Savings, a trick and two unknowns

Example 9: Jahnavi starts with ₹4000 and saves ₹650 a month; Sunita starts with ₹5050 and saves ₹500 a month. After how many months will they have the same amount?

Let m be that number of months. Jahnavi's savings are 4000 + 650m and Sunita's are 5050 + 500m, and they are equal, so 4000 + 650m = 5050 + 500m. Subtract 500m from both sides: 4000 + 150m = 5050. Subtract 4000: 150m = 1050. Divide by 150: m = 7. After 7 months they have the same amount.

Example 10: solve 28(x + 4) + 300 = 1000. There are several routes. Subtract 300 to get 28(x + 4) = 700, divide by 28 to get x + 4 = 25, so x = 21. Or simplify the LHS first: 28x + 412 = 1000, so 28x = 588 and x = 21. Or notice that 28, 300 and 1000 are all divisible by 4, divide throughout to get 7(x + 4) + 75 = 250, and carry on to x = 21 again. Look for the route with the smallest numbers.

Example 11: Riyaz's trick. Think of a number; subtract 3; multiply by 4; add 8; say the answer. Akash says 24. Let the number be x. After each step: x, then x − 3, then 4(x − 3) = 4x − 12, then 4x − 12 + 8 = 4x − 4.

So 4x − 4 = 24, that is 4(x − 1) = 24, so x − 1 = 6 and x = 7. Akash thought of 7. Try the trick with other starting numbers and look for a simple rule taking the final answer back to the start.

Example 12: Ramesh and Suresh have 60 marbles between them, and Ramesh has 30 more than Suresh. Here there are TWO unknowns. Writing Ramesh as x and Suresh as y gives x + y = 60 and x = y + 30 — but we only have a method for ONE unknown.

The way through is to use one letter for both. Let Suresh have y; then Ramesh has y + 30. The total is 60, so y + (y + 30) = 60, that is 2y + 30 = 60. One unknown, and the method applies. Use it to find both counts.