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SCERT Class 7 Maths Chapter 8 Solutions Repeated Multiplication
Class 7 Maths Chapter 8 Repeated Multiplication Questions and Answers Kerala State Syllabus
Repeated Multiplication Class 7 Questions and Answers Kerala Syllabus
Page 112
Question 1.
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:
i) 125
125 = 5 × 5 × 5 = 53
ii) 72
72 = 2 × 2 × 2 × 3 × 3
= 23 × 32
iii) 100
100 = 2 × 2 × 5 × 5
= 22 × 52
iv) 250
250 = 2 × 5 × 5 × 5
= 2 × 53
v) 3600
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 24 × 32 × 52
(vi) 10800
10800 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5
= 24 × 33 × 52
Page 115
Question 1.
Calculate the powers below as fractions:
(i) \(\left(\frac{2}{3}\right)^2\)
Answer:
\(\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)=\frac{4}{9}\)
(ii) \(\left(1 \frac{1}{2}\right)^2\)
Answer:
\(\left(1 \frac{1}{2}\right)=\left(\frac{3}{2}\right)\)
= \(\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)=\left(\frac{9}{4}\right\)
(iii) \(\left(\frac{2}{5}\right)^3\)
Answer:
\(\left(\frac{2}{5}\right)\left(\frac{2}{5}\right)\left(\frac{2}{5}\right)\)
= \(\frac{8}{125}\)
(iv) \(\left(2 \frac{1}{2}\right)^3\)
Answer:
\(\left(2 \frac{1}{2}\right)=\left(\frac{5}{2}\right)\)
\(\left(\frac{5}{2}\right)\left(\frac{5}{2}\right)\left(\frac{5}{2}\right)=\frac{125}{8}\)
Question 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 × 0.5
= 0.25
ii) (1.5)2
= (1.5)(1.5)
= 2.25
iii) (0.1)3
= (0.1)(0.1)(0.1)
= 0.001
iv) (0.01)3
= (0.01)(0.01)(0.01)
= 0.000001
Question 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 × 1.5 × 1.5 = 3.375
ii) (0.15)3 = 0.15 × 0.15 × 0.15 = 0.003375
iii) (0.015)3 = 0.015 × 0.015 × 0.015 = 0.000003375
Page 118
Question 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:
i) 72 × 162
72 = 2 × 2 × 2 × 3 × 3 = 23 × 32
162 = 3 × 3 × 3 × 3 × 2 = 34 × 2
72 × 162 = (23 × 32)(34 × 2)
= 24 × 36
ii) 225 × 135
225 = 3 × 3 × 5 × 5 = 32 × 52
135 = 3 × 3 × 3 × 5 = 33 × 51
225 × 135 = (32 × 52)( 33 × 51)
= 35 × 53
iii) 105 × 175
105 = 3 × 5 × 7 = 31 × 51 × 71
175 = 5 × 5 × 7 = 52 × 71
105 × 175 = (31 × 51 × 71)(52 × 71)
= 3 × 53 × 72
iv) 25 × 45 × 75
25 = 51 × 51
45 = 32 × 51
75 =31 × 52
25 × 45 × 75 = (51 × 51)(32 × 51)(31 × 52)
= 55 × 33
Question 2.
Write the product of the numbers from 1 to 15 as the product of powers of different primes.
Answer:
1 (no primes)
2 = 21
3 = 31
4 = 22
5 = 51
6 = 21 × 31
7 = 71
8 = 23
9 = 32
10 = 21 × 51
11 = 111
12 = 22 × 31
13 = 131
14 = 21 × 71
15 = 31 × 51
∴ The product of the numbers from 1 to 15 as the product of powers of different primes are
1 × 211 × 36 × 53 × 72 × 111 × 131
Question 3.
Consider the numbers from 1 to 25
i) Which of them are divisible by 2, but not by 4?
ii) Which of them are divisible by 4, but not by 8?
iii) Which of them are divisible by 8, but not by 16?
iv) Which of them are divisible by 16?
v) What is the highest power of 2 that divides the product of the numbers from 1 to 25 without remainder?
