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# A certain debt will be paid in 52 installments from January 1 to Decem

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A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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16 Nov 2015, 21:56
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A certain debt will be paid in 52 installments from January 1 to December 31 of a certain year. Each of the first 20 payments is to be $410; each of the remaining payments is to be$65 more than each of the first 20 payments. What is the average (arithmetic mean) payment that will be made on the debt for the year?

A. 443
B. 450
C. 465
D. 468
E. 475
[Reveal] Spoiler: OA

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Re: A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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16 Nov 2015, 22:39
Total = 410(20) + 475(32)
Number of installments = 52
Average = Total/Number of installments = 450.

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Re: A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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16 Nov 2015, 22:43
Total number of installments = 52
Payment per installment for the first 20 installments = 410

Payment per installment for the remaining 32 installments = 410+65=475

Average = (20*410 + 32*475)/52 =(8200+15200)/52=23400/52 = 450

Alternatively , we can proceed as -
65 is difference in value between payment per installment in first 20 installments and the payment per installment for the remaining 32 installments.
At this point we can eliminate option A and E .
(Option A= 443 - If the number of installments were in equal proportion , then average would have been = (410+475)/2=442.5
Option E = 475 - It would be true if all the installments were of 475 )

Number of installments are in the ratio = 20/32 = 5/8
Total number of parts = 5+8 = 13
We need to divide 65 into 13 parts , and the parts will be 25 and 40 . And the distances will be in inverse proportion
Therefore 410 +40 = 450

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A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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18 Nov 2015, 04:00
The most efficient way I found was to split the difference into its own entity.

By that,

$$\frac{(20*410 + 32*475)}{52}$$ can be rewritten as $$\frac{(20*410 + (32*410 + 32*65))}{52}$$

This can be simplified much more easily compared to the original expression
$$\frac{(52*410 + 32*65)}{52} = 410 + \frac{32*65}{52}$$
$$= 410 + 40 =450$$

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Re: A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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19 Apr 2016, 11:48
$$\frac{(20*410 +32*475)}{52}$$ = 450
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Re: A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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05 May 2017, 22:13
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Re: A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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10 May 2017, 00:40
I have tried something as below.

32 * 65 = 2080

$$\frac{2080}{52}$$ = 40 [ additional amount paid per month]

average = 410 + 40 = 450/ month

Explanation

Let's say $410 is the average amount paid for 52 weeks. We know that for 20 week installment =$ 410.
Next 32 weeks installment exceeded $65. So additional amount paid over$410 = 32 * 65 = $2080 => $$\frac{2080}{52}$$= 40 [additional amount paid over$ 410]
=> 410 + 40 = 450

Ans: Option B
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Re: A certain debt will be paid in 52 installments from January 1 to Decem [#permalink]

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16 May 2017, 17:18
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Expert's post
Bunuel wrote:
A certain debt will be paid in 52 installments from January 1 to December 31 of a certain year. Each of the first 20 payments is to be $410; each of the remaining payments is to be$65 more than each of the first 20 payments. What is the average (arithmetic mean) payment that will be made on the debt for the year?

A. 443
B. 450
C. 465
D. 468
E. 475

We are given that 52 total payments have to be made and that the first 20 payments are 410 dollars per payment. Thus, 20 x 410 = 8200 dollars was paid in those 20 payments.

Thus, after those 20 payments, there are 52 - 20 = 32 payments left to be made. If each one is 65 dollars more than each of the first 20 payments, each of the 32 remaining payments will be 410 + 65 = 475 dollars, and the total amount of these 32 payments is:

32 x 475 = 10,450

Thus, the average of the 52 payments is:

average = (8,200 + 15,200)/52 = 23,400/52 = 450 dollars

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Re: A certain debt will be paid in 52 installments from January 1 to Decem   [#permalink] 16 May 2017, 17:18
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