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01. Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 3 B. 1 : 1
C. 3 : 4 D. 4 : 3

Answer and Explanation

Answer: 4 : 3

Explanation:
5% of A + 4% of B = 2/3  (6% of A + 8% of B)

5/100  A + 4/100  B = 2/3   (6/100  A + 8/100 B)  

1/20  A + 1/25  B = 1/25  A + 4/75  B
   
1/20 - 1/25   A = 4/75 - 1/25   B
  

1/100  A = 1/100  B

A/B = 100/75 = 4/3 .

Required ratio = 4 : 3

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02. 1 litre of water is evaporated from 6 litres of sugar solution containing 4% of sugar. Find the percentage of sugar remaining in the solution
A. 4% B. 4.8% 
C. 5.4%  D. 6.5%

Answer and Explanation

Answer: 4.8% 

Explanation:
Let there be x litres of sugar in solution.
x / 6 × 100 = 4
x = 24/100 = 0.24 lit
Percentage of sugar in remaining solution = 0.24 / 5 × 100 = 4.8%

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03. In an examination the maximum marks is 500. A scores 10% less than B, B scores 25% more than C and C scores 20% less than D. if a scored 260, what percentage of the maximum marks did D get?
A. 64% B. 72%
C. 77% D. 80%

Answer and Explanation

Answer: 80%

Explanation:
Let D get x%. Then C gets 0.8x%, B gets 1.25(0.8)x% and A gets (0.9 × 1.25 × 0.8 × x%) i.e. 0.9x%
360/500 × 100 = 0.9x
x = 80%

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04. The population of a town decreases at the rate of 5% every year. If the population of the town 2 years from now would be 902500, then what is the population of the town at present?
A. 1000000 B. 995000
C. 814500 D. 947625

Answer and Explanation

Answer: 1000000

Explanation:
Explanation:
Population after 2 years = Present Population ( 1 – r1/100) (1 – r2/100)
P(1 – 5/100) (1 – 5/100) = 902500
P = 1000000

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05. A candidate secured 20% marks in an examination and failed by 10 marks. Another secured 42% marks and got 1 mark more than the marks required to pass the examination. Determine the maximum possible marks in the test?
A. 20 B. 50
C. 21 D. 45

Answer and Explanation

Answer: 50

Explanation:
Let the maximum marks be x. 20 / 100 x + 10 = 42 / 100 x -1
11 = 22 / 100 x
x = 50 marks.

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06. If price of sugar falls by 10%, how many quintals can be bought for the same money which was sufficient to buy 18 quintals at the higher price?
A. 22 B. 21 
C. 19 D. 20

Answer and Explanation

Answer: 20

Explanation:
Percentage increase in consumption = (x / 100 – x) ×100
= 10 / 90 × 100 = 11 1/9%
Quantity of sugar that can be bought for new price =
18 + [((100 / 9) / 100) × 18] = 18 + 2 = 20 quintals

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07. The length of a rectangle is decreased by 5 % and its breadth is increased by 5%. What is the percentage change in the area of the rectangle?
A. 20%  B. 25%
C. 2.5% D. 0.25%

Answer and Explanation

Answer: 0.25%

Explanation:
Let the sides of the rectangle be l1 and b1.
Initial area of the rectangle , A1 = l1b1
Find the area of the rectangle, A2 = (0.95 l1)(1.05)b1 = 0.9975 l1b1.
Change in the area = l1b1 – 0.9975 l1b1 = 0.0025 l1b1
Percentage change = (0.0025 l1b1) / l1b1 × 100 = 0.25%

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08. A’s salary is 20% below B’s salary. By how much percent is B’s salary above A?
A. 16 1/3%  B. 25% 
C. 20% D. 16 2/3%

Answer and Explanation

Answer: 25% 

Explanation:
 Let B’s salary is Rs. 100. A’s salary is 20% below than B’s salary. Hence, salary of A = Rs. 80. B’s salary is (20 / 80 × 100) percentage above A’s salary, i.e 25% above A’s salary.

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09. If 60% of the students in a school are boys and the number of girls is 812, how many students are there in the school?
A. 1218  B. 1355 
C. 2842 D. 2030

Answer and Explanation

Answer: 2030

Explanation:
Let the total number of students be x. If 60% are boys then 40% are girls. 40/100 x =812
x = (812 × 100) / 40 = 2030.

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10. Find the single discount which is equivalent to successive discounts of 20%, 15% and 10%.
A. 38.8%  B. 61.2% 
C. 30% D. 51.8%

Answer and Explanation

Answer: 38.8% 

Explanation:
Let marked Price = Rs. 100
1st discount = 100 × 20 /100 = Rs. 20
2nd discount = (100 – 20) × 15 / 100 = 80 × 15 /100 = Rs. 12
3rd discount = (80 – 12) × 10/100 = 68/10 = Rs. 6.8
Single discount = Sum of all the discounts = 20 + 12 + 6.8 = 38.8%

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