Bank Math Master

Question: A water tank has two taps (Tap-1 and Tap-2): Tap-1 can fill a tank in 6 hours and Tap- 2 can empty the tank in 12 hours. How long will they take to fill the tank if both taps are opened simultaneously but Tap-2 is closed after 6 hours?

Solution:
Tap-1, in 1 hour it fills = 1/6 part
Tap-2 in 1 hour it empties = 1/12 part

When both taps are open, in 1 hrs it fills = (1/6 - 1/12) part
= (2 -1)/12 part
= 1/12 part
When both taps are open, in 6 hrs it fills = {(1/12) × 6} part
= 1/2 part
∴ Remaining = (1 - 1/2) = 1/2 part

As Tap-2 is closed after 6 hours
∴ Tap-1, 1 part will be filled in = 6 hours
Remaining 1/2 part will be filled in = 6 × 1/2 = 3 hours

∴ Total time required = 6 + 3 = 9 hours

Question: A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2 cm, then the area of the triangle is-

Solution:
Let, the side of the square be a cm
Then, its diagonal √2a
√2a = 12√2
⇒ a = 12

∴ Perimeter of the square = 4a
= 4 × 12
= 48 cm

and also perimeter of the equilateral triangle = 48 cm

∴ Each side of the triangle = 48/3
= 16 cm

Area of the triangle = (√3/4) × (16)² cm²
= (√3/4) × 256 cm²
= 64√3 cm²

Question: A does 20% less work than B. If A can complete a piece of work in 15/2 hours, then B can do it in-

Solution:
Ratio of times taken by A and B = 100/80 = 5/4
Let
B takes x days to do the work

ATQ,
5/4 = (15/2)/x
⇒ 5/4 = (15/2)/x
⇒ 5x = (15 × 4)/2
⇒ 5x = 30
∴ x = 6 hours

Question: The length of a ractangle is 20% more than its breadth. What will be the ratio of the area of a rectangle to that of a square whose side is equal to the breadth of the rectangle?

Solution:
Let,
breadth be x metres
∴ length = (120% of x) metres
= 120x/100 metres
= 6x/5 metres

∴ the area of the rectangle = (6x/5 × x) m²

∴ the area of the square = (x × x) m²

∴ The ratio = (6x/5 × x) : (x × x)
= 6 : 5

Question: If the area of the trapezium, whose parallel sides are 6 cm and 10 cm is 64 sq. cm, what will be the distance between the parallel sides?

Solution:
Given,
Parallel sides of a trapezium = 6 cm, and 10 cm

We know,
Area of trapezium = (1/2)(sum of the parallel sides) × distance between the parallel sides
64 = (1/2)(6 + 10) × distance
⇒ 64 = 8 × distance
⇒ distance = 64/8
∴ distance = 8 cm

So, the distance between the parallel lines of trapezium = 8 cm.

Question: One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the faster pipe alone will be able to fill the tank in:

Solution:
Let,
the slower pipe alone fill the tank in x minutes.
Then, Faster pipe alone will fill it in x/4 minutes.

ATQ,
(1/x) + (4/x) = 1/36
⇒ 5/x = 1/36 
∴ x = 180

The slower pipe alone fill the tank in 180 minutes.
the faster pipe alon will be able to fill the tank in = (180 ÷ 4) minutes.
= 45 minutes

Question: If the volume of a sphere is divided by its surface area, the result is 27 cm. the radius of the sphere is-

Solution:
Let,
the radius of the sphere is r cm
∴ the volume of a sphere = (4/3)πr³
∴ the surface area of a sphere = 4πr²

ATQ,
{(4/3)πr³}/(4πr²) = 27
⇒ r/3 = 27
∴ r = 81

∴ the radius of the sphere is 81 cm

Question: A can complete a piece of work in 18 days, B in 20 days and C in 30 days. B and C together start the work and are forced to leave after 2 days. The time taken by A alone to complete the remaining work is-

Solution:
Here,
B's 1 day's work = 1/20 part
C's 1 day's work = 1/30 part

∴ (B + C)'s 1 day's work = (1/20 + 1/30) part
∴ (B+C)'s 2 day's work = {(1/20 + 1/30) × 2} part
= [{(3+2)/60} × 2] part
= (5/60 × 2) part
= 1/6 part

Remaining work = (1 -1/6) part
= 5/6 part

A's one day's work = 1/18 part
∴ The time taken by A alone to complete the remaining work is = (5/6)/(1/18) days
= 15 days

Question: 55 men can finish a work in 42 days. How many additional men must be engaged to complete the work 9 days earlier?

Solution:
Here,
Days needed = (42 - 9) = 33 days

To complete the work in 42 days, it requires 55 men
To complete the work in 1 days, it requires (55 × 42) men
To complete the work in 33 days, it requires (55× 42)/33 men
= 70 men

∴ additional men required = (70 - 55) men
= 15 men

Question: 15 buckets of water fill a tank when the capacity of each bucket is 15 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 25 liters?

Solution:
The capacity of the tank = (15 x 15) litres
= 225 litres

The capacity of each bucket = 25 litres
Number of buckets needed = (225 ÷ 25) = 9

Question: How many bricks need to build a wall 8 m long, 6 m high and 11 cm thick, where as each bricks measuring 25 cm × 60 cm × 11 cm?

Solution:
Given,
Wall long = 8 m = 800 cm
Wall thick = 11 cm
Wall high = 6 m = 600 cm

∴ Volume of the wall = (800 × 600 × 11) cm³

∴ Volume of the brick = 25 cm × 11 cm × 60 cm

∴ bricks need to build the wall = Volume of the wall ÷ Volume of the brick
= (800 × 600 × 11) ÷ (25 × 11 × 60)
= 320

Question: If A can do 1/4 of a work in 3 days and B can do 1/9 of the same work in 4 days, how much will A get if both work together and paid Tk 800 in all?

Solution:
Whole work is done by A in (3 × 4) = 12 days
∴ A's 1 day's work = 1/12 part
Whole work is done by B in (4 × 9) = 36 days
∴ B's 1 day's work = 1/36 part

A's 1 day's work : B's 1 day's work
= A's wages : B's wages
= 1/12 : 1/36
= 3 : 1

∴ A's share = (800 × 3/4) Tk
= 600 Tk

Question: A pipe can fill a cistern in 16 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the cistern completely?

Solution:
Time is taken to fill half of the tank = (1/2) × 16 = 8 hrs

In 1 hour 1 pipe can fill = 1/16 part
∴ In 1 hour 4 pipe can fill = {4 × (1/16)} part
= 1/4 part

4 pipe can fill 1/4 part in 1 hour
∴ 4 pipe can fill 1 part in 4 hour
∴ 4 pipe can fill 1/2 part in (4 × 1/2) hour
= 2 hours

∴ Total time = (8 + 2) = 10 hours.