To Find: The correct symbol (>, 5%. (ii) 5/10 __ 50 % Convert the fraction to a percentage: 5/10 = 1/2. We know that 1/2 is equivalent to 50%. Therefore, 5/10 = 50%. (iii) 3/11 __ 61 % To compare, let's convert the fraction to a percentage. 3/11 × 100% = 300/11% ≈ 27.27%. Since 27.27% 5% (ii) 5/10…

Q5 (ii): 5/10 __ 50 %

To Find: The correct symbol (>, 5%. (ii) 5/10 __ 50 % Convert the fraction to a percentage: 5/10 = 1/2. We know that 1/2 is equivalent to 50%. Therefore, 5/10 = 50%. (iii) 3/11 __ 61 % To compare, let's convert the fraction to a percentage. 3/11 × 100% = 300/11% ≈ 27.27%. Since 27.27% 5% (ii) 5/10…

Q5 (iii): 3/11 __ 61 %

To Find: The correct symbol (>, 5%. (ii) 5/10 __ 50 % Convert the fraction to a percentage: 5/10 = 1/2. We know that 1/2 is equivalent to 50%. Therefore, 5/10 = 50%. (iii) 3/11 __ 61 % To compare, let's convert the fraction to a percentage. 3/11 × 100% = 300/11% ≈ 27.27%. Since 27.27% 5% (ii) 5/10…

Q1 (i): 20% of __ is 40. (Answer: 200)

To Find: The missing numbers in each statement. Solution: (i) Let the missing number be x. 20% of x is 40. 0.20 × x = 40 x = 40/0.20 = 400/2 = 200 (ii) Let the missing percentage be y. y% of 150 is 30. y/100 × 150 = 30 y × 1.5 = 30 y = 30/1.5 = 300/15 = 20 So, it is 20%. (iii) Let the missing…

Q1 (ii): __% of 150 is 30.

To Find: The missing numbers in each statement. Solution: (i) Let the missing number be x. 20% of x is 40. 0.20 × x = 40 x = 40/0.20 = 400/2 = 200 (ii) Let the missing percentage be y. y% of 150 is 30. y/100 × 150 = 30 y × 1.5 = 30 y = 30/1.5 = 300/15 = 20 So, it is 20%. (iii) Let the missing…

Q1 (iii): 15% of 200 is __.

To Find: The missing numbers in each statement. Solution: (i) Let the missing number be x. 20% of x is 40. 0.20 × x = 40 x = 40/0.20 = 400/2 = 200 (ii) Let the missing percentage be y. y% of 150 is 30. y/100 × 150 = 30 y × 1.5 = 30 y = 30/1.5 = 300/15 = 20 So, it is 20%. (iii) Let the missing…

Fractions in Disguise Class 8 Mathematics NCERT Solutions

Q1Figure it Out 1

Express the following fractions as percentages.

(i)

$\frac{3}{5}$

(ii)

$\frac{7}{14}$

(iii)

$\frac{9}{20}$

(iv)

$\frac{72}{150}$

(v)

$\frac{1}{3}$

(vi)

$\frac{5}{11}$

Solution

To Find: The percentage equivalent for the given fractions.

Formula: To convert a fraction to a percentage, we multiply the fraction by 100. Percentage = Fraction

\times 100\%

Solution:

(i)

\frac{3}{5} = \frac{3}{5} \times 100\% = 3 \times 20\% = 60\%

(ii)

\frac{7}{14} = \frac{1}{2} = \frac{1}{2} \times 100\% = 50\%

(iii)

\frac{9}{20} = \frac{9}{20} \times 100\% = 9 \times 5\% = 45\%

(iv)

\frac{72}{150} = \frac{72}{150} \times 100\% = \frac{72 \times 2}{3}\% = 24 \times 2\% = 48\%

(v)

\frac{1}{3} = \frac{1}{3} \times 100\% = 33.33...\%

. This can also be written as

33\frac{1}{3}\%

.

(vi)

\frac{5}{11} = \frac{5}{11} \times 100\% = \frac{500}{11}\% = 45.45...\%

. This can also be written as

45\frac{5}{11}\%

.

Final Answer:

(i)

60\%

(ii)

50\%

(iii)

45\%

(iv)

48\%

(v)

33.33...\%

or

33\frac{1}{3}\%

(vi)

45.45...\%

or

45\frac{5}{11}\%

Q2Figure it Out 1

Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?

(i)

$10 \%$

(ii)

$15 \%$

(iii)

$25 \%$

(iv)

$60 \%$

(v)

$40 \%$

(vi)

None of these

Solution

Given: Total number of marbles = 25 Number of white marbles = 15

To Find: The percentage of white marbles.

Solution: First, we find the fraction of white marbles. Fraction of white marbles =

\frac{\text{Number of white marbles}}{\text{Total number of marbles}} = \frac{15}{25}

To convert this fraction to a percentage, we multiply by 100. Percentage of white marbles =

\frac{15}{25} \times 100\%

Simplify the fraction:

\frac{15}{25} = \frac{3}{5}

Percentage =

\frac{3}{5} \times 100\% = 3 \times 20\% = 60\%

This corresponds to option (iv).

Final Answer: The correct option is (iv)

60\%

.

Q3Figure it Out 1

In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?

Solution

Given: Total number of students = 80 Number of students who walk = 15

To Find: The percentage of students who come by walking.

Solution: First, find the fraction of students who walk. Fraction =

\frac{\text{Number of students who walk}}{\text{Total number of students}} = \frac{15}{80}

Simplify the fraction:

\frac{15}{80} = \frac{15 \div 5}{80 \div 5} = \frac{3}{16}

Now, convert the fraction to a percentage by multiplying by 100. Percentage =

\frac{3}{16} \times 100\% = \frac{300}{16}\%

\frac{300}{16} = \frac{150}{8} = \frac{75}{4} = 18.75

So, the percentage of students who come by walking is

18.75\%

.

Final Answer:

18.75\%

of the students come by walking.

Q5Figure it Out 1

Pairs of quantities are shown below. Identify and write appropriate symbols ' >', '<', '=' in the blanks. Try to do it without calculations.

(i)

$50 \%$ __ $5 \%$

(ii)

$\frac{5}{10}$ __ $50 \%$

(iii)

$\frac{3}{11}$ __ $61 \%$

(iv)

$30 \%$ __ $\frac{1}{3}$

Solution

To Find: The correct symbol (>, <, or =) for each pair.

Solution:

(i)

50 \%

__

5 \%

50 parts out of 100 is greater than 5 parts out of 100. Therefore,

50\% > 5\%

.

(ii)

\frac{5}{10}

__

50 \%

Convert the fraction to a percentage:

\frac{5}{10} = \frac{1}{2}

. We know that

\frac{1}{2}

is equivalent to

50\%

. Therefore,

\frac{5}{10} = 50\%

.

(iii)

\frac{3}{11}

__

61 \%

To compare, let's convert the fraction to a percentage.

\frac{3}{11} \times 100\% = \frac{300}{11}\% \approx 27.27\%

. Since

27.27\% < 61\%

, we have

\frac{3}{11} < 61\%

. Alternatively, without calculation, we know

\frac{3}{11}

is less than

\frac{1}{3}

(which is

\approx 33\%

), and

61\%

is much larger. Therefore,

\frac{3}{11} < 61\%

.

(iv)

30 \%

__

\frac{1}{3}

Convert

\frac{1}{3}

to a percentage:

\frac{1}{3} \times 100\% = 33.33...\%

. Since

30\% < 33.33...\%

, we have

30\% < \frac{1}{3}

.

Final Answer:

(i)

50\% > 5\%

(ii)

\frac{5}{10} = 50\%

(iii)

\frac{3}{11} < 61\%

(iv)

30\% < \frac{1}{3}

Q1Figure it Out 2

Find the missing numbers. The first problem has been worked out.

(i)

20% of __ is 40. (Answer: 200)

(ii)

__% of 150 is 30.

(iii)

15% of 200 is __.

(iv)

120% of 50 is __.

Solution

To Find: The missing numbers in each statement.

Solution:

(i)

Let the missing number be

x

.

20\%

of

x

is 40.

0.20 \times x = 40

x = \frac{40}{0.20} = \frac{400}{2} = 200

(ii)

Let the missing percentage be

y

.

y\%

of 150 is 30.

\frac{y}{100} \times 150 = 30

y \times 1.5 = 30

y = \frac{30}{1.5} = \frac{300}{15} = 20

So, it is

20\%

.

(iii)

Let the missing number be

z

.

15\%

of 200 is

z

.

z = \frac{15}{100} \times 200

z = 15 \times 2 = 30

(iv)

Let the missing number be

w

.

120\%

of 50 is

w

.

w = \frac{120}{100} \times 50

w = 1.2 \times 50 = 60

Final Answer:

(i)

200

(ii)

20

(iii)

30

(iv)

60

Q2Figure it Out 2

Find the value of the following and also draw their bar models.

(i)

$25 \%$ of 160

(ii)

$16 \%$ of 250

(iii)

$62 \%$ of 360

(iv)

$140 \%$ of 40

(v)

$1 \%$ of 1 hour

(vi)

$7 \%$ of 10 kg

Solution

To Find: The value of each expression.

Solution:

(i)

25 \%

of 160

25\% = \frac{1}{4}

. Value =

\frac{1}{4} \times 160 = 40

. Bar model description: A bar representing the whole (160) is divided into 4 equal parts. One of these parts is shaded, representing

25\%

or 40.

(ii)

16 \%

of 250 Value =

\frac{16}{100} \times 250 = \frac{16 \times 2.5}{1} = 40

. Bar model description: A bar representing 250. A small portion,

16\%

, is shaded. This is a bit less than one-fifth of the bar.

(iii)

62 \%

of 360 Value =

\frac{62}{100} \times 360 = 0.62 \times 360 = 223.2

. Bar model description: A bar representing 360. A portion slightly larger than half (

62\%

) is shaded.

