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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeThe present age of Sarita is two less than the total age of her two sons and one daughter. Four years hence, the age of younger son and the age of daughter is 5:6 and daughter is 8 years younger than the older son. What will be the average age of Sarita and her two sons, 12 years hence if the age of Sarita 9 years ago is 36 years?
Correct
Answer: 4) 38 years
Explanation:
4 years hence, let the age of younger son = 5x and the age of daughter = 6x
Age of older son = 6x+8
Present age of younger son = 5x4, Present age of older son = 6x+84 = 6x+4 and Present age of daughter = 6x4
Given, the age of Sarita 9 years ago = 36 years
Present age of Sarita = 36+9 = 45 years
ATQ, the present age of Sarita is two less than the total age of her two sons and one daughter.
45 = 5x4 + 6x+4 + 6x42
45+2 = 5x4 + 6x+4 + 6x4
47 = 17x 4
17x = 51
x = 3
Age of Sarita 12 years hence = 45+12 = 57
Age of her younger son 12 years hence = 5x4+12 = 5x+8 = 5*3+8 = 23
Age of her older son 12 years hence = 6x+4+12 = 6x+16 = 6*3+16 = 34
Age of her daughter 12 years hence = 6x4+12 = 6x+8 = 6*3+6 = 24
Required average = (57+23+34)/3 = 114/3 = 38 yearsIncorrect
Answer: 4) 38 years
Explanation:
4 years hence, let the age of younger son = 5x and the age of daughter = 6x
Age of older son = 6x+8
Present age of younger son = 5x4, Present age of older son = 6x+84 = 6x+4 and Present age of daughter = 6x4
Given, the age of Sarita 9 years ago = 36 years
Present age of Sarita = 36+9 = 45 years
ATQ, the present age of Sarita is two less than the total age of her two sons and one daughter.
45 = 5x4 + 6x+4 + 6x42
45+2 = 5x4 + 6x+4 + 6x4
47 = 17x 4
17x = 51
x = 3
Age of Sarita 12 years hence = 45+12 = 57
Age of her younger son 12 years hence = 5x4+12 = 5x+8 = 5*3+8 = 23
Age of her older son 12 years hence = 6x+4+12 = 6x+16 = 6*3+16 = 34
Age of her daughter 12 years hence = 6x4+12 = 6x+8 = 6*3+6 = 24
Required average = (57+23+34)/3 = 114/3 = 38 years 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeSumit bought a combined total of 15 mouse and keyboards. He marked up the mouse by 25% on the cost price, while each keyboard was marked up by Rs. 200. He was able to sell 80% of the mouse and 3 keyboards and make a profit of Rs. 4200. The remaining mouse and 2 keyboards could not be sold by him. Find his overall loss if he gets no return on unsold items and it is known that a keyboard costs 400% of a mouse.
