- Staff Selection Commission Mathematics - Percentage (1999-2017) Part 4

# Staff Selection Commission Mathematics - Percentage (1999-2017)

### TYPE–VII (Contd.)

Ans .

1218

1. Explanation :

Percentage of boys = 60%

∴ Percentage of girls = 40%

Boys : Girls = 60 : 40 = 3 : 2

Number of girls = 812

∴ Number of boys = $$\frac{3}{2} \times 812$$ = 1218

Ans .

75.5%

1. Explanation :

Marks scored by A :

First subject $$→$$ $$\frac{900 \times 72}{100}$$ = 648

Second subject $$→$$ $$\frac{700 \times 80}{100}$$ = 560

Total marks scored = 648 + 560 = 1208

Total maximum marks = 900 + 700 = 1600

\ Required per cent = $$\frac{1208}{1600} \times 100$$ = 75.5%

Ans .

40%

1. Explanation :

Percentage of failures either in 1 subject or both subjects = (35 + 45 – 20)% = 60%

∴ Percentage of the successful = (100– 60)% = 40%

Ans .

80%

1. Explanation :

Total marks of 50 students = 50 × 70 = 3500

Total marks of 25 students = 25 × 60 = 1500

Total marks of 24 students = 24 × 80 = 1920

∴ Marks obtained by last student = 3500 – 1500 – 1920

= 80 i.e, 80%

Ans .

42 and 33

1. Explanation :

Let marks obtained by the first student be x.

∴ Marks obtained by the second student = x – 9

According to the question,

x = 56% of (x + x – 9)

x = $$\frac{(2x - 9) \times 56}{100}$$

$$→$$ 100x = 112x – 504

$$→$$ 112x – 100x = 504

$$→$$ 12x = 504

$$→$$ x = $$\frac{504}{12}$$ = 42

∴ Marks obtained by the second student = 42 – 9 = 33

Ans .

360 marks

1. Explanation :

Maximum marks of examination = x (let)

According to the question,

25% of x = 47 + 43

$$\frac{x \times 25}{100}$$ = 90

$$\frac{x}{4}$$ = 90

$$→$$ x = 4 × 90 = 360

### TYPE–VIII

Ans .

4% decreased

1. Explanation :

Change in his salary = 20 - 20 - $$\frac{20 \times 20}{100}$$ %

= -$$\frac{400}{100}$$% = -4%

i.e. 4% decrease

Ans .

8% increase

1. Explanation :

Net % change = A + B + $$\frac{AB}{100}$$ %

Here, A = 20%, B = – 10%

∴ Net % change

= 20 - 10 - $$\frac{200}{100}$$

= 10 – 2= 8%

+ ve sign shows increase

Ans .

1% decrease

1. Explanation :

Required change = $$\frac{(10)^2}{100}$$ %decrease

= 1% decrease

Ans .

19%

1. Explanation :

A single equivalent reduction to reduction series of x%, y%

= x + y - $$\frac{xy}{100}$$ %

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

= 20 - 1 % = 19%

Ans .

10% decrease

1. Explanation :

The net change in price = -25 +20 - $$\frac{-25 \times 20}{100}$$ % = (–25 + 20 – 5)% = – 10% Negative sign shows decrease.

Ans .

$$2 \frac{1}{2}$$% decrease

1. Explanation :

Let the price of the article be 100 and the daily sale be 100

units.

∴ Revenue day = 100 × 100 = 10000

New receipts = 75 × 130 = 9750

Decrease = (10000 – 9750) = 250

∴ % decrease = $$\frac{2500}{10000} \times 100$$ = $$2 \frac{1}{2}$$% decrease

Ans .

21%

1. Explanation :

Single equivalent percentage increase in price = 10 + 10 + $$\frac{10 \times 10}{100}$$ % = 21%

Ans .

25

1. Explanation :

Effective increase percentage

= 10 + 20 + $$\frac{10 \times 20}{100}$$ % = 32%

$$\frac{x \times 132}{100}$$ = 33

$$→$$ x = $$\frac{33 \times 100}{132}$$ = 25

Ans .

44%

1. Explanation :

Effective percentage increase = 20 + 20 + $$\frac{20 \times 20}{100}$$ % = 44%

Ans .

1% decrease

1. Explanation :

Net change = 10 - 10 - $$\frac{10 \times 10}{100}$$ %

= –1% = 1% decrease

Ans .

32%

1. Explanation :

Required percentage increase = 10 + 20 + $$\frac{10 \times 20}{100}$$ % = 32%

Ans .