Answer:
i) The numbers from 1 to 25 that are divisible by 2 but not by 4 are:
2, 6, 10, 14, 18, 22
ii) The numbers from 1 to 25 that are divisible by 4 but not by 8 are:
4, 12, 20
iii) The numbers from 1 to 25 that are divisible by 8 but not by 16 are:
8, 24
iv) The numbers from 1 to 25 that are divisible by 16 are:
16
v) Numbers that are divisible by 2 from 1 to 25 are:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Thus, the total numbers that can be divisible by 2 are 12.
Number that are divisible by 4 from 1 to 25 are:
4, 8, 12, 16, 20, 24
Thus, the total number numbers that can be divisible by 4 are 6.
Number that are divisible by 8 from 1 to 25 are:
8, 16, 24
Thus, the total number numbers that can be divisible by 8 are 3.
Number that are divisible by 16 from 1 to 25 are:
16
Thus, the total number numbers that can be divisible by 16 are 1.
From this, the highest power of 2 = 12 + 6 + 3 + 1 = 22
So, the highest power of 2 that divides the product of number from 1 to 25 is 222.
Question 4.
Consider the product of the numbers from 1 to 25
i) What is the highest power of 5 which divides this product without remainder?
ii) And the highest power of 10 dividing this product without remainder?
iii) How many zeros does this product end with?
Answer:
i) The numbers divisible by 5 are 5, 10, 15, 20, 25 that is 5 number.
The number divisible by 25 is 25 that is 1 number.
Thus, the total contribution from multiple of 5 = 5 + 1 = 6
Thus, the highest power of 5 that divides the product between 1 to 25 is 56.
ii) Since 10 = 2 × 5. We need to count how many times both 2 and 5 appear as factors in the product.
Thus, the smallest count between the 2 and 5 will give us the highest powers of 10.
The highest power of 2, which divides the product of numbers from 1 to 25 is 22 and
the highest power of 5, which divides the product of numbers from 1 to 25 is 6.
Thus, the highest power of 10 = smallest count of the highest power of 2 and 5
Therefore, the highest power of 10 dividing the product from 1 to 25 is 106.
iii) Since we found that 106 divides the product of numbers from 1 to 25.
This means that the product ends with 6 zeros.
Page 122
Question 1.
Calculate the following quotients:
i) 512 ÷ 64
ii) 3125 ÷ 125
iii) 243 ÷ 27
iv) 1125 ÷ 45
Answer:
i) 512 ÷ 64
512 = 29 = 26 × 23
64 = 26 = 23 × 23
512 ÷ 64 = ( 26 × 23) ÷ (23 × 23)
= 26 ÷ 23
= 26-3
= 23
= 8
ii) 3125 ÷ 125
3125 = 55 = 52 × 53
125 = 53 = 52 × 51
3125 ÷ 125 = (52 × 53) ÷ (52 × 51)
= 53 ÷ 51
= 53-1
= 52
=25
iii) 243 ÷ 27
243 = 35 = 32 × 33
27 = 33 = 32 × 31
243 ÷ 27 = (32 × 33) – (32 × 31)
= 33 – 31
= 33-1
= 32
= 9
iv) 1125 ÷ 45
1125 = 53 × 32
45 = 51 × 32
1125 ÷ 45 = (53 × 32) ÷ (51 × 32)
= 53-1
= 52
= 25
Question 2.
i) Write half of 210 as a power of 2.
ii) Write one-third of 312 as a power of 3.
Answer:
i) Half of 210 = 210 ÷ 21
= 210-1
= 29
ii) One-third of 312 = 312 ÷ 31
= 311
There is another way for doing this type of problems, and it can be stated as a general principle, using algebra:
\(\frac{x^m}{x^n}=\frac{1}{x^{n-m}}\), for all natural numbers x ≠ 0 and for all natural numbers m < n
For example, let’s factorize the numerator and denominator, and calculate 64 ÷ 512.
Page 123
Question 1.
Can’t you simplify the fractions below like this?
i) \(\frac{27}{243}\)
Answer:
ii) \(\frac{125}{3125}\)
Answer:
iii) \(\frac{48}{64}\)
Answer:
iv) \(\frac{54}{81}\)
Answer:
Page 126
Question 1.
