- Staff Selection Commission Mathematics - Percentage (1999-2017) Part 3

Staff Selection Commission Mathematics - Percentage (1999-2017)


TYPE–IV (Contd.)





Ans .

48,900


  1. Explanation :

    Women = \(\frac{43}{83} \times 311250\) = 161250

    Men = 311250 – 161250 = 150000

    ∴ Total number of literate persons

    = \(\frac{161250 \times 8}{100}\) + 150000\(\times \frac{24}{100}\)

    = 12900 + 36000 = 48900







Ans .

500


  1. Explanation :

    7x – 5x = 200

    \(→\) 2x = 200 \(→\) x = 100

    ∴ Price of a pair of shoes

    = 5x = 5 × 100 = 500







Ans .

4 : 3


  1. Explanation :

    \(x \times \frac{15}{100}\) = \(y \times \frac{20}{100}\)

    \(→\) x × 15 = y × 20

    \(→\) \(\frac{x}{y}\) = \(\frac{20}{15}\) = \(\frac{4}{3}\) = 4 : 3







Ans .

72


  1. Explanation :

    Boys in school = 2x

    Girls = 3x

    Students who are not scholarship holders :

    Boys \(→\) \(\frac{2x \times 75}{100}\) = \(\frac{6x}{4}\)

    Girls \(→\) \(\frac{3x \times 70}{100}\) = \(\frac{21x}{10}\)

    Total students who do not hold scholarship = \(\frac{6x}{4}\) + \(\frac{21x}{10}\)

    = \(\frac{72x}{20}\) = \(\frac{18x}{5}\)

    ∴ Required percentage = \(\frac{\frac{18x}{5}}{5x} \times 100\) = 72%







Ans .

4 : 3


  1. Explanation :

    Numbers \(→\) A and B

    ∴ \(\frac{A \times 5}{100}\) + \(\frac{B \times 4}{100}\)

    = \(\frac{2}{3}\) \((\frac{A \times 6}{100} + \frac{B \times 8}{100})\)

    \(→\) 5A + 4B = \(\frac{12A + 16B}{3}\)

    \(→\) 15A + 12B = 12A + 16B

    \(→\) 15A – 12A = 16B – 12B

    \(→\) 3A = 4B

    \(→\) \(\frac{A}{B}\) = \(\frac{4}{3}\)







Ans .

3 : 4


  1. Explanation :

    Acid – I

    60

    60 – 40 = 20



    Acid – II

    25

    40 – 25 = 15



    ∴ Required ratio = 15 : 20 = 3 : 4







Ans .

3 : 5


  1. Explanation :

    50% of x = 30% of y

    \(→\)\(\frac{x \times 50}{100}\) = \(\frac{y \times 30}{100}\)

    \(→\)\(\frac{x}{y}\) = \(\frac{30}{50}\) = \(\frac{3}{5}\)







Ans .

23 : 27


  1. Explanation :

    Boys in the village = 3x

    Girls in the village = 2x

    Villagers who appeared in the examination

    = \(\frac{3x \times 30}{100}\) + \(\frac{2x \times 70}{100}\)

    = \(\frac{9x}{10}\) + \(\frac{14x}{10}\)

    \(\frac{23x}{10}\)

    Villagers who did not appear in the examination

    = \(\frac{3x \times 70}{100}\) + \(\frac{2x \times 30}{100}\)

    = \(\frac{21x}{10}\) + \(\frac{6x}{10}\)

    \(\frac{27x}{10}\)

    Required ratio = \(\frac{23x}{10}\) : \(\frac{27x}{10}\)

    = 23 : 27







Ans .

1 : 4


  1. Explanation :

    C.P. of 1 litre of milk= Rs. 100

    ∴ Mixture sold for Rs. 125 = \(\frac{125}{100}\) = \(\frac{5}{4}\) litre

    ∴ Quantity of water = \(\frac{5}{4}\) - 1 = \(\frac{1}{4}\) litre

    ∴ Required ratio = \(\frac{1}{4}\) : 1 = 1 : 4







Ans .

75%


  1. Explanation :

    Percentage of syrup = \(\frac{3}{4} \times 100\) = 75%







Ans .

