Healthforceontario Interview Questions, Process, and Tips

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D

9 AND 3

The formula to find number of diagonals (D) given total number of vertices or sides (N) is

N * (N – 3)

D = ———–

2

Using the formula, we get
1325 * 2 = N * (N – 3)
N2 – 3N – 2650 = 0

Solving the quadratic equation, we get N = 53 or -50

It is obvious that answer is 53 as number of vertices can not be negative.

Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so on……

Hence the series is 0, 0, 0, 2, 5, 9, 14, …….. (as diagram with 1,2 or 3 vertices will have 0 diagonals).

Using the series one can arrive to the formula given above.

16

put 1 red marble in one jar and all the rest (99) in the
other.
This gives you 50% from the first jar (if they pick that
jar they will get red 100% of the time). For the other jar
the chances are 49/99 or 49.494949%. Divide that by 2 and
its 24.7474%. Total odds are 50% plus 24.7474% = 74.7474%

8

21

So we consider the 2nd statement first. We can form an equation out of it.
14x-6=13y+3=9z+3
Using this, we can understand that the multiple of 14 and the multiple of 13 and 9 must have a difference of 9. The easiest way to ensure that is multiplying it by 9
14*9=126
13*9=117
If the 5th farmer gives 3 apples to the 4th farmer, they would have 123 and 120 apples respectively. However, we also know that the 2nd farmer has 117 apples (13*9=117, and this is a multiple of 9) if the 5th farmer gives 3 apples too the 2nd farmer, the 3rd, 4th and 5th farmers would have 120 apples each.
Now that we got 120, we should check if the first part of the question makes sense along with it. The equation would be
7a+1=11b-1=120
We know that 11*11=121 and 7*17=119. When we add 1 to 119 and subtract 1 from 121, we get 120 for each. In this way, all the farmers have 120 apples each.
Therefore, the 3rd farmer had a yield of 11 per tree and the 4th farmer had a yield of 9 per tree.

GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC

b