SOLUTION: 61) In Anna's purse there are 11 coins consisting of nickels and dimes.If the total value of these coins is $0.95,how many of each type of coin are there?
#61) A bank customer w
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Question 44695This question is from textbook beginning algebra
: 61) In Anna's purse there are 11 coins consisting of nickels and dimes.If the total value of these coins is $0.95,how many of each type of coin are there?
#61) A bank customer withdraws $200 from her account in $10 and $20 bills.she requests twice as many $20 as $10 bills.How many of each kind of bill does she receive?
This question is from textbook beginning algebra
Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
#60) In Anna's purse there are 11 coins consisting of nickels and dimes.If the total value of these coins is $0.95,how many of each type of coin are there?
Let number of nickels be "x"
Then number of dimes is "11-x"
Value of the nickels is 5x cents
Value of the dimes is 10(11-x)=110-10x cents
EQUATION:
value + value = 95 cents
5x+110-10x=95
-5x=-15
x=3 (number of nickels)
11-x=8 (number of dimes)
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#61) A bank customer withdraws $200 from her account in $10 and $20 bills.she requests twice as many $20 as $10 bills.How many of each kind of bill does she receive?
Let the number of $10 bills be "x"
Then number of $20 bills = 2x
Value of the $10 bills is 10x dollars
Value of the $20 bills is 20(2x)=40x dollars
EQUATION:
value + value = $200
10x + 40x = 200
50x = 200
x=4 (number of $10 bills)
2x=8 (number of $20 bills)
Cheers,
Stan H.
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
X+Y=11 & 5X+10Y=95 OR X=11-Y THUS 5(11-Y)+10Y=95 OR 55-5Y+10Y=95 OR 5Y=40 OR
Y=40/5 OR Y=8 THUS THERE ARE 8 DIMES FOR 80 CENTS THEN THERE ARE X+8=11 OR X=3 NICKELS FOR 15 CENTS FOR 80+15=95 OR 95=95.
10*X+20*2X=200 OR 10X+40X=200 OR 50X=200 OR X=200/50 OR X=4 TEN DOLLAR BILLS
THEN 2*4=8 20 DOLLAR BILLS
PROOF 4*10+8*20=200 OR 40+160=200 OR 200=200
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