Bank Math Master

Question: A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

Solution:
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = {3 - (3x/8) + x} litres

Quantity of syrup in new mixture = (5 - 5x/8) litres

ATQ,
{3 - (3x/8) + x}  = (5 - 5x/8)
⇒ 5x + 24 = 40 - 5x
⇒ 10x = 16
∴ x = 8/5

So, part of the mixture replaced = (8/5) × (1/8) = 1/5

Question: A person crosses a 1200 m long street in 10 minutes. What is his speed in km per hour?

Solution:
Speed = 1200 meters/10 minutes
= (1200 × 60)/(10× 1000) km/hr
= 7.2 km/hr

Question: In a mixture 120 liters, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then estimate the quantity of water in liter to be further added in the mixture.

Solution:
Sum of the given ratio = 2 +1=3

Quantity of milk = 120 × (2/3) = 80 liters
Hence, quantity of water will be
= 120 - 80 = 40 liters

Assume,
x liters water needs to be added.
ATQ,
80/(x + 40) = 1/2
⇒ 160 = x + 40
∴ x = 120

∴ 120 liters water needs to be added.

Question: A boatman goes 1 km against the current of the stream in 1/2 hour and goes 1 km along the current in 20 minutes. How long will it take to go 5 km in stationary water?

Solution:
Speed upstream = 1/(1/2) = 2 km/hr

Speed downstream = 1/(20/60) = 3 km/hr

Speed in still water = (1/2)(3 + 2) = 2.5 km/hr

Time taken to travel 5 km in still water = 5/2.5 hr
= 2 hr

Question: Tea worth Tk. 126 per kg and Tk. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Tk. 153 per kg, the price of the third variety per kg will be:

Solution:
let, price of third variety x tk per kg 

126y + 135 y + x × 2y = 153 (y + y + 2y)
⇒ 126y + 135y + 2xy = 153 × 4y
⇒ 126 + 135 + 2x = 612
⇒ 2x + 261 = 612
⇒ 2x = 351
∴ x = 175.5 tk

Question: A train 200 m long passed a pole in 20 seconds. How long will it take to pass a platform 610 m long?

Solution:
Speed = 200/20 = 10 m/s
∴ Required time = (200 + 610)/10
= 81 sec

Question: The cost of Type 1 rice is Tk. 15 per kg and Type 2 rice is Tk. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is:

Solution:
Let, 
Quantity of type 1 rice is 2x kg.
Quantity of type 2 rice is 3x kg.
The price per kg of the mixed variety of rice is y taka

∴ Total price of type 1 rice is 15 × 2x = 30x Taka
∴ Total price of type 2 rice is 20 × 3x = 60x Taka

ATQ,
30x + 60x = y(2x + 3x)
⇒ 90x = y × 5x
⇒ y = (90x)/(5x)
∴ y = 18

Question: Speed of a boat in standing water is 18 kmph and the speed of the stream is 3 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is:

Solution:
Speed of a boat in standing water = 18 kmph
The speed of the stream = 3 kmph

∴ Speed upstream
= (18 - 3) kmph
= 15 kmph.

∴ Speed downstream
=  (18 + 3) kmph
= 21 kmph.

Total time taken
= (105/15 + 105/21) hours
= 7 + 5 hours
= 12 hours

Question: An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 100 minutes, it must travel at a speed of:

Solution:
Total Distance = (240 × 5) = 1200 km.
Time = 100 minutes = 100/60 hr
= 5/3 hr

We know that,
Speed = Distance/Time
 ∴ Required speed = 1200/(5/3) km/hr
= 720 km/hr.

Question: In what ratio must a grocer mix two varieties of tea worth Tk. 60 a kg and Tk. 65 a kg so that by selling the mixture at Tk. 68.20 a kg he may gain 10%?

Solution:
Quantity of Tk. 60 tea is x kg.
Quantity of Tk. 65 tea is y kg

S.P. of 1 kg of the mixture = Tk. 68.20,
Gain = 10%.
C.P of 1 kg of the mixture = Tk. (100/110 × 68.20) = Tk. 62

ATQ,
60x + 65y = (x + y)62
⇒ 60x + 65y = 62x + 62y
⇒ 62x - 60x = 65y - 62y
⇒ 2x = 3y
∴ x/y = 3/2

Question: A man can row three-quarters of a kilometre against the stream in 11.25 minutes and down the stream in 7.5 minutes. The speed (in km/hr) of the man in still water is:

Solution:
Three - quarters of a kilometer = (3 × 1000)/4 meters = 750 meters

11.25 minutes = 11.25 × 60 = 675 sec
7.5 minutes = 7.5 × 60 = 450sec

Rate upstream = 750/ 675 m/sec
= 10/9 m/sec

Rate downstream =750/450 m/sec
 = 5/3 m/sec


∴Rate in still water = (1/2){(10/9) + (5/3)} m/sec
= (1/2){(10 + 15)/9}
= (1/2)(25/9)
= 25/18
= (25/18)(18/5) km/hr.
= 5 km/hr

Question: A train 200 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 16 seconds. The speed of the train is:

Solution:
Speed of the train relative to man = 200/16 m/sec
= 12.5 m/sec
= (12.5 × 3600)/1000 km/hr
= 45 km/hr 

Let,
The speed of the train be x km/hr.
∴ Relative speed = (x - 5) km/hr.
⇒ x - 5 = 45         
∴ x = 50 km/hr.

