 – 2, *3 + 3, *4 – 4, *5 + 5,..
The answer is, 1784
Question 11 of 20
11. Question
A sold an article for Rs. 1272 and earned a profit of 6 %. At what price should it have been sold so as to earn a profit of 14 %?
Correct
Answer c
Let us take CP of an article be Rs. X
SP = x* 106/100 = 1272
X= 1200
Required SP = 1200* 114/100 = 1368
Shortcut:
(100+ Profit %)/SP1 = (100 + Profit %)/SP2
(100+ 6)/1272 = (100 + 14)/SP2
(106/1272) = (114/SP2)
SP2 = 114*(1272/106) = Rs. 1368
Incorrect
Answer c
Let us take CP of an article be Rs. X
SP = x* 106/100 = 1272
X= 1200
Required SP = 1200* 114/100 = 1368
Shortcut:
(100+ Profit %)/SP1 = (100 + Profit %)/SP2
(100+ 6)/1272 = (100 + 14)/SP2
(106/1272) = (114/SP2)
SP2 = 114*(1272/106) = Rs. 1368
Question 12 of 20
12. Question
The difference between the two numbers is 540. 32 % of one number is equal to 48 % of another number. Find the two numbers.
Correct
Answer a
Let the two numbers be x and y,
X – Y = 540
(32/100)X = (48/100)Y
(X/Y) = 3/2
X : y = 3 : 2
1’s = 540
The two numbers be,
= > 3’s = (540*3) = 1620
= > 2’s = (540*2) = 1080
Incorrect
Answer a
Let the two numbers be x and y,
X – Y = 540
(32/100)X = (48/100)Y
(X/Y) = 3/2
X : y = 3 : 2
1’s = 540
The two numbers be,
= > 3’s = (540*3) = 1620
= > 2’s = (540*2) = 1080
Question 13 of 20
13. Question
At present, Priyanka is 8 years older than Renuka. The ratio of the present ages of Priyanka to Myna is 4: 5. At present Renuka is 16 years younger than Myna. What is Renuka’s present age?
Correct
Answer b
Priyanka = Renuka + 8
The ratio of the present ages of Priyanka to Myna = 4: 5 (4x, 5x)
Renuka = Myna – 16
Renuka’s present age = 5x – 16
Priyanka = Renuka + 8
4x = 5x – 16 + 8
8 = 5x – 4x
X = 8
Present age of Renuka = 5x – 16 = 40 – 16 = 24 years
Incorrect
Answer b
Priyanka = Renuka + 8
The ratio of the present ages of Priyanka to Myna = 4: 5 (4x, 5x)
Renuka = Myna – 16
Renuka’s present age = 5x – 16
Priyanka = Renuka + 8
4x = 5x – 16 + 8
8 = 5x – 4x
X = 8
Present age of Renuka = 5x – 16 = 40 – 16 = 24 years
Question 14 of 20
14. Question
C is 25% less efficient than A. A and B together can finish a piece of work in 3 days. B and C together can do it in 4 days. In how many days can A alone finish the same piece of work?
Correct
Answer d
(A + B)’s one day work = (1/3)
(B + C)’s one day work = (1/4)
(A – C)’s one day work = (1/3) – (1/4) = (1/12)
C is 25% less efficient than A
Efficiency = > A: C = 100: 75 = 4: 3
Day ratio = > A: C = 3: 4
1’s = 12
A alone can finish the whole work in 36 days.
Incorrect
Answer d
(A + B)’s one day work = (1/3)
(B + C)’s one day work = (1/4)
(A – C)’s one day work = (1/3) – (1/4) = (1/12)
C is 25% less efficient than A
Efficiency = > A: C = 100: 75 = 4: 3
Day ratio = > A: C = 3: 4
1’s = 12
A alone can finish the whole work in 36 days.