Correct
Answer: 5) Rs. 9000
Explanation:
Total keyboard sold = 3 and mouse sold = 80% of 15 = 12
Total profit obtained by selling 3 keyboards = 3*200 = 600
Total profit obtained by selling 12 mouse = 4200600 = 3600
The profit obtained by selling each mouse = 3600/12 = 300
Cost price of each mouse = 300*(100/25) = 1200
Selling price of each mouse = 1200+300 = 1500
Selling price of 12 mouse = 12*1500 = 18000
Total cost price of 15 mouse = 15*1200 = 18000
Cost price of each keyboard = 400% of 1200 = 4800
Total cost price of 5 keyboards = 5*4800 = 24000
Selling price of each keyboard = 4800+200 = 5000
Total selling price of 3 keyboard = 3*5000 = 15000
Total cost price of 15 mouse and 5 keyboards = 18000+24000= 42000
Total selling price of 12 mouse and 3 keyboards = 18000+15000 = 33000
Loss = Cost price – Selling price
Loss = 4200033000
Loss = Rs. 9000Incorrect
Answer: 5) Rs. 9000
Explanation:
Total keyboard sold = 3 and mouse sold = 80% of 15 = 12
Total profit obtained by selling 3 keyboards = 3*200 = 600
Total profit obtained by selling 12 mouse = 4200600 = 3600
The profit obtained by selling each mouse = 3600/12 = 300
Cost price of each mouse = 300*(100/25) = 1200
Selling price of each mouse = 1200+300 = 1500
Selling price of 12 mouse = 12*1500 = 18000
Total cost price of 15 mouse = 15*1200 = 18000
Cost price of each keyboard = 400% of 1200 = 4800
Total cost price of 5 keyboards = 5*4800 = 24000
Selling price of each keyboard = 4800+200 = 5000
Total selling price of 3 keyboard = 3*5000 = 15000
Total cost price of 15 mouse and 5 keyboards = 18000+24000= 42000
Total selling price of 12 mouse and 3 keyboards = 18000+15000 = 33000
Loss = Cost price – Selling price
Loss = 4200033000
Loss = Rs. 9000 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeA, B and C started a business by investing in the ratio of 5:4:8 respectively. A manages the business and for his activity, he gets an amount of Rs. 50000 at the end of the year. If at the end of year total profit obtained by them are in the ratio of 5:2:4 then what is the profit B got at the end of the year?
Correct
Answer: 3) Rs. 40000
Explanation:
The ratio of investment of A, B and C is 5:4:8.
Ratio of profit of A, B and C = 5: 2: 4
5: 4: 8
(5: 2: 4) * 2
5: 4: 8
(10:4: 8)
Amount received by A for managing the business = 50000
10 – 5 — 50000
5—— 50000
1 ——10000
Profit obtained by B =
4 ——4*1 ——4*10000 = Rs. 40000Incorrect
Answer: 3) Rs. 40000
Explanation:
The ratio of investment of A, B and C is 5:4:8.
Ratio of profit of A, B and C = 5: 2: 4
5: 4: 8
(5: 2: 4) * 2
5: 4: 8
(10:4: 8)
Amount received by A for managing the business = 50000
10 – 5 — 50000
5—— 50000
1 ——10000
Profit obtained by B =
4 ——4*1 ——4*10000 = Rs. 40000 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeA sum of money was invested for 6 years in Scheme A, which offered simple interest at a rate of 12% per annum. The amount received from Scheme A after 6 years was then invested for two years in Scheme B, which offers compound interest (compounded annual) at a rate of 5% per annum. If the interest received from Scheme B was Rs. 5289, what was the sum invested in Scheme A?
Correct
Answer: 1) Rs. 30000
Explanation:
Let the amount invested in scheme A = 100x
Interest obtained in 6 years = (100x*12*6)/100 = 72x
Amount received = 100x+72x = 172x
Amount invested in compound interest for 2 years = 172x
Interest received in 2 years = 172x*(10.25/100) {since 5+5+5*5/100 = 10.25%}
17.63x
17.63x = 5289
x = 5289/17.63
x = 300
Amount invested in scheme A = 100x = 100*300 = Rs. 30000Incorrect
Answer: 1) Rs. 30000
Explanation:
Let the amount invested in scheme A = 100x
Interest obtained in 6 years = (100x*12*6)/100 = 72x
Amount received = 100x+72x = 172x
Amount invested in compound interest for 2 years = 172x
Interest received in 2 years = 172x*(10.25/100) {since 5+5+5*5/100 = 10.25%}
17.63x
17.63x = 5289
x = 5289/17.63
x = 300
Amount invested in scheme A = 100x = 100*300 = Rs. 30000 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeA tank is connected with 8 taps some of them fill the tank and others that are emptying taps. Each of tap that fill the tank can fill it in 4 hr while each of the emptying tap can empty it in 6 hr. If all the taps kept open it take 2 hr to fill 25% of the tank. How many of the taps are emptying taps.
