**Ans . **

7.00

**Explanation :**Original price of article = x per kg.

New price = \(\frac{79x}{100}\)per kg

∴ \(\frac{100}{\frac{79x}{100}}\) - \(\frac{79x}{100}\) = 3

\(→\) 79x = 700 \(→\) x = \(\frac{700}{79}\)

∴ New price = \(\frac{79x}{100}\) = \(\frac{79}{100} \times \frac{700}{79}\)

= 7 per kg

**Ans . **

Rs. 4

**Explanation :**Let the original price of sugar be Rs. x per kg.

Reduced price = Rs. \(\frac{80x}{100}\) = Rs. \(\frac{4x}{5}\)per kg.

According to the question,

\(\frac{160}{\frac{4x}{5}}\) - \(\frac{4x}{5}\) = 8

\(→\) 8x = 40

\(→\) x = \(\frac{40}{8}\) = 5 per kg.

Reduced Price = Rs. \(\frac{4 \times 5}{5}\) = Rs. 4 per kg

**Ans . **

12% decrease

**Explanation :**Required percentage change = 10 - 20 + \(\frac{10 \times -20}{100}\) % = -12%

Negative sign shows decrease.

**Ans . **

\( 37 \frac{1}{2}\)%

**Explanation :**Required per cent = \(\frac{x}{100 + x} \times 100\)

where x = 60% = \(\frac{60}{160} \times 100\) = \(\frac{75}{2}\) = \( 37 \frac{1}{2}\)%

**Ans . **

12 kg.

**Explanation :**Let original price of sugar be Rs. x per kg.

New price = Rs.\(\frac{120x}{100}\) = Rs.\(\frac{6x}{5}\)per kg

According to the question,

\(\frac{50}{x}\) - \(\frac{50}{\frac{6x}{5}}\) = 2

\(→\) 6x = 25

\(→\) x = Rs \(\frac{25}{6}\)kg

∴ Required quantity of sugar = \(\frac{50}{x}\)kg

= \(\frac{50}{\frac{25}{6}}\)kg = \(\frac{50 \times 6}{25}\) = 12kg.

**Ans . **

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

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

= \(\frac{10}{100-10} \times 100\) = \( 11 \frac{1}{9}\)%

**Ans . **

25%

**Explanation :**Required percentage increase = \(\frac{x}{100-x} \times 100\)

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

**Ans . **

Rs. 5 per kg

**Explanation :**Original price of sugar = Rs. x/kg. (let)

New price = Rs. \(\frac{120x}{100}\) per kg.

= Rs. \(\frac{6x}{5}\) per kg.

According to the question,

\(\frac{120}{x}\) - \(\frac{120}{\frac{6x}{5}}\) = 4

\(→\) x = Rs. 5 per kg.

**Ans . **

25%

**Explanation :**Original price of building = Rs. 100 (let)

∴ Its price in 2001 = Rs. 80

Its price in 2002 = Rs. 60

Required percentage decrease = \(\frac{80-60}{80} \times 100\) = 25%

**Ans . **

1700

**Explanation :**Percentage of boys = 100% – 70% = 30%

Let total no. of students be x

\ According to question,

30% of x = 510

x = \(\frac{510}{30} \times 100\) = 1700

**Ans . **

1458

**Explanation :**40% of students = 972

∴ 60% of students = \(\frac{972}{40} \times 60\) = 1458

**Ans . **

1176

**Explanation :**Number of boys = \(\frac{70}{30} \times 504\) = 1176

**Ans . **

95

**Explanation :**Required sum = 0.5% of 19000 = \(19000 \times \frac{0.5}{100}\) = 95

**Ans . **

9 m

**Explanation :**Remaining height = 192 - \(\frac{125}{2}\)% of 192

= 192 – 120 = 72 m

∴ Required distance (distance covered in second hour) then,

= \(\frac{25}{2}\)% of 72

= \(\frac{25 \times 72}{2 \times 100}\) = 9m

**Ans . **

52 kgs.

