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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (1-5): The line graph given below shows the % decrease and % increase in the profit of Company A and Company B respectively. Read the data carefully and answer the following questions.
Note: Profit = (Income – Expenditure)
If the profit of company A and Company B is same in the year 2014 then the profit of Company A in the year 2016 is what % of the profit of company B in the year 2015?
Correct
Answer: 2) 51%
Explanation:
2014: Let the profit of company A = profit of company B = 100x
Profit of company A in 2015 = 100x*{(100-15)/100} = 100x*(85/100) = 85x
Profit of company A in 2016 = 85x*{(100-25)/100} = 85x*(75/100) = 63.75x
Profit of company B in 2015 = 100x*{(100+25)/100} = 100x*(125/100) = 125x
Required percentage = (63.75x/125x)*100 = 51%Incorrect
Answer: 2) 51%
Explanation:
2014: Let the profit of company A = profit of company B = 100x
Profit of company A in 2015 = 100x*{(100-15)/100} = 100x*(85/100) = 85x
Profit of company A in 2016 = 85x*{(100-25)/100} = 85x*(75/100) = 63.75x
Profit of company B in 2015 = 100x*{(100+25)/100} = 100x*(125/100) = 125x
Required percentage = (63.75x/125x)*100 = 51% -
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections (1-5): The line graph given below shows the % decrease and % increase in the profit of Company A and Company B respectively. Read the data carefully and answer the following questions.
Note: Profit = (Income – Expenditure)
If the ratio between profit earned by company A and company B in the year 2015 and 2016 is 4:5 respectively then find the ratio of profit earned by company A in the year 2016 and company B in the year 2017.
Correct
Answer: 5) 4:9
Explanation:
Let the profit of company A in 2015 = 400x and profit of company B in 2016 = 500x
Profit of company A in 2016 = 400x*{(100-25)/100} = 400x*(75/100) = 300x
Profit of company B in 2017 = 500x*{(100+35)/100} = 675x
Required ratio = (300x): (675x) = 4:9Incorrect
Answer: 5) 4:9
Explanation:
Let the profit of company A in 2015 = 400x and profit of company B in 2016 = 500x
Profit of company A in 2016 = 400x*{(100-25)/100} = 400x*(75/100) = 300x
Profit of company B in 2017 = 500x*{(100+35)/100} = 675x
Required ratio = (300x): (675x) = 4:9 -
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections (1-5): The line graph given below shows the % decrease and % increase in the profit of Company A and Company B respectively. Read the data carefully and answer the following questions.
Note: Profit = (Income – Expenditure)
If the expenditure of company A and B in 2017 are equal and the ratio of expenditure and income of A and B in 2017 are 2:3 and 5:6 respectively. Find the ratio between the income of company A and company B in 2018 respectively if the expenditures of 2017 and 2018 are same for both the company.
Correct
Answer: 5) 285:248
Explanation:
Let the expenditure of company A and company B in 2017 = 100x
Income of company A in 2017 = 100x*(3/2) = 150x
Income of company B in 2017 = 100x*(6/5) = 120x
Profit of Company A in 2017 = 150x-100x = 50x
Profit of Company B in 2017 = 120x – 100x = 20x
Profit of company A in 2018 = 50x*{(100-15)/100} = 50x*(85/100) = 42.5x
Profit of company B in 2018 = 20x*{(100+20)/100} = 20x*(120/100) = 24x
Income = Profit + Expenditure
Income of company A in 2018 = 42.5x + 100x = 142.5x
Income of company B in 2018 = 24x + 100x = 124x
Required ratio = (142.5/124) = 285/248 = 285:248Incorrect
Answer: 5) 285:248
Explanation:
Let the expenditure of company A and company B in 2017 = 100x
Income of company A in 2017 = 100x*(3/2) = 150x
Income of company B in 2017 = 100x*(6/5) = 120x
Profit of Company A in 2017 = 150x-100x = 50x
Profit of Company B in 2017 = 120x – 100x = 20x
Profit of company A in 2018 = 50x*{(100-15)/100} = 50x*(85/100) = 42.5x
Profit of company B in 2018 = 20x*{(100+20)/100} = 20x*(120/100) = 24x
Income = Profit + Expenditure
Income of company A in 2018 = 42.5x + 100x = 142.5x
Income of company B in 2018 = 24x + 100x = 124x
Required ratio = (142.5/124) = 285/248 = 285:248 -
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections (1-5): The line graph given below shows the % decrease and % increase in the profit of Company A and Company B respectively. Read the data carefully and answer the following questions.
Note: Profit = (Income – Expenditure)
If the profit of company A in 2015 is twice the profit of company B in 2014 then the profit of company A in 2016 is approximately what percent of profit of company B in 2017.
