Fractions in Disguise

Looking at the bar model diagram showing the equivalence between $\dfrac{3}{4}$ and 75%, can you tell what percentage of the colour was made using yellow?

Answer

Given:

The red paint makes up $\dfrac{3}{4}$ of the mixture, which is equal to 75%.

The whole colour (the complete mixture) represents 100%.

Since the colour is made by mixing only red paint and yellow paint:

Percentage of yellow = (Total percentage) – (Percentage of red)

= 100% – 75%

= 25%

Hence, 25% of the colour was made using yellow.

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

Answer

We have:

$\phantom{=} 24$

Hence, 24% expressed as a fraction is $\dfrac{6}{25}$.

Express the following fractions as percentages.

(i) $\dfrac{3}{5}$

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

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

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

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

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

Answer

To express a fraction as a percentage, we multiply the fraction by 100.

(i) $\dfrac{3}{5}$

$\phantom{=} \dfrac{3}{5} \times 100 \\[1em] = \dfrac{3}{\cancel{5}} \times \overset{20}{\cancel{100}} \\[1em] = 3 \times 20 \\[1em] = 60$

Hence, $\mathbf{\dfrac{3}{5}}$ = 60%

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

$\phantom{=} \dfrac{7}{14} \times 100 \\[1em] = \dfrac{\overset{1}{\cancel{7}}}{\underset{2}{\cancel{14}}} \times 100 \\[1em] = \dfrac{1}{\cancel{2}} \times \overset{50}{\cancel{100}} \\[1em] = 50$

Hence, $\mathbf{\dfrac{7}{14}}$ = 50%

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

$\phantom{=} \dfrac{9}{20} \times 100 \\[1em] = \dfrac{9}{\cancel{20}} \times \overset{5}{\cancel{100}} \\[1em] = 9 \times 5 \\[1em] = 45$

Hence, $\mathbf{\dfrac{9}{20}}$ = 45%

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

$\phantom{=} \dfrac{72}{150} \times 100 \\[1em] = \dfrac{72}{\underset{3}{\cancel{150}}} \times \overset{2}{\cancel{100}} \\[1em] = \dfrac{\overset{24}{\cancel{72}}}{\cancel{3}} \times 2 \\[1em] = 24 \times 2 \\[1em] = 48$

Hence, $\mathbf{\mathbf{\dfrac{72}{150}}}$ = 48%

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

$\phantom{=} \dfrac{1}{3} \times 100 \\[1em] = \dfrac{100}{3} \\[1em] = 33\dfrac{1}{3}$

Hence, $\mathbf{\dfrac{1}{3}}$ = $33\dfrac{1}{3}$% (≈ 33.33%)

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

$\phantom{=} \dfrac{5}{11} \times 100 \\[1em] = \dfrac{500}{11} \\[1em] = 45\dfrac{5}{11}$

Hence, $\mathbf{\dfrac{5}{11}}$ = $45\dfrac{5}{11}$% (≈ 45.45%)

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

1. 10%
2. 15%
3. 25%
4. 60%
5. 40%
6. None of these

Answer

Given:

Total number of marbles = 25

Number of white marbles = 15

Fraction of white marbles = $\dfrac{15}{25}$

Percentage of white marbles = (Fraction of white marbles) × 100

$= \dfrac{15}{25} \times 100 \\[1em] = \dfrac{15}{\cancel{25}} \times \overset{4}{\cancel{100}} \\[1em] = 15 \times 4 \\[1em] = 60$

Hence, 60% of her marbles are white, and option 4 is the correct option.

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

Answer

Given:

Total number of students = 80

Number of students who walk = 15

Fraction of students who walk = $\dfrac{15}{80}$

Percentage of students who walk = (Fraction of students who walk) × 100

$= \dfrac{15}{80} \times 100 \\[1em] = \dfrac{15}{\underset{4}{\cancel{80}}} \times \overset{5}{\cancel{100}} \\[1em] = \dfrac{15}{4} \times 5 \\[1em] = \dfrac{75}{4} \\[1em] = 18.75$

Hence, 18.75% of the students come to school by walking.

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.

Answer

The total distance from Start to Finish represents 100% of the race. The fraction of this distance each runner has covered is read off and matched to the nearest given percentage.

Reading the positions from the picture:

A has covered a little less than two-fifths of the track ⇒ approximately 38%

B is just past the middle of the track ⇒ approximately 55%

C has covered close to three-quarters of the track ⇒ approximately 72%

D is very near the Finish line ⇒ approximately 93%

Hence, A → 38%, B → 55%, C → 72% and D → 93%.

(The options 20% and 84% do not match any runner.)

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

(i) 50% ............... 5%

(ii) $\dfrac{5}{10}$ ............... 50%

(iii) $\dfrac{3}{11}$ ............... 61%

(iv) 30% ............... $\dfrac{1}{3}$

Answer

(i) 50% > 5%

50 parts out of 100 is clearly more than 5 parts out of 100.

(ii) $\dfrac{5}{10}$ = 50%

$\dfrac{5}{10} = \dfrac{1}{2}$, which is the same as 50%.

(iii) $\dfrac{3}{11}$ < 61%

In $\dfrac{3}{11}$, the numerator 3 is less than half of the denominator 11, so $\dfrac{3}{11}$ is less than 50%, and therefore less than 61%.

(iv) 30% < $\dfrac{1}{3}$

$\dfrac{1}{3} = 33\dfrac{1}{3}$, which is greater than 30%.

Madhu and Madhav each ate biscuits of a different variety. Madhu's biscuits had 25% sugar, while Madhav's had 35% sugar. Suppose Madhu ate 120 g of biscuits and Madhav ate 95 g of biscuits. Who consumed more sugar? Try to find out.

Answer

Given:

Madhu ate 120 g of biscuits having 25% sugar.

Madhav ate 95 g of biscuits having 35% sugar.

The weight of sugar is proportional to the weight of the biscuits consumed.

Sugar eaten by Madhu = 25% of 120 g

$= \dfrac{25}{100} \times 120 \text{ g} \\[1em] = \dfrac{\overset{1}{\cancel{25}}}{\underset{4}{\cancel{100}}} \times 120 \text{ g} \\[1em] = \dfrac{1}{\cancel{4}} \times \overset{30}{\cancel {120}} \text{ g} \\[1em] = 30 \text{ g}$

Sugar eaten by Madhav = 35% of 95 g

$= \dfrac{35}{100} \times 95 \text{ g} \\[1em] = \dfrac{\overset{7}{\cancel{35}}}{\underset{20}{\cancel{100}}} \times 95 \text{ g} \\[1em] = \dfrac{7}{\underset{4}{\cancel{20}}} \times \overset{19}{\cancel{95}} \text{ g} \\[1em] = \dfrac{7}{4} \times 19 \text{ g} \\[1em] = \dfrac{7 \times 19}{4} \text{ g} \\[1em] = \dfrac{133}{4} \text{ g} \\[1em] = 33.25 \text{ g}$

Comparing the two amounts, 33.25 g > 30 g.

Hence, Madhav consumed more sugar (33.25 g) than Madhu (30 g).

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.

Answer

25% of a value is one-fourth of the value.

10% of a value is one-tenth of the value.

20% of a value is twice the 10% value.

5% of a value is half the 10% value.

Using these observations, we get:

Using the understanding that 20% of a value is double 10% of the same value,

(i) mentally calculate how much 40% of the values in the table above would be.

(ii) mentally calculate how much 15% of the values in the table would be.

(iii) Suppose you have to mentally calculate the following percentages of some value: 75%, 90%, 70%, 55%. How would you do it? Discuss.

Answer

(i) 40% of a value is double 20% of the same value because

⇒ 40% = 2 × 20%

Therefore, we double the values in the 20% row:

Hence, the values of 40% are 40, 80, 20, 32, 4, 14 and 114.8 respectively.

(ii)

We observe that (10% of a value) + (5% of the same value) = 15% of that value, since 10 parts and 5 parts together make 15 parts out of 100.

⇒ 15% of a value = (10% of the value) + (5% of the value)

Adding the 10% row and the 5% row of the table:

Hence, the values of 15% are 15, 30, 7.5, 12, 1.5, 5.25 and 43.05 respectively.

(iii)

To calculate these percentages mentally, we can express them using simpler percentages.

(a) 75%:

75% = 50% + 25%

So, 75% of a value = half of the value + one-fourth of the value.