(iv)

140 \%

of 40 Value =

\frac{140}{100} \times 40 = 1.4 \times 40 = 56

. Bar model description: A bar representing the original amount 40 (which is

100\%

). An additional segment, representing

40\%

of 40 (which is 16), is attached to it. The total length of the combined bar represents 56.

(v)

1 \%

of 1 hour 1 hour = 60 minutes. Value =

\frac{1}{100} \times 60 \text{ minutes} = 0.6 \text{ minutes}

. To convert to seconds:

0.6 \times 60 = 36

seconds. Bar model description: A bar representing 60 minutes. A very thin slice (

1\%

) is shaded, representing 36 seconds.

(vi)

7 \%

of 10 kg Value =

\frac{7}{100} \times 10 \text{ kg} = 0.7 \text{ kg}

. To convert to grams:

0.7 \times 1000 = 700

grams. Bar model description: A bar representing 10 kg. A small slice (

7\%

) is shaded, representing 0.7 kg or 700 g.

Final Answer:

(i)

40

(ii)

40

(iii)

223.2

(iv)

56

(v)

0.6 minutes or 36 seconds

(vi)

0.7 kg or 700 g

Q3Figure it Out 2

Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up $\frac{3}{4}$ of the deep orange paint?

Solution

Given: Total volume of deep orange paint = 60 ml Fraction of red paint in the mixture =

\frac{3}{4}

To Find: The amount of red paint used.

Solution: The amount of red paint is

\frac{3}{4}

of the total mixture. Amount of red paint =

\frac{3}{4} \times 60 \text{ ml}

Amount of red paint =

3 \times \frac{60}{4} \text{ ml} = 3 \times 15 \text{ ml} = 45 \text{ ml}

Final Answer: Surya used 45 ml of red paint.

Q4Figure it Out 2

Pairs of quantities are shown below. Identify and write appropriate symbols ' >', '<', '=' in the boxes. Visualising or estimating can help. Compute only if necessary or for verification.

(i)

$50 \%$ of $510 \square 50 \%$ of 515

(ii)

$37 \%$ of $148 \square 73 \%$ of 148

(iii)

$29 \%$ of $43 \square 92 \%$ of 110

(iv)

$30 \%$ of $40 \square 40 \%$ of 50

(v)

$45 \%$ of $200 \square 10 \%$ of 490

(vi)

$30 \%$ of $80 \square 24 \%$ of 64

Solution

To Find: The correct symbol (>, <, or =) for each pair.

Solution:

(i)

50 \%

of

510 \square 50 \%

of 515 Since the percentage is the same (

50\%

), the result depends on the base number. As

510 < 515

,

50\%

of 510 will be less than

50\%

of 515. Answer:

<

(ii)

37 \%

of

148 \square 73 \%

of 148 Since the base number is the same (148), the result depends on the percentage. As

37\% < 73\%

,

37\%

of 148 will be less than

73\%

of 148. Answer:

<

(iii)

29 \%

of

43 \square 92 \%

of 110 Estimate: LHS is approx.

30\%

of 40, which is 12. RHS is approx.

90\%

of 110, which is 99. Clearly, LHS is much smaller than RHS. Calculation: LHS =

0.29 \times 43 = 12.47

. RHS =

0.92 \times 110 = 101.2

. Answer:

<

(iv)

30 \%

of

40 \square 40 \%

of 50 LHS =

0.30 \times 40 = 12

. RHS =

0.40 \times 50 = 20

. Since

12 < 20

. Answer:

<

(v)

45 \%

of

200 \square 10 \%

of 490 LHS =

0.45 \times 200 = 90

. RHS =

0.10 \times 490 = 49

. Since

90 > 49

. Answer:

>

(vi)

30 \%

of

80 \square 24 \%

of 64 LHS =

0.30 \times 80 = 24

. RHS =

0.24 \times 64 = 15.36

. Since

24 > 15.36

. Answer:

>

Final Answer:

(i)

<

(ii)

<

(iii)

<

(iv)

<

(v)

>

(vi)

>

Q5Figure it Out 2

Fill in the blanks appropriately:

(i)

$30 \%$ of $k$ is $70, 60 \%$ of $k$ is , $90 \%$ of $k$ is , $120 \%$ of $k$ is .

(ii)

$100 \%$ of $m$ is $215, 10 \%$ of $m$ is , 1% of $m$ is , 6% of $m$ is .

(iii)

$90 \%$ of $n$ is $270, 9 \%$ of $n$ is , $18 \%$ of $n$ is , 100% of $n$ is .

(iv)

Make 2 more such questions and challenge your peers.

Solution

To Find: The values for the blanks.

Solution:

(i)

Given:

30\%

of

k = 70

.

$60\%$ of $k$ is $2 \times (30\% \text{ of } k) = 2 \times 70 = 140$ .
$90\%$ of $k$ is $3 \times (30\% \text{ of } k) = 3 \times 70 = 210$ .
$120\%$ of $k$ is $4 \times (30\% \text{ of } k) = 4 \times 70 = 280$ .

(ii)

Given:

100\%

of

m = 215

. This means

m=215

.

$10\%$ of $m = \frac{10}{100} \times 215 = 0.1 \times 215 = 21.5$ .
$1\%$ of $m = \frac{1}{100} \times 215 = 0.01 \times 215 = 2.15$ .
$6\%$ of $m = 6 \times (1\% \text{ of } m) = 6 \times 2.15 = 12.9$ .

(iii)

Given:

90\%

of

n = 270

.

From this, $10\%$ of $n = \frac{270}{9} = 30$ . Therefore, $100\%$ of $n = 10 \times 30 = 300$ .
$9\%$ of $n = \frac{9}{100} \times 300 = 27$ . (Alternatively, $9\%$ is one-tenth of $90\%$ , so $270/10=27$ ).
$18\%$ of $n = 2 \times (9\% \text{ of } n) = 2 \times 27 = 54$ .
$100\%$ of $n = 300$ .

(iv)

This part asks to create questions, which is not a solvable problem.

Final Answer:

(i)

140, 210, 280

(ii)

21.5, 2.15, 12.9

(iii)

27, 54, 300

Q6Figure it Out 2

Fill in the blanks:

(i)

3 is ____ % of 300.

(ii)

____ is $40 \%$ of 4.

(iii)

40 is $80 \%$ of ____ .

Solution

To Find: The values for the blanks.

Solution:

(i)

Let

x

be the percentage. We need to find what percentage of 300 is 3.

x = \frac{3}{300} \times 100 = \frac{1}{100} \times 100 = 1

So, 3 is

1\%

of 300.

(ii)

We need to find

40\%

of 4. Value =

\frac{40}{100} \times 4 = 0.4 \times 4 = 1.6

. So, 1.6 is

40\%

of 4.

(iii)

Let

y

be the number. 40 is

80\%

of

y

.

40 = \frac{80}{100} \times y

40 = 0.8 \times y

y = \frac{40}{0.8} = \frac{400}{8} = 50

So, 40 is

80\%

of 50.

Final Answer:

(i)

1

(ii)

1.6

(iii)

50

Q7Figure it Out 2

Is $10 \%$ of a day longer than $1 \%$ of a week? Create such questions and challenge your peers.

Solution

To Find: Which quantity is larger:

10\%

of a day or

1\%

of a week.

Solution: First, let's convert both quantities to the same unit, for example, hours.

A day has 24 hours.

10\%

of a day =

10\% \times 24

hours

= \frac{10}{100} \times 24 = 0.1 \times 24 = 2.4

hours.

A week has 7 days, and each day has 24 hours. Total hours in a week =

7 \times 24 = 168

hours.

1\%

of a week =

1\% \times 168

hours

= \frac{1}{100} \times 168 = 1.68

hours.

Now, we compare the two values:

2.4

hours versus

1.68

hours. Since

2.4 > 1.68

,

10\%

of a day is longer than

1\%

of a week.

Final Answer: Yes,

10\%

of a day is longer than

1\%

of a week.

Q8Figure it Out 2

Mariam's farm has a peculiar bull. One day she gave the bull 2 units of fodder and the bull ate 1 unit. The next day, she gave the bull 3 units of fodder and the bull ate 2 units. The day after, she gave the bull 4 units and the bull ate 3 units. This continued, and on the 99th day she gave the bull 100 units and the bull ate 99 units. Represent these quantities as percentages. This task can be distributed among the class. What do you observe?

Solution

To Find: The percentage of fodder eaten each day and to observe the pattern.

Solution: We calculate the percentage of fodder eaten for each given day. Percentage eaten =

\frac{\text{Fodder eaten}}{\text{Fodder given}} \times 100\%

Day 1: Gave 2, ate 1. Percentage =

\frac{1}{2} \times 100\% = 50\%

.

Day 2: Gave 3, ate 2. Percentage =

\frac{2}{3} \times 100\% \approx 66.67\%

.

Day 3: Gave 4, ate 3. Percentage =

\frac{3}{4} \times 100\% = 75\%

.

This pattern continues. On day 'n', the bull is given n+1 units and eats n units. So, on the

k^{th}

day, the bull is given

k+1

units and eats

k

units. The percentage eaten on the

k^{th}

day is

\frac{k}{k+1} \times 100\%

.

On the 99th day: Gave 100, ate 99. Percentage =

\frac{99}{100} \times 100\% = 99\%

.

Observation: The percentage of fodder eaten by the bull increases every day. The fraction of fodder eaten is of the form

\frac{n}{n+1}

. As the number of days (

n

) increases, this fraction gets closer and closer to 1, and therefore the percentage gets closer and closer to

100\%

. The bull becomes more 'efficient' in its eating over time.

Final Answer: Day 1:

50\%

Day 2:

66.67\%

Day 3:

75\%

... Day 99:

99\%

The observation is that the percentage of fodder eaten increases each day and approaches

100\%

.

