ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: 40 litre of mixture of milk and water contains 25% of water. When 10 liters of water is added, what will be the percentage of milk in the final mixture?

Solution:
25% of 40 litre = 10 litre. 
10 liter water is present in the mixture. 
When 10 liters more water is added in the mixture then the final mixture contains 20 liters water and 30 liters milk. 
% of milk in final mixture= (30/50) × 100 = 60%

Question: A vendor bought 6 oranges for Tk. 10 and sold them at 4 for Tk. 6. Find his loss or gain percent.

Solution:
Suppose, number of oranges bought = LCM of 6 and 4 = 12
∴ CP of 12 oranges = (10/6) × 12 = Tk. 20
and SP of 12 oranges = (6/4) × 12 = Tk. 18
∴ Loss = 20 - 18 = 2

∴ Loss% = (2/20) × 100% = 10%

Question: Applied to a bill for Tk. 100000 the difference between a discount of 40% and two successive discounts of 36% and 4% is-

Solution:
40% of Tk. 100000 = Tk. 40000

36% of Tk. 100000 = 36000
4% of 36000 = Tk. 2560.
Therefore, two successive discounts on Tk 100000 = 36000 + 2560 = Tk. 38560

Difference between a discount of 40% and two successive discounts of 36% and 4%
= 40000 - 38560
=  1440

Question: In Live MCQ's class there are 12 boys and 18 girls. Write the ratio of girls : boys in its simplest form.

Solution:
The question asks for the ratio girls:boys so girls must be first and boys second. It also asks for the answer in its simplest form.

Number of girl : Number of boys = 18 : 12 = 3 : 2

The ratio 18 : 12 represents the number of girls to boys but is not in its simplest form. Ratios are typically expressed in the simplest form by dividing both terms by their greatest common divisor.
Here’s the simplification process:
Original Ratio= 18 : 12
Greatest Common Divisor (GCD): The GCD of 18 and 12 is 6.
Simplify: Divide both terms by 6
18/6 : 12/6 = 3 : 2

Question: Pure ghee costs Tk.100 per kg. A shopkeeper mixes vegetable oil costing Tk. 50 per kg and sells the mixture at Tk. 96 per kg, making a profit of 20%. In what ratio does he mix the pure ghee with the vegetable oil.

Solution:
Cost price of 1 kg of mixure = Tk. (100/120) × 96 = Tk. 80 = mean price
Cost of 1 kg of pure ghee = Tk. 100 = dearer quantity.  
Cost of 1 kg of vegetable = Tk. 50 = cheaper quantity. 
And, the mean price = m = Tk. 20

Therefore,
(Dearer quantity) : (Cheaper quantity) = (m - c) : (d - m) = (80 - 50) : (100 - 80) = 30 : 20 = 3 : 2 

∴ Required ratio = 3 : 2

Question: In order to obtain an income of Tk. 650 from 10% stock at Tk. 97, one must make an investment of-

Solution:
To obtain Tk. 10, investment = Tk. 97.
To obtain Tk. 650

investment = Tk. (97/10) × 650
= Tk. 6305

Question: In school, three rows are formed of the students. First row has 50% more students than the second row and the third row has 50% less students than the second row. If the total number of students in all the rows is 900, then find the number of students in the third row.

Solution:
Let the number of students in the second row is 100. 
Then number of students in first row is 150 and number of students in third row is 50. 
Total students = 150 + 100 + 50 = 300 
Given = 900 
Then 900/300 = 3 
Third row has 50 × 3 =150 students.

Question: By selling 33 meters of cloth, one gains the selling price of 11 meters. Find the gain percent.