(b) 90%:

90% = 100% − 10%

So, 90% of a value = the whole value − one-tenth of the value.

(c) 70%:

70% = 50% + 20%

So, 70% of a value = half of the value + one-fifth of the value.

(d) 55%:

55% = 50% + 5%

So, 55% of a value = half of the value + one-twentieth of the value.

Hence, we can mentally calculate these percentages by breaking them into simpler percentages such as 50%, 25%, 20%, 10% and 5%, and then combining the results.

To find 10% of a quantity, what decimal value should be multiplied?

Answer

A percentage is a fraction with denominator 100, which can be written as a decimal.

10% = $\dfrac{10}{100}$

= $\dfrac{1}{10}$

= 0.1

So, to find 10% of a quantity, we multiply the quantity by 0.1.

Hence, the decimal value to be multiplied is 0.1.

Complete the following table:

Answer

Each percentage z% is written as the fraction $\dfrac{z}{100}$, which is then expressed as a decimal.

For example:

25% = $\dfrac{25}{100}$ = 0.25

1% = $\dfrac{1}{100}$ = 0.01

43% = $\dfrac{43}{100}$ = 0.43

Completing the table in the same way:

Kishanlal aims to achieve a daily sales of at least ₹5000. In the next two days, he made ₹5000 and ₹6000 respectively. What percentage of his target are these values?

(i) On Days 5 and 6 his sales were ₹7800 and ₹9550 respectively. Calculate the percentage of the target (₹5000) achieved on these days.

(ii) 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.

Answer

Given:

Target sales = ₹5000

Percentage of target achieved = $\dfrac{\text{Sales made}}{\text{Target}} \times 100$

For sales of ₹5000:

= $\dfrac{5000}{5000} \times 100$

= 1 × 100

= 100%

For sales of ₹6000:

$= \dfrac{6000}{5000} \times 100 \\[1em] = \dfrac{6}{5} \\[1em] = \dfrac{6 \times 100}{5} \\[1em] = \dfrac{600}{5} \\[1em]$

$= 120$%

∴ ₹5000 is 100% of the target and ₹6000 is 120% of the target.

(i)

Given:

Target sales = ₹5000

Percentage of target achieved = $\dfrac{\text{Sales made}}{\text{Target}} \times 100$

For Day 5 (sales of ₹7800):

$= \dfrac{7800}{5000} \times 100 \\[1em] = \dfrac{78}{\underset{1}{\cancel{50}}} \times {\overset{2}{\cancel{100}}} \\[1em] = \dfrac{78}{1} \times 2 \\[1em]$

$= 156$%

For Day 6 (sales of ₹9550):

$= \dfrac{9550}{5000} \times 100 \\[1em] = \dfrac{955}{{\underset{5}{\cancel{500}}}} \times {\overset{1}{\cancel{100}}} \\[1em] = \dfrac{955}{5} \times 1 \\ \\[1em] = \dfrac{955}{5} \\[1em]$

$= 191$%

Hence, on Day 5 he achieved 156% of his target and on Day 6 he achieved 191% of his target.

(ii)

Given:

Target sales = ₹5000

Sales made = (Percentage of target achieved) × Target

For Day 7 (150% of target):

$= \dfrac{150}{100} \times 5000 \\[1em] = \dfrac{{\overset{3}{\cancel{150}}}}{{\underset{2}{\cancel{100}}}} \times 5000 \\[1em] = \dfrac{3}{2} \times 5000 \\[1em] = \dfrac{3 \times 5000}{2} \\[1em] = \dfrac{15000}{2} \\[1em] = ₹7500$

For Day 8 (210% of target):

$= \dfrac{210}{100} \times 5000 \\[1em] = \dfrac{21}{10} \times 5000 \\[1em] = \dfrac{21 \times 5000}{10} \\[1em] = 21 \times 500 \\[1em] = ₹10500$

∴ The sales made were ₹7500 on Day 7 and ₹10500 on Day 8.

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

Answer

Each percentage z% is written as the fraction $\dfrac{z}{100}$ and then as a decimal.

For example:

90% = $\dfrac{90}{100}$ = 0.9

250% = $\dfrac{250}{100}$ = 2.5

28.9% = $\dfrac{28.9}{100} = \dfrac{289}{1000}$ = 0.289

Completing the table in the same way:

The approximate locations of these values are marked on the diagram below (where 100% corresponds to 1, 200% to 2, and so on):

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

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

Answer

In each bar model, the strip representing 100% is divided into a number of equal parts, so every part stands for an equal share of 100%.

(i)

The strip is divided into 5 equal parts.

Each part = $\dfrac{100}{5} = 20$%

In the second strip, 4 parts make up 80% and this equals 60.

Hence 1 part (20%) = 15, and the whole strip (5 parts = 100%) = 75.

(ii)

Each strip is divided into 10 equal parts.

Each part = $\dfrac{100}{10} = 10$%

⇒ In the first strip, the single marked part = 10%.

In the second strip, the whole = 90 (100%) and the marked portion is 6 parts = 60%.

60% of 90

= $\dfrac{60}{100} \times 90$

= 54

(iii)

Each strip is divided into 4 equal parts.

Each part = $\dfrac{100}{4} = 25$%

⇒ In the first strip, the single marked part = 25%.

In the second strip, the whole = 140 (100%) and the marked portion is 3 parts = 75%.

75% of 140

= $\dfrac{75}{100} \times 140$

= 105

Hence, the missing numbers are — (ii) 10% and 54; (iii) 25% and 105.

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

Answer

(i) 25% of 160

$= \dfrac{25}{100} \times 160 \\[1em] = \dfrac{\overset{1}{\cancel{25}}}{\underset{4}{\cancel{100}}} \times 160 \\[1em] = \dfrac{1}{\cancel{4}} \times \overset{40}{\cancel{160}} \\[1em] = 40$

Hence, 25% of 160 = 40.

(ii) 16% of 250

$= \dfrac{16}{100} \times 250 \\[1em] = \dfrac{16}{\underset{2}{\cancel{100}}} \times \overset{5}{\cancel{250}} \\[1em] = \dfrac{\overset{8}{\cancel{16}}}{\cancel{2}} \times 5 \\[1em] = 8 \times 5 \\[1em] = 40$

Hence, 16% of 250 = 40.

(iii) 62% of 360

$= \dfrac{62}{100} \times 360 \\[1em] = \dfrac{62 \times 360}{100} \\[1em] = \dfrac{22320}{100} \\[1em] = 223.2$

Hence, 62% of 360 = 223.2.

(iv) 140% of 40

$= \dfrac{140}{100} \times 40 \\[1em] = \dfrac{14}{10} \times 40 \\[1em] = 14 \times 4 \\[1em] = 56$

Hence, 140% of 40 = 56.

(v) 1% of 1 hour

1 hour = 60 minutes.

1% of 1 hour

$= \dfrac{1}{100} \times 60 \text{ minutes} \\[1em] = \dfrac{60}{100} \text{ minutes} \\[1em] = 0.6 \text{ minutes} \\[1em] = 36 \text{ seconds} \quad \text{[Since } 0.6 \times 60 = 36\text{]}$

Hence, 1% of 1 hour = 36 seconds.

(vi) 7% of 10 kg

10 kg = 10000 g.

7% of 10 kg

$= \dfrac{7}{100} \times 10000 \text{ g} \\[1em] = 7 \times 100 \text{ g} \\[1em] = 700 \text{ g} \\[1em] = \dfrac{700}{1000} \text{ kg} \\[1em] = 0.7 \text{ kg}$

Hence, 7% of 10 kg = 700 g (i.e., 0.7 kg).

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

Answer

Given:

Total deep orange paint = 60 ml

Red paint = $\dfrac{3}{4}$ of the deep orange paint

$\phantom{=} \text{Red paint} = \dfrac{3}{4} \text{ of } 60 \text{ ml} \\[1em] = \dfrac{3}{4} \times 60 \text{ ml} \\[1em] = \dfrac{3}{{\cancel{4}}} \times {\overset{15}{\cancel{60}}} \text{ ml} \\[1em] = 3 \times 15 \text{ ml} \\[1em] = 45 \text{ ml}$

Hence, Surya used 45 ml of red paint.

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

Answer

(i) 50% of 510 $\quad$ < $\quad$ 50% of 515

Since the percentage is the same in both cases, the quantity with the larger base will be larger.

As 510 < 515,

50% of 510 < 50% of 515.

