ANSWERs for figure it out Class 8 Mathematics – NCERT (Ganita Prakash) Part 2 Chapter 1: FRACTIONS IN DISGUISE

ANSWERs for figure it out Class 8 Mathematics – NCERT (Ganita Prakash) Part 2 

 Chapter 1: FRACTIONS IN DISGUISE

1.1 Fractions as Percentages

The symbol ʻ%ʼ means per cent.

The word ‘per cent’ is derived from the Latin phrase ‘per centum’, meaning ‘by the hundred’ or ‘out of hundred’.

Thus, percentages are simply fractions where the denominator is 100.

Expressing Fractions as Percentages

Example 1: 

Surya wants to use a deep orange colour to capture the sunset. He mixes some red paint and yellow paint to make this colour. The red paint makes up 3 4 of this mixture. What percentage of the colour is made with red?

 3/4 is 3 out of every 4. 

That is, 6 out of every 8 (equivalent fraction). 

That is, 30 out of every 40. 

That is, 75 out of every 100. This means 75%

Example 2: 

Surya won some prize money in a contest. He wants to save 2 5 of the money to purchase a new canvas. Express this quantity as a percentage.

A fraction is of a unit, while a percentage is per 100. Therefore, to express a fraction as a percentage, we can just multiply the fraction by 100

Example 3: 

Given a percentage, can you express it as a fraction? For example, express 24% as a fraction.

A percentage is a fraction, 24% is the same as \( \frac{24}{100} \) . 

other equivalent forms of \( \frac{24}{100} \)  = \( \frac{12}{50} \)  =  \( \frac{6}{25} \)  = \( \frac{48}{200} \) .

A percentage, z%, can be expressed by any of the fractions that are equivalent to \( \frac{z}{100} \).

Figure it Out Page 3 PART 2

1. Express the following fractions as percentages. 

(i) \( \frac{5}{8} \)

(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:

$$\text{Percentage}=\text{Fraction}\times 100$$

(i) $\frac{5}{8}$

$$\frac{5}{8}\times 100=62.5\mathrm{\%}$$

(i) $\frac{3}{5}$

$$\frac{3}{5}\times 100=60\mathrm{\%}$$

(ii) $\frac{7}{14}$

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

(iii) $\frac{9}{20}$

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

(iv) $\frac{72}{150}$

$$\frac{72}{150} = \frac{24}{50} = \frac{12}{25}$$
$$\frac{12}{25} \times 100 = 48\%$$

(v) $\dfrac{1}{3}$

$$\dfrac{1}{3} \times 100 = 33.33\% \ (\text{approximately})$$

(vi) $\dfrac{5}{11}$

$$\dfrac{5}{11} \times 100 = 45.45\% \ (\text{approximately})$$
2. Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white? 

(i) 10% (ii) 15% (iv) 60% (v) 40% (iii) 25% (vi) None of these

​ $$\frac{15}{25} \times 100 = 60\%$$

Correct answer: 60% (Option iv)

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

$$\frac{15}{80} = \frac{3}{16}$$ $$\frac{3}{16} \times 100 = 18.75\%$$

Answer: 18.75%

4. A group of friends is participating in a long-distance run. The positions of each of them after 15 minutes are shown in the following picture. Match (among the given options) what percentage of the race each of them has approximately completed. 

Position A

• A is just a little ahead of the start.

• Clearly less than 25% of the total distance.

 Best match: 20%

Position B

• B is around the middle, but still before halfway.

• Slightly less than 50%.

•  Best match: 38%

Position C

• C is past the halfway point, but not close to the finish.

• Roughly around 70–75%.

Best match: 72%

 Position D

• D is very close to the finish line, but not exactly at the end.

• So it must be more than 90%, but not 100%.

 Best match: 93%

Final Matching Answer

Position	Percentage
A	20%
B	38%
C	72%
D	93%

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

 (i) 50% ____ 5%(ii) 5 /10 ____ 50%(iii) 3/ 11 _____ 61% (iv) 30% ____ 1/3

(i) 50% ___5%

$$50\mathrm{\%}=0.5$$

5% = 0.05

50% > 5%

(ii) $\frac{5}{10}$ ___ 50%

$\frac{5}{10} =$  50% 

Answer $\frac{5}{10}$= 50% 

(iii) $\frac{1}{3}$ ___ 61%

$$\frac{1}{3} \approx 33.3\%$$
$$\frac{3}{11}<61\mathrm{\%}$$

(iv) 30% ___ $\frac{1}{3}$

$$30\% < 33.3\%$$
$$30\% < \frac{1}{3}$$

Try to calculate (without using pen and paper) the indicated percentages of the values shown in the table below. Write your answers in the table.

The FDP Trio — Fractions, Decimals, and Percentages

Example 2: 

We can find 50% of a value by multiplying 1 2 with the value. Will multiplying the value by 0.5 also give the answer for 50% of the value?

Yes, since 1/2 = 0.5

50% = 50 /100 = 1/2 = 0.5

50% of 24 = 12

0.5 x 24 = 12

Complete the following table

Example 3: 

The maximum marks in a test are 75. If students score 80% 

or above in the test, they get an A grade. How much should Zubin score 

at least to get an A grade?

Example 4:

 To prepare a particular millet kanji (porridge), suppose the ratio of millet to water to be mixed for boiling is 2:7. What percentage does the millet constitute in this mixture? If 500 ml of the mixture is to be made, how much millet should be used?