12


  1. Explanation :

    Let the numbers be 5x and 4x respectively

    According to the question ,

    \(5x \times \frac{40}{100}\) = 12

    \(→\) 2x = 12 \(→\) x = 6

    ∴ 4x 50% = 4 × 6 × \(\frac{1}{2}\) = 12







Ans .

9 : 2


  1. Explanation :

    According to the question,

    \(x \times \frac{10}{100}\) = \(3 \times y \times \frac{15}{100}\)

    \(→\) 10x = 45y

    \(→\) \(\frac{x}{y}\) = \(\frac{45}{10}\) = \(\frac{9}{2}\)







Ans .

110%


  1. Explanation :

    Required per cent = \(\frac{11}{10} \times 100\) = 110%







Ans .

78


  1. Explanation :

    Let the number of students

    in school be 100.

    Boys \(→\) 60

    Girls \(→\) 40

    Students who do not hold scholarship :

    Boys \(→\) \(\frac{60 \times 80}{100}\) = 48

    Girls \(→\) \(\frac{40 \times 75}{100}\) = 30

    Required answer = 48 + 30

    = 78 i.e., 78%







Ans .

5 : 7


  1. Explanation :

    According to the question,

    A × 35% = B × 25%

    \(→\) \(\frac{A}{B}\) =\(\frac{25}{35}\) =\(\frac{5}{7}\)




TYPE – V





Ans .

150 litres


  1. Explanation :

    Glycerine in mixture = 40 litres

    Water = 10 litres

    Let x litres of pure glycerine is mixed with the mixture.

    ∴ \(\frac{40 + x}{50+x}\) = \(\frac{95}{100}\) = \(\frac{19}{20}\)

    \(→\) 800 + 20x = 950 + 19x

    \(→\) x = 950 – 800 = 150 litres







Ans .

\( 33 \frac{1}{3}\)%


  1. Explanation :

    Alcohol in original solution = \(\frac{40}{100} \times 5\) = 2 litres

    Water in original solution = 3 litres

    On adding 1 litre water, water becomes 4 litres.

    Now, 6 litres of solution contains 2 litres of alcohol.

    ∴ 100 litres of solution contains = \(\frac{2}{6} \times 100\)

    = \(\frac{100}{3}\) = \( 33 \frac{1}{3}\)% alcohol







Ans .

10.5%


  1. Explanation :

    In 12 litres salt solution,

    Salt = \(\frac{7 \times 12}{100}\) = 0.84 units

    Water = \(\frac{93 \times 12}{100}\) = 1116. units

    After evaporation,

    Percentage of salt = \(\frac{0.84}{8} \times 100\) = 10.5%







Ans .

20 litres


  1. Explanation :

    In 60 litres of solution, Water

    = \(\frac{60 \times 20}{100}\) = 12 litres

    On adding x litres of water, \(\frac{12 + x}{60+x} \times 100\) = 40

    \(→\) 60 + 5x = 120 + 2x

    \(→\) 3x = 60

    \(→\) x = 20 litres







Ans .

100 gm


  1. Explanation :

    Sugar in original solution

    = \(\frac{75 \times 30}{100}\) = 22.5 gm

    Let x gm of sugar be mixed.

    ∴\(\frac{22.5 + x}{75 + x} \times 100\) = 70

    \(→\) 2250 + 100x = 75 × 70 + 70x

    \(→\) 2250 + 100x = 5250 + 70x

    \(→\) 30x = 5250 – 2250 = 3000

    \(→\) x = \(\frac{3000}{30}\) = 100 gm







Ans .

60%


  1. Explanation :

    In 30% alcohol solution,

    Alcohol = \(\frac{30}{100} \times 6\) = 1.8 litres

    Water = 4.2 litres

    On mixing 1 litre of pure alcohol,

    Percentage of water = \(\frac{4.2}{7} \times 100\) = 60%







Ans .

266.67


  1. Explanation :

    In 4 kg of ore, iron = 0.9 kg.

    ∴ Quantity of ore for 60 kg of iron = \(\frac{60 \times 4}{0.9}\) = 266.67 kg







Ans .

60 ml


  1. Explanation :

    Let x ml of water be added.

    ∴ \(\frac{20 + x}{100 + x} \times 100\) = 50

    \(→\) 40 + 2x = 100 + x

    \(→\) x = 60 ml







Ans .