(ii) 37% of 148 $\quad$ < $\quad$ 73% of 148

Since the base is the same in both cases, the quantity with the larger percentage will be larger.

As 37% < 73%,

37% of 148 < 73% of 148.

(iii) 29% of 43 $\quad$ < $\quad$ 92% of 110

29% of 43 is less than one-third of 43, whereas 92% of 110 is almost the whole of 110.

∴ 29% of 43 < 92% of 110.

(iv) 30% of 40 $\quad$ < $\quad$ 40% of 50

30% of 40 is less than one-third of 40, while 40% of 50 is nearly half of 50.

∴ 30% of 40 < 40% of 50.

(v) 45% of 200 $\quad$ > $\quad$ 10% of 490

45% of 200 is nearly half of 200, whereas 10% of 490 is one-tenth of 490.

∴ 45% of 200 > 10% of 490.

(vi) 30% of 80 $\quad$ > $\quad$ 24% of 64

30% of 80 is greater than one-fourth of 80, while 24% of 64 is slightly less than one-fourth of 64.

∴ 30% of 80 > 24% of 64.

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.

Answer

(i) Given 30% of k = 70. We use multiples of 30%.

60% of k = 2 × (30% of k) = 2 × 70 = 140

90% of k = 3 × (30% of k) = 3 × 70 = 210

120% of k = 4 × (30% of k) = 4 × 70 = 280

⇒ 60% of k is 140, 90% of k is 210, 120% of k is 280.

(ii) Given 100% of m = 215.

10% of m = $\dfrac{215}{10}$ = 21.5

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

6% of m = 6 × (1% of m) = 6 × 2.15 = 12.9

⇒ 10% of m is 21.5, 1% of m is 2.15, 6% of m is 12.9.

(iii) Given 90% of n = 270.

9% of n = 90% of n ÷ 10

= $\dfrac{270}{10}$

= 27

18% of n = 2 × (9% of n) = 2 × 27 = 54

10% of n = $\dfrac{270}{9}$ = 30 ⇒ 100% of n = 10 × 30 = 300

⇒ 9% of n is 27, 18% of n is 54, 100% of n is 300.

(iv) This is an open-ended activity. A possible pair of such questions you can pose to your peers:

⇒ 40% of p is 24, 80% of p is ..........., 20% of p is ..........., 100% of p is ........... (Answers: 48, 12, 60.)

⇒ 50% of q is 35, 25% of q is ..........., 150% of q is ..........., 100% of q is ........... (Answers: 17.5, 105, 70.)

Fill in the blanks:

(i) 3 is ............... % of 300.

(ii) ............... is 40% of 4.

(iii) 40 is 80% of ................ .

Answer

(i) Let 3 be x% of 300.

x% of 300 = 3

$\Rightarrow x = \dfrac{3}{300} \times {100} \\[1em] = \dfrac{3}{\underset{3}{\cancel{300}}} \times \overset{1}{\cancel{100}} \\[1em] = \dfrac{3}{3} \\[1em] = 1$

⇒ 3 is 1% of 300.

(ii) Required value = 40% of 4.

40% of 4

$= \dfrac{40}{100} \times 4 \\[1em] = \dfrac{\overset{4}{\cancel{40}}}{\underset{5}{\cancel{100}}} \times 4 \\[1em] = \dfrac{2}{5} \times 4 \\[1em] = \dfrac{8}{5} \\[1em] = 1.6$

⇒ 1.6 is 40% of 4.

(iii) Let 40 be 80% of y.

80% of y = 40

$\Rightarrow \dfrac{80}{100} \times y = 40 \\[1em] \Rightarrow y = 40 \times \dfrac{100}{80} \\[1em] \Rightarrow y = 40 \times \dfrac{\overset{5}{\cancel{100}}}{\underset{4}{\cancel{80}}} \\[1em] = \overset{10}{\cancel{40}} \times \dfrac{5}{\cancel{4}} \\[1em] = 50$

⇒ 40 is 80% of 50.

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

Answer

We compare the two durations by converting each to the same unit.

A day = 24 hours, so:

10% of a day = $\dfrac{10}{100} \times 24$ hours

= 2.4 hours = 144 minutes

A week = 7 days = 7 × 24 = 168 hours, so:

1% of a week = $\dfrac{1}{100} \times 168$ hours

= 1.68 hours = 100.8 minutes

⇒ Since 144 minutes > 100.8 minutes, 10% of a day is longer than 1% of a week.

A similar question you can pose to your peers: Is 5% of an hour longer than 10% of 10 minutes? (5% of 60 min = 3 min; 10% of 10 min = 1 min; so yes, 5% of an hour is longer.)

Hence, 10% of a day (144 minutes) is longer than 1% of a week (100.8 minutes).

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?

Answer

On any day, if the bull is given a certain amount of fodder, the percentage eaten is:

$\phantom{=} \text{Percentage eaten} = \dfrac{\text{fodder eaten}}{\text{fodder given}} \times 100$

On the n^th day she gives $(n+1)$ units and the bull eats $n$ units, so the percentage eaten = $\dfrac{n}{n+1} \times 100$.

A few values are:

$\text{Day 1: } \dfrac{1}{2} \times 100 = 50$%

$\text{Day 2: } \dfrac{2}{3} \times 100 = 66.67$%

$\text{Day 3: } \dfrac{3}{4} \times 100 = 75$%

$\text{Day 4: } \dfrac{4}{5} \times 100 = 80$%

$\text{Day 5: } \dfrac{5}{6} \times 100 = 83.33$%

$\vdots \\[1em]$

$\text{Day 99: } \dfrac{99}{100} \times 100 = 99$%

⇒ Although the bull leaves exactly 1 unit uneaten every single day, the percentage of fodder it eats keeps increasing day after day and gets closer and closer to 100% (without ever reaching it). This is because the uneaten part as a percentage, $\dfrac{1}{n+1} \times 100$, keeps shrinking as the amount of fodder grows.

Hence, the percentage eaten rises steadily towards 100%, even though the bull always leaves 1 unit uneaten.

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?

Answer

Given:

Time to pick 20% of the plantation = 18 days

Time to pick the entire (100%) plantation = ?

Since 100% is 5 times 20%, the time taken is also 5 times (at the same rate of work).

Time for 100% = $\dfrac{100}{20} \times 18$ days

= 5 × 18 days

= 90 days

⇒ The assumption that the rate of work stays the same is necessary because only then is the time taken directly proportional to the area picked. If the workers worked faster or slower (due to fatigue, weather, more or fewer workers, etc.), the time would not scale in this simple way and the answer would change.

Hence, the workers will take 90 days to pick the entire plantation.

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?

Answer

Given:

Total training time = 90 minutes

warm up : play : cool down = 10% : 80% : 10%

Warm up = 10% of 90 = $\dfrac{10}{100} \times 90$ = 9 min

Play = 80% of 90 = $\dfrac{80}{100} \times 90$ = 72 min

Cool down = 10% of 90 = $\dfrac{10}{100} \times 90$ = 9 min

⇒ Check: 9 + 72 + 9 = 90 minutes. $\quad$[Verified]

∴ Warm up = 9 minutes, Play = 72 minutes, and Cool down = 9 minutes.

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.

Answer

Taking the worldwide population as about 8.2 billion:

People in the Northern Hemisphere = 90% of 8.2 billion

= $\dfrac{90}{100} \times 8.2$ billion

= 0.9 × 8.2 billion

= 7.38 billion

⇒ About 7.38 billion (roughly 7.4 billion) people live in the Northern Hemisphere.

Hence, approximately 7.38 billion people live in the Northern Hemisphere.

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?

Answer

(i) The proportion (percentage) of each ingredient depends only on the recipe, not on the number of people. Changing from 4 people to 8 people increases the total quantity, but the share of each ingredient stays the same.

⇒ For 8 people, the proportions remain Rava: 40%, Sugar: 40%, and Ghee: 20%.

Hence, for 8 people, the proportions remain Rava: 40%, Sugar: 40%, and Ghee: 20%.

(ii) Total weight of ingredients = 2 kg = 2000 g.

Rava = 40% of 2000 = $\dfrac{40}{100}$ × 2000 = 800 g

Sugar = 40% of 2000 = $\dfrac{40}{100}$ × 2000 = 800 g

Ghee = 20% of 2000 = $\dfrac{20}{100}$ × 2000 = 400 g

⇒ Check: 800 + 800 + 400 = 2000 g = 2 kg. $\quad$[Verified]

Hence, Rava = 800 g, Sugar = 800 g, and Ghee = 400 g.

