Part II, Chapter 1Fractions in Disguise

To express a fraction as a percentage, multiply it by 100.

(i) \(\dfrac35\times100=60\) → 60%

(ii) \(\dfrac{7}{14}\times100=\dfrac12\times100=50\) → 50%

(iii) \(\dfrac{9}{20}\times100=9\times5=45\) → 45%

(iv) \(\dfrac{72}{150}\times100=\dfrac{72\times2}{3}=48\) → 48%

(v) \(\dfrac13\times100=\dfrac{100}{3}=33\dfrac13\) → 33.33…% (about 33.3%)

(vi) \(\dfrac{5}{11}\times100=\dfrac{500}{11}=45\dfrac{5}{11}\) → about 45.45%

Fraction of white marbles \(=\dfrac{15}{25}=\dfrac35\).

Percentage \(=\dfrac35\times100=60\).

Fraction who walk \(=\dfrac{15}{80}=\dfrac{3}{16}\).

Percentage \(=\dfrac{15}{80}\times100=\dfrac{1500}{80}=18.75\).

Take the whole track from Start to Finish as \(100\%\) and see what part of it each friend has covered.

A has covered a little less than \(\frac13\) of the track, about 32%.

The nearest option is 38% (20% is much further away).

B has gone just past the middle, about 53%.

The nearest option is 55%.

C has covered about \(\frac34\) of the track, about 75%.

The nearest option is 72%.

D is almost at the Finish, about 94%.

The nearest option is 93%.

The options 20% and 84% are not needed.

(i) 50% is ten times 5%, so \(50\%\gt5\%\).

(ii) \(\dfrac{5}{10}=\dfrac12=50\%\), so \(\dfrac{5}{10}=50\%\).

(iii) \(\dfrac{3}{11}\) is less than \(\dfrac12\) (since \(3\) is less than half of \(11\)), so it is less than 50%, which is already less than 61%.

So \(\dfrac{3}{11}\lt61\%\).

(iv) \(\dfrac13=33.33\ldots\%\), which is more than 30%.

So \(30\%\lt\dfrac13\).

In each pair, both bars are cut into equal parts.

Every part is the same fraction of the whole, so: value of 1 part \(=\) total \(\div\) number of parts, and percentage of 1 part \(=100\%\div\) number of parts.

(i) (worked out) 5 equal parts.

Each part \(=100\%\div5=20\%\), and in value \(75\div5=15\).

The marked part on the left is 1 part \(=20\%\).

On the right, 4 parts \(=4\times15=60\).

(ii) 10 equal parts, total 90.

Each part \(=100\%\div10=10\%\), and in value \(90\div10=9\).

The marked part on the left is 1 part, so the missing number is 10%.

On the right, 6 parts \(=6\times9=\) 54.

(iii) 4 equal parts, total 140.

Each part \(=100\%\div4=25\%\), and in value \(140\div4=35\).

The marked part on the left is 1 part, so the missing number is 25%.

On the right, 3 parts \(=3\times35=\) 105.

(i) \(\dfrac{25}{100}\times160=\dfrac14\times160=40\)

(ii) \(\dfrac{16}{100}\times250=16\times2.5=40\)

(iii) \(\dfrac{62}{100}\times360=62\times3.6=223.2\)

(iv) \(\dfrac{140}{100}\times40=1.4\times40=56\)

(v) 1 hour \(=60\) minutes, so \(\dfrac{1}{100}\times60=0.6\) minutes \(=36\) seconds

(vi) \(\dfrac{7}{100}\times10=0.7\) kg \(=700\) g

Red paint \(=\dfrac34\) of 60 ml.

\(\dfrac34\times60=\dfrac{180}{4}=45\) ml.

(Check: \(\frac34=75\%\) and 75% of 60 \(=45\).)

(i) Same percentage, and \(510\lt515\), so 50% of 510 \(\lt\) 50% of 515.

(Check: \(255\lt257.5\).)