The ratio of millet to the volume of the mixture is 2:9. 

In other words, in one unit of the mixture, millet occupies 2 /9 units and water occupies 7/ 9 units.

The percentage (i.e., in 100 such units) of millet in the mixture is 2/9 x 100 = 22.22%.

 The percentage of water in the mixture will be 100 – 22.22 = 77.78%. 

A mixture with 22.22% millet means 100 ml mixture will have 22.22 ml millet. 

 Therefore, 500 ml with 22.22% millet will have 5 × 22.22 = 111.1 ml of millet.

Example 5: 

A cyclist cycles from Delhi to Agra and completes 40% of the journey. If he has covered 92 km, how many more kilometres does he have to travel to reach Agra?

Example 6: 

Kishanlal recently opened a garment shop. He aims to achieve a daily sales of at least ₹5000. The sales on the first 2 days were ₹2000 and ₹3500. What percentage of his target did he achieve? 

It is 40% on Day 1 and 70% on Day 2. Another way of saying it is — he was 60% short of his target on Day 1 and 30% short of his target on Day 2.

In the next two days, he made ₹5000 and ₹6000 respectively. What percentage of his target are these values?What percentage of the target was achieved on Day 4?

His target is ₹5000, and he made ₹5000 on Day 3 — this is 100%. 

On Day 4, he made ₹6000, which is 1000 more than his target.

1000 is 20% of 5000. Therefore, 6000, (5000 + 1000) is 100% + 20% = 120% of 5000. It can also be computed as 6000 / 5000 × 100 = 6 / 5 × 100 = 120%. This means he achieved 120% of his target, i.e., 20% more than his target.

On Days 5 and 6 his sales were ₹7800 and ₹9550 respectively. Calculate the percentage of the target achieved on these days.

On Day 7, he achieved 150% of his target. On Day 8, he achieved 210% of his target. Find the sales made on these days.

Suppose on some day, he made ₹2500. This can be expressed as “He achieved 1 2 of his target” or “He achieved 50% of his target” or “He achieved 0.5 of his target”. On some other day, he made ₹10,000. We can say “He achieved twice/double/2 times his target” or “He achieved 200% of his target”

Complete the table below. Mark the approximate locations in the following diagram.

Example 7: 

A farmer harvested 260 kg of wheat last year. This year, they harvested 650 kg of wheat. What percentage of last year’s harvest is this year’s harvest? 

This year’s harvest = \( \frac{650}{260} \) × 100 = 250% of last year’s harvest. 250% indicates that it is 2.5 times the original value

Figure it Out Page 12

Estimate first before making any computations to solve the following questions. Try different methods including mental computations. 

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

(i) The whole bar = 100%

It is divided into 5 equal parts

One part = 20%

Next bar : - The full bar (100%) = 75

Bar divided into 5 equal parts

100%

┌───┬───┬───┬───┬───┐

│20% │20% │20%  │20%│20%  │

└───┴───┴───┴───┴───┘

Value of 1 part= 75/5​= 15  The arrow shows the four bars

 Missing value = 15 x 4 = 60

(ii) The bar is divided into 10 equal parts

Whole bar = 100%

Each part= (100%)/10=10%  

100%

┌─┬─┬─┬─┬─┬─┬─┬─┬─┬─┐

│   │   │   │   │  │   │   │   │   │  │

└─┴─┴─┴─┴─┴─┴─┴─┴─┴─┘

Each part = 10%

Missing value = 10%

Next bar : - The full bar (100%) = 90

Bar divided into 10 equal parts

90

┌─┬─┬─┬─┬─┬─┬─┬─┬─┬─┐

│9 │9 │9│9 │9 │9 │9 │9 │9│9 │

└─┴─┴─┴─┴─┴─┴─┴─┴─┴─┘

Value of 1 part= 90/10​= 9  The arrow shows the six parts

 Missing value = 9 x 6 = 54

(iii) Left diagram

Bar divided into 4 equal parts

Whole = 100%

100%

┌────┬────┬────┬────┐

│25%    │25%    │25%    │25%    │

└────┴────┴────┴────┘

Each part= (100%)/4 = 25% Arrow indicates one part

Missing value = 25%

Right diagram

Whole bar (100%) = 140

Divided into 4 equal parts

140

┌────┬────┬────┬────┐

│35       │35       │35        │35       │

└────┴────┴────┴────┘

Each part= ​ (140 %)/4 =35 Arrow shows the three parts

 Missing value = 3 x 35 = 105

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

 (i) 25% of 160 (ii) 16% of 250 (iv) 140% of 40 (v) 1% of 1 hour (iii) 62% of 360 (vi) 7% of 10 kg 

(i) 25% of 160

$$25\% = \frac{1}{4} \Rightarrow \frac{1}{4} \times 160 = 40$$

160

┌────┬────┬────┬────┐

│40        │40       │40       │40       

└────┴────┴────┴────┘

25% = 40

(ii) 16% of 250

$$16\% = \frac{16}{100} \Rightarrow \frac{16}{100} \times 250 = 40$$

10 kg

┌──────────────────────────────┐

│  7%     |                                                                |

= 0.7 kg  |                                                               │

│            |                                                                 |    

 = 700 g |                                                                │

└──────────────────────────────┘

3. Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up 3 4 of the deep orange paint? 

