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Re: If the least common multiple of 2x and y is 400 and the greatest commo [#permalink]
ktzsikka wrote:
If the least common multiple of 2x and y is 400 and the greatest common divisor of 6x and 3y is 30, what is the value of \(\frac{x*y}{2} \) ?

A. 100

B. 500

C. 1000

D. 1500

E. 2000


A number is the product of it's LCM and HCF that's 2x and y equals 400*10

=> x*y = 2000
=>x*y/2 = 1000

Therefore IMO C
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Re: If the least common multiple of 2x and y is 400 and the greatest commo [#permalink]
Here LCM (2x,y) = 400 and HCF (2x,y) = 10

Hence 2x * y = 4000

=> x*y/2 = 4000/4 = 1000

Option C
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Re: If the least common multiple of 2x and y is 400 and the greatest commo [#permalink]
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Given that LCM of 2x and y is 400 and GCD of 6x and 3y is 30 and we need to find the value of \(\frac{x*y}{2}\)

Lets solve the Problem using Two Methods:

Method 1:

LCM of 2x and y = 400

If we multiply 2x and y by 3 to get 6x and 3y then their LCM = (LCM of 2x and y) * 3 = 400*3 = 1200

LCM of two numbers * GCD of two numbers = Product of the two numbers

=> LCM(6x,3y) * GCD(6x,3y) = 6x * 3y
=> 1200 * 30 = 18 *xy
=> xy = \(\frac{1200*30}{18}\) = 2000

=> \(\frac{x*y}{2}\) = \(\frac{2000}{2}\) = 1000

So, Answer will be C

Method 2:

GCD of 6x and 3y = 30

If we divide 6x and 3y by 3 to get 2x and y then their GCD = (GCD of 6x and 3y) / 3 = \(\frac{30}{3}\) = 10

LCM of two numbers * GCD of two numbers = Product of the two numbers

=> LCM(2x,y) * GCD(2x,y) = 2x * y
=> 400 * 10 = 2xy
=> xy = 2000

=> \(\frac{x*y}{2}\) = \(\frac{2000}{2}\) = 1000

So, Answer will be C
Hope it helps!

To learn more about LCM and GCD watch the following videos



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Re: If the least common multiple of 2x and y is 400 and the greatest commo [#permalink]
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Re: If the least common multiple of 2x and y is 400 and the greatest commo [#permalink]
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