Class 8 Maths

# Square Roots

## Exercise 6.3 Part 2

Question 5: For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252

Answer: By prime factorisation we get,
252 = 2 x 2 x 3 x 3 x 7
Here, 2 and 3 are in pairs but 7 needs a pair. Thus, 7 can become pair after multiplying 252 with 7.
So, 252 will become a perfect square when multiplied by 7.

(ii) 180

Answer: By prime factorisation, we get, 180 = 3 x 3 x 2 x 2 x 5
Here, 3 and 2 are in pair but 5 needs a pair to make 180 a perfect square.
180 needs to be multiplied by 5 to become a perfect square.

(iii) 1008

Answer: By prime factorisation of 1008, we get
1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
Here, 2 and 3 are in pair, but 7 needs a pair to make 1008 a perfect square.
Thus, 1008 needs to be multiplied by 7 to become a perfect square

(iv) 2028

Answer: By prime factorisation of 2028, we get
2028 = 2 x 2 x 3 x 13 x 13
Here, 2 and 13 are in pair, but 3 needs a pair to make 2028 a perfect square.
Thus, 2028 needs to be multiplied by 3 to become a perfect square.

(v) 1458

Answer: By prime factorisation of 1458, we get
1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3
Here, 3 are in pair, but 2 needs a pair to make 1458 a perfect square.
So, 1458 needs to be multiplied by 2 to become a perfect square.

(vi) 768

Answer: By prime factorisation of 768, we get
768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
Here, 2 are in pair, but 3 needs a pair to make 768 a perfect square.
So, 768 needs to be multiplied by 3 to become a perfect square.

Question 6: For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252

Answer: By prime factorisation of 768, we get
252 = 2 x 2 x 3 x 3 x 7
Here, 2 and 3 are in pair, but 7 has no pair, which can be eliminated after dividing 768 by 7.
Hence, 252 needs to be divided by 7 to become a perfect square

(ii) 2925

Answer: By prime factorisation of 2925, we get
2925 = 5 x 5 x 3 x 3 x 13
Here, 5 and 3 are in pair, but 7 has no pair, which can be eliminated after dividing 2925 by 13.
Thus, 2925 needs to be divided by 13 to become a perfect square

(iii) 396

Answer: By prime factorisation of 396, we get
396 = 2 x 2 x 3 x 3 x 11
Here, 2 and 3 are in pair, but 11 needs another 11 to make a pair. Thus, 11 can be eliminated after dividing 396 by 11.
Thus, 396 needs to be divided by 11 to become a perfect square

(iv) 2645

Answer: By prime factorisation of 2645, we get
2645 = 5 x 23 x 23
Here, 23 is in pair, but 5 needs a pair. Thus, 5 can be eliminated after dividing 2645 by 5
Hence, 2645 needs to be divided by 5 to become a perfect square.

(v) 2800

Answer: By prime factorisation of 2800, we get
2800 = 2 x 2 x 7 x 10 x 10
Since, only 7 is not in pair, thus, by eliminating 7, from 2800, it can be a perfect square.
Hence, 2800 needs to be divided by 7 to become a perfect square.

(vi) 1620

Answer: By prime factorisation of 1620, we get
1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5
Here, 2 and 3 are in pair, but 5 is not in pair. Thus, by eliminating 5 from 1620, obtained number will be a perfect square.
Hence, 1620 needs to be divided by 5 to become a perfect square

Question 7: The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Answer: We need to calculate the square root of 2401 to get the solution.
By prime factorisation of 2401, we get
2401 = 7 xx 7 xx 7 xx 7

Or, 2401=7^2xx7^2

Or, sqrt(2401)=sqrt(7^2xx7^2)

=7xx7=49

There are 49 students, each contributing 49 rupees

Question 8: 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Answer: To calculate the number of rows and number of plants, we need to find the square root of 2025.
By prime factorisation of 2025, we get
2025 = 5 xx 5 xx 3 xx 3 xx 3 xx 3

Or, 2025=5^2xx3^2xx3^2

Or, sqrt(2025)=sqrt(5^2xx3^2xx3^2)

=5xx3xx3=45

There are 45 rows with 45 plants in each of them.

Question 9: Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Answer: Let us find LCM of 4, 9 and 10
4 = 2 x 2
9 = 3 x 3
10 = 5 x 2
So, LCM = 2 2 x 3 2 x 5 = 180
Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a perfect square.
The Required number = 180 x 5 = 900

Question 10: Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Answer: 8=2xx2xx2
15=3xx5
20=2xx2xx5
So, LCM =2^3xx3xx5=120
As 2, 3 and 5 are not in pairs in LCM’s factor so we need to multiply 120 by product of 2, 3 and 5 to make it a perfect square.
Required Number =120xx2xx3xx5=3600