• Total deep orange paint = 60 ml

• Red paint = $\frac{3}{4}$ of deep orange paint

Calculation:

$$\frac{3}{4} \times 60 = 45$$

Answer:
 Surya used 45 ml of red paint.

4. 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 ______ 50% of 515 

(ii) 37% of 148  ______73% of 148 

 (iii) 29% of 43 ______ 92% of 110 

(iv) 30% of 40  ______40% of 50 

(v) 45% of 200  ______10% of 490 

(vi) 30% of 80  ______24% of 64 

(i) 50% of 510 ⬜ 50% of 515

• 50% of 510 = 255

• 50% of 515 = 257.5

$$255 < 257.5$$

Answer:

$$50\% \text{ of } 510 < 50\% \text{ of } 515$$

(ii) 37% of 148 ⬜ 73% of 148

Same number, but 73% > 37%

Answer:

$$37\% \text{ of } 148 < 73\% \text{ of } 148$$

 5. 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. 

(i) 30% of k = 70

$$k = \frac{70 \times 100}{30} = \frac{700}{3}$$Now,

60% of k = $2\times 70=140$
90% of k = $3\times 70=210$
120% of k = $4\times 70=280$

Answers:
60% = 140, 90% = 210, 120% = 280

(ii)

100% of m = 215

• 10% of m = $\frac{215}{10} = 21.5$
  

• 1% of m = $\frac{215}{100} = 2.15$
  

• 6% of m = $6 \times 2.15 = 12.9$
  

Answers:
10% = 21.5, 1% = 2.15, 6% = 12.9

(iii)

90% of n = 270

$$1\% = \frac{270}{90} = 3$$

9% of n = 9×3=27
18% of n = 18×3=54
100% of n = 100×3=300
Answers:
9% = 27, 18% = 54, 100% = 300

(iv) Two sample questions

1. If 25% of a number is 50, find 100% of the number.

2. If 40% of a number is 120, find 10% and 80% of the number.

6. Fill in the blanks: 

(i) 3 is ____ % of 300. 

(ii) _____ is 40% of 4. 

(iii) 40 is 80% of _____. 

(i) 3 is ___% of 300

$$\frac{3}{300} \times 100 = 1\%$$

Answer: 1%

(ii) ___ is 40% of 4

$$40\% \text{ of } 4 = \frac{40}{100} \times 4 = 1.6$$

Answer: 1.6

(iii) 40 is 80% of ___

$$\text{Whole} = \frac{40 \times 100}{80} = 50$$

Answer: 50

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

Is 10% of a day longer than 1% of a week?

• 1 day = 24 hours

$$10\% \text{ of 1 day} = 0.10 \times 24 = 2.4 \text{ hours}$$

• 1 week = 7 days = 168 hours

$$1\% \text{ of 1 week} = 0.01 \times 168 = 1.68 \text{ hours}$$

Since, 2.4>1.68
Yes, 10% of a day is longer than 1% of a week.

Create a similar challenge question

Example:
Is 5% of a year longer than 10% of a month?
(Students can compare by converting into days)

8. 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?

Mariam gave the bull:

• Day 1: 2 units → ate 1 → ate $\frac{1}{2} = 50\%$

• Day 2: 3 units → ate 2 → ate $\frac{2}{3} \approx 66.7\%$

• Day 3: 4 units → ate 3 → ate $\frac{3}{4} = 75\%$

• …

• Day 99: 100 units → ate 99 → ate $\frac{99}{100} = 99\%$

Observation:
As days increase,

• the % eaten increases

• from 50% → 66.7% → 75% → … → 99%

So the bull eats a larger percentage each day as the total given increases.

9. 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? 

Workers take 18 days to finish 20% of the plantation.

If 20% takes 18 days, then full (100%) takes:

$$\text{Days} = 18 \times \frac{100}{20} = 18 \times 5 = 90$$

Answer: 90 days

10. 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? 

Total training = 90 minutes, ratio:

• Warm-up = 10%

• Play = 80%

• Cool-down = 10%

$$10\% \text{ of } 90 = 9\text{ min}$$
$$80\% \text{ of } 90 = 72\text{ min}$$

Time allocated:

• Warm-up = 9 min

• Play = 72 min

• Cool-down = 9 min

11. 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. 

Current estimated world population (as of January 1, 2026) is approximately 8,266,435,578 people.

If about 90% of the world’s population lives in the Northern Hemisphere, then:

$$90\% \text{ of } 8,\!266,\!435,\!578 = 0.90 \times 8,\!266,\!435,\!578 \approx 7,\!439,\!791,\! (≈ 7.44 \text{ billion})$$

Approximate number of people in the Northern Hemisphere: ~7.44 billion.

12. 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? 

