Q) Aarush bought 2 pencils and 3 chocolates for Rs. 11 and Tanish bought 1 pencil and 2 chocolates for Rs. 7 from the same shop. Represent this situation in the form of a pair of linear equations. Find the price of 1 pencil and 1 chocolate, graphically.

(Q 33 – 30/2/2 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s consider price of 1 pencil is X and price of 1 chocolate is Y.

Next, by given condition “Aarush bought 2 pencils and 3 chocolates for Rs. 11”

∴ we can write it as 2 X + 3 Y = 11 …………. (i)

Step 2: Next condition is: 2. “Tanish bought 1 pencil and 2 chocolates for Rs. 7”

∴ we can write it as X + 2 Y = 7 ………… (ii)

These 2 linear equations represent the situation

Step 3: Next, Let’s find values of X and Y by solving these 2 eqiations:

By multiplying equation (ii) by 2, we get:

2X + 4 Y = 14 …………… (iii)

Now, we subtract equation (i) from equation (iii), we get:

(2 X + 4 y) – (2 X + 3 Y) = 14 – 11

∴ 2 X + 4 Y – 2 X – 3 Y = 3

∴ Y = 3

Step 4: Now, we substitute value of Y in equation (ii), we get:

X + 2 Y = 7

∴ X + 2 (3) = 7

∴ X + 6 = 7

∴ X = 7 – 6 = 1

Therefore, price of one pencil is Rs. 1 and the price of one chocolate is Rs. 3.

Step 5:  tNext, we plot these two equations:

It is clear that the two linear equation intersect each other at point (1, 3). Hence, this is the solution of these equations.

Here X = 1 => price of one pencil is Rs. 1 and Y = 3 => the price of one chocolate is Rs. 3.

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