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In a box, there are red and blue pens in the ratio of \(4:3\). If I were to remove a red pen and two blue pens from the box, this ratio would become \(3:2\). How many red pens are there currently in the box?

 Feb 5, 2021
 #1
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Let the  number of red pens  = R  and blue pens =  B

 

So

 

B / R  = 3  /4   ⇒    B =  (3/4) R

 

In the second scenario we  have  that

 

(R - 1)  / ( B - 2)  = 3 /2

 

Sub in for B

 

(R  -1)  / ( (3/4)R -2 ) =  3/2      cross-multiply

 

2 ( R  -1)  = 3  ( (3/4) R  - 2)      simplify

 

2R  -2  =  (9/4)R   - 6        rearrange  as

 

 

6  - 2   = (9/4)R  - 2R

 

4  =   (9/4)R - (8/4)R

 

4 =   (1/4) R      multiply through  by 4

 

16 =   R

 

Check 

 

Initially  :  R / B   =    16   / (3/4)R =    16  /12  =  4 /3

 

Then  :    (16 - 1) / ( 12 -2)   =     15  /10     =    3   /  2

 

 

 

cool cool cool

 Feb 5, 2021

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