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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Dr. Kramer plans to invest $20,000 in an account paying 6% i
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21 Mar 2020, 22:23
21 Mar 2020, 22:23
Expert Reply
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Question Stats:
45% (01:26) correct
54% (04:16) wrong based on 11 sessions
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Dr. Kramer plans to invest $20,000 in an account paying 6% interest annually. How much more must she invest at the same time at 3% so that her total annual income during the first year is 4% of her entire investment?
(A) $32,000
(B) $36,000
(C) $40,000
(D) $47,000
(E) $49,000
(A) $32,000
(B) $36,000
(C) $40,000
(D) $47,000
(E) $49,000
Re: Dr. Kramer plans to invest $20,000 in an account paying 6% i
[#permalink]
27 Mar 2020, 11:32
27 Mar 2020, 11:32
2
Expert Reply
Total amount = x
Total annual income = 4x/100
Initial amount = 20000
Income generated = 6/100 * 20000 = 1200
Extra amount to invest = y
Income generated = 3y/100
y + 20000 = x ------- equation 1
y = x - 20000 ---- y expressed in terms of x
\(1200 + \frac{3y}{100} = \frac{4x}{100}\) ----- equation 2
\(1200 + \frac{3(x-20000)}{100 }= \frac{4x}{100} \)------ substitute the value of y
\(\frac{(120000 + 3x - 60000)}{100}= \frac{4x}{100}\)
x = 60000
the additional amount to invest = 60000 - 20000 = 40000
Answer is C.
Total annual income = 4x/100
Initial amount = 20000
Income generated = 6/100 * 20000 = 1200
Extra amount to invest = y
Income generated = 3y/100
y + 20000 = x ------- equation 1
y = x - 20000 ---- y expressed in terms of x
\(1200 + \frac{3y}{100} = \frac{4x}{100}\) ----- equation 2
\(1200 + \frac{3(x-20000)}{100 }= \frac{4x}{100} \)------ substitute the value of y
\(\frac{(120000 + 3x - 60000)}{100}= \frac{4x}{100}\)
x = 60000
the additional amount to invest = 60000 - 20000 = 40000
Answer is C.
gmatclubot
Re: Dr. Kramer plans to invest $20,000 in an account paying 6% i [#permalink]
27 Mar 2020, 11:32
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