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# ratios

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Hi good people!,

This problem I bet is simple and straight forward, but I cannot wrap my brain around this..please help:

A rectangle has a length that is 16cm more than it's breadth. If the length and breadth are in the ratio 4:3, determine the length and breadth of the rectangle..

I did this bet get stuck:

$${length \over Breadth}={4 \over 3}={x+16 \over y}$$

that gives:

$$4y=3x+48$$

Now...I'm stuck.. Apr 23, 2018
edited by juriemagic  Apr 23, 2018
edited by juriemagic  Apr 23, 2018

#1
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Let the width of the rectangle = W

Then the length of the rectangle = W + 16

[W + 16] / W = 4 / 3, solve for W                 cross multiply

[W + 16] = 4 / 3W                                        Subtract W frm both sides

16 = 4 / 3W - W

16 = 1 / 3W                                                  divide both sides by 1/3

W = 16 / (1/3)

W = 16 x 3

W = 48 cm - the width of the rectangle

48 + 16 = 64 cm - the length of the rectangle. So that:

64:48 = 4:3

Apr 23, 2018
edited by Guest  Apr 23, 2018
#3
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Hi guest,

sorry for replying only now, thanx a lot,you have helped me see my error..thanx again..

juriemagic  Apr 23, 2018
#2
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Hi Juriemagic,

Your y is equal to x.

The length and bredth are     x and  x+16

Apr 23, 2018
#4
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Hi Melody,

yep, I realised my error...thanx for your help!!

juriemagic  Apr 23, 2018
#5
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Here's a solution.

l:b=4:3, so that's 4x:3x

So, x=16

4x=64, and 3x=48

Thus, 64:48=4:3

Apr 23, 2018
#6
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Thanx Tertre, I do appreciate...

juriemagic  Apr 25, 2018