Solution:
Gain = (SP of 33 m) - (CP of 33 m) = SP of 11m
⇒ SP of 22m = CP of 33m

Let
CP of each meter be Tk 1.
Then,
CP of 22 m = Tk. 22.
CP of 33 m = Tk. 33

Hence, SP of 22m = Tk. 33
∴ Gain = 33 - 22 = Tk. 11

∴ %Gain = (11/22) × 100
= 50%

Question: A discount of 15% on one article is the same as discount of 20% on a second article. The costs of the two articles can be-

Solution:
Let the prices of two articles be X and Y
From the question 15X/100 = 20Y/100
⇒ 15X = 20Y
⇒ X/Y = 20/15

Thus the ratio of prices of two articles is = 20 : 15 = 4 : 3
Any two amounts in the ratio 4 : 3 will satisfy the condition.
In the above instance, Tk. 80 and Tk. 60 is the answer.

Question: Paint Pro makes pink paint by mixing red paint and white paint in the ratio 3 : 4. Colour Co makes pink paint by mixing red paint and white paint in the ratio 5 : 7. Which company uses a higher proportion of red paint in their mixture?

Solution:
The proportion of red paint for Paint Pro is 3/7
The proportion of red paint for Colour Co is 5/12

We can compare fractions by putting them over a common denominator using equivalent fractions
3/7 = 36/84
5/12 = 35/84

3​/7 is a bigger fraction so Paint Pro uses a higher proportion of red paint.

Question: A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Tk. 175, how much must C pay as his share of rent?

Solution:
A : B : C = (10 × 7) : (12 × 5) : (15 × 3)
= 70 : 60 : 45
= 14 : 12 : 9

∴ C's rent = 175 × (9/35)
= Tk. 45

Question: A number is divided into two parts in such a way that 60% of 1st part is 2 more than the 40% of 2nd part and 60% of the 2nd part is 1 more than the 40% of 1st part. Find the number.

Solution:
Let,
Two part of the number be a, b respectively
∴ The number = a + b

ATQ,
60% of a  -  40% of b = 2 
⇒ (60/100)a - (40/100)b = 2 
∴ 6a - 4b = 20…………(1)

and
60% of b - 40% of a = 1 
⇒ (60/100)b - (40/100)a = 1 
∴ 6b -  4a = 10………….(2)

From (1) × 2 + (2) × 3 we get,
12a - 8b + 18b - 12a = 40 + 30
⇒ 10b  = 70
∴ b = 7 

Put b in 1, we get
6a - 28 = 20
⇒ 6a = 48
∴ a = 8 

Hence, the number is (7 + 8) = 15 

Question: Radha’s salary is 50% more than Seeta’s salary. Radha got a raise of 40% on her salary while Seeta got a raise of 30% on her salary. By what percent is Radha’s salary more than Seeta’s?

Solution:
Let Seeta’s salary = Tk. 100 
Then Radha’s salary = 100 × 50% + 100 = 150

Radha got a raise of 40% then 150 × 40% + 150 = 210 
Seeta got a raise of 30% then 100 × 30% + 100 = 130 

Now Radha salary more than Seeta salary by 210 - 130 = 80 
Percent = (80/130) × 100 = 61.53%

Question: A man bought cookies at 3 for Tk. 1. How many for Tk. 1 should he sell to make a profit a 50%.

Solution:
CP of 3 cookies = Tk. 1
SP of 3 cookies = 150% of Tk. 1 = 3/2

For Tk. 3/2, the man sells 3 cookies.
Hence for Tk. 1, number of cookies sold = (3 × 2)/3 = 2

Question: A man bought 20 shares of Tk. 50 at 5tk discount, the rate of dividend being 13.5 . The rate of interest obtained is-

Solution:
Investment = Tk. [20 × (50 - 5)] = Tk. 900.
Face value = Tk. (50 × 20) = Tk. 1000.

 Dividend = (13.5/100) × 1000
= Tk. 135

Interest obtained = (135/900) × 100%
= 15%

Question: The students in Parimal’s class walk, cycle or drive to school in the ratio 2 : 1 : 4. If 8 students walk, how many students are there in Parimal’s class altogether?

