Quantitative Aptitude — Notes

1. Number System

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Is 918,082 divisible by 11?

1. Number the digits from the right: 2 (1st), 8 (2nd), 0 (3rd), 8 (4th), 1 (5th), 9 (6th).
2. Odd places: 2 + 0 + 1 = 3. Even places: 8 + 8 + 9 = 25.
3. Difference = 3 − 25 = −22, which is a multiple of 11.

Answer: Yes, 918,082 is divisible by 11 (918,082 = 11 × 83,462).

Example 2 (Easy). Find the unit digit of 7⁹⁵ × 3⁵⁸.

1. 7⁹⁵: 95 ÷ 4 leaves 3, so it ends like 7³, which is 3.
2. 3⁵⁸: 58 ÷ 4 leaves 2, so it ends like 3², which is 9.
3. Multiply the unit digits: 3 × 9 = 27, so the unit digit is 7.

Answer: 7.

Example 3 (Medium). Find the remainder when 2¹⁰⁰ is divided by 7.

1. Look for a small power of 2 that leaves remainder 1: 2³ = 8 = 7 + 1. Yes.
2. Write 100 = 3 × 33 + 1, so 2¹⁰⁰ = (2³)³³ × 2.
3. (2³)³³ leaves 1³³ = 1. Then 1 × 2 = 2.

Answer: 2.

Example 4 (Medium). How many factors does 360 have, and what is their sum?

1. Prime factorise: 360 = 2³ × 3² × 5¹.
2. Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24. Why add 1? Each prime can appear 0, 1, ... p times, which is p + 1 choices.
3. Sum = (1+2+4+8)(1+3+9)(1+5) = 15 × 13 × 6 = 1,170.

Answer: 24 factors, sum 1,170.

Example 5 (Medium). How many zeros are at the end of 125!?

1. Each zero needs a pair 2 × 5. There are many more 2s than 5s, so count the 5s.
2. ⌊125/5⌋ = 25, ⌊125/25⌋ = 5, ⌊125/125⌋ = 1.
3. Total = 25 + 5 + 1 = 31.

Answer: 31. Faster: keep dividing by 5 and adding the quotients: 125 → 25 → 5 → 1, sum 31.

Example 6 (Hard). A number divided by 357 leaves remainder 39. What is the remainder when the same number is divided by 17?

1. Number = 357k + 39. Since 357 = 17 × 21, the part 357k is a multiple of 17.
2. So only 39 matters: 39 = 2 × 17 + 5.

Answer: 5. This works only because 17 divides 357. If the new divisor is not a factor of the old one, the answer cannot be fixed.

Practice set

2. HCF & LCM

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Find the HCF and LCM of 36, 84 and 120.

1. 36 = 2² × 3², 84 = 2² × 3 × 7, 120 = 2³ × 3 × 5.
2. HCF: common primes are 2 and 3; lowest powers 2² and 3¹. HCF = 4 × 3 = 12.
3. LCM: all primes at highest powers: 2³ × 3² × 5 × 7 = 8 × 9 × 35 = 2,520.

Answer: HCF 12, LCM 2,520.

Example 2 (Easy). Find the HCF of 1071 and 462 by the division method.

1071 = 2 × 462 + 147
 462 = 3 × 147 + 21
 147 = 7 × 21  + 0    -> last divisor is 21

Answer: 21. Why it works: any number dividing both 1071 and 462 also divides 1071 − 2 × 462 = 147, and so on down, so the HCF never changes as the numbers shrink.

Example 3 (Medium). The HCF of two numbers is 12 and their LCM is 336. If one number is 48, find the other.

1. Use HCF × LCM = product: 12 × 336 = 48 × other.
2. Other = 4,032 ÷ 48 = 84.

Answer: 84. Check: HCF(48, 84) = 12 and LCM(48, 84) = 336.

Example 4 (Medium). Find the HCF and LCM of 2/3, 8/9 and 16/81.

1. Numerators 2, 8, 16: HCF 2, LCM 16. Denominators 3, 9, 81: HCF 3, LCM 81.
2. HCF = 2/81. LCM = 16/3.

Answer: HCF 2/81, LCM 16/3.

Example 5 (Medium). Four bells toll at intervals of 6, 8, 12 and 15 minutes. They toll together at 9:00 AM. When will they next toll together, and how many times will they toll together in 6 hours (counting 9:00 AM)?

1. LCM(6, 8, 12, 15) = 2³ × 3 × 5 = 120 minutes = 2 hours.
2. Next time together: 9:00 AM + 2 hours = 11:00 AM.
3. In 6 hours = 360 minutes: 360 ÷ 120 = 3 times after the start, plus the start = 4.

Answer: 11:00 AM; 4 times.

Example 6 (Hard). Find the greatest number that divides 43, 91 and 183 leaving the same remainder each time.

1. If all three leave the same remainder r, then the differences have no remainder at all.
2. Differences: 91 − 43 = 48, 183 − 91 = 92, 183 − 43 = 140.
3. HCF(48, 92, 140) = 4.

Answer: 4 (each number leaves remainder 3).

Example 7 (Hard). Find the smallest number which, when divided by 12, 15, 20 and 54, leaves remainder 8 in each case.

1. LCM(12, 15, 20, 54) = 2² × 3³ × 5 = 540.
2. Add the common remainder: 540 + 8 = 548.

Answer: 548.

Practice set

3. Simplification & Approximation

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Simplify 18 ÷ 3 × 2 + 4 − 6.

1. Division and multiplication go left to right: 18 ÷ 3 = 6, then 6 × 2 = 12.
2. Then addition and subtraction left to right: 12 + 4 − 6 = 10.

Answer: 10. (Doing 3 × 2 first would give 18 ÷ 6 = 3, and the wrong answer 1.)

Example 2 (Medium). Simplify 36 − [18 − {14 − (15 − 4 ÷ 2 × 2)}].

1. Innermost ( ): 4 ÷ 2 × 2 = 2 × 2 = 4, so 15 − 4 = 11.
2. { }: 14 − 11 = 3.
3. [ ]: 18 − 3 = 15.
4. Finally 36 − 15 = 21.

Answer: 21.

Example 3 (Easy). Simplify 48 ÷ 1/4 of 16.

1. "Of" comes before division: 1/4 of 16 = 4.
2. 48 ÷ 4 = 12.

Answer: 12. (If you divided first you would get 48 × 4 × 16, a huge wrong number.)

Example 4 (Easy). Find 0.25 × 0.4 + 1.2 ÷ 0.3.

1. 0.25 × 0.4 = 0.1.
2. 1.2 ÷ 0.3 = 12 ÷ 3 = 4.
3. 0.1 + 4 = 4.1.

Answer: 4.1.

Example 5 (Medium). Find the approximate value of 39.97% of 649.8 + 25.02 × 11.99.

1. Round: 40% of 650 = 260.
2. Round: 25 × 12 = 300.
3. Estimate = 260 + 300 = 560.

Answer: about 560 (exact value 559.71).

Example 6 (Hard). Simplify 1 + 1 / (1 + 1 / (1 + 1/2)).

1. Start at the bottom: 1 + 1/2 = 3/2.
2. 1 ÷ 3/2 = 2/3, so the middle is 1 + 2/3 = 5/3.
3. 1 ÷ 5/3 = 3/5, so the whole thing is 1 + 3/5 = 8/5.

Answer: 8/5. For continued fractions, always start at the bottom and climb up.

Practice set

4. Squares, Cubes, Roots & Indices

Key formulas & rules

Squares 1 to 30

Cubes 1 to 15

Shortcuts

Solved examples

Example 1 (Easy). Simplify 2⁵ × 2³ ÷ 2⁴.

1. Same base, so add and subtract the powers: 5 + 3 − 4 = 4.
2. 2⁴ = 16.

Answer: 16.

Example 2 (Easy). Find 27^2/3 and 16^−3/4.

1. 27^2/3: cube root first (it keeps numbers small): ∛27 = 3, then 3² = 9.
2. 16^−3/4 = 1/16^3/4. Fourth root of 16 = 2, then 2³ = 8. So 1/8.

Answer: 9 and 1/8.

Example 3 (Medium). Find √5776.