Given proportions:

Rava = 40%
Sugar = 40%
Ghee = 20%
(i) For 8 people
8 is double 4, so ingredients double too:
Rava = 40% of total doubled → stays 40%
Sugar = 40%
Ghee = 20%
Proportions remain the same: 40%, 40%, 20%

(ii) If total = 2 kg
Convert % to weight:
Rava =40% of 2=0.40×2=0.8 kg
Sugar =40% of 2=0.8 kg
Ghee =20% of 2=0.4 kg
Ingredient weights:
Rava = 0.8 kg
Sugar = 0.8 kg
Ghee = 0.4 kg

Example 1: 

Eesha scored 42 marks out of 50 on an English test and 70 marks out of 80 in a Science test. Since she lost only 8 marks in English but 10 marks in Science, she thinks she has done better at English. Reema does not agree! She argues that since Eesha has scored more marks in Science, she has done better at Science. Vishu thinks we cannot compare the scores because the maximum marks are different. Who do you think is correct? If the maximum marks are the same, the comparison becomes easier, isn’t it?

 English score as a percentage =  \( \frac{42}{50} \)​ × 100 = 84% 

Science score as a percentage = \( \frac{70}{80} \)​ × 100 = 87.5%.

 The Science score (as a percentage) is higher than the English score (as a percentage). 

So, Eesha has scored better on the Science test 

Example 2: 

Madhu and Madhav recently learnt about the importance of reading labels on processed food before purchase. They are at a shop to buy badam drink mix. They are looking at two products and wondering which has a larger share of badam. Can you figure it out? Which product uses a smaller proportion of food chemicals?

DEF’s sugar content as a percentage of total weight = 99 150 × 100 = 66%

Product 1: DEF Badam Mix

Total weight = 150 g

Ingredient	Weight (g)
Sugar	99 g
Milk solids	30 g
Badam powder	12 g
Food chemicals	9 g

Product 2: Zacni Badam Mix

Total weight = 400 g

Ingredient	Weight (g)
Sugar	227 g
Milk solids	64 g
Badam powder	40 g
Food chemicals	24 g

Percentage= \( \frac{Part}{Total} \) ​×100

DEF Badam Mix (Total = 150 g)

• Sugar

$$\frac{99}{150} \times 100 = 66\%$$

• Milk solids

$$\frac{30}{150} \times 100 = 20\%$$

• Badam powder

$$\frac{12}{150} \times 100 = 8\%$$

• Food chemicals

$$\frac{9}{150} \times 100 = 6\%$$

$$66 + 20 + 8 + 6 = 100\%$$

Zacni Badam Mix (Total = 400 g)

• Sugar

$$\frac{227}{400} \times 100 = 56.75\%$$

• Milk solids

$$\frac{64}{400} \times 100 = 16\%$$

• Badam powder

$$\frac{40}{400} \times 100 = 10\%$$

• Food chemicals

$$\frac{24}{400} \times 100 = 6\%$$

Check:

$$56.75 + 16 + 10 + 6 = 88.75\%$$

Product	Sugar	Milk Solids	Badam Powder	Food Chemicals
DEF	66%	20%	8%	6%
Zacni	56.75%	16%	10%	6%

Which product has a larger share of badam?

• DEF: 8%

• Zacni: 10%

Zacni has a larger share of badam powder.

Which product uses a smaller proportion of food chemicals?

• DEF: 6%

• Zacni: 6%

Both products use the same proportion of food chemicals (6%).

Percentages help us compare products fairly, even when packet sizes are different.

Reading labels carefully helps us make healthier choices.

Example 3: 

Do the following two statements mean the same thing? (i) The population of this state in 1991 is 165% of that in 1961. (ii) The population of this state has increased by 65% from 1961 to 1991. Yes, both mean the same. Suppose p is the population of the state in 1961 and q is the population of the state in 1991.

Statement A implies,

q = 165% of p,   

\( \frac{165}{100} \) p = 1.65p

Statement B implies,

 q = p + 65% of p 

q = p + 0.65 × p = 1.65p 

The population of the state in 1991 is 1.65 times that in 1961

Figure it Out Page - 19 

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

Cost Price (CP) = ₹75

Selling Price (SP) = ₹110

Profit = SP − CP = 110 − 75 = ₹35

Profit % = \( \frac{35}{75} \)  x 100 = 46 \( \frac{2}{3} \)%

Answer: 46⅔% profit

2. 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? 

Cost of materials = ₹475

Profit margin required = 50%

$$50\% \text{ of } 475 = \frac{1}{2} \times 475 = 237.5$$

Selling Price = 475 + 237.5 = ₹712.50

Answer: He should sell the chair for ₹712.50

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? 

Revenue (Sales) = ₹2.5 crore
Profit margin = 25%

Profit = 25% of expenditure
So, Revenue = Cost + Profit = 125% of cost

$$\text{Cost} = \frac{2.5}{1.25} = 2 \text{ crore}$$

Answer: Total expenditure = ₹2 crore

4. 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? 

Marked Price = ₹300

Discount = 25%

$$25\% \text{ of } 300 = 75$$

Amount to pay = 300 − 75 = ₹225

Answer: Anwar pays ₹225

5. 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% (iv) 66.66% (v) 140% (iii) 60% (vi) 160.66%

Old price = ₹60
New price = ₹100

Increase = 100 − 60 = 40

$$\text{Percentage increase} = \frac{40}{60} \times 100 = 66.66\%$$

Correct option: (iv) 66.66%

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? 

Discount = 15%
So, SP = 85% of original price

$$85\% = 4,40,000$$ 

$$\text{Original Price} = \frac{4,40,000 \times 100}{85} \approx ₹5,17,647$$

Answer: Original price ≈ ₹5,17,650 (approx.)

4. 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? 

