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?
Answers
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
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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