(ii) Same number (148), and \(37\%\lt73\%\), so 37% of 148 \(\lt\) 73% of 148.

(iii) 29% of 43 is less than 30% of 43, which is less than 13.

92% of 110 is more than 90% of 110, which is 99.

So \(29\%\text{ of }43\lt92\%\text{ of }110\).

(Check: \(12.47\lt101.2\).)

(iv) 30% of 40 \(=12\);

40% of 50 \(=20\).

So \(12\lt20\).

(v) 45% of 200 \(=90\);

10% of 490 \(=49\).

So \(90\gt49\).

(vi) 30% of 80 \(=24\);

24% of 64 \(=15.36\).

So \(24\gt15.36\).

(i) 60%, 90% and 120% are 2, 3 and 4 times 30%.

So 60% of \(k=2\times70=140\);

90% of \(k=3\times70=210\);

120% of \(k=4\times70=280\).

(ii) 10% of \(m=215\div10=21.5\);

1% of \(m=215\div100=2.15\);

6% of \(m=6\times2.15=12.9\).

(iii) 9% is one-tenth of 90%, so 9% of \(n=270\div10=27\).

18% is twice 9%, so 18% of \(n=2\times27=54\).

100% of \(n\): 10% of \(n=270\div9=30\), so 100% of \(n=300\).

(iv) Two more such questions: “40% of \(p\) is 90, 20% of \(p\) is ____, 60% of \(p\) is ____, 100% of \(p\) is ____” (answers: 45, 135, 225);

“50% of \(q\) is 36, 10% of \(q\) is ____, 25% of \(q\) is ____, 100% of \(q\) is ____” (answers: 7.2, 18, 72).

(i) \(\dfrac{3}{300}\times100=1\), so 3 is 1% of 300.

(ii) 40% of \(4=\dfrac{40}{100}\times4=1.6\), so 1.6 is 40% of 4.

(iii) 80% of the number is 40, so 10% of it is \(40\div8=5\), and 100% of it is \(5\times10=50\).

So 40 is 80% of 50.

10% of a day \(=\dfrac{10}{100}\times24\) hours \(=2.4\) hours \(=144\) minutes.

1% of a week \(=\dfrac{1}{100}\times7\) days \(=0.07\) day \(=0.07\times24\) hours \(=1.68\) hours \(=100.8\) minutes.

\(144\gt100.8\), so 10% of a day is longer.

A question to challenge your peers: “Is 5% of an hour longer than 10% of 4 minutes?” (5% of 60 min = 3 min;

10% of 4 min = 0.4 min, so yes.)

20% of the plantation takes 18 days.

So 10% takes \(18\div2=9\) days, and 100% takes \(9\times10=90\) days.

(By proportion: \(20:18::100:x\), so \(x=\dfrac{18\times100}{20}=90\).)

Why the assumption is needed: the working time is proportional to the area picked only if the workers pick at the same speed all the time.

If they speed up, slow down or are fewer on some days, the time for the full plantation would change.

Total number of parts \(=10+80+10=100\).

Time for each part \(=90\div100=0.9\) minutes.

Warm up \(=10\times0.9=9\) minutes.

Play \(=80\times0.9=72\) minutes.

Cool down \(=10\times0.9=9\) minutes.

Check: \(9+72+9=90\) minutes.

This year is 2025.

The textbook gives the world population in 2025 as about 8.2 billion (Figure it Out, last set, Question 2).

90% of \(8.2\) billion \(=\dfrac{90}{100}\times8.2=7.38\) billion.

If you use a different value for this year’s world population, multiply that value by 0.9 in the same way.

(i) For 8 people every ingredient is doubled, so the total is doubled too.

Each ingredient’s share of the total does not change: Rava 40%, Sugar 40%, Ghee 20%.

(ii) Rava : Sugar : Ghee \(=40:40:20\).

Total number of parts \(=40+40+20=100\).

Total weight \(=2\) kg \(=2000\) g.