Q9Figure it Out 2

Workers in a coffee plantation take 18 days to pick coffee berries in $20\%$ of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?

Solution

Given: Time taken to pick

20\%

of the plantation = 18 days.

To Find:

Time to complete the work for the entire ( $100\%$ ) plantation.
Why the assumption of a constant rate of work is necessary.

Solution:

Let's find the time required to pick the entire plantation. The entire plantation is $100\%$ . We can find how many '20%' blocks are in '100%'. $\frac{100\%}{20\%} = 5$ . So, the total work is 5 times the work done in 18 days. Assuming the rate of work is constant, the total time required will be: Total time = $5 \times (\text{time for } 20\%) = 5 \times 18 \text{ days} = 90$ days.
The assumption that the rate of work stays the same is necessary because many factors could change the rate of work over a long period like 90 days. These factors include:

Number of workers: The number of workers might change.
Worker efficiency: Workers might get tired and work slower, or become more skilled and work faster.
Weather conditions: Bad weather could slow down or stop the work.
Plantation yield: The density of coffee berries might not be uniform across the plantation. Some areas might have more berries and take longer to pick. Without this assumption, we cannot use a simple proportion to calculate the total time.

Final Answer: It will take 90 days to complete the picking work for the entire plantation. The assumption of a constant rate of work is necessary because it ensures a direct proportional relationship between the area picked and the time taken, ignoring real-world variables that could affect the work speed.

Q10Figure it Out 2

The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is $10 \% : 80 \% : 10 \%$ . If he wants to conduct a training of 90 minutes. How long should each activity be done?

Solution

Given: Total training time = 90 minutes Ratio of activities (warm up : play : cool down) =

10\% : 80\% : 10\%

To Find: The duration of each activity.

Solution: We calculate the time for each activity by taking its percentage of the total time.

Warm up duration:

10\%

of 90 minutes =

\frac{10}{100} \times 90 = 0.1 \times 90 = 9

minutes.

Play duration:

80\%

of 90 minutes =

\frac{80}{100} \times 90 = 0.8 \times 90 = 72

minutes.

Cool down duration:

10\%

of 90 minutes =

\frac{10}{100} \times 90 = 0.1 \times 90 = 9

minutes.

Check: The sum of durations should be 90 minutes.

9 + 72 + 9 = 90

minutes. The calculation is correct.

Final Answer:

Warm up: 9 minutes
Play: 72 minutes
Cool down: 9 minutes

Q11Figure it Out 2

An estimated $90\%$ of the world's population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year's worldwide population.

Solution

Given: Percentage of world population in the Northern Hemisphere =

90\%

To Find: The approximate number of people living in the Northern Hemisphere.

Solution: The question requires using the current worldwide population. As of the early 2020s, the world population is approximately 8 billion people. Let's use this value for calculation. Worldwide population

\approx 8,000,000,000

Number of people in the Northern Hemisphere =

90\%

of 8 billion

= \frac{90}{100} \times 8,000,000,000

= 0.9 \times 8,000,000,000

= 7,200,000,000

This is 7.2 billion people.

Final Answer: Based on a world population of approximately 8 billion, the number of people living in the Northern Hemisphere is approximately 7.2 billion.

Q12Figure it Out 2

A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions-Rava: 40%, Sugar: 40%, and Ghee: 20%.

(i)

If you want to make halwa for 8 people, what is the proportion of each of the above ingredients?

(ii)

If the total weight of the ingredients is 2 kg , how much rava, sugar and ghee are present?

Solution

Given: Recipe proportions for 4 people: Rava 40%, Sugar 40%, Ghee 20%.

Solution: (i) Proportions for 8 people: Making halwa for 8 people means doubling the quantity of the recipe for 4 people. However, the proportions or the percentage composition of the ingredients remain the same to maintain the taste and texture of the dish. The absolute amount of each ingredient will double, but the ratio between them stays constant.

Final Answer (i): The proportions remain the same: Rava: 40%, Sugar: 40%, and Ghee: 20%.

(ii) Weight of ingredients for a 2 kg mixture: Total weight of ingredients = 2 kg = 2000 g. We need to calculate the weight of each ingredient based on its percentage.

Weight of Rava:

40\%

of 2000 g =

\frac{40}{100} \times 2000 = 0.4 \times 2000 = 800

g.

Weight of Sugar:

40\%

of 2000 g =

\frac{40}{100} \times 2000 = 0.4 \times 2000 = 800

g.

Weight of Ghee:

20\%

of 2000 g =

\frac{20}{100} \times 2000 = 0.2 \times 2000 = 400

g.

Check:

800 \text{ g} + 800 \text{ g} + 400 \text{ g} = 2000 \text{ g} = 2

kg. The calculation is correct.

Final Answer (ii):

Rava: 800 g
Sugar: 800 g
Ghee: 400 g

Q1Figure it Out 3

If a shopkeeper buys a geometry box for ₹75 and sells it for ₹110, what is his profit margin with respect to the cost?

Solution

Given: Cost Price (CP) of the geometry box = ₹75 Selling Price (SP) of the geometry box = ₹110

To Find: The profit margin (profit percentage) with respect to the cost.

Solution: First, calculate the profit. Profit = Selling Price - Cost Price Profit = ₹110 - ₹75 = ₹35

Now, calculate the profit percentage with respect to the cost price. Profit Percentage =

\frac{\text{Profit}}{\text{Cost Price}} \times 100\%

Profit Percentage =

\frac{35}{75} \times 100\%

Simplify the fraction:

\frac{35}{75} = \frac{7}{15}

Profit Percentage =

\frac{7}{15} \times 100\% = \frac{700}{15}\% = \frac{140}{3}\% \approx 46.67\%

Final Answer: The shopkeeper's profit margin is

46.67\%

(or

46\frac{2}{3}\%

).

Q2Figure it Out 3

I am a carpenter and I make chairs. The cost of materials for a chair is ₹475 and I want to have a profit margin of 50%. At what price should I sell a chair?

Solution

Given: Cost Price (CP) of materials for a chair = ₹475 Desired profit margin = 50%

To Find: The selling price (SP) of the chair.

Solution: The profit margin is calculated on the cost price. Profit =

50\%

of Cost Price Profit =

\frac{50}{100} \times 475 = 0.5 \times 475 = ₹237.50

The selling price is the sum of the cost price and the profit. Selling Price = Cost Price + Profit Selling Price = ₹475 + ₹237.50 = ₹712.50

Final Answer: The carpenter should sell the chair for ₹712.50.

Q3Figure it Out 3

Samson bought a car for ₹4,40,000 after getting a 15% discount from the car dealer. What was the original price of the car?

Solution

Given: Selling Price (SP) of the car = ₹4,40,000 Discount = 15%

To Find: The original price (Marked Price, MP) of the car.

Let: The original price be

MP

.

Equation: A 15% discount means the selling price is

100\% - 15\% = 85\%

of the original price. SP =

85\%

of MP SP =

0.85 \times MP

Solution: We are given SP = ₹4,40,000.

4,40,000 = 0.85 \times MP

MP = \frac{4,40,000}{0.85}

MP = \frac{44,000,000}{85} \approx 517,647.06

Final Answer: The original price of the car was approximately ₹5,17,647.

Q3Figure it Out 3

The total sales of a company (also called revenue) was ₹2.5 crore last year. They had a healthy profit margin of 25%. What was the total expenditure (costs) of the company last year?

Solution

Given: Total Sales (Revenue or Selling Price, SP) = ₹2.5 crore Profit Margin = 25% Profit margin is usually calculated with respect to the cost.

To Find: The total expenditure (Cost Price, CP) of the company.

Let: The total expenditure be

C

.

Equation: Profit =

25\%

of Cost =

0.25 \times C

Selling Price = Cost Price + Profit SP =

C + 0.25 \times C = 1.25 \times C

Solution: We are given SP = ₹2.5 crore.

2.5 \text{ crore} = 1.25 \times C

C = \frac{2.5 \text{ crore}}{1.25}

C = 2 \text{ crore}

Final Answer: The total expenditure of the company last year was ₹2 crore.

Q4Figure it Out 3

A clothing shop offers a 25% discount on all shirts. If the original price of a shirt is ₹300, how much will Anwar have to pay to buy this shirt?

Solution

Given: Original Price (Marked Price, MP) of the shirt = ₹300 Discount = 25%

To Find: The price Anwar has to pay (Selling Price, SP).

Solution: First, calculate the discount amount. Discount Amount =

25\%

of Original Price Discount Amount =

\frac{25}{100} \times 300 = 0.25 \times 300 = ₹75

The price to pay is the original price minus the discount. Selling Price = Original Price - Discount Amount Selling Price = ₹300 - ₹75 = ₹225

Alternatively, a 25% discount means the paying price is

100\% - 25\% = 75\%

of the original price. Selling Price =

75\%

of ₹300 =

0.75 \times 300 = ₹225

.

Final Answer: Anwar will have to pay ₹225 to buy the shirt.

Q4Figure it Out 3

1600 people voted in an election and the winner got 500 votes. What percent of the total votes did the winner get? Can you guess the minimum number of candidates who stood for the election?

Solution

Given: Total votes = 1600 Votes for the winner = 500

Part 1: Percentage of votes for the winner

To Find: The percentage of total votes the winner received.

Solution: Percentage =

\frac{\text{Votes for winner}}{\text{Total votes}} \times 100\%

Percentage =

\frac{500}{1600} \times 100\% = \frac{5}{16} \times 100\% = \frac{500}{16}\% = \frac{125}{4}\% = 31.25\%

Part 2: Minimum number of candidates

To Find: The minimum number of candidates.

Reasoning: The winner received 500 votes. This means every other candidate must have received fewer than 500 votes. The number of remaining votes is

1600 - 500 = 1100

. These 1100 votes must be distributed among the other candidates such that no single candidate gets 500 or more votes. Let's test the minimum number of other candidates needed.

