10D + 25Q = 170 and D = Q + 3Substitute: 10(Q + 3) + 25Q = 170ie 10Q + 30 + 25Q = 170ie 35Q = 140so Q = 4, making D = 7
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).
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!
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.
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)
Work in cents... 10D + 25Q = 480 and D = Q + 6Substitute: 10(Q + 6) + 25Q = 48010Q + 60 + 25Q = 48035Q = 420Q = 12, making D = 18
10D + 25Q = 170 and D = Q + 3Substitute: 10(Q + 3) + 25Q = 170ie 10Q + 30 + 25Q = 170ie 35Q = 140so Q = 4, making D = 7
Work in cents...10D + 25Q = 1405 and D = 100 - QSubstitute: 10(100 - Q) + 25Q = 1405ie 1000 - 10Q + 25Q = 1405ie 15Q = 405so Q = 27, making D = 73
Work in cents...10D + 25Q = 405 and D = Q - 5Substitute: 10(Q - 5) + 25Q = 405ie 10Q - 50 + 25Q = 405ie 35Q = 455so Q = 13, making D = 8
Work in cents...10D + 25Q = 675 and D = Q + 8Substitute: 10(Q + 8) + 25Q = 675ie 10Q + 80 + 25Q = 675ie 35Q = 595so Q = 17, making D = 25
Multi-Channel Transmission System Operator Maintainer
Work in cents...10D + 25Q = 850 and D = 55 - QSubstitute: 10(55 - Q) + 25Q = 850ie 550 - 10Q + 25Q = 850ie 15Q = 300so Q = 20, making D = 35
d = 2q - 18; 10d + 25q = 945 10(2q - 18) + 25q = 945 20q - 180 + 25q = 945 45q = 1125 q = 25 You have 32 dimes ($3.20) and 25 quarters ($6.25)
25 questions on this test. it is the 25th question on the test. funny huh?
5
16 quarts
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).