IMO E

This is my approach:
let the no of $10 cheques be X and $50 cheques be Y
Then, 10X+50Y = 1500
For the max amount to be en cashed, Y>X
Y=X+1
Therefore,10X +50(X+1) = 7n (Since 7 Cheques are en cashed)
=> 60X+50 = 7n

putting X=5, 60X+50 = 350 which a multiple of 7.

Hope it helps
hence total value en cashed = 10*5 + 50*6 =350
Amount of lost cheques = 1500-350 = 1150

grandh01 wrote:A tourist purchased a total of $1,500
worth of traveler's checks in $10 and
$50 denominations, During the trip
the tourist cashed 7 checks and then
lost all of the rest. If the number of
$10 checks cashed was one more or
one less than the number of $50
checks cashed, what is the minimum
possible value of the checks that were
lost?
(A) $1,430
(B) $1,310
(C) $1,290
(D) $1,270
(E) $1,150

Let # of $10 checks cashed = a
and # of $50 checks cashed = b
Value of the checks that were cashed = $10*a + $50*b

To find the minimum possible value of the checks that were lost , we have to find the maximum possible value of the checks that were cashed.

According to the question

Total # of checks cashed = 7
=> a+b = 7

And
|a - b| = 1

Hence
either a = 3 and b = 4

Value of the checks cashed = 10*3 + 50*4 = $230

or a = 4 and b = 3

Value of the checks cashed = 10*4 + 50*3 = $190

Clearly, we can see that the maximum value of the checks cashed is = $230
So the minimum value of the checks lost = $1500 - $230 = [spoiler]$1270[/spoiler]

The correct answer is D