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 (DEF and Zacni) and wondering which has a larger share of badam. Can you figure it out? Which product uses a smaller proportion of food chemicals?

It is easier to compare the proportions of the ingredients if we convert them into percentages. For example, DEF's sugar content as a percentage of total weight = $\dfrac{99}{150} \times 100 = 66$%.

Complete this table by calculating the percentages to answer the questions:

Check if the percentages of each product add up to 100.

Answer

To express each ingredient as a percentage of the total weight, we use:

Percentage of an ingredient $= \dfrac{\text{Weight of the ingredient}}{\text{Total weight}} \times 100$

For DEF (Total weight = 150 g):

Milk Solids = $\dfrac{30}{150}$ × 100 = 20%

Badam Powder = $\dfrac{12}{150}$ × 100 = 8%

Food Chemicals = $\dfrac{9}{150}$ × 100 = 6%

For Zacni (Total weight = 400 g):

Sugar = $\dfrac{272}{400}$ × 100 = 68%

Milk Solids = $\dfrac{64}{400}$ × 100 = 16%

Badam Powder = $\dfrac{40}{400}$ × 100 = 10%

Food Chemicals = $\dfrac{24}{400}$ × 100 = 6%

The completed table is:

Comparing the badam powder content ⇒ Zacni has 10% while DEF has 8%, so Zacni has the larger share of badam.

Comparing the food chemical content ⇒ both DEF and Zacni contain 6%, so they use the same proportion of food chemicals.

Hence, Zacni has the larger share of badam powder (10% as against DEF's 8%), and both products use an equal proportion of food chemicals (6% each).

Adding the percentages of all the ingredients of each product:

For DEF:

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

For Zacni:

68% + 16% + 10% + 6% = 100%

Hence, the percentages of each product add up to 100%.

The journey of a sweater from the manufacturer to the customer shows the cost price (CP), the marked price (MP), and the selling price (SP) at each step:

Kishanlal (the retailer) made a percentage profit of $\dfrac{130}{300} \times 100 = 43.3$%. Find the profit percentage of the wholesaler and the manufacturer.

Answer

The percentage profit is calculated with respect to the cost price (the price at which the goods were bought).

Percentage profit $= \dfrac{\text{Profit}}{\text{Cost Price}} \times 100$, where Profit = Selling Price − Cost Price.

Manufacturer:

Cost Price = ₹230, Selling Price = ₹253

Profit = ₹253 - ₹230 = ₹23

Percentage profit = $\dfrac{23}{230} \times 100 = 10$%

Wholesaler:

Cost Price = ₹253, Selling Price = ₹300

Profit = ₹300 - ₹253 = ₹47

Percentage profit = $\dfrac{47}{253} \times 100 = 18.58$%

Hence, the manufacturer made a percentage profit of 10% and the wholesaler made a percentage profit of about 18.58%.

(i) Shambhavi owns a stationery shop. She procures 200 page notebooks at ₹36 per book. She sells them with a profit margin of 20%. Find the selling price.

(ii) She sells crayon boxes at ₹50 per box with a profit margin of 25%. How much did Shambhavi buy them from the wholesaler?

Answer

(i)

Given:

Cost Price of one notebook = ₹36

Profit margin = 20% (of the cost price)

Selling Price = ?

The selling price is the cost price together with the profit:

Profit = 20% of ₹36 = $\dfrac{20}{100}$ × 36 = ₹7.20

Selling Price = Cost Price + Profit

= ₹36 + ₹7.20

= ₹43.20

Hence, the selling price of each notebook is ₹43.20.

(ii)

Given:

Selling Price of one crayon box = ₹50

Profit margin = 25% of the cost price

Therefore, Selling Price = 125% of Cost Price

125% of CP = ₹50

$\dfrac{125}{100}$ × CP = ₹50

CP = ₹50 × $\dfrac{100}{125}$

= ₹40

Hence, Shambhavi bought each crayon box from the wholesaler for ₹40.

Raghu had purchased rice at ₹35 per kg. To clear his old stock, he sells 10 kg rice for ₹300, incurring a percentage loss of $\dfrac{50}{350} \times 100 = 14.28$%. Could we have just calculated the loss percentage per kg instead? Would it be the same?

Answer

Yes, the loss percentage can also be calculated per kilogram.

Working it out per kg:

Cost Price per kg = ₹35

Selling Price per kg = $\dfrac{300}{10} = ₹30$

Loss per kg = ₹35 - ₹30 = ₹5

Percentage loss = $\dfrac{5}{35} \times 100 = 14.28$%

This is exactly the loss percentage obtained for 10 kg:

$\dfrac{50}{350} \times 100 = 14.28$%

This happens because the percentage loss is a ratio (loss compared to cost price), and this ratio stays the same whether we consider 1 kg or 10 kg.

∴ Yes, the loss percentage calculated per kg is 14.28%, which is the same as the loss percentage for 10 kg, because the loss percentage does not depend on the quantity.

Due to heavy rains, Snehal could not transport strawberries to Hyderabad from his farm in Panchgani. He sells some of his stock at ₹80 per kg with a 12% loss. What is the cost price?

Answer

Given:

Selling Price per kg = ₹80

Loss = 12% of the cost price

Cost Price = ?

Since the loss is 12% of the cost price, the selling price is (100% − 12%) = 88% of the cost price.

Let the cost price be ₹CP. Then:

88% of CP = ₹80

$\dfrac{88}{100}$ × CP = ₹80

CP = ₹80 × $\dfrac{100}{88}$

= ₹$\dfrac{1000}{11}$

= ₹90.91

∴ The cost price of the strawberries is ₹$\mathbf{\dfrac{1000}{11}}$ ≈ ₹90.91 per kg.

A utensil store is offering a 35% discount on the cooker with an MRP ₹1800. What is the selling price? If the cost price was ₹900, what is the percentage profit made after the sale?

Answer

Given:

Marked Price (MRP) = ₹1800

Discount = 35% (of the MRP)

Cost Price = ₹900

Finding the Selling Price:

A 35% discount means the selling price is (100% − 35%) = 65% of the MRP.

Selling Price = 65% of ₹1800

$= \dfrac{65}{100} \times 1800 \\[1em] = ₹1170$

Finding the percentage profit:

Profit = Selling Price - Cost Price

= ₹1170 - ₹900

= ₹270

Percentage profit = $\dfrac{270}{900} \times 100 = 30$%

Hence, the selling price of the cooker is ₹1170, and the percentage profit made after the sale is 30%.

Check if the calculations are correct in the bill shown.

Answer

Amount of CFL bulbs = 3 × ₹150.00 = ₹450.00

So, the subtotal is ₹450.

CGST at 9%:

$\dfrac{9}{100} \times 450 = ₹40.50$

SGST at 9%:

$\dfrac{9}{100} \times 450 = ₹40.50$

Total bill:

Sub Total + CGST + SGST

= ₹450.00 + ₹40.50 + ₹40.50

= ₹531.00

Yes, all the calculations in the bill are correct.

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

Answer

Given:

Cost Price (CP) = ₹75

Selling Price (SP) = ₹110

Since SP > CP, there is a profit.

Profit = SP − CP

= ₹(110 − 75)

= ₹35

Profit margin with respect to the cost is calculated as:

$\phantom{=} \text{Profit margin} = \dfrac{\text{Profit}}{\text{CP}} \times 100 \\[1em] = \dfrac{{35}}{75} \times 100 \\[1em] = \dfrac{\overset{7}{\cancel{35}}}{\underset{15}{\cancel{75}}} \times 100 \\[1em] = \dfrac{7}{15} \times 100 \\[1em] = \dfrac{700}{15} \\[1em] = \dfrac{140}{3} \\[1em]$

= $46\dfrac{2}{3}$%

Hence, the profit margin with respect to the cost is $46\dfrac{2}{3}$% (≈ 46.67%).

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?

Answer

Given:

Cost Price (CP) = ₹475

Profit margin = 50% (with respect to the cost)

Selling Price = ?

Profit = 50% of CP

$\phantom{=} \text{Profit} = \dfrac{50}{100} \times 475 \\[1em] = \dfrac{\overset{1}{\cancel{50}}}{\underset{2}{\cancel{100}}} \times 475 \\[1em] = \dfrac{1}{2} \times 475 \\[1em] = ₹\dfrac{475}{2} \\[1em] = ₹237.50$

Now,

Selling Price = CP + Profit

= ₹(475 + 237.50)

= ₹712.50

Hence, the chair should be sold at ₹712.50.