If there is 1 other candidate, they would have 1100 votes, making them the winner, which contradicts the problem.
If there are 2 other candidates, the 1100 votes are split between them. To prevent one from being a winner, their votes must be less than 500. If we split 1100 into two parts, at least one part must be $\geq 1100/2 = 550$ . This person would be the winner. So, 2 other candidates is not possible.
If there are 3 other candidates, can we split 1100 votes into 3 parts, each less than 500? Yes. For example, 400, 400, and 300. ( $400+400+300 = 1100$ ). In this case, the candidate with 500 votes is the winner. So, we need at least 3 other candidates besides the winner. Minimum total candidates = 1 (winner) + 3 (others) = 4.

Final Answer: The winner got

31.25\%

of the total votes. The minimum number of candidates who stood for the election is 4.

Q5Figure it Out 3

The price of 1 kg of rice was ₹38 in 2024. It is ₹42 in 2025. What is the rate of inflation? (Inflation is the percentage increase in prices.)

Solution

Given: Original Price (in 2024) = ₹38 New Price (in 2025) = ₹42

To Find: The rate of inflation (percentage increase).

Formula: Rate of Inflation =

\frac{\text{Increase in Price}}{\text{Original Price}} \times 100\%

Solution: First, find the increase in price. Increase in Price = New Price - Original Price = ₹42 - ₹38 = ₹4

Now, calculate the rate of inflation. Rate of Inflation =

\frac{4}{38} \times 100\% = \frac{2}{19} \times 100\% = \frac{200}{19}\% \approx 10.53\%

Final Answer: The rate of inflation is approximately

10.53\%

.

Q5Figure it Out 3

The petrol price in 2015 was ₹60 and ₹100 in 2025. What is the percentage increase in the price of petrol?

(i)

50%

(ii)

40%

(iii)

60%

(iv)

66.66%

(v)

140%

(vi)

160.66%

Solution

Given: Original Price (in 2015) = ₹60 New Price (in 2025) = ₹100

To Find: The percentage increase in the price.

Formula: Percentage Increase =

\frac{\text{Amount of Increase}}{\text{Original Amount}} \times 100\%

Solution: First, find the amount of increase. Amount of Increase = New Price - Original Price = ₹100 - ₹60 = ₹40

Now, calculate the percentage increase. Percentage Increase =

\frac{40}{60} \times 100\% = \frac{2}{3} \times 100\% \approx 66.66\%

This corresponds to option (iv).

Final Answer: The correct option is (iv) 66.66%.

Q6Figure it Out 3

A number increased by $20\%$ becomes 90. What is the number?

Solution

Given: A number increased by

20\%

is 90.

To Find: The original number.

Let: The original number be

x

.

Equation: The number increased by

20\%

can be written as

x + 20\% \text{ of } x

.

x + \frac{20}{100}x = 90

x + 0.2x = 90

1.2x = 90

Solution:

x = \frac{90}{1.2} = \frac{900}{12}

x = \frac{300}{4} = 75

Final Answer: The number is 75.

Q7Figure it Out 3

A milkman sold two buffaloes for ₹80,000 each. On one of them, he made a profit of $5 \%$ and on the other a loss of $10 \%$ . Find his overall profit or loss.

Solution

Given: Selling Price (SP) of each buffalo = ₹80,000 Profit on first buffalo = 5% Loss on second buffalo = 10%

To Find: The overall profit or loss.

Solution: First, we need to find the Cost Price (CP) for each buffalo.

For the first buffalo (5% profit): Let its cost price be

CP_1

. SP =

CP_1 + 5\% \text{ of } CP_1 = 1.05 \times CP_1

80,000 = 1.05 \times CP_1

CP_1 = \frac{80,000}{1.05} \approx ₹76,190.48

Profit on this buffalo =

SP - CP_1 = 80,000 - 76,190.48 = ₹3,809.52

For the second buffalo (10% loss): Let its cost price be

CP_2

. SP =

CP_2 - 10\% \text{ of } CP_2 = 0.90 \times CP_2

80,000 = 0.90 \times CP_2

CP_2 = \frac{80,000}{0.90} \approx ₹88,888.89

Loss on this buffalo =

CP_2 - SP = 88,888.89 - 80,000 = ₹8,888.89

Overall Profit or Loss: Total Selling Price = ₹80,000 + ₹80,000 = ₹1,60,000 Total Cost Price =

CP_1 + CP_2 = ₹76,190.48 + ₹88,888.89 = ₹1,65,079.37

Since Total CP > Total SP, there is an overall loss. Overall Loss = Total CP - Total SP Overall Loss = ₹1,65,079.37 - ₹1,60,000 = ₹5,079.37

Final Answer: The milkman had an overall loss of approximately ₹5,079.37.

Q8Figure it Out 3

The population of elephants in a national park increased by $5 \%$ in the last decade. If the population of the elephants last decade is $p$ , the population now is

(i)

$p \times 0.5$

(ii)

$p \times 0.05$

(iii)

$p \times 1.5$

(iv)

$p \times 1.05$

(v)

$p+1.50$

Solution

Given: Population last decade =

p

Increase in population = 5%

To Find: The expression for the current population.

Solution: The new population is the original population plus the increase. Increase =

5\%

of

p = \frac{5}{100} \times p = 0.05p

New Population = Original Population + Increase New Population =

p + 0.05p

Factoring out

p

, we get: New Population =

p(1 + 0.05) = p(1.05) = p \times 1.05

This matches option (iv).

Final Answer: The correct option is (iv)

p \times 1.05

.

Q9Figure it Out 3

Which of the following statement(s) mean the same as - "The demand for cameras has fallen by $85\%$ in the last decade"?

(i)

The demand now is 85% of the demand a decade ago.

(ii)

The demand a decade ago was 85% of the demand now.

(iii)

The demand now is 15% of the demand a decade ago.

(iv)

The demand a decade ago was 15% of the demand now.

(v)

The demand a decade ago was 185% of the demand now.

(vi)

The demand now is 185% of the demand a decade ago.

Solution

Given: The demand for cameras has fallen by

85\%

in the last decade.

To Find: The equivalent statement(s).

Let:

D_{old}

be the demand a decade ago.

D_{now}

be the demand now.

Analysis: A fall of

85\%

means the current demand is the original demand minus the fall. Fall =

85\%

of

D_{old} = 0.85 \times D_{old}

D_{now} = D_{old} - (85\% \text{ of } D_{old})

D_{now} = D_{old} - 0.85 \times D_{old}

D_{now} = D_{old}(1 - 0.85)

D_{now} = 0.15 \times D_{old}

This means "The demand now is

15\%

of the demand a decade ago." This matches statement (iii).

Let's check other statements:

(i)

False. The demand is

15\%

, not

85\%

. A fall of

85\%

is not the same as being

85\%

.

(iii)

True. As derived above.

(ii)

, (iv), (v), (vi) are incorrect as they reverse the relationship or use incorrect percentages.

Final Answer: The correct statement is (iii) The demand now is 15% of the demand a decade ago.

Q1Figure it Out 4

Bank of Yahapur offers an interest of $10\%$ p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually.

Solution

Given: Principal (P) = ₹20,000 Rate of interest (r) =

10\%

p.a. = 0.10 Time (t) = 2 years

To Find: The final amount with and without compounding.

Solution:

1. Without Compounding (Simple Interest): Formula: Amount (A) =

P(1 + rt)

A = 20,000(1 + 0.10 \times 2)

A = 20,000(1 + 0.20)

A = 20,000(1.2) = ₹24,000

2. With Compounding Annually: Formula: Amount (A) =

P(1 + r)^t

A = 20,000(1 + 0.10)^2

A = 20,000(1.1)^2

A = 20,000(1.21) = ₹24,200

Comparison: The amount received with compounding (₹24,200) is ₹200 more than the amount received without compounding (₹24,000).

Final Answer: Without compounding, one gets ₹24,000. With compounding, one gets ₹24,200.

Q2Figure it Out 4

Bank of Wahapur offers an interest of $5\%$ p.a. Compare how much one gets if one deposits ₹20,000 for a period of 4 years with compounding and without compounding annually.

Solution

Given: Principal (P) = ₹20,000 Rate of interest (r) =

5\%

p.a. = 0.05 Time (t) = 4 years

To Find: The final amount with and without compounding.

Solution:

1. Without Compounding (Simple Interest): Formula: Amount (A) =

P(1 + rt)

A = 20,000(1 + 0.05 \times 4)

A = 20,000(1 + 0.20)

A = 20,000(1.2) = ₹24,000

2. With Compounding Annually: Formula: Amount (A) =

P(1 + r)^t

A = 20,000(1 + 0.05)^4

A = 20,000(1.05)^4

Calculation for

(1.05)^4

:

(1.05)^2 = 1.1025

.

(1.05)^4 = (1.1025)^2 = 1.21550625

.

A = 20,000(1.21550625) = ₹24,310.125

Comparison: The amount received with compounding (₹24,310.13) is ₹310.13 more than the amount received without compounding (₹24,000).

Final Answer: Without compounding, one gets ₹24,000. With compounding, one gets ₹24,310.13 (rounded to two decimal places).

Q3Figure it Out 4

Do you observe anything interesting in the solutions of the two questions above? Share and discuss.

Solution

Observation: In both Question 1 (10% for 2 years) and Question 2 (5% for 4 years), the total simple interest earned is the same.

In Q1, total simple interest rate is $r \times t = 10\% \times 2 = 20\%$ . Amount = $20000(1.2) = ₹24,000$ .
In Q2, total simple interest rate is $r \times t = 5\% \times 4 = 20\%$ . Amount = $20000(1.2) = ₹24,000$ . This is because the product of rate and time ( $rt$ ) is the same in both cases ( $0.10 \times 2 = 0.20$ and $0.05 \times 4 = 0.20$ ).

