A collection of 30 coins worth $5.50 consists of nickels,dimes and quarters. There are twice as many dimes as nickels. Howmany quarters are there?
sittesf
Answered question
2022-08-08
A collection of 30 coins worth $5.50 consists of nickels,dimes and quarters. There are twice as many dimes as nickels. Howmany quarters are there?
Answer & Explanation
vibrerentb
Beginner2022-08-09Added 21 answers
Let n=number of nickels, d=number of dimes, q=number ofquarters. 2n=d by the second statement for. So n+d+q=n+2n+q= 3n+q=30. [Quantity function] 5n+10d+25q=5n+20n+25q= 25n+25q=550 [Value function] n+q=22. Subtract the quantity function from the rewritten value function toget: 2n=8 n=4. Then q=18. This mean there are 18 quarters.
darcybabe98ub
Beginner2022-08-10Added 6 answers
To solve this problem, you need to make three different equationswith three different variables, and then use the combination methodto solve. x=nickels y=dimes z=quarters Here are the equations: x+y+z=30 (all thecoins added together equal 30 coins in all) 0.05x+0.10y+0.25z=5.50(this is the dollar amount of all the coins added together equals$5.50) 2x=y (there are twice asmany dimes as nickels) Rewrite the second equation as 5x+10y+25z=550 (multiply everythingby 100) and rewrite the third equation as 2x-y=0. Now your equations are: x+y+z=30 5x+10y+25z=550 2x-y=0 Now use the combination method to get rid of the variable z inorder to isolate the variables x and y. 5x+10y+25z=550 -25x-25y-25z=-750 (multiply everything by -25) Add both of these equations together to get: -20x-15y=-200 Then, add 2x-y=0 to this equation in order to isolate the variablex. -20x-15y=-200 -30x+15y=0 (multiply everything by -15) Add these two equations together and get: x=4 This is the key that will allow you to solve theproblem. Plug x=4 into the equation 2x=y. 2(4)=y y=8. Then, plug x=4 and y=8 into x+y+z=30. 4+8+z=30 12+z=30 z=18. The answer to your question: There are 18 quarters.