Question

Ten percent of the red balls were added to twenty percent of the blue balls and the total was 24. If three times the number of red balls exceed the number of blue balls by 20, find the number of red balls and blue balls.​

Answer 1

Answer:

suppose, number of red balls is r and blue ball is b.

r/10 + b/5 = 24    so, r + 2b = 240

3r-20=b    so, 3r-b=20

solving those equation we get,  r=40  and b=100

Step-by-step explanation:

Answer 2

The number of red balls is 40 and number of blue balls is 100

Solution:

Let the red balls be r

Let the blue balls be b

Ten percent of the red balls were added to twenty percent of the blue balls and the total was 24

10 % of r + 20 % of b = 24

$\frac{10}{100} \times r + \frac{20}{100} \times b = 24\\\\10r + 20b = 2400\\\\r + 2b = 240 ----- eqn\ 1$

Three times the number of red balls exceed the number of blue balls by 20

3r - b = 20 ----- eqn 2

Multiply eqn 2 by 2

6r - 2b = 40 ------ eqn 3

Add eqn1  and eqn 3

r + 2b = 240

6r - 2b = 40

---------------------

7r = 280

r = 40

Substitute r = 40 in eqn 1

40 + 2b = 240

2b = 200

b = 100

Thus number of red balls is 40 and number of blue balls is 100

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