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Quantitative Aptitude0%

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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Directions (1-5): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

2x^{2} + 3x – 2 = 0, 8y^{2} – 10y + 3 = 0

Correct

Explanation:
2x^{2} + 3x – 2 = 0
2x^{2} + 4x – x – 2 = 0
Gives x = -2, 1/2
8y^{2} – 10y + 3 = 0
8y^{2} – 4y – 6y + 3 = 0
Gives y = 1/2, 3/4
Put on number line
-2… 1/2 … 3/4

Incorrect

Explanation:
2x^{2} + 3x – 2 = 0
2x^{2} + 4x – x – 2 = 0
Gives x = -2, 1/2
8y^{2} – 10y + 3 = 0
8y^{2} – 4y – 6y + 3 = 0
Gives y = 1/2, 3/4
Put on number line
-2… 1/2 … 3/4

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

Directions (1-5): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

3x^{2} – 32x – 11 = 0, y^{2} – 14y + 33 = 0

Correct

Explanation:
3x^{2} – 32x – 11 = 0
3x^{2} – 33x + x – 11 = 0
Gives x = -1/3, 11
y^{2} – 14y + 33 = 0
y^{2} – 11y – 3y + 33 = 0
Gives y = 3, 11
Put on number line
-1/3… 3… 11
When y = 3, y < x (11) and y > x (-1/3) – here relation cant be determined.

Incorrect

Explanation:
3x^{2} – 32x – 11 = 0
3x^{2} – 33x + x – 11 = 0
Gives x = -1/3, 11
y^{2} – 14y + 33 = 0
y^{2} – 11y – 3y + 33 = 0
Gives y = 3, 11
Put on number line
-1/3… 3… 11
When y = 3, y < x (11) and y > x (-1/3) – here relation cant be determined.

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

Directions (1-5): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

4x^{2} – 23x – 6 = 0, 4y^{2} + 9y + 2 = 0

Correct

Explanation:
4x^{2} – 23x – 6 = 0
4x^{2} – 24x + x – 6 = 0
Gives x = -1/4, 6
4y^{2} + 9y + 2 = 0
4y^{2} +8y + y + 2 = 0
Gives y = -2, -1/4
Put on number line
-2… -1/4… 6

Incorrect

Explanation:
4x^{2} – 23x – 6 = 0
4x^{2} – 24x + x – 6 = 0
Gives x = -1/4, 6
4y^{2} + 9y + 2 = 0
4y^{2} +8y + y + 2 = 0
Gives y = -2, -1/4
Put on number line
-2… -1/4… 6

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Directions (1-5): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

x^{2} – 8x + 7 = 0, 3y^{2} + 20y + 12 = 0

Correct

Explanation:
x^{2} – 8x + 7 = 0
x^{2} – x – 7x + 7 = 0
Gives x = -1/4, 6
3y^{2} + 20y + 12 = 0
3y^{2} + 18y + 2y + 12 = 0
Gives y = -6, -2/3
Put on number line
-6… -2/3… 1…. 7

Incorrect

Explanation:
x^{2} – 8x + 7 = 0
x^{2} – x – 7x + 7 = 0
Gives x = -1/4, 6
3y^{2} + 20y + 12 = 0
3y^{2} + 18y + 2y + 12 = 0
Gives y = -6, -2/3
Put on number line
-6… -2/3… 1…. 7

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

Directions (1-5): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

2x^{2} – 13x + 6 = 0, 2y^{2} – 33y + 130 = 0

Correct

Explanation:
2x^{2} – 13x + 6 = 0
2x^{2} – 13x + 6 = 0
Gives x = 1/2, 6
2y^{2} – 33y + 130 = 0
2y^{2} – 20y – 13y + 130 = 0
Gives y= 13/2, 10
Put on number line
1/2…. 6… 13/2… 10

Incorrect

Explanation:
2x^{2} – 13x + 6 = 0
2x^{2} – 13x + 6 = 0
Gives x = 1/2, 6
2y^{2} – 33y + 130 = 0
2y^{2} – 20y – 13y + 130 = 0
Gives y= 13/2, 10
Put on number line
1/2…. 6… 13/2… 10

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

A person has a monthly salary of Rs 10,000 out of which he spends Rs 6,000. After a month, He got a salary hike by 10% and so he increases his expenditure by 20%. Find the percentage change in his monthly savings?

Correct

Explanation:
Savings = 10,000 – 6000 = 4000
New Income becomes = (110/100)*10,000 = 11,000
Exp. becomes = (120/100)*6,000 = 7,200
So savings now = 11000 – 7200 = 3800
So % decrease in savings = (4000-3800)/4000 * 100 = 5%

Incorrect

Explanation:
Savings = 10,000 – 6000 = 4000
New Income becomes = (110/100)*10,000 = 11,000
Exp. becomes = (120/100)*6,000 = 7,200
So savings now = 11000 – 7200 = 3800
So % decrease in savings = (4000-3800)/4000 * 100 = 5%

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

There are 3 red balls, 5 blue balls and 2 white balls in a bag from which 2 balls are chosen randomly. What is the probability that at most one ball is blue?

Correct

Explanation:
All blue prob = 5C2/10C2 = 2/9
So atmost 1 = 1 – 2/9 = 7/9

Incorrect

Explanation:
All blue prob = 5C2/10C2 = 2/9
So atmost 1 = 1 – 2/9 = 7/9

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

The ratio of the ages of A and B is 2 : 3. Also the ratio of age of A 4 years from now and that of C’s present age is 7 : 12. If 9 years ago, C was 39 years old, what was the age of B 5 years ago?

Correct

Explanation:
C’s present age = 39+9 = 48
So (A+4)/48 = 7/12
So A = 24
And 24/B = 2/3
So B = 36

Incorrect

Explanation:
C’s present age = 39+9 = 48
So (A+4)/48 = 7/12
So A = 24
And 24/B = 2/3
So B = 36

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

The perimeter of a rectangular plot is 130 m. Find the cost of gardening 1.5 m broad boundary around rectangular plot if the cost of gardening is Rs 16 per m^{2}.

Correct

Explanation:
Given 2(l+b) = 130
1.5 m broad boundary means increase in l and b by 3 m
So area of the boundary will be [(l+3)(b+3) – lb] = 3(l+b) + 9 = 3*130/2 + 9 = 204 m^{2}
So cost of gardening = 204*16

Incorrect

Explanation:
Given 2(l+b) = 130
1.5 m broad boundary means increase in l and b by 3 m
So area of the boundary will be [(l+3)(b+3) – lb] = 3(l+b) + 9 = 3*130/2 + 9 = 204 m^{2}
So cost of gardening = 204*16

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

A and B started a business by investing Rs 5,000 and Rs 7,000 respectively. After 4 months A added Rs 1000 while B withdrew Rs 1000. At the same time C joined them with an investment of Rs 10,000. If a total profit received after a year is Rs 7,280, what is the difference in the shares of A and C?