Kerala Syllabus Class 7 Mathematics - Unit 8 Repeated Multiplication - Questions and Answers | Teaching Manual 

Questions and Answers for Class 7 Mathematics - Unit 8 Repeated Multiplication - Study Notes | Text Books Solution STD 7 - Maths: Unit 8 Repeated Multiplication - Questions and Answers | ഗണിതം - ആവർത്തന ഗുണനം - ചോദ്യോത്തരങ്ങൾ 

ഏഴാം ക്ലാസ്സ്‌  Mathematics - Unit 8 Repeated Multiplication എന്ന പാഠം ആസ്പദമാക്കി തയ്യാറാക്കിയ ചോദ്യോത്തരങ്ങള്‍. ഈ അധ്യായത്തിന്റെ Teachers Handbook, Teaching Manual എന്നിവ ഡൗൺലോഡ് ചെയ്യാനുള്ള ലിങ്ക് ചോദ്യോത്തരങ്ങളുടെ അവസാനം നൽകിയിട്ടുണ്ട്. പുതിയ അപ്‌ഡേറ്റുകൾക്കായി ഞങ്ങളുടെ Telegram Channel ൽ ജോയിൻ ചെയ്യുക.

Kerala Syllabus STD 7 Maths: Unit 8 Repeated Multiplication - Textbook Solutions

♦ Textbook Activities (Textbook Page No: 112)
♦ Write each number below either as a power of a single prime or as a product of powers of different primes:
(i) 125    (ii) 72       (iii) 100 
(iv) 250  (v) 3600   (vi) 10800
Answer:
♦ Textbook Activities (Textbook Page No: 115)
Now try these problems:
(1) Calculate the powers below as fractions:
(i) (⅔)²      
(ii) (1½)²
(iii) (⅖)³
(iv) (2½)³
Answer:
(2) Calculate the powers below in decimal form:
(i) (0.5)²
(ii) (1.5)²
(iii) (0.1)³
(iv) (0.01)³
Answer:
(i) (0.5)² = 0.5 x 0.5 = 0.25 (2 digits after the decimal point in the product)
(ii) (1.5)² = 1.5 x 1.5 = 2.25 (2 digits after the decimal point in the product)
(iii) (0.1)³ = 0.1 x 0.1 x 0.1 = 0.001 (3 digits after the decimal point in the product)
(iv) (0.01)³ = 0.01 x 0.01 x 0.01 = 0.000001 (6 digits after the decimal point in the product)

(3) Using 15³ = 3375, calculate the powers below:
(i) (1.5)³
(ii) (0.15)³
(iii) (0.015)³
Answer:
(i) (1.5)³ = 1.5 x 1.5 x 1.5 = 3.375 (3 digits after the decimal point in the product)
(ii) (0.15)³ = 0.15 x 0.15 x 0.15 = 0.003375 (6 digits after the decimal point in the product)
(iii) (0.015)³ = 0.015 x 0.015 x 0.015 = 0.000003375 (9 digits after the decimal point in the product)

♦ Textbook Activities (Textbook Page No: 118, 119)
Now try these problems:
(1) Write each product below as the product of powers of different primes:
(i) 72 × 162 
(ii) 225 × 135 
(iii) 105 × 175 
(iv) 25 × 45 × 75
Answer:
(2) Write the product of the numbers from 1 to 15 as the product of powers of different primes
Answer:
Among the natural numbers from 1 to 15, the prime numbers 2, 3, 5, 7, 11 and 13 can be written as such. The remaining numbers can be factorized as:
4 = 2²
6 = 2 x 3
8 = 2³
9 = 3²
10 = 2 x 5
12 = 2 x 2 x 3 = 2² x 3
14 = 2 x 7
15 = 3 x 5
Then, 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 x 15 
= 2 x 3 x 2² x 5 x (2 x 3) x 7 x 2³ x 3² x (2 x 5) x 11 x (2² x 3) x 13 x (2 x 7) x (3 x 5)
= 2¹¹ x 3⁶ x 5³ x 7² x 11 x 13

(3) Consider the numbers from 1 to 25
(i)  Which of them are divisible by 2, but not by 4?
Answer:
The numbers that are divisible by 2 are the multiples of 2
2, 4, 6, 8, 12, 14, 16, 18, 20, 22, 24
The numbers that are not divisible by 4 are not multiples of 4
They are 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25.
So the numbers that are divisible by 2, but not by are those belonging to both the groups.
They are 2, 6, 10, 14, 18, 22.