44% increase

1. Explanation :

Required effect = 80 - 20 - $$\frac{80 \times 20}{100}$$ % = 44%

Positive sign shows increase.

Ans .

decreased by 1%

1. Explanation :

Net effect = 10 - 10 - $$\frac{10 \times 10}{100}$$ % = -1%

Negative sign shows decrease

Ans .

10% decrease

1. Explanation :

Required per cent effect = 20 - 25 - $$\frac{20 \times 25}{100}$$ % = -10%

Ans .

21

1. Explanation :

Percentage effect = 10 + 10 + $$\frac{10 \times 10}{100}$$ % = 21%

Ans .

$$\frac{8}{15}$$

1. Explanation :

Original fraction = $$\frac{x}{y}$$

∴ $$\frac{\frac{120x}{100}x}{y \times \frac{80}{100}}$$ = $$\frac{4}{5}$$

∴ $$\frac{120x}{80y}$$ = $$\frac{4}{5}$$

∴ $$\frac{6x}{4y}$$ = $$\frac{4}{5}$$

∴ $$\frac{x}{y}$$ = $$\frac{8}{15}$$

Ans .

$$\frac{95}{48}$$

1. Explanation :

Let original fraction be

$$\frac{x}{y}$$

According to the question,

$$\frac{\frac{120}{100}x}{\frac{95y}{100}}$$ = \frac{5}{2}

$$\frac{120x}{95y}$$ = \frac{5}{2}

$$\frac{x}{y}$$ = \frac{95}{48}

Ans .

decreased by $$6 \frac{1}{4}$$%

1. Explanation :

Effective value = x + y + $$\frac{x \times y}{100}$$ %

= 25 - 25 - $$\frac{25 \times 25}{100}$$ % = decreased by $$6 \frac{1}{4}$$%

Ans .

240

1. Explanation :

Larger number = x and smaller

number = 520 – x

$$\frac{96x}{100}$$ = $$\frac{520-x}{100} \times 112$$

$$→$$ 96x = 520 × 112 – 112x

$$→$$ 112x + 96x = 520 × 112

$$→$$ 208x = 520 × 112

$$→$$ x = $$\frac{520 \times 112}{208}$$ = 280

∴ Smaller number

= 520 – 280 = 240

Ans .

400

1. Explanation :

If the original price of article be x, then

$$x \times \frac{80}{100}\times \frac{130}{100}$$ = 416

$$→$$ x = $$\frac{416 \times 100\times 100}{80 \times 130}$$ = 400

Ans .

80

1. Explanation :

If the number be x, then $$x \times \frac{245}{200}$$ = 98

$$→$$ x = $$\frac{98 \times 200}{245}$$ = 80

Ans .

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

1. Explanation :

Let the original price be 100

New price after 10% decrease = 90

In order to restore the price to its original value,

it must be increased by 10% increase = $$\frac{10}{90} \times 100$$ = $$\frac{100}{9}$$ = $$11 \frac{1}{9}$$%

Ans .

25%

1. Explanation :

Clearly, 75% of the number = 225

∴ Number = $$\frac{225 \times 100}{75}$$ = 300

Again, 125% of 300 = 375

Hence, the number should be increased by 25%

Ans .

$$30 \frac{5}{9}$$%

1. Explanation :

Let the number be 100.

After 20% increase, number = 120

After 20% increase of 120, number = $$120 \times \frac{120}{100}$$ = 144

∴ Per cent decrease

= $$\frac{44}{144} \times 100$$

= $$\frac{275}{9}$$ = $$30 \frac{5}{9}$$%

Ans .

$$\frac{25}{4}$$%

1. Explanation :

Let the original number of employees be 100 and wages per

Total wages = (100 × 100) = 10000

New number of employees = 125

New wages per head = 75

Total new wages = (125 × 75) = 9375

Decrease = (10000 – 9375) = 625

∴ Percentage decrease = $$\frac{625}{10000} \times 100$$

= $$\frac{625}{100}$$ = $$\frac{25}{4}$$%

Ans .

5000

1. Explanation :

Let original number be x

$$\frac{90}{100}x \times \frac{110}{100}$$ = x - 50

$$\frac{99x}{100}$$ = x-50

x - $$\frac{99x}{100}$$ = 50

$$\frac{x}{100}$$ = 50

$$→$$ x = 5000

Ans .