Question 15 of 20
15. Question
A bag contains 3 red marbles, 7 green marbles and 6 pink marbles. If three marbles are taken at random, than what is the probability that 2 marbles are green?
Correct
Answer c
Total probability = 16C_{3}
Required probability = 7C_{2} and 9C_{1}
The probability that 2 marbles are green,
= > [7C_{2} and 9C_{1}] / 16C_{3}
= > (21*9) / (5*7*16)
= > 27/80
Incorrect
Answer c
Total probability = 16C_{3}
Required probability = 7C_{2} and 9C_{1}
The probability that 2 marbles are green,
= > [7C_{2} and 9C_{1}] / 16C_{3}
= > (21*9) / (5*7*16)
= > 27/80
Question 16 of 20
16. Question
Directions (Q. 16 – 20) Study the following information carefully and answer the given questions:
Following pie chart shows the total number of mobile phones sold by a shopkeeper in different months and the table shows the ratio of Lenovo and Acer mobile phones sold in the given months.
Total number of mobiles sold in the month of Jan and Feb together is approximately what percentage of the total number of Acer mobiles sold in the same months?
Correct
Answer a
Total number of mobiles sold in the month of Jan and Feb together
= > 56000*(22/100) + 56000*(25/100)
= > 12320 + 14000 = 26320
Total number of Acer mobiles sold in the month of Jan and Feb together
= > 56000*(22/100)*(1/4) + 56000*(25/100)*(2/5)
= > 3080 + 5600 = 8680
Required % = (26320/8680)*100 = 303 %
Incorrect
Answer a
Total number of mobiles sold in the month of Jan and Feb together
= > 56000*(22/100) + 56000*(25/100)
= > 12320 + 14000 = 26320
Total number of Acer mobiles sold in the month of Jan and Feb together
= > 56000*(22/100)*(1/4) + 56000*(25/100)*(2/5)
= > 3080 + 5600 = 8680
Required % = (26320/8680)*100 = 303 %
Question 17 of 20
17. Question
Directions (Q. 16 – 20) Study the following information carefully and answer the given questions:
Following pie chart shows the total number of mobile phones sold by a shopkeeper in different months and the table shows the ratio of Lenovo and Acer mobile phones sold in the given months.
Find the difference between the total number of Acer mobiles sold in the month of Mar to that of May?
Correct
Answer d
The total number of Acer mobiles sold in the month of Mar
= > 56000*(18/100)*(1/3) = 3360
The total number of Acer mobiles sold in the month of May
= > 56000*(20/100)*(3/5) = 6720
Required difference = 6720 – 3360 = 3360
Incorrect
Answer d
The total number of Acer mobiles sold in the month of Mar
= > 56000*(18/100)*(1/3) = 3360
The total number of Acer mobiles sold in the month of May
= > 56000*(20/100)*(3/5) = 6720
Required difference = 6720 – 3360 = 3360
Question 18 of 20
18. Question
Directions (Q. 16 – 20) Study the following information carefully and answer the given questions:
Following pie chart shows the total number of mobile phones sold by a shopkeeper in different months and the table shows the ratio of Lenovo and Acer mobile phones sold in the given months.
Find the ratio between the total number of mobiles sold in the month of Jan and Mar together to that of Apr and May together?
Correct
Answer c
The total number of Lenovo mobiles sold in the month of Jan and Mar together
= > 56000*[(22 + 18)/100]
The total number of mobiles sold in the month of Apr and May together
= > 56000*[(15 + 20)/100]
Required ratio = [(56000/100)*40]: [(56000/100)*35]
= > 40: 35 = 8: 7
Incorrect
Answer c
The total number of Lenovo mobiles sold in the month of Jan and Mar together
= > 56000*[(22 + 18)/100]
The total number of mobiles sold in the month of Apr and May together
= > 56000*[(15 + 20)/100]
Required ratio = [(56000/100)*40]: [(56000/100)*35]
= > 40: 35 = 8: 7
Question 19 of 20
19. Question
Directions (Q. 16 – 20) Study the following information carefully and answer the given questions:
Following pie chart shows the total number of mobile phones sold by a shopkeeper in different months and the table shows the ratio of Lenovo and Acer mobile phones sold in the given months.