Calculate the products below in head:
i) 52 × 42
ii) 53 × 63
iii) 253 × 43
iv) 1252 × 82
Answer:
i) 52 × 42 = (5 × 4)2
= 202
= 400
ii) 53 × 63 = (5 × 6)3
= 303
= 27000
iii) 253 × 43 = (25 × 4)3
= (100)3
= 1000000
iv) 1252 × 82 = (125 × 8)2
= (1000)2
= 1000000
Question 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 = (3 × 5)3
= 32 × 52
ii) 303 = (2 × 3 × 5)3
= 23 × 33 × 53
iii) 122 × 212
122 = (2 × 2 × 3)2
= 22 × 22 × 32
212 = (7 × 3)2
= 72 × 32
122 × 212 = 22 × 22 × 32 × 72 × 32
= 24 × 34 × 72
iv) 122 × 213
122 = (2 × 2 × 3)2 = 22 × 22 × 32
213 = (7 × 3)3 = 73 × 33
122 × 213 = 22 × 22 × 32 × 73 × 33
= 24 × 35 × 73
Class 7 Maths Chapter 8 Kerala Syllabus Repeated Multiplication Questions and Answers
Question 1.
Write each number below either as a power of a single prime or as a product of powers of different primes.
i) 3125
ii) 200
iii) 1600
Answer:
i) 3125 = 5 × 5 × 5 × 5 × 5 = 5s
ii) 200 = 2 × 2 × 2 × 5 × 5 = 233 × 52
iii) 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 26 × 52
Question 2.
Calculate the powers below as fractions:
i) \(\left(\frac{3}{2}\right)^3\)
Answer:
\(\left(\frac{3}{2}\right)^3=\left(\frac{3}{2}\right) \times\left(\frac{3}{2}\right) \times\left(\frac{3}{2}\right)\)
= \(\left(\frac{27}{8}\right)\)
ii) \(\left(\frac{3}{5}\right)^2\)
Answer:
\(\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}\right) \times\left(\frac{3}{5}\right)\)
= \(\left(\frac{9}{25}\right)\)
iii) \(\left(2 \frac{3}{2}\right)^2\)
Answer:
\(\left(2 \frac{3}{2}\right)^2=\left(\frac{7}{2}\right)^2\)
= \(\left(\frac{49}{4}\right)\)
Question 3.
Write each product below as the product of powers of different primes:
i) 75 × 45
ii) 96 × 144
iii) 72 × 175
Answer:
i) 75 × 45
75 = 3 × 52
45 = 32 × 5
5 × 45 = (3 × 52) × (32 × 5)
= 33 × 53
ii) 96 × 144
96 = 25 × 31
144 = 122 = (22 × 3)2 = 24 × 32
96 × 144 = (25 × 31) × (24 × 32)
= 29 × 33
iii) 72 × 175
72 = 8 × 9 = 23 × 32
175 = 25 × 7 = 52 × 71
72 × 175 = (23 × 32) × (52 × 71)
Question 4.
Calculate the following quotients:
(i) \(\frac{1440}{120}\)
(ii) \(\frac{729}{27}\)
Answer:
i) 1440 = 122 × 10
= (22 × 3)2 × (21 × 51)
= 25 × 322 × 51
120 = 12 × 10 = (22 × 3) × (2 × 5)
120 = 23 × 31 × 51
= \(\frac{1440}{120}=\frac{2^5 \times 3^2 \times 5^1}{2^3 \times 3^1 \times 5^1}\)
= 25-3 x 32-1 x 51-1
= 22 × 31 × 50
=4 × 3 × 1
= 12
(ii) \(\frac{729}{27}\)
729 = 36
27 = 33
\(\frac{729}{27}=\frac{3^6}{3^3}\) = 36-3 = 33 = 27
Question 5.