5%

1. Explanation :

Let the income be x and the rate of income tax be y %

According to the question,

$$\frac{xy \times 1.19}{100}$$ - $$\frac{xy}{100}$$ = $$(x - \frac{xy}{100}) \times \frac{1}{100}$$

$$→$$ 1.19 xy – xy = - $$\frac{xy}{100}$$

$$→$$ 0.19 y = - $$\frac{y}{100}$$

$$→$$ $$\frac{y}{100}$$ + 0.19y = 1 $$→$$ y$$\frac{1+19}{100}$$ = 1

$$→$$ = $$\frac{100}{20}$$ = 5%

Ans .

50%

1. Explanation :

Man’s income = 100 (let)

Expenditure = 75

Savings = 25

New income = $$\frac{100 \times 120}{100}$$ = 120

New expenditure = $$\frac{75 \times 110}{100}$$ = 82.5

Savings = 120 – 82.5 = 37.5

Increase in savings = 37.5 – 25 = 12.5

\ Increase per cent = $$\frac{12.5}{25} \times 100$$ = 50%

Ans .

33.1%

1. Explanation :

Single equivalent increase for 10% and 10% = 10 + 10 + $$\frac{10 \times 10}{100}$$ % = 21%

Again, single equivalent increase for 21% and 10% = 21 + 10 + $$\frac{21 \times 10}{100}$$ % = 33.1%

Ans .

increased by 8.9%

1. Explanation :

Increase in first year = 10%

Decrease in 2nd year = 10%

Effective result = 10 - 10 - $$\frac{10 \times 10}{100}$$ % = –1%

Increase in 3rd year = 10%

∴ Effective result = 10 - 1 - $$\frac{10 \times 1}{100}$$%

= (9 – 0.1)% = 8.9% (increase)

Ans .

80

1. Explanation :

Let the number be x.

∴ (20 + 25)% of x = 36

$$\frac{45x}{100}$$ = 36

$$→$$ x = $$\frac{36 \times 100}{45}$$ = 80

Ans .

500

1. Explanation :

Effective percentage = -20 + 20 - $$\frac{20 \times 20}{100}$$ % = -4%

If the number be x, then 4% of x = 20

$$x \times \frac{4}{100}$$ = 20

x = $$\frac{20 \times 100}{4}$$ = 500

Ans .

$$\frac{100x}{100+x}$$%

1. Explanation :

$$\frac{B \times 30}{100}$$ $$→$$ increased value $$→$$ P + $$\frac{Px}{100}$$

= P$$\frac{100+x}{100}$$

∴ Required answer =$$\frac{x}{100+x} \times 100$$

Ans .

28%

1. Explanation :

Effective percentage decrease = x + y + $$\frac{x \times y}{100}$$ %

= - 10 - 20 + $$\frac{-10 \times -20}{100}$$ %

= (– 30 + 2)% = –28%

Ans .

0

1. Explanation :

Cost of edible oil = 100 per kg.

Consumption = 1 kg.

Again, New price = 125 per kg.

Consumption = 0.8 kg.

Expenditure = Rs. (125 × 0.8) = Rs. 100

Percentage effect = 25 - 20 - $$\frac{20 \times 25}{100}$$ % = 0%

### TYPE - IX

Ans .

27500

1. Explanation :

Let the total number of votes be 100.

Number of uncast votes = 8

∴ Number of votes polled = 92

Number of votes obtained by the winner = 48

∴ Number of votes obtained by the loser = 92 – 48 = 44

If the difference of win be 4 votes, total voters = 100

∴ When the difference be 1100 votes, total voters

= $$\frac{100}{4} \times 1100$$ = 27500

Ans .

16800

1. Explanation :

Let the total number of voters enrolled be x.

Number of votes polled = = 75% of x = $$\frac{3x}{4}$$

Number of valid votes = $$\frac{3x}{4}$$ - $$\frac{2}{100} \times \frac{3x}{4}$$

= $$\frac{3x}{4}$$ - $$\frac{2}{100} \times \frac{3x}{4}$$

= $$\frac{147x}{200}$$

Now, 75% of = $$\frac{147x}{200}$$ = 9261

or $$\frac{3}{4}$$ of $$\frac{147x}{200}$$ = 9261

or x = $$\frac{9261 \times 4 \times 200}{3 \times 147}$$ = 16800

Ans .

42,000

1. Explanation :

Difference of percentage of

votes = 60% – 40% = 20%

∴ 20% of total votes = 14000

= $$\frac{14000}{20} \times 60$$ = 42000

Ans .

