GMATBaumgartner wrote:
Tourist purchased a total of 30 travelers checks in $50 and $100 denominations. The total worth of the travelers checks is $1800. How many checks of $50 denominations can he spend so that average amount (arithmetic mean) of the remaining travelers checks is $80?
A. 4
B. 12
C. 15
D. 20
E. 24
I present an alternate and fast method to do such problems. These are essentially "mixture" problems and can be done by using alligation approach.
Refer following figure:
Attachment:
Moneybill-1.JPG [ 30.34 KiB | Viewed 10613 times ]
Total checks are 30. Total value is 1800. This means average value is (1800/30) = 60.
Once we have individual values, we can get the ratios.
We find that 100$ checks and 50$ checks are in the ratio of 1:4.
say 100$ checks are x in no. then 50$ checks are 4x in number
x+4x = 30, so 5x = 30; x = 6.
100$ --> 6 in Number
50$--> 24 in Number
Now let us come to second part. Refer following figure
Attachment:
Moneybill-2.JPG [ 28.84 KiB | Viewed 10616 times ]
Revised average = 80
We find that 100$ checks and 50$ checks are in the ratio of 3:2
Now we know that 100$ checks are 6 in number.
3: 2 is same is 6:4.
hence 50$ checks will have to be 4 in Number.
Originally 50$ checks were 24 in number. So 20 checks will be given away.
D is the answer.