1000 ml


  1. Explanation :

    In 1 litre i.e. 1000 ml of mixture,

    Alcohol = 700 ml.

    Water = 300 ml.

    Let x ml of alcohol is mixed.

    \(\frac{300}{1000 + x} \times 100\) = 15

    \(→\) 1000 + x = 2000

    \(→\) x = 1000 ml.







Ans .

\( 24 \frac{2}{7}\)%


  1. Explanation :

    In 10 litres of first type of liquid,

    Water = \(\frac{1}{5} \times 10\) = 2 litres

    In 4 litres of second type of liquid,

    Water = \(4 \times \frac{35}{100}\) = \(\frac{7}{5}\) litres

    Total amount of water = \(2 + \frac{7}{5}\) = \(\frac{17}{5}\) litres

    Required percentage = \(\frac{\frac{17}{5}}{14} \times 100\) = \(\frac{170}{7}\) = \( 24 \frac{2}{7}\)%







Ans .

5 litres


  1. Explanation :

    Water content in 40 litres of mixture

    = \(40 \times \frac{10}{100}\) = 4 litres

    ∴ Milk content = 40 – 4 = 36 litres

    Let x litres of water is mixed.

    Then,\(\frac{4 +x }{40+x}\) = \(\frac{20}{100}\) = \(\frac{1}{5}\)

    \(→\) 20 + 5x = 40 + x

    \(→\) 4x = 20 \(→\) x = 5 litres







Ans .

100 ml


  1. Explanation :

    Alcohol = \(\frac{15}{100} \times 400\)ml = 60 ml.

    Water = 340 ml.

    Let x ml of alcohol be added.

    Then,\(\frac{60 + x}{400 + x} \times 100\) = 32

    \(\frac{60 + x}{400 + x}\) = \(\frac{32}{100}\)= \(\frac{8}{25}\)

    or 1500 + 25x = 3200 + 8x

    or 17x = 1700

    or x = 100 ml







Ans .

150 gm


  1. Explanation :

    Initial quantity of gold = \(\frac{50 \times 80}{100}\) = 40 gm

    Let 'x ' gm be mixed.

    (40+x) = \((50 + x) \times \frac{95}{100}\)

    (40+x) = \((50 + x) \times \frac{19}{20}\)

    \(→\) 800 + 20x = 950 + 19x

    \(→\) x = 150 gm







Ans .

40 litres


  1. Explanation :

    In 200 litres of mixture,

    Quantity of milk = \(\frac{85}{100} \times 200\) = 170 litres

    Quantity of water = 30 litres

    Let the quantity of additional milk added be x litres.

    According to the question

    \(\frac{170+x}{200+x} \times 100\) = 87.5

    \(→\) (170 +x) × 100 = 17500 + 87.5x

    \(→\) 100x – 87.5x = 17500 – 17000

    \(→\) 12.5x = 500

    \(→\) x = \(\frac{500}{12.5}\) = 40 litres







Ans .

1 : 3


  1. Explanation :

    Let x litres of first mixture is mixed with y litres of the second mixture.

    According to the question,

    \(\frac{x \times \frac{30}{100} + y \times \frac{50}{100}}{x \times \frac{70}{100} + y \times \frac{50}{100}}\) = \(\frac{45}{55}\)

    \(\frac{0.3x + 0.5y}{0.7x + 0.5y}\) = \(\frac{9}{11}\)

    \(→\) 6.3x + 4.5y = 3.3x + 5.5y

    \(→\) 6.3x – 3.3x = 5.5y – 4.5y

    \(→\) 3x = y

    \(→\) \(\frac{x}{y}\) = 1:3







Ans .

2 : 3


  1. Explanation :

    ∴ Required ratio = 10 : 15 = 2 : 3







Ans .

\( 16 \frac{2}{3}\)%


  1. Explanation :

    Alcohol = \(15 \times \frac{1}{5}\) = 3 litres

    Water = \(15 \times \frac{4}{5}\) = 12 litres

    ∴ Required percentage

    = \(\frac{3}{15 + 3} \times 100\)

    = \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%







Ans .

575 kg


  1. Explanation :

    ∵ 12 kg copper is contained

    in 100 kg of alloy

    69 kg copper is contained in

    \(\frac{100}{12} \times 69\)= 575 kg of alloy







Ans .

