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If Arnold has a 1.70 in dimes and quarters. He has 3 more dimes then quarters. How many of each coin does he have?
10D + 25Q = 170 and D = Q + 3Substitute: 10(Q + 3) + 25Q = 170ie 10Q + 30 + 25Q = 170ie 35Q = 140so Q = 4, making D = 7
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).
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!
You have 1000 coins to make 199.90 how many dimes and quarters?
This is a system of equations. Let d= number of dimes. q= number of quarters. d+q=1000 -> 10d + 10q = 10000 .10d+.25q=199.90-> -10d+(-25q)=-19990 Multiply each equation to get one of the variables to drop out. Add the equations. (-15q)=(-9990) q=666 1000-666=334 There are 666 quarters and 334 dimes.
The machine only takes dimes and quarters and it has 52 coins worth 8.95 how many of each domination of coins are contained?
Given: 10d + 25q = 895 cents d + q = 52 coins Multiply everything in the second equasion by 10 (to get rid of the dimes). 10d + 10q = 520 Subtract this 3rd equation from the 1st equation: 10d + 25q = 895 10d + 10d = 520 Giving you: 15q = 375 Divide both sides by 15 and you get q = 25 This means you have 25 quarters ($6.25) and 27 dimes ($2.70) CHECK: 25 quarters + 27 dimes = 52 coins (CORRECT) $6.25 + $2.70 = $8.95 (CORRECT)
