Question

Rajan and Manu went to a market. Rajan bought 3 erasers and 5 pens for Rs 105 and Manu bought 4 erasers and 6 pens for Rs 130. What is the price of one eraser?

A) Rs 25 B) Rs 20 C) Rs 10 D) Rs 19

Answer 1

✬ C.P of Eraser = Rs 10 ✬

Step-by-step explanation:

Given:

• Rajan bought 3 erasers and 5 pens for Rs 105.
• Manu bought 4 erasers and 6 pens for Rs 130.

To Find:

• What is the cost price of one eraser?

Solution: Let C.P of one eraser and one pen be Rs x & y respectively.

• In Rajan's case •

➮ 3 erasers and 5 pens = Rs 105

➮ (3x + 5y) = 105

➮ 3x = 105 – 5y

➮ x = (105 – 5y/3) ........(1)

• In Manu's case •

➮ 4 erasers and 6 pens = Rs 130

➮ (4x + 6y) = 130........(2)

Now, put the value of x from equation 1 in equation 2.

$\implies{\rm }$ 4x + 6y = 130

$\implies{\rm }$ 4(105 – 5y/3) + 6y = 130

$\implies{\rm }$ 420 – 20y/3 + 6y = 130

$\implies{\rm }$ 420 – 20y + 18y/3 = 130

$\implies{\rm }$ 420 – 2y = 130 $\times$ 3

$\implies{\rm }$ 420 – 2y = 390

$\implies{\rm }$ –2y = 390 – 420

$\implies{\rm }$ –2y = –30

$\implies{\rm }$ y = 30/2 = 15

Hence, Price of one pen is y = Rs 15

➟ Price of one eraser is x

➟ x = (105 – 5y/3)

➟ x = (105 – 75/3) = 30/3 = Rs 10

Answer 2

Given that ,

• The price of 3 erasers and 5 pens is Rs 105
• The price of 4 erasers and 6 pens is Rs 130

Let , the prices of one eraser and pen be " x " and " y "

According to the question ,

3x + 5y = 105 -- (i)

4x + 6y = 130 --- (ii)

Multiply eq (i) by 4 and eq (ii) by 3 , we obtain

12x + 20y = 420 --- (iii)

and

12x + 18y = 390 --- (iv)

Subtract eq (iv) from (iii) , we get

2y = 30

y = 15

Substitute the value of y = 15 in eq (i) , we get

3x + 5(15) = 105

3x + 75 = 105

3x = 30

x = 10

$\therefore \sf \underline{The \: price \: of \: one \: eraser \: will \: be \: Rs \: 10}$