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28 May 2019, 03:55
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In a developing country, the price of a stock is directly proportional to the reciprocal of the inflation in the country, which is in turn directly proportional to the local diesel prices. The price of the stock is 100 currency units when the inflation is 10 units and when the inflation is 12 units, the local price of the diesel is 60 currency units. By how many currency units should the local diesel price fall so that the price of the stock, which is currently at 200 currency units, increases by 25 percent?
A. 5
B. 10
C. 25
D. 30
E. 40
source: e-GMAT Scholaranium
A. 5
B. 10
C. 25
D. 30
E. 40
source: e-GMAT Scholaranium
28 May 2019, 09:49
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CP=k/I
100=k/10, or k=1000
CP=1000/I......(1)
I=x*DP
12=x*60, or x=1/5
I=DP/5 .......(2)
From (1) and (2)
CP=5000/DP
or DP= 5000/CP
When CP=200, DP=25
When CP=200*1.25=250, DP=20
Number of currency units should the local diesel price fall=25-20=5
100=k/10, or k=1000
CP=1000/I......(1)
I=x*DP
12=x*60, or x=1/5
I=DP/5 .......(2)
From (1) and (2)
CP=5000/DP
or DP= 5000/CP
When CP=200, DP=25
When CP=200*1.25=250, DP=20
Number of currency units should the local diesel price fall=25-20=5
31 Mar 2020, 01:59
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Gmatprep998
In a developing country, the price of a stock is directly proportional to the reciprocal of the inflation in the country, which is in turn directly proportional to the local diesel prices. The price of the stock is 100 currency units when the inflation is 10 units and when the inflation is 12 units, the local price of the diesel is 60 currency units. By how many currency units should the local diesel price fall so that the price of the stock, which is currently at 200 currency units, increases by 25 percent?
A. 5
B. 10
C. 25
D. 30
E. 40
source: e-GMAT Scholaranium
Let S, I, and D be the stock price, inflation, and diesel price, respectively.A. 5
B. 10
C. 25
D. 30
E. 40
source: e-GMAT Scholaranium
- • \(S = \frac{K}{I}\)…(i), where K is the proportionality constant.
• When \(I = 10, S =100\)
- o \(100 = \frac{K}{10}\)
o \(K = 1000\)
- o When \(D = 60, I = 12\)
o \(12 = M*60\)
o \(M = \frac{1}{5}\)
- o \(S = \frac{K}{M*D}\)
- o Required \(S = 200*1.25 = 240\)
• When S is 200
- o \(200 = \frac{1000}{\frac{1}{5}*D}\)
o \(D = 25\)
- o \(240 = \frac{1000}{\frac{1}{5}*D}\)
o \(D = \frac{5000}{240} = 20\)(approx.)
