CBSE Class 8 Mathematics Algebra Play Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. State the rule by which the boxes of a number pyramid are filled.
Answer:
Each box holds the sum of the two numbers written directly below it, so the pyramid is built upwards from the bottom row.

Teacher's Note:
1. Going down, the rule becomes subtraction.
2. Only the two touching boxes below are added.

Question 2. A friend thinks of a number x, doubles it and then adds 4. Write the expression she now has.
Answer:
Doubling gives 2x; adding 4 gives 2x + 4.

Teacher's Note:
1. Write each step as an expression in turn.
2. Do not write 2(x + 4); the 4 is added after doubling.

Question 3. In the trick “think of a number, double it, add 4, divide by 2, subtract the number you thought of”, why is the answer the same for every starting number?
Answer:
In algebra the steps end with (x + 2) − x = 2. The letter-number x cancels out, so the answer is 2 whatever number is chosen.

Teacher's Note:
1. If the starting number disappears, the answer cannot depend on it.
2. This is the key idea behind every such trick.

Question 4. The bottom row of a three-row number pyramid is 7, 5, 2. Write the two numbers of its middle row.
Answer:
7 + 5 = 12 and 5 + 2 = 7, so the middle row is 12 and 7.

Teacher's Note:
1. Add each pair of neighbours.
2. The top would then be 12 + 7 = 19.

Question 5. Write the expression for the number at the top of a three-row pyramid whose bottom row is a, b, c.
Answer:
Middle row: a + b and b + c. Top: (a + b) + (b + c) = a + 2b + c.

Teacher's Note:
1. The middle number feeds both middle bricks, so it is counted twice.
2. Remember the coefficients 1, 2, 1.

Question 6. The top-left number of a 2 × 2 block taken from a calendar page is a. Write the other three numbers of the block in terms of a.
Answer:
a + 1 (to its right), a + 7 (just below it) and a + 8 (diagonally below right).

Teacher's Note:
1. One step right adds 1; one row down adds 7.
2. Diagonally down-right adds 1 + 7 = 8.

Question 7. Write, in terms of a, the sum of the four numbers of a 2 × 2 block of a calendar page whose top-left number is a.
Answer:
Sum = a + (a + 1) + (a + 7) + (a + 8) = 4a + 16.

Teacher's Note:
1. Add the letter terms and the number terms separately.
2. 4a + 16 = 4(a + 4), always a multiple of 4.

Question 8. Using the digits 2, 3 and 8 once each, write the arrangement of a two-digit number times a one-digit number that gives the largest product, and give that product.
Answer:
Use the largest digit 8 as the multiplier and write 3, 2 in decreasing order: 32 × 8 = 256.

Teacher's Note:
1. The next best, 82 × 3 = 246, is smaller.
2. Biggest digit multiplies; the others go in decreasing order.

Question 9. Reverse the digits of 62 and subtract the smaller number from the larger one. Is the difference divisible by 9?
Answer:
Reversing 62 gives 26, and 62 − 26 = 36. Yes, 36 = 9 × 4.

Teacher's Note:
1. The quotient 4 is the difference of the digits, 6 − 2.
2. This always happens with reverse-and-subtract.

Question 10. Add 45 and the number obtained by reversing its digits. Which number of the chapter’s trick divides this sum?
Answer:
45 + 54 = 99, which is divisible by 11 (99 = 11 × 9).

Teacher's Note:
1. A number plus its reverse is 11 × (sum of digits).
2. Here 4 + 5 = 9, so the sum is 11 × 9.

Question 11. Write 1001 as a product of three prime numbers.
Answer:
1001 = 7 × 11 × 13.

Teacher's Note:
1. Check: 7 × 11 = 77 and 77 × 13 = 1001.
2. This explains the repeating-number trick.

Question 12. According to the Summary of this chapter, algebra is an indispensable tool for doing what?
Answer:
For justifying mathematical statements, that is, showing why a claim is true for every number.

Teacher's Note:
1. Examples can only suggest a pattern.
2. Algebra proves it for all numbers at once.

Question 13. A friend follows the date trick of the chapter and tells you the answer 870. Which date did she choose?
Answer:
870 − 165 = 705 = 100M + D, so M = 7 and D = 05. The date was 5 July.

Teacher's Note:
1. Always subtract 165 first.
2. The last two digits give the day.

Question 14. In the trick “double it, add 4, divide by 2, subtract the number you thought of”, what number should be added in place of 4 so that the answer is always 5?
Answer:
Add 10: (2x + 10) ÷ 2 − x = (x + 5) − x = 5. In general, adding 2k makes the answer k.

Teacher's Note:
1. The added number is halved, so add twice the wanted answer.
2. Test with any number to be sure.

Question 15. A three-row number pyramid has 48 at the top, and the two end numbers of its bottom row are 9 and 11. Find the middle number of the bottom row.
Answer:
Let the middle number be b. Then 9 + 2b + 11 = 48, so 2b = 28 and b = 14.
Check: 9 + 14 = 23, 14 + 11 = 25 and 23 + 25 = 48.

Teacher's Note:
1. Use the formula top = a + 2b + c.
2. A quick check confirms the answer.

Question 16. Using algebra, show that the sum of a two-digit number and the number obtained by reversing its digits is a multiple of 11.
Answer:
(10a + b) + (10b + a) = 11a + 11b = 11(a + b), which is a multiple of 11.

Teacher's Note:
1. Write each number in expanded form.
2. The quotient is the sum of the digits.

Question 17. The four numbers of a 2 × 2 block of a calendar page add up to 100. Write the four numbers.
Answer:
4a + 16 = 100 gives 4a = 84 and a = 21. The numbers are 21, 22, 28 and 29.

