Model 1 Basic Pipes & Cisterns problems Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (d)

Part of the cistern filled in 1 hour = $1/8$

Part of the cistern emptied in 1 hour = $1/16$

When both the taps are opened simultaneously, part of cistern filled in 1 hour

= $1/8 - 1/16 = {2 - 1}/16 = 1/16$

Hence, the cistern will be filled in 16 hours.

Using Rule 7,

Here, x = 8, y = 16

Required time = ${xy}/{y- x}$

= ${8 × 16}/{16 - 8}$ = 16 hours

Answer: (c)

Part of the tank filled by pipes A and B in 1 minute

= $1/36 + 1/45 = {5 + 4}/180$

= $9/180 = 1/20$

Part of the tank filled by these pipes in 7 minutes

= $7/20$

Remaining unfilled part

= $1 - 7/20 = {20 - 7}/20 = 13/20$

When all three pipes are opened.

= $1/20 - 1/30 = {3 - 2}/60 = 1/60$

Time taken in filling $13/20$ part

= $13/20$ × 60 = 39 minutes

Required time = 39 + 7 = 46 minutes

Answer: (c)

If pipe y be closed after x minutes, then

$18/24 + x/32$ = 1

$x/32 = 1 - 18/24 = 1- 3/4 = 1/4$

$x = 32/4$ = 8 minutes

x = 24, y = 32, t = 18

Required time = $[y(1 –{t/x})]$ minutes

= $[32(1 - {18/24})]$ minutes

= $[32(1 - {3/4})] = 32 × 1/4$ = 8 minutes

Answer: (a)

Using Rule 2,

Part of the cistern filled in 1 hour

= $1/3 + 1/4 - 1/2$

[Cistern filled by 1st pipe + Cistern filled by 2nd pipe - Cistern emptied by 3rd pipe]

= ${4 + 3 - 6}/12 = 1/12$

Hence, the cistern will be filled in 12 hours.

Answer: (b)

Using Rule 1,

If the slower pipe fills the tank in x hours, then

$1/x + 1/{x - 10} = 1/12$

${x -10 + x}/{x(x - 10)} = 1/12$

$x^2 - 10x = 24x - 120$

$x^2 - 34x + 120$ = 0

$x^2 - 30x - 4x + 120$ = 0

$x (x - 30) - 4 (x - 30)$ = 0

$(x - 4) (x - 30)$ = 0

$x$ = 30 because $x ≠ 4$

Required time = 30 - 10 = 20 hours