CBSE Class 8 Mathematics Part I Chapter 5 Questions and Answers

Class 8 Maths Ganita Prakash

Class 8 Maths Ganita Prakash Chapter 5 Number Play NCERT Solutions

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Textbook Page 113

Q. Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and ‘–’ signs in between the numbers. How many different possibilities exist? Write all of them.

Eight such expressions are possible. You can use the diagram below to systematically list all the possibilities.

Evaluate each expression and write the result next to it. Do you notice anything interesting?

Solution:

1. 3 + 4 + 5 + 6 = 18
2. 3 + 4 + 5 − 6 = 6
3. 3 + 4 − 5 + 6 = 8
4. 3 + 4 − 5 − 6 = −4
5. 3 − 4 + 5 + 6 = 10
6. 3 − 4 + 5 − 6 = −2
7. 3 − 4 − 5 + 6 = 0
8. 3 − 4 − 5 − 6 = −12

Observations:

• (i) The results of all expressions are even numbers.
• (ii) The results can be positive, negative, or zero.

Q. Now, take four other consecutive numbers. Place the ‘+’ and ‘–’ signs as you have done before. Find out the results of each expression. What do you observe?

Solution:
Let the four consecutive numbers be 7, 8, 9, 10.

1. 7 + 8 + 9 + 10 = 34
2. 7 + 8 + 9 − 10 = 14
3. 7 + 8 − 9 + 10 = 16
4. 7 + 8 − 9 − 10 = −4
5. 7 − 8 + 9 + 10 = 18
6. 7 − 8 + 9 − 10 = −2
7. 7 − 8 − 9 + 10 = 0
8. 7 − 8 − 9 − 10 = −2

Observations:
The result of every expression is an even number. The results may be positive, negative, or zero, but they are always even.

Q. Repeat this for one more set of 4 consecutive numbers. Share your findings.

Solution:
Let the four consecutive numbers be 12, 13, 14, 15.

1. 12 + 13 + 14 + 15 = 54
2. 12 + 13 + 14 − 15 = 24
3. 12 + 13 − 14 + 15 = 26
4. 12 + 13 − 14 − 15 = −4
5. 12 − 13 + 14 + 15 = 28
6. 12 − 13 + 14 − 15 = −2
7. 12 − 13 − 14 + 15 = 0
8. 12 − 13 − 14 − 15 = −30

Findings:
When four consecutive numbers are chosen, no matter how the ‘+’ and ‘–’ signs are placed between them, the resulting expressions are always even.

Textbook Page 114

Q. Replace any negative sign in the expression a + b – c – d with a positive sign and find the difference between the two numbers.

Solution:
Given expression: a + b – c – d
Replacing –c by c, we get: a + b + c – d
Difference: a + b – c – d – (a + b + c – d)
= a + b – c – d – a – b – c + d
= –2c.

Q. What do you conclude from this observation?

Solution:
The difference between the two numbers is even. So either both are even, or both are odd.

Textbook Page 115

Q. We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even.

Solution:

• (i) 43 + 37: 43 = odd, 37 = odd. Since odd + odd = even ∴ 43 + 37 = even.
• (ii) 672 – 348: 672 = even, 348 = even. Since even – even = even ∴ 672 – 348 = even.
• (iii) 4 × 347 × 3: 4 = even. Since even × anything = even ∴ 4 × 347 × 3 = even.
• (iv) 708 – 477: 708 = even, 477 = odd. Since even – odd = odd ∴ 708 – 477 = odd.
• (v) 809 + 214: 809 = odd, 214 = even. Since odd + even = odd ∴ 809 + 214 = odd.
• (vi) 119 × 303: 119 = odd, 303 = odd. Since odd × odd = odd ∴ 119 × 303 = odd.
• (vii) 543 – 479: 543 = odd, 479 = odd. Since odd – odd = even ∴ 543 – 479 = even.
• (viii) 513³: 513 = odd. odd³ = odd × odd × odd = odd ∴ 513³ = odd.

Q. Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.

Solution:

• (i) 2a + 2b: 2a = even, 2b = even. Since even + even = even ∴ 2a + 2b is always even.
• (ii) 3g + 5h: If g and h are odd, 3g and 5h are odd. If g and h are even, 3g and 5h are even. ∴ 3g + 5h can be even or odd depending upon g and h.
• (iii) 4m + 2n: 4m = even, 2n = even. Since even + even = even ∴ 4m + 2n is always even.
• (iv) 2u − 4v: 2u = even, 4v = even. Since even – even = even ∴ 2u − 4v is always even.
• (v) 13k − 5k: 13k – 5k = 8k. Since 8k is a multiple of 2 ∴ 13k − 5k is always even.
• (vi) 6m − 3n: 6m = even. If n is odd, 3n is odd, making the result odd. ∴ 6m − 3n can be even or odd depending on n.
• (vii) x² + 2: If x is odd, x² + 2 is odd. ∴ x² + 2 can be even or odd depending on x.
• (viii) b² + 1: If b is even, b² + 1 is odd. ∴ b² + 1 can be even or odd depending on b.
• (ix) 4k × 3j: 4k × 3j = 12kj. Since 12kj is a multiple of 2 ∴ 4k × 3j is always even.

