Proportion and Dividing a Quantity in a Ratio
समानुपात और किसी राशि का अनुपात में विभाजन
Proportion questions in NTPC are formula hits (third, fourth and mean proportional), while division questions test whether you can turn "shares in a ratio" into "value of one part". The 2025-26 papers added two-step versions: proportionals chained together, shares with a given difference and pairwise sums such as (x + y) : (y + z) : (z + x).
1. Proportion and the Three Proportionals
a : b :: c : d means a/b = c/d, so a × d = b × c (product of extremes = product of means)
| Term | Definition | Formula | Example |
|---|---|---|---|
| Fourth proportional to a, b, c | d in a : b :: c : d | d = bc/a | 4, 6, 8 → 6 × 8/4 = 12 |
| Third proportional to a, b | c in a : b :: b : c | c = b²/a | 12, 18 → 324/12 = 27 |
| Mean proportional between a, b | x in a : x :: x : b | x = √(ab) | 12, 27 → √324 = 18 |
| Continued proportion a, b, c | a : b = b : c | b² = ac | 12, 18, 27 |
| Missing term | Any one unknown | Cross multiply | 0.6 : x :: 4.5 : 12 → x = 0.6 × 12/4.5 = 1.6 |
✗ Third proportional to 12 and 18 = 12 × 12/18 = 8 | ✓ Third proportional is b²/a = 18 × 18/12 = 27; the ORDER matters (8 is the third proportional to 18 and 12)
2. Dividing a Quantity in a Given Ratio
Share of A = (A's ratio term / sum of ratio terms) × total
| Situation | Method | Example |
|---|---|---|
| Whole ratio | Value of one part = total ÷ sum of terms | ₹1,800 in 2 : 3 : 4 → 200 per part → 400, 600, 800 |
| Fractional ratio | Multiply by the LCM of denominators first | 1/2 : 1/3 : 1/4 → × 12 → 6 : 4 : 3 |
| Difference of two shares given | One part = difference ÷ difference of terms | R − Q = 1,350 in 9 : 4 : 7 : 5 → 3 parts = 1,350 |
| One share given | One part = share ÷ its term | Girls 1,375 at 8 : 11 → 125 per part |
| Sum and difference of two numbers given | Larger = (S + D)/2, smaller = (S − D)/2 | S = 24, D = 3.6 → 13.8 : 10.2 = 23 : 17 |
3. Changing a Ratio by Adding or Subtracting
If x is added to both terms of a : b to give c : d, then (a + x)/(b + x) = c/d; for unknown numbers in ratio a : b, write them as ak and bk
To make four numbers p, q, r, s proportional by adding x to each: (p + x)(s + x) = (q + x)(r + x). The x² terms cancel, leaving a linear equation. Shortcut: x = (qr − ps)/(p + s − q − r).
4. Pairwise Sums in a Ratio
If (x + y) : (y + z) : (z + x) = p : q : r, then the three pair-sums add up to 2(x + y + z) = (p + q + r) parts
So one part = 2(x + y + z)/(p + q + r). Then z = (x + y + z) − (x + y), and so on.
हिंदी नोट: तीसरा समानुपाती b²/a और चौथा समानुपाती bc/a होता है, जबकि मध्यानुपाती √(ab) होता है। संख्याओं का क्रम बदलने से उत्तर बदल जाता है, इसलिए प्रश्न में दिए क्रम पर ही काम कीजिए।
Exam Pointer: Verified NTPC patterns: third, fourth and mean proportional, including chained versions (January 2021, June 2025 Graduate CBT-1, May 2026 UG CBT-1), dividing a sum in a ratio or with a given difference of shares (March 2021, June 2025), one share known or ratio from sum and difference (December 2020, June 2022 CBT-2), subtracting a number to change the ratio (April 2016) and pairwise sums (March 2026 Graduate CBT-1).
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Third, fourth and mean proportional, and missing terms
[PYQ: NTPC UG CBT-1 9-May-2026 Shift-3 | NTPC Graduate CBT-1 20-Jun-2025 Shift-1 | NTPC CBT-1 20-Jan-2021 Shift-1]
EXAM LEVEL
Q. Find the third proportional to 12 and 18, and the mean proportional between 12 and 27.
Third proportional = 18²/12 = 324/12 = 27. Mean proportional = √(12 × 27) = √324 = 18. Together 12, 18, 27 form a continued proportion (12 : 18 = 18 : 27 = 2 : 3), which is why the two answers are linked.
Answer: 27 and 18
EXAMATLAS LEVEL
Q. x is the mean proportional between 4 and 36, and y is the third proportional to x and 18. Find the fourth proportional to x, y and 20.
x = √(4 × 36) = √144 = 12. y = 18²/12 = 27. Fourth proportional to 12, 27, 20 = 27 × 20/12 = 45. Check: 12 : 27 = 20 : 45 = 4 : 9.
