Sharing a whole in a given ratio

What to assume they know

• Simplest form, and using it to test whether two ratios are proportional — reducing a ratio to its simplest form
• What a ratio claims, and why it is not a difference — reading a ratio as a claim about matched amounts
• Division with and without remainder, and multiplying a whole number by a fraction
• Adding the terms of a ratio
• Reading a fraction of a quantity, as in two fifths of 30
• Working with money, weights and volumes in mixed units

What they should be able to do

• Divide a given quantity between two shares in a stated ratio, showing the group size as a separate step
• Explain why the number of groups is the sum of the terms of the ratio
• Verify an answer by reducing the ratio of the two shares back to the given ratio
• Use the general form — m × x / (m + n) and n × x / (m + n) — and say what each factor is counting
• Recover the amounts in a mixture from its total and its ratio
• Change a mixture to a new ratio by adding to one ingredient only, identifying first which quantity is unchanged
• Show that a ratio by itself does not determine the shares, using two problems with the same ratio and different wholes
• Notice when a sharing problem needs an assumption that the question does not state

Where it usually goes wrong

• "Sharing in the ratio 3 : 1 means giving one person 3 and the other 1." That is one round of dealing, not the answer, unless the whole happens to be 4.
• "The ratio tells me the shares." It does not until the whole is named. The chapter puts 3 : 1 on 12 counters and on ₹4,000 within three pages.
• "Divide by the bigger term." The divisor is the sum of the terms, because that is how many groups the whole contains.
• "Divide by 2, because there are two people." Only for 1 : 1. The two people do not get equal groups; they get equal-sized groups in unequal numbers.
• "3 : 1 means three quarters and one quarter, so the ratio is a pair of fractions." The fractions come out of the ratio and are not the ratio. Keep the group step visible or students will guess the wrong denominator.
• "To change a mixture from 3 : 1 to 5 : 2, change both ingredients." The problem says only cement is added, so the sand is the anchor. Choosing the wrong anchor is the commonest error on Example 12.
• "Adding to a mixture is the same as adding to both terms of a ratio." Adding yellow paint adds to one term only. This is where this topic and the ratio-is-not-a-difference topic meet.
• "The answer must come out in whole objects." For counters and cups of rice it does; for a length it need not — a height of 152 cm split 4 : 6 gives 60.8 cm and 91.2 cm, and the answer is still correct.

Questions to check understanding

• Divide a stated quantity in a stated two-term ratio, showing the group size
• Recover both ingredient amounts from a mixture's total and its ratio
• Given one share and the ratio, find the other share and the whole
• Adjust a mixture to a new ratio by adding to one ingredient, and state which quantity was held fixed
• Share a profit or a bill in the ratio of two contributions given in unreduced form
• Verify a proposed split by checking both the total and the reduced ratio
• Explain why two problems with the same ratio have different answers
• Identify the missing information in an under-specified sharing problem

Examples worth working on the board

Values marked verified are worked out here or an added count on the printed page; the chapter works several of these itself, and that is said where it does.

