L.C.M. of 4, 9 and 10 is 180. Prime factors of 180 = 2 x 2 x 3 x 3 x 5 Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square. ∴ 180 x 5 = 900 Hence, the smallest square number which is divisible by 4, 9 and 10 is 900. Class 8 Maths Chapter 6 Exercise 6.3Read more
L.C.M. of 4, 9 and 10 is 180.
Prime factors of 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 180 must be multiplied
by 5 to make it a perfect square.
∴ 180 x 5 = 900
Hence, the smallest square number which is divisible by 4, 9 and 10
is 900.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
Here, Number of plants = 2025 Let the number of rows of planted plants be x. And each row contains number of plants = x According to question, x²=2025 ⇒ x = √2025= √3x3x3x3x5x5 ⇒ x = 3x3x5= 45 Hence, each row contains 45 plants. Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video for more answersRead more
Here, Number of plants = 2025
Let the number of rows of planted plants be x.
And each row contains number of plants = x
According to question,
x²=2025
⇒ x = √2025= √3x3x3x3x5x5
⇒ x = 3x3x5= 45
Hence, each row contains 45 plants.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
Here, Donated money = ₹ 2401 Let the number of students be x. Therefore donated money = xXx According to question, x²=2401 ⇒ x = √2401 = √7x7x7x7 ⇒ x = 7x7=49 Hence, the number of students is 49. Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video for more answers vist to: https://www.tiwariacadeRead more
Here, Donated money = ₹ 2401
Let the number of students be x.
Therefore donated money = xXx
According to question,
x²=2401
⇒ x = √2401 = √7x7x7x7
⇒ x = 7×7=49
Hence, the number of students is 49.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
(i) 252 = 2 x 2 x 3 x 3 x 7 Here, prime factor 7 has no pair. Therefore 252 must be divided by 7 to make it a perfect square. ∴ 252 ÷ 7 = 36 And √36 = 2 x 3 = 6 (ii) 2925 = 3 x 3 x 5 x 5 x 13 Here, prime factor 13 has no pair. Therefore 2925 must be divided by 13 to make it a perfect square. ∴ 2925Read more
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be
divided by 7 to make it a perfect square.
∴ 252 ÷ 7 = 36
And √36 = 2 x 3 = 6
(ii) 2925 = 3 x 3 x 5 x 5 x 13
Here, prime factor 13 has no pair.
Therefore 2925 must be divided by 13
to make it a perfect square.
∴ 2925 ÷ 13 = 225
And √225 = 3 x 5 = 15
(iii) 396 = 2 x 2 x 3 x 3 x 11
Here, prime factor 11 has no pair. Therefore 396 must be
divided by 11 to make it a perfect square.
∴ 396 ÷ 11 = 36
And √36 = 2 x 3 = 6
(iv) 2645 = 5 x 23 x 23
Here, prime factor 5 has no pair. Therefore 2645 must be
divided by 5 to make it a perfect square.
∴ 2645 ÷ 5 = 529
And √529 = 23 x 23 = 23
(v) 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7
Here, prime factor 7 has no pair. Therefore 2800 must be
divided by 7 to make it a perfect square.
∴ 2800 ÷ 7 = 400
And √400 = 2 x 2 x 5 = 20
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
(i) 252 = 2 x 2 x 3 x 3 x 7 Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square. ∴ 252 x 7 = 1764 and √1764 = 2x3x7=42 (ii) 180 = 2 x 2 x 3 x 3 x 5 Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square. ∴ 1Read more
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be
multiplied by 7 to make it a perfect square.
∴ 252 x 7 = 1764
and √1764 = 2x3x7=42
(ii) 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 180 must be
multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900
and √900= 2x3x5=30
(iii) 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 1008 must be
multiplied by 7 to make it a perfect square.
∴ 1008 x 7 = 7056
and √7056 = 2x2x3x7 =84
(iv) 2028 = 2 x 2 x 3 x 13 x 13
Here, prime factor 3 has no pair. Therefore 2028 must be
multiplied by 3 to make it a perfect square.
∴ 2028 x 3 = 6084
And √6080 = √2x2x3x3x13x13 =78
(v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3
Here, prime factor 2 has no pair. Therefore 1458 must be
multiplied by 2 to make it a perfect square.
∴ 1458 x 2 = 2916
and √2916 = 2x3x3x3 = 54
(vi) 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
Here, prime factor 3 has no pair. Therefore 768 must be
multiplied by 3 to make it a perfect square.
