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# If x is the number on the number line between 5 and 15 that

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If x is the number on the number line between 5 and 15 that [#permalink]  26 Aug 2018, 03:32
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Question Stats:

45% (01:08) correct 54% (01:06) wrong based on 31 sessions
If x is the number on the number line between 5 and 15 that is twice as far from 5 as from 15, then x is

A. $$5 \frac{2}{3}$$

B. $$10$$

C. $$11 \frac{2}{3}$$

D. $$12 \frac{1}{2}$$

E. $$13 \frac{1}{3}$$
[Reveal] Spoiler: OA

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VP
Joined: 20 Apr 2016
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WE: Engineering (Energy and Utilities)
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Re: If x is the number on the number line between 5 and 15 that [#permalink]  26 Aug 2018, 04:35
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Carcass wrote:
If x is the number on the number line between 5 and 15 that is twice as far from 5 as from 15, then x is

A. $$5 \frac{2}{3}$$

B. $$10$$

C. $$11 \frac{2}{3}$$

D. $$12 \frac{1}{2}$$

E. $$13 \frac{1}{3}$$

The distance from $$5$$ = $$2x$$

and the distance from $$15 = x$$

The distance between $$15$$ and$$5$$= $$15 - 5 = 10$$

That means $$2x + x = 10$$

or $$3x = 10$$

or $$x = \frac{10}{3}$$

or $$2x = \frac{20}{3}$$

Now we can calulate the value of $$x$$ in $$2$$ways

i.e.$$5 + 2x = 5 + \frac{20}{3} = \frac{35}{3} = 11{\frac{2}{3}}$$

or $$15 - x = 15 - \frac{10}{3} = \frac{35}{3}= 11{\frac{2}{3}}$$
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Re: If x is the number on the number line between 5 and 15 that   [#permalink] 26 Aug 2018, 04:35
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