- 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

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 :

= $$\frac{10}{100 + 10} \times 100$$

= $$\frac{10}{110} \times 100$$

$$\frac{100}{11}$$% = $$9 \frac{1}{11}$$%

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