Why the number 2028 is not a perfect square number find the smallest number with which 2028 is to be multiplied so that it become perfect square?

2028 can be factorised as follows.

2028 = 2 × 2 × 3 × 13 × 13

Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a perfect square. Therefore, 2028 has to be multiplied with 3 to obtain a perfect square.

Therefore, 2028 × 3 = 6084 is a perfect square.

2028 × 3 = 6084 = 2 × 2 × 3 × 3 × 13 × 13

∴ `sqrt[6084]` = 2 x 3 x 13  = 78

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 7Here, 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.

Thus, Answer = 7

[ii] 180

Answer: By prime factorisation, we get, 180 = 3 x 3 x 2 x 2 x 5Here, 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.

Thus, Answer = 5

[iii] 1008

Answer: By prime factorisation of 1008, we get1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7Here, 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

Hence, Answer = 7

[iv] 2028

Answer: By prime factorisation of 2028, we get2028 = 2 x 2 x 3 x 13 x 13Here, 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.

Hence, Answer = 3

[v] 1458

Answer: By prime factorisation of 1458, we get1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3Here, 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.

Therefore, Answer = 2

[vi] 768

Answer: By prime factorisation of 768, we get768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3Here, 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.

Hence, Answer = 3

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 get252 = 2 x 2 x 3 x 3 x 7Here, 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

Thus, Answer = 7

[ii] 2925

Answer: By prime factorisation of 2925, we get2925 = 5 x 5 x 3 x 3 x 13Here, 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

Thus, Answer = 13

[iii] 396

Answer: By prime factorisation of 396, we get396 = 2 x 2 x 3 x 3 x 11Here, 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

Hence, Answer = 11

[iv] 2645

Answer: By prime factorisation of 2645, we get2645 = 5 x 23 x 23Here, 23 is in pair, but 5 needs a pair. Thus, 5 can be eliminated after dividing 2645 by 5Hence, 2645 needs to be divided by 5 to become a perfect square.

Therefore, Answer = 5

[v] 2800

Answer: By prime factorisation of 2800, we get2800 = 2 x 2 x 7 x 10 x 10Since, 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.

Therefore, Answer = 7

[vi] 1620

Answer: By prime factorisation of 1620, we get1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5Here, 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

Thus, Answer = 5

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
Thus, Answer = 49

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.
Thus, Answer = 45

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 104 = 2 x 29 = 3 x 310 = 5 x 2

So, LCM = 2 2 x 3 2 x 5 = 180

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`

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Solution:

We have to find the smallest whole number by which the number should be multiplied so as to get a perfect square number

To get a perfect square, each factor of the given number must be paired.

[i] 252

Hence, prime factor 7 does not have its pair. If 7 gets a pair, then the number becomes a perfect square. Therefore, 252 has to be multiplied by 7 to get a perfect square.

So, perfect square is 252 × 7 = 1764

1764 = 2 × 2 × 3 × 3 × 7 × 7

Thus, √1764 = 2 × 3 × 7 = 42

[ii] 180

Hence, prime factor 5 does not have its pair. If 5 gets a pair, then the number becomes a perfect square. Therefore, 180 has to be multiplied by 5 to get a perfect square.

So, perfect square is 180 × 5 = 900

900 = 2 × 2 × 3 × 3 × 5 × 5

Thus, √900 = 2 × 3 × 5 = 30

[iii] 1008

Hence, prime factor 7 does not have its pair. If 7 gets a pair, then the number becomes a perfect square. Therefore, 1008 has to be multiplied by 7 to get a perfect square.

So, perfect square is 1008 × 7 = 7056

7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

Thus, √7056 = 2 × 2 × 3 × 7 = 84

[iv] 2028

Hence, prime factor 3 does not have its pair. If 3 gets a pair, then the number becomes a perfect square. Therefore, 2028 has to be multiplied by 3 to get a perfect square.

So, perfect square is 2028 × 3 = 6084

6084 = 2 × 2 × 13 × 13 × 3 × 3

Thus, √6084 = 2 × 13 × 3 = 78

[v] 1458

Hence, prime factor 2 does not have its pair. If 2 gets a pair, then the number becomes a perfect square. Therefore, 1458 has to be multiplied by 2 to get a perfect square.

So, perfect square is 1458 × 2 = 2916

2916 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2

Thus, √2916 = 3 × 3 × 3 × 2 = 54

[vi] 768

Hence, prime factor 3 does not have its pair. If 3 gets a pair, then the number becomes a perfect square. Therefore, 768 has to be multiplied by 3 to get a perfect square.

So, perfect square is 768 × 3 = 2304

2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

Thus, √2304 = 2 × 2 × 2 × 2 × 3 = 48

☛ Check: NCERT Solutions for Class 8 Maths Chapter 6

Video Solution:

NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.3 Question 5

Summary:

For each of the following numbers, [i] 252 [ii] 180 [iii] 1008 [iv] 2028 [v] 1458 [vi] 768 the smallest whole number by which it should be multiplied so as to get a perfect square number and the square root of the square number so obtained are as follows: [i] 7; √1764 = 42 [ii] 5; √900 = 30 [iii] 7; √7056 = 84 [iv] 3; √6084 = 78 [v] 2; √2916 = 54 and [vi] 3; √2304 = 48

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