Aham56 wrote:
Let p,n,d and q be the number of pennies, nickels, dimes and quarters.
(p + 5n + 10d +25q)/(p+n+d+q) = 20
Adding one more quarter makes the average 21, so:
[p+5n+10d+25 (q+1)]/[p+n+d+q+1] = 21
Solving gives us p+n+q+d = 4 and so p + 5n + 10d + 25q = 80 which can be the case only with p = 0, n =1, d =0 and q =3. Adding another quarter makes the total 105/5 to give an average of 21.
So, I arrived at A as the answer.
However, I could not do this within two minutes and am interested in knowing how to get faster.
Hey! I also "struggled" at first, but I think I found a method as to how to tackle these types of questions:
First of all,
forget all those p + n + d... equations (they are way too complicated) instead, we have two easy equations:
T = Total amount of money
X = Number of pennies
I) \(\frac{T}{X}\) = 20ct
II)\(\frac{T+25}{X+1}\) = 21ct
Now, first solve I) for T. We get: T = 20x
Then, insert 20x into the second equation and solve for x.
You will get x = 4, and consequently, T = 20*4 = 80.
So now you know, that in the first equation your total amount of money is 80ct and to get to that number, you have x= 4 pennies.
Now, try to think of the different pennies that, combined, will amount to 80ct.
(Tip: Always start with the "largest" pennies!)How many quarters "fit" into 80ct?
--> 3, because 3 * 25ct = 75ct.
What is left?
--> 1 penny, that needs to be a nickel (5ct).
So: There are no dimes!
But why start with the largest pennies you may ask? Well, try this out:
Let´s take 2 quarters = 50ct, so we still have 2 pennies left.
However, we only have nickels (5ct) and dimes (10ct) that are the next largest pennies.
2 dimes are 20ct and 50ct + 20ct =/= 80ct.
I think with these types of questions you can "trust" the GMAC that it will be relatively easy to come up with the different tyes of pennies, as long as you try to "fill out" the sum of 80ct with the largest possible pennies first. (I hope you can understand what I mean)
Hope this helped!