Toronto east general hospital Interview Questions, Process, and Tips

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3 hours ago.

Thin candle melts 3/4 in 3 hours leaving 1/4
Where as in the same time thick candle melts 3/6 leaving 3/6 which is 1/2. Now thick candle is exactly twice than the thin candle.

Or via modeling:

We need to find time at which the length of the thin candle is half the thick candle. Let x be the time. Thin candle melts at 1/4 an hour and thick candle melts at 1/6 an hour. In x hours they melt at x/4 and x/6 respectively. What’s left will be 1 – x/4 and 1 – x/6. We need to find x at which :

2 * (1 – (x/4)) = 1 – (x/6)

This equation results in x = 3

A Roman was born the first day of the 35th year before Christ means before the start of year 0, 35 years has been past and died the first day of the 35th year after Christ means 35th year is about to start at his death so only 34 years has been completed so total years = 69

Take 5 pills from jar 1, 4 pills from jar2,…. and 1 pill
from jar 5 and put altogether in the scale. the ideal
weight should be (1+2+3+4+5)x10= 150gms. for eg if the jar2
is contaminated then the weight will be 146 gms. so
depending on the amount of weight loss we can identify the
contaminated jar.

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros

6

6.25

Actually
carrots=> $12
celery => $3

5c2*4c1/9c3 +5c3/9c3=25/42

1500

brownish grey

2/3 * 15 miles = 10 miles then

time for 10 miles travel is t1 = 10/40 = 1/4

then remaning distance is 5 miles

time for 5 miles travel is t2 = 5/60 = 1/12

t1+ t2= 1/3 in seconds * 60 = 20 min