1. ## Pencil Mania

I know the answer of this sum but would be glad to get an explanation to how this is done:

Box P has 5 more pencils than Box Q. How many pencils must be moved from Box P to Box Q, such that Box P has 9 fewer pencils than Box Q now?

a) 4
b) 5
c) 6
d) 7

2. $\frac{5+9}{2}$

3. Hello, MoniMini!

Box P has 5 more pencils than Box Q.
How many pencils must be moved from Box P to Box Q,
so that Box P has 9 fewer pencils than Box Q now?

. . a) 4 . . b) 5 . . c) 6 . . d) 7

Let $\,q$ = number of pencils in Box Q.
Then there are $q + 5$ pencils in Box P.

Suppose $\,x$ pencils are moved from Box P to Box Q.
. . Then Box P will have: $q + 5 - x$ pencils. .[1]
. . And Box Q will have: $q + x$ pencils. .[2]

We want [1] to be 9 less than [2].

. . . . $q + 5 - x \;=\;(q + x) - 9$

Got it?