# Math Help - problem simultaneous equations

1. ## problem simultaneous equations

Hey i need help, i dont know the working outs.

a shopkeeper sold his entire stock of shirts and ties in a sale for $10 000. The shirts were 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. how many of each did he sell in the sale?

answer- 120 shirts and 300 ties

attempt.
shirts -x ties - y
x+y= 10 000

3x= $100 y=$20
if he had sold only half the shirts and two-thirds of the ties he would have received $6000 ? shirts x/2 ties 2y/3 ?? so x/2 + 2y/3= 6000 ?? 3- a party was organized for thirty people at which they could have either a hamburger or a pizza. if there was five times as many hamburgers as pizzas calculate the number of each. x be hamburger, y be pizza 5x.. answer- 5 pizzas and 25 hamburgers thankss 2. Originally Posted by fresh_ Hey i need help, i dont know the working outs. a shopkeeper sold his entire stock of shirts and ties in a sale for$10 000. The shirts were 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 $6000. how many of each did he sell in the sale? How about we say S for the total number of shirts and T for the total number of ties. And let s be the cost of one shirt and t be the cost of one tie. So S*s + T*t = 10 000. 3s = 100, t = 20. So if we rewrite 3s = 100 as s = 100/3, and filling this and t = 20 into the first equation we have: 100/3 * S + 20*T = 10 000. Now, if we sold half the shirts and two-thirds of the ties is if we sell 1/2 * S and 2/3 * T, and from selling this we get 6000. So 1/2 * S * s + 2/3 * T * t = 6000. or, filling in s = 100/3 and t = 20 as before: 1/2 * S * 100/3 + 2/3 * T * 20 = 6000. So there you have your two simultaneous equations: 100/3 * S + 20*T = 10000 100/6 * S + 40/3 * T = 6000 Solving for S and T will give you the total amount of shirts and ties. And the question is how many of each did he sell in his sale. Well, he sold all of them. So S is the number of shirts he sold, and T is the number of ties he sold... (And just to check with your given answers so as I don't get it wrong and look a complete fool: 100/3 * 120 + 20*300 = 4000 + 6000 = 10000. 100/6*120 + 40/3*300 = 2000 + 4000 = 6000. Phew...) So yes - your method was fine! You got your simultaneous equations, and you just had to finish it out... ( Note to self: read entire post in future...) 3. ## Answer 3*x - Sold shirts (3*x because 3 shirts=100$ and profit from shirts=soldshirts/3*100)
y - Sold ties

$\frac {3*x}{3}*100+y*20=10000$

$100x+20y=10000$

$20(5x+y)=10000$

$5x+y=\frac {10000}{20}$

$5x+y=500$

$y=500-5x$

If he had sold half his shirts- $\frac {3*x}{2}$
two-thirds of the ties - $2*\frac {y}{3}$
he would have received 6000\$ so:

$\frac {\frac {3*x}{3}}{2}*100+2*\frac {y}{3}*20=6000$

$\frac {x}{2}*100+2*\frac {y}{3}*20=6000$

$y=500-5x$ from above

$\frac {x}{2}*100+2*\frac {500-5x}{3}*20=6000$

$\frac {x}{2}*100+\frac {40(500-5x)}{3}=6000$

$\frac {3*100x+2*40(500-5x)}{6}=6000$

$300x+80(500-5x)=36000$

$300x+40000-400x=36000$

$-100x=-4000$

$x=40$

But the num. of shirts= $3*x=40*3=120$

$y=500-5x=500-40*5=500-200=300$

So we have 120 shirts and 300 ties

4. Originally Posted by fresh_
3- a party was organized for thirty people at which they could have either a hamburger or a pizza. if there was five times as many hamburgers as pizzas calculate the number of each.
x be hamburger, y be pizza
5x..

answer- 5 pizzas and 25 hamburgers

What fool would order a hamburger instead of a pizza?

Ok - 30 people. 5 times as many hamnburgers as pizzas.
So, for every one Pizza ordered there are 5 hamburgers.
If we let H = total number of hamburgers and P = total number of pizzas, we know H = 5P, right?
And then the total amount of 'foodstuffs' purchased is 30 because there are 30 people present. So H + P = 30.
But we know H = 5P.
So
5P + P = 30.
6P = 30.
P = 5, so H = 5*P = 25.
So we have: 5 pizzas and 25 hamburgers...

5. thanks guys , you guys rock !