Help on SAT math question

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Discus: SAT/ACT Tests and Test Preparation: June 2004 Archive: Help on SAT math question
By Ajoshi05 (Ajoshi05) on Monday, May 31, 2004 - 02:20 pm: Edit

Hey, I have a math question from MAY 2002. Here it is:

Ina bag of marbles, 1/2 of the marbles are red, 1/4 of them are green, and 1/5 of them are blue. If the remaining 2 marbles are white, what is the number of green marbles in the bag?
A. 4
B. 5
C. 8
D. 10
E. 40

I definitely have probs doing prbability crap, so somebody help me please. Thanx!!

By Justperfect (Justperfect) on Monday, May 31, 2004 - 02:27 pm: Edit

the first thing i did was find a number that all go into which is 40/40 which is one of the answers (e) i then found that all of the question fit the number 40. i then saw that i needed 1/4 which was green divided and found out the answer was D10. E>total marbles D>answer C>blue marbles

By Haithman (Haithman) on Monday, May 31, 2004 - 02:28 pm: Edit

Let x eqaul the total number of marbles in the bag.

1/2x + 1/4x + 1/5x + 2 = x
Multiply both sides by 20 to get rid of the fractions.

(20) 1/2x + 1/4x + 1/5x + 2 = x (20)
10x + 5x + 4x + 40 = 20x
Add common numbers, so 10x+5x+4x = 19x
So 19x + 40 = 20x
Subtract 19x from both sides
So 40 = x
Now we denoted x to be the total number of marbles in the bag, so we know there are 40 marbles in the bag (of all colors).
So to find the amount of green, we look back in the problem and it says that 1/4 are green. So to find green, just plug 40 into x of 1/4x = green. So 1/4(40)= 10 which equals the green. So there are 10 green in the bag, or answer choice D.

Hope that helped!

By Optimizerdad (Optimizerdad) on Monday, May 31, 2004 - 02:28 pm: Edit

This isn't a probability question, just algebra.

Let T = total #marbles in bag
R = #red marbles
G = #green marbles
B = #blue marbles
W = #white marbles
Then R = 0.5T, G = 0.25T, B = 0.2T
and W = T - R - G - B
= T - 0.5T - 0.25T - 0.2T
= 0.05T
W also = 2, so 0.05T = 2 and T = 40
Finally, G = 0.25T = 10. Answer is (D)

By Haithman (Haithman) on Monday, May 31, 2004 - 02:30 pm: Edit

Yeah good method perfect. My way was just more mathematical. Either way would work though.

By Drusba (Drusba) on Monday, May 31, 2004 - 02:32 pm: Edit

This is not probability. You simply need to determine the total number of marbles in the bag and multiply by 1/4. Simply add up the total that is a color other than white: 1/2 + 1/4 + 1/5 = 19/20. The two white ones thus equals 1/20 of the total. The total is Y and 2 (the white ones) = 1/20 x Y, meaning Y =40. 1/4 of that = 10 green marbles

By Ajoshi05 (Ajoshi05) on Monday, May 31, 2004 - 10:10 pm: Edit

Cool, everyone got 40. That's what i got at first and then checked my answer and saw that it was 10. I was like wtf! So, i guess we all proved collegeboard wrong, RIGHT ON!! Anyway, thanx for the responses.

By Haithman (Haithman) on Tuesday, June 01, 2004 - 02:38 am: Edit

No, the answer to the problem is 10. There are 40 total marbles, but 10 green (which it asks for.) So sadly, we have yet to prove them wrong.

By Nightflarer (Nightflarer) on Tuesday, June 01, 2004 - 03:07 pm: Edit

Some of you'll do way too much work
Here are the steps:
1. Add up the fractions for red, blue and green.
2. So 19/20 of all the marbles are red, blue or green.
3. Therefore, white must be 1/20 which means 1/20 of the marbles is 2.
4. Multiply 20 by 2 and get 40 total marbles.
5. Divide 40 by 4 to get 10 green marbles, the answer.

By Mhawk177 (Mhawk177) on Tuesday, June 01, 2004 - 03:11 pm: Edit

Go for the common denominator of 20... so you get 10/20 + 5/20 + 4/20 = 19/20.. leaving the 2 white to make up 1/20. 2= 1/20 of the bag.. therefore 40 would be the total. Green is 1/4 and 1/4 of 40= 10 D. being the answer

By Haithman (Haithman) on Tuesday, June 01, 2004 - 03:20 pm: Edit

Well my method looks long, but I can do it equally fast in me head...Do whatever way works for you.

By Ajoshi05 (Ajoshi05) on Tuesday, June 01, 2004 - 10:39 pm: Edit

yeah, that's what i meant. SHUT UP!! Man, maybe i was drunk when i looked at that question and did this conversation. Anyway, i have another one from the october 2003 test. here it is:

x+y = 5x+3y
How many different ordered pairs(x,y) are solutions to the equation above?
A. None
B. One
C. Two
D. Four
E. More than four

Ok, now you can answer this for me and make me feel more like an idiot, please. Thanx!

By Qwert271 (Qwert271) on Tuesday, June 01, 2004 - 10:56 pm: Edit

Take an x and a y away from both sides to get 4x + 2y = 0. If you put any random number in for x, you get a number for y. This will work for any and every number you put in for x, so there are infinite solutions. E.

For example: for the equation 4x + 2y = 0
4(1) + 2(2) = 0
4(2) + 2(4) = 0
4(3) + 2(6) = 0
4(4) + 2(8) = 0
4(5) + 2(10) = 0 etc. indefinitely.

Just as a side note: equations with only single order exponents i.e. x, y, not x^2 or x^3, will always be lines, which have infinite points.

By Ajoshi05 (Ajoshi05) on Tuesday, June 01, 2004 - 11:08 pm: Edit

kool, thanx a lot!! I really needed to know how to do that.

By Haithman (Haithman) on Wednesday, June 02, 2004 - 12:55 am: Edit



Good Luck, and lets OWN this test!

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