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Math Expert
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Director
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VP
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Re: When the cost of an article increases by $170, a trader increases his [#permalink]
Old C.P = c
Old S.P = s
given
\(\frac{s-c}{c }= \frac{20}{100}\)
or
5s = 6c ----- eq (1)

new \(SP = s^#\)
new \(CP = c^#\)
given
\( s^# - c^#/ c^# = 15/100\)
or
\(20 s^# = 23 c^#\) ---- eq (2)

also \( s^# = 11/10 s\)
and \(c^# = c +170\)
substitute back in equation 2 from 1 and given above props

\(\frac{20*11}{ 10 } * 6c = 23 (c+170)\)
17 c = 23 *170*5
or c = 1150
and \(c^# = c +170\)

Thus \(c^# = 1150+170 = 1320\)
E
Intern
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When the cost of an article increases by $170, a trader increases his [#permalink]
According to question -

the Old SP can be written as Old SP = 120/100 of Old CP
&
the new SP can be written as New SP = 120/100 * 110/100 of CP i.e. SP = 132/100 of Old CP


Also New SP (can be written as) New SP = 115/100 of New CP i.e. 115/100 * (Old CP + 170)

So equation will be 115/100 * (Old CP+170) = 132/100 * Old CP. It will be equals to Old CP = 1150


So the New CP = Old CP + 170 = 1320

Option E is the Answer­
GMAT Club Bot
When the cost of an article increases by $170, a trader increases his [#permalink]
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