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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 –
4x2 + 23x + 15 = 0, 2y2 – 11y + 14 = 0
Correct
Explanation:
4x2 + 23x + 15 = 0
4x2 + 20x + 3x + 15 = 0
Gives x = -5, -3/4
2y2 – 11y + 14 = 0
2y2 – 4y – 7y + 14 = 0
Gives y = 2, 7/2
Put on number line
-5… -3/4 … 2….. 7/2
Incorrect
Explanation:
4x2 + 23x + 15 = 0
4x2 + 20x + 3x + 15 = 0
Gives x = -5, -3/4
2y2 – 11y + 14 = 0
2y2 – 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 –
3x2 – 2x – 8 = 0, 3y2 – 14y – 24 = 0
Correct
Explanation:
3x2 – 2x – 8 = 0
3x2 – 6x + 4x – 8 = 0
Gives x = 2, -4/3
3y2 – 14y – 24 = 0
3y2 – 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:
3x2 – 2x – 8 = 0
3x2 – 6x + 4x – 8 = 0
Gives x = 2, -4/3
3y2 – 14y – 24 = 0
3y2 – 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 –
2x2 – 3x – 9 = 0, 6y2 + 23y + 21 = 0
Correct
Explanation:
2x2 – 3x – 9 = 0
2x2 – 6x + 3x – 9 = 0
Gives x = -3/2, 3
6y2 + 23y + 21 = 0
6y2 + 9y + 14y + 21 = 0
Gives y = -3/2, -7/3
Put on number line
-7/3… -3/2… 3
Incorrect
Explanation:
2x2 – 3x – 9 = 0
2x2 – 6x + 3x – 9 = 0
Gives x = -3/2, 3
6y2 + 23y + 21 = 0
6y2 + 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 –
12x2 – 11x + 2 = 0, 3y2 – 25y + 52 = 0
Correct
Explanation:
12x2 – 11x + 2 = 0
12x2 – 3x – 8x + 2 = 0
Gives x = 1/4, 2/3
3y2 – 25y + 52 = 0
3y2 – 12y – 13y + 52 = 0
Gives y = 13/3, 4
Put on number line
1/4…. 2/3… 4….. 13/3
Incorrect
Explanation:
12x2 – 11x + 2 = 0
12x2 – 3x – 8x + 2 = 0
Gives x = 1/4, 2/3
3y2 – 25y + 52 = 0
3y2 – 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 –
2x2 + 13x + 21 = 0, 3y2 + 7y – 6 = 0
Correct
Explanation:
2x2 + 13x + 21 = 0
2x2 + 6x + 7x + 21 = 0
Gives x = -7/2, -3
3y2 + 7y – 6 = 0
3y2 + 9y – 2y – 6 = 0
Gives y= -3, 2/3
Put on number line
-7/2…. -3… 2/3
Incorrect
Explanation:
2x2 + 13x + 21 = 0
2x2 + 6x + 7x + 21 = 0
Gives x = -7/2, -3
3y2 + 7y – 6 = 0
3y2 + 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 = ᴨr2h = 12320
CSA = 2ᴨrh = 1760
Divide both equations, r = 14 m
So diameter = 28 m, h = 20 m
Incorrect
Explanation:
Vol = ᴨr2h = 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
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