Find the total number of Acer mobiles sold in the month of Jan, Feb and Apr together?
Correct
Answer b
The total number of Acer mobiles sold in the month of Jan, Feb and Apr together
Directions (Q. 16 – 20) Study the following information carefully and answer the given questions:
Following pie chart shows the total number of mobile phones sold by a shopkeeper in different months and the table shows the ratio of Lenovo and Acer mobile phones sold in the given months.
The total number of mobiles sold in the month of Feb is what percentage of the total number of mobiles sold in the months of May?
Correct
Answer e
Required % = (25/20)*100 = 125 %
Incorrect
Answer e
Required % = (25/20)*100 = 125 %
Click “Start Quiz” to attend these Questions and view Solutions
Directions (Q. 6 – 10): What value should come in the place of question mark (?) in the following number series?
56, 55, 102, 279, 1052,?
3127
4283
3759
4841
5135
72, 144, 576, 3456,? , 276480
34560
38954
27648
25172
21566
2400, 1295, 624, ? , 80, 15
255
376
328
214
357
72000, 14400, 1440, 96, ?
12.4
24.6
8.2
4.8
16.4
74, 75, 148, 447, ? , 8925
2453
1784
2194
1592
2876
A sold an article for Rs. 1272 and earned a profit of 6 %. At what price should it have been sold so as to earn a profit of 14 %?
Rs. 1456
Rs. 1224
Rs. 1368
Rs. 1512
None of these
The difference between the two numbers is 540. 32 % of one number is equal to 48 % of another number. Find the two numbers.
1620 and 1080
1510 and 970
1480 and 940
1350 and 810
1730 and 1190
At present, Priyanka is 8 years older than Renuka. The ratio of the present ages of Priyanka to Myna is 4: 5. At present Renuka is 16 years younger than Myna. What is Renuka’s present age?
28 years
24 years
20 years
32 years
None of these
C is 25% less efficient than A. A and B together can finish a piece of work in 3 days. B and C together can do it in 4 days. In how many days can A alone finish the same piece of work?
30 days
24 days
28 days
36 days
None of these
A bag contains 3 red marbles, 7 green marbles and 6 pink marbles. If three marbles are taken at random, than what is the probability that 2 marbles are green?
13/45
21/53
27/80
11/67
None of these
Directions (Q. 16 – 20) Study the following information carefully and answer the given questions:
Following pie chart shows the total number of mobile phones sold by a shopkeeper in different months and the table shows the ratio of Lenovo and Acer mobile phones sold in the given months.
Total number of mobiles sold in the month of Jan and Feb together is approximately what percentage of the total number of Acer mobiles sold in the same months?
303 %
240 %
190 %
150 %
280 %
Find the difference between the total number of Acer mobiles sold in the month of Mar to that of May?
5180
4620
4230
3360
None of these
Find the ratio between the total number of mobiles sold in the month of Jan and Mar together to that of Apr and May together?
3 : 5
5 : 9
8 : 7
11 : 15
None of these
Find the total number of Acer mobiles sold in the month of Jan, Feb and Apr together?
13560
12880
14920
15310
None of these
The total number of mobiles sold in the month of Feb is what percentage of the total number of mobiles sold in the months of May?
