Percentage - Exercise 9(C)

Increase :

(i) 375 by 4%

(ii) 500 by 3.4%

(iii) 70 by 140%

(iv) 48 by $12\dfrac{1}{2}$%

(v) 90 by 2%

Answer

(i) 375 by 4%

Given:

Increase = 4% of 375

$= \left( 375 \times \dfrac{4}{100} \right) \\[1em] = \left( 15 \times \dfrac{4}{4} \right) \quad \text{[Dividing 375 and 100 by 25]} \\[1em] = \left( 15 \times \dfrac{1}{1} \right) \\[1em] = 15$

Increased value = 375 + 15 = 390

∴ Increased value is 390

(ii) 500 by 3.4%

Increase = 3.4% of 500

$= \left( 500 \times \dfrac{3.4}{100} \right) \\[1em] = \left( 5 \times \dfrac{3.4}{1} \right) \quad \text{[Dividing 500 and 100 by 100]} \\[1em] = 5 \times 3.4 \\[1em] = 17$

Increased value = 500 + 17 = 517

∴ Increased value is 517

(iii) 70 by 140%

Increase = 140% of 70

$= \left( 70 \times \dfrac{140}{100} \right) \\[1em] = \left( 7 \times \dfrac{140}{10} \right) \quad \text{[Dividing 70 and 100 by 10]} \\[1em] = \left( 7 \times \dfrac{14}{1} \right) \quad \text{[Dividing 140 and 10 by 10]} \\[1em] = 7 \times 14 \\[1em] = 98$

Increased value = 70 + 98 = 168

∴ Increased value is 168

(iv) 48 by $12\dfrac{1}{2}$%

First, convert to an improper fraction: $12\dfrac{1}{2}$% = $\dfrac{25}{2}$%.

Increase = $\dfrac{25}{2}$% $\text{ of } 48$

$= \left( 48 \times \dfrac{25}{2 \times 100} \right) \\[1em] = \left( 48 \times \dfrac{1}{2 \times 4} \right) \quad \text{[Dividing 25 and 100 by 25]} \\[1em] = 48 \times \dfrac{1}{8} \\[1em] = 6 \times \dfrac{1}{1} \quad \text{[Dividing 48 and 8 by 8]} \\[1em] = 6$

Increased value = 48 + 6 = 54

∴ Increased value is 54

(v) 90 by 2%

Increase = 2% of 90

$= \left( 90 \times \dfrac{2}{100} \right) \\[1em] = \left( 9 \times \dfrac{2}{10} \right) \quad \text{[Dividing 90 and 100 by 10]} \\[1em] = \dfrac{18}{10} \\[1em] = 1.8$

Increased value = 90 + 1.8 = 91.8

∴ Increased value is 91.8

Decrease :

(i) 70 by 40%

(ii) 340 by 35%

(iii) 65 by 4%

(iv) 36 by $16\dfrac{2}{3}$%

(v) 260 by 1.5%

Answer

(i) 70 by 40%

Decrease = 40% of 70

$= \left( 70 \times \dfrac{40}{100} \right) \\[1em] = \left( 7 \times \dfrac{40}{10} \right) \quad \text{[Dividing 70 and 100 by 10]} \\[1em] = \left( 7 \times \dfrac{4}{1} \right) \quad \text{[Dividing 40 and 10 by 10]} \\[1em] = 28$

Decreased value = 70 - 28 = 42

∴ Decreased value is 42

(ii) 340 by 35%

Decrease = 35% of 340

$= \left( 340 \times \dfrac{35}{100} \right) \\[1em] = \left( 34 \times \dfrac{35}{10} \right) \quad \text{[Dividing 340 and 100 by 10]} \\[1em] = \left( 34 \times \dfrac{7}{2} \right) \quad \text{[Dividing 35 and 10 by 5]} \\[1em] = \left( 17 \times \dfrac{7}{1} \right) \quad \text{[Dividing 34 and 2 by 2]} \\[1em] = 17 \times 7 \\[1em] = 119$