Total votes = 1600
Votes for winner = 500

$$\text{Percentage} = \frac{500}{1600} \times 100 = 31.25\%$$

Minimum number of candidates?
Remaining votes = 1100
If there were only 2 candidates, the other would get 1100 votes → winner impossible.

So, minimum = 3 candidates

Answer:

• Winner got 31.25% votes

• Minimum candidates = 3

5. 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.) 

Old price = ₹38
New price = ₹42

Increase = 4

$$\text{Inflation rate} = \frac{4}{38} \times 100 \approx 10.53\%$$

Answer: Inflation ≈ 10.5%

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

A number increased by 20% becomes 90

$$120\% = 90$$
$$\text{Original number} = \frac{90 \times 100}{120} = 75$$

Answer: The number is 75

7. 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. 

SP of each buffalo = ₹80,000

First buffalo (5% profit)

$$\text{CP} = \frac{80,000}{1.05} \approx 76,190$$

Second buffalo (10% loss)

$$\text{CP} = \frac{80,000}{0.90} \approx 88,889$$

Total CP ≈ 1,65,079
Total SP = 1,60,000

Loss = 5,079

Loss % = \( \frac{5079}{165079} \) x 100 approx 3.08%

Answer: Overall loss of about 3%

8. 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 × 0.5 (iv) p × 1.05 (ii) (v) p × 0.05 p + 1.50 (iii) p × 1.5 

Population last decade = p
Increase = 5%

$$\text{New population} = p \times 1.05$$

Correct option: (iv) p × 1.05

9. 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.

If demand fell by 85%, only 15% remains.

Correct statements:

(iii) Demand now is 15% of the demand a decade ago
(iv) Demand a decade ago was 15% of the demand now  (wrong logic)

Correct meanings:

• (iii)  only correct. All others are incorrect.

Answer: Only (iii)

Example 7: 

If one deposits ₹6000 in the bank, what is the amount after 3 years?

That depends on the choice of FD. There are two possibilities: 1. Option 1: The interest is paid out regularly (for example, every year). The principal amount is returned after the maturity period

2. Option 2: The interest gained every time (say after each year) is added back to the FD, thus increasing the principal amount for the subsequent period. After the maturity period, the entire amount is returned. This phenomenon is called compounding.

Example 8: 

What percent is the total amount received with respect to the amount deposited in both the options?

This can be calculated by finding total amount received amount deposited × 100. 

Without Compounding 

 7800 /6000 × 100 = 130% = 1.3. 

The total amount received = 6000 × (1 + 0.1 + 0.1 + 0.1) = 6000 × 1.3. 

The percentage gain over 3 years is 30%. 

 With Compounding

 7986 /6000 × 100 = 133.1% = 1.331. 

The total amount received = 6000 × 1.1 × 1.1 × 1.1. = 6000 × 1.331 

The percentage gain over 3 years is 33.1%

Figure it Out page no 22-23

1. 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.

1. Bank of Yahapur (10% p.a., 2 years, ₹20,000)

(A) Without compounding (Simple Interest)

Formula:

$$\text{SI} = \frac{P \times R \times T}{100}$$ $$= \frac{20000 \times 10 \times 2}{100} = 4000$$

Amount received = 20,000 + 4,000 = ₹24,000

(B) With compounding annually

Year 1:

$$10\% \text{ of } 20000 = 2000 \Rightarrow \text{Amount} = 22000$$

Year 2:

$$10\% \text{ of } 22000 = 2200$$

Final amount = 22,000 + 2,200 = ₹24,200

Method	Amount
Without compounding	₹24,000
With compounding	₹24,200

₹200 more with compounding

2. 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.

2. Bank of Wahapur (5% p.a., 4 years, ₹20,000)

(A) Without compounding

$$\text{SI} = \frac{20000 \times 5 \times 4}{100} = 4000$$

Amount = 20,000 + 4,000 = ₹24,000

(B) With compounding annually

Year-wise:

Year	Amount (₹)
Start	20,000
After 1 year	21,000
After 2 years	22,050
After 3 years	23,152.50
After 4 years	24,310.13

Method	Amount
Without compounding	₹24,000
With compounding	₹24,310.13

 3. Do you observe anything interesting in the solutions of the two questions above? Share and discuss. Let us try to generalise the pattern observed in each of the options. 

Observations

1. Compounding always gives more amount than simple interest.

2. Difference increases with time.

3. Interest is earned on interest itself in compounding.

Generalisation

If

• Principal = $P$

• Rate = $R\mathrm{\%}$

• Time = $t$ years

Without compounding (Simple Interest):

$$\text{Amount} = P + \frac{PRT}{100}$$

With compounding (annually):

$$\text{Amount} = P \left(1 + \frac{R}{100}\right)^t$$

Example 9: 

What is the amount we get back if we invest ₹6000 at an interest rate of 10% p.a. for ‘t’ years?