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?

Answer

Given:

Revenue (total sales) = ₹2.5 crore = ₹2,50,00,000

Profit margin = 25%

For a company, the profit margin is calculated with respect to the revenue (sales amount).

Profit = 25% of Revenue

$\phantom{=} \text{Profit} = \dfrac{25}{100} \times 2,50,00,000 \\[1em] = \dfrac{1}{4} \times 2,50,00,000 \\[1em] = ₹62,50,000$

Now,

Expenditure = Revenue − Profit

= ₹(2,50,00,000 − 62,50,000)

= ₹1,87,50,000

Hence, the total expenditure (cost) of the company last year was ₹1,87,50,000 (i.e. ₹1.875 crore).

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?

Answer

Given:

Original (marked) price = ₹300

Discount = 25%

Amount to be paid = ?

Discount = 25% of original price

$\phantom{=} \text{Discount} = \dfrac{25}{100} \times 300 \\[1em] = \dfrac{\overset{1}{\cancel{25}}}{\underset{4}{\cancel{100}}} \times 300 \\[1em] = \dfrac{1}{{\cancel{4}}} \times \overset{75}{\cancel{300}} \\[1em] = ₹75$

Now,

Amount to be paid = Original price − Discount

= ₹(300 − 75)

= ₹225

Hence, Anwar will have to pay ₹225 to buy the shirt.

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

1. 50%
2. 40%
3. 60%
4. 66.66%
5. 140%
6. 160.66%

Answer

Given:

Price in 2015 (original) = ₹60

Price in 2025 = ₹100

Increase in price = ₹(100 − 60) = ₹40

The percentage increase is taken with respect to the original price:

$\phantom{=} \text{Percentage increase} = \dfrac{\text{Increase}}{\text{Original price}} \times 100 \\[1em] = \dfrac{40}{60} \times 100 \\[1em] = \dfrac{\overset{2}{\cancel{40}}}{\underset{3}{\cancel{60}}} \times 100 \\[1em] = \dfrac{2}{3} \times 100 \\[1em] = \dfrac{200}{3} \\[1em]$

= 66 $\dfrac{2}{3}$% $\approx 66.66$%

The percentage increase in the price of petrol is approximately 66.66%.

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?

Answer

Given:

Price paid after discount = ₹4,40,000

Discount = 15%

Original price = ?

Let the original price be ₹x.

Since a 15% discount was given, the price paid is (100 − 15)% = 85% of the original price.

$\phantom{\Rightarrow} 85$% $\text{ of } x = 4,40,000$

$\Rightarrow \dfrac{85}{100} \times x = 4,40,000 \\[1em] \Rightarrow x = 4,40,000 \times \dfrac{100}{85} \\[1em] \Rightarrow x = \dfrac{4,40,000 \times 100}{85} \\[1em] \Rightarrow x = ₹\dfrac{88,00,000}{17} \\[1em] \Rightarrow x ≈ ₹5,17,647.06$

Hence, the original price of the car was ₹5,17,647.06.

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?

Answer

Given:

Total votes = 1600

Votes received by the winner = 500

Percentage of votes received by the winner:

$\phantom{=} \dfrac{\text{Winner's votes}}{\text{Total votes}} \times 100 \\[1em] = \dfrac{500}{1600} \times 100 \\[1em] = \dfrac{5}{16} \times 100 \\[1em] = \dfrac{500}{16}$

$= 31.25$%

So, the winner got 31.25% of the total votes.

Minimum number of candidates:

Remaining votes (other than the winner's) = 1600 − 500 = 1100

For the winner to actually win, every other candidate must get fewer votes than the winner, i.e. at most 499 votes each.

To make the number of other candidates as small as possible, give each of them the maximum possible (499 votes):

⇒ With 2 other candidates: maximum they can hold = 499 + 499 = 998, which is less than 1100. Not enough.

⇒ With 3 other candidates: they can hold up to 3 × 499 = 1497 votes, which is enough to share the 1100 remaining votes (for example, 499 + 499 + 102).

So at least 3 other candidates are needed, in addition to the winner.

Hence, the winner got 31.25% of the total votes, and the minimum number of candidates who stood for the election is 4 (the winner and 3 others).

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.)

Answer

Given:

Price in 2024 (original) = ₹38

Price in 2025 = ₹42

Increase in price = ₹(42 − 38) = ₹4

The rate of inflation is the percentage increase with respect to the original price:

$\phantom{=} \text{Inflation} = \dfrac{\text{Increase}}{\text{Original price}} \times 100 \\[1em] = \dfrac{4}{38} \times 100 \\[1em] = \dfrac{2}{19} \times 100 \\[1em] = \dfrac{200}{19} \\[1em]$

= 10 $\dfrac{10}{19}$% $\approx 10.53$%

Hence, the rate of inflation is $10\dfrac{10}{19}$% (≈ 10.53%).

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

Answer

Given:

The number increased by 20% becomes 90.

Let the number be x.

When the number is increased by 20%, it becomes (100 + 20)% = 120% of the number.

120% of x = 90

$\Rightarrow \dfrac{120}{100} \times x = 90 \\[1em] \Rightarrow \dfrac{6}{5} \times x = 90 \\[1em] \Rightarrow x = 90 \times \dfrac{5}{6} \\[1em] \Rightarrow x = \dfrac{450}{6} \\[1em] \Rightarrow x = 75$

Hence, the number is 75.

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.

Answer

Given:

Selling Price of each buffalo = ₹80,000

First buffalo (5% profit):

Here the Selling Price is (100 + 5)% = 105% of its Cost Price.

$\phantom{\Rightarrow} 105$% $\text{ of } CP_1 = 80,000$

$\Rightarrow \dfrac{105}{100} \times CP_1 = 80,000 \\[1em] \Rightarrow CP_1 = 80,000 \times \dfrac{100}{105} \\[1em] \Rightarrow CP_1 = ₹\dfrac{16,00,000}{21} \\[1em] \Rightarrow CP_1 ≈ ₹76,190.48$

Second buffalo (10% loss):

Here the Selling Price is (100 − 10)% = 90% of its Cost Price.

$\phantom{\Rightarrow} 90$% $\text{ of } CP_2 = 80,000$

$\Rightarrow \dfrac{90}{100} \times CP_2 = 80,000 \\[1em] \Rightarrow CP_2 = 80,000 \times \dfrac{100}{90} \\[1em] \Rightarrow CP_2 = ₹\dfrac{8,00,000}{9} \\[1em] \Rightarrow CP_2 ≈ ₹88,888.89$

Now,

Total Cost Price = $CP_1 + CP_2$

$\phantom{=} = ₹\left(\dfrac{16,00,000}{21} + \dfrac{8,00,000}{9}\right) \\[1em] = ₹\dfrac{48,00,000 + 56,00,000}{63} \\[1em] = ₹\dfrac{1,04,00,000}{63} \\[1em] ≈ ₹1,65,079.37$

Total Selling Price = ₹(80,000 + 80,000) = ₹1,60,000

Since Total Cost Price > Total Selling Price, there is an overall loss.

$\phantom{=} \text{Loss} = \text{Total CP} − \text{Total SP} \\[1em] = ₹\left(\dfrac{1,04,00,000}{63} − 1,60,000\right) \\[1em] = ₹\dfrac{1,04,00,000 − 1,00,80,000}{63} \\[1em] = ₹\dfrac{3,20,000}{63} \\[1em] ≈ ₹5,079.37$

Hence, the milkman had an overall loss of approximately ₹5,079.37.

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

1. p × 0.5
2. p × 0.05
3. p × 1.5
4. p × 1.05
5. p + 1.50

Answer

Given:

Population last decade = p

Increase = 5%

When the population increases by 5%, the present population becomes (100 + 5)% = 105% of the earlier population.

Population now = 105% of p

= $\dfrac{105}{100}$ × p

= 1.05 × p

= p × 1.05

Hence, the population now is p × 1.05, and option 4 is the correct option.

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

1. The demand now is 85% of the demand a decade ago.
2. The demand a decade ago was 85% of the demand now.
3. The demand now is 15% of the demand a decade ago.
4. The demand a decade ago was 15% of the demand now.
5. The demand a decade ago was 185% of the demand now.
6. The demand now is 185% of the demand a decade ago.