Correct
Answer: 4) 3
Explanation:
Let the number of taps that fill the tank are x and the number of emptying taps are 8x
Given, each tap can fill the tank in 4 hr
Therefore, tank filled by each tap in 1 hour = ¼
Tank filled by x taps in 1 hour = x*(1/4) = x/4
Tank filled by x taps in 2 hours = 2*(x/4) = 2x/4 = x/4
Also each emptying tap can empty the tank in 6 hours
Tank emptied by one emptying tap in 1 hour = 1/6
Tank emptied by (8x) emptying tap in 1 hour = (8x)/6
And tank emptied by (8x) emptying tap in 2 hours = 2*{(8x)/6} = (8x)/3
ATQ,
As the tank gets 25% filled in 2 hours
x/4 – (8x)/3 = 25% of 1
{3x – 4*(8x)}/12 = .25
(3x32+4x) = 3
7x 32 = 3
7x = 35
x = 5
Therefore, emptying taps = 8x = 85 = 3Incorrect
Answer: 4) 3
Explanation:
Let the number of taps that fill the tank are x and the number of emptying taps are 8x
Given, each tap can fill the tank in 4 hr
Therefore, tank filled by each tap in 1 hour = ¼
Tank filled by x taps in 1 hour = x*(1/4) = x/4
Tank filled by x taps in 2 hours = 2*(x/4) = 2x/4 = x/4
Also each emptying tap can empty the tank in 6 hours
Tank emptied by one emptying tap in 1 hour = 1/6
Tank emptied by (8x) emptying tap in 1 hour = (8x)/6
And tank emptied by (8x) emptying tap in 2 hours = 2*{(8x)/6} = (8x)/3
ATQ,
As the tank gets 25% filled in 2 hours
x/4 – (8x)/3 = 25% of 1
{3x – 4*(8x)}/12 = .25
(3x32+4x) = 3
7x 32 = 3
7x = 35
x = 5
Therefore, emptying taps = 8x = 85 = 3 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudePritam sold two items: A and B. The ratio of the loss incurred by selling item A to the profit gained by selling item B is 2:3. The cost price of A is 60% of the cost price of B and the selling price of B is two times the selling price of A. If the sum of the selling price of A and cost price of B is Rs. 270, what is the cost price of B?
Correct
Answer: 5) Rs. 175
Explanation:
Let the loss incurred by selling item A = 2x
And profit gained by selling item B = 3x
Let the cost price of item B = 100y
And the cost price of item A = 60% of 100x = 60y
Selling price of item B = 100y+3x
And selling price of item A = 60y2x
ATQ,
100y+3x = 2*(60y2x)
100y+3x = 120y4x
3x+4x = 120y100y
7x = 20y
Also,
60y2x+100y = 270
21x2x+35x = 270 (7x = 20y)
54x = 270
x = 270/54
x = 5
Cost price of item B = 100y = 35x = 35*5 = 175 (7x=20y)Incorrect
Answer: 5) Rs. 175
Explanation:
Let the loss incurred by selling item A = 2x
And profit gained by selling item B = 3x
Let the cost price of item B = 100y
And the cost price of item A = 60% of 100x = 60y
Selling price of item B = 100y+3x
And selling price of item A = 60y2x
ATQ,
100y+3x = 2*(60y2x)
100y+3x = 120y4x
3x+4x = 120y100y
7x = 20y
Also,
60y2x+100y = 270
21x2x+35x = 270 (7x = 20y)
54x = 270
x = 270/54
x = 5
Cost price of item B = 100y = 35x = 35*5 = 175 (7x=20y) 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudePipe P and pipe Q together can fill a tank in 24 minutes. With pipe R they can fill the tank in 20 minutes. Pipe P and pipe R together can fill the tank in 40 minutes. Find the time taken by pipe Q and pipe R to fill the tank.
