Answer :
He sold his entire stock for $10000.
Let the number of shirts be x.
Let the number of ties be y.
The shirt is at 3 for $100. Therefore, a shirt would cost $(100/3)
The tie cost $20 each.
Therefore, Multiplying by the number of shirts and ties:
100x / 3 + 20y = 10000 .............(i).
If he had sold half the shirts and two-thirds the ties, the equation becomes:
100/3 * (x/2) + 20 * (2y / 3) = 6000
50x / 3 + 40y / 3 = 6000 ..............(ii)
Multiply equation (ii) by 2.
2*(50x / 3 + 40y / 3 ) = 2* 6000
100x / 3 + 80y / 3 = 12000 .........(iii).
Equation (i) Minus (iii) . Inorder to eliminate x.
100x / 3 - 100x / 3 + 20y - 80y / 3 = 10000 - 12000
-20y / 3 = - 2000.
y = 3 * 2000 / 20
y = 300.
Substituting into equation (i).
100x / 3 + 20y = 10000
100x / 3 + 20*300 = 10000
100x / 3 + 6000 = 10000
100x / 3 = 10000 - 6000
100x / 3 = 4000
x = 4000 * 3 / 100
x = 120.
So he actually sold 120 shirts and 300 ties. Cheers.
Let the number of shirts be x.
Let the number of ties be y.
The shirt is at 3 for $100. Therefore, a shirt would cost $(100/3)
The tie cost $20 each.
Therefore, Multiplying by the number of shirts and ties:
100x / 3 + 20y = 10000 .............(i).
If he had sold half the shirts and two-thirds the ties, the equation becomes:
100/3 * (x/2) + 20 * (2y / 3) = 6000
50x / 3 + 40y / 3 = 6000 ..............(ii)
Multiply equation (ii) by 2.
2*(50x / 3 + 40y / 3 ) = 2* 6000
100x / 3 + 80y / 3 = 12000 .........(iii).
Equation (i) Minus (iii) . Inorder to eliminate x.
100x / 3 - 100x / 3 + 20y - 80y / 3 = 10000 - 12000
-20y / 3 = - 2000.
y = 3 * 2000 / 20
y = 300.
Substituting into equation (i).
100x / 3 + 20y = 10000
100x / 3 + 20*300 = 10000
100x / 3 + 6000 = 10000
100x / 3 = 10000 - 6000
100x / 3 = 4000
x = 4000 * 3 / 100
x = 120.
So he actually sold 120 shirts and 300 ties. Cheers.
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 = [tex]\rm \dfrac{100x}{3}[/tex]
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.
[tex]\rm \dfrac{Price \ of \ each \ set \ of \ shirt}{Number \ of \ sets \ of \ shirts } + Price \ of \ each\ ties = Total \ sale \ of \ shirt \ and \ ties \\\\\dfrac{100x}{3} + 20y = 10000\\\\33.33 x+ 20y = 10000[/tex]
And he had sold only half the shirts and two-thirds of the ties he would have received $6000.
[tex]\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}[/tex]
[tex]= \rm Two \ third \times Price \ of \ each\ ties = \dfrac{2}{3} \times 20y = \dfrac{40y}{3}[/tex]
Then,
The required equation is,
[tex]\rm\dfrac{50x }{3} + \dfrac{40y }{3} = 6000[/tex]
Multiply equation (ii) by 2,
[tex]\rm 2\times [\dfrac{50x }{3} ] + \dfrac{40y }{3} = 2\times 6000\\\\\dfrac{100x }{ 3 } + \dfrac{80y }{3} = 12000[/tex]
Subtract equation (i) from (ii) in order to eliminate x.
[tex]\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.[/tex]
Substitute the value of y in equation 1,
[tex]\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[/tex]
Hence, He sells 300 shirts and 120 ties in the sale.
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https://brainly.com/question/15894203
