MC02 · Percentage, Ratio & Proportion

Question 1 of 20

A salary is increased by 10% and then decreased by 10%. The net percentage change is: (CDS PYQ)

A No change B 1% increase C 1% decrease D 2% decrease

1% decrease. Successive % formula: net = a+b+ab/100 = 10+(−10)+(10×−10)/100 = −1%. A net 1% decrease. Verify: 100→×1.1→110→×0.9→99. Net = −1% ✓. Classic CDS trap — students assume up 10% then down 10% cancel out.

Question 2 of 20

What percentage of 250 is 75? (CDS PYQ)

A 25% B 30% C 35% D 40%

30%. (75/250)×100 = 30%. Quick: 25% of 250=62.5; 30% of 250=75 ✓. Formula: "A is what % of B?" = (A/B)×100.

Question 3 of 20

In an election, one candidate gets 55% of total valid votes and wins by 1800 votes. The total valid votes are: (CDS PYQ)

A 9000 B 12000 C 18000 D 20000

18000. Winning margin = (55−45)% = 10% of total = 1800. Total = 1800/10% = 18000. Formula: Margin = (2W−100)% × Total. Election questions appear in almost every CDS paper.

Question 4 of 20

If A:B = 2:3 and B:C = 4:5, then A:B:C equals: (CDS PYQ)

A 8:12:15 B 6:9:12 C 2:4:5 D 4:6:10

8:12:15. Make B common: LCM(3,4)=12. A:B = 2:3 = 8:12; B:C = 4:5 = 12:15. So A:B:C = 8:12:15. Verify: A:B = 8:12 = 2:3 ✓; B:C = 12:15 = 4:5 ✓.

Question 5 of 20

A number is increased by 20% and then decreased by 20%. Compared to the original, the final number is: (CDS PYQ)

A The same B 4% less C 4% more D 2% less

4% less. Net = 20+(−20)+(20×−20)/100 = −4%. Verify: 100→120→96. Net = −4% ✓. Symmetrical percentage changes always yield a net loss, because the decrease applies to a larger base.

Question 6 of 20

If 15% of A equals 20% of B, then A:B equals: (CDS PYQ)

A 3:4 B 4:3 C 15:20 D 1:1

4:3. 0.15A = 0.20B → A/B = 0.20/0.15 = 4:3. A is larger because the same amount is a smaller percentage of A.

Question 7 of 20

In a mixture of 60 litres with milk:water = 2:1, how much water must be added to make the ratio 1:2? (CDS PYQ)

A 20 litres B 40 litres C 60 litres D 80 litres

60 litres. Milk=40L, water=20L. Target: milk:water=1:2, so for 40L milk, water needed=80L. Additional water = 80−20 = 60L. Verify: 40:80 = 1:2 ✓. Keep the unchanged component (milk) fixed and solve for the other.

Question 8 of 20

A shopkeeper marks goods 40% above cost price and gives a 25% discount. His profit percentage is: (CDS PYQ)

A 5% B 10% C 12% D 15%

5%. Successive: 40+(−25)+(40×−25)/100 = 15−10 = 5%. Verify: CP=100, MP=140, SP=140×0.75=105. Profit = 5% ✓.

Question 9 of 20

If A is 25% more than B, B is what percent less than A? (CDS PYQ)

A 20% B 25% C 16.67% D 12.5%

20%. A=1.25B → B=A/1.25=0.8A → B is 20% less than A. Formula: if X is p% more than Y, then Y is [p/(100+p)]×100 % less than X = 25/125×100 = 20%.

Question 10 of 20

The mean proportion of 4 and 16 is: (CDS PYQ)

A 6 B 8 C 10 D 12

8. Mean proportion = √(a×b) = √(4×16) = √64 = 8. Verify: 4:8 = 8:16 = 1:2 ✓. Do not confuse with arithmetic mean (10) or harmonic mean (6.4).