Decreased value = 340 - 119 = 221

∴ Decreased value is 221

(iii) 65 by 4%

Decrease = 4% of 65

$= \left( 65 \times \dfrac{4}{100} \right) \\[1em] = \left( 13 \times \dfrac{4}{20} \right) \quad \text{[Dividing 65 and 100 by 5]} \\[1em] = \left( 13 \times \dfrac{1}{5} \right) \quad \text{[Dividing 4 and 20 by 5]} \\[1em] = \dfrac{13}{5} \\[1em] = 2.6$

Decreased value = 65 - 2.6 = 62.4

∴ Decreased value is 62.4

(iv) 36 by $16\dfrac{2}{3}$%

Convert to an improper fraction:

$16\dfrac{2}{3}$% = $\dfrac{50}{3}$%

Decrease = $\dfrac{50}{3}$% of 36

$= \left( 36 \times \dfrac{50}{3 \times 100} \right) \\[1em] = \left( 36 \times \dfrac{1}{3 \times 2} \right) \quad \text{[Dividing 50 and 100 by 50]} \\[1em] = 36 \times \dfrac{1}{6} \quad \text{[Dividing 36 and 6 by 6]} \\[1em] = 6$

Decreased value = 36 - 6 = 30

∴ Decreased value is 30

(v) 260 by 1.5%

Decrease = 1.5% of 260

$= \left( 260 \times \dfrac{1.5}{100} \right) \\[1em] = 2.6 \times 1.5 \quad \text{[Dividing 260 and 100 by 100]} \\[1em] = 3.9$

Decreased value = 260 - 3.9 = 256.1

∴ Decreased value is 256.1

Find a number :

(i) Which when increased by 10% becomes 66.

(ii) Which when increased by 120% becomes 77.

(iii) Which when increased by 2.5% becomes 246.

Answer

(i) Which when increased by 10% becomes 66.

Let the required number be x.

Increase = 10% of x

$= x \times \dfrac{10}{100} \\[1em] =\dfrac{10}{100}x \\[1em] = \dfrac{x}{10}$

Increased number = $x + \dfrac{x}{10} = \dfrac{11}{10}x$

$\therefore \dfrac{11}{10}x = 66 \\[1em] \Rightarrow x = \dfrac{66 \times 10}{11} \\[1em] \Rightarrow x = \dfrac{6 \times 10}{1} \\[1em] \Rightarrow x = 60 \\[1em]$

Hence, the required number is 60

(ii) Which when increased by 120% becomes 77.

Let the required number be x.

Increase = 120% of x

$= x \times \dfrac{120}{100} \\[1em] = \dfrac{6}{5}x \\[1em]$

Increased number = $x + \dfrac{6}{5}x = \dfrac{11}{5}x$

$\therefore \dfrac{11}{5}x = 77 \\[1em] \Rightarrow x = \dfrac{77 \times 5}{11} \\[1em] \Rightarrow x = \dfrac{7 \times 5}{1} \\[1em] \Rightarrow x = 35 \\[1em]$

Hence, the required number is 35

(iii) Which when increased by 2.5% becomes 246.

Let the required number be x.

Increase = 2.5% of x

$= x \times \dfrac{2.5}{100} \\[1em] = x \times \dfrac{2.5 \times 10}{100 \times 10} \\[1em] = \dfrac{25}{1000}x$

Increased number = $x + \dfrac{25}{1000}x = \dfrac{1025}{1000}x$

$\therefore \dfrac{1025}{1000}x = 246 \\[1em] \Rightarrow x = \dfrac{246 \times 1000}{1025} \\[1em] \Rightarrow x = \dfrac{246 \times 40}{41} \\[1em] \Rightarrow x = \dfrac{6 \times 40}{1} \\[1em] \Rightarrow x = 240 \\[1em]$

Hence, the required number is 240.

Find a number :

(i) Which when decreased by 35% becomes 52.

(ii) Which when decreased by 8% becomes 115.

(iii) Which when decreased by 3.4% becomes 483.

Answer

(i) Which when decreased by 35% becomes 52.

Let the required number be x.

