Word Problems/Maths Problem sum
a) John had 320 one-dollar coins and 50-cent coins in the ratio of 7:1. After his mother gave him thrice as many one-dollar coins as 50-cent coins, the ratio of one-dollar coins to 50-cent coins he had became 4:1. How many 50-cent coins did his mother give him?
b) Eric has 2/3 as many erasers as Fred. If Eric gives 8 erasers to fred, he will have 2/5 as many erasers as Fred. how many erasers does eric have at first?
(a) It is given that D + F = 320, which means that = 320 - F.
Since there are 7 dollars for each 50 cent piece, it is known that D/F = 7.
Since D = 320 - F, putting that into D/F = 7 gives (320-F)/F = 7.
To get the answer, multiply both sides by F, add F to both sides,
and then divide both sides by 8.
That gives 320 - F = 7F, going to 320 = 8F, an then F = 40.
Since D = 7F, and F = 40, that means D = 7*40 = 280.
Checking the answer midway gives gives the number of coins is 40 + 280 = 320, which matches.
The ratio if 280/40 = 7, which matches.
Now if we add on three dollars for every half dollar, we get a ratio of 4:1.
This means our new equation should be (280+3x)/(40+x) = 4.
Multiply this out, subtract 3x from both sides, subtract 160 from both sides, and that's it.
That is, 280 + 3x = 160 + 4x. This goes to 120 = x.
Since we added 4 times as many dollars, that means we added 360 of them.
For dollars, we have 280+360 = 640 and for fifty cent pieces, we have 40 + 120 = 160.
Noting that 640/160 = 4, that is also seen to be correct.
(b) It is given that E = 2F/3.
It is also given that E - 8 = 2(F+8)/5.
Putting the 1st equation into the 2nd gives 2F/3 - 8 = 2(F+8)/5.
To get rid of fractions, multiply by 15. This gives 10F - 120 = 6F + 48.
Adding 120 - 6F to both sides gives 4F = 168.
That means F = 42.
From there, using E = 2F/3, it can be seen the E = 28.
Checking the 2nd equation gives 28 - 8 = 2(42+8)/5.
That turns into 20 = 2(50)/5 = 20, so that must be right.