Simplify ( 6a^-2bc^-3/4ab^-3c² ) ÷ ( 5a^-3 b²c^-1/3ab^-2c³ )

Effective annual rate = 100(1 + {(4/4) /100})⁴ - 100
= (101/100)⁴ x 100 - 100
= 104.0604 - 100
= 4.0604%

Time at which they meet at starting point
= LCM of 2, 4 and 5.5
= 44 h

S_d = 12 km/h
S_u = 5 km/h
Speed of current = (12 - 5)/2 = 3.5 km/h

Let the number be a and b.
Then, (a + b) = (a² -b²) / (a-b)
= 135/5
= 27

Given expression = (20 ÷ 5) ÷ 2 + (16 ÷ 8) x 2 + (10 ÷ 5) x (3 ÷ 2)
= (4 / 2) + (2 x 2) + 2 x (3 / 2)
= 2 + 4 + 3
= 9

According to the question,
a + b + c = 11 x 3 = 33 ......(i)
c + d + e = 17 x 3 = 51 ...........(ii)
e + f = 22 x 2 = 44 ........... (iii)
e + c = 17 x 2 = 34 ..........(iv)
From Eqs. (ii) and (iv), we get
34 + d = 51
or d = 17 ....(v)

Now, by adding Eqs. (i), (iii) and (v), we get
a + b + c + d + e + f = 33 + 44 + 17 = 94

? Average of a, b, c, d, e and f = 94/6
= 47 / 3
= 15²/₃

? 22 men do the work in 16 days.
? 1 man will do the work in 16 x 22 days.
? 32 men will do the job in = (16 x 22)/32 days
? Required number of days = (16 x 22)/32 = 11 days

Let the sum be ? P .
Then, SI = 4P - P = ? 3P
(P x 12 x T)/100 = 3P
? T = (3 x 100)/12 = 25 yr

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Let total distance be y km.
Then , (2y / 3) / 4 + ( y / 3) / 5 = 42/60
? y/6 + y/ 15 = 7/10
? 5y + 2y = 21
? y = 3
? Required distance = 3 km.