Weight of each part \(=2000\div100=20\) g.

Rava \(=40\times20=800\) g.

Sugar \(=40\times20=800\) g.

Ghee \(=20\times20=400\) g.

Profit \(=110-75=35\).

Profit margin with respect to cost \(=\dfrac{35}{75}\times100=\dfrac{140}{3}\approx46.67\).

Profit \(=50\%\text{ of }475=\dfrac{50}{100}\times475=237.50\).

Selling price \(=475+237.50=712.50\).

(Or SP \(=150\%\) of 475 \(=1.5\times475=712.50\).)

The profit margin of a company is calculated on its revenue (sales).

Revenue = ₹2.5 crore.

Profit \(=25\%\text{ of }2.5=\dfrac{25}{100}\times2.5=0.625\) crore.

Expenditure \(=\) revenue \(-\) profit

\(=2.5-0.625\)

\(=1.875\) crore

(Check: \(\dfrac{0.625}{2.5}\times100=25\%\).)

Shortcut

Profit is 25% of revenue, so the costs are the remaining \(100\%-25\%=75\%\) of revenue.

Expenditure \(=75\%\text{ of }2.5\)

\(=0.75\times2.5\)

\(=1.875\) crore

Note: the textbook does not say whether the 25% margin is taken on the revenue or on the cost. We take it on the revenue, as the question says. If it is taken on the cost, the revenue is 125% of the cost, so the cost \(=\dfrac{2.5}{1.25}=2\) crore.

Discount \(=25\%\text{ of }300=\dfrac14\times300=75\).

Anwar pays \(300-75=225\).

(Or he pays 75% of 300 \(=225\).)

Increase \(=100-60=40\).

Percentage increase \(=\dfrac{\text{increase}}{\text{original price}}\times100=\dfrac{40}{60}\times100=66.66\ldots\).

After a 15% discount, Samson paid \(100\%-15\%=85\%\) of the original price.

85% of the original price \(=4{,}40{,}000\).

So 1% \(=\dfrac{4{,}40{,}000}{85}\), and 100% \(=\dfrac{4{,}40{,}000}{85}\times100\approx5{,}17{,}647.06\).

Winner’s percentage \(=\dfrac{500}{1600}\times100=31.25\).

Remaining votes \(=1600-500=1100\), shared by the other candidates.

Each of them must have got fewer than 500 votes, otherwise the winner would not have won.

With 1 other candidate, that person would get all 1100 votes, which is more than 500.

Not possible.

With 2 other candidates, one of them must get at least 550 votes (since 1100 ÷ 2 = 550).

Not possible.

With 3 other candidates, the votes can be, for example, 400, 400 and 300 (total 1100), each less than 500.

Possible.

Increase in price \(=42-38=4\).

Inflation \(=\dfrac{4}{38}\times100=\dfrac{200}{19}\approx10.53\).

Let the number be \(x\).

Number increased by 20% \(=x+20\%\text{ of }x=x+\dfrac{20}{100}x=1.2x\).

It becomes 90, so \(1.2x=90\).

\(x=\dfrac{90}{1.2}=\dfrac{900}{12}=75\).

(Check: 20% of 75 \(=15\), and \(75+15=90\).)

First buffalo (5% profit): SP \(=105\%\) of CP, so CP \(=\dfrac{80{,}000}{105}\times100\approx76{,}190.48\).

Second buffalo (10% loss): SP \(=90\%\) of CP, so CP \(=\dfrac{80{,}000}{90}\times100\approx88{,}888.89\).

Total cost price \(\approx76{,}190.48+88{,}888.89=1{,}65{,}079.37\).

Total selling price \(=2\times80{,}000=1{,}60{,}000\).

Since the selling price is less than the cost price, there is a loss: \(1{,}65{,}079.37-1{,}60{,}000\approx5{,}079.37\).

Loss percentage \(=\dfrac{5{,}079.37}{1{,}65{,}079.37}\times100\approx3.08\).

