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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (15): 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….. 7/2Incorrect
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 pointsCategory: Quantitative AptitudeDirections (15): 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 pointsCategory: Quantitative AptitudeDirections (15): 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… 3Incorrect
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 pointsCategory: Quantitative AptitudeDirections (15): 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….. 13/3Incorrect
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 pointsCategory: Quantitative AptitudeDirections (15): 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… 2/3Incorrect
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 pointsCategory: Quantitative AptitudeIf 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 = ((10025)/100) * 24 = 18
Now to have loss of 10% (2515), he must sell for a rupee:
Candies = 18 * (100/(10010) = 20
So for 5 rupees, 5*20 = 100 candiesIncorrect
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 = ((10025)/100) * 24 = 18
Now to have loss of 10% (2515), he must sell for a rupee:
Candies = 18 * (100/(10010) = 20
So for 5 rupees, 5*20 = 100 candies 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeA 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) * 32520Incorrect
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 pointsCategory: Quantitative AptitudeA, 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 3/5Incorrect
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 pointsCategory: Quantitative AptitudeThe 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 mIncorrect
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 pointsCategory: Quantitative AptitudeA 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+5800Incorrect
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
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