Proportional Reasoning–1

 HEGP107 — Proportional Reasoning–1

Complete Answers to the Mathematical Problems

This guide answers the mathematical problems and “Figure it Out” exercises in the attached Class 8 chapter.

 

1. Ratios and similar images

The image dimensions are:

 

Image	Width	Height	Width : height in simplest form
A	60 mm	40 mm	(3:2)
B	40 mm	20 mm	(2:1)
C	30 mm	20 mm	(3:2)
D	90 mm	60 mm	(3:2)
E	60 mm	60 mm	(1:1)

Therefore, A, C, and D are similar because their width-to-height ratios are equal. B and E are not similar to them.

 

The ratio of width to height is unchanged when both dimensions are multiplied by the same factor. For example,

 

[ 60:40=30:20=90:60=3:2. ]

 

2. Coffee and milk ratios

The regular coffee ratio is (300:600=1:2). Compare each mixture with this ratio.

 

Coffee decoction	Milk	Simplest ratio	Result
300 mL	600 mL	(1:2)	Regular
150 mL	500 mL	(3:10)	Lighter
200 mL	400 mL	(1:2)	Regular
24 mL	56 mL	(3:7)	Stronger
100 mL	300 mL	(1:3)	Lighter

The coffee is stronger when the amount of decoction is larger relative to the amount of milk, and lighter when it is smaller relative to the milk.

 

3. Rule of Three and direct proportion

Mid-day meal

For 120 students, the cook uses 15 kg of rice. For 80 students,

 

[ 120:15::80:x. ]

 

Thus,

 

[ x=\frac{15\times80}{120}=\boxed{10\text{ kg}}. ]

 

Car journey

The car travels 90 km in 150 minutes. Four hours equals 240 minutes. Therefore,

 

[ 150:90::240:x. ]

 

[ x=\frac{240\times90}{150}=\boxed{144\text{ km}}. ]

 

Tea prices

Himachal tea costs ₹200 for 200 g, so the price per kilogram is

 

[ \frac{1000}{200}\times200=\boxed{₹1000}. ]

 

Meghalaya tea costs ₹800 per kilogram. Therefore, the Himachal tea is more expensive. The weight-to-price ratios are

 

[200:200=1:1,
\qquad1000:800=5:4,]

 

so the ratios are not proportional.

 

4. Additional proportional-reasoning problems

Earth’s weekly travel

The Earth travels approximately 940 million kilometres in 52 weeks. Therefore,

 

[ \frac{940,000,000}{52}\approx\boxed{18,076,923\text{ km per week}}. ]

 

Mason’s bricks

The total length of the outer and inner walls shown in the diagram is 108 ft. Since 10 ft of wall requires 1,450 bricks,

 

[ 10:1450::108:x. ]

 

Thus,

 

[ x=\frac{108\times1450}{10}=\boxed{15,660\text{ bricks}}. ]

 

Motorcycle speed and travel time

At 50 km/h, the journey takes 2 hours, so the distance is

 

[ 50\times2=100\text{ km}. ]

 

At 75 km/h, the time is

 

[ \frac{100}{75}=\frac43\text{ hours}=\boxed{1\text{ hour }20\text{ minutes}}. ]

 

This is an inverse-proportion problem: increasing speed decreases the travel time. Therefore, it cannot be written as (50:2::75:x) as a direct proportion.

 

5. Figure it Out: ratios and proportions

Question 1. Identify true proportions.

Two ratios (a:b) and (c:d) are proportional if (ad=bc).

 

Statement	Check	Answer
(4:7::12:21)	(4\times21=7\times12=84)	True
(8:3::24:6)	(8\times6\ne3\times24)	False
(7:12::12:7)	(7\times7\ne12\times12)	False
(21:6::35:10)	(21\times10=6\times35=210)	True
(12:18::28:12)	(12\times12\ne18\times28)	False
(24:8::9:3)	(24\times3=8\times9=72)	True

The true statements are (i), (iv), and (vi).

 

Question 2. Give three ratios proportional to (4:9).

Multiplying both terms by 2, 3, and 4 gives

 

[ \boxed{8:18,\quad12:27,\quad16:36}. ]

 

Question 3. Complete ratios proportional to (18:24).

Since

 

[ 18:24=3:4, ]

 

we obtain

 

[ \boxed{3:4,\quad12:16,\quad20:\frac{80}{3},\quad27:36}. ]

 

The third missing term is (80/3=26\frac23). If the exercise intends whole-number terms only, 20 cannot be paired with a whole-number second term while preserving exactly the ratio (18:24).

 

Question 4. Similar rectangles

After measuring the corresponding sides and allowing for rotation, the similar pairs are approximately

 

[ \boxed{A\text{ and }D,\qquad C\text{ and }E}. ]

 

Rectangle B has a different width-to-height ratio. Two rectangles are similar when the ratios of corresponding sides are equal.