An increase of 5% means the new population is the old population plus 5% of it:

\(p+5\%\text{ of }p=p+0.05p=1.05p=p\times1.05\).

Let the demand a decade ago be 100.

A fall of 85% means the demand has dropped by 85, so the demand now is \(100-85=15\).

So the demand now is 15% of the demand a decade ago, which is statement (iii).

The other statements are not the same.

(i) says the demand now is 85, which is a fall of only 15, so (i) is false.

(ii) says \(100=85\%\) of the demand now, so the demand now \(=\dfrac{100}{0.85}\approx117.6\), an increase, so (ii) is false.

(iv) says \(100=15\%\) of the demand now, so the demand now \(=\dfrac{100}{0.15}\approx667\), a big increase, so (iv) is false.

(v) says \(100=185\%\) of the demand now, so the demand now \(=\dfrac{100}{1.85}\approx54\), a fall of about 46%, so (v) is false.

(vi) says the demand now is 185, which is an increase of 85%, so (vi) is false.

Without compounding: interest each year \(=10\%\text{ of }20{,}000=2000\).

For 2 years: \(2\times2000=4000\).

Amount \(=20{,}000+4000=24{,}000\).

With compounding: each year the amount becomes 110% of the amount at the start of that year.

Difference \(=24{,}200-24{,}000=200\).

Without compounding: interest each year \(=5\%\text{ of }20{,}000=1000\).

For 4 years: \(4\times1000=4000\).

Amount \(=20{,}000+4000=24{,}000\).

With compounding: each year the amount is multiplied by 1.05.

Difference \(\approx24{,}310.13-24{,}000=310.13\).

Without compounding both banks give the same amount, ₹24,000.

This is because the total interest rate is the same: \(10\%\times2=20\%\) and \(5\%\times4=20\%\).

With compounding the amounts are different: ₹24,200 for Bank of Yahapur and about ₹24,310.13 for Bank of Wahapur.

Even though the rate multiplied by the number of years is 20% in both, the 5% bank gives more with compounding,

because interest is added back and earns more interest in each of its 4 years.

In both cases, compounding gives more than not compounding.

Without compounding, the interest each year is 6% of \(p\), that is \(p\times0.06\).

For 4 years the interest is \(p\times0.06\times4\).

Total amount \(=\) principal \(+\) interest \(=p+(p\times0.06\times4)=p(1+0.24)=1.24p\).

This matches option (vii).

The other expressions do not give \(1.24p\):

(i) \(=24p\),

(ii) \(=2.4p\),

(iii) \(=0.024p\),

(iv) \(=0.0024p\),

(v) \(=6.4p\),

(vi) \(=4.24p\).

None of these is \(p\) plus the interest.

Without compounding: interest each year \(=7\%\text{ of }50{,}000=3500\).

For 3 years: \(3\times3500=10{,}500\).

With compounding: multiply by 1.07 each year.

Interest with compounding \(=61{,}252.15-50{,}000=11{,}252.15\).

Extra interest \(=11{,}252.15-10{,}500=752.15\).

Giridhar (no compounding): interest each year \(=12\%\text{ of }12{,}500=1500\).

For 3 years: \(3\times1500=4500\).

Raghava (compounded): multiply by 1.1 each year.

Raghava’s interest \(=16{,}637.50-12{,}500=4137.50\).

Difference \(=4500-4137.50=362.50\).

Each year the population becomes 103% of the previous year’s, so we multiply by 1.03 each year.

Each hour the count is multiplied by 1.025.

After 1 hour: \(506{,}000\times1.025=518{,}650\).

After 2 hours: \(518{,}650\times1.025=531{,}616.25\).

250% of 50 lakh \(=\dfrac{250}{100}\times50=2.5\times50=125\) lakh.

World population \(=8.2\) billion \(=8.2\times10^{9}\).

Percentage share \(=\dfrac{\text{population of the country}}{\text{world population}}\times100\).

