A woman buys toys worth 2490 which include dolls worth 29 each and balls worth 33 each. How of each did she buy.
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