SAT math..another problem





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Discus: What Are My Chances?: January 2003 Archive: SAT math..another problem
By mathz on Monday, January 20, 2003 - 12:58 am: Edit

the numerator and denominator of a fraction are in the ratio 2:3.
If 6 is subtracted from the numerator, the result will be a fraction that has a value 2/3 of the original fraction. What is the numerator of the original fraction??


answer: 18...but HOW!?

By hey on Monday, January 20, 2003 - 01:13 am: Edit

Let n be the numerator and d be the denominator.

Then n/d = 2/3 ==> d = 3n / 2

When you subtract 6 from n, you get n-6.

So (n-6)/d = (2/3) (2/3) = (4/9)

Plugging in d from the first equation, you get

(n-6) / (3n/2) = 4/9

Since (n-6) / (3n/2) = 2(n-6) / (3n),

(2n - 12) / 3n = 4/9

Bringing 3n to the right side, you get

2n - 12 = (4/9) (3n) = (4/3 n)

Putting all the n's on the left and numbers on the right, you get

2n - (4/3)n = 12

So (2/3)n = 12

So n = 18.

By mathz on Monday, January 20, 2003 - 01:28 am: Edit

thanks..again!!
for some random reason i can never SEE it; it seems so easy once you see the answer.


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