Question

Luke has blue and red balls. Every day, he wins 2 blue balls and loses 3 red balls. After 5 days, he has the same amount of blue and red balls. After 9 days, he has twice as many blue balls as red balls. How many red balls did he have at the beginning?

Answer 1

Answer:

let the number of red balls be x

let the number of blue balls be y

after 5 days

there are y+10 blue balls and x-15 red balls

so

y+10 = x-15

y+35 = x

after 9 days

there are y+18 blue balls and x-27 red balls

y+18 = 2*(x-27)

y+18 = 2x -54

y+72 = 2x

subtituting

x - 35+72 = 2x

37 = x

Answer 2

The number of red balls at the beginning is 47.

GIVEN

balls lost every day = 3 red balls

Balls won everyday = 2 blue balls

Days after the Number of blue and red balls are equal = 5

After 9 days, blue balls = 2(Red balls).

TO FIND

Number of Red balls at the beginning.

SOLUTION

We can simply solve the above problem as follows-

Let,

The initial number of blue balls = B

The initial number of red balls = R

ATQ,

Number of balls won every day = 2 blue balls

Number of blue balls After 5 days = B + 10

Number of balls lost everyday = 3 Red balls

So,

Number of Red balls after 5 days = R - 15

After 5 days, the number of blue and red balls is the same.

So,

B + 10 = R - 15.

B = R - 15 - 10

B = R - 25 (Equation 1 )

The number of Blue balls after 9 days = B + 18.

The number of Red balls after 9 days = R - 27.

It is given,

The number of blue balls = 2 red balls.

So,

B + 18 = 2(R-27)

B + 18 = 2R - 54. (Equation 2)

Putting the Value of B from equation 1 in equation 2

R-25 +18 = 2R - 54

-25+18+54 = 2R-R

47 = R

Hence, The number of red balls at the beginning is 47.

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