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Bunuel
The total income of Mr. Teng in the years 2003, 2004, and 2005 was $36,400. His income increased by 20% each year. What was his income in 2005?

(A) 5,600
(B) 8,800
(C) 10,000
(D) 12,000
(E) 14,400


suppose,

income in 2003 = 100

income in 2004 = 120

income in 2005 = 144.

Total = 364.

when total income 364, 2005's income is 144

when total income 1, 2005's income is 144/364

when total income 36400, 2005's income is = 144/364 * 36400 = 14400.

E is the correct answer.
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Bunuel
The total income of Mr. Teng in the years 2003, 2004, and 2005 was $36,400. His income increased by 20% each year. What was his income in 2005?

(A) 5,600
(B) 8,800
(C) 10,000
(D) 12,000
(E) 14,400

income 2003 = x
2004 = 1.2x
2005 = 1.44 x
so total 3.64 x = 36400
x= 10,000
2005 income = 10,000*1.4 ; 14,400
IMO E
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Bunuel
The total income of Mr. Teng in the years 2003, 2004, and 2005 was $36,400. His income increased by 20% each year. What was his income in 2005?

(A) 5,600
(B) 8,800
(C) 10,000
(D) 12,000
(E) 14,400

We can let n = his income in 2003; thus:

n + 1.2n + (1.2)^2 * n = 36,400

2.2n + 1.44n = 36,400

3.64n = 36,400

n = 36,400/3.64 = 3,640,000/364 = 10,000

Hence, his income in 2005 was (1.2)^2*10000 = 14,400.

Alternate Solution:

Notice that Mr. Teng’s income in 2005 was the greatest among the three years 2003, 2004 and 2005. Thus, his income in 2005 must be greater than the 3-year average of 36400/3 = 12133 1/3 . Only answer choice E is greater than 12133 1/3.

Answer: E
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2003: X
2004: 1.2X
2005: 1.2^2X = 1.44X

Thus, we have x + 1.2x + 1.44x = 36,000 which gets us 3.64x = 36,000, and x = 10,000.

To find year 2005, simply do 1.44x or 10,000(1.44) = 14,400 to get to our answer.
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