China's amphibian aircraft, which is stated to be the world's largest, has successfully passed engine tests and is expected to make its maiden flight in the first half of the year 2017.

Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms.

As speed is same, area of cross sections should be same. 

Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12 

i.e πR2 = 6πR12   

Here R = R and R1 = 12/2 = 6   

⇒R2 = 6×6×6   

R = 14.696   

=> D = X = 14.696 * 2 = 29.3938 cm.

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

Given the company has 8 SD + 6 SRP + 7 IN = 21 cars + 'x' small cars.
Given SD makes 1/4 th of the total fleet.
=> 8 = 1/4 (21 + x)
=> 32 = 21 + x
=> x = 11
Therefore, the company has 11 small cars.

The statement mentions that chances of heart ailments are greatly reduced by a regular half-hour exercise. So, a follows. However, it talks of only reducing the probability which does not mean that persons involved in sedentary jobs shall definitely suffer from heart ailments. So, b does not follow.

Hence, concusion a only follows the given statement.

If there are total 100 candidates,
40 candidates are really capable and
60 candidates are incapable.
Of the candidates who are really capable = 80%
40*(80/100)=32 pass the test and
of the incapable= 30%
60*(30/100)=18 pass the test.
Total candidates who passed the test = 32+18 = 50
The proportion of the really capable students who can pass the test to the total students who can pass = 32/50 = 16/25.

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.