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A box contains balls of two colours, black and white. A total of 28 balls were taken out from the box constituting of 1/5 of the black balls and 1/10 of the white balls. When 3/4 of the remaining black balls and 2/3 of the remaining white balls were also taken out, only 50 balls were left in the box. How many total balls were initially there in the box?

A box contains balls of two colours, black and white. A total of 28 balls were taken out from the box constituting of 1/5 of the black balls and 1/10 of the white balls. When 3/4 of the remaining black balls and 2/3 of the remaining white balls were also taken out, only 50 balls were left in the box. How many total balls were initially there in the box? Correct Answer 195

Let the no. of black and white balls in the box be ‘x’ and ‘y’ respectively

When 28 balls were taken out,

⇒ x/5 + y/10 = 28

⇒ 2x + y = 280       ----(1)

No. of black balls remaining = x – x/5 = 4x/5

No. of white balls remaining = y – y/10 = 9y/10

When more balls were taken out,

No. of black balls taken out = (3/4) × (4x/5) = 3x/5

No. of black balls remaining = 4x/5 – 3x/5 = x/5

No. of white balls taken out = (2/3) × (9y/10) = 3y/5

No. of white balls remaining = 9y/10 – 3y/5 = 3y/10

But, 50 balls were remaining in the box,

⇒ x/5 + 3y/10 = 50

⇒ 2x + 3y = 500      ----(2)

Subtracting eq. (1) from (2),

⇒ 2x + 3y – 2x – y = 500 – 280

⇒ 2y = 220

⇒ y = 220/2 = 110

Substituting in (1),

⇒ x = (280 – 110)/2 = 85

∴ Total no. of balls in box = x + y = 85 + 110 = 195

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