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14 Apr 2008, 20:21
Kudos
Bookmarks
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.
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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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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14 Apr 2008, 20:48
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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
2) Total Paid = 3*2662 = 7986
X + X*10/100 = 7986
1.1X = 7986
=> X = 7260
14 Apr 2008, 20:54
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Value
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
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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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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