Then, find the formula for total interest gained for both:

• (a) Without compounding (Simple Interest)

• (b) With compounding (Compound Interest, annually)

• (A) Without Compounding (Simple Interest)

  Given:
  Principal $P=6000$ Rate $r = 10\% = 0.10$ Time $t$ years

  Amount after $t$ years:

  $$A = P + P \times r \times t$$
  $$A = 6000 + 6000 \times 0.10 \times t$$
  $$\boxed{A = 6000(1 + 0.10t)}$$

  Total Interest gained:

  $$\boxed{I = P \times r \times t = 6000 \times 0.10 \times t = 600t}$$Observation:

  Interest increases by a fixed amount every year → linear growth

• (B) With Compounding (Compound Interest – Annually)

  Amount after $t$ years:

  $$A = P(1 + r)^t$$
  $$A = 6000(1.10)^t$$

  Total Interest gained:

  $$\boxed{I = A - P = 6000(1.10)^t - 6000}$$

  👉 Observation:
  Interest is earned on interest + principal → exponential growth

Suppose we want to know the expression/formula to find the total interest amount gained at the end of the maturity period. What would be the formula for each of the two options?

Generalised Formula 

Type of Interest	Amount Formula	Interest Formula
Simple Interest	( A = P(1 + rt) )	( I = Prt )
Compound Interest	( A = P(1 + r)^t )	( I = P[(1+r)^t - 1] )

Key Continuity Idea 

Without compounding: growth is linear

• With compounding: growth is exponential

• For small $t$, both are close

• As $t$ increases, compound interest becomes much larger

This links directly to population growth, bacteria growth, inflation, etc.

Figure it Out page 23

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 × 6 × 4 

(ii) p × 0.6 × 4

(iii) p × 0.6 /100 × 4

(iv) p × 0.06 /100 × 4 

(v) p × 1.6 × 4

(vi) p × 1.06 × 4

 (vii) p + (p × 0.06 × 4)

4. Jasmine invests ₹p for 4 years at 6% (No compounding)

Correct expression:

$$\text{Amount} = p + \frac{p \times 6 \times 4}{100}$$

Correct option:

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

5. 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? 

Without compounding

$$\text{SI} = \frac{50000 \times 7 \times 3}{100} = 10500$$

Method	Interest
Without compounding	₹10,500
With compounding	₹11,250

₹750 more with compounding

 6. 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? 

Giridhar

$$\text{SI} = \frac{12500 \times 12 \times 3}{100} = 4500$$

Without compounding

$$1000 + 100t = 2000 \Rightarrow t = 10 \text{ years}$$

8. 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? 

$$A = 1.5 (1.03)^3 \approx 1.64 \text{ crore}$$

Expected population ≈ 1.64 crore

9. 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.

Initial = 5,06,000
Rate = 2.5% per hour
Time = 2 hours

$$A = 506000 (1.025)^2 \approx 5,31,616$$

Final bacteria ≈ 5,31,616

Example 10: 

A TV is bought at a price of ₹21,000. After 1 year, the value of the TV depreciates by 5%. Find the value of the TV after one year.

The amount of reduction in the value is 5% of 21,000 = 0.05 × 21,000 = 1050. The current value is 21,000 – 1050 = 19,950.

The value of the TV after 1 year will be 95% of the current value = 95% of 21,000 = 0.95 × 21,000 = 19,950.

The value of the TV after 1 year will be ₹19,950.

Example 11: 

The population of a village was observed to be reducing by about 10% every decade. If the current population is 1250, what is the expected population after 3 decades?

The population 1 decade later will be 0.9 times the population of the current decade. Therefore, the population after 1 decade will be 1250 × 0.9. 

The population after 2 decades will be 1250 × 0.9 × 0.9. 

The population after 3 decades will be 1250 × 0.9 × 0.9 × 0.9 = 911.25

 

First decade’s decrease = 0.1 × 1250 = 125.

Population after 1 decade = 1250 – 125 = 1125.

Second decade’s decrease = 0.1 × 1125 = 112.5 ≅ 112. 

Population after 2 decades = 1125 – 112 = 1013 

Third decade’s population decrease = 0.1 × 1013 = 101.3 ≅ 101. 

Population after 3 decades = 1013 – 101 = 912.

Rounding off, we can say that the expected population after 3 decades will be around 910.

A provision store is offering a stock clearance sale. Customers can choose one of the two options — 20% discount or ₹50 discount—for any purchase above ₹150. Which option would you choose if you want to: 

(i) buy items worth ₹180 

(ii) buy items worth ₹225 

(iii) buy items worth ₹300

Example 12: A bakery called Cakely is offering a 30% + 20% discount on all cakes. Another bakery called Cakify is offering a 50% discount on all cakes. Would you rather choose Cakely or Cakify if you want the cheaper cost?

30% + 20% is the same as 50%, the usage of 30% + 20% in shopping means compounding.

 Suppose you want to buy a cake worth ₹200. 

Cakely’s 30% + 20% 

→ Applying the 30% discount 

→ the price of cake is ₹200 – ₹60 = ₹140.

 Applying the 20% discount on ₹140

 → the price of cake is ₹140 – ₹28 = ₹112. 

Cakify’s 50% → The 50% discount makes the price of the cake ₹100

Example 13: After Surbhi bought cookware from the wholesaler, she kept a profit margin of 50% on all the products. To clear off the remaining stock, she thought she would offer a 50% discount and come out without any loss. (i) Do you think she didn’t make any loss? (ii) If she had sold goods (originally) for ₹12,000 after discount, how much loss did she incur? What is the percentage loss? (iii) What should have been the percentage discount offered so that she sold the goods at the price she had bought (i.e., no profit or loss)?

(i) This means the selling price is 3 /4 of the price the goods were bought at, i.e., a 25% loss. (ii) If she had sold goods worth ₹12,000, 

 0.75x = 12,000

 x = 16,000. 

