He sells 300 shirts and 120 ties in the sale.
Given that,
A shopkeeper sold his entire stock of shirts and ties in a sale for $10000.
The shirt was priced at 3 for $100 and the ties $20 each.
If he had sold only half the shirts and two-thirds of the ties he would have received $6000.
We have to determine,
How many did he sell in the sale?
According to the question,
Let the number of shirts be x.
And the number of ties be y.
The shirt was priced at 3 for $100 = @@MATH_TOKEN_0@@
The ties are $20 each = 20y
A shopkeeper sold his entire stock of shirts and ties in a sale for $10000.
The shirt was priced at 3 for $100 and the ties $20 each.
@@MATH_TOKEN_1@@
And he had sold only half the shirts and two-thirds of the ties he would have received $6000.
\(\rm = \dfrac{Price \ of \ each \ set \ of \ shirt \times he \ had \ sold \ half \ shirts}{Number \ of \ sets \ of \ shirts }\\\\= \dfrac{100x \times \dfrac{1}{2}}{3}\\\\= \dfrac{50x}{3}\)
\(= \rm Two \ third \times Price \ of \ each\ ties = \dfrac{2}{3} \times 20y = \dfrac{40y}{3}\)
Then,
The required equation is,
\(\rm\dfrac{50x }{3} + \dfrac{40y }{3} = 6000\)
Multiply equation (ii) by 2,
\(\rm 2\times [\dfrac{50x }{3} ] + \dfrac{40y }{3} = 2\times 6000\\\\\dfrac{100x }{ 3 } + \dfrac{80y }{3} = 12000\)
Subtract equation (i) from (ii) in order to eliminate x.
\(\rm \dfrac{100x }{ 3 }- \dfrac{100x }{ 3 } + 20y - \dfrac{80y }{3 }= 10000 - 12000\\\\ \dfrac{ 60y -80y }{3 } = - 2000\\\\ \dfrac{ -20y }{3 } = - 2000\\\\ y = \dfrac{3 \times2000 }{20}\\\\ y=\dfrac{6000}{20}\\\\ y = 300.\)
Substitute the value of y in equation 1,
\(\dfrac{100x }{3} + 20y = 10000\\\\ \dfrac{100x }{3} + 20\times300 = 10000\\\\\dfrac{100x }{3} + 6000 = 10000 \\\\ \dfrac{100x }{3} = 10000 - 6000\\\\ \dfrac{100x }{3} = 4000\\\\ x =\dfrac{ 4000 \times 3 }{ 100}\\\\x =\dfrac{12000}{10} \\\\x = 120\)
Hence, He sells 300 shirts and 120 ties in the sale.
For more details refer to the link given below.
https://brainly.com/question/15894203
