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9. Ratios and Proportion

CSS GSA — Paper 7, Part-II, Section VI, Sub-topic 9 (Ratios and Proportion)

Past Paper Weightage

  • Very common in word problems (dividing amounts, scaling recipes/maps, mixture problems) — high frequency.

Ratios

  • A ratio a:b compares two quantities of the same kind — can be simplified like a fraction
a:b=aba : b = \frac{a}{b}
  • Dividing a quantity in a given ratio: split total into (sum of ratio parts), then multiply by each part

Proportion

  • A proportion states two ratios are equal:
a:b=c:dorab=cda : b = c : d \quad \text{or} \quad \frac{a}{b} = \frac{c}{d}
  • Cross-multiplication:
ab=cd  ⟹  ad=bc\frac{a}{b} = \frac{c}{d} \implies ad = bc
  • Direct proportion: as one quantity increases, the other increases at the same rate
y=kxy = kx
  • Inverse proportion: as one quantity increases, the other decreases proportionally
y=kxi.e.xy=constanty = \frac{k}{x} \quad \text{i.e.} \quad xy = \text{constant}

(e.g., more workers → less time for a fixed job)

Worked Example

  • Divide Rs. 6000 in the ratio 2:3:5 → total parts = 10; shares = 1200, 1800, 3000
  • If 6 workers finish a job in 10 days, how long for 15 workers? (inverse proportion) → 6×10 = 15×x → x = 4 days

Past Paper Style Numericals (Solved)

  1. Two numbers are in the ratio 3:5. If their sum is 96, find the numbers.
    • Step 1: The ratio 3:5 means the total is divided into 3+5 = 8 equal parts.
    • Step 2: Find the value of 1 part: 96 ÷ 8 = 12.
    • Step 3: Multiply each ratio number by this part-value: first number = 3×12 = 36; second number = 5×12 = 60. Answer: 36 and 60.
  2. A sum of money is divided among A, B, C in the ratio 2:3:4. If C gets Rs. 800 more than A, find the total sum.
    • Step 1: The difference between C's share (4 parts) and A's share (2 parts) is 4−2 = 2 parts.
    • Step 2: This 2-part difference equals Rs. 800, so 1 part = 800÷2 = Rs. 400.
    • Step 3: Total parts = 2+3+4 = 9, so total sum = 9 × 400 = Rs. 3,600.
  3. If a:b = 2:3 and b:c = 4:5, find a:b:c.
    • Step 1: The two ratios share "b", but with different values (3 in the first, 4 in the second) — find the LCM of 3 and 4, which is 12.
    • Step 2: Scale the first ratio (a:b=2:3) so b=12: multiply both sides by 4 → a:b = 8:12. Scale the second ratio (b:c=4:5) so b=12: multiply both sides by 3 → b:c = 12:15.
    • Step 3: Now both ratios share b=12, so combine directly: a:b:c = 8:12:15.
  4. If 15 men can build a wall in 12 days, how many men are needed to build it in 9 days?
    • Step 1: This is inverse proportion — fewer days means more men are needed. The total "work units" (men × days) stays constant: 15 × 12 = 180 man-days.
    • Step 2: Set up the equation for the new scenario: (new men) × 9 = 180.
    • Step 3: Solve: new men = 180 ÷ 9 = 20 men.

Rates

  • Rate: how much of one quantity per one unit of another (km per hour, rupees per kg, litres per minute); a unit rate is the rate per 1
  • Speed, distance, time
    • Speed = distance ÷ time; distance = speed × time; time = distance ÷ speed
    • km/h to m/s: multiply by 5/18; m/s to km/h: multiply by 18/5 (72 km/h = 20 m/s)
  • Average speed = total distance ÷ total time, not the average of the two speeds
    • Example: 60 km at 30 km/h (2 h) and 60 km at 60 km/h (1 h): average = 120 ÷ 3 = 40 km/h (not 45)
  • Relative speed: objects moving in the same direction → subtract speeds; opposite directions → add speeds
    • Example: two trains at 60 and 40 km/h towards each other close in at 100 km/h
  • Best buy: compare unit prices
    • Example: 2 kg for Rs. 500 is Rs. 250/kg; 5 kg for Rs. 1,150 is Rs. 230/kg, so the second is cheaper
  • Rates of work and flow (jobs per day, litres per minute) are covered in the Work, Time and Pipes chapter
  • Past-paper link: CSS 2025 Q8(a), 2021 Q6(d) and 2024 Q8(d) are rate problems (answered in the Work chapter)

MCQ Traps

  • Worker-time-work problems are usually inverse proportion (more workers = less time), while speed-distance at constant time is direct — read the scenario carefully before setting up the equation
  • A ratio must compare quantities of the same unit — convert units first if needed

Self-Test Questions

  1. Divide 450 in the ratio 4:5.
  2. If 8 pipes fill a tank in 15 hours, how long will 12 identical pipes take?
  3. If a:b = 3:4 and b:c = 2:5, find a:b:c.

