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The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the numbers is
- a)3
- b)4
- c)1
- d)2
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The HCF and product of two numbers are 15 and 6300 respectively. The n...
Let the number be 15x and 15y, where x and y are co –prime.
∴ 15x × 15y = 6300
So, two pairs are (7, 4) and (14, 2)
∴ 15x × 15y = 6300
So, two pairs are (7, 4) and (14, 2)
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The HCF and product of two numbers are 15 and 6300 respectively. The n...
Given information:
- HCF of two numbers = 15
- Product of two numbers = 6300
Solution:
Finding the numbers:
Let the two numbers be x and y.
Given that the HCF of x and y is 15, we can write:
x = 15a
y = 15b
where a and b are co-prime numbers.
Given that the product of x and y is 6300, we have:
x * y = 6300
Substitute the values of x and y:
15a * 15b = 6300
225ab = 6300
ab = 28
Now we need to find the number of pairs (a, b) that satisfy the condition ab = 28.
Finding possible pairs:
The factors of 28 are:
1, 2, 4, 7, 14, 28
Since a and b are co-prime, the possible pairs are:
(1, 28), (2, 14), (4, 7)
Therefore, there are 3 possible pairs of numbers that satisfy the given conditions.
Final answer:
The number of possible pairs of numbers is 3.
- HCF of two numbers = 15
- Product of two numbers = 6300
Solution:
Finding the numbers:
Let the two numbers be x and y.
Given that the HCF of x and y is 15, we can write:
x = 15a
y = 15b
where a and b are co-prime numbers.
Given that the product of x and y is 6300, we have:
x * y = 6300
Substitute the values of x and y:
15a * 15b = 6300
225ab = 6300
ab = 28
Now we need to find the number of pairs (a, b) that satisfy the condition ab = 28.
Finding possible pairs:
The factors of 28 are:
1, 2, 4, 7, 14, 28
Since a and b are co-prime, the possible pairs are:
(1, 28), (2, 14), (4, 7)
Therefore, there are 3 possible pairs of numbers that satisfy the given conditions.
Final answer:
The number of possible pairs of numbers is 3.
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The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the numbers isa)3b)4c)1d)2Correct answer is option 'D'. Can you explain this answer?
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