**Ans . **

48,900

**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

**Explanation :**7x – 5x = 200

\(→\) 2x = 200 \(→\) x = 100

∴ Price of a pair of shoes

= 5x = 5 × 100 = 500

**Ans . **

4 : 3

**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

**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

**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

**Explanation :**Acid – I

60

60 – 40 = 20

Acid – II

25

40 – 25 = 15

∴ Required ratio = 15 : 20 = 3 : 4

**Ans . **

3 : 5

**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

**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

**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%

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

**Ans . **

12

**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

**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%

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

**Ans . **

78

**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

**Explanation :**According to the question,

A × 35% = B × 25%

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

**Ans . **

150 litres

**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 . **

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

**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%

**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

**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

**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%

**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

**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

**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

**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 . **

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

**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

**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

**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

**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

**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

**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

**Explanation :**∴ Required ratio = 10 : 15 = 2 : 3

**Ans . **

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

**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

**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

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

**Ans . **

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

**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

**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

**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%

**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%

**Ans . **

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

**Explanation :**Required percentage decrease

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

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

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

**Ans . **

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

**Explanation :**Required answer

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

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

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

**Ans . **

20%

**Explanation :**Required reduction in consumption

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

where x = 25

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

**Ans . **

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

**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

**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 . **

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

**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 . **

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

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

**Ans . **

20%

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

**Ans . **

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

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

**Ans . **

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

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

**Ans . **

20%

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

Percentage decrease in consumption

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

**Ans . **

4% decrease

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

**Ans . **

37.5%

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

**Ans . **

20%

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

**Ans . **

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

**Explanation :**Percentage decrease in the consumption of petrol

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

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

**Ans . **

44.4%

**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%

**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

**Explanation :**According to question,

40% \(→\) 220 + 20 or 40% \(→\) 240

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

**Ans . **

500

**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

**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

**Explanation :**Let marks obtained by Supriyo = x

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

**Ans . **

250

**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%

**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

**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

**Explanation :**Let the maximum marks be x.

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

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

**Ans . **

550

**Explanation :**Let maximum marks be x,

then,

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

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

**Ans . **

67%

**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%

**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%

**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

**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%

**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%

**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

**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%

**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%

**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%

**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

**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%

**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%

**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

**Explanation :**Clearly, 75 candidates qualify

∴ 75% of appearing candidates = 450

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

**Ans . **

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

**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

**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

**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

**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

**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%

**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%

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

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

**Ans . **

3700

**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

**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 . **

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

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

**Ans . **

450

**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 . **

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

**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

**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

**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

**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

**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

**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

**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%

**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

**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%

**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

**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