Difference between revisions of "Euclid 2018"
(→Problem) |
|||
| Line 4: | Line 4: | ||
b) Given <math>\frac{a}{6}</math> + <math>\frac{6}{18}</math> = 1, find the value of a | b) Given <math>\frac{a}{6}</math> + <math>\frac{6}{18}</math> = 1, find the value of a | ||
| − | c) The total cost of one chocolate bar and two identical packs of gums is | + | c) The total cost of one chocolate bar and two identical packs of gums is 4.15 dollars. One chocolate bar costs 1 dollar more than one pack of gum. Find the price of a chocolate bar. |
| + | |||
| + | == Solution == | ||
| + | a) Plug x = 11 back to the original equation | ||
| + | |||
| + | (11) + (11+1) + (11+2) + (11+3) = <math>\boxed{50}</math> | ||
| + | |||
| + | b)By simplifcation: | ||
| + | |||
| + | <math>\frac{3a}{18}</math> + <math>\frac{6}{18}</math> = 1 | ||
| + | |||
| + | <math>\frac{3a+6}{18}</math> = 1 | ||
| + | |||
| + | <math>3a + 6 = 18</math> | ||
| + | |||
| + | <math>3a = 12</math> | ||
| + | |||
| + | <math>a = 4</math> | ||
| + | |||
| + | Therefore a = <math>\boxed{4}</math> | ||
| + | |||
| + | c) Set One chocolate bar as x + 1 and one piece of gum as x dollars. We get: | ||
| + | <math>2(x) + (x + 1) = 4.15</math> | ||
| + | <math>3x + 1 = 4.15</math> | ||
| + | <math>x = 1.05</math> | ||
| + | |||
| + | Therefore one piece of chocolate bar is 1 + 1.05 = <math>\boxed{2.05}</math> | ||
== Video Solution == | == Video Solution == | ||
Revision as of 14:40, 11 April 2021
Problem
a) Given x = 11, find (x) + (x+1) + (x+2) + (x+3)
b) Given 
+ 
= 1, find the value of a
c) The total cost of one chocolate bar and two identical packs of gums is 4.15 dollars. One chocolate bar costs 1 dollar more than one pack of gum. Find the price of a chocolate bar.
Solution
a) Plug x = 11 back to the original equation
(11) + (11+1) + (11+2) + (11+3) = 
b)By simplifcation:
+
= 1
+ = 1
= 1
Therefore a = 
c) Set One chocolate bar as x + 1 and one piece of gum as x dollars. We get:
Therefore one piece of chocolate bar is 1 + 1.05 = 
