Welcome to Online Quants Test in AffairsCloud.com. We are starting IBPS PO Prelim 2016 Course and we are creating sample questions in Quants section, type of which will be asked in IBPS PO 2016 !!!

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading...

You must sign in or sign up to start the quiz.

You have to finish following quiz, to start this quiz:

Results

0 of 10 questions answered correctly

Your time:

Time has elapsed

You have reached 0 of 0 points, (0)

Categories

Quantitative Aptitude0%

1

2

3

4

5

6

7

8

9

10

Answered

Review

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 –

4x^{2} + 23x + 15 = 0, 2y^{2} – 11y + 14 = 0

Correct

Explanation:
4x^{2} + 23x + 15 = 0
4x^{2} + 20x + 3x + 15 = 0
Gives x = -5, -3/4
2y^{2} – 11y + 14 = 0
2y^{2} – 4y – 7y + 14 = 0
Gives y = 2, 7/2
Put on number line
-5… -3/4 … 2….. 7/2

Incorrect

Explanation:
4x^{2} + 23x + 15 = 0
4x^{2} + 20x + 3x + 15 = 0
Gives x = -5, -3/4
2y^{2} – 11y + 14 = 0
2y^{2} – 4y – 7y + 14 = 0
Gives y = 2, 7/2
Put on number line
-5… -3/4 … 2….. 7/2

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} – 2x – 8 = 0, 3y^{2} – 14y – 24 = 0

Correct

Explanation:
3x^{2} – 2x – 8 = 0
3x^{2} – 6x + 4x – 8 = 0
Gives x = 2, -4/3
3y^{2} – 14y – 24 = 0
3y^{2} – 18y + 4y – 24 = 0
Gives y = -4/3, 6
Put on number line
-4/3… 2… 6
When x = 2, x < y (6) and x > y (-4/3) – here relation cant be determined.

Incorrect

Explanation:
3x^{2} – 2x – 8 = 0
3x^{2} – 6x + 4x – 8 = 0
Gives x = 2, -4/3
3y^{2} – 14y – 24 = 0
3y^{2} – 18y + 4y – 24 = 0
Gives y = -4/3, 6
Put on number line
-4/3… 2… 6
When x = 2, x < y (6) and x > y (-4/3) – 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 –

2x^{2} – 3x – 9 = 0, 6y^{2} + 23y + 21 = 0

Correct

Explanation:
2x^{2} – 3x – 9 = 0
2x^{2} – 6x + 3x – 9 = 0
Gives x = -3/2, 3
6y^{2} + 23y + 21 = 0
6y^{2} + 9y + 14y + 21 = 0
Gives y = -3/2, -7/3
Put on number line
-7/3… -3/2… 3

Incorrect

Explanation:
2x^{2} – 3x – 9 = 0
2x^{2} – 6x + 3x – 9 = 0
Gives x = -3/2, 3
6y^{2} + 23y + 21 = 0
6y^{2} + 9y + 14y + 21 = 0
Gives y = -3/2, -7/3
Put on number line
-7/3… -3/2… 3

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 –

12x^{2} – 11x + 2 = 0, 3y^{2} – 25y + 52 = 0

Correct

Explanation:
12x^{2} – 11x + 2 = 0
12x^{2} – 3x – 8x + 2 = 0
Gives x = 1/4, 2/3
3y^{2} – 25y + 52 = 0
3y^{2} – 12y – 13y + 52 = 0
Gives y = 13/3, 4
Put on number line
1/4…. 2/3… 4….. 13/3

Incorrect

Explanation:
12x^{2} – 11x + 2 = 0
12x^{2} – 3x – 8x + 2 = 0
Gives x = 1/4, 2/3
3y^{2} – 25y + 52 = 0
3y^{2} – 12y – 13y + 52 = 0
Gives y = 13/3, 4
Put on number line
1/4…. 2/3… 4….. 13/3

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 + 21 = 0, 3y^{2} + 7y – 6 = 0

Correct

Explanation:
2x^{2} + 13x + 21 = 0
2x^{2} + 6x + 7x + 21 = 0
Gives x = -7/2, -3
3y^{2} + 7y – 6 = 0
3y^{2} + 9y – 2y – 6 = 0
Gives y= -3, 2/3
Put on number line
-7/2…. -3… 2/3

Incorrect

Explanation:
2x^{2} + 13x + 21 = 0
2x^{2} + 6x + 7x + 21 = 0
Gives x = -7/2, -3
3y^{2} + 7y – 6 = 0
3y^{2} + 9y – 2y – 6 = 0
Gives y= -3, 2/3
Put on number line
-7/2…. -3… 2/3

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

If a man loses 25% by selling 24 candies for a rupee, how many candies should he sell for five rupees in order to reduce loss by 15%?