1. Last digit 6, so the root ends in 4 or 6.
2. First two digits 57: the largest square below 57 is 49 = 7². Tens digit is 7.
3. 75² = 5,625, and 5,776 is bigger, so take the larger candidate: 76.

Answer: 76.

Example 4 (Medium). Find the cube root of 185,193.

1. Last digit 3, so the root ends in 7 (3 and 7 swap).
2. Drop the last three digits: 185. Largest cube below 185 is 125 = 5³. Tens digit is 5.

Answer: 57 (57³ = 185,193).

Example 5 (Medium). Rationalise 1/(√5 − √3).

1. Multiply top and bottom by the conjugate √5 + √3.
2. Bottom: (√5)² − (√3)² = 5 − 3 = 2.

Answer: (√5 + √3)/2.

Example 6 (Hard). Solve 2^x = 4^{x − 3}.

1. Make the bases equal: 4^{x − 3} = (2²)^{x − 3} = 2^{2x − 6}.
2. So x = 2x − 6, which gives x = 6.

Answer: x = 6. Check: 2⁶ = 64 and 4³ = 64.

Example 7 (Easy). Is 347 × 29 = 10,063 correct? Check with digital roots.

1. 347: 3 + 4 + 7 = 14 → 5. 29: 2 + 9 = 11 → 2.
2. 5 × 2 = 10 → 1.
3. 10,063: 1 + 0 + 0 + 6 + 3 = 10 → 1. They match.

Answer: Yes, it passes the check. (A match does not prove it right, but a mismatch always proves it wrong.)

Practice set

5. Speed Maths: Calculation Tricks

Key formulas & rules

Percentage–fraction table (learn by heart)

Shortcuts

Solved examples

Example 1 (Easy). Find 4,862 × 5 and 128 × 25.

1. 4,862 × 10 = 48,620; half of it is 24,310.
2. 128 × 100 = 12,800; a quarter of it is 3,200.

Answer: 24,310 and 3,200.

Example 2 (Easy). Find 72 × 11, 87 × 11 and 352 × 11.

1. 72: 7 | 7+2 | 2 = 792.
2. 87: 8 | 15 | 7. The middle is 15, so write 5 and carry 1: 9 | 5 | 7 = 957.
3. 352: 3 | 8 | 7 | 2 = 3,872.

Answer: 792, 957, 3,872.

Example 3 (Easy). Find 65² and 115².

1. 65: 6 × 7 = 42, then 25: 4,225.
2. 115: 11 × 12 = 132, then 25: 13,225.

Answer: 4,225 and 13,225.

Example 4 (Medium). Find 97 × 94, 106 × 112 and 98 × 103.

1. 97 × 94: distances −3 and −6. Left = 97 − 6 = 91, right = (−3)(−6) = 18. Answer 9,118.
2. 106 × 112: distances +6, +12. Left = 106 + 12 = 118, right = 72. Answer 11,872.
3. 98 × 103: distances −2, +3. Left = 98 + 3 = 101, right = −6. So 10,100 − 6 = 10,094.

Answer: 9,118; 11,872; 10,094.

Example 5 (Medium). Find 468 × 99.

1. 468 × 100 = 46,800.
2. Subtract one 468: 46,800 − 468 = 46,332.

Answer: 46,332.

Example 6 (Hard). Find 37.5% of 816 and 16.67% of 1,266.

1. 37.5% = 3/8. 816 ÷ 8 = 102, × 3 = 306.
2. 16.67% = 1/6. 1,266 ÷ 6 = 211.

Answer: 306 and 211.

Practice set

Tables to memorise: tables 1–30, squares 1–100, cubes 1–50, roots, fractions and factorials are all on the Tables & Speed Tricks page. Drill them in Calculation Practice.

6. Percentages

Key formulas & rules

Learn this table by heart. It turns "divide by 16" into "take 6.25%", which is much faster in your head.

Shortcuts

Solved examples

Example 1 (Easy). Find 35% of 240.

1. 10% of 240 = 24 (move the decimal one place).
2. 30% = 3 × 24 = 72. 5% = half of 10% = 12.
3. 35% = 72 + 12 = 84.

Answer: 84.

Example 2 (Easy). A shirt's price goes from ₹80 to ₹92. Find the percentage increase.

1. Change = 92 − 80 = ₹12.
2. Divide by the OLD price: 12 / 80 = 0.15.
3. 0.15 × 100 = 15%.

Answer: 15% increase.

Example 3 (Medium). A price is raised by 20% and then reduced by 10%. What is the net change?

1. Take the price as 100 (100 is the easiest base for percent questions).
2. After +20%: 100 × 1.2 = 120.
3. After −10% of 120 (not of 100): 120 − 12 = 108.
4. 108 vs 100 → net +8%.

Answer: 8% increase. Shortcut: 20 − 10 + (20 × −10)/100 = 8.

Example 4 (Medium). A's salary is 25% more than B's. By what percent is B's salary less than A's?

1. Let B = 100. Then A = 125.
2. Difference = 25. Now the base is A (we compare B with A), so divide by 125.
3. 25 / 125 × 100 = 20%.

Answer: 20%. Shortcut: 25% = 1/4, so the answer is 1/5 = 20%.

Example 5 (Medium). A town has 50,000 people and grows 10% every year. What is the population after 2 years? A machine worth ₹80,000 loses 10% of its value each year. What is it worth after 2 years?

1. Growth: 50,000 × 1.1 × 1.1 = 50,000 × 1.21 = 60,500.
2. Depreciation: 80,000 × 0.9 × 0.9 = 80,000 × 0.81 = 64,800.

Answers: 60,500 people and ₹64,800. Note the growth is 21% in two years, not 20% (the formula a + b + ab/100 gives 10 + 10 + 1 = 21).

Example 6 (Hard). The price of sugar rises by 25%. By what percent must a family cut its sugar use so that its spending stays the same?

1. Spending = price × quantity. Let old price = 100 and old quantity = 100, so spending = 10,000.
2. New price = 125. To keep spending 10,000, new quantity = 10,000 / 125 = 80.
3. Quantity falls from 100 to 80, a cut of 20%.

Answer: 20%. Shortcut: r / (100 + r) × 100 = 25/125 × 100 = 20%.

Practice set

7. Profit, Loss & Discount

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). An article bought for ₹500 is sold for ₹575. Find the profit %.

1. Profit = 575 − 500 = ₹75.
2. Profit % = 75 / 500 × 100 = 15%.

Answer: 15%.

Example 2 (Easy). A fan sold for ₹540 gives a loss of 10%. Find its cost price.

1. 10% loss means SP is 90% of CP: 0.9 × CP = 540.
2. CP = 540 / 0.9 = ₹600.

Answer: ₹600. (Not 540 + 54 = 594 — the 10% is of CP, not of SP.)

Example 3 (Medium). The marked price of a watch is ₹1,200. Two successive discounts of 20% and 10% are given. Find the SP and the single equivalent discount.

1. After 20% off: 1,200 × 0.8 = 960.
2. After 10% off 960: 960 × 0.9 = ₹864.
3. Total discount = 1,200 − 864 = 336 = 28% of 1,200.

Answer: SP ₹864, single discount 28%. Shortcut: 20 + 10 − 200/100 = 28%.

Example 4 (Medium). A shopkeeper marks goods 40% above cost and gives a 15% discount. What is his profit %?

1. Let CP = 100. MP = 140.
2. SP = 140 × 0.85 = 119.
3. Profit = 19 on a CP of 100 → 19%.

Answer: 19%. Shortcut: m − d − md/100 = 40 − 15 − 6 = 19.

Example 5 (Medium). A dishonest grocer claims to sell at cost price but uses a 900 g weight instead of 1 kg. Find his gain %.

1. He charges for 1,000 g but gives only 900 g. He gains the value of 100 g.
2. His real cost is for 900 g, so gain % = 100 / 900 × 100 = 11.11%.

Answer: 11 1/9 % (11.11%). Divide by what he actually gives (900), not by 1,000.

Example 6 (Hard). (i) A shop offers "Buy 4, get 1 free". What is the effective discount? (ii) Two phones are sold for ₹990 each, one at 10% profit and the other at 10% loss. Find the overall result.

