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1) The cost prices of 3 kinds of tea are 10$, 12$,16$ per

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Manager
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1) The cost prices of 3 kinds of tea are 10$, 12$,16$ per [#permalink] New post 14 Apr 2008, 19:21
1) The cost prices of 3 kinds of tea are 10$, 12$,16$ per bag. In what proportion should they be mixed so that the price of the mixture may be 14$ per bag?
1) 2:3:4
2) 1:2:3
3) 1:1:3
4) 1:1:2
5) 1:1:1

2) A loan is discharged in three equal installments of 2662$ each. If the rate of interest is 10%, Find the amount of loan?


Please explain the procedure for both.
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Re: Mixture/S.I. [#permalink] New post 14 Apr 2008, 19:48
1) Multiply first one by 10, second one by 12, and third one by 16. Add all the three and check if it is divisible by 3 or not. Only Answer 3 will satisfy the answer.

2) Total Paid = 3*2662 = 7986
X + X*10/100 = 7986
1.1X = 7986
=> X = 7260
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Re: Mixture/S.I. [#permalink] New post 14 Apr 2008, 19:54
Value wrote:
1) The cost prices of 3 kinds of tea are 10$, 12$,16$ per bag. In what proportion should they be mixed so that the price of the mixture may be 14$ per bag?

1) 2:3:4
2) 1:2:3
3) 1:1:3
4) 1:1:2
5) 1:1:1


1) if price of the mixture is $14, then sum of the mixture must be 126. lets check whether the sum is 126 or not.
with the ratio of 2:3:4, the actual sum is (2x10 + 3x12 + 4x16) = 120. so not suff.


2) if price of the mixture is $14 and the tees are in the ratio of 1:2:3, then sum of the mixture must be 84. lets check whether the sum is 84 or not.
with the ratio of 1:2:3, the actual sum is (1x10 + 2x12 + 3x16) = 82. so not suff.

3) similarly the sum must be 70.
actual sum = 10x1 + 12x1 + 16x3 = 70.

bingo...

4) cannot be
5) cannot be

so C.


Value wrote:
2) A loan is discharged in three equal installments of 2662$ each. If the rate of interest is 10%, Find the amount of loan?

Please explain the procedure for both.


this more about PV concept.

assuming no advance payment, the loan amt = 2662/1.1 + 266/(1.1)^2 + 266/(1.1)^2 = 2420 + 2200 + 2000 = 6620

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Re: Mixture/S.I.   [#permalink] 14 Apr 2008, 19:54
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