# Please help with easy maths questions?

1.) p:p + q = 3:5. Find 3p+2q: 2q-p

2.) How many numbers greater than 40000 can you make using these following numbers?

4, 8, 3, 1, 7

This is extracurricular competition maths. Please help, I am in a maths competition next week.

### 5 Answers

- RealProLv 73 weeks agoFavorite Answer
1)

I think parentheses are key.p : (p + q) = 3 : 5You know this is p / (p + q) = 3 / 5 so get rid of the denominators.3(p + q) = 5pWe have two variables so why not just express one in terms of the other?3q = 2pp = (3/2)q(3p + 2q) / (2q - p)= (9q/2 + 2q) / (2q - 3q/2)= (13q/2) / (1q/2) = 13 = 13/1(3p + 2q) : (2q - p) = 13 : 1Of course, if algebraic work need not be shown, then you take p = 3 and q = 2 and put the numbers in.2)For example (4*8*3*1*7)^(4+8+3+1+7)? You can come up with infinitely many of those.Assuming each digit must be used exactly once, then you have 3 choices for the first digit (4, 7, 8), 4 choices for the 2nd digit, 3 for the 3rd, 2 for the 4th and 1 for the 5th.3 * 4 * 3 * 2 * 1 = 3 * 4! = 72 different numbers.

- KrishnamurthyLv 73 weeks ago
1.)

p : p + q = 3 : 5.

3p + 2q : 2q - p = 13 : 1

2.)

One can make more than 12 numbers (48317, 43178, 41783, 47831, ....)

greater than 40,000 using the numbers? 4, 8, 3, 1, and 7.

- az_lenderLv 73 weeks ago
(2) How many times are you allowed to use each digit? If only once, then the answer is

3*24 = 72. The "3" accounts for the fact that the first digit must be 4, 7, or 8. The "24" is the number of ways of arranging the remaining 4 digits.

- JOHNLv 73 weeks ago
(p + q)/p= 1 + q/p = 5/3

q/p = 5/3 – 1 = 2/3

(3p+2q)/(2q-p) = (3 + 2(q/p))/((2q/p) – 1)

= (3 + 4/3)/(4/3 -1) = 13/3 x 3/1 = 13.

13 : 1.

4, 8, 3, 1, 7

Numbers > 40000 begin with 4, 8, 7.

Those beginning with a 4: 4! numbers

So 3 x 4! numbers in all.

= 3 x 4 x 3 x 2 x 1 = 72 numbers in all.

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- Anonymous3 weeks ago
< Please help, I am in a maths competition >

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