**Explanation :**Water in 100 kg fresh fruit = 68%

Water in dry fruit = 20%

Decrease = 48%

∴ Dry fruit obtained = 100 – 48 = 52 kg

**Ans . **

33%

**Explanation :**The net tax rate = 30 + \(30 \times \frac{10}{100}\) % = 33%

**Ans . **

180

**Explanation :**Let z have x

∴ Money with Y = \(\frac{3}{2}\)x and Money with X = 3x

∴ 3x + \(\frac{3x}{2}\) + x = \(3 \times 110\)

\(→\) \(\frac{6x+3x+2x}{2}\) = 330

\(→\) 11 x = 2 × 330

x = \(\frac{2 \times 330}{11}\) = 60

∴ Money with X = 3x = (3×60) = 180

**Ans . **

\( 83 \frac{1}{3}\)%

**Explanation :**If a number is x% more than other, then the other number is less than the first number by

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

\ Required answer = \(\frac{500}{100+500} \times 100\) = \( 83 \frac{1}{3}\)%

**Ans . **

48.8%

**Explanation :**Let x = 10 and y = 10

∴ x\(y^2\) = 10 × 10 × 10=1000 units

Decreasing values of x and y by 20%,

Expression = x\(y^2\)

= 8 × 8 × 8

= 512

Decrease= 1000–512 = 488 units

Percentage decrease = \(\frac{488}{1000} \times 100\) = 48.8%

**Ans . **

88%

**Explanation :**If two numbers are respectively x% and y% more than a third number, the first as a per cent of

second is = \(\frac{100+x}{100+y} \times 100\) = \(\frac{110}{125} \times 100\) = 88%

**Ans . **

140

**Explanation :**Let sum of money be x.

\(\frac{11}{2}\)% of x = 220

\(→\) x = \(\frac{220 \times 200}{11}\) = 4000

\( 3 \frac{1}{2}\)% of 4000 = \(\frac{7}{2} \times \frac{4000}{100}\) = 140

**Ans . **

3000

**Explanation :**Let the total number of workers in the factory be x

\(x \times \frac{60}{100} \times \frac{75}{100}\) = 1350

\(→\) x = \(\frac{1350 \times 100^2}{60 \times 75}\) = 3000

**Ans . **

10%

**Explanation :**Let the third number = 100.

∴ First number = 70

Second number = 63

∴ Required per cent = \(\frac{70-63}{70} \times 100\) = 10%

**Ans . **

160%

**Explanation :**Let Rani’s weight be x kg.

∴ Meena’s weight = 4x kg.

Tara’s weight = \(\frac{5x}{4}\)kg

∴ Required percentage = \(\frac{4x}{\frac{5x}{2}} \times 100\) = 160%

**Ans . **

570

**Explanation :**Number of people who have the saving habit = \(\frac{2500 \times 60}{100}\)

∴ Number of shareholders = (100 – 62)% of 1500

= \(\frac{1500 \times 38}{100}\) = 570

**Ans . **

8000

**Explanation :**Let the original price of the article be x.

According to the question,

5832 = \(x(1-\frac{1}{100})^3\)

5832 = \(x(\frac{9}{10})^3\)

x = 8000

**Ans . **

9%

**Explanation :**Increase in AC = 6%

∴ Increased AC = \(\frac{106}{100} \times 3\) = 3.18 cm

∴ Decreased CB = 5 – 3.18 = 1.82 cm.

∴ Decrease = 2 – 1.82 = 0.18 cm

∴ Percentage decrease = \(\frac{0.18}{2} \times 100\) = 9%

**Ans . **

\( 91 \frac{2}{3}\)%

**Explanation :**∵ 24= 100% ∴22 = \(\frac{100}{24} \times 22\) = \( 91 \frac{2}{3}\)%

**Ans . **

40%

**Explanation :**Let the number of boys be x and that of girls be y.

Then, 71x + 73y = 71.8 (x + y)

\(→\) 71.8x – 71x = 73y – 71.8y

\(→\) 0.8x = 1.2y

\(\frac{x}{y}\) = \(\frac{1.2}{0.8}\) = \(\frac{12}{8}\) = \(\frac{3}{2}\)

∴ \(\frac{x}{y}\) + 1 = \(\frac{3}{2}\) + 1

\(\frac{x + y}{y}\) = \(\frac{5}{2}\)

∴ Percentage of girls = \(\frac{y}{x+y} \times 100\) = \(\frac{2}{5} \times 100\) = 40%

**Ans . **

50%

**Explanation :**Let the number of books in shelf B be 100.

∴ Number of books in shelf A = 80

On transferring 25% i.e \(\frac{1}{4}\) of books of shelf A to shelf B.

