Percentages

1) Find the value 7\({1 \over {12}}\)% of 2400.

Show Solution
7 \({1 \over {12}}\) % of 2400 =  \({85 \over {12}}\) % of 2400

= \({85 \over {12*100}}\) * 2400 = 170.     Hence (a).

2) Which of the following is the largest number?

Show Solution
20% of 850 = 170

25% of 700 = 175

18% of 900 = 162

20% of 750 = 150.  Hence (b).

3) In a Bus, there are 60 passengers. out of which 16.66% are children. Number of male passengers in the Bus are 50% more than the female passengers. Find the number of male passengers in the Bus.

Show Solution
Total number of passengers in Bus = 60

16.66% are children   =  \({1 \over {6}}\) * 60 = 10

  Number of male passengers   in the Bus are 50% more than the female passengers
  • M = 1.5F = 50

  • \({M \over {F}}\) =  \({3 \over {2}}\)   = 50

  • 3x + 2x = 50  à x =10


  • Number of male passengers =  3x  à 30 .  Hence (d).

    4) Sam’s monthly income is Rs.70, 000; he spends 35% for the living expenses and 18% for income tax. 10% for Car loan. How much does he save per month?

    Show Solution
    Sam’ s monthly income = Rs.70, 000

    The monthly expenses are 35%, 18% and 10%.
  • (35 + 18 + 10)% of 70000 = 63% of 70000 


  • =  \({63 \over {100}}\) * 70000 

    = Rs.44 , 100
      Savings = Income – Expenditure
    = 70000 – 44100 = Rs.25 , 900 .   Hence (c).

    5)

    A number is reduced by 15% of its present value is 612. What was \({2 \over {3}}\)rd of its original value?

    Show Solution

    Let the original value be x. And it’s reduced by 15% 85% of x= 612 à x = 720 \({2 \over {3}}\)rd of 720 = 480 Hence (b).

    6) The population of a city at the end of the year 2012 was 1, 90,000. It decreased by 10% in the year 2013 and increased by 16% in the year 2014. What is the population of the city in the beginning of the year 2015?

    Show Solution
    Population of city = 190000 * \({90 \over {100}}\) *  \({116 \over {100}}\)  

    = 198360 .  Hence (c).

    7)

    What percentages of numbers between 1 and 95 both exclusive have 3 or 7 in the unit digit?(approximately)

    Show Solution

    Numbers between 1 and 95 both exclusive have 3 or 7 in the unit digit are 13, 17, 23, 27, …93 à 17 such numbers Percentage = \({17 \over {93}}\) * 100 = 18%. Hence (b).

    8) A scored 26% marks in Machines and failed by 23 marks. B scored 38% marks and scored 49 marks more than the pass marks. Find the maximum mark in the subject.

    Show Solution
      Let the maximum marks in Machines be n.
    Pass mark = 26% of n + 23 = 38% of n – 49
    12% of n = 72
      n = 600.   Hence (c).

    9) A number is mistakenly divided by 4 instead of being multiplied by 4. Find the percentage change in the result due to this mistake.

    Show Solution
      Let us assume the number is 100

    Initial value = 100 * 4 = 400

    Final value = 100/4 = 25

    % change = \({400-25 \over {400}}\) * 100 = 93.75% .  Hence (d).

    10) The price of Refined oil has increased by 14.28%. By what percentage must a family reduce its consumption so as the family able to balance their budget?

    Show Solution
    Let the initial price of Refined oil be Rs.100

    Price increased by 14.28% = Rs.114.28

    After increase the quantity of Refined oil bought for Rs.100 = \({1 \over {114.28}}\) * 100 =  \({7 \over {8}}\)

    %  Reduction in consumption = 1 -  \({7 \over {8}}\) * 100 = 12.5%.  Hence (a).

    11) In a school, Boys and Girls ratio was 8: 9 in the academic year 2016. After a year Boys and Girls strength of the school increased by 10% and decreased by 18% respectively. The total students in the year 2017 are 6472. Find the Boys in the school for the year 2017.