Decrease = 35% of x

$= x \times \dfrac{35}{100} \\[1em] = x \times \dfrac{7}{20} \\[1em] = \dfrac{7}{20}x$

Decreased number = $x - \dfrac{7}{20}x = \dfrac{13}{20}x$

$\therefore \dfrac{13}{20}x = 52 \\[1em] \Rightarrow x = \dfrac{52 \times 20}{13} \\[1em] \Rightarrow x = \dfrac{4 \times 20}{1} \\[1em] \Rightarrow x = 80$

Hence, the required number is 80.

(ii) Which when decreased by 8% becomes 115.

Let the required number be x.

Decrease = 8% of x

$= x \times \dfrac{8}{100} \\[1em] = x \times \dfrac{2}{25} \\[1em] = \dfrac{2}{25}x$

Decreased number = $x - \dfrac{2}{25}x = \dfrac{23}{25}x$

$\therefore \dfrac{23}{25}x = 115 \\[1em] \Rightarrow x = \dfrac{115 \times 25}{23} \\[1em] \Rightarrow x = \dfrac{5 \times 25}{1} \\[1em] \Rightarrow x = 125$

Hence, the required number is 125.

(iii) Which when decreased by 3.4% becomes 483.

Let the required number be x.

Decrease = 3.4% of x

$= x \times \dfrac{3.4}{100} \\[1em] = x \times \dfrac{3.4 \times 10}{100 \times 10} \\[1em] = x \times \dfrac{34}{1000} \\[1em] = \dfrac{17}{500}x$

Decreased number = $x - \dfrac{17}{500}x = \dfrac{483}{500}x$

$\therefore \dfrac{483}{500}x = 483 \\[1em] \Rightarrow x = \dfrac{483 \times 500}{483} \\[1em] \Rightarrow x = \dfrac{1 \times 500}{1} \\[1em] \Rightarrow x = 500$

Hence, the required number is 500.

By what number must a given number be multiplied to increase it by 12%?

Answer

Let the number be x.

Increase in its value = 12% of x

$= x \times \dfrac{12}{100} \\[1em] = \dfrac{12}{100}x \\[1em] = \dfrac{3}{25}x$

∴ Increased value = $\Big(x + \dfrac{3}{25}x\Big) = \dfrac{28}{25}x$

Hence, for an increase of 12%, the given number should be multiplied by $\dfrac{28}{25}$.

By what number must a given number be multiplied to decrease it by 30%?

Answer

Let the number be x.

Decrease in its value = 30% of x

$= x \times \dfrac{30}{100} \\[1em] = \dfrac{3}{10}x$

∴ Decreased value = $\Big(x - \dfrac{3}{10}x\Big) = \dfrac{7}{10}x$

Hence, for an decrease of 30%, the given number should be multiplied by $\dfrac{7}{10}$.

The price of a fan increases from ₹ 3260 to ₹ 3749. Find the increase per cent in its price.

Answer

Given:

Original Price = ₹ 3260

New Price = ₹ 3749.

Increase in price = New Price - Original Price

Substituting the values in above, we get:

Increase in price = ₹ 3749 - ₹ 3260 = ₹ 489

Increase % =

$\left( \dfrac{\text{Increase in price}}{\text{Original Price}} \times 100 \right)$%

= $\left( \dfrac{489}{3260} \times 100 \right)$%

= $\left( \dfrac{489}{815} \times 25 \right)$%

= $\left( \dfrac{489}{163} \times 5 \right)$%

= $\left( \dfrac{3}{1} \times 5 \right)$%

= 15%

Hence, the increase percentage in its price = 15%.

The monthly salary of Mr Rakesh is ₹ 32500. After deducting the provident fund, he gets ₹ 29900 per month. What per cent of the salary is deducted as provided fund?

Answer

Given:

Total Salary = ₹ 32500

Net Salary = ₹ 29900.

Provident fund deduction = Total Salary - Net Salary

Substituting the values in above, we get:

Provident fund deduction = ₹ 32500 - ₹ 29900 = ₹ 2600

Deduction % =

$\left( \dfrac{\text{Provident fund deduction}}{\text{Total Salary}} \times 100 \right)$%

= $\left( \dfrac{2600}{32500} \times 100 \right)$%

= $\left( \dfrac{2600}{325} \times 1 \right)$%

= $\left( \dfrac{104}{13} \right)$%

= 8%

Percentage of the salary deducted as provided fund = 8%

A car was purchased last years for ₹ 415000. Now, its value is ₹ 356900. At what rate is the car depreciating?

