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4350 is invested into three parts at simple interest so that

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4350 is invested into three parts at simple interest so that [#permalink]

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11 May 2008, 03:09
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$4350 is invested into three parts at simple interest so that the amounts received after 1, 2 and 3 years respectively in each part are equal. Find the amount invested for 3 years, if the rate of interest is 10% p.a. (1)$1320
(2)$1430 (3)$1540
(4)$1560 (5)$1650

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Re: PS: Simple Interest question [#permalink]

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11 May 2008, 04:59
let x,y,z are the amount invested for 3,2 and 1 years respectively.
x+x*3*0.1 = y + y*2*0.1 = z + z*1*0.1
=> 1.3x=1.2y=1.1z

x+y+z =4350
substitute z and y in terms of x
4.31x/1.32=435
x = 1332.25
Re: PS: Simple Interest question   [#permalink] 11 May 2008, 04:59
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