2 : 3


  1. Explanation :

    ∴ Required ratio = \(\frac{1}{10}\) : \(\frac{3}{20}\) = 2 : 3







Ans .

\( 16 \frac{2}{3}\)%




  1. Explanation :

    In 20 litres of mixture,

    Alcohol \(→\) \(\frac{20 \times 20}{100}\) = 4 litres

    Water \(→\) 20 - 4 = 16 litres

    On adding 4 litres of water,

    Quantity of water \(→\) 16 + 4 = 20 litres

    Quantity of mixture = 24 litres

    ∴ Required per cent = \(\frac{4}{24} \times 100\) = \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%







Ans .

60 gram


  1. Explanation :

    In 300 gm of solution,

    Sugar = \(\frac{300 \times 40}{100}\) = 120 gm.

    Let x gm of sugar be mixed.

    According to the question,

    \(\frac{120 + x}{300 + x}\) = \(\frac{1}{2}\)

    \(→\) 240 + 2x = 300 + x

    \(→\) 2x – x = 300 – 240

    \(→\) x = 60 gm.







Ans .

45


  1. Explanation :

    Quantity of sugar in the solution

    = \(\frac{3 \times 60}{100}\) = 1.8 units

    On adding 1 litre of water,

    ∴ Required percent = \(\frac{1.8}{4} \times 100\) = 45%







Ans .

16%


  1. Explanation :

    In 32 litres of solution,

    Alcohol = \(\frac{32 \times 20}{100}\) = 6.4 litres

    Water = 32 – 6.4 = 25.6 litres

    On adding 8 litres of water,

    Required percent = \(\frac{6.4}{40} \times 100\) = 16%




TYPE – VI





Ans .

\( 16 \frac{2}{3}\)%


  1. Explanation :

    Required percentage decrease

    = \(\frac{Increase}{100 + Increase} \times 100\)

    = \(\frac{20}{100 + 20} \times 100\)

    = \(\frac{100}{6} \) = \( 16 \frac{2}{3}\)%







Ans .

\( 9 \frac{1}{11}\)%


  1. Explanation :

    Required answer

    = \(\frac{10}{100 + 10} \times 100\)

    = \(\frac{10}{110} \times 100\)

    \(\frac{100}{11}\)% = \( 9 \frac{1}{11}\)%







Ans .

20%


  1. Explanation :

    Required reduction in consumption

    = \(\frac{x}{100 + x} \times 100\)%

    where x = 25

    = \(\frac{25}{100 + 25} \times 100\)% = 20%







Ans .

\( 16 \frac{2}{3}\)%


  1. Explanation :

    Reduction in consumption = \(\frac{R}{100 + R} \times 100\)% = \(\frac{20}{120} \times 100\)% = \(\frac{50}{3}\)% = \( 16 \frac{2}{3}\)%







Ans .

4% decrease


  1. Explanation :

    Let the CP of each article = 100 and consumption = 100 units

    Initial expenditure = (100 × 100) = 10000

    New price of article = 80

    Consumption = 120 units

    Expenditure = (120 × 80) = 9600

    Decrease = (10000 – 9600) = 400

    ∴ Percentage decrease = \(\frac{400 \times 100}{10000}\) = 4%







Ans .

\( 13 \frac{1}{23}\)%


  1. Explanation :

    If the price of a commodity increases by R%, then reduction

    in consumption, not to increase the expenditure is given by

    \(\frac{R}{100 + R} \times 100\) %

    =\(\frac{15}{100 + 15} \times 100\) %

    =\(\frac{15}{115} \times 100\) %

    =\(\frac{300}{23}\)

    \( 13 \frac{1}{23}\)%







Ans .

\(\frac{1}{3}\)


  1. Explanation :

    Required fractional decrease = \(\frac{R}{100 + R}\) = \(\frac{50}{100 + 50}\) = \(\frac{1}{3}\)







Ans .

20%


  1. Explanation :

    Percentage decrease = \(\frac{25}{125} \times 100\) = 20%







Ans .

\( 66 \frac{2}{3}\)%


  1. Explanation :

    Required increase percent = \(\frac{40}{100 - 40} \times 100\) = \(\frac{40}{60} \times 100\) = \( 66 \frac{2}{3}\)%







Ans .

