Question

The numerator and denominator of a fraction is in the ratio 2:3. If 6 are subtracted from the numerator the value of the fraction becomes 2/3 of the original fraction. The numerator of the original fraction is,

• A. 16
• B. 21
• C. 18
• D. 30

Hint:

Let numerator = 2x and denominator = 3x, then set (2x-6)/3x equal to two-thirds of 2/3 and solve for x.

Answer 1

The Correct Option is C

Step 1: Set up the fraction directly using the ratio.
Since numerator : denominator = 2 : 3, write numerator = 2k and denominator = 3k for a common multiplier k, so the fraction always reduces to 2/3 regardless of k's value.
Step 2: Compute the target fraction after the change.
The problem states that after subtracting 6 from the numerator, the new fraction becomes 2/3 of the original value. Since the original fraction is 2/3, the new value must be (2/3) multiplied by (2/3), which is 4/9.
Step 3: Compare the original fraction 2/3 with the target fraction 4/9 by finding a common denominator.
Convert 2/3 to ninths: 2/3 = 6/9. The target fraction is 4/9. So the numerator, when both fractions are written over a denominator of 9, drops from 6 to 4, a drop of 2 out of every 9 parts of the denominator.
Step 4: Scale this drop back up to the actual denominator, 3k.
The denominator 3k corresponds to 9 parts when scaled to ninths, so each "part" equals 3k/9 = k/3. The numerator drop of 2 parts, in actual terms, equals 2 x (k/3) = 2k/3. The problem says the actual drop in the numerator is exactly 6, since 6 is subtracted, so set 2k/3 = 6.
Step 5: Solve for k.
2k/3 = 6 leads to 2k = 18, so k = 9.
Step 6: Compute the original numerator.
Numerator = 2k = 2 x 9 = 18, matching option (3). Denominator = 3k = 27, and indeed 18/27 reduces to 2/3, and (18-6)/27 = 12/27 = 4/9 = (2/3) of 2/3, confirming the solution is consistent.
The numerator of the original fraction is 18. \[\boxed{18}\]