Question 33

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waecmaths question: 

if $x+0.4y=3$ and $y=\tfrac{1}{2}x$ find the value of $(x+y)$

Option A: 

$1\tfrac{1}{4}$

Option B: 

$2\tfrac{1}{2}$

Option C: 

$3\tfrac{3}{4}$

Option D: 

5

waecmaths solution: 

$\begin{align}  & x+0.4y=3---(i) \\ & y=\tfrac{1}{2}x---(ii) \\ & \text{Substitute }\tfrac{1}{2}x\text{ for }y\text{ in }(i) \\ & x+0.4(\tfrac{1}{2}x)=3 \\ & x+0.2x=3 \\ & 1.2x=3 \\ & x=\frac{3}{1.2}=2.5 \\ & y=\tfrac{1}{2}x=\tfrac{1}{2}(2.5)=1.25 \\ & x+y=2.5+1.25=3.75 \\ & x+y=3\tfrac{75}{100}=3\tfrac{3}{4} \\\end{align}$

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