Past-Paper MCQs (FPSC)

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The FPSC MCQ papers you shared (2024 and 2025) had no direct ratio question.

Past-Paper Questions with Answers (FPSC)

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Every question below appeared in the FPSC papers you shared; the year and question number are in each title. Answers are worked out by me.

  • CSS 2025 Q6(c): Divide Rs. 600 among A, B and C so that Rs. 40 more than 2/5 of A's share, Rs. 20 more than 2/7 of B's share and Rs. 10 more than 9/17 of C's share are equal. (5)
    1. Let the common value be k: 2A/5 + 40 = 2B/7 + 20 = 9C/17 + 10 = k
    2. A = 5(k − 40)/2; B = 7(k − 20)/2; C = 17(k − 10)/9
    3. A + B + C = 600 → 2.5k − 100 + 3.5k − 70 + (17/9)k − 170/9 = 600
    4. (71/9)k = 7100/9 → k = 100
    5. A = 150, B = 280, C = 170 (check: 150 + 280 + 170 = 600)
  • CSS 2024 Q7(d): Distribute Rs. 4,320 among Zain, Aslam and Ashraf so that if Zain gets 2 parts, Aslam gets 3 and Ashraf 7. (5)
    1. Total parts = 2 + 3 + 7 = 12; each part = 4,320 ÷ 12 = Rs. 360
    2. Zain = 720, Aslam = 1,080, Ashraf = 2,520
  • CSS 2024 Q7(b): Hassan has one-third of Ali's pocket money; Ali has 5 times Akbar's; Akbar has 3 times Nasir's; Shahbaz gets as much as Nasir and Ali together. Total Rs. 8,000. Find each. (5)
    1. Nasir = n; Akbar = 3n; Ali = 15n; Hassan = 5n; Shahbaz = n + 15n = 16n
    2. Total = n + 3n + 15n + 5n + 16n = 40n = 8,000 → n = 200
    3. Nasir 200, Akbar 600, Ali 3,000, Hassan 1,000, Shahbaz 3,200
  • CSS 2019 Q7(a): Moiz and Mair share Rs. 2,000 in the ratio 1:4. Moiz shares his part among himself, his wife and son in the ratio 4:5:1. How much more does his wife get than his son? (5)
    1. Moiz's part = 1/5 × 2,000 = Rs. 400
    2. Parts 4 + 5 + 1 = 10, so each part = 40: Moiz 160, wife 200, son 40
    3. Wife − son = 200 − 40 = Rs. 160
  • CSS 2017 Q8(A): Divide Rs. 500 among Arham, Mariam and Sarim so that Arham gets 2/3 of Mariam's share and Mariam gets 1/4 of Sarim's. (5)
    1. Let Mariam = 3 parts; then Arham = 2 parts and Sarim = 12 parts
    2. Total parts = 17, so each part = 500 ÷ 17 ≈ 29.41
    3. Arham ≈ Rs. 58.82, Mariam ≈ Rs. 88.24, Sarim ≈ Rs. 352.94
  • CSS 2017 Q8(B): A school has food for 52 days. After 20 days, 400 students arrive and the food lasts only 24 more days. How many students are there actually? (5)
    1. Food left after 20 days would last 32 more days for N students
    2. Food is fixed, so N × 32 = (N + 400) × 24 (more students, fewer days: inverse proportion)
    3. 32N = 24N + 9,600 → 8N = 9,600
    4. N = 1,200 students
  • CSS 2018 Q6(C): Tahir starts with Rs. 15,000. After 5 months Umar joins with Rs. 30,000; at the start of the 9th month Usman joins with Rs. 45,000. Annual profit Rs. 406,000. Share of each? (5)
    1. Capital × months: Tahir 15,000 × 12 = 180,000; Umar 30,000 × 7 = 210,000; Usman 45,000 × 4 = 180,000
    2. Ratio 180 : 210 : 180 = 6 : 7 : 6 (total 19 parts)
    3. Tahir ≈ Rs. 128,210.53; Umar ≈ Rs. 149,578.95; Usman ≈ Rs. 128,210.53
    • The figures do not divide exactly, so give rounded values
  • CSS 2020 Q8(c): 50 men build 20 km of road in 40 days. How long will 70 men take for the same road? (5)
    1. Total work = 50 × 40 = 2,000 man-days
    2. Days for 70 men = 2,000 ÷ 70 ≈ 28.6 days

CSS Exam Perspective

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  • Most repeated areas: sharing money in a ratio, chained ratios, inverse proportion, partnership profit
  • Tip: turn every ratio into 'parts', find the value of one part, then multiply