Germany: \(\dfrac{8.3\times10^{7}}{8.2\times10^{9}}\times100\approx1.01\).

Share \(\approx1\%\).

India: \(\dfrac{1.46\times10^{9}}{8.2\times10^{9}}\times100\approx17.8\).

Share \(\approx18\%\).

Bangladesh: \(\dfrac{1.75\times10^{8}}{8.2\times10^{9}}\times100\approx2.1\).

Share \(\approx2\%\).

USA: \(\dfrac{3.47\times10^{8}}{8.2\times10^{9}}\times100\approx4.2\).

Share \(\approx4\%\).

Let the price of the phone be \(p\).

GST \(=18\%\text{ of }p=\dfrac{18}{100}p=0.18p\)

Final price \(=\) price \(+\) GST

\(=p+0.18p\)

\(=1.18p\)

So adding 18% GST means multiplying the price by 1.18.

Now \(p=8250\).

Final price \(=1.18\times8250=9735\)

Also \(p+0.18p=8250+0.18\times8250=8250+1485=9735\).

Options (v) and (vi) both give 9735.

The others do not give 9735.

(i) \(8250+18=8268\)

(ii) \(8250+1800=10050\)

(iii) \(8250+\dfrac{18}{100}=8250.18\)

(iv) \(8250\times18=148500\)

(vii) \(1.8\times8250=14850\)

Month 1: +5% means multiply by 1.05.

Month 2: −2% means multiply by 0.98.

Month 3: −3% means multiply by 0.97.

Each change is applied to the previous month’s population.

Population after three months \(=p\times1.05\times0.98\times0.97\).

So (ii) is true.

\(1.05\times0.98=1.029\) and \(1.029\times0.97=0.99813\).

So the population is \(0.99813\,p\), which is a little less than \(p\).

So (vi) is true.

(iv) is false, because \(0.99813\,p\ne p\).

(v) is false, because \(0.99813\,p\lt p\).

(i) is false, because it multiplies by the percentages themselves (0.05, 0.02, 0.03).

(iii) is false, because it adds and subtracts the percentages as plain numbers.

Let the cost price be ₹100.

Price with 35% profit margin \(=100+35=135\).

30% discount on 135 \(=\dfrac{30}{100}\times135=40.50\).

New selling price \(=135-40.50=94.50\).

The selling price (₹94.50) is less than the cost price (₹100), so he makes a loss of \(100-94.50=5.50\), which is 5.5% of the cost price.

Reason: the discount is taken on the higher price (₹135), not on the cost price.

The discount of ₹40.50 is more than the profit margin of ₹35.

\(40.50\gt35\), so the profit is wiped out and a loss remains.

Take the distance between two neighbouring dots as 1 unit.

The whole square is 8 units by 8 units.

Area of the whole square \(=8\times8=64\) square units.

E is a right triangle with base 4 units and height 4 units.

Area of E \(=\dfrac12\times4\times4=8\) square units.

Percentage \(=\dfrac{8}{64}\times100=12.5\).

Check with the other regions

A \(=4\times4=16\)

B \(=4\times6=24\)

C \(=4\times2=8\)

D \(=\) the other half of a \(4\times4\) square \(=8\)

E \(=8\)

Total \(=16+24+8+8+8=64\) ✓

5% of 40 \(=\dfrac{5}{100}\times40=2\) and 40% of 5 \(=\dfrac{40}{100}\times5=2\).

25% of 12 \(=\dfrac{25}{100}\times12=3\) and 12% of 25 \(=\dfrac{12}{100}\times25=3\).

15% of 60 \(=\dfrac{15}{100}\times60=9\) and 60% of 15 \(=\dfrac{60}{100}\times15=9\).

We notice that in each pair the two answers are equal.

General statement: \(x\%\) of \(y\) is always equal to \(y\%\) of \(x\).

