Question

Hari bought 5 tables and 13 chairs for 11,850, He sold the
table at a profit of 15% and chairs at a loss of 10%. If his total
gain is 315, then the difference between the cost price of one
table and one chair is:

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The difference between the cost price of one  table and one chair is Rs 750

Solution:

Let the cost of 1 table be Rs. x

Let the cost of 1 chair be Rs. y

Hari bought 5 tables and 13 chairs for 11,850

Therefore,

5x + 13y = 11850  ------- eqn 1

Cost Price of 5 tables = 5x  

Cost Price of 13 Chairs = 13y

He sold the  table at a profit of 15% and chairs at a loss of 10%

Selling price of 5 Tables = cp of 5 tables + 15 % of cp of 5 tables

Selling price of 5 Tables = 5x + 15 % of 5x

$\text{Selling price of 5 Tables } = 5x + \frac{15}{100} \times 5x\\\\\text{Selling price of 5 Tables } = 5x + \frac{3x}{4}\\\\\text{Selling price of 5 Tables } = \frac{23x}{4}$

He sold chairs at a loss of 10%

Therefore,

Selling price of 13 chairs = cp of 13 chairs - 10 % of cp of 13 chairs

Selling price of 13 chairs = 13y - 10 % of 13y

$\text{Selling price of 13 chairs } = 13y - \frac{10}{100} \times 13y\\\\\text{Selling price of 13 chairs } = 13y - \frac{13y}{10}\\\\\text{Selling price of 13 chairs } = \frac{130y-13y}{10}\\\\\text{Selling price of 13 chairs } = \frac{117y}{10}$

His total  gain is 315

SP - CP = Profit

Therefore,

$(\frac{23x}{4} + \frac{117y}{10}) - (5x + 13y) = 315\\\\\frac{23x}{4} - 5x + \frac{117y}{10} - 13y = 315\\\\\frac{3x}{4} - \frac{13y}{10} = 315\\\\30x - 52y = 315 \times 40\\\\30x - 52y = 12600 ----- eqn 2$

Solving eqn 1 and eqn 2 by elimination,

x = 1200

y = 450

The difference between the cost price of one  table and one chair is:

x - y = 1200 - 450 = 750

Thus the difference between the cost price of one  table and one chair is Rs 750

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