1. (i) You get 5 items and pay for 4. Free = 1 of 5 = 20% discount.
2. (ii) CP of first = 990 / 1.1 = 900. CP of second = 990 / 0.9 = 1,100.
3. Total CP = 2,000, total SP = 1,980. Loss = 20 = 1% of 2,000.

Answers: (i) 20%, (ii) 1% loss. Shortcut for (ii): x²/100 = 100/100 = 1% loss.

Practice set

8. Simple & Compound Interest

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Find the SI on ₹5,000 at 8% per year for 3 years.

1. SI = 5,000 × 8 × 3 / 100 = 1,200.

Answer: ₹1,200.

Example 2 (Easy). Find the CI on ₹10,000 at 10% per year for 2 years.

1. Year 1: 10,000 → 11,000.
2. Year 2: 11,000 × 1.1 = 12,100.
3. CI = 12,100 − 10,000 = 2,100.

Answer: ₹2,100. Shortcut: 21% of 10,000.

Example 3 (Medium). Find the CI on ₹8,000 for 1 year at 10% per year, compounded half-yearly.

1. Half-yearly: rate = 10/2 = 5% per half-year, time = 2 half-years.
2. Amount = 8,000 × 1.05 × 1.05 = 8,000 × 1.1025 = 8,820.
3. CI = 820.

Answer: ₹820 (more than the ₹800 you would get with yearly compounding).

Example 4 (Medium). Find the difference between CI and SI on ₹15,000 for 2 years at 6%.

1. The extra in CI is just the interest on year 1's interest.
2. Difference = P(R/100)² = 15,000 × 0.06 × 0.06 = 15,000 × 0.0036 = 54.

Answer: ₹54.

Example 5 (Hard). Find CI − SI on ₹8,000 for 3 years at 5%.

1. Formula: P(R/100)²(3 + R/100) = 8,000 × 0.0025 × 3.05.
2. 8,000 × 0.0025 = 20. 20 × 3.05 = 61.
3. Check: CI = 8,000 × 1.05³ − 8,000 = 9,261 − 8,000 = 1,261. SI = 1,200. Difference 61.

Answer: ₹61.

Example 6 (Hard). (i) A sum doubles in 5 years at CI. In how many years will it become 8 times? (ii) A sum doubles in 8 years at SI. Find the rate.

1. (i) 8 = 2³. Each doubling takes 5 years, so 3 doublings take 15 years.
2. (ii) Doubling at SI means interest = P in 8 years. R = 100 / 8 = 12.5%.

Answers: (i) 15 years, (ii) 12.5%.

Practice set

9. Ratio & Proportion

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Divide ₹1,200 between Ravi and Sita in the ratio 3 : 5.

1. Total parts = 3 + 5 = 8.
2. One part = 1,200 / 8 = ₹150.
3. Ravi = 3 × 150 = ₹450. Sita = 5 × 150 = ₹750.

Answer: ₹450 and ₹750.

Example 2 (Easy). A : B = 2 : 3 and B : C = 4 : 5. Find A : B : C.

1. B is 3 in the first ratio and 4 in the second. Make both 12 (LCM of 3 and 4).
2. A : B = 2 : 3 = 8 : 12. B : C = 4 : 5 = 12 : 15.
3. So A : B : C = 8 : 12 : 15.

Answer: 8 : 12 : 15.

Example 3 (Medium). Find (i) the mean proportional of 9 and 16, (ii) the third proportional to 4 and 6, (iii) the fourth proportional to 3, 5 and 12.

1. (i) √(9 × 16) = √144 = 12. (Check: 9 : 12 = 12 : 16.)
2. (ii) b²/a = 36/4 = 9. (Check: 4 : 6 = 6 : 9.)
3. (iii) bc/a = 5 × 12 / 3 = 20. (Check: 3 : 5 = 12 : 20.)

Answers: 12, 9, 20.

Example 4 (Medium). Two numbers are in the ratio 3 : 5. If 9 is added to each, the ratio becomes 3 : 4. Find the numbers.

1. Let the numbers be 3x and 5x.
2. (3x + 9) / (5x + 9) = 3/4 → 4(3x + 9) = 3(5x + 9) → 12x + 36 = 15x + 27.
3. 3x = 9, so x = 3. Numbers = 9 and 15.

Answer: 9 and 15. Check: 18 : 24 = 3 : 4.

Example 5 (Medium). A bag has ₹1, 50 paise and 25 paise coins in the ratio 2 : 3 : 4. The total value is ₹180. How many coins of each kind?

1. Let the counts be 2x, 3x, 4x.
2. Value = 2x × 1 + 3x × 0.5 + 4x × 0.25 = 2x + 1.5x + x = 4.5x.
3. 4.5x = 180 → x = 40. Coins: 80, 120, 160.

Answer: 80, 120 and 160 coins. The trap is dividing ₹180 in 2 : 3 : 4 directly — that is the ratio of counts, not of values.

Example 6 (Hard). The incomes of A and B are in the ratio 4 : 3 and their expenses are in the ratio 3 : 2. Each saves ₹6,000. Find A's income.

1. Incomes 4x, 3x. Expenses 3y, 2y. Savings = income − expense.
2. 4x − 3y = 6,000 and 3x − 2y = 6,000.
3. Multiply the first by 2 and the second by 3: 8x − 6y = 12,000 and 9x − 6y = 18,000. Subtract: x = 6,000.
4. A's income = 4 × 6,000 = ₹24,000. (Then y = 6,000, expenses 18,000 and 12,000.)

Answer: ₹24,000.

Practice set

10. Partnership

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). A invests ₹20,000 and B invests ₹30,000 for a year. The profit is ₹15,000. Find each share.

1. Same time, so ratio = 20,000 : 30,000 = 2 : 3.
2. 5 parts = 15,000, one part = 3,000.
3. A = 6,000, B = 9,000.

Answer: A ₹6,000, B ₹9,000.

Example 2 (Easy). A invests ₹40,000 for 12 months and B invests ₹60,000 for 8 months. Find the profit-sharing ratio.

1. A: 40,000 × 12 = 4,80,000. B: 60,000 × 8 = 4,80,000.
2. Ratio = 1 : 1.

Answer: 1 : 1. More money for less time can equal less money for more time.

Example 3 (Medium). A starts a business with ₹50,000. After 4 months B joins with ₹60,000. At the end of the year the profit is ₹27,000. Find each share.

1. A is in for 12 months, B for 12 − 4 = 8 months.
2. A: 50,000 × 12 = 6,00,000. B: 60,000 × 8 = 4,80,000.
3. Ratio = 600 : 480 = 5 : 4. Total 9 parts, one part = 3,000.
4. A = 15,000, B = 12,000.

Answer: A ₹15,000, B ₹12,000.

Example 4 (Medium). A, B and C invest in the ratio 3 : 4 : 5 for times in the ratio 4 : 3 : 2. The profit is ₹17,000. Find C's share.

1. Multiply term by term: 3 × 4 : 4 × 3 : 5 × 2 = 12 : 12 : 10.
2. Total 34 parts, one part = 500.
3. C = 10 × 500 = 5,000.

Answer: ₹5,000.

Example 5 (Hard). A invests ₹10,000 and withdraws ₹2,000 after 6 months. B invests ₹12,000 for the whole year. Profit is ₹14,000. Find A's share.

1. A: 10,000 × 6 + 8,000 × 6 = 60,000 + 48,000 = 1,08,000.
2. B: 12,000 × 12 = 1,44,000.
3. Ratio = 108 : 144 = 3 : 4. One part = 14,000 / 7 = 2,000.
4. A = 6,000.

Answer: ₹6,000.

Example 6 (Hard). A and B invest in the ratio 3 : 2. A is the working partner and first takes 10% of the profit as salary. The profit is ₹50,000. Find A's total earning.

1. Salary = 10% of 50,000 = 5,000.
2. Remaining 45,000 split 3 : 2 → A gets 27,000, B gets 18,000.
3. A's total = 5,000 + 27,000 = 32,000.

Answer: ₹32,000.

Practice set

11. Averages

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Find the average of 12, 15, 18, 21 and 24.

1. They are equally spaced (step 3), so the average is the middle number.
2. Middle = 18. (Check: sum = 90, 90 / 5 = 18.)

Answer: 18.

Example 2 (Easy). Class A has 30 students with an average of 60. Class B has 20 students with an average of 70. Find the combined average.

