Question 48858: Good evening.I would like to ask you about this two problem, i try alday i cannot get the answer. Please help in this problem.
(1) A parking meter takes only dimes and quarters. IF the meter held $7.55 at the end of one day, and the number of dimes was 8 more than 20 times the number of quarters, then the total number of coins is how much?
(2) The number of dimes in Gina's coin collection is 4 less than the number of nickels, and then number of quarters is 3 times the number of dimes. If the value of her collection is $7.40, how many nickels does she have?
Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1) A parking meter takes only dimes and quarters. IF the meter held $7.55 at the end of one day, and the number of dimes was 8 more than 20 times the number of quarters, then the total number of coins is how much?
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Let number of quarters be "x"; value of those quarters is 25x cents
Number of dimes = 20x+8 ; value of those dimes is 10(20x+8)=200x+80 cents
EQUATION:
value of quarters + value of dimes = 755 cents
25x+200x+80=755
225x=675
x=3 (number of quarters)
20x+8 = 68 (number of dimes)
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(2) The number of dimes in Gina's coin collection is 4 less than the number of nickels, and then number of quarters is 3 times the number of dimes. If the value of her collection is $7.40, how many nickels does she have?
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Let number of nickels be "x" ; value of those nickels is 5x cents
Number of dimes is x-4 ;value of those dimes is 10(x-4)=10x-40 cents
Number of quarters is 3(x-4) ; value is 25(3x-12)= 75x-300 cents
EQUATION:
value of nickels + value of dimes + value of quarters = 740 cents
5x+10x-40+75x-300=740
Solve this equation for x.
Then 5x will be the number of nickels.
Cheers,
Stan H.
Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! 755=(8+20q)10+25q or 755=80+200q+25q or 755-80=225q or 675=225q or q=3 thus
there are 3 quarters & 20*3+8=68 dimes thus 3*25+68*10=775 or 75+680=775 or
775=775
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