However, the amounts from compounding are different.

In Q1, the final amount is $₹24,200$ . The effective growth factor is $(1.1)^2 = 1.21$ .
In Q2, the final amount is $₹24,310.13$ . The effective growth factor is $(1.05)^4 \approx 1.2155$ .

Conclusion: Even though the total simple interest is the same, the compound interest is greater in the second case (lower rate for a longer period). This shows that for the same overall simple interest rate (

r \times t

), compounding more frequently (or over more periods) results in a larger final amount. The effect of compounding becomes more powerful over longer periods of time, even with a smaller interest rate.

Q4Figure it Out 4

Jasmine invests amount 'p' for 4 years at an interest of $6\%$ p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done?

(i)

$p \times 6 \times 4$

(ii)

$p \times 0.6 \times 4$

(iii)

$p \times \frac{0.6}{100} \times 4$

(iv)

$p \times \frac{0.06}{100} \times 4$

(v)

$p \times 1.6 \times 4$

(vi)

$p \times 1.06 \times 4$

(vii)

$p + (p \times 0.06 \times 4)$

Solution

Given: Principal =

p

Time (t) = 4 years Rate (r) =

6\%

p.a. = 0.06 Compounding is not done (Simple Interest).

To Find: The correct expression for the total amount.

Formula: For simple interest, the total amount (A) is given by:

A = \text{Principal} + \text{Simple Interest}

A = P + (P \times r \times t)

Which can also be written as

A = P(1 + rt)

.

Solution: Substituting the given values into the formula:

A = p + (p \times 0.06 \times 4)

Now, let's check the given options:

(i)

p \times 6 \times 4

: This is incorrect as it uses

r=6

instead of

r=0.06

and calculates only interest, not amount.

(ii)

p \times 0.6 \times 4

: Incorrect rate.

(iii)

p \times \frac{0.6}{100} \times 4

: Incorrect rate.

(iv)

p \times \frac{0.06}{100} \times 4

: Incorrect rate.

(v)

p \times 1.6 \times 4

: Incorrect formula.

(vi)

p \times 1.06 \times 4

: Incorrect formula.

(vii)

p + (p \times 0.06 \times 4)

: This matches our derived formula for the total amount.

Final Answer: The correct expression is (vii)

p + (p \times 0.06 \times 4)

.

Q5Figure it Out 4

The post office offers an interest of $7\%$ p.a. How much interest would one get if one invests ₹50,000 for 3 years without compounding? How much more would one get if it was compounded?

Solution

Given: Principal (P) = ₹50,000 Rate (r) =

7\%

p.a. = 0.07 Time (t) = 3 years

Part 1: Interest without compounding (Simple Interest)

Formula: Simple Interest (SI) =

P \times r \times t

Solution: SI =

50,000 \times 0.07 \times 3

SI =

50,000 \times 0.21 = ₹10,500

Part 2: How much more with compounding

Formula: Amount with compounding (A) =

P(1 + r)^t

Compound Interest (CI) =

A - P

Solution:

A = 50,000(1 + 0.07)^3 = 50,000(1.07)^3

(1.07)^3 = 1.225043

A = 50,000 \times 1.225043 = ₹61,252.15

CI = ₹61,252.15 - ₹50,000 = ₹11,252.15

Difference = Compound Interest - Simple Interest Difference = ₹11,252.15 - ₹10,500 = ₹752.15

Final Answer: One would get ₹10,500 interest without compounding. If it was compounded, one would get ₹752.15 more.

Q6Figure it Out 4

Giridhar borrows a loan of ₹ 12,500 at $12\%$ per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at $10\%$ per annum, compounded annually. Who pays more interest and by how much?

Solution

Given: Principal (P) = ₹12,500 Time (t) = 3 years

Giridhar's Loan: Rate (

r_G

) =

12\%

p.a. = 0.12, without compounding (simple interest).

Raghava's Loan: Rate (

r_R

) =

10\%

p.a. = 0.10, compounded annually.

To Find: Who pays more interest and by how much.

Solution:

Interest paid by Giridhar (Simple Interest):

I_G = P \times r_G \times t

I_G = 12,500 \times 0.12 \times 3

I_G = 12,500 \times 0.36 = ₹4,500

Interest paid by Raghava (Compound Interest): Amount

A_R = P(1 + r_R)^t

A_R = 12,500(1 + 0.10)^3 = 12,500(1.1)^3

A_R = 12,500(1.331) = ₹16,637.50

Interest

I_R = A_R - P = 16,637.50 - 12,500 = ₹4,137.50

Comparison: Giridhar's interest = ₹4,500 Raghava's interest = ₹4,137.50 Since ₹4,500 > ₹4,137.50, Giridhar pays more interest.

Difference: Difference =

I_G - I_R = ₹4,500 - ₹4,137.50 = ₹362.50

Final Answer: Giridhar pays more interest by ₹362.50.

Q7Figure it Out 4

Consider an amount ₹1000. If this grows at $10\%$ p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth and not-compounding an example of linear growth?

Solution

Given: Principal (P) = ₹1000 Rate (r) =

10\%

p.a. = 0.10 Target Amount (A) = Double the principal = ₹2000

Part 1: Time to double without compounding (Simple Interest)

Formula:

A = P(1 + rt)

Solution:

2000 = 1000(1 + 0.10 \times t)

2 = 1 + 0.1t

1 = 0.1t

t = \frac{1}{0.1} = 10

years.

Part 2: Time to double with compounding

Formula:

A = P(1 + r)^t

Solution:

2000 = 1000(1 + 0.10)^t

2 = (1.1)^t

To solve for t, we can use logarithms:

t = \log_{1.1}(2) = \frac{\ln(2)}{\ln(1.1)} \approx \frac{0.693}{0.0953} \approx 7.27

years. So, it takes about 7.3 years to double with compounding.

Part 3: Type of growth

Analysis:

Without compounding (Simple Interest): The formula for the amount is $A(t) = P + (Pr)t$ . This is a linear equation of the form $y = mx + c$ , where $A$ is $y$ , $t$ is $x$ , the slope $m$ is $Pr$ (a constant), and the y-intercept $c$ is $P$ . The amount grows by a fixed amount ( $Pr$ ) each year. This is linear growth.
With compounding: The formula is $A(t) = P(1+r)^t$ . This is an exponential equation of the form $y = ab^x$ , where $A$ is $y$ and $t$ is $x$ . The amount is multiplied by a fixed factor ( $1+r$ ) each year. This is exponential growth.

Final Answer:

Without compounding, it will take 10 years to double.
With compounding, it will take approximately 7.3 years to double.
Yes, compounding is an example of exponential growth, and not-compounding (simple interest) is an example of linear growth.

Q8Figure it Out 4

The population of a city is rising by about $3\%$ every year. If the current population is 1.5 crore, what is the expected population after 3 years?

Solution

Given: Current Population (P) = 1.5 crore = 15,000,000 Rate of increase (r) =

3\%

per year = 0.03 Time (t) = 3 years

To Find: The expected population after 3 years.

Analysis: Population growth is a compounding process because the growth in each year is based on the new, larger population from the previous year.

Formula: Expected Population =

P(1 + r)^t

Solution: Expected Population =

1.5 \text{ crore} \times (1 + 0.03)^3

=

1.5 \text{ crore} \times (1.03)^3

Calculate

(1.03)^3

:

(1.03)^3 = 1.092727

Expected Population =

1.5 \times 1.092727

crore =

1.6390905

crore

In numbers, this is

16,390,905

.

Final Answer: The expected population after 3 years is approximately 1.64 crore (or 16,390,905).

Q9Figure it Out 4

In a laboratory, the number of bacteria in a certain experiment increases at the rate of $2.5\%$ per hour. Find the number of bacteria at the end of 2 hours if the initial count is 5,06,000.

Solution

Given: Initial count of bacteria (P) = 5,06,000 Rate of increase (r) =

2.5\%

per hour = 0.025 Time (t) = 2 hours

To Find: The number of bacteria at the end of 2 hours.

Analysis: The growth of bacteria is a compounding process.

Formula: Final Count =

P(1 + r)^t

Solution: Final Count =

5,06,000 \times (1 + 0.025)^2

=

5,06,000 \times (1.025)^2

Calculate

(1.025)^2

:

(1.025)^2 = 1.050625

Final Count =

5,06,000 \times 1.050625

=

531,616.25

Since the number of bacteria must be an integer, we can round the result.

Final Answer: The number of bacteria at the end of 2 hours will be approximately 531,616.

Q1Figure it Out 5

The population of Bengaluru in 2025 is about $250\%$ of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?

Solution

Given: Population in 2000 = 50 lakhs Population in 2025 =

250\%

of the population in 2000.

To Find: The population in 2025.

Solution: Population in 2025 =

\frac{250}{100} \times (\text{Population in 2000})

=

2.5 \times 50

lakhs = 125 lakhs

125 lakhs can also be written as 1.25 crore or 12,500,000.

Final Answer: The population of Bengaluru in 2025 is 125 lakhs.

Q2Figure it Out 5

The population of the world in 2025 is about 8.2 billion. The populations of some countries in 2025 are given. Match them with their approximate percentage share of the worldwide population. China: 1.425 billion, India: 1.428 billion, USA: 0.339 billion, Nigeria: 0.223 billion Percentages: 17.41%, 2.72%, 17.38%, 4.13%

Solution

Given: World population = 8.2 billion Country populations: China (1.425b), India (1.428b), USA (0.339b), Nigeria (0.223b)

To Find: The percentage share for each country and match it with the given percentages.

