Percentage Questions Answers

• 29. Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first

  1. 8%
  2. 9%
  3. 10%
  4. 11%

  Answer And Explanation

  Answer: Option C

  Explanation:

  Let the third number is x.
  then first number = (100-30)% of x
  = 70% of x = 7x/10
  
  Second number is (63x/100)
  
  Difference = 7x/10 - 63x/100 = 7x/10
  
  So required percentage is, difference is what percent of first number
  
  => (7x/100 * 10/7x * 100 )% = 10%

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• 30. Due to an increase in 30% in the price of eggs, 3 eggs less are available for Rs. 7.80. Find the present rate of eggs per dozen.

  1. Rs. 9.36
  2. Rs. 10.36
  3. Rs. 11.36
  4. Rs. 12.36

  Answer And Explanation

  Answer: Option A

  Explanation:

  Let the original price per egg be Rs x
  Then increased price will be,
  

  $$\begin{aligned} \left(\frac{130}{100}x\right) \\ => \frac{7.80}{x}-\frac{7.80}{\frac{130}{100}x} = 3\\ => \frac{7.80}{x}-\frac{780}{130x} = 3 \\ => 390x = 234 \\ => x = 0.6 \\ \text{Actual price was Rs 0.6} \\ \text{Present price per dozen will be} \\ Rs.\left(12*\frac{130}{100}*0.6 \right) \\ = Rs. 9.36 \end{aligned}$$

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• 31. In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.

  1. 40%
  2. 42%
  3. 44%
  4. 46%

  Answer And Explanation

  Answer: Option C

  Explanation:

  Failed in mathematics, n(A) = 34
  Failed in English, n(B) = 42
  

  $$\begin{aligned} n(A\cup B) = n(A)+n(B)-n(A\cap B) \\ = 34+42-20 = 56 \\ \text{Failed in either or both subjects are 56} \\ \text{Percentage passed = }(100-56)\% \\ = 44\% \end{aligned}$$
  

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• 32. Out of 450 students of a school, 325 play football, 175 play cricket and 50 neither play football nor cricket. How many students play both football and cricket ?

  1. 75
  2. 100
  3. 125
  4. 150

  Answer And Explanation

  Answer: Option B

  Explanation:

  Students who play cricket, n(A) = 325
  Students who play football, n(B) = 175
  Total students who play either or both games,
  

  $$\begin{aligned} = n(A\cup B) = 450-50 = 400\\ \text{Required Number}, n(A \cap B) \\ = n(A)+n(B)-n(A\cup B) \\ = 325 + 175 - 400 = 100 \end{aligned}$$

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• 33. In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ?

  1. 27
  2. 26
  3. 25
  4. 24

  Answer And Explanation

  Answer: Option D

  Explanation:

  $$\begin{aligned} n(A) = \left(\frac{60}{100}*96\right) = \frac{288}{5} \\ n(B) = \left(\frac{30}{100}*96\right) = \frac{144}{5} \\ n(A\cap B) = \left(\frac{15}{100}*96\right) = \frac{72}{5} \\ \text{People who have either or both lunch} \\ n(A\cup B) = \frac{288}{5}+\frac{144}{5}-\frac{72}{5} \\ = \frac{360}{5} = 72 \end{aligned}$$
  
  So People who do no have either lunch were = 96 -72
  = 24

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• 34. Teacher took exam for English, average for the entire class was 80 marks. If we say that 10% of the students scored 95 marks and 20% scored 90 marks then calcualte average marks of the remaining students of the class.
  

  1. 60
  2. 70
  3. 75
  4. 80

  Answer And Explanation

  Answer: Option C

  Explanation:

  Lets assume that total number of students in class is 100 and required average be x.
  Then from the given statement we can calculate :
  (10 * 95) + (20 * 90) + (70 * x) = (100 * 80)
  
  => 70x = 8000 - (950 + 1800) = 5250
  
  => x = 75.

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