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Darion has at least one quarter, one dime, one nickel, and o
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29 Jul 2020, 10:37
29 Jul 2020, 10:37
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Question Stats:
70% (02:46) correct
29% (02:44) wrong based on 81 sessions
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Darion has at least one quarter, one dime, one nickel, and one penny in his pocket. He has three times as many pennies as nickels, three times as many nickels as dimes, and three times as many dimes as quarters. If he has 120 coins, how much money does he have?
A. $ 1.20
B. $ 3.81
C. $ 4.06
D. $ 4.07
E. $ 7.75
A. $ 1.20
B. $ 3.81
C. $ 4.06
D. $ 4.07
E. $ 7.75
Re: Darion has at least one quarter, one dime, one nickel, and o
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29 Jul 2020, 11:41
29 Jul 2020, 11:41
1
Let us denote # of pennies as P, dimes as D, nickels as N, and quarters as Q.
three times as many dimes as quarters => D = 3Q
three times as many nickels as dimes => N = 3D = 9Q
three times as many pennies as nickels => P = 3N = 27Q
total number of coins = Q + 3Q + 9Q + 27Q = 40Q
=> 120 = 40Q
=> Q = 3
money in quarters = 3/4 = 0.75
money in dimes = 9/10 = 0.9
money in nickels = 27/20 = 1.35
money in pennies = 81/100 = 0.81
total money = 3.81
So, the answer is B.
three times as many dimes as quarters => D = 3Q
three times as many nickels as dimes => N = 3D = 9Q
three times as many pennies as nickels => P = 3N = 27Q
total number of coins = Q + 3Q + 9Q + 27Q = 40Q
=> 120 = 40Q
=> Q = 3
money in quarters = 3/4 = 0.75
money in dimes = 9/10 = 0.9
money in nickels = 27/20 = 1.35
money in pennies = 81/100 = 0.81
total money = 3.81
So, the answer is B.
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Re: Darion has at least one quarter, one dime, one nickel, and o
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03 Aug 2020, 05:19
03 Aug 2020, 05:19
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Carcass wrote:
Darion has at least one quarter, one dime, one nickel, and one penny in his pocket. He has three times as many pennies as nickels, three times as many nickels as dimes, and three times as many dimes as quarters. If he has 120 coins, how much money does he have?
A. $ 1.20
B. $ 3.81
C. $ 4.06
D. $ 4.07
E. $ 7.75
A. $ 1.20
B. $ 3.81
C. $ 4.06
D. $ 4.07
E. $ 7.75
NOTE: The official test-makers will not assume that you know the values of quarters, nickels, dimes and pennies.
1 quarter = $0.25
1 dime = $0.10
1 nickel = $0.05
1 penny = $0.01
Let's start with the smallest value, which happens to be the number of quarters
Let x = the number of quarters in Darion's pocket
He has three times as many dimes as quarters
So, 3x = the number of dimes
He has three times as many nickels as dimes
So, 3(3x) = the number of nickels
In other words, 9x) = the number of nickels
He has three times as many pennies as nickels
So, 3(9x) = the number of pennies
In other words, 27x) = the number of pennies
He has 120 coins
In other words: (number of quarters) + (number of dimes) + (number of nickels) + (number of pennies) = 120
Substitute to get: x + 3x + 9x + 27x = 120
Simplify: 40x = 120
Solve: x = 3
When we plug x = 3 into all of the above expressions, we find that there are:
3 quarters ($0.75)
9 dimes ($0.90)
27 nickels ($1.35)
81 pennies ($0.81)
How much money does he have?
$0.75 + $0.90 + $1.35 + $0.81 = $3.81
Answer: B
Cheers,
Brent
