Ratio & Proportion • Topic 2 of 5

Continued Ratio

A continued (or combined) ratio links three or more quantities, such as a:b:c. The constant headache is that two separate ratios usually share a common term written with different numbers — for example a:b = 2:3 and b:c = 4:5. Here b is 3 in the first ratio and 4 in the second, so you must scale each ratio to make the shared term equal: take the LCM of 3 and 4 (which is 12). Then a:b = 8:12 and b:c = 12:15, giving a:b:c = 8:12:15. This LCM-bridging trick is the heart of every continued-ratio question and feeds straight into partnership and mixture problems. Once combined, treat the whole chain with a single common multiplier k (8k, 12k, 15k) so a total or a difference becomes one equation. Watch the order: a:b:c is not the same as c:b:a.

✅ Solved examples

1. If a:b = 2:3 and b:c = 4:5, find a:b:c.
Make b common. LCM(3,4) = 12. a:b = 8:12, b:c = 12:15. So a:b:c = 8:12:15.
2. A:B = 3:4 and B:C = 6:7. Find A:B:C.
B is 4 then 6; LCM = 12. A:B = 9:12, B:C = 12:14. So A:B:C = 9:12:14.
3. P:Q = 2:3, Q:R = 4:5, R:S = 6:7. Find P:S.
Multiply the chain: P/S = (2/3)(4/5)(6/7) = 48/105 = 16/35. So P:S = 16:35.
4. If a:b:c = 2:3:5 and a + b + c = 200, find c.
Sum of parts = 10, total = 200 ⇒ k = 20. c = 5k = 100.

✏️ Practice — try these, take hints as needed

1. If x:y = 3:5 and y:z = 10:13, find x:y:z.
Make y common.
y is 5 then 10; scale first ratio by 2.
x:y = 6:10.
6 : 10 : 13
2. A:B = 5:6 and B:C = 8:9. Find A:B:C.
B is 6 then 8; LCM = 24.
A:B = 20:24, B:C = 24:27.
Chain them.
20 : 24 : 27
3. If P:Q = 1:2, Q:R = 3:4, find P:R.
Multiply P/Q × Q/R.
(1/2)(3/4).
= 3/8.
3 : 8
4. a:b:c = 4:5:6 and the total is 225. Find b.
Sum of parts = 15.
k = 225/15.
b = 5k.
75
5. If a:b = 2:3, b:c = 5:7 and a = 20, find c.
Combine: b common, LCM(3,5)=15 ⇒ a:b:c = 10:15:21.
a = 10k = 20 ⇒ k = 2.
c = 21k.
42

📝 Topic test — 8 questions

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