Question

Q. The cost of 4 pens and 8 pencils rupees is 84. If one pen costs rupees 12 more than the cost of one pencil, find the price of one pen and one pencil separately.​

Answer 1

Answer:

$\large{\underline{\boxed{\sf Cost \: of \: one\: pen = Rs. 15}}}$

$\large{\underline{\boxed{\sf Cost \: of \: one \: pencil = Rs. 3}}}$

Step-by-step explanation:

Let the cost of one pen be ₹x and cost of a pencil be ₹y.

According to the question ;

The cost of 4 pens and 8 pencils is ₹84.

⇒ 4x + 8y = 84 $\quad$ ... (i)

One pen cost ₹12 more than the cost of a pencil.

⇒ x = 12 + y $\quad$ ... (ii)

Substituting the value of (ii), in (i) -

⇒ 4x + 8y = 84

⇒ 4(12 + y) + 8y = 84

⇒ 48 + 4y + 8y = 84

⇒ 48 + 12y = 84

⇒ 12y = 84 - 48

⇒ 12y = 36

⇒ y = $\sf \dfrac{36}{12}$

⇒ y = 3

Putting the value of y in (ii),

⇒ x = 12 + y

⇒ x = 12 + 3

⇒ x = 15

Hence, the cost of one pen is ₹15, and the cost of a pencil is ₹3.

Answer 2

Answer:-

Cost of 1 pen = ₹15

Cost of 1 pencil = ₹3

Given :-

Cost of 4 pens and 8 pencil is ₹84.

One pen costs is ₹ 12 more than cost of one pencil.

To find :-

The price of one pen and pencil separately.

Solution:-

Let the cost of one pen be ₹x and one pencil be ₹y.

A/Q

→$\mathsf{4x + 8y = 84}$

• Take 4 as common.

→$\mathsf{4 (x + 2y ) = 84}$

→$\mathsf{ x + 2y = \dfrac{84}{4}}$

→$\mathsf{x+2y = 21}------1)$

→$\mathsf{ x = y + 12} --------2)$

• Put the value of x in eq. 1

→$\mathsf{x + 2y = 21 }$

→$\mathsf{y + 12 +2y = 21}$

→$\mathsf{ 3y = 21 -12}$

→$\mathsf{3y = 9}$

→$\mathsf{ y = \dfrac{9}{3}}$

→$\mathsf{ y = 3}$

• put the value of y in eq. 2

→$\mathsf{ x = y +12 }$

→$\mathsf{x = 3+12}$

→$\mathsf{ x = 15}$

hence,

Cost of one pen is ₹15 and cost of one pencil is ₹3.