CBSE Class 8 Mathematics Algebra Play Short Answer Questions with Answers

Short Answer Questions

Question 1. A ‘Think of a Number’ trick has these steps: think of a number, triple it, add 9, divide by 3, and finally subtract the number you first thought of. Taking the starting number as x, write the expression obtained after each step and state the value the trick always ends with.
Answer:
Think of a number: x
Triple it: 3x
Add 9: 3x + 9
Divide by 3: x + 3
Subtract the original number: (x + 3) − x = 3
The letter-number x cancels in the last step, so the trick always ends at 3.
In simple words: Tripling and then dividing by 3 undoes itself, so only the 9 ÷ 3 = 3 is left.

Teacher's Note:
1. Divide every term when dividing an expression: (3x + 9) ÷ 3 = x + 3.
2. The answer is the added number divided by the multiplier.

Question 2. Priya thinks of the date 15/08. She multiplies the month by 5, adds 6, multiplies by 4, adds 9, multiplies by 5, and finally adds the day. Show all the steps and write the number Priya announces.
Answer:
Here the month is M = 8 and the day is D = 15.
Multiply the month by 5: 8 × 5 = 40
Add 6: 46
Multiply by 4: 184
Add 9: 193
Multiply by 5: 965
Add the day: 965 + 15 = 980
Priya announces 980. Check: 100M + 165 + D = 800 + 165 + 15 = 980.
In simple words: Following the steps with month 8 and day 15 gives 980.

Teacher's Note:
1. 15/08 means day 15 of month 8.
2. Use 100M + 165 + D as a check.

Question 3. In a number pyramid every brick holds the sum of the two bricks directly below it. The bottom row of a three-row pyramid is 7, 9, 4.
(i) Write the two numbers in the middle row.
(ii) Write the number at the top.
Answer:
(i) Middle row: 7 + 9 = 16 and 9 + 4 = 13.
(ii) Top brick: 16 + 13 = 29.
The rule a + 2b + c gives the same: 7 + 18 + 4 = 29. The middle number is counted twice because it feeds both bricks of the middle row.
In simple words: Add the neighbouring pairs to get 16 and 13, then add those to get 29.

Teacher's Note:
1. Check the top with a + 2b + c.
2. Keep the bricks in order from left to right.

Question 4. Devansh picks a 2 × 2 grid of dates from the August 2025 calendar page and tells his sister that the four dates add up to 56. Taking the top-left date as a, write the sum in terms of a and find the four dates.
Answer:
The dates are a, a + 1 in one week and a + 7, a + 8 in the next.
Sum = a + (a + 1) + (a + 7) + (a + 8) = 4a + 16.
4a + 16 = 56 gives 4a = 40 and a = 10.
The four dates are 10, 11, 17 and 18.
In simple words: The sum is 4a + 16, so 4a + 16 = 56 gives a = 10 and the dates 10, 11, 17, 18.

Teacher's Note:
1. Check: 10 + 11 + 17 + 18 = 56.
2. On the August 2025 page, 10 and 11 are Sunday and Monday.

Teacher's Note:
1. Start with the row that has only one kind of shape.
2. Substitute back to check every row.

Question 6. The digits 2, 4 and 7 are to be written in the boxes of □□ × □, using each digit once.
(i) Which arrangement gives the largest product, and what is that product?
(ii) Which digit is used as the multiplier?
Answer:
(i) The largest product is 42 × 7 = 294.
(ii) The multiplier is 7, the largest of the three digits.
The other two digits, 4 and 2, are written in decreasing order. The next best arrangement, 72 × 4, gives only 288, and the smallest, 47 × 2, gives 94.
In simple words: Put the biggest digit 7 as the multiplier and make 42 from the rest: 42 × 7 = 294.

Teacher's Note:
1. Compare 42 × 7 = 294 with 72 × 4 = 288.
2. The rule: largest digit multiplies, others in decreasing order.

Question 7. Start with the two-digit number 83, reverse its digits and subtract the smaller number from the larger one. Divide the result by 9 and write the quotient. What does this quotient tell you about the two digits of 83?
Answer:
Reversing 83 gives 38.
Difference: 83 − 38 = 45
Divide by 9: 45 ÷ 9 = 5
The quotient 5 is exactly the difference of the two digits, 8 − 3 = 5. This happens because the difference always equals 9(a − b), where a and b are the digits.
In simple words: 83 − 38 = 45 and 45 ÷ 9 = 5, which is the gap between the digits 8 and 3.

Teacher's Note:
1. The difference of a number and its reverse is 9 × (digit difference).
2. Try another number to see the same pattern.

