Proportional Reasoning-1 · Lesson 5 of 8
Sharing, but Not Equally!
“Divide a known whole into parts in a specified ratio, derive the general sharing formula and track how adding one component changes a mixture.”
• Distinguish a part-to-part ratio from the whole. • Find one equal ratio part and calculate each share. • Derive shares mx/(m+n) and nx/(m+n). • Solve profit-sharing and mixture problems. • Track which amount stays fixed when one ingredient is added.
Suppose you and a friend have twelve counters to share. Equal sharing gives six each, but what if your friend should receive three counters for every one you receive? The ratio describes how the two shares compare, not how many counters either person immediately takes. To use all twelve counters, you need repeated groups of that same pattern. Finding the size of one ratio part will turn this hands-on idea into a method for any suitable total.
This lesson begins with counters and then applies the same reasoning to money, recipes and mixtures. After dividing the whole, always check both the total and the ratio.
7.5 Sharing, but Not Equally!
In a ratio 3 : 1, imagine three equal-sized groups for the first share and one equal-sized group for the second. There are four groups in total. Sharing twelve counters gives 12 ÷ 4 = 3 counters per group.
For a larger total, counting out small batches is slow. To share 42 counters in the ratio 4 : 3, first add the ratio terms: there are seven groups. Each group contains 42 ÷ 7 = 6 counters. Four groups give 24, while three groups give 18.
Problem
Divide 42 counters in the ratio 4 : 3.
Generalise the same process. If a total x is divided in the ratio m : n, the whole contains m + n equal parts. One part is x/(m+n). Multiply this by m for the first share and by n for the second.
The denominator is the sum of the ratio terms because it represents the whole. Using m/n × x for the first share confuses “first compared with second” with “first compared with the whole.”
Problem
Prashanti invests ₹75,000 and Bhuvan ₹25,000. They agree to divide ₹4,000 profit in their investment ratio. Find each share.
Problem
Rice : urad dal = 2 : 1. How many cups of each make six cups of the mixture in this model?
A changing-mixture problem has two stages. First recover the actual amounts from the original total and ratio. Then identify what stays unchanged when an ingredient is added. The new total is usually different from the old total.
For the sand-and-cement example, adding cement leaves the amount of sand fixed. The new ratio determines the required total cement, after which you subtract the cement already present to find the amount to add.
Problem
A 40 kg mixture contains sand and cement in the ratio 3 : 1. How much cement must be added to make the ratio 5 : 2?
Figure it Out
Problem
A mathematical mixture model uses acid : water = 1 : 5 and total volume 240 mL. Find the component quantities.
Problem
Blue : yellow = 3 : 5 in 40 mL of paint. Find the two amounts and the ratio after adding 20 mL yellow.
Problem
A full bucket contains red : yellow paint in the ratio 3 : 5. Add one equally sized full bucket of yellow paint. Find the new ratio.
Money and measured amounts can often use fractional shares, subject to practical precision. Whole counters cannot be split if the activity forbids breaking them. For instance, ten counters cannot be shared exactly in the ratio 2 : 1 using only whole counters because one part would be 10/3 counters.
Quiz
When dividing a total in the ratio 3 : 5, how many equal parts make the whole?
Divide 42 in the ratio 4 : 3. What are the shares?
Which expression is the first share of total x in ratio m : n?
A 40 kg sand–cement mixture is 3 : 1. What stays fixed when only cement is added?
Why does adding one equal bucket of yellow change 3 : 5 paint to 3 : 13?
Practice Problems
- Divide ₹4500 in the ratio 2 : 3.
- A 240 mL mixture has components in the ratio 1 : 5. Find the amounts.
- A 40 mL blue-and-yellow mixture has ratio 3 : 5. Find the new ratio after adding 20 mL yellow.
- How much cement is added to 40 kg of a 3 : 1 sand–cement mixture to change the ratio to 5 : 2?
- An original full bucket has red : yellow = 3 : 5. Add another equally sized full bucket of yellow. Find the new ratio and each colour’s fraction of the final whole.
1. Total parts = 5; one part = ₹4500/5 = ₹900. 2. Shares are ₹1800 and ₹2700. Their sum is ₹4500.
1. One part = 240/6 = 40 mL. 2. The amounts are 40 mL and 200 mL.
1. The original amounts are 15 mL blue and 25 mL yellow. 2. After adding yellow they are 15 mL and 45 mL, giving 1 : 3.
1. Sand is 30 kg and stays fixed; original cement is 10 kg. 2. The new cement must be 2/5 × 30 = 12 kg. 3. Add 12 − 10 = 2 kg, not 12 kg.
1. In original bucket units, red = 3/8 and yellow becomes 13/8. Their ratio is 3 : 13. 2. The final whole is two buckets, or 16 original ratio parts. 3. Red is 3/16 and yellow is 13/16 of the final whole. These fractions add to 1.
Key Takeaways
• A ratio compares parts; adding its terms identifies the whole’s number of equal parts. • Find one part first, then multiply to obtain each share. • The two shares must both add to the total and simplify to the requested ratio. • When one component is added, track the unchanged amount before using the new ratio. • Required final amount and amount to add are different quantities. • A full added container must be expressed in the same units as the original ratio parts.