Question 11 of 20

Two numbers are in ratio 7:11. If 7 is added to each, the new ratio is 2:3. The smaller number is: (CDS PYQ)

A 42 B 49 C 56 D 77

49. Let 7k and 11k. (7k+7)/(11k+7)=2/3 → 21k+21=22k+14 → k=7. Numbers=49 and 77. Smaller=49. Verify: (49+7)/(77+7)=56/84=2:3 ✓.

Question 12 of 20

In what ratio must tea at Rs 60/kg be mixed with tea at Rs 80/kg to get a blend costing Rs 72/kg? (CDS PYQ)

A 1:2 B 2:3 C 3:2 D 4:3

2:3. Alligation: (80−72):(72−60) = 8:12 = 2:3. So cheaper:dearer = 2:3. Verify: (2×60+3×80)/5 = (120+240)/5 = 72 ✓. Alligation is the fastest method for all mixing questions.

Question 13 of 20

A person spends 75% of income. If income rises 20% and expenditure rises 10%, the percentage increase in savings is: (CDS PYQ)

A 40% B 50% C 60% D 70%

50%. Let income=100, savings=25, expenditure=75. New income=120, expenditure=82.5. New savings=37.5. Increase=(37.5−25)/25×100=50%.

Question 14 of 20

If x% of y equals y% of z, then: (CDS PYQ)

A x = y B x = z C x > z D x + z = y

x = z. xy/100 = yz/100 → xy = yz → x = z (since y≠0). A neat algebraic identity — the percentages are equal regardless of the common base y.

Question 15 of 20

Two numbers are in ratio 5:6. If 8 is subtracted from each, the ratio becomes 4:5. The larger number is: (CDS PYQ)

A 36 B 42 C 48 D 54

48. Let 5k and 6k. (5k−8)/(6k−8)=4/5 → 25k−40=24k−32 → k=8. Numbers=40 and 48. Larger=48. Verify: (40−8)/(48−8)=32/40=4:5 ✓.

Question 16 of 20

If p:q = 3:4 and q:r = 6:7, then p:r equals: (CDS PYQ)

A 7:8 B 8:9 C 9:14 D 18:14

9:14. p/r = (p/q)×(q/r) = (3/4)×(6/7) = 18/28 = 9:14. Method: multiply corresponding ratios. Always simplify by dividing by GCF.

Question 17 of 20

A train covers a distance in 50 minutes. If speed is increased by 25%, the time taken is: (CDS PYQ)

A 40 minutes B 42 minutes C 45 minutes D 48 minutes

40 minutes. Speed and time are inversely proportional. Speed increases by 25% → time × 4/5. New time = 50×4/5 = 40 min. Formula: T₂ = T₁ × S₁/S₂ = 50 × 1/1.25 = 40 min.

Question 18 of 20

If A:B = 3:4 and B:C = 8:9, then A:C is: (CDS PYQ)

A 1:2 B 2:3 C 1:3 D 3:8

2:3. p/r = (3/4)×(8/9) = 24/36 = 2:3. Confirm: A:B:C — make B common: LCM(4,8)=8. A:B=6:8; B:C=8:9. A:C=6:9=2:3 ✓.

Question 19 of 20

The ratio of two numbers is 3:5. If 10 is added to each, the ratio becomes 5:7. The smaller number is: (CDS PYQ)

A 10 B 15 C 20 D 25

15. Let 3k and 5k. (3k+10)/(5k+10)=5/7 → 21k+70=25k+50 → 4k=20 → k=5. Numbers=15 and 25. Smaller=15. Verify: (15+10)/(25+10)=25/35=5:7 ✓.

Question 20 of 20

A sum of money is divided among A, B, C in the ratio 3:4:5. If C gets Rs 1800 more than A, the total sum is: (CDS PYQ)

A Rs 5400 B Rs 7200 C Rs 9000 D Rs 10800

Rs 9000. C−A = (5−3)=2 parts = Rs 1800 → 1 part = Rs 900. Total = (3+4+5)×900 = 12×900 = Rs 10800. Hmm: 12×900=10800 → option D. Let me recheck: 2 parts=1800 → 1 part=900; total=12 parts=10800. Answer D: Rs 10800.