Chapter 4 · Exercises 4 & 5Textbook pages 80 and 82

Exercises 4 & 5 – Animated Solutions വ്യത്യാസ ഗുണനം

Eight questions, all built on the identities of page 9: (x − 1)(y − 1) = xy − (x + y) + 1, (x + 1)(y − 1) = xy − (x − y) − 1 and their friends. Read each question, try it in your notebook (use the hint if you are stuck), then press Next ▶ to watch the solution grow step by step. The in-text practice products from pages 77 and 81 are at the end.

📖 Tools you will use

Neighbours above: (x + 1)(y + 1) = xy + (x + y) + 1; below: (x − 1)(y − 1) = xy − (x + y) + 1. With 2 or 3 in place of 1: (x + a)(y + a) = xy + a(x + y) + a².
One up, one down (x the bigger): (x + 1)(y − 1) = xy − (x − y) − 1 and (x − 1)(y + 1) = xy + (x − y) − 1.
Add / subtract two such equations to get xy and x + y (or x − y).
Class 8: (x − y)² = (x + y)² − 4xy and (x + y)² = (x − y)² + 4xy; then x = ½(sum + difference), y = ½(sum − difference).

Exercise 4 (page 80)

Question 1 · A rectangle 3 cm smaller each way

Q1The perimeter of a rectangle is 40 cm and its area is 70 sq cm. Find the area of the rectangle whose length and breadth are each 3 cm less.

(ഒരു ചതുരത്തിന്റെ ചുറ്റളവ് 40 സെന്റിമീറ്ററും, പരപ്പളവ് 70 ചതുരശ്ര സെന്റിമീറ്ററുമാണ്. നീളവും വീതിയും ഇതിനേക്കാൾ 3 സെന്റിമീറ്റർ കുറവായ ചതുരത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.)

You do not need the length and breadth themselves. The perimeter tells you x + y; the area tells you xy. Expand (x − 3)(y − 3) and see which two things appear.

🎬 Solution animation

Answer: x + y = 20, xy = 70, so (x − 3)(y − 3) = xy − 3(x + y) + 9 = 70 − 60 + 9 = 19 sq cm.

💡 Why we never found x and y

The sides are 10 + √30 ≈ 15.5 cm and 10 − √30 ≈ 4.5 cm – not nice numbers at all. The identity needs only their sum and product, which the question gives directly. That is the whole point of identities: work with the combination you know.

Question 2 · One metre less, one metre more

Q2If the length and breadth of a rectangle are each decreased by 1 m, the area becomes 741 sq m; if each is increased by 1 m, the area becomes 861 sq m. (i) What is the area of the rectangle? (ii) What is its perimeter? (iii) What are its length and breadth?

Write both areas with the identities. One equation has + (x + y), the other − (x + y). Adding and subtracting them separates xy from x + y.

🎬 Solution animation

Answer: adding gives 2xy + 2 = 1602 ⇒ xy = 800 sq m; subtracting gives 2(x + y) = 120 ⇒ x + y = 60 ⇒ perimeter 120 m; (x − y)² = 60² − 4 × 800 = 400 ⇒ x − y = 20 ⇒ length 40 m, breadth 20 m. Check: 39 × 19 = 741 ✓, 41 × 21 = 861 ✓.

Question 3 · Products of the neighbours

Q3Two numbers were each increased by 1 and multiplied: the product was 1271. Each was decreased by 1 and multiplied: the product was 1131. (i) What is the product of the numbers? (ii) What is their sum? (iii) What are the numbers?

Exactly the method of the textbook's 525 / 437 example (page 9 of this guide): add, subtract, then (x − y)² = (x + y)² − 4xy.

🎬 Solution animation

Answer: (i) xy = ½(1271 + 1131 − 2) = 1200; (ii) x + y = ½(1271 − 1131) = 70; (iii) (x − y)² = 4900 − 4800 = 100 ⇒ x − y = 10 ⇒ 40 and 30. Check: 41 × 31 = 1271 ✓, 39 × 29 = 1131 ✓.

Question 4 · Odd numbers and their odd neighbours

Q4For two odd numbers, the product of the odd numbers just ahead of each is 285, and the product of the odd numbers just behind each is 165. What are the numbers?

The odd neighbours of an odd number x are x + 2 and x − 2, not x ± 1. So the identities have 2 in place of 1: (x + 2)(y + 2) = xy + 2(x + y) + 4.

🎬 Solution animation

Answer: (x + 2)(y + 2) = xy + 2(x + y) + 4 = 285 and (x − 2)(y − 2) = xy − 2(x + y) + 4 = 165. Add: 2xy + 8 = 450 ⇒ xy = 221. Subtract: 4(x + y) = 120 ⇒ x + y = 30. (x − y)² = 900 − 884 = 16 ⇒ x − y = 4 ⇒ the numbers are 17 and 13. Check: 19 × 15 = 285 ✓, 15 × 11 = 165 ✓.

⚠️ Reading the Malayalam

The textbook says തൊട്ടുമുമ്പിലുള്ള ("just in front") for the product 285 and തൊട്ടുപുറകിലുള്ള ("just behind") for 165. On page 78 the book used the same words with (x + 1)(y + 1) = 525 as the "in front" product, so "in front" means the next numbers (x + 2, y + 2) and "behind" the previous ones (x − 2, y − 2). The larger product belongs to the larger neighbours, which confirms the reading.