Question 8. Rahul tries a number trick with four different starting numbers, gets the same answer every time and declares, “This trick works for every number.” Give two reasons why algebra is needed before such a claim can be accepted.
Answer:
Reason 1: There are endlessly many numbers, so testing four of them cannot show that the trick works for the fifth or the five-hundredth. Examples can only suggest a pattern, never settle it.
Reason 2: Algebra uses one letter-number for every possible starting number at once, so a single piece of working covers all cases and also shows why the starting number cancels out.
In simple words: A few examples cannot cover all numbers, but one algebra working covers every number and shows why.

Teacher's Note:
1. Checking examples convinces; algebra proves.
2. This is what the chapter means by justifying a statement.

Question 9. For the trick “think of a number, double it, add 4, divide by 2, subtract the original number”, the answer is always 2.
(i) Change exactly one step so that the answer is always 5.
(ii) Justify your new trick using algebra.
Answer:
(i) Replace “add 4” by “add 10”. The steps are: think of a number, double it, add 10, divide by 2, subtract the original number.
(ii) With starting number x: x → 2x → 2x + 10 → x + 5 → (x + 5) − x = 5.
Since x cancels, the answer is always 5. In general, adding 2k makes the trick end at k.
In simple words: Adding 10 instead of 4 makes the halved part 5, so the trick always ends at 5.

Teacher's Note:
1. The number added is halved later, so add double the wanted answer.
2. Test it with a number such as 3: 6, 16, 8, 5.

Question 10. In the date trick, the steps turn a date into the number 100M + 165 + D, where M is the month and D is the day. Two friends announce 869 and 1295. Find the date each friend had thought of.
Answer:
Subtract 165 and read the last two digits as the day.
869 − 165 = 704, so M = 7 and D = 04: the date is 4 July.
1295 − 165 = 1130, so M = 11 and D = 30: the date is 30 November.
The day is never more than 31, so it always fills exactly the last two digits, and the rest is the month.
In simple words: Take away 165 from each number; 704 means 4 July and 1130 means 30 November.

Teacher's Note:
1. Read 04 as day 4.
2. Check: 700 + 165 + 4 = 869.

Question 11. In a three-row number pyramid the top brick is 54, the left brick of the middle row is 25 and the last brick of the bottom row is 9. Taking the middle brick of the bottom row as c, form an equation and find every missing number of the pyramid.
Answer:
Let the bottom row be a, c, 9 and the middle row be 25 and b.
25 + b = 54, so b = 29.
c + 9 = b, that is c + 9 = 29, so c = 20.
a + c = 25, so a = 25 − 20 = 5.
The pyramid is: bottom 5, 20, 9; middle 25, 29; top 54. Check: a + 2c + 9 = 5 + 40 + 9 = 54.
In simple words: The right middle brick is 54 − 25 = 29, so the middle bottom is 20 and the left bottom is 5.

Teacher's Note:
1. Here the middle bottom brick is called c, so the top is a + 2c + 9.
2. Work downwards by subtraction.

Question 12. A number pyramid has four rows and its bottom row is a, b, c, d.
(i) Write the expression for the number at the top in terms of a, b, c, d.
(ii) Use it to find the top number when the bottom row is 2, 5, 3, 6, without building the pyramid.
Answer:
(i) Second row: a + b, b + c, c + d; third row: a + 2b + c, b + 2c + d; so the top is a + 3b + 3c + d.
(ii) With a = 2, b = 5, c = 3, d = 6: top = 2 + 15 + 9 + 6 = 32.
The two inner numbers are multiplied by 3 because more paths lead from them to the top.
In simple words: The top of a four-row pyramid is a + 3b + 3c + d, which gives 32 here.

Teacher's Note:
1. Check by building: 7, 8, 9; 15, 17; 32.
2. The coefficients 1, 3, 3, 1 are symmetric.

Question 13. From a calendar page, three dates lying one below the other in the same column are chosen (for example 4, 11, 18). Taking the first date as a, show by algebra that their sum is always three times the middle date, and find the three dates if the sum is 54.
Answer:
Dates in a column differ by 7, so the three dates are a, a + 7 and a + 14.
Sum = a + (a + 7) + (a + 14) = 3a + 21 = 3(a + 7).
a + 7 is the middle date, so the sum is always three times the middle date.
If the sum is 54, the middle date is 54 ÷ 3 = 18, so the dates are 11, 18 and 25.
In simple words: The three dates are a, a + 7, a + 14, whose sum is 3(a + 7), three times the middle one; for 54 they are 11, 18, 25.

Teacher's Note:
1. Factorise 3a + 21 to see the middle date.
2. Check: 11 + 18 + 25 = 54.

Teacher's Note:
1. Compare rows that differ by a single shape.
2. Then substitute one value at a time.

Question 15. Using the digits 1, 6 and 8 once each in □□ × □, Ira writes 81 × 6 and Jaya writes 61 × 8. Expand both products as a sum of two terms and explain which arrangement is larger and why.
Answer:
81 × 6 = (8 × 10 + 1) × 6 = 480 + 6 = 486
61 × 8 = (6 × 10 + 1) × 8 = 480 + 8 = 488
The first terms are equal, since both are 10 × 6 × 8. Only the second terms differ, and 1 × 8 > 1 × 6.
So Jaya’s arrangement 61 × 8 is larger, by 2.
In simple words: Both products share 480; Jaya adds 1 × 8 while Ira adds only 1 × 6, so 61 × 8 wins.