 

Question 5. Draw larger and smaller rectangles with the same ratio.

Yes. If the given rectangle has width-to-height ratio (w:h), then any rectangles with dimensions

 

[ 2w\times2h,\qquad \frac12w\times\frac12h ]

 

have the same ratio and are similar. Different students may draw different-sized rectangles; they are not wrong if the ratio is preserved.

 

Question 6. Brick-wall ratios

For pattern (a), the number of grey bricks in one repeating block is 9 and the number of coloured bricks is 6. Thus,

 

[ 9:6=\boxed{3:2}. ]

 

For pattern (b), there are 16 grey bricks and 12 coloured bricks in one block. Thus,

 

[ 16:12=\boxed{4:3}. ]

 

Question 7. Human-figure ratios

This is a measurement activity. If a student measures head, torso, arm, and leg lengths as (h,t,a,l), the required ratios are

 

[ \boxed{h:t,\qquad t:a,\qquad t:l}. ]

 

A proportional drawing should multiply all corresponding body lengths by the same scale factor.

 

5. Shampoo-price activity

The sample prices are not proportional to volume. For example,

 

[6:180=1:30,
\qquad2:154=1:77,]

 

so the volume ratio and price ratio are not equal. The larger bottles may offer better value to customers, while larger containers may reduce packaging per millilitre and therefore reduce environmental impact. The exact recommendation should be based on the price-per-millilitre and packaging data collected from the market.

 

6. Sharing quantities in ratios

Twelve counters

Equal sharing gives

 

[ 12\div2=6 ]

 

for each person, so the ratio is

 

[ 6:6=\boxed{1:1}. ]

 

If one person receives 5 counters, the other receives 7, so the ratio is

 

[ \boxed{5:7}. ]

 

Sharing 12 counters in the ratio (3:1):

 

[3+1=4,
\qquad12\div4=3.]

 

The shares are

 

[3\times3=\boxed{9},
\qquad1\times3=\boxed{3}.]

 

Forty-two counters in the ratio (4:3)

[4+3=7,
\qquad42\div7=6.]

 

The shares are

 

[4\times6=\boxed{24},
\qquad3\times6=\boxed{18}.]

 

In general, dividing (x) in the ratio (m:n) gives

 

[ \boxed{\frac{mx}{m+n}\quad\text{and}\quad\frac{nx}{m+n}}. ]

 

7. Profit sharing and mixtures

Profit sharing

The investments are ₹75,000 and ₹25,000, so the ratio is

 

[ 75000:25000=3:1. ]

 

The ₹4,000 profit is divided into 4 parts:

 

[ 4000\div4=1000. ]

 

Therefore, Prashanti receives

 

[ 3\times1000=\boxed{₹3000}, ]

 

and Bhuvan receives

 

[ 1\times1000=\boxed{₹1000}. ]

 

Sand and cement mixture

The original 40 kg mixture has ratio 3:1. Thus,

 

[\text{sand}=\frac34\times40=30\text{ kg},
\qquad
\text{cement}=\frac14\times40=10\text{ kg}.]

 

For a new ratio of 5:2, 30 kg of sand requires

 

[ \frac25\times30=12\text{ kg cement}. ]

 

Already there are 10 kg, so the amount to add is

 

[ \boxed{2\text{ kg of cement}}. ]

 

8. Figure it Out: sharing and mixtures

Question 1. Divide ₹4,500 in the ratio (2:3).

[2+3=5,
\qquad4500\div5=900.]

 

The parts are

 

[ \boxed{₹1800\text{ and }₹2700}. ]

 

Question 2. Acid and water in a 1:5 mixture of 240 mL

[1+5=6,
\qquad240\div6=40.]

 

Acid is

 

[ \boxed{40\text{ mL}}, ]

 

and water is

 

[ \boxed{200\text{ mL}}. ]

 

Question 3. Blue and yellow paint in the ratio 3:5

For 40 mL,

 

[3+5=8,
\qquad40\div8=5.]

 

Blue paint is (3\times5=\boxed{15\text{ mL}}), and yellow paint is (5\times5=\boxed{25\text{ mL}}). After adding 20 mL yellow, the amounts are 15 mL and 45 mL. The new ratio is

 

[ 15:45=\boxed{1:3}. ]

 

Question 4. Rice and urad dal in the ratio 2:1 for 6 cups

[2+1=3,
\qquad6\div3=2.]