52%

1. Explanation :

Let total employees = 100

∴ Required percentage = $$\frac{40 \times 40}{100}$$ + $$\frac{60 \times 60}{100}$$

= 16 + 36 = 52%

Ans .

1490

1. Explanation :

$$x \times \frac{60-40}{100}$$ = 298

$$→$$ x ×$$\frac{1}{5}$$ = 298

$$→$$ x = 298 × 5 = 1490

Ans .

56056

1. Explanation :

Number of valid votes = $$104000 \times \frac{98}{100}$$ = 101920

∴ Valid votes received by the candidate = $$\frac{101920 \times 55}{100}$$ = 56056

Ans .

8000

1. Explanation :

If the number of votes polled be x, then $$\frac{x \times 20}{100}$$ = 1600

$$→$$ x = $$\frac{1600 \times 100}{20}$$ = 8000

Ans .

8640

1. Explanation :

Vote percentage of third candidate

= 100 – 40 – 36 = 24%

∴ Votes got by third candidate = $$\frac{36000 \times 24}{100}$$ = 8640

Ans .

3,00,000

1. Explanation :

∴ (57 – 43)% of x = 42000

$$x \times \frac{14}{100}$$ = 42000

x = $$\frac{42000 \times 100}{14}$$ = 300000

Ans .

700

1. Explanation :

Total number of votes polled = x

$$\frac{x \times 84}{100}$$ - $$\frac{x \times 16}{100}$$ = 476

$$\frac{68x}{100}$$ = 476

$$→$$ x = $$\frac{476 \times 100}{68}$$ = 700

Ans .

30000

1. Explanation :

Number of valid votes = x (let)

\ (62 – 38)% of x = 7200

$$x \times \frac{24}{100}$$ = 7200

x = $$\frac{7200 \times 100}{24}$$ = 30000

Ans .

600

1. Explanation :

Total number of votes polled = x (let)

According to the question

$$\frac{x \times 62}{100}$$ - $$\frac{x \times (100 - 62)}{100}$$ = 144

$$\frac{x \times 62}{100}$$ - $$\frac{x \times 38}{100}$$ = 144

x = 600

Ans .

6200

1. Explanation :

Total voters in the list = x

Votes got by the winner = $$\frac{47x}{100}$$

Votes got by the loser = x – $$\frac{x}{10}$$ - 60 - $$\frac{47x}{100}$$

= $$\frac{9x}{10}$$ - 60 - $$\frac{47x}{100}$$

= $$\frac{43x}{100}$$ - 60

According to the question,

$$\frac{47x}{100}$$ - $$\frac{43x}{100}$$ + 60 = 308

x = 6200

Ans .

1490

1. Explanation :

According to the question,

(60 – 40)% of x = 298

$$→$$ $$x \times \frac{20}{100}$$ = 298

$$→$$ $$\frac{x}{5}$$ = 298

$$→$$ x = 298 × 5 = 1490

### TYPE–X

Ans .

217800

1. Explanation :

Required population after two years = = 180000 $$(1+ \frac{10}{100})^2$$

= $$180000 \times \frac{11}{10}\times \frac{11}{100}$$ = 217800

Ans .

48.8%

1. Explanation :

If the present worth of the equipment be 100, then its price after 3 years

= $$100 \times (\frac{80}{100})^3$$ = 51.2

∴ Depriciation = 48.8%

Ans .

68921

1. Explanation :

P = P0 $$(1 + \frac{R}{100})^T$$

P = 64000 $$(1 + \frac{5}{200})^3$$

P = 64000 $$( \frac{41}{40})^3$$

P = $$\frac{64000 \times 41 \times 41 \times 41}{40 \times 40 \times 40}$$

P = 68921

Ans .

10000

1. Explanation :

Suppose the value of property two years ago was x

According to question

∴ x$$(1-\frac{10}{100})^2$$ = 8100

∴ x$$(\frac{90}{100})^2$$ = 8100

$$→$$ x = 10000

Ans .

56,700

1. Explanation :

Let the present population be P.

∴ P = 62500$$(1 - \frac{4}{100})^2$$

P = 57600

Ans .

54,080

1. Explanation :

Required population = $$( 1 + \frac{4}{100})^2$$ = 54,080

Ans .

Let the man’s annual salary in 2006 be x.

∴ $$\frac{110x}{100}$$ = 880000

x = = 800000

1. Explanation :

Ans .

62500

1. Explanation :

Population of the village two years ago = $$\frac{P}{(1+\frac{R}{100})^2}$$

= $$\frac{67600}{(1+\frac{4}{100})^2}$$

= 62500

Ans .

