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): Pie-chart given bellows shows the income of five different persons and bar graph shows the percentage distribution of their income on different things. Study the question carefully and answer them.
Q1. Who among the following spend maximum amount on food?
D
E
B
C
A
Solution:
Alternate,
By seeing the chart and graph it can be easily concluded that ‘D’ spend maximum amount on food as he has maximum income among five persons and he spend maximum on food as compare to others
Q2. Amount spend by ‘E’ on furniture is what percent more than amount spend by ‘D’ on Transportation?
70%
45%
80%
65%
60%
Solution:
Q3. Find the average amount spend by A, B and C on furniture?
4622
4626
4262
4266
4662
Solution:
Q4. ‘D’ buy only three type of food X, Y and Z and amount spend on buying X, Y and Z is in the ratio 5 : 7 : 8. What is the difference between amount spend on buying Z type food to amount spend on buying X type food.
2520
1680
8400
1260
2100
Solution:
Q5. Find the ratio of amount spend by ‘A’ and ‘B’ together on food to the amount spend by ‘C’ and ‘D’ together on furniture?
295 : 277
277 : 295
311 : 301
301 : 305
301 : 310
Solution:
Q6. Quantity I: Percentage mark-up above cost price of an article so as to gain 33% after allowing the customer a discount of 5%.
Quantity II: Percentage of dancers under 25 years out of a group of 20 singers and 40 dancer if 20% of the singers are under 25 years old and 40% of the entire group is under 25 years.
Quantity I > Quantity II
Quantity I < Quantity II
Quantity I ≥ Quantity II
Quantity I ≤ Quantity II
Quantity I = Quantity II or No relation
Solution:
Q7. Quantity I: Value of fifth number when Average of five numbers is 61. The average of 1st and 3rd number is 69 and average of second and fourth number is 69.
Quantity II: No. of boys in the class. The average age of all students of a class is 18 years. The average age of boys of the class is 20 years and that of the girls is 15 years. The no. of girls in the class is 20.
Note: Compare the magnitudes of quantities.
Note: Compare the magnitudes of quantities.
Quantity I > Quantity II
Quantity I < Quantity II
Quantity I ≥ Quantity II
Quantity I ≤ Quantity II
Quantity I = Quantity II or No relation
Solution:
Q8. A 20 litres mixture contains milk and water in the respective ratio of 3 : 2. Then 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of milk and water in the resultant mixture respectively ?
17 : 3
9 : 1
4 : 17
5 : 3
3 : 14
Solution:
Q9. On walking at 3/4 of his usual speed a man reaches his office 20 minutes late. What is the usual time taken by him in reaching his office ?
75 minutes
60 minutes
40 minutes
30 minutes
None of these
Solution:
Q10. A wholesaler blends two varieties of tea, one costing Rs 60 per kilo and another costing Rs 105 per kilo. The respective ratio of quantities they were mixed in was 7 : 2. If he sold the mixed variety of Rs 100 per kilo, how much was his profit percentage?
Solution:
Directions (11-15): In each of the following questions two equations are given. You have to solve the equations and
Give answer –
Give answer –
Q11. I. 8x² + 26x + 15 = 0
II. 4y² + 24y + 35 = 0
if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:
I. 8x² + 26x + 15 = 0
⇒ 8x² + 20x + 6x + 15 = 0
⇒ 4x (2x + 5) + 3(2x + 5) = 0
⇒ (2x + 5) (4x + 3) = 0
⇒ x = – 5/2, –3/4
II. 4y² + 24y + 35 = 0
⇒ 4y² + 10y + 14y + 35 = 0
⇒ 2y (2y + 5) + 7 (2y + 5) = 0
⇒ (2y + 5) (2y + 7) = 0
⇒ y = –5/2, –7/2
x ≥ y
Q12. I. 2x² + 9x + 9 = 0
II. 2y² + 17y + 36 = 0
if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:
I. 2x² + 9x + 9 = 0
⇒ 2x² + 6x + 3x + 9 = 0
⇒ (x + 3) (2x + 3) = 0
⇒ x = –3, –3/2
II. 2y² + 17y + 36 = 0
⇒ 2y² + 8y + 9y + 36 = 0
⇒ (y + 4) (2y + 9) = 0
y = – 4, –9/2
x > y
Q13. I. 5x² + 29x + 20 = 0
II. 25y² + 25y + 6 = 0
if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:
I. 5x² + 29 + 20 = 0
⇒ 5x² + 25x + 4x + 20 = 0
⇒ (x + 5) (5x + 4) = 0
⇒ x = –5, –4/5
II. 25y² + 25y + 6 = 0
⇒ 25y² + 15y + 10y + 6 = 0
⇒ (5y + 3) (5y + 2) = 0
⇒ y = – 3/5, –2/5
y > x
Q14. I. 3x² – 16x + 21 = 0
II. 3y² – 28y + 65 = 0
if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:
I. 3x² – 16x + 21 = 0
⇒ 3x² – 9x – 7x + 21 = 0
⇒ (x – 3) (3x – 7) = 0
⇒ x = 3, 7/3
II. 3y² – 28y + 65 = 0
⇒ 3y² – 15y – 13y + 65 = 0
⇒ (y – 5) (3y – 13) = 0
⇒ y = 5, 13/3
y > x
Q15. I. 8x²-26x+15=0
II. 2y²-17y+30=0
if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:
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