Question

Q1. Two friends, Akash & Beenu had some candies
each. One of them had 15 candies more than the other.
The candies with Akash was 60% of the total candies
with them. How many candies did each have?
A. 40, 25
B. 47, 32
C. 45, 30
D. 49, 34​

Answer 1

Answer

Option C) 45, 30

$\rule{200}1$

Explanation

Two friends, Akash and Beenu had some candies each. One of them had 15 candies more than the other.

Let us assume that Beenu has (x) number of candies.

So, Akash has (x + 15) number of candies.

Now,

Total number of candies that both of them had = x + x + 15 = (2x + 15) candies

The candies with Akash was 60% of the total candies with them.

According to question,

⇒ x + 15 = 60/100 (2x + 15)

⇒ x + 15 = 3/5 (2x + 15)

Cross-multiply them

⇒ 5(x + 15) = 3(2x + 15)

⇒ 5x + 75 = 6x + 45

⇒ 5x - 6x = 45 - 75

⇒ -x = -30

⇒ x = 30

Therefore,

Beenu has 30 candies and Akash has (30 + 15) = 45 candies

Answer 2

Answer:

$\large{\underline{\boxed{\mathfrak\blue{\: \: \: Answer:-C)45,30 \: \: \: }}}}$

$\rule{100}{1}$

Given:-

• Candies of one of them = Candies of other + 15

• Candies with Akash = 60% of total candies.

To find:-

• Candies of Akash and Beenu = ?

As Akash has 60% of total candies,so Akash has more candies than Beenu.

Let number of candies of Beenu = y & number of candies of Akash = y + 15

$\therefore$ Total candies = y + y + 15

$\therefore$ Total candies = 2y + 15

A/Q,

$\: \: \: \: \: \: \dashrightarrow\sf y + 15 = 60\% \: of \: (2y + 15) \\\\ \: \: \: \: \: \: \dashrightarrow\sf y + 15 = \dfrac{60}{100}\times (2y + 15) \\\\ \: \: \: \: \: \: \dashrightarrow\sf y + 15 = \dfrac{3(2y + 15)}{5} \\\\ \: \: \: \: \: \:\dashrightarrow\sf y + 15 = \dfrac{6y + 45}{5} \\\\ \: \: \: \: \: \:\dashrightarrow\sf 5(y + 15) = 6y + 45 \\\\ \: \: \: \: \: \:\dashrightarrow\sf 5y + 75 = 6y + 45 \\\\ \: \: \: \: \: \:\dashrightarrow\sf 6y - 5y = 75 - 45 \\\\\: \: \: \: \: \:\dashrightarrow\large{\underline{\boxed{\sf\blue{y = 30 }}}}$

$\therefore{\underline{\text{Beenu \: has \: 30 \: candies }}}$

$\rule{100}{2}$

• Putting value of x:-

$\: \: \: \: \: \: \dashrightarrow\sf Akash \: has = (30 + 15)\: candies \\\\ \: \: \: \: \: \: \dashrightarrow\large{\underline{\boxed{\sf\green{Akash \: has = 45 \: candies }}}}$

$\therefore{\underline{\text{Akash \: \& \: Beenu \: has \: 45 \: and \: 30 \: candies \: respectively }}}$