Textbook Page 116

Q. Write a few algebraic expressions which always give an even number.

Solution:
Examples of such expressions:

1. 2n (multiple of 2 = even)
2. 2n + 6 (even + even = even)
3. 4x − 10 (even – even = even)
4. 8y × 3z (even × anything = even)
5. 8a + 6b + 4c (sum of multiples of 2 = even)
6. 6m² (even × anything = even)
7. 2(p² + q²) (multiple of 2 = even)

Figure it Out – Page 122

1. The sum of four consecutive numbers is 34. What are these numbers?

Solution:
Let the four consecutive numbers be x, (x + 1), (x + 2) and (x + 3).
x + (x + 1) + (x + 2) + (x + 3) = 34
4x + 6 = 34
4x = 34 – 6
4x = 28
x = 28 / 4 = 7
Then:
x + 1 = 8
x + 2 = 9
x + 3 = 10
Therefore, the four consecutive numbers are 7, 8, 9, and 10.

2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

Solution:
If p is the greatest of five consecutive numbers, the other four numbers are obtained by subtracting 1, 2, 3, and 4 from p.
Hence, the other four numbers in terms of p are: (p – 1), (p – 2), (p – 3), and (p – 4).

3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

• (i) The sum of two even numbers is a multiple of 3.
  Solution: Sometimes true.
  Let the two even numbers be 2a and 2b. 2a + 2b = 2(a + b). The sum is even, but is a multiple of 3 only when (a + b) is a multiple of 3.
  Example: 2 + 4 = 6 (multiple of 3). Non-example: 2 + 8 = 10 (not a multiple of 3).
• (ii) If a number is not divisible by 18, then it is also not divisible by 9.
  Solution: Sometimes true.
  If a number is divisible by 18, it is also divisible by 9. However, the converse is not always true.
  Example: 45 is divisible by 9 but not by 18. 40 is neither divisible by 9 nor by 18.
• (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
  Solution: Sometimes true.
  Example: 10 and 8 are not divisible by 6, but their sum 10 + 8 = 18 is divisible by 6.
  Non-example: 15 and 10 are not divisible by 6, and their sum 15 + 10 = 25 is also not divisible by 6.
• (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
  Solution: Always true.
  Let the numbers be 6a and 9b. 6a + 9b = 3(2a + 3b). Since 2a + 3b is an integer, the sum is always a multiple of 3.
  Examples: 6 + 9 = 15; 12 + 18 = 30.
• (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
  Solution: Sometimes true.
  Let the numbers be 6a and 3b. 6a + 3b = 3(2a + b). The sum is always a multiple of 3, but is a multiple of 9 only when (2a + b) is a multiple of 3.
  Example: 6 + 3 = 9 (multiple of 9). Non-example: 12 + 3 = 15 (not a multiple of 9).

4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Solution:
Numbers leaving remainder 2 when divided by 3: 3a + 2.
Numbers leaving remainder 2 when divided by 4: 4b + 2.
Numbers leaving remainder 2 when divided by both 3 and 4 are 2 more than a common multiple of 3 and 4.
Since LCM(3, 4) = 12, the expression describing all such numbers is 12n + 2 (where n = 0, 1, 2, 3, …).
Examples:
12 × 1 + 2 = 14
12 × 2 + 2 = 26
12 × 3 + 2 = 38

5. “I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”

Solution:
The number of pebbles is less than 100.
LCM of 2, 3, and 5 = 30.
Numbers leaving a remainder of 1 when divided by 2, 3, and 5 are of the form 30k + 1.
For k = 1: 30 × 1 + 1 = 31 (not divisible by 7).
For k = 2: 30 × 2 + 1 = 61 (not divisible by 7).
For k = 3: 30 × 3 + 1 = 91 (divisible by 7: 91 ÷ 7 = 13).
Hence, the required number of pebbles is 91.

6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?

Solution:
A number that leaves a remainder of 2 when divided by 6 is of the form 6k + 2.
Let three such numbers be (6a + 2), (6b + 2), (6c + 2).
Sum = (6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6 = 6(a + b + c + 1).
Since a + b + c + 1 is an integer, the sum is divisible by 6.
Therefore, Tathagat’s claim is true.
Examples: 20 + 26 + 32 = 78 (divisible by 6); 2 + 8 + 14 = 24 (divisible by 6).

