Construct formulae and then solve............?

Question

A boy buys some packets of sweets for £1.20. If the sweets had been 3p a packet cheaper, he would have recieved 2 more packets for his money. How many packets did he buy?

Answer 1

Let N be the number of packets
Let P be the price of packets

NP = 120
(N+2)(P - 3) = 120

Solve the first equation for one of the variables:
P = 120/N

Substitute into the second:
(N + 2)(120/N - 3) = 120

Multiply out using FOIL:
(N*120/N - 3N + 2(120/N) - 6 = 120
120 - 3N + 240/N - 6 = 120
-3N + 240/N - 6 = 0

Multiply both sides by N:
-3N² + 240 - 6N = 0

Divide both sides by -3:
N² + 2N - 80 = 0

Now factor:
(N - 8)(N + 10) = 0

So the number of packets is either 8 or -10. Only one makes sense.

The boy bought 8 packets (at 15p). He could have bought 10 at 12p for the same amount.

Answer 2

Let X be the number of packets bought for 1. 20.
thus each packet cost 1.20 / X = 120/X (in pence)

If 3 pence cheaper then the cost is 120 /X - 3 each.

thus X + 2 costs
(120/X - 3 ) * (X + 2) = 120

so 120 + 240/ X - 6X - 6 = 120;

for which x is either -10 or 8 , so i pick 8, as to be the numberof packets bought

thus if X = 8 then it cost 120 / 8 = 15 pence each.

if 2 more packets then X = 10 , and the 120/10 = 12 pence three pence cheaper.

Answer 3

let x = the price of sweets
let y = no of packets purchased

therefore,

xy = 120 --------> eqn 1

from eqn 1,
x = 120/y -------> eqn 2

and,

( x - 3 )( y + 2 ) = 120 -------> eqn 3

substituting eqn 2 into eqn 3, we get,

( 120/y - 3 )( y + 2 ) = 120
120 + 240/y - 3y - 6 - 120 = 0
240/y - 3y - 6 = 0
240 - 3y^2 - 6y = 0
3y^2 + 6y - 240 = 0
y^2 + 2y - 80 = 0

solving the quad eqn using factorization,

( y + 10 )( y - 8 ) = 0

therefore,

y = -10 or y = 8

since the number of packets he bought cannot be valued negatively, therefore the value y = -10 is neglected.

so the boy bought 8 packets.

Answer 4

n is number of packets, p is price:
np = 120 = (n+2)(p - 3)
np + 2p - 3n - 6 = 120
120 + 2p - 3n - 6 = 120
2p - 3n - 6 = 0
also if np = 120, p = 120/n, so
2(120/n) - 3n - 6 = 0
80/n - n - 2 = 0
80 - n² - 2n = 0
n² + 2n - 80 = 0
(n+10)(n-8) = 0
n = 8 ....... ignoring the negative solution