1. Total A = 30 × 60 = 1,800. Total B = 20 × 70 = 1,400.
2. Combined = 3,200 / 50 = 64.

Answer: 64. Not 65 — the bigger class pulls the average toward 60.

Example 3 (Medium). The average weight of 8 people rises by 2.5 kg when one of them, weighing 65 kg, is replaced by a new person. Find the new person's weight.

1. The count stays 8. The total rose by 8 × 2.5 = 20 kg.
2. That extra 20 kg came only from the swap, so the new person is 20 kg heavier than 65.
3. New person = 85 kg.

Answer: 85 kg.

Example 4 (Medium). The average age of 10 students is 20. When the teacher's age is included, the average becomes 21. Find the teacher's age.

1. Old total = 10 × 20 = 200. New total = 11 × 21 = 231.
2. Teacher = 231 − 200 = 31.

Answer: 31 years. Shortcut: new average + (old count × rise) = 21 + 10 × 1 = 31.

Example 5 (Medium). A car goes from A to B at 60 km/h and comes back at 40 km/h. Find the average speed for the whole trip.

1. Same distance both ways, so use 2xy/(x + y).
2. 2 × 60 × 40 / 100 = 4,800 / 100 = 48.
3. Check with distance 120 km: time 2 h + 3 h = 5 h, total 240 km / 5 h = 48.

Answer: 48 km/h (not 50).

Example 6 (Hard). A batsman has a certain average after 11 innings. In the 12th innings he scores 90, which raises his average by 5. Find his new average.

1. Let the old average be x. Old total = 11x.
2. New total = 11x + 90 = 12(x + 5) = 12x + 60.
3. So x = 30. New average = 35.

Answer: 35. Shortcut: the 90 gives 5 extra to each of 12 innings (60) and still leaves the old average: 90 − 60 = 30 = old, new = 35.

Practice set

12. Mixtures & Alligation

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). In what ratio must rice at ₹40/kg be mixed with rice at ₹55/kg to get a mixture worth ₹45/kg?

1. Cheaper = 40, dearer = 55, mean = 45.
2. Cheaper : dearer = (55 − 45) : (45 − 40) = 10 : 5 = 2 : 1.

Answer: 2 : 1. Check: (2 × 40 + 1 × 55) / 3 = 135 / 3 = 45.

Example 2 (Medium). How much water must be added to 60 litres of milk costing ₹20/litre so that the mixture is worth ₹16/litre?

1. Water is the cheaper item at ₹0. Milk is dearer at ₹20. Mean = 16.
2. Water : milk = (20 − 16) : (16 − 0) = 4 : 16 = 1 : 4.
3. Milk is 60 L = 4 parts, so 1 part = 15 L of water.

Answer: 15 litres.

Example 3 (Medium). A 40% acid solution and a 70% acid solution are mixed to make 30 litres of 50% solution. How much of each is used?

1. Weaker : stronger = (70 − 50) : (50 − 40) = 20 : 10 = 2 : 1.
2. 30 L in 2 : 1 → 20 L of the 40% solution and 10 L of the 70% solution.

Answer: 20 L of 40% and 10 L of 70%. Check: 8 + 7 = 15 L acid in 30 L = 50%.

Example 4 (Medium). A vessel holds 40 litres of milk. 4 litres are taken out and replaced with water. This is done 3 times in all. How much milk is left?

1. Each round keeps (1 − 4/40) = 9/10 of the milk.
2. After 3 rounds: 40 × (9/10)³ = 40 × 0.729 = 29.16 L.

Answer: 29.16 litres. (Not 40 − 12 = 28: later rounds remove a mix, not pure milk.)

Example 5 (Hard). Two vessels have milk and water in the ratios 3 : 1 and 5 : 3. Equal amounts are poured from both into a third vessel. Find the milk : water ratio there.

1. Milk fraction: first = 3/4, second = 5/8.
2. Equal amounts, so average the fractions: (3/4 + 5/8)/2 = (6/8 + 5/8)/2 = 11/16.
3. Water fraction = 5/16. Ratio = 11 : 5.

Answer: 11 : 5.

Example 6 (Hard). A trader has 60 kg of sugar. He sells part at 8% profit and the rest at 18% profit, making 14% overall. How much did he sell at 8%?

1. Alligate the profit %: 8% part : 18% part = (18 − 14) : (14 − 8) = 4 : 6 = 2 : 3.
2. 60 kg in 2 : 3 → 24 kg at 8% and 36 kg at 18%.

Answer: 24 kg.

Practice set

13. Problems on Ages

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). A father is 3 times as old as his son. After 12 years he will be twice as old. Find their present ages.

1. Son = x, father = 3x.
2. After 12 years: 3x + 12 = 2(x + 12) = 2x + 24.
3. x = 12. Son 12, father 36.

Answer: son 12, father 36. Check: 48 = 2 × 24.

Example 2 (Medium). The ages of A and B are in the ratio 4 : 5. After 5 years the ratio will be 5 : 6. Find their present ages.

1. Ages 4x and 5x. (4x + 5)/(5x + 5) = 5/6.
2. 6(4x + 5) = 5(5x + 5) → 24x + 30 = 25x + 25 → x = 5.
3. A = 20, B = 25.

Answer: 20 and 25. Shortcut: ratio-difference trick above.

Example 3 (Medium). The sum of the ages of 5 children born 3 years apart is 50 years. Find the youngest child's age.

1. Ages: y, y + 3, y + 6, y + 9, y + 12.
2. Sum = 5y + 30 = 50 → y = 4.

Answer: 4 years. Shortcut: the middle child is the average, 50/5 = 10, so the youngest is 10 − 6 = 4.

Example 4 (Medium). Ten years ago, A was half as old as B. The ratio of their present ages is 3 : 4. Find the sum of their present ages.

1. Present ages 3x and 4x. Ten years ago: 3x − 10 and 4x − 10.
2. 3x − 10 = (4x − 10)/2 → 6x − 20 = 4x − 10 → x = 5.
3. Ages 15 and 20. Sum = 35.

Answer: 35 years. Check: ten years ago 5 and 10.

Example 5 (Hard). A is twice as old as B was when A was as old as B is now. The sum of their present ages is 63. Find A's age.

1. Let the age gap be d = A − B. "When A was as old as B is now" was d years ago.
2. At that time B was B − d. The sentence says A (now) = 2(B − d) = 2(B − (A − B)) = 4B − 2A.
3. So 3A = 4B, i.e. A : B = 4 : 3. With A + B = 63: A = 36, B = 27.
4. Check: 9 years ago A was 27 (= B now) and B was 18. A now = 36 = 2 × 18.

Answer: 36 years.

Practice set

14. Time & Work

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). A can do a job in 12 days and B in 18 days. How long will they take working together?

1. Total work = LCM(12, 18) = 36 units.
2. A does 36 ÷ 12 = 3 units a day. B does 36 ÷ 18 = 2 units a day.
3. Together: 3 + 2 = 5 units a day.
4. Time = 36 ÷ 5 = 7.2 days.

Answer: 7 1/5 days. Faster: ab/(a + b) = 12 × 18 / 30 = 216/30 = 7.2.

Example 2 (Easy). A and B together finish a job in 10 days. A alone takes 15 days. How long does B alone take?

1. Total = LCM(10, 15) = 30 units.
2. A + B do 30 ÷ 10 = 3 units a day. A alone does 30 ÷ 15 = 2 units a day.
3. So B does 3 − 2 = 1 unit a day.
4. B alone needs 30 ÷ 1 = 30 days.

Answer: 30 days.

Example 3 (Medium). A is twice as efficient as B. Together they finish a job in 14 days. How long would A take alone?

1. Efficiency means rate. Let A do 2 units a day and B do 1 unit a day.
2. Together: 3 units a day for 14 days = 42 units. That is the total work.
3. A alone: 42 ÷ 2 = 21 days.

Answer: 21 days. Notice B alone would take 42 days: twice as efficient means half the time.

Example 4 (Medium). A can do a job in 20 days and B in 30 days. They work together for 5 days, then A leaves. How many more days does B need?

1. Total = LCM(20, 30) = 60 units. A = 3 units/day, B = 2 units/day.
2. In 5 days together they do 5 × 5 = 25 units.
3. Remaining = 60 − 25 = 35 units.
4. B alone: 35 ÷ 2 = 17.5 days.