(ii)  Which of them are divisible by 4, but not by 8?
Answer:
Numbers that are divisible by 4 are: 4, 8, 12, 16, 20, 24
Numbers that are not divisible by 8 are: 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 25
So the numbers that are divisible by 4,  but not by 8, are: 4, 12, 20

(iii) Which of them are divisible by 8, but not by 16?
Answer:
Numbers that are divisible by 8 are: 8, 16, 24, 
Numbers that are not divisible by 16 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25.
So the numbers that are divisible by 8,  but not by 16, are 8 and 24 

(iv) Which of them are divisible by 16?
Answer: Only 16

(v) What is the highest power of 2 that divides the product of the numbers from 1 to 25 without remainder?
Answer:
Let's write down the product of numbers from 1 to 25
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x14 x 15 x 16 x 17 x 18 x 19 x 20 x 21 x 22 x 23 x 24 x 25
Now, let's prime factorize each of them.
1 x 2 x 3 x (2 x 2) x 5 x (2 x 3) x 7 x (2 x 2 x 2) x (3 x 3) x (2 x 5) x 11 x (2 x 2 x 3) x 13 x (2 x 7) x (3 x 5) x (2 x 2 x 2 x 2) x 17 x (2 x 3 x 3) x 19 x (2 x 2 x 5) x (3 x 7) x (2 x 11) x 23 x (2 x 2 x 2 x 3) x (5 x 5)
It is easy to see that there are twenty-two 2s in this product. Thus, we can write, 1 x 3 x 5 x 7 x 3 x 3 x ......x 5 x 5 x 2²²
Therefore, the highest power of 2 that divides the product of the numbers 1 to 25 without remainder is 2²²

(4) Consider the product of the numbers from 1 to 25
(i) What is the highest power of 5 which divides this product without a remainder?
Answer:
From question 3 (v), we can see that there are six 5s in the product of numbers from 1 to 25. Therefore, the highest power of 5 which divides the product of numbers from 1 to 25 without remainder is 5⁶.

(ii)  And the highest power of 10 dividing this product without remainder?
Answer:
First, let's prime factorize the numbers in the product.
That is,
1 x 2 x 3 x (2 x 2) x 5 x (2 x 3) x 7 x (2 x 2 x 2) x (3 x 3) x (2 x 5) x 11 x (2 x 2 x 3) x 13 x (2 x 7) x (3 x 5) x (2 x 2 x 2 x 2) x 17 x (2 x 3 x 3) x 19 x (2 x 5 x 2) x (3 x 7) x (11 x 2) x 23 x (2 x 2 x 2 x 3) x (5 x 5)
Since 10 = 2 x 5 we need to find the number of 10s that can be formed using 2s and 5s in the above product. Here, there are six 5s in the product, so six 10s can be formed.
That is, 1 x 2 x 3 x 2 x .... (2 x 3) x 10⁶
That is the highest power of 10 dividing this product without remainder is 10⁶ (6th power of 10)

(iii) How many zeros does this product end with?
Answer:
Since 2 x 5 equals 10, we can determine the number of zeroes in the product by examining the number of 10s.
From the prime factorization, we can pair 2 and 5 six times. From question 4 (ii), we saw there are six 10s in the product. Thus, the number of zeroes is 6.
♦ Textbook Activities (Textbook Page No: 122)
(1) Now try these problems:
(i) 512 ÷ 64 
(ii) 3125 ÷ 125 
(iii) 243 ÷ 27 
(iv) 1125 ÷ 45
Answer: 
2. (i) Write half of 2¹⁰ as a power of 2.
Answer:
To find half of 2¹⁰, divide it by 2.

(ii)  Write one-third of 3¹² as a power of 3.
Answer:
To find ⅓rd of 3¹², we need to find 3¹² x ⅓ 
3¹² x ⅓ = 3¹² ÷ 3 = 3¹²⁻¹ = 3¹¹

♦ Textbook Activities (Textbook Page No: 123)
Can’t you simplify the fractions below like this?
(i) ²⁷⁄₂₄₃
(ii) ¹²⁵⁄₃₁₂₅
(iii) ⁴⁸⁄₆₄
(iv) ⁵⁴⁄₈₁
Answer:
♦ Textbook Activities (Textbook Page No: 126)
Now try these problems:
(1) Calculate the products below in head:
(i) 5² x 4²
(ii) 5³ x 6³
(iii) 25³ x 4³
(iv) 125² x 8²
Answer:
(i) 5² x 4² 
 5² x 4² = (5 x 4)² = 20² = 400

(ii) 5³ x 6³
 5³ x 6³ = (5 x 6)³ = 30³ = 27000

(iii) 25³ x 4³
 25³ x 4³ = (25 x 4)³ = (100)³ = 1000000

(iv) 125² x 8²
 125² x 8² = (125 x 8)² = (1000)² = 1000000

(2) Write each number below as a product of powers of different primes:
(i) 15²
(ii) 30³
(iii) 12² × 21²
(iv) 12² × 21³
Answer:
(i) 15² 
 15² = (3 x 5)² = 3² x 5²

(ii) 30³
 30³ = (2 x 3 x 5)³ = 2³ x 3³ x 5³

(iii) 12² × 21²
 12² × 21² = (2 x 2 x 3)² x (3 x 7)²
                = (2²)² x 3² x 3² x 7²
                = 2⁴ x 3⁴ x 7²

(iv) 12² × 21³
 12² × 21³ = (2 x 2 x 3)² x (3 x 7)³
                = (2²)² x 3² x 3³ x 7³
                = 2⁴ x 3⁵ x 7³