\( 16 \frac{2}{3}\)%


  1. Explanation :

    Required percentage decrease = \(\frac{20}{100+20} \times 100\) = \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%







Ans .

20%


  1. Explanation :

    Percentage increase = \(\frac{7.50 -6 }{6} \times 100\) = 25%

    Percentage decrease in consumption

    = \(\frac{25}{125} \times 100\) = 20%







Ans .

4% decrease


  1. Explanation :

    Percentage effect = 20 - 20 + \(\frac{20 \times -20}{100}\) % = – 4% Negative sign shows decrease







Ans .

37.5%


  1. Explanation :

    Required percentage = \(\frac{60}{160} \times 100\) = \(\frac{300}{8}\) = \(\frac{75}{2}\) = 37.5%







Ans .

20%


  1. Explanation :

    Required per cent = \(\frac{25 \times 100}{125}\) = 25%







Ans .

\( 16 \frac{2}{3}\)%


  1. Explanation :

    Percentage decrease in the consumption of petrol

    = \(\frac{20}{100 + 20} \times 100\)%

    = \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%




TYPE – VII





Ans .

44.4%


  1. Explanation :

    Total candidates

    = 1000 + 800 = 1800

    The candidates who are passed

    = \(1000 \times \frac{60}{10}\) + \(800 \times \frac{50}{100}\)

    = 600 + 400 = 1000

    The number of candidates who failed = 1800 – 1000 = 800

    ∴ Required percent = \(\frac{800}{1800} \times 100\) = 44.4%







Ans .

22%


  1. Explanation :

    Let the maximum marks be x.

    According to question,

    20% of x + 5 = 30% of x –20

    \(→\) (30 –20)% of x = 25

    \(→\) x = \(\frac{25 \times 100}{10}\) = 250

    ∴ Passing marks = 20% of 250 + 5 = 55

    ∴ % Passing marks = \(\frac{55}{250} \times 100\) = 22%







Ans .

600


  1. Explanation :

    According to question,

    40% \(→\) 220 + 20 or 40% \(→\) 240

    ∴ 100% \(→\) \(\frac{240}{40} \times 100\) = 600







Ans .

500


  1. Explanation :

    Let the total marks be x.

    According to the question,

    25% of x + 40 = 33% of x

    \(→\) (33 – 25)% of x = 40

    \(→\) 8% of x = 40

    \(→\) x = \(\frac{40 \times 100}{8}\) = 500







Ans .

42, 33


  1. Explanation :

    Let the marks obtained by first student be x.

    ∴ Marks obtained by second student = x + 9

    Sum of their marks = 2x + 9

    As given,

    x + 9 = 56% of (2x + 9)

    \(→\) x + 9 = \(\frac{56}{100} \times (2x+9)\)

    \(→\) x + 9 = \(\frac{15}{25} \times (2x+9)\)

    \(→\) 25x + 225 = 28x + 126

    \(→\) 3x = 225 – 126

    \(→\) x = \(\frac{99}{3}\) = 33







Ans .

90


  1. Explanation :

    Let marks obtained by Supriyo = x

    ∴ \(\frac{9x}{10}\) = 81 \(→\) x = \(\frac{81 \times 10}{9}\) = 90







Ans .

250


  1. Explanation :

    Let the maximum marks be x.

    According to the question,

    40% of x = 90 + 10

    \(→\) x = \(\frac{100 \times 100}{40}\) = 250







Ans .

17%


  1. Explanation :

    Students passed only in Math = 65 – 30 = 35%

    Students passed only in Physics = 48 – 30 = 18%

    ∴ Total passing % = 35 + 18 + 30 = 83%

    ∴ Failed = 100 – 83 = 17%







Ans .

250


  1. Explanation :

    Let the number of students in the class be 100.

    ∴ Number of students in Biology = 72 and number of students in Maths = 44.

    ∴ Number of students opting for both subjects = 72 + 44 – 100 = 16

    ∵ When 16 students opt for both subjects, total number of students = 100

    ∴ When 40 students opt for both subjects,

    total number of students = \(\frac{100}{16} \times 40\) = 250







Ans .

500


  1. Explanation :

    Let the maximum marks be x.

    ∴ \(\frac{x \times 33}{100}\) = 125 + 40 = 165

    \(→\) x = \(\frac{165 \times 100}{33}\) = 500







Ans .

