Bank Math Master

Question: A, B, and C start together from the same place to walk around a circular path of length 12km. A walk at the rate of 4 km/h, B 3 km/h and C 3/2 km/h. They will meet together at the starting place at the end of-

Solution:
We know,
Time = Distance/speed

Time taken to complete the revolution
A = 12/4 = 3 hours
B = 12/3 = 4 hours
C = 12 × (2/3) = 8 hours

Now, required time = LCM of 3, 4, and 8
= 24 hours

Question: Mobin deposited Tk 8000 at 10% interest rate in a bank. What would be the compound interest after one year if the interest is calculated on half yearly basis?

Solution:
Compound Principal = P(1 + r)ⁿ
= 8000 × {1 + 10/(2 × 100)}^{2 × 1}
= 8000 × (21/20)²
= 8000 × {(21 × 21)/(20 × 20)}
= 8820

Compound interest = 8820 - 8000 = 820 Tk

Question: Three girls have candies in the ratio of 7 : 4 : 5. If the girl with the least number of candies has 12 candies, how many candies does the girl with the greatest number have?

Solution:
Let their amounts of candies be 7x, 4x, and 5x.

ATQ,
4x = 12
⇒ x = 3

So, the girl with the greatest number of candies has = 7 × 3
= 21 candies

Question: A juice vendor has two cans of juice. The first contains 30% water and the rest juice. The second contains 60% water. How much juice should he mix from each of the containers so as to get 18 litres of juice such that the ratio of water to juice is 1:2?

Solution:
Let the cost of 1 litre of pure juice be Tk. 1.

Juice in 1 litre mix from 1st can = 7/10 litre; C.P. of 1 litre mix. in 1st can Tk. 7/10
Juice in 1 litre mix from 2nd can = 4/10 litre; C.P. of 1 litre mix. in 2nd can Tk. 4/10
Juice in 1 litre of the final mix = 2/3 litre; C.P. of 1 litre of the final mix = Tk. 2/3

By the rule of alligation, we have:
Quantity of Cheaper/ Qunatity of Dearer = (CP of Dearer - Mean Price)/ (Mean Price - CP of Cheaper)
⇒ Quantity of 2nd can : Quantity of 1st can = {(7/10) - (2/3)} : {(2/3) - (4/10)}
= (21 - 20)/30 : (20 - 12)/30
= (1/30) : (8/30)
= 1 : 8

Thus, the quantity from the first can is = (1/9) × 18 = 2 
The quantity from the second can is= (8/9) × 18 = 16

To get 18 litres of juice with a water-to-juice ratio of 1:2, the vendor should mix 2 litres from the first can and 16 litres from the second can.

Question: If 25% of 1200 + 35% of 600 = x% of 4000, find the value of x.

Solution:
25% of 1200 + 35% of 600 = x% of 4000
⇒ (25/100) × 1200 + (35/100) × 600 = (x/100) × 4000
⇒ 300 + 210 = 40x
⇒ 40x = 510
⇒ x = 510/40
∴ x = 12.75

Question: A shopkeeper sells a watch for Tk. 8,400 and makes a profit of 25%. If the profit is reduced to 15%, what will be the new selling price?

Solution:
Let Cost Price was x
ATQ,
x + 25% of x = 8400
⇒ x + (25x/100) = 8400
⇒ x + 0.25x = 8400
⇒ 1.25x = 8400
∴ x = 6720

So, Cost Price = 6720
Now, Selling Price When profit remains at 15%,
= 6720 + 15% of 6720
= 6720 + 1008
= 7728

Question: In a school having roll strength 286, the ratio of boys and girls is 8 : 5. If 22 more girls get admitted into the school, the number of girls becomes-

Solution:
Boys : girls = 8 : 5
Let, the boys = 8x, girl = 5x

According to the 1st condition,
8x + 5x = 286
⇒ 13x = 286
⇒ x = 286/13
∴ x = 22
Boys = 8 × 22 = 176 and girls = 5 × 2 = 110

22 more girls get admitted then the number of girls become = (5x + 22)
= 110 + 22
= 132

Question: Sunny and Rajib bought the same item. Sunny paid Tk 15,000. Rajib got discount and paid Tk. 3,000 less than Sunny. How much discount did Rajib get?

Solution:
In Tk. 15000 Rajib get discount Tk. 3000
In Tk. 1 Rajib get discount Tk. (3000/15000)
In Tk. 100 Rajib get discount Tk. (3000 × 100)/15000
= Tk. 20

∴ Rajib get 20% discount.

Question: The price of a notebook is Tk. 80. If you buy more than 5 notebooks, you will get a 15% discount. How much do you have to pay if you buy 10 notebooks?

Solution:
The price of 1 notebook = 80
∴ The price of 10 notebook  = 80 × 10 = 800

If price 100 then have to be paid = 85
If price 1 then have to be paid = 85/100
If price 800 then have to be paid = (85 × 800)/100
= 680 tk

Therefore, you have to pay Tk. 680 if you buy 10 notebooks.