• Activity 3 (Part I, §7.5, p.172). Work in pairs, collect 12 countable objects — the page suggests coins, seeds or pebbles — and share them between two people in several ways. The page works the equal split as 6 and 6, giving 6 : 6 and 1 : 1 reduced, then asks what happens if one partner takes 5, and what the ratio of that split is. An illustration shows two children at a table with a scatter of counters, and beside it three circular vignettes each showing the twelve counters divided into two piles. Verified by counting the drawn counters on each vignette separately: the three vignettes show 3 and 9, then 8 and 4, then 2 and 10 — three genuine splits of twelve, each totalling twelve as drawn. None of them is the 5-and-7 split the text asks about, so a teacher must not point at a vignette when posing that question; the first vignette is the 3 : 1 answer the page reaches next.
• The dealing procedure (Part I, §7.5, p.173, worked on the page). To reach 3 : 1 from 12 counters: partner takes 3 and you take 1, leaving 8; again, leaving 4; again, leaving nothing. Three rounds, so 9 counters against 3. The page prints all three rounds and the totals.
• The chapter's sharing diagram, twelve (Part I, §7.5, p.173). An ellipse holding 12 at the top, an arrow to the label that the whole is 12; below it two boxes reading 3 and 1 with a colon between them, labelled to say that this smaller whole is 4 because 3 + 1 = 4 and that 12 is three times 4; a curved multiply-by-3 arrow on each side; and beneath, two boxes reading 9 and 3, labelled to say the parts are three times the ratio. The multiplier is set in a contrasting colour wherever it appears. This is the figure the topic is built on.
• The chapter's sharing diagram, forty-two (Part I, §7.5, p.174). The same four-tier layout with 42 at the top, 4 and 3 in the middle, a multiply-by-6 arrow, and 24 and 18 at the foot.
• Forty-two in the ratio 4 : 3 (Part I, §7.5, p.173, worked on the page). The page rejects dealing as too slow, asks for the number of groups, gives it as 4 + 3 = 7, divides 42 by 7 to get a group size of 6, and multiplies to reach 24 and 18.
• The general statement (Part I, §7.5, p.174, worked on the page). To split a total x between shares standing as m : n, the number of groups is m + n, so the group size is x / (m + n), and the two parts are m and n times that. The page closes by writing the two parts as a proportion against m : n, which is the check in section 9.
• Why the answer is unique (the argument of section 9). The two shares must add to x, and their ratio must reduce to m : n, so both are multiples of one group size g: they are mg and ng, and mg + ng = x forces g = x / (m + n). One equation, one unknown, one answer. The chapter derives the formula but does not argue that no other split could work; that argument is added here and it is what turns a recipe into a reason.
• Example 11, the food cart (Part I, §7.5, p.174, worked on the page). Prashanti invested ₹75,000 and Bhuvan ₹25,000; the first month's profit is ₹4,000, to be shared in the ratio of the investments. The page reduces 75000 : 25000 to 3 : 1, adds the terms to get 4, divides 4,000 by 4 to get 1,000, and prints the shares as ₹3,000 and ₹1,000. Put this beside the twelve counters: same ratio, same three-and-one structure, completely different amounts. That pairing is the thesis.
• Example 12, the mixture (Part I, §7.5, pp.174–175, worked on the page). A 40 kg mixture holds sand and cement in the ratio 3 : 1. The page finds the sand as three quarters of 40, that is 30 kg, and the cement as one quarter, 10 kg. It then observes that the sand does not change, sets up 5 : 2 :: 30 : ?, notes that in a 5 : 2 ratio the second term is two fifths of the first, computes two fifths of 30 as 12 kg, and concludes that 2 kg of cement must be added because 10 kg is already there. Two things to draw out: the anchor is the ingredient nobody touches, and the step that finds 12 uses the comparison inside one ratio rather than the factor between two ratios — a different move from every earlier example in the chapter, and worth naming as such.
• Five sharing problems (Part I, §7.5, p.175, exercise items 1 to 5). As printed: divide ₹4,500 in the ratio 2 : 3; acid and water mixed 1 : 5 in a 240 mL bottle, find each; blue and yellow mixed 3 : 5 to make 40 mL of green, find each, then add 20 mL of yellow and give the new ratio; rice and urad dal mixed 2 : 1 for 6 cups of mixture, find each; and one bucket of orange paint made from red and yellow in the ratio 3 : 5, to which another bucket of yellow paint is added — give the new ratio. Item 5 cannot be answered without an assumption: the second bucket's size is never stated, and the intended reading is presumably that the two buckets hold the same amount. Verified: under that reading the first bucket is 8 parts, the added yellow is another 8 parts, and the new ratio is 3 : 13. Under any other reading the answer moves. Make the assumption a talking point rather than a silent step; the same item is also the chapter's best case of the difference between adding to a ratio's terms and adding to a mixture.
• The crane's neck (Part I, p.176, exercise item 4). A crane 155 cm tall carries its neck against the remainder of its body as 4 : 6; the student is to apply the same ratio to their own height and find their neck's length. Note the trap hiding in plain sight: the crane's 155 cm is not needed at all, and the ratio is not in simplest form.

Figures to have open

• The chapter's four-tier sharing diagram (Part I p.173 and again p.174), redrawn: whole at the top, the ratio's terms in the middle, the multiplier on the arrows, the parts at the foot. Essential — it is the chapter's own summary of the method.
• A bar of the whole divided into m + n equal cells, with the first m cells shaded. An added figure, and the one that makes "groups, not objects" visible.
• Twelve counters as movable tokens for sections 1 to 4. Standard schematic; redraw as plain discs rather than reproducing the printed seeds.
• The mixture of Example 12 as a 40 kg bar in two materials, with the added cement appearing at the end while the sand band stays exactly as it was. Standard schematic and central to section 11.
• No textbook photograph or illustration is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part I, printed Chapter 7, "Proportional Reasoning-1", §7.5 "Sharing, but Not Equally!", Part I pp.172–175, including Activity 3 (Part I p.172), the two sharing diagrams (Part I pp.173–174), the general derivation (Part I p.174) and Examples 11 and 12 (Part I pp.174–175).
• Exercises: Part I p.175 items 1 to 5. Part I p.176 item 4, which is printed in the §7.6 exercise block but is a sharing problem and is carried here.
• The chapter's SUMMARY (Part I p.177) restates the two parts in the general form, and is the one summary line this topic must match exactly in meaning.
• Related within the chapter: Example 12 is the first place the chapter compares the two terms inside one ratio. Its multiplier is the same kind of number as the three sevenths in Example 7 (Part I p.164, carried in Solving a proportion problem, and the Trairasika rule of three), but Example 7's factor runs between the two ratios, which is a different move.
• Item 12 on Part I p.177, the ₹10 coin, is a sharing-in-ratio problem wrapped in a unit conversion, and is carried in Unit conversion is a proportion in disguise.