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# Rob has to make 5 payments on his student loans from college. Each pay

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Math Expert
Joined: 02 Sep 2009
Posts: 61544
Rob has to make 5 payments on his student loans from college. Each pay  [#permalink]

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23 Jan 2020, 01:22
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Difficulty:

25% (medium)

Question Stats:

93% (01:27) correct 7% (05:46) wrong based on 15 sessions

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Rob has to make 5 payments on his student loans from college. Each payment will be twice the amount of the previous payment. If the total amount he has to pay back is \$1,550, how much is Rob's first payment?

A. \$10
B. \$20
C. \$25
D. \$50
E. \$75

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Re: Rob has to make 5 payments on his student loans from college. Each pay  [#permalink]

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23 Jan 2020, 01:42
Explanation:

As per given Data
x+2x+4x+8x+16x = 1550
31x = 1550
x = 50

So First payment = x = 50\$

IMO-D
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Re: Rob has to make 5 payments on his student loans from college. Each pay  [#permalink]

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23 Jan 2020, 02:10
Question:
Rob has to make 5 payments on his student loans from college. Each payment will be twice the amount of the previous payment. If the total amount he has to pay back is \$1,550, how much is Rob's first payment?

Solution:

Assuming his first payment as \$x, we have the 5 payments as: \$x, \$2x, \$4x, \$8x, and \$16x
Thus, the sum = \$1550
=> 31x = 1550
=> x = \$50

Alternate approach:
I was trying to find an alternate logic for this (though the above method is the easiest).

Since each amount is a multiple of the first payment and their sum is 1550, the first payment must be a factor of 1550.
This rules out 20 (1550 is not divisible by 20) and 75 (1550 is not divisible by 3 which is a factor in 75).

This leaves us with 10 (too low a value, hence inadmissible), 25 and 50.

Assuming the initial value as 25, if we multiply this by 2, all other payments would become a multiple of 10 (i.e. have the units digit '0'). Thus, the sum of all values would have the units digit as 5, which doesn't match with that of 1550.

Thus, the only valid option is 50.

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Re: Rob has to make 5 payments on his student loans from college. Each pay   [#permalink] 23 Jan 2020, 02:10
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