Euclid 2018

Revision as of 15:40, 11 April 2021 by Timz2005 (talk | contribs) (Problem)

Problem

a) Given x = 11, find (x) + (x+1) + (x+2) + (x+3)

b) Given $\frac{a}{6}$ + $\frac{6}{18}$ = 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) = $\boxed{50}$

b)By simplifcation:

    $\frac{3a}{18}$ + $\frac{6}{18}$ = 1
    $\frac{3a+6}{18}$ = 1
    $3a + 6 = 18$
    $3a = 12$
    $a = 4$

Therefore a = $\boxed{4}$

c) Set One chocolate bar as x + 1 and one piece of gum as x dollars. We get:

    $2(x) + (x + 1) = 4.15$
    $3x + 1 = 4.15$
    $x = 1.05$

Therefore one piece of chocolate bar is 1 + 1.05 = $\boxed{2.05}$

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