Answer: 45
Pattern 2: Dividing a sum in a ratio, including a given difference of shares
[PYQ: NTPC Graduate CBT-1 23-Jun-2025 Shift-3 | NTPC CBT-1 14-Mar-2021 Shift-2]
EXAM LEVEL
Q. Divide ₹6,500 among A, B and C in the ratio 1/2 : 1/3 : 1/4.
Multiply the fractional ratio by the LCM 12: 6 : 4 : 3, a total of 13 parts. One part = 6,500 ÷ 13 = 500. A = ₹3,000, B = ₹2,000, C = ₹1,500. Splitting in 2 : 3 : 4 (the denominators) is the trap.
Answer: ₹3,000, ₹2,000 and ₹1,500
EXAMATLAS LEVEL
Q. A sum is divided among P, Q, R and S in the ratio 9 : 4 : 7 : 5. If R gets ₹1,350 more than Q, find the total sum and S's share.
R − Q corresponds to 7 − 4 = 3 parts, so one part = 1,350 ÷ 3 = ₹450. Total = (9 + 4 + 7 + 5) × 450 = 25 × 450 = ₹11,250. S = 5 × 450 = ₹2,250. No need to find all four shares.
Answer: Total ₹11,250; S gets ₹2,250
Pattern 3: One quantity known: the other quantity, the whole, or the ratio itself
[PYQ: NTPC CBT-2 15-Jun-2022 Shift-3 | NTPC CBT-2 12-Jun-2022 Shift-2 | NTPC CBT-1 30-Dec-2020 Shift-1]
EXAM LEVEL
Q. The ratio of boys to girls in a college is 8 : 11 and there are 1,375 girls. How many boys are there?
11 parts = 1,375, so one part = 125. Boys = 8 × 125 = 1,000.
Answer: 1,000
EXAMATLAS LEVEL
Q. The sum of two numbers is 24 and their difference is 3.6. Find the ratio of the larger number to the smaller.
Larger = (24 + 3.6)/2 = 13.8, smaller = (24 − 3.6)/2 = 10.2. Ratio = 13.8 : 10.2 = 138 : 102 = 23 : 17 (divide by 6). Shortcut: ratio = (S + D) : (S − D) = 27.6 : 20.4 = 23 : 17, no halving needed.
Answer: 23 : 17
Pattern 4: Adding or subtracting a number to change a ratio or create a proportion
[PYQ: NTPC CBT-1 27-Apr-2016 Shift-3]
EXAM LEVEL
Q. Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, the ratio becomes 12 : 23. Find the numbers.
Let them be 3k and 5k. Then 23(3k − 9) = 12(5k − 9), so 69k − 207 = 60k − 108, giving 9k = 99 and k = 11. Numbers: 33 and 55. Check: 24 : 46 = 12 : 23.
Answer: 33 and 55
EXAMATLAS LEVEL
Q. What number must be added to each of 7, 16, 43 and 79 so that the results are in proportion?
(7 + x)(79 + x) = (16 + x)(43 + x). Expand: 553 + 86x = 688 + 59x after the x² terms cancel, so 27x = 135 and x = 5. Shortcut: x = (16 × 43 − 7 × 79)/(7 + 79 − 16 − 43) = (688 − 553)/27 = 5. Check: 12, 21, 48, 84 and 12/21 = 48/84 = 4/7.
Answer: 5
Pattern 5: Pairwise sums in a ratio
[PYQ: NTPC Graduate CBT-1 24-Mar-2026 Shift-2]
EXAM LEVEL
Q. If (a + b) : (b + c) : (c + a) = 5 : 6 : 7 and a + b + c = 27, find c.
The three pair-sums add up to 2 × 27 = 54, which is 5 + 6 + 7 = 18 parts, so one part = 3. Then a + b = 15 and c = 27 − 15 = 12. (Also b + c = 18 gives b = 6 and a = 9; c + a = 21 = 7 × 3 confirms.)
Answer: 12
EXAMATLAS LEVEL
Q. If (x + y) : (y + z) : (z + x) = 7 : 8 : 9 and x + y + z = 36, find x : y : z.
Pair-sums total 72 = 24 parts, so one part = 3: x + y = 21, y + z = 24, z + x = 27. Subtract each from 36: z = 15, x = 12, y = 9. So x : y : z = 12 : 9 : 15 = 4 : 3 : 5. Treating 7 : 8 : 9 as x : y : z directly is the trap.
Answer: 4 : 3 : 5
60-Second Revision
- a : b :: c : d means ad = bc; fourth proportional bc/a, third b²/a, mean √(ab).
- Share = (own term ÷ sum of terms) × total; fractional ratios: multiply by the LCM first.
- Difference of shares: one part = difference ÷ difference of terms.
- Sum S and difference D: ratio = (S + D) : (S − D).
- Adding x to make p, q, r, s proportional: x = (qr − ps)/(p + s − q − r).
- Pairwise sums: they total 2(x + y + z); find one part, then subtract.
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