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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

The length of rectangle is 33 m and its breadth is two-third the radius of circle whose area is 924 sq. cm more than the area of rectangle. Find is the breadth of the rectangle?

Correct

Explanation:
Radius = x, so breadth = (2x/3), length = 33
So l × b = ᴨr^{2} – 924
(2x/3) * 33 = (22/7)x^{2} – 924
x = x^{2}/7 – 42
x^{2} – 7x – 294 = 0
solve, x = 21
so breadth = 2/3 * 21

Incorrect

Explanation:
Radius = x, so breadth = (2x/3), length = 33
So l × b = ᴨr^{2} – 924
(2x/3) * 33 = (22/7)x^{2} – 924
x = x^{2}/7 – 42
x^{2} – 7x – 294 = 0
solve, x = 21
so breadth = 2/3 * 21

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A and B started a business by investing Rs 7,000 and Rs 7,500 respectively. After 9 months, their friend C joined them with Rs 10500. At the same time, A and B added Rs 1,000 each. If at the end of year B and C together received Rs 12,450 as their share from the total profit, then what is the total profit?

10 men can complete a work in 6 days. 12 women can complete in 10 days and 10 children in 12 days. 1 man, 1 woman and 1 child started the work. In how many days will the work get completed if the man did twice the work as woman and also twice the work as child?

Correct

Explanation:
Man did twice work from woman and child both. This means woman and child did equal work and man twice of that. So 2x + x + x = 1, x = 1/4
Woman did 1/4 of work, child did 1/4 of work and man 2*(1/4) = 1/2 of work
10 m in 6 days, so 1 man in 60 days, so 1/2 work in 30 days.
12 w in 10 days, so 1 woman in 120 days, so 1/4 work in 30 days.
10 c in 12 days, so 1 child in 120 days, so 1/4 work in 30 days.
So 1/30 + 1/30 + 1/30

Incorrect

Explanation:
Man did twice work from woman and child both. This means woman and child did equal work and man twice of that. So 2x + x + x = 1, x = 1/4
Woman did 1/4 of work, child did 1/4 of work and man 2*(1/4) = 1/2 of work
10 m in 6 days, so 1 man in 60 days, so 1/2 work in 30 days.
12 w in 10 days, so 1 woman in 120 days, so 1/4 work in 30 days.
10 c in 12 days, so 1 child in 120 days, so 1/4 work in 30 days.
So 1/30 + 1/30 + 1/30

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

A person having salary of Rs 40,000 per month spends Rs 12,000. After a year, his salary increases by 15% so he increases his expenditure by 10%. What is the approximate percentage change in his savings?

Correct

Explanation:
Savings = 40,000 – 12000 = 28000
Income becomes = (115/100)*40,000 = 46,000
Exp. becomes = (110/100)*12,000 = 13,200
So savings now = 46000 – 13200 = 32800
So % increase in savings = (32800-28000)/28000 * 100 = 17%

Incorrect

Explanation:
Savings = 40,000 – 12000 = 28000
Income becomes = (115/100)*40,000 = 46,000
Exp. becomes = (110/100)*12,000 = 13,200
So savings now = 46000 – 13200 = 32800
So % increase in savings = (32800-28000)/28000 * 100 = 17%

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

One-third of an article is sold at 12% profit and remaining at 3% loss. If a total of Rs 800 is made as profit, what is the cost price of the article?

Correct

Explanation:
It can be calculated as:
CP = 800 *100 / [1/3 *12 + 2/3 * (-3)] = 40,000
* 2/3rd sold at 3% loss so negative sign, 1/3rd sold at 12% profit.

Incorrect

Explanation:
It can be calculated as:
CP = 800 *100 / [1/3 *12 + 2/3 * (-3)] = 40,000
* 2/3rd sold at 3% loss so negative sign, 1/3rd sold at 12% profit.