Marbles in a bag

Number Theory Level 1

A bag contains 6 6 r e d \color{#D61F06}{red} marbles, 5 5 y e l l o w \color{#CEBB00}{yellow} marbles, and 7 7 g r e e n \color{#20A900}{green} marbles. How many additional r e d \color{#D61F06}{red} marbles must be added to the 18 18 marbles already in the bag so that the probability of randomly drawing a r e d \color{#D61F06}{red} marble is 3 5 \dfrac{3}{5} ?


The answer is 12.

Anuj Shikarkhane by Anuj Shikarkhane , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Munem Shahriar
Oct 25, 2017

Suppose x x is the number additional red marbles must be added.

( x + 6 ) ( x + 18 ) = 3 5 \dfrac{(x+6)}{(x+18)} = \dfrac{3}{5}

5 x + 30 = 3 x + 54 5x+30 = 3x+54

5 x = 3 x + 24 5x = 3x+24

5 x − 3 x = 24 5x-3x = 24

2 x = 24 2x = 24

x = 24 2 x = \dfrac{24}{2}

Hence x = 12 x = \boxed{12}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Let x x be the additional red marbles. Then we have

6 + x 18 + x = 3 5 \dfrac{6+x}{18+x}=\dfrac{3}{5}

30 + 5 x = 54 + 3 x 30+5x=54+3x

2 x = 24 2x=24

x = 12 \boxed{x=12}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...