55656.A person who has a certain amount with him goes to market. He can buy 50 oranges or 40 mangoes. He retains 10% of the amount for taxi fares and buys 20 mangoes and of the balance, he purchases oranges. Number of oranges he can purchase is:
36
40
15
20
Explanation:
The person can buy 50 oranges or 40 mangoes.
Let the price of one orange be Rs. x
Total amount the person has = Rs. 50x
40 mangoes cost 50x, So one mango costs 1.25x
10% of the total amount is retained for taxi fare = 10% of 50x = 5x
20 mangoes bought for 20 x 1.25x = 25x
Money left with the person = 50x – (Taxi fare) – (Mangoes cost)
= 50x – 5x – 25x = 20x
One Orange was for Rs. x, Therefore, 20 oranges can be bought with Rs. 20 x
Thus, the person bought 20 oranges.
The person can buy 50 oranges or 40 mangoes.
Let the price of one orange be Rs. x
Total amount the person has = Rs. 50x
40 mangoes cost 50x, So one mango costs 1.25x
10% of the total amount is retained for taxi fare = 10% of 50x = 5x
20 mangoes bought for 20 x 1.25x = 25x
Money left with the person = 50x – (Taxi fare) – (Mangoes cost)
= 50x – 5x – 25x = 20x
One Orange was for Rs. x, Therefore, 20 oranges can be bought with Rs. 20 x
Thus, the person bought 20 oranges.
55657.A report consists of 20 sheets each of 55 lines and each such line consists of 65 characters. This report is reduced onto sheets each of 65 lines such that each line consists of 70 characters. The percentage reduction in number of sheets is closest to:
20%
5%
30%
35%
Explanation:
No. of Characters in one line = 65
No. of characters in one sheet = No. of lines × No. of characters per line = 55 × 65
Total number of characters = No. of sheets × No. of characters in one sheet = 20 × 55 × 65 = 71500
If the report is retyped –
New sheets have 65 lines, with 70 characters per line
No. of characters in one sheet = 65 × 70
Number of pages required = $\dfrac{Total no.of characters}{No.of characters in one sheet if retyped}$=$\dfrac{71500}{65 * 70}$=15.71
Hence, 16 pages will be required if report is retyped.
Hence, reduction of (20 – 16) = 4 pages
% reduction is = (4/20) x 100 = 20%
No. of Characters in one line = 65
No. of characters in one sheet = No. of lines × No. of characters per line = 55 × 65
Total number of characters = No. of sheets × No. of characters in one sheet = 20 × 55 × 65 = 71500
If the report is retyped –
New sheets have 65 lines, with 70 characters per line
No. of characters in one sheet = 65 × 70
Number of pages required = $\dfrac{Total no.of characters}{No.of characters in one sheet if retyped}$=$\dfrac{71500}{65 * 70}$=15.71
Hence, 16 pages will be required if report is retyped.
Hence, reduction of (20 – 16) = 4 pages
% reduction is = (4/20) x 100 = 20%
55658.Class B has 50% more students than class A. Number of girls in class A is equal to number of boys in class B. The percentage of girls is the same in both classes. What percentage of the student group are boys?
33.33%
40%
25%
60%
Explanation:
50% more than x is 1.5x. Simple, but very useful idea that might help you in solving these kinds of problems.
Let number of girls in class A = x
Let number of boys in class A = y
Total number of students = x + y
Proportion of girls = $\dfrac{x}{x+y}$
Number of boys in class B = x
Total number of students in class B = 1.5(x + y)
Proportion of girls = 1 - $\dfrac{x}{1.5(x+y)}$
Percentage of boys in the overall student community = $\dfrac{x+y}{2.5}* (x + y) * 100 = 40%$
The question is " What percentage of the student group are boys? "
40% of the student group are boys
Hence, the answer is 40%
50% more than x is 1.5x. Simple, but very useful idea that might help you in solving these kinds of problems.
Let number of girls in class A = x
Let number of boys in class A = y
Total number of students = x + y
Proportion of girls = $\dfrac{x}{x+y}$
Number of boys in class B = x
Total number of students in class B = 1.5(x + y)
Proportion of girls = 1 - $\dfrac{x}{1.5(x+y)}$
Percentage of boys in the overall student community = $\dfrac{x+y}{2.5}* (x + y) * 100 = 40%$
The question is " What percentage of the student group are boys? "
40% of the student group are boys
Hence, the answer is 40%
55659.Ramola’s monthly income is three times Ravina’s monthly income, Ravina’s monthly income is fifteen percent more that Ruchika’s monthly income. Ruchika’s monthly income is Rs. 32,000. What is Ramola’s annual income?
Rs. 1, 10, 400
Rs. 13, 24, 800
Rs. 36, 800
Rs. 52, 200
Explanation:
Ruchika’s monthly income = Rs 32000 Ravina’s monthly income = 32000 x (1 +15/100 ) = 32000 x115/100 = Rs. 36800
Ramola’s monthly income = 3 x 36800 = 110400
Ramola’s annual income = 12 x 110400 = 1324800
Ruchika’s monthly income = Rs 32000 Ravina’s monthly income = 32000 x (1 +15/100 ) = 32000 x115/100 = Rs. 36800
Ramola’s monthly income = 3 x 36800 = 110400
Ramola’s annual income = 12 x 110400 = 1324800
55660.If the numerator of a fraction is increased by 400% and the denominator is increased by 500% the resultant fraction is 10/21.What was the original fraction?
5/12
8/13
17/14
4/7
Explanation:
Let the original fraction be X/Y
After 400% increase, the numerator becomes 5X and after 500% increase, the denominator becomes 6Y.
