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Thread: Help for my 9 yo son...

  1. #1
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    Help for my 9 yo son...

    Hello there,

    I'm helping my son with a long list of math problems. At the end I was able to help him with all of them except for three questions.
    I would appreciate your guidance in solving these:

    1. In a class 5/12 were girls. If there were 4 fewer girls than boys, how many were in the class?

    2. 12 exercise books and a textbook cost $56. 6 exercise books and 2 text books $58. How many dollars was each text book?

    3. Alison shared 56 stamps with Julia so that she had 3 times as many as Julia. How many stamps did Julia received?

    Many thanks in advance.
    Jota
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  2. #2
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    Re: Help for my 9 yo son...

    1. Let b represent the number of boys and g represent the number of girls.

    The total number of students in the class is $\displaystyle \begin{align*} b + g \end{align*}$, and you know that 5/12 of this amount is how many girls there are. So

    $\displaystyle \begin{align*} \frac{5}{12} \,\left( b + g \right) &= g \\ \frac{5}{12}\,b + \frac{5}{12}\,g &= g \\ \frac{5}{12}\,b &= g - \frac{5}{12}\,g \\ \frac{5}{12}\,b &= \frac{7}{12}\,g \\ 5\,b &= 7\,g \\ b &= \frac{7}{5}\,g \end{align*}$

    We also know that there are four fewer girls than boys, so $\displaystyle \begin{align*} g + 4 &= b \end{align*}$. So we can see that

    $\displaystyle \begin{align*} g + 4 &= \frac{7}{5}\,g \\ 4 &= \frac{7}{5}\,g - g \\ 4 &= \frac{2}{5}\,g \\ 20 &= 2\,g \\ 10 &= g \end{align*}$

    So there are 10 girls, and thus 14 boys.
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  3. #3
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    Re: Help for my 9 yo son...

    1) Perhaps better for a nine year old: Since the number of girls is a whole number, the total number of students must be a multiple of 12.

    Try 12. Then the number of girls is (5/12)(12)= 5 and the number of boys is 12- 5= 7. That is 2 more than the number of girls, not 4.

    Try 24. Then the number of girls is (5/12)(24)= 10 and the number of boys is 24- 10= 14. That is 4 more than the number of girls!
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    Re: Help for my 9 yo son...

    Quote Originally Posted by antananarivo View Post
    3. Alison shared 56 stamps with Julia so that she had 3 times as many as Julia. How many stamps did Julia received?
    A....J...total
    30, 10 : 40 : Allison has 3 times as many as Julia
    ??, ?? : 56

    Can you wrap that up?
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  5. #5
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    Re: Help for my 9 yo son...

    12 exercise books and a textbook cost 56.
    6 exercise books and 2 textbooks cost 58.

    How many dollars was each text book?

    I tried to find a "non-algebra" way to do this, similar to the way I did the first problem, but it looks like the simplest thing to do is:

    Let "x" be the cost of one exercise book and "y" be the cost of one text book, in dollars. Further, I will assume that the "56" and "58" you give are in dollars, although you didn't say that.

    "12 exercise books and a textbook cost 56"
    Since each exercise book cost "x" dollars, 12 exercise books cost 12x dollars. Since each textbook cost "y" dollars one text book cost y dollars. 12 exercise books and one text book cost 12x+ y= 56

    "6 exercise books and 2 text books cost 58"
    As above 6 exercise books and 2 text books cost 6x+ 2y= 58. Solve those equations for x and y.

    The simplest thing to do, I think, is solve the first equation, 12x+ y= 56 "for y". Subtracting 12x from both sides, y= 56- 12x. Now replace the y in 6x+ 2y= 58 with that: 6x+ 2(56- 12x)= 6x+ 112- 24x= 112- 18x= 58. Subtract 112 from both sides to get -18x= -64. Can your son finish that?
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