The number of red, blue, and green marbles in a bag is in the ratio 3 : 4 : 6. If 20 red marbles, 15 blue marbles, and an unknown number of green marbles are added to the bag, the ratio of red, blue, and green marbles becomes 4 : 5 : 7. Determine the number of green marbles added.

Correct — B. Let the original counts be 3k, 4k and 6k.

Red and blue both have known additions, so use them:
(3k + 20) : (4k + 15) = 4 : 5
Cross-multiply: 5(3k + 20) = 4(4k + 15)
15k + 100 = 16k + 60, so k = 40

Original: red 120, blue 160, green 240
New red 140, new blue 175, so one new ratio unit = 140 ÷ 4 = 35

New green = 7 × 35 = 245
Green added = 245 − 240 = 5 → option (b)

Why the others are wrong

• (a)
  3 — Adding 3 makes green 243. Red 140 and blue 175 fix one ratio unit at 35, so green must be 7 × 35 = 245, and 140 : 175 : 243 is not 4 : 5 : 7.
• (c)
  7 — 7 is green's term in the new ratio, not a count of marbles. Adding 7 makes green 247, which overshoots the 245 that the ratio needs.
• (d)
  9 — Adding 9 makes green 249, above the 245 that one ratio unit of 35 demands, so the new ratio would not be 4 : 5 : 7.

When different amounts are added to each part of a ratio, the parts no longer grow by a common factor, so you cannot just rescale the ratio.

Write the old quantities as 3k, 4k and 6k with one multiplier k, and find k from a pair whose additions are both known. Red and blue qualify here, green does not.

Once k is known, the new ratio fixes a new unit (35 here), and the unknown addition is the gap between green's new and old counts.

Key facts

• Parts in the ratio a : b : c can be written ak, bk and ck for one common multiplier k.
• a : b = c : d means a × d = b × c.
• Here the counts go from 120, 160, 240 to 140, 175, 245.

Common traps

• Starting the equation with green, whose addition is unknown, which leaves two unknowns in one equation.
• Reading 7 from the new ratio as a number of marbles instead of 7 units of 35.

No directly related past PYQ was found.