Answer

Let the demand a decade ago be d.

Since the demand has fallen by 85%, the demand now is (100 − 85)% = 15% of the demand a decade ago.

Demand now = d − 85% of d

= d − $\dfrac{85}{100}$ × d

= d − 0.85d

= 0.15d

= 15% of d

So, the demand now is 15% of the demand a decade ago.

This matches statement (iii). None of the other statements describe the same relationship.

Hence, statement (iii) — "The demand now is 15% of the demand a decade ago" — means the same thing.

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.

Answer

Given:

Principal (P) = ₹20,000

Rate of interest (r) = 10% p.a. = 0.1

Time (t) = 2 years

Without compounding

The interest gained every year remains the same. So,

Amount = P + (P × r × t) = P(1 + rt)

Substituting the values, we get:

20000 × (1 + 0.1 × 2)

= 20000 × (1 + 0.2)

= 20000 × 1.2

= ₹24000

With compounding

The interest gained every year is added back to the principal. So,

Amount = P(1 + r)^t

Substituting the values, we get:

20000 × (1 + 0.1)²

= 20000 × (1.1)²

= 20000 × 1.21

= ₹24200

⇒ Extra amount with compounding = ₹24200 − ₹24000 = ₹200

Hence, without compounding one gets ₹24,000 and with compounding one gets ₹24,200, which is ₹200 more.

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.

Answer

Given:

Principal (P) = ₹20,000

Rate of interest (r) = 5% p.a. = 0.05

Time (t) = 4 years

Without compounding

Amount = P + (P × r × t) = P(1 + rt)

Substituting the values, we get:

20000 × (1 + 0.05 × 4)

= 20000 × (1 + 0.2)

= 20000 × 1.2

= ₹24000

With compounding

Amount = P(1 + r)^t

Substituting the values, we get:

20000 × (1 + 0.05)⁴

= 20000 × (1.05)⁴

= 20000 × 1.21550625

= ₹24310.125

⇒ Extra amount with compounding = ₹24310.125 − ₹24000 = ₹310.125

∴ Without compounding one gets ₹24,000 and with compounding one gets ₹24,310.125, which is ₹310.125 more.

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

Answer

In Question 1, the rate × time = 10% × 2 = 20%, and in Question 2, the rate × time = 5% × 4 = 20%. Since this product is the same in both cases, the amount without compounding is exactly ₹24,000 in both questions.

However, with compounding the two amounts are different:

⇒ Question 1 (10% for 2 years) gives ₹24,200

⇒ Question 2 (5% for 4 years) gives ₹24,310.125

Even though the total simple interest is the same, the deposit that is compounded over more periods (4 years in Question 2) grows to a larger amount than the one compounded over fewer periods (2 years in Question 1). This is because each year's interest also earns interest in the following years, and the more years there are, the more often this happens.

Hence, with the same rate × time, the amount without compounding is the same, but compounding over a greater number of periods yields a larger amount.

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?

1. p × 6 × 4

2. p × 0.6 × 4

3. $p \times \dfrac{0.6}{100} \times 4$

4. $p \times \dfrac{0.06}{100} \times 4$

5. p × 1.6 × 4

6. p × 1.06 × 4

7. p + (p × 0.06 × 4)

Answer

Given:

Principal = p

Rate of interest (r) = 6% p.a. = 0.06

Time (t) = 4 years

When compounding is not done, the total amount is:

Amount = Principal + Interest = p + (p × r × t)

Substituting the values, we get:

p + (p × 0.06 × 4)

= p + (p × 0.24)

= p(1 + 0.24)

= 1.24p

Comparing this with the given options, only option 7 matches:

⇒ p + (p × 0.06 × 4) = 1.24p

(The other options either use a wrong rate, or give only the interest instead of the total amount.)

The expression p + (p × 0.06 × 4) describes the total amount.

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?

Answer

Given:

Principal (P) = ₹50,000

Rate of interest (r) = 7% p.a. = 0.07

Time (t) = 3 years

Interest without compounding

Interest = P × r × t

Substituting the values, we get:

50000 × 0.07 × 3

= 50000 × 0.21

= ₹10500

Interest with compounding

Amount = P(1 + r)^t

Substituting the values, we get:

50000 × (1 + 0.07)³

= 50000 × (1.07)³

= 50000 × 1.225043

= ₹61252.15

⇒ Interest with compounding = ₹61252.15 − ₹50000 = ₹11252.15

⇒ Extra interest = ₹11252.15 − ₹10500 = ₹752.15

∴ Without compounding the interest is ₹10,500, and with compounding the interest is ₹11252.15, which is ₹752.15 more.

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?

Answer

Given:

Loan amount (P) = ₹12,500 (for both)

Time (t) = 3 years (for both)

Giridhar — without compounding (r = 12% = 0.12)

Interest = P × r × t

Substituting the values, we get:

12500 × 0.12 × 3

= 12500 × 0.36

= ₹4500

Raghava — with compounding (r = 10% = 0.1)

Amount = P(1 + r)^t

Substituting the values, we get:

12500 × (1 + 0.1)³

= 12500 × (1.1)³

= 12500 × 1.331

= ₹16637.50

⇒ Raghava's interest = ₹16637.50 − ₹12500 = ₹4137.50

⇒ Difference in interest = ₹4500 − ₹4137.50 = ₹362.50

Hence, Giridhar pays more interest, by ₹362.50.

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?

Answer

Given:

Principal (P) = ₹1000

Rate of interest (r) = 10% p.a. = 0.1

Amount required (doubled) = ₹2000

Without compounding

Amount = P(1 + rt)

Setting the amount equal to ₹2000:

1000 × (1 + 0.1t) = 2000

⇒ 1 + 0.1t = 2

⇒ 0.1t = 1

⇒ t = 10 years

With compounding

Amount = P(1 + r)^t = 1000 × (1.1)^t

Calculating the amount year by year:

⇒ After 7 years = 1000 × (1.1)⁷ = ₹1948.72 (not yet doubled)

⇒ After 8 years = 1000 × (1.1)⁸ = ₹2143.59 (more than doubled)

So the amount crosses ₹2000 during the 8th year, i.e., it doubles in about 8 years.

The amount without compounding follows P(1 + rt), which is a straight-line (linear) growth in t. The amount with compounding follows P(1 + r)^t, where t is an exponent — this is exponential growth.

∴ With compounding the amount doubles in about 8 years, while without compounding it takes 10 years. Yes — compounding is exponential growth and not-compounding is 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?

Answer

Given:

Current population (P) = 1.5 crore = 1,50,00,000

Rate of increase (r) = 3% per year = 0.03

Time (t) = 3 years

Since the population rises by 3% on the previous year's population, it grows by compounding. So,

Population after t years = P(1 + r)^t

Substituting the values, we get:

15000000 × (1 + 0.03)³

= 15000000 × (1.03)³

= 15000000 × 1.092727

= 16390905

Hence, the expected population after 3 years is 1,63,90,905 (about 1.64 crore).

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.

Answer

Given:

Initial count (P) = 5,06,000

Rate of increase (r) = 2.5% per hour = 0.025

Time (t) = 2 hours

Since the count increases by 2.5% on the previous hour's count, it grows by compounding. So,

Number of bacteria after t hours = P(1 + r)^t

Substituting the values, we get:

506000 × (1 + 0.025)²

= 506000 × (1.025)²

= 506000 × 1.050625

= 531616.25

Hence, the number of bacteria at the end of 2 hours is about 5,31,616.

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 (without compounding and with compounding)?

Answer

Let the principal be p, the rate of interest be r and the number of terms (years) be t.

(i) Without compounding

The principal stays the same for every term, so the total amount after the maturity period is:

Total amount = p(1 + rt)

The total interest is the amount over and above the principal:

Total interest = Total amount - p

= p(1 + rt) - p

⇒ Total interest = prt

(ii) With compounding

The interest earned each term is added back, so the total amount after the maturity period is:

Total amount = p(1 + r)^t

The total interest is:

Total interest = Total amount - p

⇒ Total interest = p(1 + r)^t - p

Hence, without compounding, total interest = prt ; with compounding, total interest = p(1 + r)^t - p.

You have won a contest. The organisers offer you two options to choose from:

Option A: You deposit ₹100 and you get back ₹300.

Option B: You deposit ₹1000 and you get back ₹1500.