Hence ,the resulting fraction becomes 5X/6Y=10/21
So,X/Y =4/7.
Let the original fraction be X/Y
After 400% increase, the numerator becomes 5X and after 500% increase, the denominator becomes 6Y.
Hence ,the resulting fraction becomes 5X/6Y=10/21
So,X/Y =4/7.
55661.Traders A and B buy two goods for Rs. 1000 and Rs. 2000 respectively. Trader A marks his goods up by x%, while trader B marks his goods up by 2x% and offers a discount of x%. If both make the same non-zero profit, find x.
25%
12.5%
37.5%
40%
Explanation:
This question is all about forming the right equations.
SP of trader A = 1000 (1 + x).
Profit of trader A = 1000 (1 + x) – 1000.
MP of trader B = 2000 (1 + 2x).
SP of trader B = 2000 (1 + 2x) (1 – x).
Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000
1000x = 2000 – 4000x2 + 4000x – 2000x – 2000
4000xx2 -1000x = 0
1000x (4x – 1) = 0
=> x = 25%
The question is "If both make the same non-zero profit, find x."
Hence, the answer is 25%.
Choice A is the correct answer.
This question is all about forming the right equations.
SP of trader A = 1000 (1 + x).
Profit of trader A = 1000 (1 + x) – 1000.
MP of trader B = 2000 (1 + 2x).
SP of trader B = 2000 (1 + 2x) (1 – x).
Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000
1000x = 2000 – 4000x2 + 4000x – 2000x – 2000
4000xx2 -1000x = 0
1000x (4x – 1) = 0
=> x = 25%
The question is "If both make the same non-zero profit, find x."
Hence, the answer is 25%.
Choice A is the correct answer.
55662.In an election between two candidates one got 52% of the total valid votes,25% of the total votes were invalid.The total number of votes was 7200,the number of valid votes that the other candidate got was.
2592
2808
2604
2716
Explanation:
Given that
Total votes=7200
Invalid votes=25% of total votes ,it means that the total valid votes are 75% of total votes.
One candidate got 52% of the total valid votes it means that other candidate got 48% of the total valid votes.
Total valid votes=$7200 \times (75/100)$ 5400 votes
Other candidate got 48% of the total valid votes that is =$5400 \times (48/100)$=2592
Given that
Total votes=7200
Invalid votes=25% of total votes ,it means that the total valid votes are 75% of total votes.
One candidate got 52% of the total valid votes it means that other candidate got 48% of the total valid votes.
Total valid votes=$7200 \times (75/100)$ 5400 votes
Other candidate got 48% of the total valid votes that is =$5400 \times (48/100)$=2592
55663.The population of a city increase at the rate of 4% per annum. There is an additional annual increase of 1% in the population due to the influx of job seekers. Find percentage increase in the population after 2 years.
10.25%
11.25%
12.25%
13.25%
Explanation:
The net annual increase = 5%
Let the initial population be 100.
Then, population after 2 years = 100×1.05×1.05 = 110.25
Therefore, % increase in population = (110.25-100) = 10.25%
The net annual increase = 5%
Let the initial population be 100.
Then, population after 2 years = 100×1.05×1.05 = 110.25
Therefore, % increase in population = (110.25-100) = 10.25%
55664.Mr Shamin’s salary increase every year by 10% in june. If there is no other increase or reduction in the salary and his salary in June 2011 was rs 22,385, what was his salary in June 2009 ?
rs 18,650
rs 18,000
rs 19,250
rs 18,500
Explanation:
Here, the salary is compounded every year.
Let salary be x.
22385=x(1+.1)^2 x=18,500
Here, the salary is compounded every year.
Let salary be x.
22385=x(1+.1)^2 x=18,500
55665.Mr. Keisham gave 40% of the money he had, to his wife. He also gave 20% of the remaining amount to each of his three sons, Half of the amount now left was spent on miscellaneous item and the remaining amount of Rs. 12,000 was deposited in the bank. How much money did Mr. Keisham have initially ?
1,000
10,000
1,00,000
10,00,000
Explanation:
Let the initial amount with Mr. Keisham be Rs. x.
Then, 1/2 [100 – (3 × 20)]% of (100 – 40)% of x = 12000.
⇒ 1/2× 40/100 × 60 /100 × x = 12000 ⇔ 3/25 x = 12000.
⇒ x=(12000 × 25)/3 = 100000.
Let the initial amount with Mr. Keisham be Rs. x.
Then, 1/2 [100 – (3 × 20)]% of (100 – 40)% of x = 12000.
⇒ 1/2× 40/100 × 60 /100 × x = 12000 ⇔ 3/25 x = 12000.
⇒ x=(12000 × 25)/3 = 100000.
- Ratio and Proportion Test 4
- Percentage Test 2
- Percentage Test 3
- Percentage Test 4
- Percentage Test 5
- Percentage Test 6
- Ratio and Proportion Test 1
- Ratio and Proportion Test 2
- Ratio and Proportion Test 3
- Percentage Test 1
- Average Test 1
- Average Test 2
- Average Test 3
- Interest Test 1
- Interest Test 2
- Interest Test 3
- Interest Test 4
- Interest Test 5
- Time and Distance Test 3
- Number system Test 2
- Partnership Test 1
- Partnership Test 2
- Profit,Loss and Discount Test 1
- Profit,Loss and Discount Test 2
- Profit,Loss and Discount Test 3
- Time and Distance Test 1
- Time and Distance Test 2
- Number System Test 1
- Time and Work Test 1
- Time and Work Test 2
- Time and Work Test 3
- Time and Work Test 4
- Time and Work Test 5
- Time and Work Test 6
- Time and Work Test 7