Write each number below as a product of powers of different primes:
i) 28
ii) 452
iii) 182 × 302
iv) 203 × 271
Answer:
i) 28 = 22 × 71
ii) 452 = (32 × 51)2 = 34 × 52
iii) 182 × 302 = (21 × 32)2 × (21 × 31 × 51)2 = 22+2 × 34+2 × 52 = 24 × 36 × 52
iv) 203 × 271 = (22 × 51)3 × (33)
= 26 × 53 × 33
Class 7 Maths Chapter 8 Notes Kerala Syllabus Repeated Multiplication
In this chapter, we will explore the concept of repeated multiplication, which is a fundamental idea in mathematics. Repeated multiplication occurs when we multiply a number by itself multiple times. In this chapter we discuss about Factors, Power of Fractions, Product of Powers, Quotient of Powers, Multiples And Powers. So,
- In the first topic ‘Factors’, we will discuss about exponents, powers and its related problems.
- And in the second topic ‘Power of fractions’, we discuss about how powers of fractions increases and decreases.
- In the third topic, we discuss about product of powers, here we discuss products and its powers.
- In the fourth topic, we discuss about Quotient of Powers, here we studied how we deal with powers in division.
- And in the last topic, Multiples and Powers, here we discuss about how we use powers in multiplication.
Understanding repeated multiplication helps us simplify complex calculations and is essential for learning about powers and roots.
Factors
Numbers can be split into products of prime numbers.
For example,128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
We can write this product in a shortened form as 27 (“read as, two to the seventh power”)
27 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Therefore, Repeated multiplication of the same number is in this form Similarly, we write repeated addition as multiplication.
For example,
5 + 5 = 2 × 5
5 × 5 = 52
5 + 5 + 5 = 3 × 5
5 × 5 × 5 = 53
5 + 5 + 5 + 5 = 4 × 5
5 × 5 × 5 × 5 = 54
The operation of multiplying a number by itself repeatedly is called exponentiation.
The number showing how many are multiplied together is called exponent.
We write the exponent in a smaller size, to the right and slightly above the number multiplied.
The number got by repeatedly multiplying a number by itself are called powers.
For example,
3 × 3 × 3 = 33 Third power of three
5 × 5 × 5 × 5 = 54 Fourth power of five
7 × 7= 72 Second power of seven
We can consider, any number as the first power of itself.
How do we split 576 as a product of primes?
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 26 × 33
Any natural number greater than 1 can be written either as the power of a prime or as the product of powers of different primes.
Powers of Fractions
We know that area of a square of side 4 meters is 4 × 4 as 42
Then what is the area of a square of side \(\frac{1}{4}\) meter?
i.e, \(\frac{1}{4} \times \frac{1}{4}=\left(\frac{1}{4}\right)^2\)
\(\frac{1}{4} \times \frac{1}{4}=\frac{1}{4 \times 4}\)
\(\left(\frac{1}{4}\right)^2=\frac{1}{4^2}\)
Similarly, what is the area of a square of sides 0.33 metres?
(0.33)2 = 0.33 × 0.33
= 0.1089 square metres
In general, the area of a square with sides of length x is x2.
Find the volume of a cube with lengths of edges 0.5 metre
(0.5)3 = 0.5 × 0.5 × 0.5
= 0.125 cubic metre
Using algebra, the volume of a cube with the length of the edges x is x3.
Then, Find the volume of a cube of edges 2\(\frac{1}{4}\) metres?
(2\(\frac{1}{4}\)) = \(\left(\frac{9}{4}\right)^3\)
= \(\frac{9}{4} \times \frac{9}{4} \times \frac{9}{4}\)
= \(\frac{9 \times 9 \times 9}{4 \times 4 \times 4}\)
= \(\frac{729}{64}\)
We can split this into quotient and reminder
\(\frac{729}{64}\) = 11\(\frac{25}{64}\)
Or use a calculator to compute
\(\frac{729}{64}\) = 11.390625
1 metre = 100 centimetres
= 102 centimetres
1 cubic metres = 100 × 100 × 100 cubic centimetres
= 106 cubic centimetres
11.390625 cubic metres = 11. 390625 × 106 cubic centimetres
= 11390625 cubic centimetres
Powers of 2
22 = 2 × 2 = 4
23 = 4 × 2 = 8
24 = 8 × 2 = 16
25 = 16 × 2 = 32
By knowing the powers of 2, we can easily compute the powers of \(\frac{1}{2}\).
Each multiplication by 2 doubles, while each multiplication by \(\frac{1}{2}\) halves.
