A plastic bag contains 12 red marbles- 15 green marbles and 5 yellow marbles- Find the number of additional red marbles that must be added to the bag- so that the probability of randomly drawing a red marble is displaystyle frac - 3 - 5 - -1328183240

Question

Toppr Admin · Accepted Answer

Let -apos-x-apos- number of additional red marbles be added to the bag-xA0-Then- the required probability will be 12-x32-x-35-xA0-x21D2-60-5x-96-3x-xA0-x21D2-2x-36-xA0-x21D2-x-18Hence- the required number of marbles is 18-

A plastic bag contains $12$ red marbles, $15$ green marbles and $5$ yellow marbles. Find the number of additional red marbles that must be added to the bag, so that the probability of randomly drawing a red marble is $\frac{3}{5}$ ?
$13$
$28$
$18$
$32$
40

Let ' $x$ ' number of additional red marbles be added to the bag.
Then, the required probability will be $\frac{12 + x}{32 + x} = \frac{3}{5}$
$\Rightarrow 60 + 5 x = 96 + 3 x$
$\Rightarrow 2 x = 36$
$\Rightarrow x = 18$
Hence, the required number of marbles is $18$ .

A plastic bag contains 12 red marbles, 15 green marbles and 5 yellow marbles. Find the number of additional red marbles that must be added to the bag, so that the probability of randomly drawing a red marble is 35?1328183240

Let 'x' number of additional red marbles be added to the bag. Then, the required probability will be 12+x32+x=35 ⇒60+5x=96+3x ⇒2x=36 ⇒x=18Hence, the required number of marbles is 18.

A plastic bag contains $12$ red marbles, $15$ green marbles and $5$ yellow marbles. Find the number of additional red marbles that must be added to the bag, so that the probability of randomly drawing a red marble is $\frac{3}{5}$ ?
$13$
$28$
$18$
$32$
40

Let ' $x$ ' number of additional red marbles be added to the bag.
Then, the required probability will be $\frac{12 + x}{32 + x} = \frac{3}{5}$
$\Rightarrow 60 + 5 x = 96 + 3 x$
$\Rightarrow 2 x = 36$
$\Rightarrow x = 18$
Hence, the required number of marbles is $18$ .