Formula: Percentage Share =

\frac{\text{Country Population}}{\text{World Population}} \times 100\%

Solution: China: Share =

\frac{1.425}{8.2} \times 100\% \approx 17.378\% \approx 17.38\%

India: Share =

\frac{1.428}{8.2} \times 100\% \approx 17.414\% \approx 17.41\%

USA: Share =

\frac{0.339}{8.2} \times 100\% \approx 4.134\% \approx 4.13\%

Nigeria: Share =

\frac{0.223}{8.2} \times 100\% \approx 2.719\% \approx 2.72\%

Matching:

China $\rightarrow 17.38\%$
India $\rightarrow 17.41\%$
USA $\rightarrow 4.13\%$
Nigeria $\rightarrow 2.72\%$

Final Answer: China:

17.38\%

India:

17.41\%

USA:

4.13\%

Nigeria:

2.72\%

Q3Figure it Out 5

The price of a mobile phone is ₹8,250. A GST of $18\%$ is added to the price. Which of the following gives the final price of the phone including the GST?

(i)

$8250+18$

(ii)

$8250+1800$

(iii)

$8250+\frac{18}{100}$

(iv)

$8250 \times 18$

(v)

$8250 \times 1.18$

(vi)

$8250+8250 \times 0.18$

(vii)

$1.8 \times 8250$

Solution

Given: Base Price of phone = ₹8,250 GST Rate = 18%

To Find: The correct expression for the final price.

Solution: The final price is the base price plus the GST amount. GST Amount =

18\%

of Base Price =

\frac{18}{100} \times 8250 = 0.18 \times 8250

.

Final Price = Base Price + GST Amount Final Price =

8250 + (0.18 \times 8250)

This can be written as

8250 + (8250 \times 0.18)

, which matches option (vi).

Alternatively, we can factor out the base price: Final Price =

8250 \times (1 + 0.18)

Final Price =

8250 \times 1.18

, which matches option (v).

Both (v) and (vi) are correct representations of the final price.

Final Answer: The correct expressions are (v)

8250 \times 1.18

and (vi)

8250+8250 \times 0.18

.

Q4Figure it Out 5

The monthly percentage change in population (compared to the previous month) of mice in a lab is given: Month 1 change was $+5\%$ , Month 2 change was $-2\%$ , and Month 3 change was $-3\%$ . Which of the following statement(s) are true? The initial population is $p$ .

(i)

The population after three months was $p \times 0.05 \times 0.02 \times 0.03$ .

(ii)

The population after three months was $p \times 1.05 \times 0.98 \times 0.97$ .

(iii)

The population after three months was $p+0.05-0.02-0.03$ .

(iv)

The population after three months was $p$ .

(v)

The population after three months was more than $p$ .

(vi)

The population after three months was less than $p$ .

Solution

Given: Initial population =

p

Month 1 change =

+5\%

Month 2 change =

-2\%

Month 3 change =

-3\%

To Find: The true statement(s).

Solution: Let's calculate the population after each month.

After Month 1: A $5\%$ increase means the new population is $105\%$ of the previous. Population = $p \times (1 + 0.05) = p \times 1.05$ .
After Month 2: A $2\%$ decrease on the new population means the next population is $98\%$ of it. Population = $(p \times 1.05) \times (1 - 0.02) = p \times 1.05 \times 0.98$ .
After Month 3: A $3\%$ decrease on the latest population means the final population is $97\%$ of it. Final Population = $(p \times 1.05 \times 0.98) \times (1 - 0.03) = p \times 1.05 \times 0.98 \times 0.97$ .

This confirms that statement (ii) is true.

Now let's evaluate the final population to check statements (v) and (vi). Final Population =

p \times (1.05 \times 0.98 \times 0.97)

= p \times (1.029 \times 0.97)

= p \times 0.99813

Since the final population is

0.99813p

, which is less than

p

, statement (vi) is true. Statement (v) is false. Statement (i) is false as it multiplies the changes, not the growth factors. Statement (iii) is false as it adds/subtracts percentages, which is incorrect for compounded changes. Statement (iv) is false as the population has changed.

Final Answer: The true statements are (ii) and (vi).

Q5Figure it Out 5

A shopkeeper initially set the price of a product with a $35\%$ profit margin. Due to poor sales, he decided to offer a $30\%$ discount on the selling price. Will he make a profit or a loss? Give reasons for your answer.

Solution

To Find: Whether the shopkeeper makes a profit or a loss.

Let: The Cost Price (CP) of the product be

x

.

Solution:

Calculate the initial Selling Price (SP): The shopkeeper sets the price with a $35\%$ profit margin on the cost. Profit = $35\%$ of $x = 0.35x$ Initial SP = CP + Profit = $x + 0.35x = 1.35x$
Calculate the final Selling Price after discount: A discount of $30\%$ is offered on the initial SP. Discount Amount = $30\%$ of Initial SP = $0.30 \times (1.35x) = 0.405x$ Final SP = Initial SP - Discount Amount = $1.35x - 0.405x = 0.945x$
Compare Final SP with CP: The shopkeeper bought the product for $x$ and sold it for $0.945x$ . Since Final SP ( $0.945x$ ) is less than CP ( $x$ ), the shopkeeper will make a loss.

Calculate the Loss: Loss = CP - Final SP =

x - 0.945x = 0.055x

Loss Percentage =

\frac{\text{Loss}}{\text{CP}} \times 100\% = \frac{0.055x}{x} \times 100\% = 5.5\%

Final Answer: The shopkeeper will make a loss. The reason is that the

30\%

discount is calculated on a higher base (the marked-up selling price), which results in a larger reduction than the initial

35\%

profit margin. The final selling price (

0.945

times the cost) is less than the cost price, resulting in a

5.5\%

loss.

Q6Figure it Out 5

What percentage of area is occupied by the region marked ' E ' in the figure?

Solution

Figure Description: The figure described is a square that is divided into four smaller, identical triangles by its two diagonals. The region marked 'E' is one of these four triangles.

To Find: The percentage of the total area occupied by region 'E'.

Solution: Since the two diagonals divide the square into four triangles of equal area, each triangle occupies an equal fraction of the total area of the square.

Fraction of area occupied by region 'E' =

\frac{1}{4}

.

To convert this fraction to a percentage, we multiply by 100. Percentage Area =

\frac{1}{4} \times 100\% = 25\%

.

Final Answer: The region marked 'E' occupies

25\%

of the total area.

Q7Figure it Out 5

What is $5\%$ of 40? What is $40\%$ of 5? What is $25\%$ of 12? What is $12\%$ of 25? What is $15\%$ of 60? What is $60\%$ of 15? What do you notice? Can you make a general statement and justify it using algebra, comparing $x\%$ of $y$ and $y\%$ of $x$ ?

Solution

Part 1: Calculations

$5\%$ of 40 = $\frac{5}{100} \times 40 = 0.05 \times 40 = 2$ .
$40\%$ of 5 = $\frac{40}{100} \times 5 = 0.40 \times 5 = 2$ .
$25\%$ of 12 = $\frac{25}{100} \times 12 = 0.25 \times 12 = 3$ .
$12\%$ of 25 = $\frac{12}{100} \times 25 = 0.12 \times 25 = 3$ .
$15\%$ of 60 = $\frac{15}{100} \times 60 = 0.15 \times 60 = 9$ .
$60\%$ of 15 = $\frac{60}{100} \times 15 = 0.60 \times 15 = 9$ .

Part 2: Observation In all the pairs of calculations, the result is the same. For example,

5\%

of 40 is equal to

40\%

of 5.

Part 3: General Statement and Justification

General Statement: For any two numbers

x

and

y

,

x\%

of

y

is always equal to

y\%

of

x

.

Justification using Algebra: Let's write the expressions for

x\%

of

y

and

y\%

of

x

.

x\%

of

y = \frac{x}{100} \times y = \frac{xy}{100}

.

y\%

of

x = \frac{y}{100} \times x = \frac{yx}{100}

.

Since the multiplication of numbers is commutative (i.e.,

xy = yx

), the two expressions are equal:

\frac{xy}{100} = \frac{yx}{100}

Therefore,

x\%

of

y = y\%

of

x

.

Hence Proved.

Q8Figure it Out 5

A school is organising an excursion for its students. 40% of them are Grade 8 students and the rest are Grade 9 students. Among these Grade 8 students, 60% are girls. [Hint: Drawing a rough diagram can help].

(i)

What percentage of the students going to the excursion are Grade 8 girls?

(ii)

If the total number of students going to the excursion is 160, how many of them are Grade 8 girls?

Solution

Given: Percentage of Grade 8 students = 40% Percentage of Grade 9 students =

100\% - 40\% = 60\%

Percentage of girls among Grade 8 students = 60%

(i) Percentage of Grade 8 girls among all students

To Find: The percentage of the total students that are Grade 8 girls.

Solution: We need to find

60\%

of the

40\%

of students. Percentage of Grade 8 girls =

60\%

of

40\%

= \frac{60}{100} \times 40\%

= 0.6 \times 40\% = 24\%

Final Answer (i):

24\%

of the students going to the excursion are Grade 8 girls.

(ii) Number of Grade 8 girls if total students are 160

Given: Total number of students = 160.

To Find: The number of Grade 8 girls.

Solution: From part (i), we know that Grade 8 girls make up

24\%

of the total students. Number of Grade 8 girls =

24\%

of 160

= \frac{24}{100} \times 160

= 0.24 \times 160 = 38.4

Since the number of students must be a whole number, the given numbers in the problem might be approximate. Based on the calculation, the result is 38.4.

Final Answer (ii): The number of Grade 8 girls is 38.4. In a real-world scenario, this would likely be rounded to 38, suggesting the percentages were approximations.

Q9Figure it Out 5

A shopkeeper sells pencils at a price such that the selling price of 3 pencils is equal to the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage?

Solution

Given: Selling Price of 3 pencils = Cost Price of 5 pencils.

To Find: Whether there is a profit or loss, and its percentage.

Let:

SP

be the selling price of one pencil.

CP

be the cost price of one pencil.