Answer: 17 1/2 days.

Example 5 (Medium). 12 men working 8 hours a day build a 240 m wall in 10 days. In how many days will 16 men working 6 hours a day build a 480 m wall?

1. Use M1 × D1 × H1 / W1 = M2 × D2 × H2 / W2.
2. 12 × 10 × 8 / 240 = 16 × D2 × 6 / 480.
3. Left side = 960/240 = 4. So 96 × D2 / 480 = 4, giving D2 = 4 × 480 / 96 = 20.

Answer: 20 days. Sense check: twice the wall needs twice the man-hours (1920), and 16 × 6 = 96 man-hours a day, so 1920 ÷ 96 = 20.

Example 6 (Hard). A can do a job in 10 days and B in 15 days. With C's help they finish it in 4 days and are paid ₹1500 in total. Find each person's share.

1. Total = LCM(10, 15, 4) = 60 units. A = 6/day, B = 4/day.
2. All three together do 60 ÷ 4 = 15 units a day, so C = 15 − 6 − 4 = 5 units a day.
3. They all work the same 4 days, so wages split in the ratio of daily work: 6 : 4 : 5 (sum 15).
4. A = 1500 × 6/15 = ₹600, B = 1500 × 4/15 = ₹400, C = 1500 × 5/15 = ₹500.

Answer: A ₹600, B ₹400, C ₹500.

Practice set

15. Pipes & Cisterns

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Pipe A fills a tank in 6 hours and pipe B in 12 hours. How long do both take together?

1. Capacity = LCM(6, 12) = 12 units. A = 2/h, B = 1/h.
2. Together 3 units/h, so 12 ÷ 3 = 4 hours.

Answer: 4 hours.

Example 2 (Easy). Pipe A fills a tank in 10 hours; pipe B empties it in 15 hours. If both are opened on an empty tank, when is it full?

1. Capacity = LCM(10, 15) = 30. A = +3/h, B = −2/h.
2. Net = +1 unit/h. Time = 30 ÷ 1 = 30 hours.

Answer: 30 hours. Faster: ab/(b − a) = 150/5 = 30.

Example 3 (Medium). A pipe fills a tank in 8 hours. Because of a leak at the bottom, it takes 10 hours. How long would the leak take to empty a full tank?

1. Capacity = LCM(8, 10) = 40. Pipe alone = 5/h. Pipe with leak = 40 ÷ 10 = 4/h (net).
2. So the leak removes 5 − 4 = 1 unit/h.
3. Leak alone: 40 ÷ 1 = 40 hours.

Answer: 40 hours. Faster: ab/(b − a) = 80/2 = 40.

Example 4 (Medium). Pipes A and B fill a tank in 12 and 15 hours, and pipe C empties it in 20 hours. All three are opened together. When is the tank full?

1. Capacity = LCM(12, 15, 20) = 60. A = +5, B = +4, C = −3 per hour (see the diagram).
2. Net = 6 units/h. Time = 60 ÷ 6 = 10 hours.

Answer: 10 hours.

Example 5 (Medium). Pipes A and B fill a tank in 20 and 30 minutes. Both are opened, and A is closed after 6 minutes. How much longer does B take to fill the tank?

1. Capacity = 60. A = 3/min, B = 2/min.
2. First 6 minutes: 6 × 5 = 30 units.
3. Left: 30 units, B alone: 30 ÷ 2 = 15 minutes.

Answer: 15 more minutes.

Example 6 (Hard). A leak can empty a full tank in 6 hours. An inlet pipe filling 4 litres a minute is opened, and now the full tank empties in 8 hours. What is the capacity?

1. Work in minutes: 6 h = 360 min, 8 h = 480 min. Let capacity = C litres.
2. Leak rate = C/360 L/min. With the inlet, net emptying = C/480 L/min.
3. Leak − inlet = net: C/360 − 4 = C/480.
4. C (1/360 − 1/480) = 4. LCM(360, 480) = 1440, so C (4 − 3)/1440 = 4, C = 5760.

Answer: 5760 litres.

Practice set

16. Time, Speed & Distance

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). Convert 72 km/h to m/s.

1. Multiply by 5/18: 72 × 5/18 = 4 × 5 = 20.

Answer: 20 m/s.

Example 2 (Easy). A car goes from A to B at 40 km/h and returns at 60 km/h. What is its average speed for the round trip?

1. The two legs have equal distance, so use 2xy/(x + y).
2. 2 × 40 × 60 / 100 = 4800/100 = 48.

Answer: 48 km/h, not 50. Check with a distance of 120 km: 3 h + 2 h = 5 h for 240 km, 240 ÷ 5 = 48.

Example 3 (Medium). Two cars 360 km apart start at the same time towards each other at 50 km/h and 70 km/h. When and where do they meet?

1. They move towards each other, so relative speed = 50 + 70 = 120 km/h.
2. Time to meet = 360 ÷ 120 = 3 hours.
3. In 3 hours the first car covers 50 × 3 = 150 km.

Answer: after 3 hours, 150 km from the first car's start.

Example 4 (Medium). A thief runs at 10 km/h. A policeman starts chasing him 30 minutes later from the same spot at 12 km/h. When does he catch the thief?

1. Head start of the thief = 10 × 1/2 = 5 km.
2. Same direction, so relative speed = 12 − 10 = 2 km/h.
3. Catch-up time = 5 ÷ 2 = 2.5 hours after the policeman starts.

Answer: 2 1/2 hours.

Example 5 (Medium). Walking at 4 km/h a student reaches school 10 minutes late. Walking at 5 km/h he is 5 minutes early. How far is the school?

1. The two times differ by 10 + 5 = 15 minutes = 1/4 hour.
2. D/4 − D/5 = 1/4, so D/20 = 1/4 and D = 5.

Answer: 5 km. Formula: (4 × 5 / 1) × 1/4 = 5.

Example 6 (Hard). Walking at 3/4 of his usual speed, a man is 20 minutes late. What is his usual time?

1. At 3/4 speed, time becomes 4/3 of usual.
2. Extra time = 4/3 T − T = T/3 = 20 minutes.
3. So T = 60 minutes.

Answer: 1 hour.

Practice set

17. Problems on Trains

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). A 150 m train runs at 54 km/h. How long does it take to pass a pole?

1. 54 km/h = 15 m/s.
2. Distance = 150 m. Time = 150 ÷ 15 = 10 s.

Answer: 10 seconds.

Example 2 (Easy). A 240 m train at 72 km/h crosses a 360 m platform. How long does it take?

1. Speed = 20 m/s. Distance = 240 + 360 = 600 m.
2. Time = 600 ÷ 20 = 30 s.

Answer: 30 seconds.

Example 3 (Medium). A train passes a pole in 15 s and a 100 m platform in 25 s. Find its length and speed.

1. The extra 10 seconds are spent covering the 100 m platform, so speed = 100 ÷ 10 = 10 m/s.
2. Length = 10 × 15 = 150 m.
3. Speed in km/h = 10 × 18/5 = 36.

Answer: 150 m, 36 km/h.

Example 4 (Medium). Trains of 120 m and 180 m run at 50 km/h and 40 km/h. How long do they take to cross each other (a) in opposite directions, (b) in the same direction?

1. Distance in both cases = 120 + 180 = 300 m.
2. Opposite: relative speed = 90 km/h = 25 m/s. Time = 300 ÷ 25 = 12 s.
3. Same direction: relative speed = 10 km/h = 25/9 m/s. Time = 300 × 9/25 = 108 s.

Answer: (a) 12 s, (b) 108 s.

Example 5 (Medium). A 200 m train passes a man walking at 6 km/h in the same direction in 20 s. Find the train's speed.

1. Relative speed = 200 ÷ 20 = 10 m/s = 36 km/h.
2. Same direction, so train speed − 6 = 36, train speed = 42 km/h.

Answer: 42 km/h.

Example 6 (Hard). Two trains start at the same time from Howrah and Patna towards each other. After meeting, they take 9 hours and 16 hours to reach their destinations. Find the ratio of their speeds.

1. Use S1 : S2 = √t2 : √t1 = √16 : √9.
2. = 4 : 3.

Answer: 4 : 3. Why: after meeting, each train covers the part the other already covered. That swap gives S1 × S1 × t1 = S2 × S2 × t2.