Correct
Answer: 1) 77%
Explanation:
Let the profit of Company A in 2015 = 200x and profit of company B in 2014 = 100x
Profit of company A in 2016 = 200x*{(100-25)/100} = 200x*(75/100) = 150x
Profit of company B in 2017 = 100x*{(100+25)/100)}*{(100+15)/100}*{(100+35)/100]
= 100x*(125/100)*(115/100)*(135/100)
= 194 (approx)
Required percentage = (150x/194x)*100 = 77.4% (approx)Incorrect
Answer: 1) 77%
Explanation:
Let the profit of Company A in 2015 = 200x and profit of company B in 2014 = 100x
Profit of company A in 2016 = 200x*{(100-25)/100} = 200x*(75/100) = 150x
Profit of company B in 2017 = 100x*{(100+25)/100)}*{(100+15)/100}*{(100+35)/100]
= 100x*(125/100)*(115/100)*(135/100)
= 194 (approx)
Required percentage = (150x/194x)*100 = 77.4% (approx) -
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections (1-5): The line graph given below shows the % decrease and % increase in the profit of Company A and Company B respectively. Read the data carefully and answer the following questions.
Note: Profit = (Income – Expenditure)
The profit of company A in 2013 is twice the profit of company B in 2013. The profit of company B will exceed the profit of company A after
Correct
Answer: 2) two years
Explanation:
Let the profit of company A in 2013 = 200x and profit of company B in 2013= 100x
After one year in 2014
The profit of company A = 200x*{(100-20)/100} = 200x*(80/100)= 160x
The profit of company B = 100x*{100+10)/100} = 100x*(110/100) = 110x
In 2015
The profit of company A = 160x*(100-15)/100 = 160x*(85/100) = 136x
The profit of company B = 110x*(100+25)/100 = 110x*(125/100) = 137.5xIncorrect
Answer: 2) two years
Explanation:
Let the profit of company A in 2013 = 200x and profit of company B in 2013= 100x
After one year in 2014
The profit of company A = 200x*{(100-20)/100} = 200x*(80/100)= 160x
The profit of company B = 100x*{100+10)/100} = 100x*(110/100) = 110x
In 2015
The profit of company A = 160x*(100-15)/100 = 160x*(85/100) = 136x
The profit of company B = 110x*(100+25)/100 = 110x*(125/100) = 137.5x -
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeSome amount was lent at 12% per annum simple interest. After one year, Rs. 17,101 is withdrawn and the remaining of the amount is repaid at 8% per annum at the end of second year. If the ratio of the first year interest to that of the second year is 15:8 then, find the amount that was lent out initially.
Correct
Answer: 2) Rs. 85505
Explanation:
Let the amount lent initially = x
ATQ,
x*(12/100) / (x-17101)*(8/100) = 15/8
4x /(x-17101) = 5/1
4x = 5x – 5*17101
x = 5*17101
x = 85505Incorrect
Answer: 2) Rs. 85505
Explanation:
Let the amount lent initially = x
ATQ,
x*(12/100) / (x-17101)*(8/100) = 15/8
4x /(x-17101) = 5/1
4x = 5x – 5*17101
x = 5*17101
x = 85505 -
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeA shopkeeper allows two successive discounts of 10% and 15% respectively on marked price of an article and customer also managed to get additional discount of Rs. 140. Shopkeeper finds that selling price of the article is 62 ½ % of marked price. Find cost price for the shopkeeper if he incurs a profit of 5% on it?
Correct
Answer: 2) Rs. 600
Explanation:
Let the marked price of the article = 200x
Two successive discount = {10+15 – 10*15/100} = 23.5%
Discount of 23.5% on 200x: 200x*{(100-23.5)/100} = 200x*{76.5/100}*100 = 153x
Selling price after addition discount of 140 = 153x-140
ATQ,
153x-140 = 62 ½% of 200x
153x-140 = 62.5% of 200x
153x-140 = 125x
153x-125x = 140
28x = 140
x = 5
Therefore, selling price = 125x = 125*5 = 625
Cost price = 625*(100/105) = 600Incorrect
Answer: 2) Rs. 600
Explanation:
Let the marked price of the article = 200x
Two successive discount = {10+15 – 10*15/100} = 23.5%
Discount of 23.5% on 200x: 200x*{(100-23.5)/100} = 200x*{76.5/100}*100 = 153x
Selling price after addition discount of 140 = 153x-140
ATQ,
153x-140 = 62 ½% of 200x
153x-140 = 62.5% of 200x
153x-140 = 125x
153x-125x = 140
28x = 140
x = 5
Therefore, selling price = 125x = 125*5 = 625
Cost price = 625*(100/105) = 600 -
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeA man invested Rs. 45000 in a scheme that offers compound interest at different rates for different years. For the first year, second year and third year the rate of interest is 10%, R% and 15% respectively. He receives Rs. 50094 at the end of 3 years as profit. Find the value of R.