Justification: \(x\%\text{ of }y=\dfrac{x}{100}\times y=\dfrac{xy}{100}\) and \(y\%\text{ of }x=\dfrac{y}{100}\times x=\dfrac{xy}{100}\).

Both are equal to \(\dfrac{xy}{100}\).

(i) Let the total number of students be 100.

Grade 8 students \(=40\% \text{ of } 100=40\).

Grade 8 girls \(=60\%\text{ of }40=\dfrac{60}{100}\times40=24\).

So 24 out of 100 students are Grade 8 girls, which is 24%.

(ii) Grade 8 students \(=40\%\text{ of }160=64\).

Grade 8 girls \(=60\%\text{ of }64=\dfrac{60}{100}\times64=38.4\).

Or directly, \(24\%\text{ of }160=38.4\).

The number of students must be a whole number, so there are about 38 Grade 8 girls.

Let the fare two years ago be ₹100.

After the 3% increase: \(100\times1.03=103\).

After the 4% increase (on ₹103): \(103\times1.04=107.12\).

The fare went from 100 to 107.12, an increase of 7.12, which is 7.12% of 100.

(It is not exactly \(3\%+4\%=7\%\) because the 4% increase is on the already increased fare.)

Let the length be 100 and the breadth be \(b\).

Area \(=100b\).

New length \(=110\).

For the same area, new breadth \(=\dfrac{100b}{110}=\dfrac{10b}{11}\).

Decrease in breadth \(=b-\dfrac{10b}{11}=\dfrac{b}{11}\).

Percentage decrease \(=\dfrac{b/11}{b}\times100=\dfrac{100}{11}=9\dfrac{1}{11}\).

From the packet: Potato 70%, Vegetable oil 24%, Salt 3%, Spices 3%.

The ingredients are in the ratio \(70:24:3:3\).

Total number of parts \(=70+24+3+3=100\).

Weight of each part \(=65\div100=0.65\) g.

Potato \(=70\times0.65=45.5\) g.

Vegetable oil \(=24\times0.65=15.6\) g.

Salt \(=3\times0.65=1.95\) g.

Spices \(=3\times0.65=1.95\) g.

Check: \(45.5+15.6+1.95+1.95=65\) g.

(i)

Shop A: pay ₹100 for 2 items, so each item costs \(100\div2=\)₹50.

Shop B: pay ₹200 for 3 items, so each item costs \(200\div3\approx\)₹66.67.

Shop C: pay ₹300 for 4 items, so each item costs \(300\div4=\)₹75.

From cheapest to costliest: Shop A, Shop B, Shop C.

(ii) Percentage discount \(=\dfrac{\text{free items}}{\text{total items received}}\times100\).

Shop A: \(\dfrac12\times100=50\%\).

Shop B: \(\dfrac13\times100\approx33.33\%\).

Shop C: \(\dfrac14\times100=25\%\).

(iii) For 4 items:

Shop A: buy 2 and get 2 free, so pay ₹200.

Shop B: buy 2 and get 1 free (3 items for ₹200), then buy the 4th item at ₹100, so pay ₹300.

Shop C: buy 3 and get 1 free, so pay ₹300.

Shop A is the cheapest for 4 items (₹200), so I would choose Shop A.

(i) Valid. The twenties have the highest percentages (26% for women, 37% for men) among all the age groups.

(ii) Valid. In every age group the bar for women is shorter than the bar for men (for example 24% against 29% for teenagers, and 26% against 37% in the twenties).

Even for children, the female bar (4%) is a little shorter than the male bar.

(iii) Not valid. The graph shows the percentage of each age group that can use computers, not how many people are in each age group.

So it tells us nothing about which age group has more people.

(iv) Not valid. In the thirties the values are 14% (women) and 25% (men).

Neither is more than a quarter (25%).

(v) Valid. For seniors the values are 2% (women) and 4% (men).

Both are less than 10%, that is, less than 1 in 10.

(vi) Not valid. In the twenties the values are 26% (women) and 37% (men), and both are less than half (50%).