What is the percentage gain each option gives? You can choose any option only once. Which option would you choose? Why?

Answer

The percentage gain is calculated as $\dfrac{\text{Gain}}{\text{Amount deposited}} \times 100$.

Option A

Gain = ₹300 − ₹100 = ₹200

Percentage gain = $\dfrac{200}{100}$ × 100

⇒ Percentage gain = 200%

Option B

Gain = ₹1500 − ₹1000 = ₹500

Percentage gain = $\dfrac{500}{1000}$ × 100

⇒ Percentage gain = 50%

Comparing the two options, Option A gives a far higher percentage gain (200%) than Option B (50%).

Since the percentage gain measures how efficiently the deposited money grows, Option A is the better deal — the money more than triples, whereas in Option B it only grows by half.

∴ Option A gives a 200% gain and Option B gives a 50% gain. Based on the percentage gain, Option A is the better choice.

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

Answer

We compare the rupee value of the 20% discount with the flat ₹50 discount for each bill amount.

(i) Bill of ₹180

20% of ₹180 = $\dfrac{20}{100} \times 180$

⇒ 20% discount = ₹36

Since ₹50 > ₹36, the flat ₹50 discount saves more.

(ii) Bill of ₹225

20% of ₹225 = $\dfrac{20}{100} \times 225$

⇒ 20% discount = ₹45

Since ₹50 > ₹45, the flat ₹50 discount saves more.

(iii) Bill of ₹300

20% of ₹300 = $\dfrac{20}{100} \times 300$

⇒ 20% discount = ₹60

Since ₹60 > ₹50, the 20% discount saves more.

The two discounts are equal when $\dfrac{20}{100} \times \text{Bill} = 50$, i.e. when the bill is ₹250.

For bills less than ₹250, choose the ₹50 discount.

For bills greater than ₹250, choose the 20% discount.

Hence, for (i) ₹180 and (ii) ₹225, choose the ₹50 discount; for (iii) ₹300, choose the 20% discount.

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?

Answer

Let the number of marbles with Arun be x.

The number of marbles with Ariba = 120% of x:

$\phantom{=} \text{Ariba's marbles} = \dfrac{120}{100} \times x \\[1em] ⇒ \text{Ariba's marbles} = \dfrac{6}{5}x$

Now Arun compares his marbles with Ariba's:

$\phantom{=} \dfrac{\text{Arun's marbles}}{\text{Ariba's marbles}} \times 100 = \dfrac{x}{\frac{6}{5}x} \times 100 \\[1em] = \dfrac{5}{6} \times 100 \\[1em] = \dfrac{500}{6} \\[1em]$

= $83\dfrac{1}{3}$%

So Arun has $83\dfrac{1}{3}$% of the marbles Ariba has (he has $16\dfrac{2}{3}$% fewer).

Hence, Arun could say, "The number of marbles I have is $83\dfrac{1}{3}$% of the marbles Ariba has."

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?

Answer

Given:

Population in 2000 = 50 lakhs

Population in 2025 = 250% of the population in 2000

$\phantom{=} \text{Population in 2025} = \dfrac{250}{100} \times 50 \text{ lakhs} \\[1em] = \dfrac{5}{2} \times 50 \text{ lakhs} \\[1em] = \dfrac{250}{2} \text{ lakhs} \\[1em] = 125 \text{ lakhs}$

Hence, the population of Bengaluru in 2025 is about 125 lakhs (i.e. 1.25 crore).

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].

1. 13%
2. 8%
3. 18%
4. 10%
5. 1%
6. 35%
7. 2%
8. 2%
9. 0.1%.

Answer

World population in 2025 ≈ 8.2 billion = 8200 million.

The percentage share of each country is $\dfrac{\text{Country's population}}{8200} \times 100$.

Germany

$\dfrac{83}{8200} \times 100 = \dfrac{8300}{8200}$

≈ 1%

India

$\dfrac{1460}{8200} \times 100 = \dfrac{146000}{8200}$

≈ 18%

Bangladesh

$\dfrac{175}{8200} \times 100 = \dfrac{17500}{8200}$

≈ 2%

USA

$\dfrac{347}{8200} \times 100 = \dfrac{34700}{8200}$

≈ 4%

∴ The approximate matches are: Germany → 1%, India → 18%, Bangladesh → 2%, and USA cannot be matched to any of the listed options.

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?

1. 8250 + 18

2. 8250 + 1800

3. $8250 + \dfrac{18}{100}$

4. 8250 × 18

5. 8250 × 1.18

6. 8250 + 8250 × 0.18

7. 1.8 × 8250

Answer

The GST of 18% is added to the price, so:

Final price = ₹8250 + 18% of ₹8250

= $₹8250 + \Big(\dfrac{18}{100} \times 8250\Big)$

= ₹8250 + 8250 × 0.18 [this is option 6]

= ₹8250 × 1.18 [this is option 5]

= ₹9735

Both option 5 and option 6 give this value. The other options are incorrect — for example, option 2 adds ₹1800 instead of 18% of ₹8250 (which is ₹1485), and option 7 multiplies by 1.8 (i.e. adds 80%) instead of 1.18.

Hence, Options 5 (8250 × 1.18) and 6 (8250 + 8250 × 0.18) give the final price of ₹9735.

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.

1. The population after three months was p × 0.05 × 0.02 × 0.03.
2. The population after three months was p × 1.05 × 0.98 × 0.97.
3. The population after three months was p + 0.05 - 0.02 - 0.03.
4. The population after three months was p.
5. The population after three months was more than p.
6. The population after three months was less than p.

Answer

A change of +5% multiplies the population by 1.05, a change of −2% multiplies it by 0.98, and a change of −3% multiplies it by 0.97. These multipliers compound month after month.

Population after 3 months = p × 1.05 × 0.98 × 0.97

⇒ Population after 3 months = p × 0.99813

⇒ Population after 3 months ≈ 0.99813p

Since the multiplier 0.99813 is less than 1, the final population is slightly less than p.

Checking the statements:

• Statement 2 is the correct compounded expression — true.
• Statement 6 says the population is less than p, which matches 0.99813 p < p — true.
• Statement 1 multiplies the decimal changes themselves, statement 3 adds the decimal changes, statement 4 claims no change, and statement 5 claims an increase — all false.

Hence, statements 2 and 6 are true.

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.

Answer

Let the cost price of the product be x.

With a 35% profit margin, the selling price (marked price) is:

Selling price = x + 35% of x

= x × 1.35

⇒ Selling price = 1.35x

A 30% discount is then offered on this selling price, so the customer pays 70% of it:

Final price = 1.35x × (1 - 0.30)

= 1.35x × 0.70

⇒ Final price = 0.945x

Since the final price 0.945x is less than the cost price x, the shopkeeper makes a loss.

Loss = x - 0.945x = 0.055x

⇒ Loss percentage = $\dfrac{0.055x}{x} \times 100 = 5.5$%

Hence, the shopkeeper makes a loss of 5.5%, because after the 30% discount the final price (0.945x) is below the cost price (x).

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

Answer

The figure is drawn on a square dot grid. Taking each grid square as 1 unit, the whole outer figure is a square of side 8 units.

Area of the whole square = 8 × 8 = 64 sq units

The square is split into five regions. The bottom-left 4 × 4 square is divided by a diagonal into two equal right triangles, D (upper-left) and E (lower-right). So region E is half of this 4 × 4 square:

$\phantom{=} \text{Area of E} = \dfrac{1}{2} \times 4 \times 4 \\[1em] ⇒ \text{Area of E} = 8 \text{ sq units}$

The percentage of the total area occupied by E is:

$\phantom{=} \dfrac{\text{Area of E}}{\text{Area of square}} \times 100 \\[1em] = \dfrac{8}{64} \times 100 \\[1em] = \dfrac{1}{8} \times 100 \\[1em]$

$= 12.5$%

Hence, the region marked 'E' occupies 12.5% of the total area.

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?

Answer

We know that, x% of y = $\dfrac{x}{100} \times y$.

Calculating each value, we get:

5% of 40 = $\dfrac{5}{100}$ × 40 = 2

40% of 5 = $\dfrac{40}{100}$ × 5 = 2

25% of 12 = $\dfrac{25}{100}$ × 12 = 3

12% of 25 = $\dfrac{12}{100}$ × 25 = 3

15% of 60 = $\dfrac{15}{100}$ × 60 = 9

60% of 15 = $\dfrac{60}{100}$ × 15 = 9

We notice that in each pair the two values are equal, i.e., x% of y = y% of x.

