Mr Roy buys potatoes at $₹ 12 / k g$ and onions at $₹ 14 / k g$ . If he buys $x k g$ of potatoes and $3 k g$ less onions than the potatoes, find the algebraic expression for his expenditure.

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Solution

Step I : To find the Cost of $x k g$ potatoes :
Given weight of potatoes $= x k g$
weight of onions $= (x - 3) k g$
Given cost of $1 k g$ potatoes $=$ $₹ 12$
Cost of $x k g$ potatoes $=$ $12 \times x = ₹ 12 x$
Step II : To find the cost of $(x - 3) k g$ onions :
Given cost of $1 k g$ onions $=$ $₹ 14$
Cost of $(x - 3) k g$ onions $=$ $14 \times (x - 3) = ₹ 14 x - 42$
Step III : To find the total expenditure of Mr Roy :
Total expenditure of Mr. Roy $= 12 x + 14 x - 42 = ₹ (26 x - 42)$
Hence, the above is the required answer.

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Mr Roy buys potatoes at ₹12/kg and onions at ₹14/kg . If he buys xkg of potatoes and 3kg less onions than the potatoes, find the algebraic expression for his expenditure.

Mr Roy buys potatoes at $₹ 12 / k g$ and onions at $₹ 14 / k g$ . If he buys $x k g$ of potatoes and $3 k g$ less onions than the potatoes, find the algebraic expression for his expenditure.