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 bag hold 3k red, 4k blue and 6k green.

After adding: red = 3k + 20, blue = 4k + 15
Red : blue must be 4 : 5, so 5(3k + 20) = 4(4k + 15)
Expand: 15k + 100 = 16k + 60, so k = 40

Red: 3 × 40 + 20 = 140 = 4 × 35
Blue: 4 × 40 + 15 = 175 = 5 × 35
So green must become 7 × 35 = 245

Green before: 6 × 40 = 240
Added: 245 − 240 = 5 → option (b)

Why the others are wrong

• (a)
  3 — 3 would make green 243. Red 140 and blue 175 fix the unit at 35, so green must be 7 × 35 = 245, and 243 is not a multiple of 7.
• (c)
  7 — 7 would make green 247. The ratio 4 : 5 : 7 with red 140 and blue 175 needs green at exactly 245, and 247 ÷ 7 is not a whole number.
• (d)
  9 — 9 would make green 249, giving 140 : 175 : 249, which is not 4 : 5 : 7. The unit 35 from red and blue forces green to 245, an addition of 5.

When a ratio changes because known amounts are added, write each part as a multiple of one unknown k and build the equation from the parts whose additions you know.

Red and blue both have stated additions, so the new red : blue ratio gives one equation in k. Solving it fixes every original count, and the unit of the new ratio (35 here) then tells you what green must become.

Key facts

• A ratio 3 : 4 : 6 means counts of 3k, 4k and 6k for some whole number k.
• Cross-multiplying (3k + 20)⁄(4k + 15) = 4⁄5 gives 5(3k + 20) = 4(4k + 15).
• Check: 140 : 175 : 245 divided by 35 is 4 : 5 : 7.

Common traps

• Building the equation on green: its addition is the unknown, so red and blue, whose additions are given, must fix k.
• Reporting green's new count (245) or its old count (240) instead of the number added.

No directly related past PYQ was found.