 

Rice is

 

[ \boxed{4\text{ cups}}, ]

 

and urad dal is

 

[ \boxed{2\text{ cups}}. ]

 

Question 5. Red and yellow paint in the ratio 3:5, then add one bucket of yellow

Take one original bucket as 8 equal units. Red is 3 units and yellow is 5 units. Adding another full bucket of yellow adds 8 units, so the new amounts are red 3 units and yellow 13 units. The new ratio is

 

[ \boxed{3:13}. ]

 

9. Unit-conversion and application problems

Question 1. Orange juice and apple juice

[ 600:900=\boxed{2:3}. ]

 

Question 2. School buses

Three buses carry 162 people, so one bus carries

 

[ 162\div3=54. ]

 

For 204 people,

 

[ 204\div54=3\frac79, ]

 

so 4 buses are required. Their total capacity is

 

[ 4\times54=216. ]

 

Thus, 4 buses are needed and

 

[ 216-204=\boxed{12\text{ seats vacant}}. ]

 

Question 3. Population density of Delhi and Mumbai

Delhi’s density is

 

[ \frac{30,000,000}{1484}\approx\boxed{20,216\text{ people/km}^2}. ]

 

Mumbai’s density is

 

[ \frac{20,000,000}{550}\approx\boxed{36,364\text{ people/km}^2}. ]

 

Therefore, Mumbai is more crowded.

 

Question 4. Crane neck ratio

The neck-to-rest-of-body ratio is 4:6, so the neck is 4 parts out of 10 parts of the total height. For a person of height (H), the neck would be

 

[ \boxed{\frac4{10}H=\frac25H}. ]

 

For the 155 cm crane, the neck is

 

[ \frac4{10}\times155=\boxed{62\text{ cm}}. ]

 

Question 5. Lilavati saffron problem

The given quantity is (2\frac12=2.5) palas for (\frac37) niskas. For 9 niskas,

 

[ 2.5:\frac37::x:9. ]

 

Thus,

 

[ x=\frac{2.5\times9}{3/7}=2.5\times21=\boxed{52.5\text{ palas}}. ]

 

Question 6. Harmain’s age

After (x) years, the ages are (1+x) and (5+x). We need

 

[ \frac{1+x}{5+x}=\frac12. ]

 

Therefore,

 

[2(1+x)=5+x
\Rightarrow x=3.]

 

Harmain’s age then is

 

[ \boxed{4\text{ years}}. ]

 

Question 7. Mass of one litre of gold

The mass ratio of equal volumes of gold and water is 37:2. If 1 L water has mass 1 kg, then

 

[ 37:2::x:1. ]

 

Thus,

 

[ x=\frac{37}{2}=\boxed{18.5\text{ kg}}. ]

 

Question 8. Manure for a 200 ft by 500 ft plot

The plot area is

 

[ 200\times500=100,000\text{ ft}^2. ]

 

One acre is 43,560 ft², and 10 tonnes of manure are needed per acre. Therefore,

 

[\text{manure}=10\times\frac{100,000}{43,560}
\approx22.9568\text{ tonnes}.]

 

The farmer should buy approximately

 

[ \boxed{22.96\text{ tonnes}}. ]

 

Question 9. Filling a 10-litre bucket

A 500 mL mug takes 15 seconds. A 10 L bucket holds 10,000 mL, which is 20 mugs. Therefore,

 

[ 20\times15=\boxed{300\text{ seconds}=5\text{ minutes}}. ]

 

Question 10. Cost of 2,400 ft² of land

One acre is 43,560 ft² and costs ₹15,00,000. Therefore,

 

[\text{cost}=15,00,000\times\frac{2400}{43560}
\approx\boxed{₹82,645.18}.]

 

Approximately, the cost is ₹82,645.

 

Question 11. Oxen and tractor ploughing

A pair of oxen takes 6 hours per acre. For 20 acres,

 

[ 20\times6=\boxed{120\text{ hours}}. ]

 

A tractor is 4 times faster, so it takes one-fourth of the time:

 

[ 120\div4=\boxed{30\text{ hours}}. ]

 

Question 12. Copper and nickel in a ₹10 coin

The mass ratio is copper:nickel = 3:1. The total mass is 7.74 g, so there are 4 equal parts:

 

[ 7.74\div4=1.935\text{ g per part}. ]

 

Copper mass:

 

[ 3\times1.935=5.805\text{ g}. ]

 

Nickel mass:

 

[ 1.935\text{ g}. ]

 

Copper costs ₹906 per kg, so its cost is

 

[ 906\times\frac{5.805}{1000}\approx₹5.26. ]

 

Nickel costs ₹1,341 per kg, so its cost is

 

[ 1341\times\frac{1.935}{1000}\approx₹2.59. ]

 

The total metal cost is

 

[ \boxed{₹7.85\text{ approximately}}. ]

 

10. Important formulas used in the chapter

For two ratios,

 

[ a:b::c:d\quad\Longleftrightarrow\quad ad=bc. ]

 

A quantity (x) divided in the ratio (m:n) gives parts

 

[ \boxed{\frac{mx}{m+n}\quad\text{and}\quad\frac{nx}{m+n}}. ]

 

For direct proportion, both quantities increase or decrease by the same factor. For inverse proportion, one quantity increases while the other decreases by the corresponding factor; for example, at fixed distance, time is inversely proportional to speed.