Question 145088: Xander and Ella pool their loose change to buy snacks on their coffee break. One day, they spend $1.85 on 1 carton of milk, 2 donuts, and 1 cup of coffee. The next day, they spend $2.30 on 3 donuts and 2 cups of coffee. The third day, they bought 1 carton of milk, 1 donut, and 2 cups of coffee and spent $1.75. On the fourth day, They have a total of $1.80 left. Is this enough to buy 2 cartons of milk and 2 donuts?
equation #1________
equation #2________
Equation #3 _______
milk cost ______
Donut Cost _____
Coffee Cost_____
4th day cost______
available $1.80
yes or no________
I am finding this question VERY hard!! Please help me :)
Found 2 solutions by faceoff57, seema230480:
Answer by faceoff57(108)   (Show Source):
You can put this solution on YOUR website! 1m + 2d + 1c = 1.85
3d + 2c = 2.30
1m + 1d + 2c = 1.75
1m + 2d + 1c = 1.85
-(1m + 1d + 2c = 1.75)
1m + 2d + 1c = 1.85
-1m - 1d - 2c = -1.75
d - c = .10
d = c + .10
3d + 2c = 2.30
3(c + .10) + 2c = 2.30
3c + .30 + 2c = 2.30
5c + .30 = 2.30
5c = 2.30 - .30
5c = 2.00
c = .40
3d + 2c = 2.30
3d + 2(.40) = 2.30
3d + .80 = 2.30
3d = 2.30 - .80
3d = 1.50
d = .50
1m + 1d + 2c = 1.75
1m + 1(.50) + 2(.40) = 1.75
1m + .50 + .80 = 1.75
1m + 1.30 = 1.75
1m = 1.75 - 1.30
1m = .45
cup of coffee = .40
donut = .50
carton of milk = .45
2 cartons of milk is .90 and 2 donuts is $1.00. The total equalling $1.90. $1.80 is not enough. They are short .10.
Answer by seema230480(3)  (Show Source):
You can put this solution on YOUR website! Let's denote Milk carton by M,
Doughnuts by D
Coffee by C
First Equation formation:
they spend $1.85 on 1 carton of milk, 2 donuts, and 1 cup of coffee
So with the above said first equation becomes: 1M+2D+1C=1.85
Second Equation formation:
next day, they spend $2.30 on 3 donuts and 2 cups of coffee.
So with the above said second equation becomes: 3D+2C=2.30
Third Equation formation:
third day, they bought 1 carton of milk, 1 donut, and 2 cups of coffee and spent $1.75
So with the above said second equation becomes:
Now we solve these 3 equations:
1M+2D+1C=1.85-------------------------->eqn 1
3D+2C=2.30 ---------------------------->eqn 2
1M+1D+2C=1.75--------------------------->eqn 3
Look for similar number of M's, D's or C's in any 2 equations. we see that Equation 1 and 3 have same numbers of M's that is "1M"
Now subtract Equation 1 from 3 or 3 from 1 whichever you find easy.
Let's say we subtract 3 from 1, so we get:
1M+2D+1C=1.85
1M+1D+2C=1.75
-----------------------
(1M-1M)+(2D-1D)+(1C-2C)=(1.85-1.75)
implies 0M+1D-1C=0.1---------------------->eqn 4
Now we will try to solve eqn 2 and eqn 4.
3D+2C=2.30 ---------------------------->eqn 2
1D-1C=0.1---------------------->eqn 4
From eqn 4 we have D=0.1+C------>eqn 5
Now putting this value of D in eqn 2, we have
3(0.1+C)+2C=2.30
implies 0.3+3C+2C=2.30
implies 5C=2.30-0.3
implies 5C=2
implies C=0.4
Now from eqn 5, D=0.1+0.4=0.5
So, D=0.5
From eqn 1, M=1.85-2D-1C
implies M=1.85-2*0.5-0.4
implies M=1.85-1-.0.4
So, M=0.45
Fourth day cost: 2M+2D=2*0.45+2*0.5
implies fourth day cost=0.9+1=1.9
So they won't be able to get 2 Dougnuts and 2 cartons of Milk with $1.8
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