Correct
Answer: 1) -12%
Explanation:
45000*(110/100)*{(100+R)/100}*(115/100) = 50094
45000*(11/10)*{(100+R)/100}*(23/20) = 50094
45*(11)*{(100+R)}*(23/20) = 50094
(100+R) = 88
R = -12%Incorrect
Answer: 1) -12%
Explanation:
45000*(110/100)*{(100+R)/100}*(115/100) = 50094
45000*(11/10)*{(100+R)/100}*(23/20) = 50094
45*(11)*{(100+R)}*(23/20) = 50094
(100+R) = 88
R = -12% -
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeMixture A and B contains milk and wine in the ratio of 1:3 and 2:3 respectively. 40% of mixture A and 10% mixture B is mixed to form another mixture C. If quantity of milk in mixture C is 15 liters and ratio of milk to wine in mixture C is 1:2, then find initial quantity of mixture B.
Correct
Answer: 1) 25 litres
Explanation:
Let the quantity of mixture A = 100x and quantity of mixture B = 100y
Quantity of milk in mixture A = 100x*(1/4) = 25x
And quantity of wine in mixture A = 100x*(3/4) = 75x
Quantity of milk in mixture B = 100y*(2/5) = 40y
And quantity of wine in mixture B = 100y*(3/5) = 60y
After mixing to form mixture C
Quantity of milk in mixture C
25x+40y = 15 (liters)
5x+8y = 3 (equation 1)
And quantity of wine in mixture C = 75x+60y
ATQ,
(25x+40y)/(75x+60y) = ½
15/(75x+60y) = ½
75x+60y = 30
5x+4y = 2 (equation 2)
Solving equation 1 and equation 2, we get
y = 0.25
Initial quantity of mixture B = 100y = 100*0.25 = 25 litersIncorrect
Answer: 1) 25 litres
Explanation:
Let the quantity of mixture A = 100x and quantity of mixture B = 100y
Quantity of milk in mixture A = 100x*(1/4) = 25x
And quantity of wine in mixture A = 100x*(3/4) = 75x
Quantity of milk in mixture B = 100y*(2/5) = 40y
And quantity of wine in mixture B = 100y*(3/5) = 60y
After mixing to form mixture C
Quantity of milk in mixture C
25x+40y = 15 (liters)
5x+8y = 3 (equation 1)
And quantity of wine in mixture C = 75x+60y
ATQ,
(25x+40y)/(75x+60y) = ½
15/(75x+60y) = ½
75x+60y = 30
5x+4y = 2 (equation 2)
Solving equation 1 and equation 2, we get
y = 0.25
Initial quantity of mixture B = 100y = 100*0.25 = 25 liters -
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeA train’s speed is 10 km/hr more than a truck. Both start from point A at the same time and reach point B 10000 m away from A at the same time. On the way, however, the train lost about 30 minutes while stopping at the stations. The speed of the truck is:
Correct
Answer: 5) 10km/hr
Explanation: `
Let the speed of truck = x km/hr and speed of train = (x + 10) km/hr
Time for which truck travels = (t+0.5) hours and time for which train travel = t hours
Distance covered by train and truck is same
Distance = Speed*time
10000 m = 10 km
x*( t+0.5) =10 (equation 1) and (x + 10)*t = 10 (equation 2)
x*( t+0.5) = (x + 10)*t
xt + 0.5x = xt + 10t
0.5x = 10t (Put in equation 2)
x(0.5x/10) + 0.5x = 10
x(5x/10) + 5x = 100
x2/2 + 5x -100 = 0
x2+ 10x -200 = 0
x2+ 20x -10x – 200 = 0
x(x+20)-10(x+20) = 0
x = 10 km/hr (Truck’s speed), -20 (speed cannot be negative)Incorrect
Answer: 5) 10km/hr
Explanation: `
Let the speed of truck = x km/hr and speed of train = (x + 10) km/hr
Time for which truck travels = (t+0.5) hours and time for which train travel = t hours
Distance covered by train and truck is same
Distance = Speed*time
10000 m = 10 km
x*( t+0.5) =10 (equation 1) and (x + 10)*t = 10 (equation 2)
x*( t+0.5) = (x + 10)*t
xt + 0.5x = xt + 10t
0.5x = 10t (Put in equation 2)
x(0.5x/10) + 0.5x = 10
x(5x/10) + 5x = 100
x2/2 + 5x -100 = 0
x2+ 10x -200 = 0
x2+ 20x -10x – 200 = 0
x(x+20)-10(x+20) = 0
x = 10 km/hr (Truck’s speed), -20 (speed cannot be negative)
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