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16 Jan 2020, 01:38
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Difficulty:

45% (medium)

Question Stats:

62% (02:14) correct 38% (01:45) wrong based on 21 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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Re: When the cost of an article increases by $170, a trader increases his [#permalink] ### Show Tags 16 Jan 2020, 02:10 1 1 Bunuel wrote: 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 Are You Up For the Challenge: 700 Level Questions 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 _________________ Director Joined: 25 Jul 2018 Posts: 549 When the cost of an article increases by$170, a trader increases his  [#permalink]

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17 Jan 2020, 11:33
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

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Re: When the cost of an article increases by $170, a trader increases his [#permalink] ### Show Tags 17 Jan 2020, 18:39 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 _________________ Keep it simple. Keep it blank Re: When the cost of an article increases by$170, a trader increases his   [#permalink] 17 Jan 2020, 18:39
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