Practice set

18. Boats & Streams

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). A boat moves at 10 km/h in still water and the stream flows at 2 km/h. Find its downstream and upstream speeds.

1. Downstream = 10 + 2 = 12 km/h.
2. Upstream = 10 − 2 = 8 km/h.

Answer: 12 km/h and 8 km/h.

Example 2 (Easy). A boat goes downstream at 16 km/h and upstream at 10 km/h. Find the boat's speed and the stream's speed.

1. Boat = (16 + 10)/2 = 13 km/h.
2. Stream = (16 − 10)/2 = 3 km/h.

Answer: boat 13 km/h, stream 3 km/h.

Example 3 (Medium). A boat covers 30 km downstream in 2 hours and 18 km upstream in 3 hours. Find the speeds of the boat and the stream.

1. D = 30 ÷ 2 = 15 km/h. U = 18 ÷ 3 = 6 km/h.
2. Boat = (15 + 6)/2 = 10.5 km/h. Stream = (15 − 6)/2 = 4.5 km/h.

Answer: 10.5 km/h and 4.5 km/h.

Example 4 (Medium). A boat's speed is 15 km/h and the stream's is 3 km/h. How long does it take to go 36 km downstream and come back?

1. D = 18 km/h, U = 12 km/h.
2. Time = 36/18 + 36/12 = 2 + 3 = 5 hours.

Answer: 5 hours.

Example 5 (Medium). A man takes three times as long to row a distance upstream as downstream. Find the ratio of his speed in still water to the stream's speed.

1. Same distance, so speeds are in inverse ratio of times: (u + v) : (u − v) = 3 : 1.
2. u + v = 3u − 3v, so 2u = 4v and u = 2v.

Answer: 2 : 1. Shortcut: (3 + 1) : (3 − 1) = 4 : 2 = 2 : 1.

Example 6 (Hard). A man rows at 5 km/h in still water. The river flows at 1 km/h. It takes him 1 hour to row to a place and come back. How far is the place?

1. D = 6 km/h, U = 4 km/h. Let the distance be d.
2. d/6 + d/4 = 1. LCM 12: (2d + 3d)/12 = 1, so 5d = 12.
3. d = 2.4 km.

Answer: 2.4 km.

Practice set

19. Races & Circular Tracks

Key formulas & rules

Shortcuts

Solved examples

Example 1 (Easy). In a 100 m race A beats B by 10 m. By how much will A beat B in a 1 km race?

1. Speed A : B = 100 : 90 = 10 : 9.
2. When A runs 1000 m, B runs 900 m.

Answer: 100 m.

Example 2 (Medium). In a 200 m race A beats B by 20 m or 5 seconds. Find A's time.

1. B runs 20 m in 5 s, so B's speed = 4 m/s.
2. B's time for 200 m = 200 ÷ 4 = 50 s.
3. A finishes 5 s earlier: 45 s.

Answer: 45 seconds.

Example 3 (Medium). A runs 1 km in 3 min 10 s and B in 3 min 20 s. By how many metres does A beat B?

1. A takes 190 s, B takes 200 s, so A wins by 10 s.
2. B's speed = 1000/200 = 5 m/s. In 10 s B covers 50 m.

Answer: 50 m.

Example 4 (Medium). In a 100 m race A beats B by 20 m and B beats C by 25 m. By how much does A beat C?

1. When A = 100, B = 80.
2. B : C = 100 : 75, so when B = 80, C = 80 × 75/100 = 60.
3. A beats C by 100 − 60 = 40 m.

Answer: 40 m, not 45.

Example 5 (Medium). A's speed is 5/3 times B's. How much start should A give B in a 200 m race so that it ends in a dead heat?

1. Speed A : B = 5 : 3. In the time A runs 200 m, B runs 200 × 3/5 = 120 m.
2. For a dead heat, B must only need 120 m, so B starts 80 m ahead.

Answer: 80 m.

Example 6 (Hard). A and B run on a 600 m circular track at 5 m/s and 3 m/s from the same point. When do they first meet (a) running in the same direction, (b) in opposite directions, (c) at the starting point?

1. Same direction: A must gain one full lap. 600 ÷ 2 = 300 s.
2. Opposite: together they must cover one lap. 600 ÷ 8 = 75 s.
3. Start point: A is back at start every 120 s, B every 200 s. Both are there together at LCM(120, 200) = 600 s.

Answer: (a) 300 s, (b) 75 s, (c) 600 s = 10 minutes.

Practice set

20. Permutations & Combinations

Key formulas & rules

Quick values worth knowing by heart: 4! = 24, 5! = 120, 6! = 720, 7! = 5040. Also 5C2 = 10, 6C2 = 15, 6C3 = 20, 8C3 = 56, 10C2 = 45, 10C3 = 120.

Shortcuts

Question patterns

Spot the key words in the question and you know the method before you start.

Solved examples

Example 1. Riya has 3 shirts, 4 trousers and 2 pairs of shoes. How many different outfits can she wear?

1. She picks a shirt and a trouser and shoes, so multiply.
2. 3 × 4 × 2 = 24.

Answer: 24 outfits.

Example 2. How many 3-digit numbers can be formed with digits 0–9 if no digit repeats?

1. Hundreds place: cannot be 0, so 9 choices (1–9).
2. Tens place: any digit except the one used, so 9 choices (0 is now allowed).
3. Units place: 8 choices left.
4. Total = 9 × 9 × 8 = 648.

Answer: 648. Faster way: all 3-digit arrangements 10P3 = 720, minus those starting with 0 (9P2 = 72) = 648.

Example 3. In how many ways can the letters of LEADER and of BANANA be arranged?

1. LEADER has 6 letters, E appears twice. Ways = 6!/2! = 720/2 = 360.
2. BANANA has 6 letters: A three times, N twice. Ways = 6!/(3! × 2!) = 720/12 = 60.
3. Why divide? Swapping the two E's gives the same word, so every word was counted 2! times.

Answer: LEADER 360, BANANA 60.

Example 4. In how many ways can 6 people sit at a round table? In how many ways can 6 different beads make a necklace?

1. In a row, 6 people give 6! = 720 orders. In a circle, the 6 rotations of each order look the same.
2. So round table = 720/6 = 5! = 120. Easy way to think: fix one person, arrange the rest.
3. A necklace can also be flipped, so clockwise and anticlockwise are the same: 120/2 = 60.

Answer: 120 seatings, 60 necklaces.

Example 5. A committee of 5 is to be chosen from 6 men and 4 women so that it has at least 2 women. In how many ways?

1. Total committees with no rule: 10C5 = 252.
2. Unwanted case "0 women": 6C5 = 6.
3. Unwanted case "1 woman": 4C1 × 6C4 = 4 × 15 = 60.
4. Wanted = 252 − 6 − 60 = 186.

Answer: 186. Check by adding cases: 2W (6 × 20 = 120) + 3W (4 × 15 = 60) + 4W (1 × 6 = 6) = 186.

Example 6. In how many ways can the letters of LEADING be arranged so that the vowels are always together?

1. LEADING has 7 different letters. Vowels: E, A, I.
2. Glue E, A, I into one block. Now we arrange 5 units: L, D, N, G, [EAI] → 5! = 120.
3. Inside the block the 3 vowels can be arranged in 3! = 6 ways.
4. Total = 120 × 6 = 720.

Answer: 720.

Practice set

21. Probability

Key formulas & rules

Sample spaces you must know

The 52-card deck

Shortcuts

Question patterns

Spot the key words in the question and you know the method before you start.

Solved examples

Example 1. Two coins are tossed. What is the probability of getting at least one head?

1. Outcomes: HH, HT, TH, TT → 4.
2. Only TT has no head. P(no head) = 1/4.
3. P(at least one head) = 1 − 1/4 = 3/4.

Answer: 3/4.

Example 2. Two dice are thrown. What is the probability that the sum is 8?

1. Total outcomes = 6 × 6 = 36.
2. Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes.
3. P = 5/36.

Answer: 5/36. Shortcut: sum 8 ≥ 7, so ways = 13 − 8 = 5.

Example 3. One card is drawn from a deck of 52. What is the probability that it is a king or a heart?

