Question

14. Cost of two pens, five pencils and seven erasers is 37.
Cost of seven pens, one eraser and two pencils is 49.
What is the cost of nine pencils and forty seven pens?
(A) 184
(B) 5276
(C) 284
(D) None of these

Answer 1

52thousand 7hundred six

Answer 2

ANSWER :

• (D) None of these.

Because, if the cost of two pens, five pencils and seven erasers is 37 and the cost of seven pens, one eraser and two pencils is 49; then cost of nine pencils and forty seven pens will be 306.

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SOLUTION :

❒ Given :-

• Cost of 2 pens, 5 pencils and 7 erasers = 37

• Cost of 7 pens, 2 pencils and 1 eraser = 49

❒ To Find :-

• Cost of 47 pen and 9 pencils = ?

❒ Calculation :-

Let,

• Cost of 1 pen be x

• Cost of 1 pencil be y

• Cost of 1 eraser be z

Then, we have to find :-

• 47x + 9y = ?

➮ According to first condition,

• 2x + 5y + 7z = 37 ———————> (1)

➮ According to second condition,

• 7x + 2y + z = 49 ———————> (2)

Now,

➮ Multiplying Eq. (2) with 7, we get :

$\bigstar \: \: \sf{ Eq. (2) \times 7 \implies \: 7 \times (7x + 2y + z) = 7 \times 49} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \implies \: 49x + 14y + 7z = 343 \: \: - - - - - > (3)}$

➮ Subtracting Eq. (1) from Eq. (3), we get :

$\bigstar \: \: \sf{Eq. (3) - Eq. (1) \implies \: (49x + 14y + 7z)-(2x + 5y + 7z) = 343 - 37} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf {\implies \: 49x + 14y + 7z-2x - 5y - 7z = 306} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \implies \: 49x - 2x + 14y - 5y + 7z - 7z = 306} \\ \\ \sf{\implies \: 47x + 9y = 306} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• As 47x + 9y = 306; hence, cost of 47 pen and 9 pencils is 306.