Bank Math Master

Question: In a zoo, there are Lions and Pigeons. If heads are counted, there are 140 and if legs are counted, there are 440. How many pigeons are there?

Solution:
Let, the pigeons have x
So, Loins has = 140 - x

ATQ,
2x + 4(140 - x) =440
⇒ 2x + 560 - 4x = 440
⇒ 2x = 120
⇒ x = 60

Question: How many kgs of wheat costing Tk 8 per kg must be mixed with 36 kg of rice costing Tk 5.4 per kg so that 20% gain may be obtained by selling the mixture at Tk 7.2 per kg?

Solution:
Cost Price = (7.2 × 100)/120 = Tk 6
Let, the quantity of wheat of Tk 8 per kg be x

ATQ,
8x + 36 × 5.4 = 6x + 216
⇒ 8x + 210 = 10x + 216
⇒ 8x + 194.4 = 10x + 216
⇒ 2x = 21.6
⇒ x = 10.8

Question: How many kgs of sugar costing Tk 9 per kg must be mixed with 27 kg of sugar costing Tk 7 per kg so that may be a gain of 10% by selling the mixture at Tk 9.24 per kg?

Solution:
Let, the sugar of Tk 9 per kg is = x kg

ATQ,
110% of (7 × 27 + 9 × x) = 9.24(27 + x)
⇒ (11/10) × (189 + 9x) = 249.48 + 9.24x
⇒ 2079 + 99x = 2494.8 + 92.4x
⇒ 6.6x = 415.8
⇒ x = 63

Question: In what ratio must water be mixed with milk to gain 20% by selling the mixture at cost price?

Solution:
Let he has to make milk to water ratio 1 : x to earn 20% profit
Let the cost price of milk be C/L

Then According to the question,
C × (x + 1) = C × (120/100)​
⇒ x + 1 = 6/5
⇒ x = 6/5 - 1
⇒ x = 1/5

So milk to water ratio = 1 : x
= 1 : 1/5
= 5 : 1

Question: The ratio of milk to water in 60 liters of a mixture is 7 : 5. The milk to be added to it make the ratio 2 : 1 is-

Solution:
Quantity of milk = 60 × (7/12) = 35 liters
Quantity of water = 60 - 35 = 25 liters

Let, x liters of milk be added to the mixture.

ATQ,
(35 + x)/25 = 2/1
⇒ 35 + x = 50
⇒ x = 50 - 35
⇒ x = 15

Question: The vessels containing water and milk in the ratio of 1 : 2 and 2 : 5 are mixed is 1 : 4 then what will be the resulting ratio?

Solution:
We can create the following ratios:
Vessel 1: water : milk = x : 2x
Vessel 2: water : milk = 2y : 5y

We can create the equation:
(x + 2x)/(2y + 5y) = 1/4
3x/7y = 1/4
12x = 7y

We see that we can let x = 7 and y = 12.
So we have:
Vessel 1: water = 7 and milk = 14
Vessel 2: water = 24 and milk = 60

Therefore, in the final mixture, the ratio of water to milk is:
(7 + 24)/(14 + 60) = 31/74

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

Question: A container filled liquid containing 4 parts of water and 6 parts of milk. How much of the mixture must be drawn off and filled with water so that the mixture contains half and half water?

Solution:
Let water = 40 liters
and milk is 60 liters.

Let, x amount taken out from the mixture.
Water = 40 - x × (2/5) + x
and milk = 60 - x × (3/5) 

ATQ,

Equate both the equation, we get x = 50/3
Now, mixture drawn ff = (50/3) / 100 = 1/6

Question: A mixture contains alcohol and water in the ratio 4 : 3. If 5 litres of water is added to the mixture, the ratio becomes 4 : 5. The quantity of alcohol in the given mixture is -

Solution:
Let the quantity of alcohol and water be 4x litres and 3x litres respectively
4x/(3x + 5) = 4/5
⇒ 20x = 4(3x+5)
⇒ 8x = 20
⇒x=2.5

Quantity of alcohol
= (4 × 2.5) litres
= 10 litres.

Question: If 4 kg of metal of which 1/3 is silver and the rest is copper, be mixed with 3 kg of metal, of which 1/4 is silver and the rest is copper, then what will be the ratio of silver to copper in the mixture?

Solution:
1st metal,
Quantity of silver = 4 ​× (1/3) = 4/3 ​kg
∴ Quantity of copper = 4 - (4/3) = 8/3 kg

2nd metal,
Quantity of silver = 3 ​× (1/4) = 3/4 ​kg
∴ Quantity of copper = 3 - 3/4 = 9/4 ​kg

In the mixture, the required ratio = (4/3 + 3/4/) : (8/3 + 9/4)
= 25/12 : 59/12
= 25 : 59

Question: Sifat traveled 2/3 as km on foot as by water and 1/3 as km on bus as by water. If he covered a total of 48 km, how many km dis he travel on foot?