Correct
Answer: 4) 30 minutes
Explanation:
Let the capacity of the tank is LCM of 24, 20 and 40 = 120 units
Tank filled by pipe P and pipe Q together in 1 hour = 120/24 = 5 units
Tank filled by P, Q, and R together in 1 hour = 120/20 = 6 units
Tank filled by P and R together in 1 hour = 120/40 = 3 units
Tank filled by Q in one hour = tank filled by P, Q and R together in one hour – tank filled by P and R together in one hour
Tank filled by Q in one hour = 6 – 3 = 3 units
Tank filled by R in one hour = tank filled by P, Q, and R together in one hour – tank filled by P and Q in one hour
Tank filled by R in one hour = 6 5 = 1 units
Then filled by Q and R together in one hour = 3+1 = 4 units
Time taken to fill the tank by Q and R together = 120/4 = 30 minutesIncorrect
Answer: 4) 30 minutes
Explanation:
Let the capacity of the tank is LCM of 24, 20 and 40 = 120 units
Tank filled by pipe P and pipe Q together in 1 hour = 120/24 = 5 units
Tank filled by P, Q, and R together in 1 hour = 120/20 = 6 units
Tank filled by P and R together in 1 hour = 120/40 = 3 units
Tank filled by Q in one hour = tank filled by P, Q and R together in one hour – tank filled by P and R together in one hour
Tank filled by Q in one hour = 6 – 3 = 3 units
Tank filled by R in one hour = tank filled by P, Q, and R together in one hour – tank filled by P and Q in one hour
Tank filled by R in one hour = 6 5 = 1 units
Then filled by Q and R together in one hour = 3+1 = 4 units
Time taken to fill the tank by Q and R together = 120/4 = 30 minutes 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections (810): Akhil is twice as efficient as Roma and can finish a piece of work in 60 hours less than Roma does.
If Akhil and Roma completed 2/5th of the work and then Roma is replaced by Mridula, the remaining work is done by Akhil and Mridula in 16 hours.
If Akhil and Roma completed 1/4th of the work and then Akhil is replaced by Yuvika, the remaining work is done by Roma and Yuvika in 30 hours.
In how many hours can Roma and Yuvika finish the work while working together?
Correct
Answer: 5) 40 hours
Explanation:
Since total work = 120 units
Work done by Roma in one hour = 1 unit
And work done by Yuvika in one hour = 2 units
Work done by Roma and Yuvika together in one hour = 3 units
Time taken by Roma and Yuvika to complete the work = 120/3 = 40 hoursCommon Explanation:
Let the efficiency of Roma = x and efficiency of Akhil = 2x
Therefore, time taken by Roma to finish the work will be 2T and time taken by Akhil to finish the work will be T
2T – T = 60 hours
T = 60 hours
Time taken by Roma to finish the work = 2T = 2*60 = 120 hours and time taken by Akhil to finish the work =T = 60 hours
Let the total work is LCM of 60 and 120 = 120 units
Work done by Akhil in one hour = 120/60 = 2 units
And work done by Roma in one hour = 120/120 = 1 unit
Work completed by Akhil and Roma = (2/5)*120 = 48units
Therefore, work left to be done by Akhil and Mridula = 12048 = 72 units
Now, Mridula and Akhil worked for 16 hours
Work done by Akhil in 16 hours = 2*16 = 32 units
Work done by Mridula in 16 hours = 7232 = 40 units
Work done by Mridula in one hour = 40/16 = 2.5 units
Again,
Work completed by Akhil and Roma = (1/4)*120 = 30 units
Therefore, work left for Roma and Yuvika = 12030 = 90units
Work done by Roma in 30 hours = 30*1 = 30units
Work done by Yuvika in 30 hours = 9030 = 60 units
Work done by Yuvika in one hour = 60/30 = 2 unitsIncorrect
Answer: 5) 40 hours
Explanation:
Since total work = 120 units
Work done by Roma in one hour = 1 unit
And work done by Yuvika in one hour = 2 units
Work done by Roma and Yuvika together in one hour = 3 units
Time taken by Roma and Yuvika to complete the work = 120/3 = 40 hoursCommon Explanation:
