Proportion and Dividing a Quantity in a Ratio
Proportion and Dividing a Quantity in a Ratio
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.