∴ 768 x 3 = 2304
and √2304 = 2x2x2x2x3 = 48
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
L.C.M. of 4, 9 and 10 is 180. Prime factors of 180 = 2 x 2 x 3 x 3 x 5 Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square. ∴ 180 x 5 = 900 Hence, the smallest square number which is divisible by 4, 9 and 10 is 900. Class 8 Maths Chapter 6 Exercise 6.3Read more
L.C.M. of 4, 9 and 10 is 180.
Prime factors of 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 180 must be multiplied
by 5 to make it a perfect square.
∴ 180 x 5 = 900
Hence, the smallest square number which is divisible by 4, 9 and 10
is 900.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
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.
Here, Number of plants = 2025 Let the number of rows of planted plants be x. And each row contains number of plants = x According to question, x²=2025 ⇒ x = √2025= √3x3x3x3x5x5 ⇒ x = 3x3x5= 45 Hence, each row contains 45 plants. Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video for more answersRead more
Here, Number of plants = 2025
Let the number of rows of planted plants be x.
And each row contains number of plants = x
According to question,
x²=2025
⇒ x = √2025= √3x3x3x3x5x5
⇒ x = 3x3x5= 45
Hence, each row contains 45 plants.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
The students of Class VIII of a school donated ₹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.
Here, Donated money = ₹ 2401 Let the number of students be x. Therefore donated money = xXx According to question, x²=2401 ⇒ x = √2401 = √7x7x7x7 ⇒ x = 7x7=49 Hence, the number of students is 49. Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video for more answers vist to: https://www.tiwariacadeRead more
Here, Donated money = ₹ 2401
Let the number of students be x.
Therefore donated money = xXx
According to question,
x²=2401
⇒ x = √2401 = √7x7x7x7
⇒ x = 7×7=49
Hence, the number of students is 49.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-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 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620
(i) 252 = 2 x 2 x 3 x 3 x 7 Here, prime factor 7 has no pair. Therefore 252 must be divided by 7 to make it a perfect square. ∴ 252 ÷ 7 = 36 And √36 = 2 x 3 = 6 (ii) 2925 = 3 x 3 x 5 x 5 x 13 Here, prime factor 13 has no pair. Therefore 2925 must be divided by 13 to make it a perfect square. ∴ 2925Read more
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be
divided by 7 to make it a perfect square.
∴ 252 ÷ 7 = 36
And √36 = 2 x 3 = 6
(ii) 2925 = 3 x 3 x 5 x 5 x 13
Here, prime factor 13 has no pair.
Therefore 2925 must be divided by 13
to make it a perfect square.
∴ 2925 ÷ 13 = 225
And √225 = 3 x 5 = 15
(iii) 396 = 2 x 2 x 3 x 3 x 11
Here, prime factor 11 has no pair. Therefore 396 must be
divided by 11 to make it a perfect square.
∴ 396 ÷ 11 = 36
And √36 = 2 x 3 = 6
(iv) 2645 = 5 x 23 x 23
Here, prime factor 5 has no pair. Therefore 2645 must be
divided by 5 to make it a perfect square.
∴ 2645 ÷ 5 = 529
And √529 = 23 x 23 = 23
(v) 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7
Here, prime factor 7 has no pair. Therefore 2800 must be
divided by 7 to make it a perfect square.
∴ 2800 ÷ 7 = 400
And √400 = 2 x 2 x 5 = 20
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
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 (ii) 180(iii) 1008 (iv) 2028 (v) 1458 (vi) 768
(i) 252 = 2 x 2 x 3 x 3 x 7 Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square. ∴ 252 x 7 = 1764 and √1764 = 2x3x7=42 (ii) 180 = 2 x 2 x 3 x 3 x 5 Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square. ∴ 1Read more
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be
multiplied by 7 to make it a perfect square.
∴ 252 x 7 = 1764
and √1764 = 2x3x7=42
(ii) 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 180 must be
multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900
and √900= 2x3x5=30
(iii) 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 1008 must be
multiplied by 7 to make it a perfect square.
∴ 1008 x 7 = 7056
and √7056 = 2x2x3x7 =84
(iv) 2028 = 2 x 2 x 3 x 13 x 13
Here, prime factor 3 has no pair. Therefore 2028 must be
multiplied by 3 to make it a perfect square.
∴ 2028 x 3 = 6084
And √6080 = √2x2x3x3x13x13 =78
(v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3
Here, prime factor 2 has no pair. Therefore 1458 must be
multiplied by 2 to make it a perfect square.
∴ 1458 x 2 = 2916
and √2916 = 2x3x3x3 = 54
(vi) 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
Here, prime factor 3 has no pair. Therefore 768 must be
multiplied by 3 to make it a perfect square.
∴ 768 x 3 = 2304
and √2304 = 2x2x2x2x3 = 48
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/