Let the efficiency of Roma = x and efficiency of Akhil = 2x
Therefore, time taken by Roma to finish the work will be 2T and time taken by Akhil to finish the work will be T
2T – T = 60 hours
T = 60 hours
Time taken by Roma to finish the work = 2T = 2*60 = 120 hours and time taken by Akhil to finish the work =T = 60 hours
Let the total work is LCM of 60 and 120 = 120 units
Work done by Akhil in one hour = 120/60 = 2 units
And work done by Roma in one hour = 120/120 = 1 unit
Work completed by Akhil and Roma = (2/5)*120 = 48units
Therefore, work left to be done by Akhil and Mridula = 12048 = 72 units
Now, Mridula and Akhil worked for 16 hours
Work done by Akhil in 16 hours = 2*16 = 32 units
Work done by Mridula in 16 hours = 7232 = 40 units
Work done by Mridula in one hour = 40/16 = 2.5 units
Again,
Work completed by Akhil and Roma = (1/4)*120 = 30 units
Therefore, work left for Roma and Yuvika = 12030 = 90units
Work done by Roma in 30 hours = 30*1 = 30units
Work done by Yuvika in 30 hours = 9030 = 60 units
Work done by Yuvika in one hour = 60/30 = 2 units 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections (810): Akhil is twice as efficient as Roma and can finish a piece of work in 60 hours less than Roma does.
If Akhil and Roma completed 2/5th of the work and then Roma is replaced by Mridula, the remaining work is done by Akhil and Mridula in 16 hours.
If Akhil and Roma completed 1/4th of the work and then Akhil is replaced by Yuvika, the remaining work is done by Roma and Yuvika in 30 hours.
Akhil and Roma worked in the first hour and Mridula and Yuvika work in the second hour. What is the ratio of work done in the first and second hour?
Correct
Answer: 1) 2:3
Explanation:
Work done by Akhil in one hour = 2 units
And work done by Roma in one hour = 1 unit
Total work done by Akhil and Roma together in one hour = 3 units
Work done by Mridula in one hour = 2.5 units
And work done by Yuvika in one hour = 2 units
Total work done by Mridula and Yuvika together in one hour = 4.5 units
Required ratio = 3:4.5 = 2:3Incorrect
Answer: 1) 2:3
Explanation:
Work done by Akhil in one hour = 2 units
And work done by Roma in one hour = 1 unit
Total work done by Akhil and Roma together in one hour = 3 units
Work done by Mridula in one hour = 2.5 units
And work done by Yuvika in one hour = 2 units
Total work done by Mridula and Yuvika together in one hour = 4.5 units
Required ratio = 3:4.5 = 2:3 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections (810): Akhil is twice as efficient as Roma and can finish a piece of work in 60 hours less than Roma does.
If Akhil and Roma completed 2/5th of the work and then Roma is replaced by Mridula, the remaining work is done by Akhil and Mridula in 16 hours.
If Akhil and Roma completed 1/4th of the work and then Akhil is replaced by Yuvika, the remaining work is done by Roma and Yuvika in 30 hours.
Akhil and Mridula worked together for 4 hours. In how much time would Yuvika have done the same amount of work if she was working alone?
Correct
Answer: 4) 9 hours
Explanation:
Work done by Akhil in one hour = 2 units
Work done by Mridula in one hour = 2.5 units
Work done by Akhil and Mridula together in one hour = 4.5 units
Work done by Akhil and Mridula together in four hours = 4*4.5 = 18 units
Time taken by Yuvika to complete 18 units work = 18/2 = 9 hoursIncorrect
Answer: 4) 9 hours
Explanation:
Work done by Akhil in one hour = 2 units
Work done by Mridula in one hour = 2.5 units
Work done by Akhil and Mridula together in one hour = 4.5 units
Work done by Akhil and Mridula together in four hours = 4*4.5 = 18 units
Time taken by Yuvika to complete 18 units work = 18/2 = 9 hours
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