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Bunuel
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When the cost of an article increases by $170, a trader increases his 16 Jan 2020, 02:38
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95% (hard)

Question Stats:

56% (03:07) correct 44% (03:06) wrong based on 167 sessions

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When the cost of an article increases by $170, a trader increases his selling price by 10%. Because of these changes his profit percentage decreases from 20% to 15%. What is the cost price of the article after the increase?

A. $980
B. $1,150
C. $1,200
D. $1,250
E. $1,320


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When the cost of an article increases by $170, a trader increases his 17 Jan 2020, 12:33
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Cost price ———x——x+170———
Selling price —1.2x——1.32x———
(——————-20%——10%———)

Now, the profit percentage is 15%.

--> \(\frac{115}{100}*(x+170) = (\frac{132}{100})*x\)

—> 115x+ 115*170 = 132x
17x = 115*170
—> x= 1150
Cost price after the increase = 1150+ 170 = 1320

The answer is E.

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GMAT Focus 1: 735 Q90 V89 DI81
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When the cost of an article increases by $170, a trader increases his 16 Jan 2020, 03:10
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Bunuel
When the cost of an article increases by $170, a trader increases his selling price by 10%. Because of these changes his profit percentage decreases from 20% to 15%. What is the cost price of the article after the increase?

A. $980
B. $1,150
C. $1,200
D. $1,250
E. $1,320


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Let the cost price and selling price initially be \(c_1\) and \(s_1\), and later after increase be \(c_2\) and \(s_2\).

profit initially =20%=\(0.2c_1=s_1-c_1......s_1=1.2c_1\)

profit later after increase =15%=\(0.15c_2=s_2-c_2.........s_2=1.15c_2\)

Now..\(s_2=1.1s_1\)..and..\(c_2=c_1+170\), so

1.1s_1=1.15(c_1+170).........1.1(1.2c_1)=1.15(c_1+170)......132c_1=115c_1=115*170.......17c_1=115*170...c_1=1150...c_2=c_1+170=1150+170=1320

E­
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When the cost of an article increases by $170, a trader increases his 17 Jan 2020, 19:39
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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
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When the cost of an article increases by $170, a trader increases his 10 Apr 2024, 09:33
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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
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When the cost of an article increases by $170, a trader increases his 30 Sep 2025, 02:42
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