Teacher's Note:
1. Check: 21 + 22 + 28 + 29 = 100.
2. The top-left number is a.

Question 18. Using the digits 2, 6 and 7 once each, find the largest product of a two-digit number and a one-digit number.
Answer:
Multiplier 7, and 6, 2 in decreasing order: 62 × 7 = 434.

Teacher's Note:
1. The next best, 72 × 6 = 432, is just smaller.
2. Follow the rule rather than guessing.

Question 19. In the reverse-and-subtract trick, what is the quotient when the difference is divided by 9? Answer in terms of the digits.
Answer:
The difference is 9(b − a), so the quotient is b − a, the difference of the two digits.

Teacher's Note:
1. For 83: 83 − 38 = 45 and 45 ÷ 9 = 5 = 8 − 3.
2. Larger digit minus smaller digit.

Question 20. The first three Virahanka-Fibonacci numbers 1, 2, 3 are written in the bottom row of a three-row pyramid. What number appears at the top, and is it a Virahanka-Fibonacci number?
Answer:
Middle row: 1 + 2 = 3 and 2 + 3 = 5. Top: 3 + 5 = 8. Yes, 8 is the next Virahanka-Fibonacci number.

Teacher's Note:
1. Every brick (3, 5, 8) belongs to the sequence.
2. The pyramid rule matches the sequence rule.

Question 21. Write the expression for the number at the top of a four-row pyramid whose bottom row is a, b, c, d.
Answer:
Top = a + 3b + 3c + d.

Teacher's Note:
1. The coefficients 1, 3, 3, 1 are symmetric.
2. Inner bricks reach the top along more paths.

Question 22. A mother is 5 times as old as her daughter. In 6 years’ time she will be 3 times as old as her daughter. Find the daughter’s present age.
Answer:
Let the daughter be d years old. Then 5d + 6 = 3(d + 6), so 5d + 6 = 3d + 18, 2d = 12 and d = 6 years (the mother is 30).

Teacher's Note:
1. Add 6 to both ages, not just one.
2. Check: in 6 years, 36 = 3 × 12.

Question 23. Naina has twice as many cows as Gauri. If Naina gives 3 of her cows to Gauri, both will have the same number. How many cows does each have?
Answer:
Let Gauri have g cows; Naina has 2g. Then 2g − 3 = g + 3, so g = 6. Gauri has 6 cows and Naina has 12.

Teacher's Note:
1. Giving away 3 lowers one count and raises the other.
2. Check: 12 − 3 = 9 = 6 + 3.

Question 24. Evaluate \( \frac{1+3+5}{7+9+11} \) and state what you notice about the answer.
Answer:
\( \frac{1+3+5}{7+9+11} \) = \( \frac{9}{27} \) = \( \frac{1}{3} \).
Every fraction of this pattern equals \( \frac{1}{3} \), because the first n odd numbers add to n², and the next n odd numbers add to 3n².

Teacher's Note:
1. The sum of the first n odd numbers is n².
2. Here 1 + 3 + 5 = 3² and the first six odd numbers add to 6² = 36.

Question 25. Explain why exactly 165 has to be subtracted from the final answer of the date trick, and how the month and the day are then read off.
Answer:
The steps turn the date into 100M + 165 + D, where 165 is a fixed number added by the steps themselves. Subtracting 165 leaves 100M + D.
As D is at most 31, the last two digits give the day and the digits in front of them give the month.

Teacher's Note:
1. 165 = (6 × 4 + 9) × 5.
2. Changing a step changes this fixed number.

Question 26. In the reverse-and-subtract trick, the two-digit number is ab with a > b. Using algebra, show that the difference is still a multiple of 9.
Answer:
The number is 10a + b and its reverse is 10b + a. As a > b, the difference is (10a + b) − (10b + a) = 9a − 9b = 9(a − b), again a multiple of 9.

Teacher's Note:
1. Always subtract the smaller number from the larger.
2. The quotient is the difference of the digits.

Question 27. Using algebra, justify that abc + bca + cab is always divisible by 37 for any three-digit number abc.
Answer:
abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111a + 111b + 111c = 111(a + b + c).
Since 111 = 3 × 37, the sum is always divisible by 37 (and by 3).

Teacher's Note:
1. Each digit appears once in each place.
2. Example: 254 + 542 + 425 = 1221 = 37 × 33.

Question 28. Explain with algebra why dividing the six-digit number abcabc by 7, then by 11, and then by 13 always returns the original three-digit number abc.
Answer:
abcabc = abc × 1000 + abc = abc × 1001, and 1001 = 7 × 11 × 13.
Dividing by 7, 11 and 13 one after another removes the factor 1001 exactly, leaving abc.

Teacher's Note:
1. Writing the digits again shifts one copy three places left.
2. Example: 538538 ÷ 1001 = 538.

Question 29. A 3 × 3 block of dates is taken from a calendar page. Using algebra, show that the sum of the nine numbers is 9 times the number at the centre of the block.
Answer:
Let the centre be m. The nine dates are m − 8, m − 7, m − 6, m − 1, m, m + 1, m + 6, m + 7, m + 8.
The added and subtracted parts cancel in pairs, so the sum is 9m.

Teacher's Note:
1. Pair each number with the one opposite it through the centre.
2. Each pair adds to 2m.

Question 30. Using algebra, show that the four numbers of a 2 × 2 block of a calendar page can never add up to 50.
Answer:
The numbers are a, a + 1, a + 7, a + 8, and their sum is 4a + 16 = 4(a + 4), always a multiple of 4.
50 is not a multiple of 4, so the sum can never be 50.

Teacher's Note:
1. 4a + 16 = 50 would give a = 8.5, not a date.
2. The multiple-of-4 test is quicker.