Question: Two equal glasses are respectively one-third and one-fourth full of milk. They are then filled up with water and the contents are mixed in a tumbler. Ratio of milk and water in tumbler is-

Solution:
Quantity of milk in tumbler = 1/4 + 1/3 = 7/12

Quantity of water in tumbler
= (1 - 1/4) + (1 -1/3) = 3/4 + 2/3 = 17/12

So, ratio of milk and water = (7/12) : (17/12)
= 7 : 17

Question: A solution contains 20% sugar by weight is made sweeter by doubling the amount of sugar. The percent of sugar, by weight, in the new solution is-

Solution: 
let, solution is 100 unit
amount of sugar = 100 × 20%
= 100 × (20/100)
= 20 unit 

by doubling, amount of sweet = 40 unit

solution now = 100 + 20 = 120 unit 

percent of sugar = (40 × 100)/120 %
= 33.33%

 

Question: A train 216 m long moving at a speed of 50 km/hr crosses a train 224 m long coming from opposite direction in 12 seconds. The speed of the second train is-

Solution:
Distance covered = (216 + 224) meter
= 440 meter.

Time = 12 seconds.

Relative speed = 440/12 = 110/3 m/s.
= (110 × 3600)/(3 × 1000) km/hr
= 132 km/hr.

Now,
50 + Speed of second train = 132 km/hr.
∴ Speed of second train = (132 - 50) km/hr.
= 82 km/hr.

Question: A mixture contains 2/5 of element A and 3/5 of element B. When 5 ml of A is added to the mixture, the proportion of B in the mixture changes to 1/5. What amount of A was originally present in the mixture before the addition was made?

Solution:
Let mixture be x ml.
A = (2x)/5 ml.
B = (3x)/5 ml.

After adding 5 ml of A to mixture, amount of B remained same. 
And the mixure be x + 5 ml.

New B = (x + 5)/5

Now,
(3x)/5 = (x + 5)/5
⇒ 15x = 5x + 25
⇒ 10x = 25
∴ x = 2.5

Original amount of A = (2 × 2.5)/5 ml = 1 ml.

Question: A 180 liter mixture of milk and water contains 20% water. How much milk, in liters must be added to the mixture will contain water and milk in the ratio of 1 : 7?

Solution:
Water in the mixtue = 180 × (20/100) liter 
= 36 liter

Milk in the mixture = 180 - 36 liter
= 144

Let,
X liter milk must be added

ATQ,
36/(144 + X) = 1/7
⇒ 144 + X = 252
⇒ X = 252 - 144
∴ X = 108 liter

Question: If a biker rides at 70 km/hr instead of 50 km/hr, he would have ride 100 km more. The actual distance travelled by biker is:

Solution:
Let,
the actual distance travelled be x km.

Then,
x/50 = (x + 100)/70
⇒ x/5 = (x + 100)/7
⇒ 7x = 5x + 500
⇒ 7x - 5x = 500
⇒ 2x = 500
⇒ x = 500/2
∴ x = 250 km

Question: There are deer and peacock in zoo. The total number of their heads is 80 and the total number of their legs is 200. How many peacocks are there?

Solution:
Let there are "X" deer and "Y" peacocks.

Total heads are 80.
∴ X + Y = 80 .................(1)

Total legs are 200. Deer has 4 legs and peacock has 2.
∴ 4X + 2Y = 200 ..................(2)

Multiplying Equation (1) by 4 and substracting with Equation (2), we get
4X + 4Y - 4X - 2Y = 320 - 200
⇒ 2Y = 120
∴ Y = 60

Putting this value of Y in Equation (1), we get
X + 60 = 80
∴ X = 20

∴ Total peacocks are Y = 60.
Hence, the correct answer is 60.

Question: A train can travel 50% faster than a car. Both start from point A at the same time and reach point B, 150 km away from A at the same time. On the way, however, the train lost about 25 minutes while stopping at the stations. The speed of the car is:

Solution: 
Let speed of the car be x kmph.
Then, speed of the train = x + (50x)/100 = x + x/2
= (3x)/2

Now
(150/x) - [150/{(3x)/2)}] = 25/60
⇒ (150/x) - {300/(3x)} = 5/12
⇒ (150/x) - (100/x) = 5/12
⇒ (150 - 100)/x = 5/12
⇒ 50/x = 5/12
⇒ 5x = 12 × 50
∴ x = 120 kmph

Question: A dishonest milkman profess to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is-

Solution:
Let, the cost price of 1 liter of milk be = 100 Tk
So, the selling price of 1 liter mixture is also = 100 Tk

Here, in 100 Tk, SP gain = 25% 
So, cost price of the mixture = (100 × 100)/125 = 80 Tk

So, water in the mixture = 100 - 80 = Tk 20