# Two grade 11 maths problems

• Jan 23rd 2008, 04:05 PM
andrew2322
1.) Tickets for a concert are available at two prices. The more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for$1800.

2.) The members of a club hire a bus for $2100. Seven members withdraw from the club and the remaining members have to pay$10 more each to cover the cost. How many members originally agreed to go on the bus?

Please explain simply and show working out, THANKS! i'd like to do more of these type problems also, so if anyone out there has some, please post them! THANKS AGAIN! :):)
• Jan 23rd 2008, 08:26 PM
earboth
Quote:

Originally Posted by andrew2322
...

2.) The members of a club hire a bus for $2100. Seven members withdraw from the club and the remaining members have to pay$10 more each to cover the cost. How many members originally agreed to go on the bus?

...

Good morning,

let x be the number of members then
(x-7) finally used the bus
let p be the original price for one ticket then
(p+10) is the new price.

With these conditions you get 2 equations:

$\displaystyle x \cdot p = 2100$ .............[1]

$\displaystyle (x-7)(p+10)=2100$ .............[2]

From [1]: $\displaystyle p=\frac{2100}x$ .............Plug in this term into the equation [2]:

$\displaystyle (x-7)\left( \frac{2100}x + 10\right)=2100$

Expand the brackets and multiply by x. You'll get a quadratic equation in x:

$\displaystyle 10x^2-70x-14700=0$ .............Solve for x. Only one of the 2 solution is valid!
• Jan 23rd 2008, 10:25 PM
andrew2322
Thanks alot
Hey thanks for posting that, thats the exact same equation that i was able to create. I tried solving that quadratic equation by dividing through by 10 then factorising, but it doesnt work... please show me how. thanks
• Jan 23rd 2008, 10:51 PM
The equation does actually factorise which becomes much easier if you take the prime factor expansion of 1470. There are however, no shortage of methods for solving quadratics that are difficult to factorise. The easiest to use is probably the quadratic formula:

if $\displaystyle ax^2+bx+c = 0$ then
$\displaystyle x = \frac {-b \pm \sqrt {b^2-4ac}}{2a}$
• Mar 21st 2009, 11:23 PM
mr fantastic
Quote:

Originally Posted by andrew2322
1.) Tickets for a concert are available at two prices. The more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for$1800.

2.) The members of a club hire a bus for $2100. Seven members withdraw from the club and the remaining members have to pay$10 more each to cover the cost. How many members originally agreed to go on the bus?

Please explain simply and show working out, THANKS! i'd like to do more of these type problems also, so if anyone out there has some, please post them! THANKS AGAIN! :):)