Teacher's Note:
1. Splitting the two-digit number by place value shows the comparison clearly.
2. The larger digit should be the multiplier.

Question 16. Take the three-digit number 254. Cycle its digits to get 542 and 425, and add all three numbers.
(i) Write the total and divide it by 37.
(ii) Using letter-numbers for the digits, explain why such a total is always divisible by 37 and by 3.
Answer:
(i) 254 + 542 + 425 = 1221, and 1221 ÷ 37 = 33.
(ii) With digits a, b, c the three numbers are 100a + 10b + c, 100b + 10c + a and 100c + 10a + b.
Each digit appears once in the hundreds, tens and units place, so the total is 111(a + b + c) = 3 × 37 × (a + b + c).
Hence 37 and 3 always divide the total exactly.
In simple words: The total is always 111 times the digit sum, and 111 = 3 × 37.

Teacher's Note:
1. For 254, 111 × (2 + 5 + 4) = 111 × 11 = 1221.
2. The quotient on dividing by 37 is 3 × (digit sum).

Question 17. A six-digit number is made by writing a three-digit number twice, for example 538 gives 538538. Using letter-numbers, explain why such a number is always divisible by 7, by 11 and by 13, and state what is left after dividing 538538 by all three.
Answer:
Let the three-digit number be n. Writing its digits twice gives n × 1000 + n = n × 1001.
Now 1001 = 7 × 11 × 13, so the six-digit number is n × 7 × 11 × 13 and each of 7, 11 and 13 divides it exactly.
For 538538: 538538 ÷ 7 = 76934, 76934 ÷ 11 = 6994, 6994 ÷ 13 = 538. What is left is the original number 538.
In simple words: Writing a number twice multiplies it by 1001 = 7 × 11 × 13, so dividing by these three gives the number back.

Teacher's Note:
1. The first copy is shifted three places, that is, multiplied by 1000.
2. The order of the divisions does not matter.

Question 18. The Virahāṅka–Fibonacci numbers 1, 2, 3, 5, 8 are written in the bottom row of a five-row number pyramid.
(i) Write the expression for the top of a five-row pyramid in terms of the bottom row a, b, c, d, e.
(ii) Use it to find the top number and say what you notice about it.
Answer:
(i) Working upwards from a, b, c, d, e, the top is a + 4b + 6c + 4d + e.
(ii) With 1, 2, 3, 5, 8: top = 1 + 8 + 18 + 20 + 8 = 55.
55 is itself a Virahāṅka–Fibonacci number. In fact every brick (3, 5, 8, 13; 8, 13, 21; 21, 34; 55) belongs to the sequence, because adding two neighbouring terms gives the next term.
In simple words: The top is a + 4b + 6c + 4d + e = 55, and 55 is again a Virahāṅka–Fibonacci number.

Teacher's Note:
1. The coefficients 1, 4, 6, 4, 1 continue the pattern 1, 3, 3, 1.
2. Build the pyramid once to check.

Question 19. A father’s present age is 4 times his son’s present age. Ten years from now the father will be only twice as old as his son. Form an equation using a letter-number and find the present age of each.
Answer:
Let the son’s present age be x years, so the father is 4x years.
After 10 years: son x + 10, father 4x + 10.
Given: 4x + 10 = 2(x + 10)
4x + 10 = 2x + 20, so 2x = 10 and x = 5.
The son is 5 years old and the father is 20 years old. Check: in 10 years they will be 15 and 30, and 30 = 2 × 15.
In simple words: Taking the son as x, 4x + 10 = 2(x + 10) gives x = 5, so the ages are 5 and 20.

Teacher's Note:
1. Both ages increase by 10.
2. Put brackets around (x + 10) before doubling.

Question 20. Evaluate \( \frac{1+3+5+7}{9+11+13+15} \) and explain, using the sum of the first n odd numbers, why a fraction built in this way always has the same value.
Answer:
\( \frac{1+3+5+7}{9+11+13+15} \) = \( \frac{16}{48} \) = \( \frac{1}{3} \).
The sum of the first n odd numbers is n². So the numerator is n², while the numerator and denominator together are the first 2n odd numbers, adding to (2n)² = 4n².
The denominator is therefore 4n² − n² = 3n², and the fraction is always \( \frac{n^2}{3n^2} \) = \( \frac{1}{3} \).
In simple words: The top is n² and the bottom is 3n², so the fraction always equals one-third.

Teacher's Note:
1. Here n = 4: numerator 16 = 4² and denominator 48 = 3 × 4².
2. This is a neat use of letter-numbers to prove a pattern.