Equation: From the given information, we can write:

3 \times SP = 5 \times CP

Solution: From the equation, we can find the ratio of SP to CP:

\frac{SP}{CP} = \frac{5}{3}

Since

\frac{5}{3} > 1

, we have

SP > CP

. This means the shopkeeper makes a profit.

To find the profit percentage, we use the formula: Profit Percentage =

\frac{SP - CP}{CP} \times 100\%

From the ratio, we have

SP = \frac{5}{3}CP

. Substitute this into the formula: Profit Percentage =

\frac{\frac{5}{3}CP - CP}{CP} \times 100\%

=

\frac{(\frac{5}{3} - 1)CP}{CP} \times 100\%

=

(\frac{5}{3} - \frac{3}{3}) \times 100\%

=

\frac{2}{3} \times 100\% \approx 66.67\%

Final Answer: The shopkeeper makes a profit of

66.67\%

(or

66\frac{2}{3}\%

).

Q10Figure it Out 5

The bus fares were increased by $3\%$ last year and by $4\%$ this year. What is the overall percentage price increase in the last 2 years?

Solution

Given: Price increase last year = 3% Price increase this year = 4%

To Find: The overall percentage price increase.

Let: The original bus fare be

P

.

Solution: These are successive percentage increases, so they compound.

After the first year (3% increase): New Fare =

P \times (1 + 0.03) = 1.03P

After the second year (4% increase on the new fare): Final Fare =

(1.03P) \times (1 + 0.04) = 1.03P \times 1.04

Final Fare =

1.0712P

Overall Increase: The overall growth factor is 1.0712. This means the final price is

107.12\%

of the original price. Overall Percentage Increase =

107.12\% - 100\% = 7.12\%

Alternatively, the increase is

1.0712P - P = 0.0712P

. As a percentage, this is

0.0712 \times 100\% = 7.12\%

. Note: Simply adding the percentages (

3\% + 4\% = 7\%

) gives an incorrect result because the second increase is applied to a larger base amount.

Final Answer: The overall percentage price increase in the last 2 years is

7.12\%

.

Q11Figure it Out 5

If the length of a rectangle is increased by $10\%$ and the area is unchanged, by what percentage (exactly) does the breadth decrease by?

Solution

Given: Increase in length = 10% Area remains unchanged.

To Find: The percentage decrease in breadth.

Let: Original length =

L

Original breadth =

B

Original Area =

A = L \times B

Solution:

Find the new length ( $L'$ ): The length is increased by $10\%$ . $L' = L + 10\% \text{ of } L = L + 0.1L = 1.1L$
Use the area information: The new area ( $A'$ ) is the same as the original area ( $A$ ). Let the new breadth be $B'$ . $A' = L' \times B' = (1.1L) \times B'$ Since $A' = A$ , we have: $L \times B = (1.1L) \times B'$
Find the new breadth ( $B'$ ) in terms of the old breadth ( $B$ ): Divide both sides by $L$ (assuming $L \neq 0$ ): $B = 1.1 B'$ $B' = \frac{B}{1.1} = \frac{1}{1.1}B = \frac{10}{11}B$
Calculate the percentage decrease in breadth: Decrease in breadth = $B - B' = B - \frac{10}{11}B = \frac{1}{11}B$ Percentage Decrease = $\frac{\text{Decrease in breadth}}{\text{Original breadth}} \times 100\%$ = $\frac{\frac{1}{11}B}{B} \times 100\% = \frac{1}{11} \times 100\% \approx 9.09\%$

Final Answer: The breadth decreases by exactly

\frac{100}{11}\%

or approximately

9.09\%

.

Q12Figure it Out 5

The percentage of ingredients in a 65 g chips packet is shown in the picture. Find out the weight each ingredient makes up in this packet. (Potatoes: 56%, Oil: 33%, Salt: 5%, Spices: 6%)

Solution

Given: Total weight of chips packet = 65 g Ingredient percentages:

Potatoes: 56%
Oil: 33%
Salt: 5%
Spices: 6%

To Find: The weight of each ingredient in grams.

Solution: We calculate the weight for each ingredient by taking its percentage of the total weight (65 g).

Weight of Potatoes:

56\%

of 65 g =

\frac{56}{100} \times 65 = 0.56 \times 65 = 36.4

g.

Weight of Oil:

33\%

of 65 g =

\frac{33}{100} \times 65 = 0.33 \times 65 = 21.45

g.

Weight of Salt:

5\%

of 65 g =

\frac{5}{100} \times 65 = 0.05 \times 65 = 3.25

g.

Weight of Spices:

6\%

of 65 g =

\frac{6}{100} \times 65 = 0.06 \times 65 = 3.9

g.

Check: The sum of the weights should be 65 g.

36.4 + 21.45 + 3.25 + 3.9 = 65.00

g.

Final Answer:

Weight of Potatoes: 36.4 g
Weight of Oil: 21.45 g
Weight of Salt: 3.25 g
Weight of Spices: 3.9 g

Q13Figure it Out 5

Three shops sell the same items at the same price. The shops offer deals as follows: Shop A: "Buy 1 and get 1 free" Shop B: "Buy 2 and get 1 free" Shop C: "Buy 3 and get 1 free" Answer the following:

(i)

If the price of one item is ₹100, what is the effective price per item in each shop? Arrange the shops from cheapest to costliest.

(ii)

For each shop, calculate the percentage discount on the items. [Hint: Compare the free items to the total items you receive.]

(iii)

Suppose you need 4 items. Which shop would you choose? Why?

Solution

Given: Price of one item = ₹100 Shop A: Buy 1, get 1 free Shop B: Buy 2, get 1 free Shop C: Buy 3, get 1 free

(i) Effective price per item and ranking

Shop A: You pay for 1 item (₹100) and get 2 items. Effective price = $\frac{₹100}{2 \text{ items}} = ₹50$ per item.
Shop B: You pay for 2 items (₹200) and get 3 items. Effective price = $\frac{₹200}{3 \text{ items}} \approx ₹66.67$ per item.
Shop C: You pay for 3 items (₹300) and get 4 items. Effective price = $\frac{₹300}{4 \text{ items}} = ₹75$ per item.

Arrangement from cheapest to costliest: Shop A, Shop B, Shop C.

(ii) Percentage discount Discount Percentage =

\frac{\text{Number of free items}}{\text{Total number of items}} \times 100\%

Shop A: You get 1 free item out of a total of 2. Discount = $\frac{1}{2} \times 100\% = 50\%$ .
Shop B: You get 1 free item out of a total of 3. Discount = $\frac{1}{3} \times 100\% \approx 33.33\%$ .
Shop C: You get 1 free item out of a total of 4. Discount = $\frac{1}{4} \times 100\% = 25\%$ .

(iii) Choice for buying 4 items

Shop A: To get 4 items, you use the "Buy 1, get 1 free" deal twice. You buy 2 items and get 2 free. Cost = $2 \times ₹100 = ₹200$ .
Shop B: To get 4 items, you use the "Buy 2, get 1 free" deal once. You pay for 2 items (₹200), get 3 items. You still need 1 more item, which you must buy at the regular price. Cost = (Cost for 3 items) + (Cost for 1 item) = $₹200 + ₹100 = ₹300$ .
Shop C: To get 4 items, you use the "Buy 3, get 1 free" deal once. You pay for 3 items. Cost = $3 \times ₹100 = ₹300$ .

Comparison of costs for 4 items:

Shop A: ₹200
Shop B: ₹300
Shop C: ₹300

Conclusion: Shop A is the cheapest option for buying 4 items.

Final Answer:

(i)

Effective prices: Shop A = ₹50, Shop B = ₹66.67, Shop C = ₹75. Cheapest to costliest: A, B, C.

(ii)

Discounts: Shop A = 50%, Shop B = 33.33%, Shop C = 25%.

(iii)

I would choose Shop A because the total cost for 4 items is ₹200, which is cheaper than Shop B and Shop C (both ₹300).

Q14Figure it Out 5

In a room of 100 people, $99\%$ are left-handed. How many left-handed people have to leave the room to bring that percentage down to $98\%$ ?

Solution

Given: Initial number of people = 100 Initial percentage of left-handed people = 99% Target percentage of left-handed people = 98%

To Find: The number of left-handed people who have to leave.

Solution:

Initial state:
- Total people = 100
- Number of left-handed (LH) people = $99\%$ of 100 = 99.
- Number of right-handed (RH) people = $100 - 99 = 1$ .
Let $x$ be the number of LH people who leave.
- The RH person stays in the room. This person is the key to solving the problem.
- After $x$ $x$ LH people leave:
  - New number of LH people = $99 - x$ .
  - New number of RH people = 1.
  - New total number of people = $(99 - x) + 1 = 100 - x$ .
Set up the equation for the target state: In the new group, the percentage of LH people is $98\%$ . This means the percentage of RH people is $100\% - 98\% = 2\%$ . The number of RH people (which is 1) must be equal to $2\%$ of the new total population. $1 = 2\% \text{ of } (100 - x)$ $1 = \frac{2}{100} \times (100 - x)$ $1 = 0.02 \times (100 - x)$
Solve for $x$ : Divide by 0.02: $100 - x = \frac{1}{0.02} = 50$ $x = 100 - 50 = 50$

Verification: If 50 LH people leave, we have:

LH people = $99 - 50 = 49$ .
RH people = 1.
Total people = $49 + 1 = 50$ .
New percentage of LH people = $\frac{49}{50} \times 100\% = 98\%$ . This is correct.

Final Answer: 50 left-handed people have to leave the room.

Q15Figure it Out 5

Look at the following graph. Based on the graph, which of the following statement(s) are valid?

(i)

People in their twenties are the most computer-literate among all age groups.

(ii)

Women lag behind in the ability to use computers across age groups.

(iii)

There are more people in their twenties than teenagers.

(iv)

More than a quarter of people in their thirties can use computers.

(v)

Less than 1 in 10 aged 60 and above can use computers.