550


  1. Explanation :

    Let maximum marks be x,

    then,

    \(\frac{36 \times x}{100}\) = 113 + 85 = 198

    \(→\) x = \(\frac{198 \times 100}{36}\) = 550







Ans .

67%


  1. Explanation :

    46% of 500 = \(\frac{500 \times 46}{100}\) = 230

    32% of 300 = \(\frac{300 \times 32}{100}\) = 96

    Required marks = 230 – 96 = 134

    Let x% of 200 = 134

    \(\frac{200 \times x}{100}\) = 134

    2x = 134

    x = \(\frac{134}{2}\) = 67%







Ans .

80%


  1. Explanation :

    A = 360;

    B = \(\frac{360 \times 100}{90}\) = 400

    C = \(\frac{400 \times 100}{125}\) = 320

    D = \(\frac{320 \times 100}{80}\) = 400

    \ Required percentage = \(\frac{400}{500} \times 100\) = 80%







Ans .

54·5%


  1. Explanation :

    Failed candidates = \(\frac{1100 \times 50}{100}\) + \(\frac{900 \times 60}{100}\)

    = 550 + 540 = 1090

    ∴ Required percentage = \(\frac{1090}{2000} \times 100\) = 54.5%







Ans .

450


  1. Explanation :

    Successful boys in English or Maths or both = 80 + 85 – 75 = 90%

    Unsuccessful boys = 10%

    ∴ Total number of boys = \(\frac{100}{10} \times 45\) = 450







Ans .

75%


  1. Explanation :

    25% of students pass in at least one subject

    i.e.; they pass in one or both subjects.

    ∴ % of students who don’t pass or fail in both subjects = (100 – 25)% = 75%







Ans .

10%


  1. Explanation :

    The percentage of students

    who pass in one or two or both subjects = 60 + 70 – 40 = 90

    ∴ Percentage of failed students = 100 – 90 = 10%







Ans .

240


  1. Explanation :

    Let total number of candidates = 100

    70 candidates passed in English and 30 failed in it.

    80 candidates passed in Maths and 20 failed in it.

    10 candidates failed in English and Maths both.

    ∴ Out of 30 failed in English, 10 failed in Maths also.

    ∴ 30 – 10 = 20 failed in English alone.

    Similarly, 20 – 10 = 10 failed in Maths alone.

    ∴ Total number of failures = 20 + 10 + 10 = 40

    ∴ 100 – 40 = 60 candidates passed in both subjects.

    Now, if 60 candidates pass, total strength = 100

    ∴ For 144 candidates, total strength = \(\frac{100}{60} \times 144\) = 240







Ans .

25%


  1. Explanation :

    Difference of percentages of maximum marks obtained by two candidates = 32% – 20% = 12%

    Difference of scores between two candidates = 30 +42 = 72

    ∴ 12% of maximum marks = 72

    ∴ Maximum marks = \(\frac{72 \times 100}{12}\) = 600

    ∴ Pass marks = 20% of 600 + 30 = 120 + 30 = 150

    ∴ Required percentage = \(\frac{150}{600} \times 100\) = 25%







Ans .

32.8%


  1. Explanation :

    Total number of students = 640 + 360 = 1000

    Number of successful boys = 60% of 640 = 384

    Number of successful girls = 80% of 360 = 288

    Total number of successful students = 384 + 288 = 672

    Number of unsuccessful students = 1000 – 672 = 328

    \ Required percentage = \(\frac{328 \times 100}{1000}\) = 32.8%







Ans .

44%


  1. Explanation :

    Let total number of students = 100

    Number of failures in Maths = 34

    Number of failures in English = 42

    Number of failures in both subjects = 20

    Number of failures in Maths or

    English or both = 34 + 42 – 20 = 56

    Number of students who passed in both subjects = 100 – 56 = 44

    The required percentage = 44%







Ans .

120


  1. Explanation :

    Difference of percentage = (40 – 30)% = 10%

    Difference of marks = 6 + 6= 12

    ∵ 10% of total marks = 12

    Total marks =\(\frac{12 \times 100}{10}\) = 120







Ans .

23%


  1. Explanation :

    Let the total number of students = 100

    ∴ Number of students who failed in Hindi or English or both = 52 + 42 – 17 = 77

    ∴ Number of students who passed in both subjects = 100 – 77 = 23%

    ∴ Required percentage = 23







Ans .

