hello,
Please solve this problem....
regards
Ket
ds13
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Let C be the cost of a cupcake and x be the no of cupcakes purchased
Let D be the cost of a doughnut and y be the no of doughnuts purchased
As per the information given in the question stem Cx+Dy=6
We are asked to determine the value of y ( i.e the no of doughnuts bought)
statement 1:
2D=3C-0.10
Cx+Dy=6(Given)
We cant determine y from the 2 equations( 4 unknowns and 2 equations)
Insufficient
statement 2:
(C+D)/2=0.35
C+D=0.70
Cx+Dy=6
y cant be determined (4 unknowns and 2 equations)
Insufficient
Combining both the equations we get
Cx+Dy=6........(1)
3C-2D=0.10....(2)
C+D=0.70.......(3)
The value of C& D can be determined from 2 and 3..
but the value of y cant be found..
Hence E is the ans...Let me know if you have any doubts...
Let D be the cost of a doughnut and y be the no of doughnuts purchased
As per the information given in the question stem Cx+Dy=6
We are asked to determine the value of y ( i.e the no of doughnuts bought)
statement 1:
2D=3C-0.10
Cx+Dy=6(Given)
We cant determine y from the 2 equations( 4 unknowns and 2 equations)
Insufficient
statement 2:
(C+D)/2=0.35
C+D=0.70
Cx+Dy=6
y cant be determined (4 unknowns and 2 equations)
Insufficient
Combining both the equations we get
Cx+Dy=6........(1)
3C-2D=0.10....(2)
C+D=0.70.......(3)
The value of C& D can be determined from 2 and 3..
but the value of y cant be found..
Hence E is the ans...Let me know if you have any doubts...
hi raju232007,
after getting value of C & D is it not possible to find x and y using trial error or trying some combination.
Please assume some real number for C & D and try it .
Am I thinking in wrong direction let me know.
ket
after getting value of C & D is it not possible to find x and y using trial error or trying some combination.
Please assume some real number for C & D and try it .
Am I thinking in wrong direction let me know.
ket
-
- Master | Next Rank: 500 Posts
- Posts: 133
- Joined: Sat Dec 29, 2007 2:43 am
- Thanked: 12 times
Yeah..It is possible to find the values of x and y using a few combination's..But the values of x and y should be considered only if we get a unique solution for both x and y...If you get more than one set of solutions the values cannot be determined...
Solving 2 and 3 we get the values of C and D as
C=0.30 and D=0.40
Now substitute the values in equation 1
0.30x+0.40y=6
30x+40y=600
We have the following 4 possibilities
x=4,y=12
x=8,y=9
x=12,y=6
x=16,y=3
So y can be any of the following values (3,6,9 and 12)
Hence the value of y cannot be determined...Hope this helps..
Solving 2 and 3 we get the values of C and D as
C=0.30 and D=0.40
Now substitute the values in equation 1
0.30x+0.40y=6
30x+40y=600
We have the following 4 possibilities
x=4,y=12
x=8,y=9
x=12,y=6
x=16,y=3
So y can be any of the following values (3,6,9 and 12)
Hence the value of y cannot be determined...Hope this helps..