B = 100 + 20 = 120

Again, on transferring \(\frac{1}{4}\) of books of shelf B to shelf A.

A = 60 + \(\frac{120}{4}\) = 90

∴ Required percentage = \(\frac{90}{180} \times 100\) = 50%

**Ans . **

178200

**Explanation :**Total revenue earned = \(9900 \times \frac{20}{100} \times 10 \) + \(9900 \times \frac{80}{100} \times 20 \)

= (19800 + 158400)

= 178200

**Ans . **

\( 77 \frac{7}{9}\)

**Explanation :**Let Tina’s weight = 1 kg

Lina’s weight = 2 kg

Neha’s weight = 1.4kg

Mina’s weight = 1.8 kg.

\(\frac{1.8x}{100}\) = 1.4

\(→\) x = \( 77 \frac{7}{9}\)

**Ans . **

37.5%

**Explanation :**Let the number of seats initially in the cinema hall be 100 and the cost of each ticket be

100.

∴ Total revenue = 100 × 100 = 10000

In second condition,

Number of seats = 125

Cost of each ticket = 110

∴ New revenue = 125 × 110 = 13750

Increase in revenue collection = . (13750 – 10000) = 3750

∴ Percentage increase = \(\frac{3750}{10000} \times 100\) = 37.5%

**Ans . **

80

**Explanation :**Total amount = x

∴ x - \(\frac{x}{5}\) - \(\frac{4x}{5} \times \frac{5}{100}\) - 120 = 1400

\(\frac{19x}{25}\) = 1520

\(→\) x = 2000

∴ Expenditure on transport = \(\frac{1}{25} \times 2000\) = 80

**Ans . **

30,000

**Explanation :**Let the total number of employees be x.

\(x \times \frac{69}{100}\) = 20700

\(→\) x = \(\frac{20700 \times 100}{69}\) = 30,000

**Ans . **

700 apples

**Explanation :**Let the fruit seller had originally x apples.

According to the question;

x – 40% of x = 420

x - \(\frac{40}{100} \times x\) = 420

\(\frac{3x}{5}\) = 420

x = 700

**Ans . **

80%

**Explanation :**Let third number = 100

First number = 120

Second number = 150

Required percentage = \(\frac{120}{150} \times 100\) = 80%

**Ans . **

\( 45 \frac{5}{11}\)%

**Explanation :**The batsman scored 3 × 4 + 8 × 6 = 60 runs by boundariesand sixes respectively.

Then, Runs scored by running

= 110 – 60 = 50

\ Required percentage = \(\frac{50}{110} \times 100\) = \(\frac{500}{11}\) =\( 45 \frac{5}{11}\)%

**Ans . **

123.75

**Explanation :**y = \(\frac{110}{100} \times 125\) = 137 5

∴ x = 90% of y

= \(\frac{90 \times 137.5}{100}\) = 123.75

**Ans . **

2.5%

**Explanation :**Error = 5.5 minutes

∴ Error per cent = \(\frac{5.5}{3 \times 60 + 40} \times 100\) = 2.5 per cent

**Ans . **

24

**Explanation :**Per cent of families having either a cow or a buffalo or both = 60 + 30 – 15 = 75

It means 25 per cent of families do not have a cow or a buffalo.