Justification using algebra:

x% of y = $\dfrac{x}{100} \times y = \dfrac{xy}{100}$

y% of x = $\dfrac{y}{100} \times x = \dfrac{xy}{100}$

Both expressions are equal to $\dfrac{xy}{100}$.

Hence, x% of y is always equal to y% of x.

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?

Answer

(i)

Grade 8 students = 40% of the total students.

Out of these Grade 8 students, 60% are girls.

So, the Grade 8 girls form 60% of 40% of the total.

Grade 8 girls = 60% of 40%

= $\dfrac{60}{100} \times \dfrac{40}{100} \times 100$%

= $\dfrac{2400}{10000} \times 100$%

$= 24$%

Hence, 24% of the students going to the excursion are Grade 8 girls.

(ii)

Total number of students = 160.

From part (i), Grade 8 girls form 24% of the total.

Number of Grade 8 girls = 24% of 160

$= \dfrac{24}{100} \times 160 \\[1em] = \dfrac{3840}{100} \\[1em] = 38.4$

∴ The number of Grade 8 girls = 38.4 (about 38 students).

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?

Answer

Given:

Selling price (SP) of 3 pencils = Cost price (CP) of 5 pencils.

Let the cost price of 1 pencil = ₹3.

Then,

CP of 5 pencils = ₹(3 × 5) = ₹15

SP of 3 pencils = CP of 5 pencils = ₹15

$\text{SP of 1 pencil} = ₹\dfrac{15}{3} = ₹5$

Since the selling price of 1 pencil (₹5) is greater than its cost price (₹3), the shopkeeper makes a profit.

Profit on 1 pencil = ₹(5 - 3) = ₹2

Profit % = $\dfrac{\text{Profit}}{\text{CP}} \times 100$

$= \dfrac{2}{3} \times 100 \\[1em] = \dfrac{200}{3} \\[1em]$

$= 66\dfrac{2}{3}$%

Hence, the shopkeeper makes a profit of $66\dfrac{2}{3}$% (≈ 66.67%).

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?

Answer

Let the original bus fare = ₹100.

Increase in the first year (3%):

Fare after 1st year = 100 + 3% of 100

= $100 \times \dfrac{103}{100}$

= ₹103

Increase in the second year (4%): $\quad$[applied on the new fare ₹103]

Fare after 2nd year = 103 + 4% of 103

= $103 \times \dfrac{104}{100}$

= ₹107.12

Overall increase = 107.12 - 100 = ₹7.12 $\quad$[on ₹100]

Hence, the overall percentage increase in the bus fare over the two years is 7.12%.

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

Answer

Let the length of the rectangle = l and the breadth = b.

Original area = l x b.

The length is increased by 10%.

New length = l + 10% of l

= $l \times \dfrac{110}{100}$

= $\dfrac{11l}{10}$

Let the new breadth = b'. Since the area is unchanged,

$\dfrac{11l}{10} \times b' = l \times b \\[1em] \Rightarrow b' = \dfrac{l \times b \times 10}{11 \times l} \\[1em] \Rightarrow b' = \dfrac{10b}{11}$

$\text{Decrease in breadth} = b - \dfrac{10b}{11} = \dfrac{b}{11}$

$\text{Percentage decrease} = \dfrac{\dfrac{b}{11}}{b} \times 100 = \dfrac{100}{11} = 9\dfrac{1}{11}$%

∴ The breadth decreases by exactly $\mathbf{9\dfrac{1}{11}}$% (≈ 9.09%).

The percentage of ingredients in a 65 g chips packet is shown in the picture (Potato: 70%, Vegetable oil: 24%, Salt: 3%, Spices: 3%). Find out the weight each ingredient makes up in this packet.

Answer

Given: Total weight of the chips packet = 65 g.

Potato = 70% of 65 g = $\dfrac{70}{100} \times 65 = 45.5$ g

Vegetable oil = 24% of 65 g = $\dfrac{24}{100} \times 65 = 15.6$ g

Salt = 3% of 65 g = $\dfrac{3}{100} \times 65 = 1.95$ g

Spices = 3% of 65 g = $\dfrac{3}{100} \times 65 = 1.95$ g

Check: 45.5 + 15.6 + 1.95 + 1.95 = 65 g.

Hence, the packet contains 45.5 g potato, 15.6 g vegetable oil, 1.95 g salt and 1.95 g spices.

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?

Answer

(i)

Price of one item = ₹100.

Shop A (Buy 1, get 1 free): Pay for 1 item (₹100) and receive 2 items.

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

Shop B (Buy 2, get 1 free): Pay for 2 items (₹200) and receive 3 items.

$\text{Effective price per item} = ₹\dfrac{200}{3} = ₹66\dfrac{2}{3} \approx ₹66.67$

Shop C (Buy 3, get 1 free): Pay for 3 items (₹300) and receive 4 items.

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

Arranging from cheapest to costliest:

Shop A (₹50) → Shop B (₹66.67) → Shop C (₹75).

(ii)

Percentage discount = $\dfrac{\text{Number of free items}}{\text{Total items received}} \times 100$.

Shop A = $\dfrac{1}{2} \times 100 = 50$%

Shop B = $\dfrac{1}{3} \times 100 = 33\dfrac{1}{3}$% ≈ 33.33%

Shop C = $\dfrac{1}{4} \times 100 = 25$%

(iii)

To get 4 items:

⇒ Shop A: Buy 2, get 2 free ⇒ 4 items for ₹200.

⇒ Shop B: Buy 3, get 1 free ⇒ 4 items for ₹300.

⇒ Shop C: Buy 3, get 1 free ⇒ 4 items for ₹300.

Hence, Shop A should be chosen, as 4 items cost only ₹200 there (₹50 per item) — the lowest cost among the three shops.

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%?

Answer

In a room of 100 people, 99% are left-handed.

Left-handed people = 99% of 100 = 99

Right-handed people = 100 - 99 = 1

Only left-handed people leave, so the number of right-handed people stays at 1.

To bring the left-handed percentage down to 98%, the right-handed people must now make up (100% − 98%) = 2% of the people remaining in the room.

Let the number of people remaining = n.

2% of n = 1

$\Rightarrow \dfrac{2}{100} \times n = 1 \\[1em] \Rightarrow n = \dfrac{100}{2} = 50$

So, 50 people remain in the room, of which 1 is right-handed and (50 − 1) = 49 are left-handed.

Left-handed people who left = 99 - 49 = 50

Hence, 50 left-handed people have to leave the room.

Look at the following graph (Ability to use computer by age and gender, 2023). Based on the graph, which of the following statement(s) are valid?

1. People in their twenties are the most computer-literate among all age groups.
2. Women lag behind in the ability to use computers across age groups.
3. There are more people in their twenties than teenagers.
4. More than a quarter of people in their thirties can use computers.
5. Less than 1 in 10 aged 60 and above can use computers.
6. Half of the people in their twenties can use computers.

Answer

The graph gives the percentage of people (Female, Male) in each age group who can use a computer:

⇒ Children: 4%, 5%

⇒ Teenage: 24%, 29%

⇒ Twenties: 26%, 37%

⇒ Thirties: 14%, 25%

⇒ Forties: 7%, 14%

⇒ Fifties: 4%, 9%

⇒ Seniors (60 and above): 2%, 4%

Examining each statement:

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

For both females (26%) and males (37%), the twenties age group shows the highest ability to use computers compared with every other age group. This statement is valid.

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

In every age group, the female percentage is lower than the male percentage (e.g., Teenage 24% < 29%, Twenties 26% < 37%, Thirties 14% < 25%, and so on). This statement is valid.

3. There are more people in their twenties than teenagers.

The graph shows only the percentage of people who can use computers within each age group — it gives no information about the actual number of people in any group. So this cannot be concluded. This statement is not valid.

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

A quarter means 25%. For people in their thirties, only 14% of females and 25% of males can use computers. The male value is exactly a quarter (not more), and the overall proportion cannot exceed 25%. This statement is not valid.

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

"1 in 10" means 10%. For seniors (60 and above), 2% of females and 4% of males can use computers — both well below 10%. This statement is valid.

6. Half of the people in their twenties can use computers.

"Half" means 50%. In the twenties, only 26% of females and 37% of males can use computers — far below 50%. This statement is not valid.

Hence, the valid statements are (1), (2) and (5).