Exercise 5 (page 82)

Question 1 · Product 713, difference 8

Q1The product of two numbers is 713 and their difference is 8. (i) What is the product of the bigger number plus 1 and the smaller number minus 1? (ii) What is the product of the bigger number minus 1 and the smaller number plus 1?

(രണ്ടു സംഖ്യകളുടെ ഗുണനഫലം 713 ഉം, വ്യത്യാസം 8 ഉം ആണ്. (i) വലിയ സംഖ്യയോട് 1 കൂട്ടിയതും, ചെറിയ സംഖ്യയിൽ നിന്ന് 1 കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം എന്താണ്? (ii) വലിയ സംഖ്യയിൽനിന്ന് 1 കുറച്ചതും, ചെറിയ സംഖ്യയോട് 1 കൂട്ടിയതും തമ്മിലുള്ള ഗുണനഫലം എന്താണ്?)

Call the bigger number x and the smaller y, so x − y = 8. Expand (x + 1)(y − 1) and (x − 1)(y + 1): only xy and x − y appear.

🎬 Solution animation

Answer: (i) (x + 1)(y − 1) = xy − (x − y) − 1 = 713 − 8 − 1 = 704. (ii) (x − 1)(y + 1) = xy + (x − y) − 1 = 713 + 8 − 1 = 720. (The numbers are 31 and 23; 32 × 22 = 704 and 30 × 24 = 720 ✓.)

Question 2 · Five less, how much more?

Q2For two numbers, the product of the bigger plus 1 and the smaller minus 1 is 5 less than the product of the numbers. By how much is the product of the bigger minus 1 and the smaller plus 1 greater than the product of the numbers?

(x + 1)(y − 1) = xy − (x − y) − 1. "5 less than xy" tells you x − y. Then use (x − 1)(y + 1) = xy + (x − y) − 1.

🎬 Solution animation

Answer: xy − (x − y) − 1 = xy − 5 ⇒ x − y = 4. Then (x − 1)(y + 1) = xy + (x − y) − 1 = xy + 3: it is 3 more than the product. (Example: 9 and 5: 45, then 10 × 4 = 40 and 8 × 6 = 48.)

Question 3 · 540 and 560

Q3For two numbers, the product of the bigger plus 1 and the smaller minus 1 is 540; the product of the bigger minus 1 and the smaller plus 1 is 560. (i) What is the product of the numbers? (ii) What is their difference? (iii) What are the numbers?

Expand both; one has − (x − y), the other + (x − y). Add to get xy, subtract to get x − y, then (x + y)² = (x − y)² + 4xy.

🎬 Solution animation

Answer: adding: 2xy − 2 = 1100 ⇒ xy = 551; subtracting: 2(x − y) = 20 ⇒ x − y = 10; (x + y)² = 100 + 4 × 551 = 2304 = 48² ⇒ x + y = 48 ⇒ the numbers are 29 and 19. Check: 30 × 18 = 540 ✓, 28 × 20 = 560 ✓.

Question 4 · The plot of land

Q4If the length of a rectangle is increased by 3 m and the breadth decreased by 2 m, the area decreases by 10 sq m. If the length is decreased by 2 m and the breadth increased by 3 m, the area increases by 30 sq m. Find the length and breadth.

Expand (x + 3)(y − 2) and (x − 2)(y + 3) with the sum × difference identity. In both, xy cancels against the original area, leaving two ordinary linear equations in x and y.

🎬 Solution animation

Answer: (x + 3)(y − 2) = xy − 10 ⇒ −2x + 3y − 6 = −10 ⇒ 2x − 3y = 4. (x − 2)(y + 3) = xy + 30 ⇒ 3x − 2y − 6 = 30 ⇒ 3x − 2y = 36. Solving: y = 12, x = 20. Length 20 m, breadth 12 m. Check: 23 × 10 = 230 = 240 − 10 ✓; 18 × 15 = 270 = 240 + 30 ✓.

Practice set: the eight products from the textbook (pages 77 and 81)

Click a product to see its picture and working. Blue is the product we want. The pictures are schematic (not to scale).

🧮 Practice products

Click a product above.

Solve your own: a neighbour-products puzzle

🧩 Make a puzzle, then solve it

secret x secret y

Choose two secret numbers (y is kept smaller than x). The puzzle shows only the two neighbour products – solve it in your notebook, then open the solution path and compare.

Check yourself

🧭 What these exercises taught

Changing both sides of a rectangle (or both numbers) by the same amount a gives (x ± a)(y ± a) = xy ± a(x + y) + a² – only the sum and product are needed.
Two neighbour products ⇒ add for xy, subtract for x + y (or x − y) ⇒ Class 8 identity for the other ⇒ the numbers.
Changing the sides by different amounts, as in Exercise 5 Q4, gives two linear equations instead.
Products near round numbers: write each factor as (round ± small) and use the four-term identity.

Next: Real-world applications – mental maths, borders, percentages and number tricks ▶