Answer

Given:

Original Value = ₹ 415000

Current Value = ₹ 356900

Depreciation amount = ₹ 415000 - ₹ 356900 = ₹ 58100

Depreciation rate =

$\left( \dfrac{\text{Depreciation amount}}{\text{Original Value}} \times 100 \right)$%

$\left( \dfrac{58100}{415000} \times 100 \right)$%

= $\dfrac{58100}{4150} \times 1$%

= $\dfrac{5810}{415}$%

= 14%

Depreciation rate of the car is 14%.

On decreasing the price of a car by 6%, its value becomes ₹ 249100. What was the original price of the car?

Answer

Given:

New value = ₹ 249100

Decrease = 6%

Let original price be x.

Since the New value represents the amount left after the decrease, we can say:

94% of the Original Price = New value

94% of x = ₹ 249100

$\dfrac{94}{100}x = ₹ 249100 \\[1em] \Rightarrow x = ₹ \dfrac{249100 \times 100}{94} \\[1em] \Rightarrow x = ₹ \dfrac{2650 \times 100}{1} \\[1em] \Rightarrow x = ₹ 265000$

The original price of the car is ₹ 265000

On increasing the salary of a man by 12%, his salary is increased by ₹ 2316. What was his original salary?

Answer

Given:

Increase % = 12%

Increase Amount = ₹ 2316

Let original salary be x.

12% of x = 2316

$\dfrac{12}{100}x = ₹ 2316 \\[1em] \Rightarrow x = ₹ \dfrac{2316 \times 100}{12} \\[1em] \Rightarrow x = ₹ \dfrac{193 \times 100}{1} \\[1em] \Rightarrow x = ₹ 19300$

The original salary is ₹ 19300.

The salary of Gopal was increased by 10% and then the increased salary was decreased by 10%. Find the net increase or decrease per cent in his original salary.

Answer

Given:

Increase of salary = 10%

Decrease of salary = 10%

Let original salary be 100.

After 10% increase: 100 + 10 = 110.

After 10% decrease on 110:

110 - (10% of 110)

$= 110 - \Big(\dfrac{10}{100} \times 110\Big) \\[1em] = 110 - \Big(\dfrac{1}{10} \times 110\Big) \\[1em] = 110 - \Big(\dfrac{1}{1} \times 11\Big) \\[1em] = 110 - 11 \\[1em] = 99 \\[1em]$

Net change = 100 - 99 = 1

Net decrease = 1%.

After deducting 4% of a bill, the amount still to be paid is ₹ 1488. How much was the original bill?

Answer

Given:

Amount still to be paid = ₹ 1488

Percentage of deduction = 4%

The original bill represents the full amount, which is 100%.

The "Amount still to be paid" is what remains after the 4% has been taken away from the 100%.

Remaining Percentage = 100% - 4% = 96%

Let the original bill be x.

Since ₹ 1488 is the amount left after the 4% deduction, it is equal to 96% of the original bill.

96% of x = ₹ 1488

$\phantom{=} \dfrac{96}{100}x = ₹ 1488 \\[1em] x = ₹ \dfrac{1488 \times 100}{96} \\[1em] x = ₹ \dfrac{1488 \times 25}{24} \\[1em] x = ₹ \dfrac{62 \times 25}{1} \\[1em] x = ₹ 1550$

The original bill is ₹ 1550.

The weight of a boy was 40 kg. But, it was wrongly measured as 42 kg. Find the error per cent.

Answer

Given:

Actual weight = 40 kg

Measured weight = 42 kg.

Error = 42 - 40 = 2 kg

$\text {Error}$% = $\left( \dfrac{\text{Error}}{\text{Actual Value}} \times 100 \right)$%

= $\dfrac{2}{40} \times 100$%

= $\dfrac{1}{20} \times 100$%

= $\dfrac{1}{1} \times 5$%

= 5%

Error percentage = 5%.