Difference between revisions of "Euclid 2018"

(Created page with "== Problem == a) Given x = 11, find (x) + (x+1) + (x+2) + (x+3) b) Given <math>\frac{a}{6}</math> + <math>\frac{6}{18}</math> = 1, find the value of a c) The total cost of o...")
 
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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 <math>4.15. One chocolate bar costs </math>1 more than one pack of gum. Find the price of a chocolate bar.
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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.
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== Solution ==
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a) Plug x = 11 back to the original equation
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(11) + (11+1) + (11+2) + (11+3) = <math>\boxed{50}</math>
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b)By simplifcation:
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    <math>\frac{3a}{18}</math> + <math>\frac{6}{18}</math> = 1
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    <math>\frac{3a+6}{18}</math> = 1
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    <math>3a + 6 = 18</math>
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    <math>3a = 12</math>
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    <math>a = 4</math>
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Therefore a = <math>\boxed{4}</math>
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c) Set One chocolate bar as x + 1 and one piece of gum as x dollars. We get:
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    <math>2(x) + (x + 1) = 4.15</math>
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    <math>3x + 1 = 4.15</math>
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    <math>x = 1.05</math>
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Therefore one piece of chocolate bar is 1 + 1.05 = <math>\boxed{2.05}</math>
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~ North America Math Contest Go Go Go
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== Video Solution ==

Latest revision as of 14:44, 11 April 2021

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 = $\boxed{2.05}$

~ North America Math Contest Go Go Go

Video Solution