1,80,500

1. Explanation :

A = $$P (1-\frac{R}{100})^T$$

A = $$200000 (1-\frac{5}{100})^2$$

= 1,80,500

Ans .

4,80,000

1. Explanation :

Value of the property 3 years ago

= $$\frac{P}{(1-\frac{R}{100})^T}$$

= $$\frac{411540}{(1-\frac{5}{100})^3}$$

= 480000

Ans .

85184

1. Explanation :

Population of town = $$P (1-\frac{R}{100})^T$$

= $$P (1-\frac{R}{100})^T$$

= $$64000 (1-\frac{10}{100})^3$$

= 85184

Ans .

6400

1. Explanation :

Present population = $$10000 (1-\frac{20}{100})^2$$ = 6400

Ans .

5

1. Explanation :

Required percent = $$\frac{1}{4} \times 3$$ + $$\frac{2}{3} \times 5$$ + $$(1-\frac{1}{4}-\frac{2}{3}) \times 11$$ = 5%

Ans .

3,150

1. Explanation :

Required price of the machine = = $$6250 (1-\frac{10}{100})(1-\frac{20}{100})(1-\frac{30}{100})$$ = 3150

Ans .

40,500

1. Explanation :

Required value = $$50000 (1-\frac{10}{100})^2$$ = 40,500

Ans .

1000

1. Explanation :

If the price of machine 3years ago be x. then 729 = $$x (1-\frac{10}{100})^3$$ 729 = $$x \times (\frac{9}{10})^3$$ $$→$$ x = 1000

Ans .

1720

1. Explanation :

Required Raman’s salary = $$\frac{100}{100+5} \times 1806$$ = $$\frac{100}{105} \times 1806$$ = 1720

Ans .

4000

1. Explanation :

∴ Men : Women = 1 : 1

∴ Number of men = $$\frac{1}{2} \times 8000$$ = 4000

Ans .

5600

1. Explanation :

Let the number of males = x

\ Number of females= 9800 – x

According to the question,

$$x \times \frac{108}{100}$$ + $$(9800-x) \times \frac{105}{100}$$ = 10858

$$→$$ 108 x + 9800 ×105 – 105x

= 1045800

$$→$$ 3x + 1029000 = 1045800

$$→$$ 3x = 1045800 – 1029000

= 16800

$$→$$ = $$\frac{16800}{3}$$ = 5600

Ans .

5120

1. Explanation :

Population in the beginning of the year = $$\frac{10000}{ (1+\frac{25}{100})^3}$$ = 5120

Ans .

111%

1. Explanation :

If the number of men be 100, then Number of women = 90

∴ Required per cent = $$\frac{100}{90} \times 100$$ = 111%

Ans .

562432

1. Explanation :

Required population = $$500000 (1+\frac{4}{100})^3$$ = 562432

Ans .

4000

1. Explanation :

If the population of village two years ago be P0 , then 4410 = $$P0 (1+\frac{1}{20})^2$$

P0 = 4000

Ans .

19,950

1. Explanation :

Value of TV after one year = 21000 × (100 – 5)% = $$\frac{21000 \times 95}{100}$$ = = Rs. 19950

Ans .

72.8 %

1. Explanation :

Single equivalent increase for 20% and 20% = 20 + 20 + $$\frac{20 \times 20}{100}$$ % = 44%

Single equivalent increase for 44% and 20% = 44 + 20 + $$\frac{44 \times 20}{100}$$ % = = 72.8%

Ans .

32.5

1. Explanation :

Population of town = 1000

Males $$→$$ 600

Females $$→$$ 400

Literate males = $$\frac{600 \times 20}{100}$$ = 20

Total literate inhabitants = $$\frac{1000 \times 25}{100}$$ = 250

∴ Literate females = 250 – 120 = 130

∴ Required percent = $$\frac{130}{400} \times 100$$ = 32.5%

Ans .

10%

1. Explanation :

If the rate of increase per annum be R%, then

A = $$P (1-\frac{R}{100})^T$$

48400 = $$40000 (1+\frac{R}{100})^2$$

R = 10% per annum

Ans .

Rs. 10,000

1. Explanation :

A = $$P (1-\frac{R}{100})^3$$

7290 = $$P (1-\frac{10}{100})^3$$

P = Rs. 10000

Ans .

8000

1. Explanation :

9261 = $$P0 (1+\frac{5}{100})^3$$

P0 = 8000

Ans .