62%


  1. Explanation :

    Let total number of students = 100

    ∴ 27 students speak none of the two languages.

    It means only 73 students speak either Hindi or English or both.

    Let x students speak both languages.

    ∴ 73 = 70 – x + x + 65 – x

    \(→\) x = 70 + 65 – 73 = 62%







Ans .

600


  1. Explanation :

    Clearly, 75 candidates qualify

    ∴ 75% of appearing candidates = 450

    ∴ Number of appearing candidates = \(\frac{450 \times 100}{75}\) = 600





Ans .

\( 68 \frac{4}{7}\)%


  1. Explanation :

    Number of students passed in first year = 75

    Number of students passed in second year = \(\frac{60 \times 75}{100}\) = 45

    Total number of passed students = 75+45=120

    Total number of appeared students =175

    ∴ Required percentage = \(\frac{120}{175} \times 100\) = \( 68 \frac{4}{7}\)%







Ans .

600


  1. Explanation :

    Let the maximum marks be x.

    According to the question,

    35% of x = 200 +10

    \(\frac{35x}{100}\) = 210

    \(→\) x = \(\frac{210 \times 100}{35}\) = 600







Ans .

50


  1. Explanation :

    Let the full marks in that examination were x.

    According to the question

    \(\frac{30x}{100}\)+ 5 = \(\frac{40x}{100}\) - 10

    \(\frac{3x}{10}\) - \(\frac{4x}{10}\) = - 10 - 5

    ∴ x = 150

    ∴ Minimum pass marks

    \(\frac{30}{100} \times 150\) + 5 = 50







Ans .

5000


  1. Explanation :

    Let the total number of candidates = x

    ∴ Number of candidates passed in English = 0.6x

    Number of candidates passed in Maths = 0.7x

    Number of candidates failed in both subjects = 0.2x

    Number of candidates passed in atleast one subject = x – 0.2x = 0.8x

    ∴ 0.6 x + 0.7x – 2500 = 0.8 x

    \(→\) 1.3x – 0.8x = 2500

    \(→\) 0.5x = 2500

    \(→\) x = \(\frac{2500}{0.5}\) = 5000







Ans .

500


  1. Explanation :

    Let the maximum marks in the

    examination = x.

    According to the question,

    \(\frac{40x}{100}\) - \(\frac{30x}{100}\) = 50

    \(→\) \(\frac{10x}{100}\) = 50

    \(→\) x = \(\frac{50 \times 100}{10}\) = 500







Ans .

8%


  1. Explanation :

    Let total candidates be 'x'

    Percentage of the candidates passing in English or Mathematics or both

    = n(E) + n(M) – n (E \({\displaystyle \cap }\) M)

    = 80 + 85 – 73 = 92

    \(→\) Percentage of candidates who failed in both the subjects

    = 100 – 92 = 8 or 8%







Ans .

50%


  1. Explanation :

    Percentage of students who failed in Maths or English or both = (25 + 35 – 10)% = 50%

    ∴ Required percentage = (100 – 50)% = 50%







Ans .

3700


  1. Explanation :

    If the total number of students be x, then

    7% of x = 259

    \(\frac{x \times 7}{100}\) = 259

    x = \(\frac{259 \times 100}{7}\) = 3700







Ans .

200


  1. Explanation :

    If the total number of students be x, then

    x = \(\frac{90x}{100}\) + \(\frac{85x}{100}\) - 150

    \(→\) 100x = 90x + 85x – 15000

    \(→\) 175x – 100x = 15000

    \(→\) 75x = 15000

    \(→\) x = 200







Ans .

\( 88 \frac{2}{3}\)%


  1. Explanation :

    Required percentage = \(\frac{40 \times 100 + 50 \times 90 + 60 \times 80}{40 + 50 + 60}\) = \( 88 \frac{2}{3}\)%







Ans .

450


  1. Explanation :

    If D gets 100 marks, then

    Marks obtained by C = 125

    Marks obtained by B = \(\frac{125 \times 90}{100}\)

    Marks obtained by A = \(\frac{125 \times 90}{100} \times \frac{125}{100}\)

    ∵ 100 = \(\frac{125 \times 125 \times 90}{10000}\)

    ∵ 320 = \(\frac{125 \times 125 \times 90 \times 320}{1000000}\) = 450 ∵







Ans .