1. Kings = 4, hearts = 13.
2. The king of hearts is in both groups, so it was counted twice. Subtract it once.
3. Favourable = 4 + 13 − 1 = 16. P = 16/52 = 4/13.

Answer: 4/13.

Example 4. A bag has 5 red and 3 blue balls. Two balls are drawn at random. Find the probability that both are red.

1. Total ways to draw 2 from 8: 8C2 = 28.
2. Ways to draw 2 red from 5: 5C2 = 10.
3. P = 10/28 = 5/14.

Answer: 5/14. One-after-another way: (5/8) × (4/7) = 20/56 = 5/14.

Example 5. A can hit a target with probability 1/2 and B with probability 2/3. Both fire once. Find the probability that the target is hit.

1. "Hit" means at least one of them hits. Use 1 − P(both miss).
2. P(A misses) = 1/2, P(B misses) = 1/3. They fire independently, so multiply: 1/2 × 1/3 = 1/6.
3. P(hit) = 1 − 1/6 = 5/6.

Answer: 5/6.

Example 6. A die is thrown 4 times. What is the probability of getting at least one six?

1. P(no six in one throw) = 5/6.
2. Throws are independent: P(no six in 4 throws) = (5/6)⁴ = 625/1296.
3. P(at least one six) = 1 − 625/1296 = 671/1296 (a little more than 1/2).

Answer: 671/1296.

Practice set

22. Linear & Quadratic Equations

Key formulas & rules

Shortcuts

Question patterns

Spot the key words in the question and you know the method before you start.

Solved examples

Example 1. Solve x + y = 10 and x − y = 4.

1. Add the two equations: 2x = 14, so x = 7.
2. Put x = 7 in the first: 7 + y = 10, so y = 3.

Answer: x = 7, y = 3. Check: 7 − 3 = 4.

Example 2. Solve 3x + 2y = 16 and 2x − y = 6.

1. From the second equation, y = 2x − 6 (substitution).
2. Put it in the first: 3x + 2(2x − 6) = 16 → 7x − 12 = 16 → x = 4.
3. y = 2(4) − 6 = 2.

Answer: x = 4, y = 2.

Example 3. 2 pens and 3 pencils cost ₹34; 3 pens and 2 pencils cost ₹41. Find the price of one pen.

1. Let pen = p, pencil = q. Then 2p + 3q = 34 and 3p + 2q = 41.
2. Add: 5p + 5q = 75 → p + q = 15.
3. Subtract (second − first): p − q = 7.
4. Add these: 2p = 22 → p = 11, and q = 4.

Answer: pen ₹11 (pencil ₹4).

Example 4. Find the roots of x² − 5x + 6 = 0 and of 2x² + 3x − 2 = 0.

1. x² − 5x + 6: two numbers with product 6 and sum −5 are −2 and −3. So (x − 2)(x − 3) = 0 → x = 2 or 3.
2. 2x² + 3x − 2: use the formula. D = 3² − 4(2)(−2) = 9 + 16 = 25, √D = 5.
3. x = (−3 ± 5)/4 → x = 2/4 = 1/2 or x = −8/4 = −2.

Answer: 2 and 3; 1/2 and −2.

Example 5. If α and β are the roots of 3x² − 12x + 7 = 0, find α² + β².

1. Sum α + β = −b/a = 12/3 = 4. Product αβ = c/a = 7/3.
2. α² + β² = (α + β)² − 2αβ = 16 − 14/3 = 34/3.

Answer: 34/3. Notice we never found α and β themselves.

Example 6. For what value of k does kx² + 4x + 1 = 0 have equal roots?

1. Equal roots means D = 0.
2. D = 4² − 4(k)(1) = 16 − 4k = 0 → k = 4.

Answer: k = 4 (the equation becomes 4x² + 4x + 1 = (2x + 1)², root −1/2 twice).

Practice set

23. Progressions: AP & GP

Key formulas & rules

Shortcuts

Question patterns

Spot the key words in the question and you know the method before you start.

Solved examples

Example 1. Find the 20th term of 3, 7, 11, 15, …

1. a = 3, d = 4, n = 20.
2. T₂₀ = 3 + 19 × 4 = 3 + 76 = 79.

Answer: 79.

Example 2. Find the sum of the first 20 terms of 5, 8, 11, …

1. a = 5, d = 3, n = 20.
2. S₂₀ = 20/2 × [2 × 5 + 19 × 3] = 10 × (10 + 57) = 670.

Answer: 670.

Example 3. How many two-digit numbers are divisible by 7, and what is their sum?

1. First is 14, last is 98, d = 7.
2. n = (98 − 14)/7 + 1 = 12 + 1 = 13.
3. Sum = 13/2 × (14 + 98) = 13 × 56 = 728.

Answer: 13 numbers, sum 728.

Example 4. In the GP 2, 6, 18, … find the 7th term and the sum of the first 6 terms.

1. a = 2, r = 3.
2. T₇ = 2 × 3⁶ = 2 × 729 = 1458.
3. S₆ = 2(3⁶ − 1)/(3 − 1) = 2 × 728/2 = 728.

Answer: 1458 and 728.

Example 5. Find the sum 16 + 8 + 4 + 2 + … to infinity.

1. a = 16, r = 1/2. Since r is between −1 and 1, the sum settles to a fixed value.
2. S∞ = 16/(1 − 1/2) = 16/(1/2) = 32.

Answer: 32.

Example 6. Find the AM, GM and HM of 4 and 16.

1. AM = (4 + 16)/2 = 10.
2. GM = √(4 × 16) = √64 = 8.
3. HM = 2 × 4 × 16/(4 + 16) = 128/20 = 6.4.
4. Check: AM ≥ GM ≥ HM (10 ≥ 8 ≥ 6.4) and GM² = AM × HM (64 = 10 × 6.4).

Answer: 10, 8, 6.4.

Practice set

24. Logarithms

Key formulas & rules

From these you can build almost any log: log 4 = 0.6020, log 6 = 0.7781, log 8 = 0.9030, log 9 = 0.9542, log 12 = 1.0791, log 15 = 1.1761 (the last digit can differ by 1 from exact tables because of rounding; options are never that close).

Number of digits: if log N = k.something, then N has k + 1 digits (for N ≥ 1). The whole-number part is called the characteristic.

Shortcuts

Question patterns

Spot the key words in the question and you know the method before you start.

Solved examples

Example 1. Find log₂ 64.

1. Ask: 2 to what power is 64? 2⁶ = 64.

Answer: 6.

Example 2. Find log 8 + log 125.

1. Product law: log 8 + log 125 = log (8 × 125) = log 1000.
2. 1000 = 10³, so log 1000 = 3.

Answer: 3.

Example 3. Given log 2 = 0.3010 and log 3 = 0.4771, find log 72.

1. 72 = 8 × 9 = 2³ × 3².
2. log 72 = 3 log 2 + 2 log 3 = 0.9030 + 0.9542 = 1.8572.

Answer: 1.8572.

Example 4. Find log₄ 8.

1. Write both as powers of 2: 4 = 2², 8 = 2³.
2. log₄ 8 = log 8/log 4 = 3 log 2/2 log 2 = 3/2.

Answer: 3/2. Check: 4^3/2 = (√4)³ = 8.

Example 5. How many digits does 2⁵⁰ have? (log 2 = 0.30103)

1. log 2⁵⁰ = 50 × 0.30103 = 15.05.
2. Characteristic is 15, so the number has 15 + 1 = 16 digits.

Answer: 16 digits. (2⁵⁰ = 1,125,899,906,842,624.)

Example 6. Solve log x + log (x − 3) = 1.

1. Product law: log [x(x − 3)] = 1, so x(x − 3) = 10¹ = 10.
2. x² − 3x − 10 = 0 → (x − 5)(x + 2) = 0 → x = 5 or x = −2.
3. Log of a negative number is not defined, so reject x = −2.

Answer: x = 5.

Practice set

25. Geometry & 2D Mensuration

Key formulas & rules

Shortcuts

Question patterns

Spot the key words in the question and you know the method before you start.

Solved examples

Example 1. A right-angled triangle has legs 6 cm and 8 cm. Find the hypotenuse and the area.

