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What are the three consecutive integers if the sum of the second and thrid exceeds a half of the first by thirty-three?
Let the three consecutive integers be n, n + 1 and n + 2. Then (n + 1) + (n + 2) = n/2 + 33 2n + 3 = n/2 + 33 : 4n + 6 = n + 66 : 3n = 60 : n = 20 The three numbers are 20, 21 and 22.
Tammy has 1 3 as many dimes as nickels 3 more quarters than nickels 17 coins in all?
Nickels = 6 Dimes = 2 Quarters = 9 (1/3N) + N + (3+N) = 17 2 1/3N = 14 N = 6
A collection of 30 coins worth 5.50 consists of nickels dimes and quarters there are twice as many dimes as nickeles how many quarters are there?
Let the number of nickels, dimes and quarters be n, d, q respectively. Then n +d + q = 30 5n + 10d + 25q = 550 But d = 2n, so: n + 2n + q = 30 => 3n + q = 30 5n + 10(2n) + 25q = 550 => 25n + 25q = 550 => n + q = 22 Which gives two simultaneous equations to solve, resulting in: n = 4, q = 18 So there are 18 quarters (plus 4 nickels and 8 dimes).
How many line segments does a N have?
Three.
If you have twice as many nickels as quarters how many quarters are there if you have 4.90?
5n + 25q = 490; n = 2q so: 10q + 25q = 490 ie 35q = 490 so q = 14 and n must be 28. 14 x 25c = $3.50, 28 x 5c = $1.40, total $4.90. ...and there you have it!