\( 68 \frac{4}{7}\)%


  1. Explanation :

    Total examinees = 80 + 60 = 140

    Total successful examinees = \(\frac{80 \times 60}{100}\) + \(\frac{60 \times 80}{100}\)

    = 48 + 48 = 96.

    ∴ Required percent = \(\frac{96}{140} \times 100\) = \(\frac{480}{7}\) = \( 68 \frac{4}{7}\)%







Ans .

78% of all students


  1. Explanation :

    a = 19%, b = 10%, c = 7%

    Passed students in both the subjects

    = 100 – (a + b – c)

    = 100 – (19 + 10 – 7)

    = 100 – 22 = 78%







Ans .

68


  1. Explanation :

    Successful students in both classes

    = \(\frac{20 \times 80}{100}\) + \(\frac{30 \times 60}{100}\)

    = 16 + 18 = 34

    \ Required percentage = \(\frac{34}{50} \times 100\) = 68%







Ans .

400


  1. Explanation :

    Failures in English = 100 – 75 = 25

    Failures in Maths = 100–60 = 40

    Failures in both subjects = 25

    Failures in English only = 25 – 25 = 0

    Failures n Maths only = 40 – 25 = 15

    Failures in one or both subjects = 25 + 15 = 40

    Percentage of successfuls = 100 – 40 = 60

    Let total students be x

    \(x \times \frac{60}{100}\) = 240

    \(→\) x = \(\frac{240 \times 100}{60}\)= 400







Ans .

400


  1. Explanation :

    Maximum marks in the examination = x (let)

    \(\frac{40x}{100}\) - \(\frac{30x}{100}\) = 12 + 28

    \(\frac{10x}{100}\) \(→\) x = 40 × 10 = 400

    \(\frac{10x}{100}\) \(→\) x = 40 × 10 = 400







Ans .

70


  1. Explanation :

    Total marks scored in all three subjects = \(\frac{300 \times 70}{100}\) = 210

    ∴ Marks scored in third subject

    = 210 – 60 – 80 = 70







Ans .

625


  1. Explanation :

    Let total marks in the exam be x.

    According to the question,

    \(\frac{x \times 36}{100}\) = 190 + 35 = 225

    \(→\) x × 36 = 225 × 100

    \(→\) x = \(\frac{225 \times 100}{36}\) = 625







Ans .

25%


  1. Explanation :

    Let the full marks of exam be x.

    According to the question,

    \(\frac{x \times 32}{100}\) - \(\frac{x \times 20}{100}\) = 30 + 42

    \(→\) \(\frac{12x}{100}\) = 72

    \(→\) x = \(\frac{72 \times 100}{12}\) = 600

    \ Minimum marks to pass = \(\frac{600 \times 20}{100}\) + 30 = 120 + 30 = 150

    \ Required percentage = \(\frac{150}{600} \times 100\) = 25%







Ans .

30000


  1. Explanation :

    Percentage of students who

    pass in one or two or both subjects = 73 + 70 – 64 = 79%

    ∴ Unsuccessful students

    \(→\) 100 – 79 = 21%

    If the total number of examinees be x, then 21% of x = 6300

    \(→\) \(x \times \frac{21}{100}\) = 6300

    \(→\) x = \(\frac{6300 \times 100}{21}\) = 30000







Ans .

12.5%


  1. Explanation :

    Let the number of students with less than 75% attendance = y

    Total students in school = x

    According to the question,

    \(\frac{x}{10}\) + \(\frac{y}{5}\) = y

    \(\frac{x}{10}\) = y - \(\frac{y}{5}\) = \(\frac{4y}{5}\)

    \(\frac{x}{2}\) = 4y

    \(\frac{y}{x}\) = \(\frac{1}{8}\)

    \(\frac{y}{x} \times 100\) = \(\frac{100}{8}\) = 12.5%







Ans .

250


  1. Explanation :

    Number of students who wear spectacles = \(\frac{1400 \times 25}{100}\) = 350

    ∴ Girls who wear spectacles = 1 -\(\frac{2}{7}\) of 350

    = \(350 \times \frac{5}{7}\) = 250