Hallie has only nickels,dimes and quarters in her pocket.If she has atleast 1 of each kind of coin and has a total of $2.75 in change,how many nickels does she have?
(1)She has a total of 21 coins,with twice as many dimes as nickels.
(2)She has $1.50 in quarters.
From the question we have,
Let x,y,z be the number of Nickels,Dimes and Quarters respectively
N(x)+D(y)+Q(z)=2.75 => 0.05(x)+0.10(y)+0.25(z)=2.75-----------------------------Equation 1
From statement 1,we get 2 more equations
x+y+z=21--------------------------------------------------------------------Equation 2
x=2y-------------------------------------------------------------------------Equation 3
Thus we have 3 equations and 3 variables and hence it can be solved.
Now my concern is what if I use another method--
0.05(x)+0.10(y)+0.25(z)=2.75
Lets substitute x=2y we got from statement 2
0.05(2y)+0.10(y)+0.25(z)=2.75 => 0.20(y)+0.25(z)=2.75---------------------Equation A
We also have from statement 1 ,x+y+z=21 again x=2(y) so we have 3(y)+z=21------Equation B
Solving Equation A and B gives us answers in decimals y=4.54 and z=7.36
The number of coins simply cannot be decimals.Please explain.I understand that one doesnt need to solve,just checking sufficiency is enough.But still,I wanted to solve this problem without using the N-Equations,N-Variables method since there are some problems out there that require solving.
Thanks people.
Dan
(1)She has a total of 21 coins,with twice as many dimes as nickels.
(2)She has $1.50 in quarters.
From the question we have,
Let x,y,z be the number of Nickels,Dimes and Quarters respectively
N(x)+D(y)+Q(z)=2.75 => 0.05(x)+0.10(y)+0.25(z)=2.75-----------------------------Equation 1
From statement 1,we get 2 more equations
x+y+z=21--------------------------------------------------------------------Equation 2
x=2y-------------------------------------------------------------------------Equation 3
Thus we have 3 equations and 3 variables and hence it can be solved.
Now my concern is what if I use another method--
0.05(x)+0.10(y)+0.25(z)=2.75
Lets substitute x=2y we got from statement 2
0.05(2y)+0.10(y)+0.25(z)=2.75 => 0.20(y)+0.25(z)=2.75---------------------Equation A
We also have from statement 1 ,x+y+z=21 again x=2(y) so we have 3(y)+z=21------Equation B
Solving Equation A and B gives us answers in decimals y=4.54 and z=7.36
The number of coins simply cannot be decimals.Please explain.I understand that one doesnt need to solve,just checking sufficiency is enough.But still,I wanted to solve this problem without using the N-Equations,N-Variables method since there are some problems out there that require solving.
Thanks people.
Dan
