Question

On selling a tea-set at 5% loss and a lemon-set at 15% gain, a crockery seller
Rs. 7. If he sells the tea-set at 5% gain and the lemon-set at 10% gain, he gains Rs. 13.
Find the actual price of the tea-set and the lemon-set.​

Answer 1

Let the cost price of the tea-set be Rs. x

Then,the cost price of the lemon-set be Rs. y

Case 1 :

When, tea -set is sold at 5 % loss and lemon-set at 15% gain. [ Given ]

According to the question,

$\sf{ \red \star \: Loss \: on \: tea \: set = \dfrac{5x}{100} = \dfrac{x}{20} }$

$\sf{ \red \star \: Gain \: on \: lemon \: set = \dfrac{15y}{100} = \dfrac{3y}{20} }$

$\therefore \sf{ \: Net \: gain = \dfrac{3y}{20} - \dfrac{x}{20} }$

$\implies \: \frac{3y}{20} - \dfrac{x}{20} = 7$

$\implies \: \dfrac{3y - x}{20} = 7$

$\implies \: 3y - x = 7 \times 20$

$\implies \: 3y - x = 140$

$\implies \: x - 3y + 140 = 0.......(i)$

Case 2 :

When, tea -set is sold at 5 % gain and lemon-set at 10% gain. [ Given ]

According to the question,

$\sf { \red\star \: Gain \: on \: tea \: set \: = \dfrac{5x}{100} = \dfrac{x}{20} }$

$\sf{ \red \star \: Gain \: on \: lemon \: set = \dfrac{10y}{100} = \dfrac{y}{10} }$

$\therefore \sf{Total \: gain \: = \dfrac{x}{20} + \dfrac{y}{10} }$

$\implies \: \dfrac{x}{20} + \dfrac{y}{10} = 13$

$\implies\dfrac{x + 2y}{20} = 13$

$\implies \: x + 2y = 13 \times 20$

$\implies \: x + 2y = 260$

$\implies \: x + 2y - 260 = 0.....(ii)$

Now,

★Subtracting equation ( i ) from ( ii )

( x + 2y - 260 ) - ( x - 3y + 140 )

⇒ -5y + 400 = 0

⇒ 400 = 5y

⇒ y = 80

★ Put the value of y in ( i )

x - 3y + 140 = 0

⇒ x - 3 × 80 + 140 = 0

⇒ x - 240 + 140 = 0

⇒ x - 100 = 0

⇒ x = 100

Hence, the actual price of the tea-set is Rs. 100 and the lemonlemon-set is Rs 80 .

Answer 2

hope it helps u dear.........