Correct

Explanation:
We have a shortcut
First find the number of candies he bought for a rupee:
Loss of 25% by selling 20 candies for a rupee
So candies bought for a rupee = ((100-25)/100) * 24 = 18
Now to have loss of 10% (25-15), he must sell for a rupee:
Candies = 18 * (100/(100-10) = 20
So for 5 rupees, 5*20 = 100 candies

Incorrect

Explanation:
We have a shortcut
First find the number of candies he bought for a rupee:
Loss of 25% by selling 20 candies for a rupee
So candies bought for a rupee = ((100-25)/100) * 24 = 18
Now to have loss of 10% (25-15), he must sell for a rupee:
Candies = 18 * (100/(100-10) = 20
So for 5 rupees, 5*20 = 100 candies

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A starts a business by investing Rs 4000. After 3 months B comes and invests Rs 5,000. After a further of 4 months C joins them by investing Rs 6,500 and A adds Rs 2000 in his share. If after a year, a total of Rs 32,520 is made as profit, what is the profit share of A?

Correct

Explanation:
4000*7 + 6000*5: 5000*9 : 6500*5
116 : 90 : 65
So A gets = 116/(116+90+65) * 32520

Incorrect

Explanation:
4000*7 + 6000*5: 5000*9 : 6500*5
116 : 90 : 65
So A gets = 116/(116+90+65) * 32520

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

A, B and C can alone complete a work in 15, 25 and 30 days respectively. They started the work with B worked for 4 days extra till the completion of work. How much of the total work did B and C did?

Correct

Explanation:
If A and C worked for x days, then B for (x+4) days
So
(1/15 + 1/30)*x + (1/25)*(x+4) = 1
Solve, x = 6
So C worked for 6 days and B for 10 days
So they did 6/30 + 10/25 work resp
So total 3/5

Incorrect

Explanation:
If A and C worked for x days, then B for (x+4) days
So
(1/15 + 1/30)*x + (1/25)*(x+4) = 1
Solve, x = 6
So C worked for 6 days and B for 10 days
So they did 6/30 + 10/25 work resp
So total 3/5

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

The volume of a right circular cylinder is 12320 cu. m and its curved surface area is 1760 sq. m. What is the diameter and height of this cylinder respectively?

Correct

Explanation:
Vol = ᴨr^{2}h = 12320
CSA = 2ᴨrh = 1760
Divide both equations, r = 14 m
So diameter = 28 m, h = 20 m

Incorrect

Explanation:
Vol = ᴨr^{2}h = 12320
CSA = 2ᴨrh = 1760
Divide both equations, r = 14 m
So diameter = 28 m, h = 20 m

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

A sum of money is invested in two schemes in parts. The first part is invested at 5% per annum for 8 years and the second part at 6% per annum for 7 years. It is found that there is a difference of Rs 44 in the interest obtained from two parts. Also if the difference for 2 years in simple interest and compound interest obtained when the first part is invested at both at 5% per annum is Rs 15.5, find the sum.

Correct

Explanation:
From: difference for 2 years in simple interest and compound interest obtained when the first part is invested at 5% per annum is Rs 15.5
Let x be 1st part invested
So x* 5*5/100*100 = 15.5
Solve x = 6200
Now from 1st part of question
6200*5*8/100 – y*6*7/100 = 44
Solve, second part = y = 5800
So total sum = 6200+5800

Incorrect

Explanation:
From: difference for 2 years in simple interest and compound interest obtained when the first part is invested at 5% per annum is Rs 15.5
Let x be 1st part invested
So x* 5*5/100*100 = 15.5
Solve x = 6200
Now from 1st part of question
6200*5*8/100 – y*6*7/100 = 44
Solve, second part = y = 5800
So total sum = 6200+5800