She lost ₹4000.

 (iii) To sell the goods at the same price, the discount offered should be 

1.5x – d × (1.5x) = x 

d = 1 /3 = 0.33. 

The discount offered should have been 33.33%.

Ariba and Arun have some marbles. Ariba says, “The number of marbles with me is 120% of the marbles Arun has”. What would be an appropriate statement Arun could make comparing the number of marbles he has with Ariba’s?

The number of marbles with me is 120% of the marbles Arun has.” 

Step 1: Assume Arun’s marbles

Let Arun have 100 marbles (this makes percentage comparison easy).

Then Ariba has:

120% of 100=120 

Step 2: Compare Arun’s marbles with Ariba’s

Arun has 100, Ariba has 120.

So Arun’s marbles as a percentage of Ariba’s:

 Appropriate statement by Arun

“The number of marbles I have is 83⅓% of the marbles Ariba has.” 

If one quantity is 120% of another,

the other is not 80% back — it must be calculated by division.

Figure it Out Page no 27 - 28 

1. 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? 

• Population in 2025 = 250% of population in 2000

• Population in 2000 = 50 lakhs

$$250\% = \frac{250}{100} = 2.5$$
$$\text{Population in 2025} = 2.5 \times 50 = 125 \text{ lakhs}$$

Answer: 125 lakhs (1.25 crore)

2. 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. [Hint: Writing these numbers in the standard form and estimating can help] 

Country	Population	% of world population
India	1.46 billion	(\approx \frac{1.46}{8.2} \times 100 \approx 18%)
USA	347 million = 0.347 bn	(\approx 4%)
Bangladesh	175 million = 0.175 bn	(\approx 2%)
Germany	83 million = 0.083 bn	(\approx 1%)

Matching:

• India → 18%

• USA → 4%

• Bangladesh → 2%

• Germany → 1%

3. 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 + 18 100 

(iv) 8250 × 18

 (v) 8250 × 1.18 

(vi) 8250 + 8250 × 0.18 

 (vii) 1.8 × 8250  

Price = ₹8250

GST = 18%

Correct expressions:

• $8250 \times 1.18$

• $8250 + 8250 \times 0.18$

Correct options: (v) and (vi)

4. 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 × 0.05 × 0.02 × 0.03. 

(ii) The population after three months was p × 1.05 × 0.98 × 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. 

Changes:

• Month 1: +5%

• Month 2: −2%

• Month 3: −3%

Correct method → successive multiplication

$$p \times 1.05 \times 0.98 \times 0.97$$

Correct statements:(ii), (vi)

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

Let CP = 100

• SP after 35% profit = 135

• Discount 30% → Selling price = $135 \times 0.7 = 94.5$

Loss = $100 − 94.5 = 5.5$

Result: Loss of 5.5%

• Since the discount is applied on a larger base, it wipes out the profit and more

Profit % and Discount % cannot be compared directly unless they are on the same base

6. What percentage of area is occupied by the region marked ‘E’ in the figure? 

Using dot-grid:

• The whole figure is drawn on a dot grid, so we can count equal unit squares.

• The outer shape is a big rectangle, divided into regions A, B, C, D, E.

• Region E is the triangular region at the bottom-left.

$$\text{Area of E} = \frac{1}{2} \times 4 = 2 \text{ unit squares}$$

$$\text{<span class="katex-display"><span class="katex"><span class="katex-display"><span class="katex"><span aria-hidden="true" class="katex-html"><span class="base"><span class="mord"><br></span></span></span></span></span></span></span>}$$

f 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?

Examples

5% of 40 = 2

40% of 5 = 2

25% of 12 = 3

12% of 25 = 3

15% of 60 = 9

60% of 15 = 9

Observation

The value of x% of y is always equal to y% of x.

Percentage is just multiplication by a fraction, and multiplication does not depend on order.

General statement x% of y = y% of x

Observation & General rule proven

$$x\% \text{ of } y = y\% \text{ of } x$$

8. 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]. D B E C (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? 

• Total students = 100%

• 40% are Grade 8

• Remaining 60% are Grade 9

• Among Grade 8, 60% are girls

•  Percentage of Grade 8 girls 

  

  
  

  • Grade 8 students = 40%

  • Girls among Grade 8 = 60% of 40%

  $$\text{Grade 8 girls} = \frac{60}{100} \times 40\% = 24\%$$

  $$\boxed{24\% \text{ of the students are Grade 8 girls}}$$
  

• Grade 8 students

  $$40\% \text{ of } 160 = \frac{40}{100} \times 160 = 64$$

  Grade 8 girls

  $$60\% \text{ of } 64 = \frac{60}{100} \times 64 = 38.4$$

  So approximately:

  $$\boxed{38 \text{ (approximately) Grade 8 girls}}$$

(i) Percentage of Grade 8 girls:

$$40\% \times 60\% = 24\%$$

(ii) Total students = 160

24%×160=38.4≈38 students

9. 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? 

• SP of 3 pencils = CP of 5 pencils

Let CP of 1 pencil = ₹1
CP of 5 = ₹5 → SP of 3 = ₹5
SP of 1 = ₹1.67

Profit = ₹0.67 per pencil

$$\text{Profit \%} = \frac{0.67}{1} \times 100 \approx 67\%$$

10. 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? 

