A large tanker can be filled by two pipes A and B in 60 and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half ?

Given, P = ? 9375, R₁ = 2% and R₂ = 4%
According to the formula,
A = P (1 + R₁/100) (1 + R₂/100)
= 9375 (1 + 2/100) (1 + 4/100)
= 9375 x (51/50) x (26/25)
= 7.5 x 51 x 26
= ? 9945

Now CI = A - P = 9945 - 9375 = ? 570

? Gain = (S.P. of 100 bananas) - (C.P. of 100 bananas)
? (S.P.of 20) = (S.P of 100) - (C.P. of 100)
? S.P.of 80 = C.P. of 100

Let C.P. of each banana = Re. 1
C.P. of 80 banana = Rs. 80
S.P. of 80 banana = Rs. 100

? Gain % = (20 / 80) x 100 % = 25%

Average of first n natural numbers = ( n + 1 ) / 2
= ( 79 + 1 ) / 2
= 40

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

∴ The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

Let us assume the first part is A then second part will be 50 - A.
According to question,
Sum of their reciprocals is 1/2.
1/A + 1/(50 - A) = 1/12
? (50 - A + A)/(50 - A)A = 1/12
? (50)/(50 - A)A = 1/12
? (50) x 12 = 1 x (50 - A)A
? 600 = (50 - A)A
? 600 = 50A - A²
? A² - 50A + 600 = 0
? A² - 30A - 20A + 600 = 0
? A(A - 30) - 20(A - 30) = 0
? (A - 30)(A - 20) = 0
? (A - 30) = 0 or (A - 20) = 0
? A = 20 , 30

Total time taken = (3/10 + 3/20 + 3/30 + 3/60) hrs.
= 3/5 hrs.

? Average speed = {12 / (3 / 5)} km/hr.
= (12 x 5 ) / 3 km/hr.
= 20 km/hr.

Given Exp. = 3.57 - 2.14
= 3 + (57 / 99) - 2 - (14 / 99)
= 1 + (57 / 99) - (14 /99)
= 1 + 43/99
= 1.43