Question

The total amount of money spent on the purchase of pencils, rubbers and pens is Rs. 45. The cost of one pencil, one rubber and two pens is Rs. 2, Rs. 3 and Rs. 8 respectively. The total number of pencils bought is greater and less than the number of pens and the number of rubbers bought respectively. How many different combinations of the number of pencils, rubbers and pens bought are possible?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C

3

Let x, y and z be the number of pencils, rubbers and pens bought. The cost of one pencil, one rubber and one pen is Rs.2, Rs.3 and Rs.4 respectively.
Now, 2x + 3y + 4z = 45.

$\therefore x=\frac{45-3y-4z}{2}$

Also,$z<\frac{45-3y-4z}{2}<y$

Or, 6z+3y $<$ 45 and 5y+4z $>45.$

Possible values of y and z that satisfy the above equation is (7, 3); (9, 2) and (11, 1) in that order. Therefore, there are 3 different possible combinations.