Question: A boat goes 6 km upstream in 24 minutes. The speed of the stream is 3 km/hr. The speed of boat in still water is

Solution:
The speed upstream is 6/24 km/min
= (6 × 60)/24 km/hr
= 15 km/hr

Let,
The speed of boat in still water is x km/hr

Here, x - 3 = 15
⇒ x = 15 + 3
∴ x = 18 

∴ The speed of the boat in still water is 18 km/hr

Question: A sum of money at simple interest amounts to Tk 960 in 5 years and to Tk 1000 in 6 years. The sum is-

Solution:
Simple interest for 1 years = 1000 - 960 tk
= 40 tk

∴ Simple interest for 5 years = 40 × 5
= 200 tk

∴ Sum = Amount after 5 years - SI for 5 years
= 960 - 200 tk
= 760 tk

Question: Jabed spends 60% of her salary and donates 20% of her salary to charity. If he is left with Tk. 2000, what is her monthly salary?

Solution:
Let Jabed's salary be x

ATQ,
x - (60% of x + 20% of x) = 2000
⇒ x - 80% of x = 2000
⇒ 20% of x = 2000
⇒ 20x/100 = 2000
⇒ 20x = 200000
⇒ x = 200000/20
⇒ x = 10000 

Therefore, Jabed's monthly salary is Tk. 10,000.

Question: In what ratio must sugar at Tk 15 per kg be mixed with sugar at Tk 25 per kg so that the mixture is worth Tk 22 per kg?

Solution:
Let the amount of sugar at Tk 15 per kg be x kg
and the amount of sugar at Tk 25 per kg be y kg.

ATQ,
15x + 25y = 22(x + y)
⇒ 15x + 25y = 22x + 22y
⇒ 7x = 3y
⇒ x/y = 3/7
⇒ x : y = 3 : 7

Question: On increasing the price of tickets for a show by 20 % the audience decreases by 30%. What is the effect on revenue?

Solution:
Effect on revenue
= percent increase - percent decrease - (percent increase × percent decrease)/100
= 20 - 30 - (20 × 30)/100
= - 10 - 600/100
= -10 - 6
= - 16 %

'-' sign shows decrease in revenue.

Question: If the price of sugar increases by 10%, by what percentage a housewife should reduce the consumption of sugar so that expense is not increased?

Solution:
Let the price be Tk. x per kg
Consumption be y kg

Expenditure = price × consumption
⇒ Expenditure = x × y

Price of sugar is increased by 10%
New price of sugar = Tk. 1.1x per kg

Let new consumption be z kg
New Expenditure = (1.1x) × z

New Expenditure = old Expenditure
⇒ (1.1x) × z = x × y
⇒ z = y/1.1

Reduction in consumption = (y - z) = y - y/1.1 = y/11

Percentage reduction in consumption = (reduction in consumption/original consumption)× 100
Percentage reduction in consumption = [(y/11)/y] × 100 = 100/11 = 9.09% ≈ 9.1%

Question: If two values are 25% and 75% of a third value, what percentage is the first value of the second value?

Solution:
Let
The third value x
∴ First value = (25x)/100 = x/4
∴ Second value = (75x)/100 = 3x/4

Now,
(First Value/Second value) × 100
= (x/4) × (4/3x) × 100
= 33.33%

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, B, and C are boxes containing marbles in the ratio 2 : 3 : 4. Total number of marbles is 90. The above ratio can be changed to 4 : 5 : 6 by transferring-

Solution:
A's share = 90 × (2/9) = 20
B's share = 90 × (3/9) = 30
C's share = 90 × (4/9) = 40

When marbles are shared in the ratio of 4 : 5 : 6

A's share = 90 × (4/15) = 24
B's share = 90 × (5/15) = 30
C's share = 90 × (6/15) = 36

Clearly,
From C (40 - 4) = 36 marbles have been transferred from to A (20 + 4) = 24

So, The above ratio can be changed to 4 : 5 : 6 by transferring 4 marbles from C to A.

Question: The distance between two places A and B is 570 km. A train starts from A at 50 km/h at 6 am and another starts from B at 80 km/h at 7 am towards each other. At what time will they meet?

Solution:
Let the two trains meet at a distance d km from place A.
Time required by the train starting from A to cover p = p/50 hr
Time taken by the other train starting from B to cover (570 - p) km = (570 - p)/80

But the first train has started 1 hr early. So, it has traveled 50 km in this 1 hr.
Therefore,
(p/50) - 1= (570 - p)/80
⇒ (p - 50)/50 = (570 - p)/80
⇒ 28500 - 50p = 80p - 4000
⇒ 130p = 32500
∴ p = 250

So, they will meet at a distance of 250 km from Place A.
So the time at which they will meet will be (250/50) = 5 hrs [after 6 am]
Hence, they will meet at 11 am.