(vi)

Half of the people in their twenties can use computers.

Solution

Graph Analysis: The graph shows the percentage of people who can use computers, broken down by age groups and gender (Male, Female, Total).

Evaluation of Statements:

(i) People in their twenties are the most computer-literate among all age groups.

By observing the 'Total' bars for each age group, the bar for the '20-29' age group is the highest. This indicates the highest percentage of computer literacy in this group.
Conclusion: Valid.

(ii) Women lag behind in the ability to use computers across age groups.

For every age group shown (10-19, 20-29, ..., 60 and above), the bar representing 'Female' is shorter than the bar representing 'Male'.
Conclusion: Valid.

(iii) There are more people in their twenties than teenagers.

The graph shows percentages of literacy within each age group, not the absolute number of people in those age groups. We cannot make any conclusion about the population size of different age groups from this graph.
Conclusion: Invalid.

(iv) More than a quarter of people in their thirties can use computers.

A quarter is $25\%$ . The 'Total' bar for the '30-39' age group is visibly higher than the $25\%$ mark on the y-axis (it appears to be around $35\%$ ).
Conclusion: Valid.

(v) Less than 1 in 10 aged 60 and above can use computers.

1 in 10 is equivalent to $10\%$ . The 'Total' bar for the '60 and above' age group is below the $10\%$ mark on the y-axis (it appears to be around $8\%$ ).
Conclusion: Valid.

(vi) Half of the people in their twenties can use computers.

Half is equivalent to $50\%$ . The 'Total' bar for the '20-29' age group is the highest, but it is slightly below the $50\%$ line (it appears to be around $48\%$ ). Therefore, it is not exactly half.
Conclusion: Invalid.

Final Answer: The valid statements are (i), (ii), (iv), and (v).

Fractions in DisguiseClass 8 Mathematics NCERT Solutions

Express the following fractions as percentages.

35\frac{3}{5}53​

714\frac{7}{14}147​

920\frac{9}{20}209​

72150\frac{72}{150}15072​

13\frac{1}{3}31​

511\frac{5}{11}115​

Solution

Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?

10%10 \%10%

15%15 \%15%

25%25 \%25%

60%60 \%60%

40%40 \%40%

None of these

Solution

In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?

Solution

Pairs of quantities are shown below. Identify and write appropriate symbols ' >', '<', '=' in the blanks. Try to do it without calculations.

50%50 \%50% __ 5%5 \%5%

510\frac{5}{10}105​ __ 50%50 \%50%

311\frac{3}{11}113​ __ 61%61 \%61%

30%30 \%30% __ 13\frac{1}{3}31​

Solution

Find the missing numbers. The first problem has been worked out.

20% of __ is 40. (Answer: 200)

__% of 150 is 30.

15% of 200 is __.

120% of 50 is __.

Solution

Find the value of the following and also draw their bar models.

25%25 \%25% of 160

16%16 \%16% of 250

62%62 \%62% of 360

140%140 \%140% of 40

1%1 \%1% of 1 hour

7%7 \%7% of 10 kg

Solution

Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up 34\frac{3}{4}43​ of the deep orange paint?

Solution

Pairs of quantities are shown below. Identify and write appropriate symbols ' >', '<', '=' in the boxes. Visualising or estimating can help. Compute only if necessary or for verification.

50%50 \%50% of 510□50%510 \square 50 \%510□50% of 515

37%37 \%37% of 148□73%148 \square 73 \%148□73% of 148

29%29 \%29% of 43□92%43 \square 92 \%43□92% of 110

30%30 \%30% of 40□40%40 \square 40 \%40□40% of 50

45%45 \%45% of 200□10%200 \square 10 \%200□10% of 490

30%30 \%30% of 80□24%80 \square 24 \%80□24% of 64

Solution

Fill in the blanks appropriately:

30%30 \%30% of kkk is 70,60%70, 60 \%70,60% of kkk is ____ , 90%90 \%90% of kkk is ____ , 120%120 \%120% of kkk is ____ .

100%100 \%100% of mmm is 215,10%215, 10 \%215,10% of mmm is ____ , 1% of mmm is ____ , 6% of mmm is ____ .

90%90 \%90% of nnn is 270,9%270, 9 \%270,9% of nnn is ____ , 18%18 \%18% of nnn is ____ , 100% of nnn is ____ .

Make 2 more such questions and challenge your peers.

Solution

Fill in the blanks:

3 is ____ % of 300.

____ is 40%40 \%40% of 4.

40 is 80%80 \%80% of ____ .

Solution

Is 10%10 \%10% of a day longer than 1%1 \%1% of a week? Create such questions and challenge your peers.

Solution

Solution

Workers in a coffee plantation take 18 days to pick coffee berries in 20%20\%20% of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?

Solution

The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is 10%:80%:10%10 \% : 80 \% : 10 \%10%:80%:10%. If he wants to conduct a training of 90 minutes. How long should each activity be done?

Solution

An estimated 90%90\%90% of the world's population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year's worldwide population.

Solution

A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions-Rava: 40%, Sugar: 40%, and Ghee: 20%.

If you want to make halwa for 8 people, what is the proportion of each of the above ingredients?

If the total weight of the ingredients is 2 kg , how much rava, sugar and ghee are present?

Solution

If a shopkeeper buys a geometry box for ₹75 and sells it for ₹110, what is his profit margin with respect to the cost?

Solution

I am a carpenter and I make chairs. The cost of materials for a chair is ₹475 and I want to have a profit margin of 50%. At what price should I sell a chair?

Solution

Samson bought a car for ₹4,40,000 after getting a 15% discount from the car dealer. What was the original price of the car?

Solution

The total sales of a company (also called revenue) was ₹2.5 crore last year. They had a healthy profit margin of 25%. What was the total expenditure (costs) of the company last year?

$\frac{3}{5}$

$\frac{7}{14}$

$\frac{9}{20}$

$\frac{72}{150}$

$\frac{1}{3}$

$\frac{5}{11}$

$10 \%$

$15 \%$

$25 \%$

$60 \%$

$40 \%$

$50 \%$ __ $5 \%$

$\frac{5}{10}$ __ $50 \%$

$\frac{3}{11}$ __ $61 \%$

$30 \%$ __ $\frac{1}{3}$

$25 \%$ of 160

$16 \%$ of 250

$62 \%$ of 360

$140 \%$ of 40

$1 \%$ of 1 hour

$7 \%$ of 10 kg

Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up $\frac{3}{4}$ of the deep orange paint?

$50 \%$ of $510 \square 50 \%$ of 515

$37 \%$ of $148 \square 73 \%$ of 148

$29 \%$ of $43 \square 92 \%$ of 110

$30 \%$ of $40 \square 40 \%$ of 50

$45 \%$ of $200 \square 10 \%$ of 490

$30 \%$ of $80 \square 24 \%$ of 64

$30 \%$ of $k$ is $70, 60 \%$ of $k$ is , $90 \%$ of $k$ is , $120 \%$ of $k$ is .

$100 \%$ of $m$ is $215, 10 \%$ of $m$ is , 1% of $m$ is , 6% of $m$ is .

$90 \%$ of $n$ is $270, 9 \%$ of $n$ is , $18 \%$ of $n$ is , 100% of $n$ is .

____ is $40 \%$ of 4.

40 is $80 \%$ of ____ .

Is $10 \%$ of a day longer than $1 \%$ of a week? Create such questions and challenge your peers.

Workers in a coffee plantation take 18 days to pick coffee berries in $20\%$ of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?

The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is $10 \% : 80 \% : 10 \%$ . If he wants to conduct a training of 90 minutes. How long should each activity be done?

An estimated $90\%$ of the world's population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year's worldwide population.

A number increased by $20\%$ becomes 90. What is the number?

A milkman sold two buffaloes for ₹80,000 each. On one of them, he made a profit of $5 \%$ and on the other a loss of $10 \%$ . Find his overall profit or loss.

The population of elephants in a national park increased by $5 \%$ in the last decade. If the population of the elephants last decade is $p$ , the population now is

$p \times 0.5$

$p \times 0.05$

$p \times 1.5$

$p \times 1.05$

$p+1.50$

Which of the following statement(s) mean the same as - "The demand for cameras has fallen by $85\%$ in the last decade"?

Bank of Yahapur offers an interest of $10\%$ p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually.

Bank of Wahapur offers an interest of $5\%$ p.a. Compare how much one gets if one deposits ₹20,000 for a period of 4 years with compounding and without compounding annually.

Jasmine invests amount 'p' for 4 years at an interest of $6\%$ p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done?

$p \times 6 \times 4$

$p \times 0.6 \times 4$

$p \times \frac{0.6}{100} \times 4$

$p \times \frac{0.06}{100} \times 4$

$p \times 1.6 \times 4$

$p \times 1.06 \times 4$

$p + (p \times 0.06 \times 4)$

The post office offers an interest of $7\%$ p.a. How much interest would one get if one invests ₹50,000 for 3 years without compounding? How much more would one get if it was compounded?

Giridhar borrows a loan of ₹ 12,500 at $12\%$ per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at $10\%$ per annum, compounded annually. Who pays more interest and by how much?

Consider an amount ₹1000. If this grows at $10\%$ p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth and not-compounding an example of linear growth?

The population of a city is rising by about $3\%$ every year. If the current population is 1.5 crore, what is the expected population after 3 years?

In a laboratory, the number of bacteria in a certain experiment increases at the rate of $2.5\%$ per hour. Find the number of bacteria at the end of 2 hours if the initial count is 5,06,000.

The population of Bengaluru in 2025 is about $250\%$ of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?

The price of a mobile phone is ₹8,250. A GST of $18\%$ is added to the price. Which of the following gives the final price of the phone including the GST?

$8250+18$

$8250+1800$

$8250+\frac{18}{100}$

$8250 \times 18$

$8250 \times 1.18$