GATE Questions

---

Given total number of passengers in the bus = 45

First average weight of 45 passengers = 52 kgs

Average weight of 5 passengers who leave bus = 48

Average weight of 5 passengers who joined the bus = 54

Therefore, the net average weight of the bus is given by

$$\begin{gathered} = 45 x 52 - 5 x 48 + 5 x 5445 \\ = 237045 \\ = 1583 \\ = 52.66 kgs. \end{gathered}$$

C.I = P1+r100t-1

594.5 = 58001+r1002-1

 594.55800+1=1+r1002

r100=1.05-1=0.05=> r = 5 % .

2nd Jan 1901 means
(1900 years and 2 day)
Now, 1600 years have 0 odd day
300 years have 1 odd day
2 days has 2 odd day
Total no. of odd days = 0 + 1 + 2 = 3 days
Hence, the day on 2nd Jan 1901 was Wednesday.

Given k : l = 4 : 3
l : m = 5 : 3

Then k : l : m = 20 : 15 : 9

The series rtus/rtus/rtus/rtus. Thus, the pattern 'rtus' is repeated.

HCF of 210 and 55 is 5

Now, 210x5 + 55P = 5

=> 1050 + 55P = 5

=> 55P = -1045

=> P = -1045/55

=> P = -19.

Tadpole is the larval stage in all the amphibians whereas larva in all the insects.