    Show Solution
    Ratio of  Boys and Girls  in 2016 = 8: 9

    Let the strength of Boys and Girls in 2016 be 8x and 9x.

    Boys strength in 2017 = 8x * \({110 \over {100}}\)  

    Girls strength in 2017 = 9 x *  \({82 \over {100}}\)

      Total  strength in 2017  =  8x *  \({110 \over {100}}\)   +  9x *  \({82 \over {100}}\) = 6472

    x = 400

    Boys strength in 2017 = 8* 400 *  \({110 \over {100}}\) = 3520.   Hence (d).

    12)

    A vendor sells 60% of the Mangoes and discards 25% of the remaining Mangoes. Next day, he sells 16.66% of the remaining, and discards 40%. On the third day, he sells only 5 Mangoes and 3 discarded. On the fourth day he discards the rest. Find the percentage of the discarded Mangoes.

    Show Solution

    Let the number of Mangoe s initially with the vendor be 100. Mango es sold on the fi rst day = 60% = 60 Mango es Mango es discarded = \({25 \over {100}}\) * 40 = 10 Mango es. ∴ Mango es left at the end of the day = 100 – (60 + 10) = 30 On the second day, Mangoes sold = 30 * \({1 \over {6}}\) = 5 and Mangoes discarded = \({40 \over {100}}\) * 25 = 10 ∴ Mango es l eft at the end of second day = 30 – (5 + 10) = 15 On the third day, Mangoes sold = 5 and Mangoes discarded = 3 ∴ Mango es l eft at the end of third day = 15 – (5 + 3) = 7 On the fourth day, he discards the rest ∴ Total number of discarded Mangoes = 10 + 10 + 3 + 7 = 30 Mango es Percentage of the Mango es discarded = \({30 \over {100}}\) *100 = 30 %. Hence (c).  

    13) In an election between two Politicians, one has got 36% of the total valid votes. 25% of the votes were invalid. If the total number of votes was 14000, find the difference between votes of two politicians

    Show Solution
    Total number of votes = 14000

    Invalid votes = 25% of 14000 = 3500

    Then, valid votes = 14000 – 3500 = 10500

    Votes received by first Politician =  36% of the total valid votes

    Votes received by second Politician =  64 % of the total valid votes

    D ifference between votes of two politicians = (64-36)%  of the total valid votes

    = 28%  of the total valid votes

    = 28% of 10500

    = 2940  Hence (a).

    14) The difference between a single discount of 30% and two successive discounts of 20% and 15% on a Shirt bill was Rs.65. Find the bill amount of a Shirt.

    Show Solution
    Let the bill amount of a Shirt be Rs.x

    SP of the Shirt with single discount 30%= x * \({70 \over {100}}\)  

    SP of the Shirt with succe s sive discount 30%= x *  \(({80 \over {100}}\) ) *  \(({85 \over {100}}\) )

    x *  \({70 \over {100}}\) -  x *  \(({80 \over {100}}\) ) *  \(({85 \over {100}}\) ) = 65
  • x = Rs.3250 .  Hence (b).

  • 15) A Project report consists of 25 sheets each of 40 lines and each such line consists of 70 characters. This report is reduced onto sheets each of 75 lines such that each line consists of 80 characters. The percentage reduction in number of sheets is closest to:

    Show Solution
    N umber of c haracters in one line = 70

    Number  o f characters in one sheet = Number  of lines * Number of  characters per line = 40 * 70 = 2800

    T otal number of characters = N umber of sheets * Number  of characters in one sheet = 25 * 2800 = 70000
     If the P roject report is retyped,  New sheets have 75 lines, with 8 0 characters per line
      Number  of characters in one sheet = 75 * 8 0
      Number of pages required be x.
    x * 75 * 80 = 70000  à x = 12 pages
    Percentage reduction = \({25-12 \over {25}}\)   = 52% .  Hence (d).