Solution:
Let,
Number of students walk = 2x
Number of student having cycle = x
Number of student driving = 4x
∴ Total number of students = 2x + x + 4x = 7x

ATQ,
2x = 8
∴ x = 4

∴ Number of students = 7 × 4 = 28

Question: A man buys Tk. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is-

Solution:
Dividend on Tk. 20 = Tk. (9/100) × 20 = Tk. 9/5

Tk. 12 is an income on Tk. 100.
∴ Tk. 9/5 is an income on = Tk. (100 × 9)/(12 × 5)
= Tk.15

Question: A number is mistakenly divided by 2 instead of being multiplied by 2. Find the percentage change in the result. 

Solution: 
Let the number be N. 
2N is the correct outcome. 
0.5N is the mistakenly calculated outcome. 

Change in the number= 2N - 0.5N = 1.5N
 
Percentage change= (1.5N/2N) × 100 = 75%

Question: Ashik buys a calculator for Tk. 600 and sells it to Neaz at 10% profit. Neaz sells it to Mahfuj for 5 % profit. Mahfuz after using it for certain time, sells it to Faysal at a loss of 20%. For how much Mahfuz sell the calculator to Faysal.

Solution:
SP for Mahfuz = 600 × (110/100) × (105/100) × (80/100)
= 600 × (924/1000)
= Tk. 554.40

Question: A trader purchases an item for Tk. 540 and sets it marked price at 20% above the cost price. He then sells it at a discount of 10% on the marked price. What is his profit percentage?

Solution:
Let the cost price be Tk. 100
Therefore the marked price is Tk. 120 (20% above CP)
Discount is Tk. 12 (10% of marked price)
Selling price is = (Tk. 120 - Tk. 12) = Tk. 108 

Therefore, profit percentage is 8%.

The given cost price of Tk. 540 is unnecessary and is only for creating confusion.

Question: Rahim is saving for a new bike which will cost Tk. 48000. Rahim earns Tk. 150000 per month. Rahim spends his money on bills, food and extras in the ratio 8 : 3 : 4. Of the money he spends on extras, he spends 80% and puts 20% into his savings account. How long will it take Rahim to save for his new bike?

Solution:
Rahim’s earnings of Tk. 150000 are divided in the ratio of 8 : 3 : 4.

The total number of shares is 8 + 3 + 4 = 15.

Each share is worth Tk. 150000 ÷ 15 = 10000

Rahim spends 4 shares on extras so 4 × 10000 = 40000

20% of 40000 = 8000

The number of months it will take Rahim is 48000 ÷ 8000 = 6

Question: The marked price of a ceiling fan is Tk. 1250 and the shopkeeper allows a discount of 6% on it. Find the selling price of the fan.

Solution:
Marked price = Tk. 1250 and discount = 6%. 

Discount = 6% of Marked Price 
= (6% of Tk. 1250) 
= Tk. {1250 × (6/100)} 
= Tk. 75 

Selling price = (Marked Price) - (discount) 
= Tk. (1250 - 75) 
= Tk. 1175. 

∴ Hence, the selling price of the fan is Tk. 1175. 

Question: How many kgs of Basmati rice costing Tk. 42 per kg should a shopkeeper mix with 25 kgs of ordinary rice costing Tk. 24 per kg so that he makes a profit of 25% on selling the mixture at Tk. 40 per kg?

Solution:
Let the amount of Basmati rice being mixed be x kgs. 
As the trader makes 25% profit by selling the mixture at Tk. 40 per kg
∴ His cost per kg of the mixture = (100 × 40)/125 =  Tk. 32 per kg

∴ (x × 42) + (25 × 24) = 32(x + 25) 
⇒ 42x + 600 = 32x + 800 
⇒ 10x = 200 
∴ x = 20

Question: The percentage of metal extracted from a mine of lead ore is 50%. Now the percentage of gold is 4/5% of the metal and the rest is lead. If the mass of ore extracted from the mine is 6000kg, the mass of the lead is- 

Solution:
Given mass of lead ore = 6000 kg 
Mass of metal is 50% of 6000 = 3000 kg 
Mass of gold is 4/5% of 3000 = (4 × 3000)/(5 × 100) = 24 kg 

Mass of lead = Mass of metal - Mass of gold
⇒ Mass of lead = 3000 - 24 = 2976 kg