7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661   (ii) 4779 – 661

Solution:
Given: 661 = 7a + 3 and 4779 = 7b + 5.
(i) 4779 + 661 = (7b + 5) + (7a + 3) = 7a + 7b + 8 = 7(a + b + 1) + 1.
Hence, the remainder is 1.
(ii) 4779 – 661 = (7b + 5) – (7a + 3) = 7(b – a) + 2.
Hence, the remainder is 2.

8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

Solution:
Divided by 3 leaves rem 2: 3a + 2 (1 less than a multiple of 3)
Divided by 4 leaves rem 3: 4b + 3 (1 less than a multiple of 4)
Divided by 5 leaves rem 4: 5c + 4 (1 less than a multiple of 5)
Since each remainder is 1 less than the divisor, the required number is 1 less than a common multiple of 3, 4, and 5.
LCM(3, 4, 5) = 60.
Hence, the smallest such number is 60 − 1 = 59.

Figure It Out – Page 126

1. Find, without dividing, whether the following numbers are divisible by 9: (i) 123, (ii) 405, (iii) 8888, (iv) 93547, (v) 358095

Solution:

• (i) 123: Sum of digits = 1 + 2 + 3 = 6. 6 is not divisible by 9 ∴ 123 is not divisible by 9.
• (ii) 405: Sum of digits = 4 + 0 + 5 = 9. 9 is divisible by 9 ∴ 405 is divisible by 9.
• (iii) 8888: Sum of digits = 8 + 8 + 8 + 8 = 32. 32 is not divisible by 9 ∴ 8888 is not divisible by 9.
• (iv) 93547: Sum of digits = 9 + 3 + 5 + 4 + 7 = 28. 28 is not divisible by 9 ∴ 93547 is not divisible by 9.
• (v) 358095: Sum of digits = 3 + 5 + 8 + 0 + 9 + 5 = 30. 30 is not divisible by 9 ∴ 358095 is not divisible by 9.

2. Find the smallest multiple of 9 with no odd digits.

Solution:
A number is divisible by 9 if its digit sum is a multiple of 9. Using only even digits (0, 2, 4, 6, 8), the smallest number whose digit sum is a multiple of 9 (sum = 18) is 288 (2 + 8 + 8 = 18).

3. Find the multiple of 9 that is closest to the number 6000.

Solution:
6000 ÷ 9 = 666.66…
Nearest multiples: 666 × 9 = 5994 and 667 × 9 = 6003.
Difference: 6000 − 5994 = 6; 6003 − 6000 = 3.
Hence, the multiple of 9 closest to 6000 is 6003.

4. How many multiples of 9 are there between the numbers 4300 and 4400?

Solution:
First multiple of 9 greater than 4300 is 4302.
Last multiple of 9 less than 4400 is 4392.
Number of multiples = [(4392 − 4302) / 9] + 1 = (90 / 9) + 1 = 10 + 1 = 11.

Figure It Out – Page 131

1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

Solution:
Digital root of the number = 5.
Digital root of 10 = 1 + 0 = 1.
Digital root of (number + 10) = 5 + 1 = 6.

2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

Solution:
Starting with 25, sequence: 25, 36, 47, 58, 69, 80, 91, 102, 113, 124, 135 …
Corresponding digital roots: 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, 9 …
Observation: The digital root repeats after every 9 steps.

3. What will be the digital root of the number 9a + 36b + 13?

Solution:
Digital root of 9a = 9.
Digital root of 36b = 9.
Digital root of 13 = 1 + 3 = 4.
Sum = 9 + 9 + 4 = 22 → 2 + 2 = 4.

4. Make conjectures by examining if there are any patterns or relations between:
(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

Solution:
Consider 14, 21, 38, 45, 52:
(i) 14 (even) → root 5 (odd); 38 (even) → root 2 (even).
Conjecture: The parity of a number and its digital root need not be the same.
(ii) Conjecture: The remainder obtained when a number is divided by 3 (or 9) is the same as the remainder obtained when its digital root is divided by 3 (or 9). (When the digital root is 9, the remainder on division by 9 is 0.)

Figure It Out – Page 132

1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.

Solution:
Sum of digits = 3 + 1 + z + 5 = 9 + z.
For 31z5 to be divisible by 9, (9 + z) must be a multiple of 9.
If z = 0: 9 + 0 = 9.
If z = 9: 9 + 9 = 18.
Hence, z = 0 or 9. Both 3105 and 3195 yield a digit sum that is a multiple of 9.

2. “I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.

