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investor purchased a bond

Expert replies
by sanju09 » Thu Apr 28, 2011 2:23 am
An investor purchased a bond for p dollars on Monday. For a certain number of days, the value of the bond increased by r percent per day. After this period of constant increase, the bond decreased the next day by q dollars and the investor decided to sell the bond that day for v dollars. When did the investor sell the bond if r = 100 √ [(v + q)/p] – 1?
A. Wednesday of the same week
B. Thursday of the same week
C. Friday of the same week
D. Monday of the next week'
E. Tuesday of the next week
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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Source: — Problem Solving |

by Anurag@Gurome » Thu Apr 28, 2011 3:01 am
sanju09 wrote:An investor purchased a bond for p dollars on Monday. For a certain number of days, the value of the bond increased by r percent per day. After this period of constant increase, the bond decreased the next day by q dollars and the investor decided to sell the bond that day for v dollars. When did the investor sell the bond if r = 100 √ [(v + q)/p] – 1?
A. Wednesday of the same week
B. Thursday of the same week
C. Friday of the same week
D. Monday of the next week'
E. Tuesday of the next week
Original price of the bond = $p
If the bond increased by r percent per day for x days, then price of the bond after x days = p(1 + r/100)^x
When the bond decreased by $q, price of the bond = p(1 + r/100)^x - q = v, which implies v + q = p(1 + r/100)^x
or (v + q)/p = (1 + r/100)^x
It is given that r = 100 √[(v + q)/p] – 1
So, (v + q)/p = (1 + √[(v + q)/p] – 1)^x
(v + q)/p = {√[(v + q)/p]}^x
[(v + q)/p]^(1/x) = [(v + q)/p]^(1/2)
x = 2, which implies the investor bought the bond on Monday, then price of the bond on Wednesday = p(1 + r/100)²
Then its price decreased by $q, so, price of the bond on Thursday = p(1 + r/100)² - q = v, which implies v + q = p(1 + r/100)² or (v + q)/p = (1 + r/100)²
√(v + q)/p = 1 + r/100
100*[√(v + q)/p - 1] = r

Hence, the investor sold the bond on Thursday of the same week.

The correct answer is B.
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by sanju09 » Fri Apr 29, 2011 1:02 am
Anurag@Gurome wrote:
sanju09 wrote:An investor purchased a bond for p dollars on Monday. For a certain number of days, the value of the bond increased by r percent per day. After this period of constant increase, the bond decreased the next day by q dollars and the investor decided to sell the bond that day for v dollars. When did the investor sell the bond if r = 100 √ [(v + q)/p] – 1?
A. Wednesday of the same week
B. Thursday of the same week
C. Friday of the same week
D. Monday of the next week'
E. Tuesday of the next week
Original price of the bond = $p
If the bond increased by r percent per day for x days, then price of the bond after x days = p(1 + r/100)^x
When the bond decreased by $q, price of the bond = p(1 + r/100)^x - q = v, which implies v + q = p(1 + r/100)^x
or (v + q)/p = (1 + r/100)^x
It is given that r = 100 √[(v + q)/p] – 1
So, (v + q)/p = (1 + √[(v + q)/p] – 1)^x
(v + q)/p = {√[(v + q)/p]}^x
[(v + q)/p]^(1/x) = [(v + q)/p]^(1/2)
x = 2, which implies the investor bought the bond on Monday, then price of the bond on Wednesday = p(1 + r/100)²
Then its price decreased by $q, so, price of the bond on Thursday = p(1 + r/100)² - q = v, which implies v + q = p(1 + r/100)² or (v + q)/p = (1 + r/100)²
√(v + q)/p = 1 + r/100
100*[√(v + q)/p - 1] = r

Hence, the investor sold the bond on Thursday of the same week.

The correct answer is B.
Nice explanation
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion