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The woman bought toys worth 2490 which included dolls worth 29 each and balls worth 33 each.
Let the number of dolls bought be D and the number of balls bought be B. Both B and D have to be integers.
33*B +29*D = 2490
Using trial and error the equation holds for B = 28 and D = 54
The woman bought 28 balls and 54 dolls
Cost of 1 set of Art-kit + 1 set of Tool-kit = 33 + 29 = 62$
Total Amount = 2490$
Let us assumed equal number (x) of each kind of toys has been purchased,
Then 29x + 33x = 2490,
=>62x = 2490 => x = 2490/62 => x = 40 and Remainder(R)= 10
Here the remainder(R=10) indicates that one of the item should be > 40 and the other kind < 40.
The difference in prices of two kinds of toys à 33-29=4 shows that
on increasing one kind of item(Tool-kit) by 1 and decreasing the other kind(Art-kit) we save 4$.
Therefore on decreasing No. of Art-kit by 12 and increasing the other toy (Tool-kit) by the same number (12) we save 4*12=48$ and adding the remaining 10$ the total savings becomes 48+10 = 58$ and with 58$, two(2) more Too-kit(29$ each) could be bought.
This makes number of Tool-kit bought = 40+12+2 = 54 and the
Number of Art-kit set bought = 40 – 12 = 28
Number of Art-kit set = 28 and Number of Tool-kit = 54
She bought 28 set of Art-kit and 54 set of tool-kit <-- Answer
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