Solution:
First number = 12a + 8; Second number = 12b – 4.
Sum = 12(a + b) + 4.
If a = 1, b = 2: 12(3) + 4 = 40 (multiple of 8).
If a = 2, b = 4: 12(6) + 4 = 76 (not a multiple of 8).
Therefore, Snehal’s claim is incorrect.

3. When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

Solution:
Sum = 3a + 3b = 3(a + b). It will be a multiple of 6 only when (a + b) is even.
If a + b is even, the sum is a multiple of 6.
If a + b is odd, the sum is not a multiple of 6.
Generalisation: The sum of two multiples of 3 is a multiple of 6 when both are even multiples of 3 or both are odd multiples of 3.

4. Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”.
(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Solution:
(i) Yes, true. Reversing digits does not change their sum, so divisibility by 9 is preserved.
(ii) Yes. Any permutation/shuffle of digits keeps the same digit sum, so every such number remains a multiple of 9.

5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

Solution:
Divisible by 18 implies it must be divisible by 2 (so b is even: 0, 2, 4, 6, 8) and 9.
Digit sum = 4 + 8 + a + 2 + 3 + b = 17 + a + b.
17 + a + b must be a multiple of 9 → a + b = 1 or 10.
Possible pairs (a, b): (1, 0), (2, 8), (4, 6), (6, 4), (8, 2).

6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

Solution:
Divisible by 44 implies divisible by 4 and 11.
Divisible by 4 → q8 is a multiple of 4 → q = 0, 2, 4, 6, 8.
Divisible by 11 → (3 + 7 + 8) − (p + q) = 18 − (p + q) is a multiple of 11 → p + q = 7.
Possible pairs (p, q): (7, 0), (5, 2), (3, 4), (1, 6).

7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?

Solution:
Sets: (2, 3, 4), (14, 15, 16), (26, 27, 28), (38, 39, 40)...
All such sets are of the form: 12n + 2, 12n + 3, 12n + 4 (n = 0, 1, 2, …).
They occur periodically every 12 numbers.

8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

Solution:
45,000 is divisible by 4 (ends in 00) and 9 (sum = 9), so 45,000 is divisible by 36.
Adding 36 repeatedly gives the next multiples:
45,036; 45,072; 45,108; 45,144; and 45,180.

9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.

Solution:
Difference between consecutive even numbers is 2.
The five numbers are: 5p − 4, 5p − 2, 5p, 5p + 2, 5p + 4.

10. Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.

Solution:
Consider 210015:
Ends in 5 (divisible by 5) and sum of digits = 9 (divisible by 3), so it is divisible by 15.
Reversed number is 510012: ends in 2 (divisible by 2) and sum of digits = 9 (divisible by 3), so it is divisible by 6.

11. Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.

Solution:
Let any multiple of 11 be 11n. Doubling gives 2 × (11n) = 11 × (2n), which is always a multiple of 11.
Therefore, Deepak’s conjecture is false. Every multiple of 11 remains a multiple of 11 when doubled.

12. Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning.

• (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9: Always true (6a × 3b = 18ab = 9 × 2ab).
• (ii) The sum of three consecutive even numbers will be divisible by 6: Always true (2n + 2n + 2 + 2n + 4 = 6n + 6 = 6(n + 1)).
• (iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6: Always true (Digit sum is identical, and the last digit f remains even).
• (iv) 8(7b – 3) – 4(11b + 1) is a multiple of 12: Sometimes true (Simplifies to 12b − 28 = 12(b − 2) − 4, leaving a remainder of 8 or 4).

13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

Solution:
The sum of three numbers is divisible by 3 if and only if:
(i) All three numbers leave the same remainder when divided by 3, or
(ii) The three numbers leave remainders 0, 1, and 2 (in any order) when divided by 3.

14. Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

Solution:

• (i) Yes, one of two consecutive integers is always even, so the product is always a multiple of 2.
• (ii) Not always a multiple of 6 (e.g., 4 × 5 = 20 is not a multiple of 6, whereas 2 × 3 = 6 is).
• (iii) The product of 4 consecutive integers is always a multiple of 24 (e.g., 2 × 3 × 4 × 5 = 120 = 24 × 5).
• (iv) The product of 5 consecutive integers is always a multiple of 120 (e.g., 1 × 2 × 3 × 4 × 5 = 120).

15. Solve the cryptarithms: (i) EF × E = GGG   (ii) WOW × 5 = MEOW

Solution:
(i) 37 × 3 = 111 → E = 3, F = 7, G = 1.
(ii) 515 × 5 = 2575 → W = 5, O = 1, M = 2, E = 7.

16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

Solution:
Multiples of 32 are a subset of multiples of 8, which in turn are a subset of multiples of 4 (Multiples of 32 ⊆ Multiples of 8 ⊆ Multiples of 4).
Thus, Venn diagram (iv) correctly captures this concentric relationship.