• Increase in fare last year = 3%

• Increase in fare this year = 4%

Successive percentage increases → use multiplication, not addition.
Let the original bus fare be ₹100

After 1st year (3% increase)

$$100+3\mathrm{\%}\text{ of }100=100\times 1.03=103$$

After 2nd year (4% increase)

$$$$103+4% of 103=103×1.04=107.12103 + 4\% \text{ of } 103 = 103 \times 1.04 = 107.12

Overall Increase

$$\text{Increase}=107.12-100=\mathrm{7.12\%}\mathrm{<br>}\mathrm{<br>}$$

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

Area of a rectangle:

$$\text{Area} = \text{Length} \times \text{Breadth}$$

If area is constant, then:

• When length increases,

• breadth must decrease
  so that the product remains the same.

This is inverse proportionality.

10% increase means:

$$\text{New length} = 1 + \frac{10}{100} = 1.1 \text{ times original length}$$

So,

$$L \rightarrow 1.1L$$

Since area is unchanged:

$$L \times B = (1.1L) \times B'$$

Cancel $L$:

$$B = 1.1B'$$
$$B' = \frac{B}{1.1}$$

👉 This is why you wrote:

$$\text{Breadth} = \frac{1}{1.1}$$

So the new breadth is $\frac{1}{1.1}$ times the original breadth.

Original breadth = 1

New breadth = $\frac{1}{1.1}$

Decrease:

$$1 - \frac{1}{1.1}$$

Convert 1 as $\frac{1.1}{1.1}$:

$$\frac{1.1 - 1}{1.1} = \frac{0.1}{1.1}$$

Step 5: Convert decrease into percentage

Breadth decreases by 9.09%​

12. 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. 

Convert each percentage to grams:

$$\text{Ingredient mass} = \% \times 65$$

Total weight of chips packet = 65 g

Percentage composition:

$$\text{Weight of ingredient} = \frac{\text{Percentage}}{100} \times \text{Total weight}$$

Ingredient	Percentage	Weight g
Potato	70%	70% of 65 = 45.5 g
Vegetable oil	24%	24% of 65 = 15.6 g
Salt	3%	3% of 65 = 1.95 g
Spices	3%	3% of 65 = 1.95 g
Total   100% 65g

13. 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?

Three shops sell the same item at the same price ₹100 each.
They offer the following deals:

• Shop A: Buy 1 and get 1 free

• Shop B: Buy 2 and get 1 free

• Shop C: Buy 3 and get 1 free

(i) Effective price per item

Shop A: Buy 1 Get 1 Free

• Pay for 1 item = ₹100

• Total items received = 2

$$\text{Effective price per item} = \frac{100}{2} = ₹50$$

Shop B: Buy 2 Get 1 Free

• Pay for 2 items = ₹200

• Total items received = 3

$$\text{Effective price per item} = \frac{200}{3} \approx ₹66.67$$

Shop C: Buy 3 Get 1 Free

• Pay for 3 items = ₹300

• Total items received = 4

$$\text{Effective price per item} = \frac{300}{4} = ₹75$$

Shop	Effective price per item
A	₹50
B	₹66.67
C	₹75

(ii) Percentage discount

$$\text{Discount \%} = \frac{\text{Free items}}{\text{Total items}} \times 100$$

Shop	Discount %
A	(\frac{1}{2} \times 100 = 50%)
B	(\frac{1}{3} \times 100 = 33.33%)
C	(\frac{1}{4} \times 100 = 25%)

(iii) If you need 4 items, which shop is best?

• Shop A: Buy 2 → Get 2 free → Pay ₹200

• Shop B: Buy 4 → Get 1 free → Pay ₹400

• Shop C: Buy 4 → Get 1 free → Pay ₹300

Best choice: Shop A (lowest cost)

14. 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%? 

Total people in the room = 100

Left-handed people = 99% of 100 = 99
Right-handed people = 1

Let x left-handed people leave

Then:
Left-handed people remaining = 99 − x
Total people remaining = 100 − x

New percentage condition

$$\frac{\text{Left-handed people}}{\text{Total people}} = 98\%$$
$$\frac{99-x}{100-x}=\frac{98}{100}$$

9−x100−x=4950\frac{99 - x}{100 - x} = \frac{49}{50}

Cross multiply:

$$50(99 - x) = 49(100 - x)$$
$$4950 - 50x = 4900 - 49x$$
$$4950 - 4900 = 50x - 49x$$
$$50 = x$$

After 50 left-handed people leave:

• Left-handed = 49

• Right-handed = 1

• Total = 50

$$\frac{49}{50} \times 100 = 98\%$$

 15. 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

The graph shows percentage of people who can use computers by:

• Age group

• Gender (Male / Female)

(i) People in their twenties are the most computer-literate

True
Both male and female percentages are highest in the twenties group.

(ii) Women lag behind men in all age groups

 True
In every age group, the male bar is higher than the female bar.

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

 False
The graph shows percentage, not population size.
We cannot compare how many people are there.

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

 False
Combined (male + female) is below 25%.

(v) Less than 1 in 10 aged 60+ can use computers

True
Seniors show around 4–6%, which is less than 10%.

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

False
Male + female together ≈ 37%, not 50%.

Final Correct Statements

(i), (ii), (v)