Ratio and Proportion
Ratio questions test whether you can keep track of parts versus actual values — a distinction CLAT exploits constantly in its data-based quant passages.
10 questions · 5 minutes · instant scoring
What this topic actually tests
A ratio compares two or more quantities in terms of equal 'parts' rather than absolute values, and the fastest CLAT technique is to treat the ratio as a multiplier x rather than converting to fractions or decimals. If ages are in ratio 4:5, write them as 4x and 5x — one unknown instead of two — and use whatever condition the question gives (a sum, a difference, or a value after some years) to solve for x directly. Once x is known, every actual value follows by simple multiplication. This is faster than cross-multiplying fractions and avoids arithmetic errors with awkward numbers. For proportion questions (a:b = c:d), remember that cross-multiplication (ad = bc) is the standard tool, and for combining two ratios that share a common term (a:b and b:c) you must scale both ratios so the shared term matches before combining into a:b:c — for example, a:b = 2:3 and b:c = 4:5 must be scaled to a:b = 8:12 and b:c = 12:15 (matching b at 12) to give a:b:c = 8:12:15. Worked example: a sum of Rs 60,000 is split among three partners in ratio 3:4:5. Total parts = 3+4+5 = 12, so each part = 60,000/12 = 5,000, and the middle partner's share = 4 x 5,000 = Rs 20,000. This 'total parts' method — divide the whole by the sum of ratio parts, then multiply by the relevant part — is the single fastest way to solve nearly every direct-division ratio question on the exam, and it generalises directly to mixture and alligation problems.
The common trap on this topic
The most common ratio trap is treating ratio parts as if they were already actual values. If a ratio is 3:5 and a question states that the smaller value is 30, students sometimes assume the larger is simply 5 more or otherwise misapply the difference instead of recognising that each 'part' equals 30/3 = 10, making the larger value 5 x 10 = 50. A second frequent trap arises in 'ratio changes after an operation' questions (adding or subtracting a fixed amount from each term, or the passage of years for ages): because the operation is not proportional, you cannot simply apply it to the ratio numbers directly — you must first express the values as multiples of x, apply the stated operation to those expressions, and only then set up the new ratio equation. Skipping the algebraic setup and eyeballing the answer is a major source of wrong answers under time pressure. A third trap is losing track of which quantity is being asked for after combining two ratios into a three-term ratio (a:b:c) — always re-read the question to confirm whether it wants a single term, a two-term sub-ratio, or a combined expression, since the combined ratio calculation is itself only an intermediate step.
Take the micro-test
Two numbers are in the ratio 3:5. If 10 is added to each number, the ratio becomes 5:7. Find the two numbers.
A sum of money is divided among A, B, and C in the ratio 2:3:5. If C's share exceeds A's share by Rs 3,000, find B's share.
How much water is present in the original 50-litre mixture?
If 10 litres of water is now added to this mixture, what is the new ratio of milk to water?
The ratio of A's age to B's age is 4:5. Five years ago, the ratio of their ages was 3:4. Find A's present age.
If a:b = 2:3 and b:c = 4:5, find a:b:c.
Find the third proportional to 4 and 12.
Find the mean proportional between 9 and 16.
A, B, and C invest in a business in the ratio 3:4:5. If the total profit at the end of the year is Rs 60,000, find B's share of the profit.
If x:y = 3:4, find the value of (3x + 2y) : (2x - y).
FAQ
What is the fastest way to solve "divide X among people in ratio a:b:c" questions?
Add the parts (a+b+c), divide the total amount by that sum to get the value of one part, then multiply by whichever part corresponds to the share you need — this is faster than setting up algebraic equations.
How do I combine two ratios that share a common term, like a:b and b:c?
Scale each ratio so that the shared term (b) has the same value in both, then write the combined ratio a:b:c using those scaled numbers.
Why do ratio and proportion questions often appear alongside averages and mixtures on CLAT?
Mixture and alligation problems are really ratio problems in disguise — the ratio in which two quantities must be combined to reach a target average is found using the same parts-based reasoning used in direct ratio-sharing questions.
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