Solution:
Let, he traveled x km by water, 2x/3 km on foot and x/3 km bus

ATQ,
x + (2x/3) + (x/3) = 48
⇒ (3x + 2x + x)/3 = 48
⇒ 6x = 144
⇒ x = 24

So, traveled (2 × 24)/3 = 16 km on foot.

Question: A train covers a distance 10 km in 12 minutes. If its speed is decreased by 10 km/hr, the time taken by it to cover the same distance will be-

Solution:
Old speed = (10 × 60)/12 = 50 kmph
New speed = 50 - 10 = 40 kmph

Time taken to go 10km at 40kmph = 10/40 hr
= (10 × 60)/40 min
= 15 min

Question: A truck cover a distance of 480 meters in 1 min whereas a bus covers a distance of 54 km in 45 min. The ratio of their speed- 

Solution:
Speed of truck = 480/60 = 8 m/s
speed of bus = (54 × 1000)/(45 × 60) = 20 m/s

Ratio = 8 : 20 = 2 : 5

Question: A man completes a certain journey in 8 hours. He covers one-third distance at 60 kmph and the rest at 80 kmph. The length of the journey-

Solution:
Let, the length of the journey be x km
he covers x/3 km  at 60 kmph
he covers (x - x/3) = 2x/3 at 80 kmph

ATQ,
(x/3)/60 + (2x/3)/80 = 8
x/180 + x/120 = 8
5x/360 = 8
x = 576 km

Question: A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of total time, to cover the remaining distance in the remaining time, his speed in kmph must be-

Solution:
Remaining distance = 6 - 3 = 3 km
remaining time = 1/3 × 45
= 15 min
= 15/60 hr
= 1/4 hr

Required speed = 3/(1/4) kmph
= 12 kmph

Question: With a uniform speed, a car covers a distance in 8 hours. Had the speed been increased by 4 kmph the same distance could have been covered in 7 hours and 30 minutes. What is the distance covered?

Solution:
Let, the distance = x km

ATq,
x/7.5 - x/8 = 4
⇒ 10x/75 - x/8 = 4
⇒ 2x/15 - x/8 = 4
⇒ x/120 = 4
⇒ x = 480

Question: A train 150 m long, takes 30 seconds to cross a bridge 500 m long. How much time will the train take to cross a platform 370 m long?

Solution:
ট্রেন + ব্রিজের মোট দূরত্ব = ১৫০ + ৫০০ = ৬৫০ মি
ট্রেন + প্লাটফর্ম = ১৫০ + ৩৭০ = ৫২০ মি

৬৫০ মি অতিক্রম করে ৩০ সেকেন্ডে
১ মি অতিক্রম করে ৩০/৬৫০ সেকেন্ডে
৫২০ মি অতিক্রম করে (৩০ × ৫২০)/৬৫০ সেকেন্ডে
= ২৪ সেকেন্ডে

Question: A train 800 meters long is running at a speed 78 kmph. If it crosses a tunnel in 1 min, then the length of the tunnel is-

Solution:
৩৬০০ সেকেন্ডে অতিক্রম করে ৭৮০০০ মি
১ সেকেন্ডে অতিক্রম করে ৭৮০০০/৩৬০০ মি
৬০ সেকেন্ডে অতিক্রম করে (৭৮০০০ × ৬০)/৩৬০০ মি
= ১৩০০ মি

টানেলের দৈর্ঘ্য = ১৩০০ - ৮০০ = ৫০০ মি

Question: A train running at 7/11 of its own speed reached a place in 22 hours. How much time could be saved if the train would have run at its own speed?

Solution:
New speed = 7/11 of the usual speed
New time = 11/7 of the usual time

So, 11/7 of usual time  = 22 hr
Ususal time = 22 × (7/11) = 14 hr

So, the time would be saved = 22 - 14 = 8 hr

Question: A train is driven at the speed of 100 kmph and stops for 10 minutes at the end of every 150 km. To cover a distance of 1000 km, it will be-

Solution:
Here, time taken to 150 km at 100 kmph  = 1 hour 30 min + 10 min
= 1 hr 40 min
= 1 hr + 2/3 hr
= 5/3 hr

time taken to cover (150 × 6) = 900 km in (5/3 × 900) = 10 hrs
Remaining 100 km is covered in 1 hr

So, total time = 10 + 1 = 11 hr