SBI PO Quantitative Aptitude Quiz
Here, we are providing SBI PO Mains Study plan as there is left not enough time to deal in details. The questions asked in the quantitative aptitude section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions
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Directions (1-5): Answer the following question based on the information given below.
Every year, a survey of 1000 people is conducted by the World Health Organization (WHO). WHO found that in the year 2005, 2006, 2007, 2008 and 2009 the percentage of people affected by malaria were 30%, 40%, 30%, 20% and 45% respectively. WHO also found that every year out of the affected people 60% were students, 10% were house-wives and 30% were drivers.The number of house-wives, students and drivers were in the ratio 20 : 11 : 9, every year.
Q1. In the year 2007, find the number of house-wives affected by malaria?
60
30
50
110
150
Solution:
In the year 2007, 30% of the population was affected by malaria out of which 10% were house-wives.
∴ The number of house-wives affected by malaria in the year 2007 = 10% of 30% of 1000 = 0.1 × 0.3 × 1000 = 30
Q2. In the year 2009, find the number of drivers who were not affected by malaria?
110
125
415
90
190
Solution:
The number of house-wives, students and drivers were in the ratio 20 : 11 : 9 in each year.
Let the common factor be x.
Also, every year 1000 people were surveyed.
∴ 20x + 11x + 9x = 1000
∴ x = 25
∴ The total number of house-wives, students and drivers was 500, 275 and 225 respectively.
Now, in the year 2009, 45% of the total population was affected by malaria.
45% of 1000 = 450
Out of the 450 affected people, 30% were drivers.
30% of 450 = 135
Hence, the numbers of drivers who were not affected by malaria in the year 2009 = 225 − 135 = 90
Q3. What is the difference between the number of students affected and not affected by malaria in the year 2006?
205
35
200
240
420
Solution:
Total population of students for each year = 275
In the year 2006, the numbers of students affected by malaria = 60% of 40% of 1000 = 0.6 × 0.4 × 1000 = 240 students
∴ The number of students not affected by malaria = 275 − 240 = 35
∴ Difference between the two = 240 − 35 = 205
Q4. Find the ratio of the number of house-wives affected by malaria in the year 2005 to that affected by malaria in the year 2008.
5: 3
9: 4
3: 2
2: 1
4: 3
Solution:
The number of house-wives affected by malaria in the year 2005 = 10% of 30% of 1000 = 0.1 × 0.3 × 1000 = 30
The number of house-wives affected by malaria in the year 2008 = 10% of 20% of 1000 = 0.1 × 0.2 × 1000 = 20
The required ratio = 30 : 20 = 3 : 2
Q5. Which year had the maximum number of students not affected by malaria?
2005
2006
2007
2008
2009
Solution:
Total number of students in each year = 275
The number of students affected by malaria in the year 2005 = 60% of 30% of 1000 = 180
∴ The number of students not affected by malaria = 275 − 180 = 95
The number of students affected by malaria in the year 2006 = 60% of 40% of 1000 = 240
∴ The number of students not affected by malaria = 275 − 240 = 35
The number of students affected by malaria in the year 2007 = 60% of 30% of 1000 = 180
∴ The number of students not affected by malaria = 275 − 180 = 95
The number of students affected by malaria in the year 2008 = 60% of 20% of 1000 = 120
∴ The number of students not affected by malaria = 275 − 120 = 155
The number of students affected by malaria in the year 2009 = 60% of 45% of 1000 = 270
∴ The number of students not affected by malaria = 275 − 270 = 5
Thus, 2008 had the maximum number of students not affected by malaria.
Directions (6-10): Calculate the approximate value of given questions:
Q6. (13.96)² – (15.03)² + (18.09)² – 32.65 = ?
223
262
334
354
201
Solution:
? ≃ 196 – 225 + 324 – 33
≃ 262
Q7.
9
11
2
5
13
Solution:
Q8. 9228.789 – 5021.832 + 1496.989 = ?
6500
6004
6314
5704
5104
Solution:
? ≃ 10,726 – 5022 = 5704
Q9. 1001 ÷ 49 × 99 – 1299 = ?
700
600
900
250
400
Solution:
Q10. 7999.99 + 72 × 49.99 = ?
12000
12600
12500
11600
11000
Solution:
? ≃ 8000 + 72 × 50
≃ 11,600
Q11. Two dice are thrown simultaneously. Find the probability of getting a sum of two numbers greater than 6.
3/4
23/36
5/12
7/12
11/36
Solution:
Here to find required probability we will find opposite probability and then subtract it from 1.
i.e. Required probability = 1 – P (sum ≤ 6)
∴ favorable cases for (sum ≤ 6) = (1,1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1),
Possible cases = 6 × 6 = 36
∴ Required probability =
Q12. A milkman mixes a certain amount of water in the pure milk and makes
profit by selling this milk at cost price of pure milk. Find the ratio of milk to water in the mixture.
profit by selling this milk at cost price of pure milk. Find the ratio of milk to water in the mixture.
5 : 1
4 : 1
1 : 6
6 : 1
3 : 1
Solution:
Q13. Find the probability of selecting 2 green balls out of 4 green and 3 red balls.
6/7
4/7
3/7
2/7
1/7
Solution:
Q14. In a 64 litre mixture of milk and water the ratio of water to milk is 3 : 5. If x litre water is mixed in this mixture, so the ratio of water to milk becomes 5 : 8. Find the value of x.
Solution:
Q15. There are two mixtures i.e. P and Q of spirit and water. The mixture P contains ratio of spirit to water as 5 : 8 and Q contains ratio of water to spirit as 4 : 5. 26 litres of mixture P and some quantity of Q is taken out and these two quantities are mixed together to form a new mixture. Ratio of spirit to water in this new mixture is 4 : 5. Find the quantity which was taken out from Q.
9 ℓ
16 ℓ
14 ℓ
12 ℓ
18 ℓ
Solution:
let quantity which was taken out from mixture Q be x lit.
Final quantity of spirit in new mixture
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