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Math Help - Several SAT Math Questions

  1. #1
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    Several SAT Math Questions

    Hi. I can't figure where to put SAT Math Questions. The test is on Sat. and we have a few questions we can't yet figure out how to do:

    Q3. A 0, 3 or 5 will be placed in each square of the grid. The sum in each row is equal to the number to the right of that row and the sum in each column is equal to the number below the column. What will row X be when the boxes are all filled out correctly. (I hope you can figure out the picture.

    Row Sum
    W 0 5
    X 0 13
    Y 0 0
    Z 3 18
    Sum 15 8 3 10











    Answer Choices:
    A) 5,3,0,5
    B) 5,0,3,5
    C) 3,5,0,5
    D) 0,3,5,5
    E) 5,3,5,0

    Correct answer: (A) 5,3,0,5

    How did they get the correct answer. We eliminate B, D, and E because we know the 3rd column of row X is a 0. However, both A and C look possible as far as I can tell.

    Best is if you can answer the question by showing us a method for converting the info into math equations.

    Q1. There are just enough microscopes (variable name M), test tubes (T) and calculators (C) so that every 3 students have to share a microscope, every 4 students have to share a test tube and every 5 students had to share a calculator. The sum of microscopes, test tubes and calculators is 94. How many students are there?

    Correct answer 120 students

    So far, I've got M+T+C=94. How do we turn "3 students to a calculator", "4 to a microscope", etc. into a math statement to solve this problem

    Thank you for helping.
    Last edited by mathDad; May 3rd 2012 at 08:07 PM.
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  2. #2
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    Re: Several SAT Math Questions

    On the first question, I'm assuming you mean A and C. It can't be C because there are no combinations of the digits 0,3, and 5 that you could add to three to make the total in the column equal 15.
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  3. #3
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    Re: Several SAT Math Questions

    Quote Originally Posted by Bean View Post
    On the first question, I'm assuming you mean A and C.
    Yup. I fixed the OP to be correct.
    Quote Originally Posted by Bean View Post
    It can't be C because there are no combinations of the digits 0,3, and 5 that you could add to three to make the total in the column equal 15.
    You are talking about column 1. Good observation.

    I'm also curious if there is a way to convert this question into math equations which can then be solved using all the regular rules of math that we all know (and love).
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    Re: Several SAT Math Questions

    We'll solve it a piece at a time. I'd suggest filling it in as we go along.

    Row Y must be all zeros to get that total.

    The remaining three cells in the first column must all be 5; this is the only way to get a total of 15 at the bottom of the column (given that one cell must be zero). So the first cell in column X must be 5.

    Since the first cell in row W is now 5, the remaining cells have to be zero in order for the final row total to be 5.

    Now rows W and Y both have zeros in the fourth column. Since the column total is 10, the only possibility is that the fourth cells in both X and Z must be the maximum allowed value of 5.

    The only blanks left are the second boxes in rows X and Z. The second Z box must be 5 (but this is unimportant to us). The second X box has to be three in order to get a total of 13, since the other boxes are 5, 0, and 5.

    This gives us a final order for row X as 5, 3, 0, 5, which is the answer for A.
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  5. #5
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    Re: Several SAT Math Questions

    Q1) We have to find a number into which 3, 4, and 5 evenly divide. They share no prime roots, so we multiply 3 x 4 x 5 to get 60. The answer must therefore be a multiple of 60.

    If they divided evenly at 60, there would have to be 20 microscopes, 15 test tubes, and 12 calculators. But 20 + 15 + 12 = 47, which is not the right answer.

    However, 47 is exactly half of the required 94. We therefore increase everything by a factor of 2, thus doubling 60 students to 120 students.
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  6. #6
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    Re: Several SAT Math Questions

    Quote Originally Posted by ssuag View Post
    We'll solve it a piece at a time. I'd suggest filling it in as we go along.

    Row Y must be all zeros to get that total.

    The remaining three cells in the first column must all be 5; this is the only way to get a total of 15 at the bottom of the column (given that one cell must be zero). So the first cell in column X must be 5.

    Since the first cell in row W is now 5, the remaining cells have to be zero in order for the final row total to be 5.

    Now rows W and Y both have zeros in the fourth column. Since the column total is 10, the only possibility is that the fourth cells in both X and Z must be the maximum allowed value of 5.

    The only blanks left are the second boxes in rows X and Z. The second Z box must be 5 (but this is unimportant to us). The second X box has to be three in order to get a total of 13, since the other boxes are 5, 0, and 5.

    This gives us a final order for row X as 5, 3, 0, 5, which is the answer for A.
    So the trick seems to be to look for restricted rows/columns.

    Is that the best way to do this type of problem? Is there a way to convert a problem like this into equations? For example, "if the red and green marbles are four times the wood marbles", we translate that to "r+g=4w". Is there some way to translate this question into math?
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  7. #7
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    Re: Several SAT Math Questions

    Quote Originally Posted by ssuag View Post
    Q1) We have to find a number into which 3, 4, and 5 evenly divide. They share no prime roots, so we multiply 3 x 4 x 5 to get 60. The answer must therefore be a multiple of 60.

    If they divided evenly at 60, there would have to be 20 microscopes, 15 test tubes, and 12 calculators. But 20 + 15 + 12 = 47, which is not the right answer.

    However, 47 is exactly half of the required 94. We therefore increase everything by a factor of 2, thus doubling 60 students to 120 students.
    It sounds like you're converting the question into
    \frac m3 + \frac t4 + \frac c5 = 1.
    Then solving for m+t+c.

    If I'm right, can you explain how you got from the question to that equation? When I look closely enough at it, I do see something like 1 microscope for 3 students plus 1 test tube for 4 students, etc. If that's the case, why is it set equal to 1?
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