Mathematics: Solved Word Problems (Average, Profit & Loss, Interest, Ratio, Speed, Work)

• Average = sum ÷ number of items. Sum = average × number.
• Average of first n natural numbers = (n + 1) ÷ 2.
• An increase of x% followed by a decrease of x% gives a net decrease of x²/100 %.

Solved example: the average of 5 numbers is 20. One number, 30, is removed. Sum = 100, remaining sum = 70, new average = 70 ÷ 4 = 17.5.

• Profit = SP – CP; Profit % = profit ÷ CP × 100 (always on cost price).
• SP = CP × (100 + profit%) ÷ 100; CP = SP × 100 ÷ (100 – loss%).
• Discount is always on the marked price.

Solved example: a fan is sold at a 10% loss for Rs 900. CP = 900 × 100 ÷ 90 = Rs 1,000.

• Simple interest = P × R × T ÷ 100.
• Amount (compound) = P(1 + R/100)ᵀ; CI = amount – P.

Solved example: CI on Rs 10,000 at 10% for 2 years. Amount = 10,000 × 1.1 × 1.1 = 12,100, so CI = Rs 2,100 (SI would be Rs 2,000).

• To divide an amount in the ratio a : b, the parts are amount × a/(a + b) and amount × b/(a + b).
• In partnership, profit is shared in the ratio of capital × time.

Solved example: 40 litres of a mixture has milk and water in the ratio 3 : 1 (30 L milk, 10 L water). To make the ratio 3 : 2 the water must be 20 L, so add 10 litres of water.

• Speed = distance ÷ time. km/h to m/s: multiply by 5/18; m/s to km/h: multiply by 18/5.
• A train crossing a pole covers its own length; crossing a platform covers train + platform length.
• Average speed for equal distances at x and y km/h = 2xy ÷ (x + y).
• Downstream speed = boat + stream; upstream speed = boat – stream.

Solved example: a 200 m train at 72 km/h (= 20 m/s) crosses a pole in 200 ÷ 20 = 10 seconds.

• If A finishes a job in a days, A's one-day work = 1/a.
• Together: 1/a + 1/b; time = ab ÷ (a + b).
• A filling pipe adds, a leak or emptying pipe subtracts.

Solved example: pipe A fills a tank in 6 h and pipe B empties it in 9 h. Net per hour = 1/6 – 1/9 = 1/18, so the tank fills in 18 hours.

• Events that repeat together (bells, lights) meet again after the LCM of their intervals.
• HCF × LCM = product of the two numbers.
• Ages: let the unknown age be x and form one equation from each statement.
• Arrangements of n different objects = n! (3! = 6). Selections of r from n: ⁿCᵣ = n! ÷ (r!(n – r)!), e.g. ⁵C₂ = 10.
• Probability = favourable outcomes ÷ total outcomes, between 0 and 1.

Solved example: bells ring every 6, 8 and 12 minutes. LCM = 24 minutes, so they ring together again after 24 minutes.