6125

1. Explanation :

Original population of village = x (let)

According to the question,

$$x \times \frac{95}{100} \times \frac{80}{100}$$ = 4655

x = 6125

Ans .

9600

1. Explanation :

In the village,

Females = 3000

Males = 9000 – 3000 = 6000

After respective increases,

Population of village = $$3000 \times \frac{105}{100}$$ + $$\frac{6000 \times 107.5}{100}$$

= 3150 + 6450 = 9600

Ans .

60

1. Explanation :

Let the population of the city be 100.

Total illiterate people = 40

Poor people = 60

Rich people = 40

Illiterate rich people = $$\frac{40 \times 10}{100}$$ = 4

∴ Illiterate poor people = 40 – 4 = 36

∴ Required per cent = $$\frac{36}{60} \times 100$$% = 60%

Ans .

31200

1. Explanation :

Population of city after two years = $$P (1+\frac{R1}{100})(1+\frac{R1}{100})$$ = $$20000 (1+\frac{20}{100})(1+\frac{30}{100})$$ = $$20000 \times \frac{120}{100} \times \frac{130}{100}$$ = 31200

Ans .

4100

1. Explanation :

Let the total population of village be x.

According to the question,

$$\frac{x \times 14}{100}$$ = 574

$$→$$ x = $$\frac{574 \times 100}{14}$$ = 4100

### TYPE–XI

Ans .

13.50

1. Explanation :

10 per cent of 837 = $$\frac{10}{100} \times 837$$ = 83.7

∴ Reduced per kg price = $$\frac{83.7}{62}$$ = 13.50

Ans .

18

1. Explanation :

Let the original price of sugar = x /kg.

Reduced price of sugar = 80% of x

= $$\frac{x \times 80}{100}$$ = $$\frac{4 \times x}{5}$$

∴ $$\frac{36}{\frac{4x}{5}}$$ - $$\frac{36}{x}$$ = $$\frac{1}{2}$$

$$→$$ x = 9 × 2 = 18/Kg

Ans .

18

1. Explanation :

Reduced price of 6.2kg of sugar

= 10% of 1116

= 111.6

∴ Reduced price per kg = $$\frac{111.6}{62}$$ = 18

Ans .

30

1. Explanation :

Let the original price of sugar be x/kg.

∴ New price = $$\frac{9x}{10}$$

$$\frac{270}{\frac{9x}{10}}$$ - $$\frac{270}{x}$$ = 1

$$→$$ x = 30 /kg

Ans .

12.96

1. Explanation :

Let the original price of apples be x/ dozen

∴ New price = $$\frac{4x}{5}$$ /dozen

$$\frac{54}{\frac{4x}{5}}$$ - $$\frac{54}{x}$$ = $$\frac{10}{2}$$

$$\frac{4x}{5}$$ = 12.96

Ans .

36

1. Explanation :

The original price of 1 egg = x

Present price = $$\frac{3}{2}x$$

$$\frac{24}{x}$$ - $$\frac{24}{\frac{3x}{2}}$$ = 4

$$→$$ x = 2

∴ Present price of eggs per dozen = $$12 \times \frac{3}{2} \times 2$$ = 36

Ans .

16

1. Explanation :

Original price of wheat = x /kg.

New price of wheat = $$\frac{4x}{5}$$

$$\frac{320}{\frac{4x}{5}}$$ - $$\frac{320}{x}$$ = 5

$$→$$ x = 16

Ans .

28.80

1. Explanation :

Original rate = x per egg

New rate = $$\frac{6x}{5}$$ per egg

$$\frac{24}{x}$$ - $$\frac{24 \times 5}{6x}$$ = 2

x = 2

∴ New rate = $$\frac{12}{5}$$ per egg.

∴ Rate per dozen of eggs = $$\frac{12}{5} \times 12$$= 28.80

Ans .

15

1. Explanation :

Let original price of rice per kg = x (let)

∴ New price of rice per kg = $$\frac{3x}{4}$$

$$\frac{600}{\frac{3x}{4}}$$ - $$\frac{600}{x}$$ = 10

$$→$$ x = 20

∴ New price = $$\frac{3 \times 20}{4}$$ = 15/kg

Ans .

20%

1. Explanation :

Percentage decrease = $$\frac{0.25}{1.25} \times 100$$ = 20%

Ans .

30

1. Explanation :

If the number be x, then x – 15 = $$\frac{4x}{5}$$

$$→$$ 5x – 75 = 4x $$→$$ x = 75

∴ 40% of 75 = $$\frac{75 \times 40}{100}$$ = 30