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7.4B HW Answers
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2. Jose has $2. 45 in nickels and dimes
2. Jose has $2.45 in nickels and dimes. He has ten more nickels than dimes. How many of each does he have? 0.05n d = 2.45 5n + 10d = 245 25 n = d + 10 = = 5(d + 10) + 10d = 245 5d d = 245 15d + 50 = 245 d = 13 15d = 195
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+ (–10) 0.10d + 0.25q = 9.65 d + q = 59 -10d–10q = -590 15q = 375
3. A jar filled with only dimes and quarters contains a total of 59 coins. The value of all the coins in the jar is $ How many quarters are in the jar? 0.10d q = 9.65 10d + 25q = 965 + (–10) d + q = 59 -10d–10q = -590 15q = 375 q = 25
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4. Ernie has $3. 65 in nickels and dimes
4. Ernie has $3.65 in nickels and dimes. He has 13 more nickels than dimes. How many of each coin does he have? 0.05n d = 3.65 5n + 10d = 365 33 n = d + 13 = = 5(d + 13) + 10d = 365 5d d = 365 15d + 65 = 365 d = 20 15d = 300
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+ (–5) 0.05n + 0.25q = 9.65 5n + 25q = 965 n + q = 65 -5n – 5q = -325
5. A jar filled with only nickels and quarters contains a total of 65 coins. The value of all the coins in the jar is $ How many quarters are in the jar? 0.05n q = 9.65 5n + 25q = 965 + (–5) n + q = 65 -5n – 5q = -325 20q = 640 q = 32
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6. Tickets to a local movie were sold at $6. 00 for adults and $4
6. Tickets to a local movie were sold at $6.00 for adults and $4.50 for students. If 210 tickets were sold for a total of $1,065, how many student tickets were sold? 6a s = 1065 600a + 450s = + (–600) a + s = 210 -600a–600s = -150s = s = 130
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+ 25a + 15c = 8415 25a + 15c = 8415 (–25) a + c = 441 -10c = -2610
7. Tickets to a play were sold for $25 for adults and $15 for children. If 441 tickets were sold for a total of $8415, how many children tickets were sold? 25a + 15c = 8415 25a + 15c = 8415 + (–25) a + c = 441 -25a –25c = -10c = -2610 c = 261
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8. The Steele Canyon ASB sold advance tickets to the dance for $4 per ticket. Anyone who attended and purchased their ticket at the door had to pay $5 a ticket. A total of 480 students attended the dance and ASB earned $ How many students bought tickets at the door? 4a + 5d = 2100 4a + 5d = 2100 + (–4) a + d = 480 -4a – 4d = -1920 d = 180
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+ 3a + 1s = 470 3a + s = 470 (–1) a + s = 292 -a – s = -292 2a = 178
9. Altogether 292 tickets were sold for a high school basketball game. An adult ticket costs $3 and a student ticket costs $1. Ticket sales were $470. How many of each type of ticket were sold? 3a + 1s = 470 3a + s = 470 + (–1) a + s = 292 -a – s = -292 2a = 178 89 + s = 292 a = 89 s = 203
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