75 %
90 %
100 %
115 %
125 %
Answers:
Answer d
(4/8) * (8/16) * (31/65) * 325 = x
X = 155/4 = 39
X = 40
2. Answer a
(14)^{2} – (13)^{2} + (4325 + 80) ÷ x = 225
196 – 169 + (4405/x) = 225
(4405/x) = 225 – 27
(4405/x) = 198
X = 4405/198
X = (4400/200) = 22
3. Answer c
(52/100)*2600 + (6/11)*20350 – (14/100)*2600 = x
X = 1352 + 11100 – 364
X = 12088 = 12090
4. Answer e
(4503/19)*50 = (x/25)*50
237 = x/25
X = 237*25
X = 5925
5. Answer b
40 % of 1540 + 44 % of 1050 = (20/100)*x
(40/100)*1540 + (44/100)*1050 = x/5
616 + 462 = (x/5)
1078*5 = x
X = 5390
Direction (6-10)
6. Answer e
The pattern is, *1 – 1^{3}, *2 – 2^{3}, *3 – 3^{3}, *4 – 4^{3}, *5 – 5^{3},..
The answer is, 5135
7. Answer c
The pattern is, *2, *4, *6, *8, *10
The answer is, 27648
8. Answer a
The pattern is, 7^{4}– 1, 6^{4}– 1, 5^{4}– 1, 4^{4} – 1, 3^{4} – 1, 2^{4} – 1….
The answer is, 255
9. Answer d
The pattern is, ÷ 5, ÷ 10, ÷ 15, ÷ 20,..
The answer is, 4.8
10. Answer b
The pattern is, *1 + 1, *2 – 2, *3 + 3, *4 – 4, *5 + 5,..
The answer is, 1784
11. Answer c
Let us take CP of an article be Rs. X
SP = x* 106/100 = 1272
X= 1200
Required SP = 1200* 114/100 = 1368
Shortcut:
(100+ Profit %)/SP1 = (100 + Profit %)/SP2
(100+ 6)/1272 = (100 + 14)/SP2
(106/1272) = (114/SP2)
SP2 = 114*(1272/106) = Rs. 1368
12. Answer a
Let the two numbers be x and y,
X – Y = 540
(32/100)X = (48/100)Y
(X/Y) = 3/2
X : y = 3 : 2
1’s = 540
The two numbers be,
= > 3’s = (540*3) = 1620
= > 2’s = (540*2) = 1080
13. Answer b
Priyanka = Renuka + 8
The ratio of the present ages of Priyanka to Myna = 4: 5 (4x, 5x)
Renuka = Myna – 16
Renuka’s present age = 5x – 16
Priyanka = Renuka + 8
4x = 5x – 16 + 8
8 = 5x – 4x
X = 8
Present age of Renuka = 5x – 16 = 40 – 16 = 24 years
14. Answer d
(A + B)’s one day work = (1/3)
(B + C)’s one day work = (1/4)
(A – C)’s one day work = (1/3) – (1/4) = (1/12)
C is 25% less efficient than A
Efficiency = > A: C = 100: 75 = 4: 3
Day ratio = > A: C = 3: 4
1’s = 12
A alone can finish the whole work in 36 days.
Direction (16-20)
16. Answer a
Total number of mobiles sold in the month of Jan and Feb together
= > 56000*(22/100) + 56000*(25/100)
= > 12320 + 14000 = 26320
Total number of Acer mobiles sold in the month of Jan and Feb together
= > 56000*(22/100)*(1/4) + 56000*(25/100)*(2/5)
= > 3080 + 5600 = 8680
Required % = (26320/8680)*100 = 303 %
17. Answer d
The total number of Acer mobiles sold in the month of Mar
= > 56000*(18/100)*(1/3) = 3360
The total number of Acer mobiles sold in the month of May
= > 56000*(20/100)*(3/5) = 6720
Required difference = 6720 – 3360 = 3360
18. Answer c
The total number of Lenovo mobiles sold in the month of Jan and Mar together
= > 56000*[(22 + 18)/100]
The total number of mobiles sold in the month of Apr and May together
= > 56000*[(15 + 20)/100]
Required ratio = [(56000/100)*40]: [(56000/100)*35]
= > 40: 35 = 8: 7
19. Answer b
The total number of Acer mobiles sold in the month of Jan, Feb and Apr together