Question

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

Answer 1

Given :

Luke wins 2 blue balls and loses 3 red ones daily .

After 5 days, he has the same amount of blue and red balls  .

After 9 days, he has twice as many red balls as blue balls .

To find :

The number of red balls in beginning .

Solution :

Let initially , the number of blue balls be x and the number of red balls be y .

after 5 days ,

x + 10 = y - 15          ....(i)

after 9 days ,

x + 18 = 2 * ( y - 27 )    ...(ii)

subtracting (i) by (ii)

8 = y - 39

=> y = 47

The number of red balls in beginning is 47 .

Answer 2

Step-by-step explanation:

Given that:Luke has blue and red balls Every day he wins 2 blue balls and loses 3 red ones,After 5 days he has the same amount of blue and red balls .After 9 days, he has twice as many red balls as blue balls.

To find:How many red balls did he have at the beginning?

Solution: Let us assume that in the starting Luke has x blue balls and y red balls

since, he wins everyday 2 blue balls;after 1 day (x+2) blue balls

since, he loses everyday 3 red balls;after 1 day (y-3) red balls

Case1:

$x + 10 = y - 15 \\ \\\bold{ x - y = - 25} \: \: \: \: ...eq1$

Case2:

$(x + 18) = 2(y - 27) \\ \\ x + 18 = 2y - 54 \\ \\ x - 2y = - 54 -18 \\ \\ \bold{x -2y = - 72} \: \: \: \: ...eq2$

Now subtract both eq1 and eq2

$x - y = - 25 \\ x - 2y = -72\\ ( - ) \: ( + ) \: \: \: ( + ) \\ - - - - - - \\ \bold{y =47}$

Luke has 47 red balls in the beginning.

Hope it helps you.