ব্যাখ্যা
a = 8; b = 9; C = 12
Capacity of a tank
= a × b × c/(c-a)
= (8 × 9 × 12)/4
= 216 Litre.
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স · তারিখ অনির্ধারিত · ১৭ প্রশ্ন
a = 8; b = 9; C = 12
Capacity of a tank
= a × b × c/(c-a)
= (8 × 9 × 12)/4
= 216 Litre.
(A +B)'s 20 day's work = 20 × 1/30 = 2/3
Remaining work = 1 − 2/3 = 1/3
Now,1/ 3 work is done by A in 20 days
∴ The whole work will be done by A
in 20×3 = 60days
A opened 2 hours early to B
In 2 hours A can do 3 × 2 = 6 unit work
Remaining work = 24 - 6 = 18
A + B can do it in
= 18/5 hours
= 3 hours 36 minutes
∴ Tank will be full in 9 A.M. + 3 hours 36 minutes = 12.36 P.M.
Work done by the C pipe in 1 minute
= 1/15 − (1/20 + 1/24)
= (1/15 − 11/120)
= −1/40 [−ve means emptying]
∴ Volume of 1/ 40 part = 3 gallons.
Volume of whole = (3 × 40) gallons = 120 gallons.
If a tank has 4x liters of total capacity and it holds 3x liters of water and if 30 liters of water is taken out, the tank becomes empty.
It means 3x liters of water is taken out
3x = 30 liters
x = 10 liters
Capacity of tank
= 4x = 4 × 10 = 40 liters
Net part filled in 1 hour
= (1/x − 1/y)
= (y−x)/xy
∴ The tank will be filled in
= xy/(y−x) hours
Apply formula of
M1D1H1/W1= M2D2H2/W2
Let 'P' pumps are required to empty the reservoir.
(12pumps×6hours×15days)/1 reservoir
= (P×9hours×12days)/1 reservoir
P = 10 pumps
Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours
∴ 1/ x + 1/ x+6 = 1/ 4
⇒ x+6+x/ x(x+6) = 1/ 4
⇒ x² − 2x−24 = 0
⇒ (x−6)(x+4) = 0
⇒ x = 6 [neglecting the negative value of x].
Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.
∴ 1/x + 2/x + 4/x = 1/5
⇒ 7/x = 1/5
⇒ x = 35 hours
(m1×h1×t1)/w1 = (m2×h2×t2)/w2
9 taps × 20 min = t taps × 15 min
So, t = 12 taps
Let the daily wages of A, B and C be Tk. 5x, Tk. 6x and Tk. 4x respectively.
Then, ratio of their amounts
= (5×6):(6×4):(4x9)
= 30:24:36
= 5:4:6
∴ A's amount
= Tk. (1800 × 5/15)
= Tk. 600
A's 1 day work = 1/18
B's 1 day work = 1/9 [because B take half time than A]
So, (A + B)'s one day work
= 1/18 + 1/9
= (1+2)/18
= 1/6
2 men can do a work in x days
1 men can do a work in (2 × x) days
y women can do a work in 3 days
1 women can do a work in 3y days
1 man : 1 woman
Days - 2x : 3y
Efficiency - 3y : 2x
From M1D1 = M2D2
⇒ M1D1 / M2D2 = 5/6
⇒ (x−1)(x+1)/(x+1)(x+2) = 5/6
⇒ (x−1)/(x+2) = 5/6
⇒ 6x−6 = 5x+10
⇒ x = 16
According to the question,
Let, D is number of days
(639×12×5)/1 road = (30×6×D)/1 road
⇒ D = 213 days
B's 10day's work = 1/15 × 10 = 2/3
Remaining work =
(1−2)/3 = 1/3
Now,
1/18 work is done by A in 1 day
∴ 1/3 work is done by A in 18 × 1/3 = 6 days
Working 5 hours a day, A can complete the work in 8 days i.e.
= 5 × 8 = 40 hours
Working 6 hours a day, B can complete the work in 10 days i.e.
= 6 × 10 = 60 hours
(A + B)'s 1 hour's work,
= 1/40 + 1/60 = (3+2)/ 120
= 5/120
= 1/24
Hence, A and B can complete the work in 24 hours i.e. they require 3 days to complete the work.