In general, Powers of numbers greater than 1 steadily increase. Powers of numbers greater than 0 and less than 1 steadily decrease. All powers of 1 remain 1 and all powers of 0 remain 0.
Multiples And Powers
Let’s see how we can add 2 times 4 and 2 times 6.
We know,
2 times 4 = 4 + 4
2 times 6 = 6 + 6
Adding this we get,
(2 times 4) + (2 times 6) = (4 + 6) + (4 + 6)
= 10 + 10 = 2 times 10
In short, we can write,
(2 × 4) + (2 × 6) = 2 × (4 + 6)
Using power, we can do the same as:
2nd power of 4 = 4 × 4
2nd power of 6 = 6 × 6
Multiplying,
(2nd power of 4) × (2nd power of 6)
= (4 × 6) × (4 × 6)
= 2nd power of (4 + 6)
That is,
42 × 62 = (4 × 6)2
In general,
The product of the same powers of two numbers is equal to the same power of the product of these numbers
And by using algebra,
xnyn = (xy)n for all numbers x, y and for all natural numbers
Now let’s check what is 5 times 2 times 7?
We can calculate like this:
5 times 2 is 5 × 2 = 10
10 times 7 = 10 × 7 = 70
Also, we can calculate like this:
2 times 7 = 2 × 7 = 14
5 times 14 is 5 × 14 = 70
Let’s see how the second computation works:
2 times 7 = 7 + 7 = 14
5 times 14 is (7 + 7) (7+ 7) (7+ 7) (7+ 7) (7+ 7) = 10 times 7
Thus, we can write the computation as:
5 × (2 × 7) = (5 × 2) × 7
Using power, we can do the same as:
That is 5th power of 7 = 7 × 7
5th power of (7 × 7) = (7 × 7)(7 × 7)(7 × 7)(7 × 7)(7 × 7)
= 10th power of 7
In short using exponent we can write this as,
(7 × 7)5 = (72)5 = 710
Thus, we get the relation, that is,
In computing a power of a power of a nmnber the exponents should be multiplied
This can be written using algebra as an equation.
(xm)n = xmn for all numbers x and for all natural numbers m and n
Products Of Powers
Let’s check how we can find 32 × 34
We know that
32 = 3 × 3
34 = 3 × 3 × 3
Multiplying
Thus 32 × 34 = 32+4 = 36
In general, we can say:
In multiplying two powers of a number, the exponents should be added
Using algebra, we can write it as follows:
xm × xn = xm+n for all numbers x and all natural numbers m and n
We should note two things here:
(i) The product of two powers of the same number is a power of that number
(ii) The exponent of the product is the sum of the exponents of the numbers multiplied
For example, compute 52 × 53 × 544
52 × 53 × 54 = (52 × 53) × 544
= 55 × 54
= 59
This method is also used for prime factorization of a product.
For example consider, 96 × 144
96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3
144 = 2 × 2 × 2 × 2 × 3 × 3 = 24 × 32
and then compute the product as
96 × 144 = (25 × 3) × (24 × 32)
= (25 × 24) × (3 × 32)
= 29 × 33
Quotient Of Powers
Now let’s see how we can find the quotient using the powers.
For that consider an example, 288 ÷ 36
First factorize the numbers 288 and 36
288 = 25 × 32
36 = 22 × 32
We can remove common factors. So that we get,
(25 × 32) ÷ (22 × 32) = 25 ÷ 22
Now we can do the division easily
25 ÷ 22 = 25-2 = 23
Write all these steps together we get:
288 ÷ 36 = (25 × 32) ÷ (22 × 32)
= 25 ÷ 22
= 23
= 8
As a general principle we can state this as:
In dividing the larger power of a non-zero number by a smaller power of the same number, the exponents should be subtracted
And we can state this using algebra like this:
\(\frac{x^m}{x^n}\) = xm-n, for all numbers x ≠ 0 and for all natural numbers m > n
- The operation of multiplying a number by itself repeatedly is called exponentiation.
- The number showing how many are multiplied together is called exponent.
- The number got by repeatedly multiplying a number by itself are called powers.
- Any natural number greater than 1 can be written either as the power of a prime or as the product of powers of different primes.