1. 6, 8 is the 3, 4 triplet doubled, so hypotenuse = 10 cm (check: 36 + 64 = 100).
2. Area = ½ × 6 × 8 = 24 cm².

Answer: 10 cm, 24 cm².

Example 2. Find the area of a triangle with sides 13, 14 and 15 cm.

1. s = (13 + 14 + 15)/2 = 21.
2. Area = √[21 × 8 × 7 × 6] = √7056 = 84 cm².

Answer: 84 cm².

Example 3. Find the area and circumference of a circle of radius 7 cm.

1. Area = (22/7) × 7 × 7 = 154 cm².
2. Circumference = 2 × (22/7) × 7 = 44 cm.

Answer: 154 cm², 44 cm.

Example 4. A field is 40 m × 30 m. A path 2.5 m wide runs around it on the outside. Find the area of the path and the cost of paving it at ₹4 per m².

1. Outer rectangle = (40 + 5) × (30 + 5) = 45 × 35 = 1575 m². (Width is added on both sides, so +2 × 2.5.)
2. Field = 40 × 30 = 1200 m².
3. Path = 1575 − 1200 = 375 m². Cost = 375 × 4 = ₹1500.

Answer: 375 m², ₹1500. Shortcut: 2w(l + b + 2w) = 5 × (40 + 30 + 5) = 375.

Example 5. The diagonals of a rhombus are 16 cm and 12 cm. Find its area and perimeter.

1. Area = ½ × 16 × 12 = 96 cm².
2. Diagonals bisect each other at 90°, giving right triangles with legs 8 and 6. Side = √(64 + 36) = 10 cm.
3. Perimeter = 4 × 10 = 40 cm.

Answer: 96 cm², 40 cm.

Example 6. A trapezium has parallel sides 10 cm and 18 cm, 7 cm apart. Find its area.

1. Area = ½ × (10 + 18) × 7 = 14 × 7 = 98 cm².

Answer: 98 cm².

Practice set

26. 3D Mensuration

Key formulas & rules

Shortcuts

Question patterns

Spot the key words in the question and you know the method before you start.

Solved examples

Example 1. A cube has edge 5 cm. Find its volume, total surface area and diagonal.

1. Volume = 5³ = 125 cm³.
2. TSA = 6 × 25 = 150 cm².
3. Diagonal = 5√3 ≈ 8.66 cm.

Answer: 125 cm³, 150 cm², 5√3 cm.

Example 2. A cuboid is 12 cm × 4 cm × 3 cm. Find its volume, TSA and longest diagonal.

1. Volume = 12 × 4 × 3 = 144 cm³.
2. TSA = 2(48 + 12 + 36) = 2 × 96 = 192 cm².
3. Diagonal = √(144 + 16 + 9) = √169 = 13 cm.

Answer: 144 cm³, 192 cm², 13 cm.

Example 3. A cylinder has radius 7 cm and height 10 cm. Find its volume, CSA and TSA.

1. Volume = (22/7) × 49 × 10 = 1540 cm³.
2. CSA = 2 × (22/7) × 7 × 10 = 440 cm².
3. TSA = CSA + 2πr² = 440 + 2 × 154 = 748 cm².

Answer: 1540 cm³, 440 cm², 748 cm².

Example 4. A cone has radius 7 cm and height 24 cm. Find its slant height, volume and total surface area.

1. l = √(7² + 24²) = √625 = 25 cm (7, 24, 25 triplet).
2. Volume = ⅓ × (22/7) × 49 × 24 = 1232 cm³.
3. CSA = (22/7) × 7 × 25 = 550 cm². Base = 154 cm². TSA = 704 cm².

Answer: 25 cm, 1232 cm³, 704 cm².

Example 5. A sphere of radius 6 cm is melted and recast into a cone of base radius 6 cm. Find the height of the cone.

1. Volume is the same: (4/3)π × 6³ = ⅓π × 6² × h.
2. Cancel π and multiply both sides by 3: 4 × 216 = 36h → 864 = 36h.
3. h = 24 cm.

Answer: 24 cm. Also: a sphere of radius 4 cm gives (4/1)³ = 64 spheres of radius 1 cm.

Example 6. A bucket is a frustum with top radius 10 cm, bottom radius 4 cm and depth 8 cm. Find its volume in terms of π.

1. V = (πh/3)(R² + Rr + r²) = (8π/3)(100 + 40 + 16).
2. = (8π/3) × 156 = 416π cm³ ≈ 1307 cm³.

Answer: 416π cm³.

Practice set

27. Data Interpretation

Key formulas & rules

Shortcuts

Solved examples

Chart 1 — Table: students enrolled in three courses at a coaching institute

Example 1. By what percent did Maths enrolment grow from 2021 to 2024?

1. Change = 210 − 120 = 90.
2. Percent = 90/120 × 100 = 75%.

Answer: 75%.

Example 2. In which year did the total enrolment grow by the highest percentage over the previous year?

1. 2022: 60/270 ≈ 22.2%. 2023: 60/330 ≈ 18.2%. 2024: 70/390 ≈ 17.9%.
2. The same increase of 60 on a smaller base (270) gives the bigger percent.

Answer: 2022. Note that 2024 has the biggest absolute rise (70) but not the biggest percent rise.

Chart 2 — Bar chart: company revenue

Example 3. In which year was the percentage growth in revenue over the previous year the highest?

1. 2020: 15/40 = 37.5%. 2021: fell (−5/55 ≈ −9.1%). 2022: 20/50 = 40%. 2023: 15/70 ≈ 21.4%. 2024: 15/85 ≈ 17.6%.
2. Highest is 40%.

Answer: 2022. Shortcut: look for a big jump from a short bar. 50 → 70 stands out.

Example 4. What is the average yearly revenue over the six years?

1. Sum = 40 + 55 + 50 + 70 + 85 + 100 = 400.
2. Average = 400/6 ≈ ₹66.67 crore.

Answer: ₹66.67 crore.

Chart 3 — Line chart: sales of two shops

Example 5. In how many months did Shop A sell more than Shop B?

1. Compare month by month: Jan 30 < 50, Feb 45 > 40, Mar 40 < 55, Apr 60 > 45, May 55 < 65, Jun 70 > 60.
2. A is ahead in Feb, Apr and Jun.

Answer: 3 months. On the chart, count where the solid line is above the dashed line.

Chart 4 — Pie chart: family budget

Example 6. How much does the family spend on rent, and what is the central angle for food?

1. Rent = 25% of 60,000 = ₹15,000.
2. Food angle = 20 × 3.6 = 72°.

Answer: ₹15,000 and 72°.

Practice set

Questions 1–3 use the table (Chart 1).

Questions 4–6 use the bar chart (Chart 2).

Questions 7–9 use the line chart (Chart 3).

Questions 10–12 use the pie chart (Chart 4). Monthly income ₹60,000.

Formula Sheet

Part 1 — Number foundations (1–5)

Part 2 — Arithmetic (6–13)

Part 3 — Time, work & motion (14–19)

Part 4 — Counting, algebra & shapes (20–27)

Time-saving tips

Test Strategy & Learning Path

How to attempt a 60-question / 60-minute test

1. Pass 1 (about 30 minutes): go through every question. If you can see the method in 10 seconds and finish in under a minute, solve it. Otherwise mark it and move on. Aim for 30–35 correct answers here.
2. Pass 2 (about 20 minutes): return to the marked questions. Pick those where you know the method but the calculation is long. Use options, approximation and elimination.
3. Pass 3 (about 8 minutes): the hardest remaining ones. Eliminate 2 options and guess if there is no negative marking. If there is (for example −1/4 per wrong), guess only when you have removed at least 2 options.
4. Final 2 minutes: make sure every answer you meant to mark is actually marked.
5. The 90-second rule: if you are stuck for 90 seconds, leave it. One stubborn question can cost you three easy ones.
6. DI sets: read the chart once carefully (30–45 seconds), then answer all its questions together. Do not jump between sets.

Which topics to learn first

Learn in this order. Each group is used by the groups after it, and the first two groups carry most of the marks.

8-week plan

Plan for about 90 minutes a day: 30 minutes learning a topic, 45 minutes practice, 15 minutes revising mistakes. One rest or mock day per week.

Speed targets

By the end of week 8, you should be able to solve a standard question in these times. Measure yourself with a stopwatch.