∴ Required number of families = 25% of 96 =\(96 \times \frac{25}{100}\) = 24

**Ans . **

3,850

**Explanation :**Let the amount invested at 6% = x

∴ Amount invested at 5% = (10000 – x )

According to the question

\(\frac{(10000 - x) \times 5}{100}\) - \(\frac{x \times 6}{100}\) = 76.50

\(→\) 50000 – 5x – 6x = 7650

\(→\) 50000 – 11x = 7650

\(→\) 11x = 50000 – 7650 = 42350

x = 3850

**Ans . **

3

**Explanation :**Kites of 20 are available for 19.

Hence, discount = 5%

If one gets kites of 20 for 18, discount = 10%

∴ Required answer

20 kites \(→\) 2 kites

27 kites \(→\) \(\frac{2}{20} \times 27\) = 3

**Ans . **

135

**Explanation :**First part = x and second part = y.

According to question

\(\frac{x \times 80}{100}\) = \(\frac{y \times 60}{100}\) + 3

\(→\) 4x – 3y = 15 ...(i)

Again

\(→\) 8y = 9x + 60

\(→\) 8y – 9x = 60 ...(ii)

By equation (i) × 8 + (ii) × 3,

32x – 24y = 120

24y – 27x = 180

5x = 300 \(→\) x = 60

From equation (i)

4 × 60– 3y = 15

\(→\) 3y = 240 – 15 = 225

\(→\) y = \(\frac{225}{3}\) = 75

∴ x + y = 60 + 75 = 135

**Ans . **

\( 11 \frac{31}{54}\)%

**Explanation :**Value of 100 stock = 108

∵ Income on investing 108 = \(\frac{25}{2}\)

∴ Income on investment of 27000

= \(\frac{25}{2 \times 108} \times 27000\)

= 3125

∴ Gain per cent = \(\frac{3125}{27000} \times 100\) = \( 11 \frac{31}{54}\)%

**Ans . **

4764

**Explanation :**Required mass of lead = \(8000 \times \frac{60}{100} \times (1-\frac{3}{400})\)

= \(8000 \times \frac{60}{100} \times (\frac{397}{400})\) = 4764 kg.

**Ans . **

\(\frac{2}{9}\)

**Explanation :**According to the question \(x + y = (x^2 + y^2) \times \frac{1}{5} \)

Again,

\(x + y = (x^2 - y^2) \times \frac{1}{4} \)

\(\frac{x^2 + y^2}{5} = \frac{x^2 - y^2}{4}\)

\(→\)\( 5x^2 -4x^2 = 5y^2 + 4y^2\)

\(→\)\( x^2 = 9y^2\)

\(→\) x = 3y

\(\frac{x + y}{x^2} = \frac{x^2 + y^2}{5x^2} = \frac{2}{9}\)

**Ans . **

200

**Explanation :**Let the total number of eggs bought be x.

10% of eggs are rotten.

\ Remaining eggs = \(\frac{90x}{100}\) = \(\frac{9x}{10}\)

After giving 80% of eggs to the neighbour,

Remaining eggs = \(\frac{9x}{50}\)

According to the question

\(\frac{9x}{50}\) = 36

\(→\) 9x = 36 × 50

\(→\) x = 200

**Ans . **

Rs. 4,00,000

**Explanation :**Amount with man in the

beginning = Rs. x (let).

Amount given to son and daughter = 80%.

Remaining amount = 20% of x = Rs.\(\frac{x}{5}\)

Remaining amount after donations to trust = \(\frac{x}{5} \times \frac{20}{100}\) = Rs. \(\frac{x}{25}\)

∴ \(\frac{x}{25}\) = 16000

\(→\) x = 16000 × 25 = Rs. 400000

**Ans . **

Rs. 40000

**Explanation :**Let the business man’s present earning be Rs. x.

According to the question,

\(x \times \frac{125}{100} \times \frac{96}{100} \times \frac{125}{100} \times \frac{96}{100} \times \frac{125}{100}\) = 72000

\(x \times \frac{5}{4} \times \frac{24}{25} \times \frac{5}{4} \times \frac{24}{25} \times \frac{5}{4}\) = 72000

\(x \times \frac{9}{5}\) = 100

x = Rs. 40000

**Ans . **

48025

**Explanation :**Number of blood cells in first 6 hours = \(40000 (1+\frac{10}{100})^2(1-\frac{10}{100})(1+\frac{5}{100})^2\) = 48024.9 = 48025

**Ans . **

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

**Explanation :**Let 100 pairs of shoes be bought for Rs. 100.

New budget = Rs. 160

New price = Rs. 1.20 pair of shoes

∴ Number of shoes bought = \(\frac{160}{1.2}\) = \( 133 \frac{1}{3}\)

∴ Percentage increase = \( 33 \frac{1}{33}\)%

**Ans . **

84

**Explanation :**Average of set A = \(\frac{27 + 28 + 30 + 32 + 33}{5}\) = \(\frac{150}{5}\) = 30

Case II,

New average = \(\frac{30 \times 130}{100}\) = 39

∴ 150 + k = 39 × 6 = 234

\(→\) k = 234 – 150 = 84

**Ans . **

Rs. 128

**Explanation :**Initial amount with the man = Rs. x (let).

Remaining amount after first bet = Rs. \(\frac{x}{4}\)

Remaining amount after second bet = Rs.\(\frac{x}{16}\)

Remaining amount after third bet = Rs.\(\frac{x}{64}\)

∴ \(\frac{x}{64}\) = 2

x = 128

**Ans . **

1000000

**Explanation :**Initial number of soldiers in the army = x

According to the question

\(x \times (\frac{90}{100})^3\) = 729000

\(→\) x = = 1000000

**Ans . **

16

**Explanation :**Required percentage decrease = \(\frac{25-21}{25} \times 100\) = \(\frac{400}{25}\) = 16%

**Ans . **

= 11088

**Explanation :**Required number of workers = \(8000 \times \frac{105}{100} \times \frac{110}{100} \times \frac{120}{100}\)

= 11088

**Ans . **

2000

**Explanation :**1% = 100 basis points

∴ 82.5% = 8250 basis points and 62.5% = 6250 basis points

∴ Required difference

= 8250 – 6250

= 2000 basis points

**Ans . **

\( 22 \frac{8}{11}\)%

**Explanation :**Percentage increase in sales = \(\frac{51300 - 41800}{41800} \times 100\)

= \(\frac{9500}{418}\) = \( 22 \frac{8}{11}\)%

**Ans . **

7

**Explanation :**Defective parts of 120 machine parts = \(\frac{120 \times 5}{100}\) = 6

Defective parts of 80 machine parts = \(\frac{80 \times 10}{100}\) = 8

Total defective parts = 6 + 8 = 14

∴ Required percent = \(\frac{14}{200} \times 100\) = 7%

**Ans . **

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

**Explanation :**Error = (1.55 – 1.5) metre = 0.55 metre

∴ Error per cent = \(\frac{0.05}{15} \times 100\) = \(\frac{50}{15}\) = \( 3 \frac{1}{3}\)%

**Ans . **

36500

**Explanation :**Duty payment :

Laptop \(→\) Rs \(\frac{210000 \times 10}{100}\) = Rs. 21000

Mobile phone \(→\) Rs \(\frac{100000 \times 8}{100}\) = Rs. 8000

Television set \(→\) Rs \(\frac{150000 \times 5}{100}\) = Rs. 7500

Total Duty Payment

= Rs. (21000 + 8000 + 7500)

= Rs. 36500

**Ans . **

Rs. 1600

**Explanation :**Initial amount with the person = Rs. x (let)

After an expense of \(\frac{15}{2}\)%

Remaining amount = \((100 - \frac{15}{2})% of x.\)

= \((\frac{185}{2})% of x.\) = Rs. \(\frac{185x}{200}\)

After an expense of 75% of it,

Remaining amount = \(\frac{37x}{40 \times 4}\) = \(\frac{37x}{160}\)

According to the question,

\(\frac{37x}{160}\) = 370

\(→\) 37x = 370 × 160

\(→\) x = Rs. 1600

**Ans . **

60

**Explanation :**Let the number of matches played between India and Pakistan in the first case be x.

Number of wins by Pakistan = \(\frac{60x}{100}\) = \(\frac{3x}{5}\)

\(\frac{\frac{3x}{5}}{x+30}\) = \(\frac{30}{100}\)

\(→\) 2x = x + 30

\(→\) x = 30

∴ Total number of matches = 30 + 30 = 60