ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: In order to obtain an income of Tk. 600 from 10% stock at Tk. 120, one must make an investment of-

Solution:
To obtain Tk. 10,
investment = Tk. 120
To obtain Tk. 600
investment = Tk. (120/10) × 600
= Tk. 7200

Question: In your wallet, there are Tk 1000, Tk 500, and Tk 200 notes in the ratio 3 : 7 : 5. The total amount of money in the wallet is Tk 22,500. Find the number of each note respectively.

Solution:
Let,
The number of Tk 1000 notes = 3x
The number of Tk 500 notes = 7x
The number of Tk 200 notes = 5x

According to the question,
(1000 × 3x) + (500 × 7x) + (200 × 5x) = 22500
⇒ 3000x + 3500x + 1000x = 22500
⇒ 7500x = 22500
⇒ x = 22500/7500
∴ x = 3

∴ Number of Tk 1000 notes = 3 × 3 = 9
∴ Number of Tk 500 notes = 7 × 3 = 21
∴ Number of Tk 200 notes = 5 × 3 = 15

Question: The ratio of the present ages of Riyad and his father is 4 : 7. The father’s age at the time of Riyad’s birth was 18 years. Find the father’s present age.

Solution:
Present ratio is 7 : 4
Let actual ages are 7x and 4x.

∴ 7x - 4x = 18
⇒ 3x = 18
∴ x = 6

Hence the father’s present age = 7 × 6 = 42 years

Question: Two numbers are 30% and 20% less than a third number respectively. The ratio of first two numbers is-

Solution:
Given that,
Two numbers are 30% and 20% less than a third number respectively.

Let the third number be 100.

Now,
First number = 100 - 30% of 100
= 100 - 30
∴ First number = 70
And
Second number = 100 - 20% of 100
= 100 - 20
∴ Second number = 80

∴ Ratio of the first two numbers = First number/Second number
= 70/80
= 7/8

∴ The ratio of the first two numbers is 7 : 8.

Question: P, Q, and R opened a shop together. P invests four times as much as Q, and Q invests one-fifth as much as R. After one year, the total profit is Tk 18,000. What is Q's share of the profit?

Solution:
Let R's investment = x taka
Then Q's investment = 1/5 of x = x/5
And P's investment = 4 × (x/5) = 4x/5

So the ratio of investments (or profit) is:
P : Q : R
= 4x/5 : x/5 : x
= 4 : 1 : 5

Total ratio = 4 + 1 + 5 = 10

Total profit = Tk 18,000

∴ Q's share = (1/10) × 18000 = Tk 1800

Question: A vessel contains 108 litres of milk and water in the ratio of 5 : 4. If 20 Liters of milk and 36 liter water is added to the mixture then difference between milk and water in mixture is Y. Find the value of 7Y?

Solution:
Given that,
Total volume of mixture = 108 liters
Ratio of milk to water = 5 : 4
Milk added = 20 liters
Water added = 36 liters

Now,
Total ratio of milk and water = 5 + 4 = 9
Milk in the mixture = (108 liters) × (5/9) = 60 liters
Water in the mixture = (108 liters) × (4/9) = 48 liters

After adding milk and water:
New amount of milk = 60 + 20 = 80 liters
New amount of water = 48 + 36 = 84 liters

∴ Difference between milk and water in the mixture = 84 - 80 = 4 liters
∴ Y = 4
⇒ 7Y = 7 × 4 = 28

∴ The value of 7Y is 28.

Question: How much water be mixed in 36 litre of milk worth Tk. 4.80 per litre, so that value of mixture is Tk. 3.60 per litre?

Solution:
Here,
Milk = Tk. 4.80
Mixture = Tk. 3.60
Water = Tk. 0

ATQ,
Ratio of Milk and water will be = 3.60 : 1.20
= 360 : 120
= 18 : 6

∴ In 18 litres milk water added = 6 litres
∴ In 1 litre milk water added = 6/18 litres
∴ In 36 litres milk water added = {(6/18) × 36} litres
= 12 litres

Question: The perimeter of a rectangle is 72 cm. If the ratio of the lengths of two adjacent sides is 7 : 5, find the lengths of these sides.

Solution:
Perimeter of a rectangle = 2(Length + Breadth)
Also Length :  Breadth = 7 : 5
Let actual values are 7x and 5x.

Hence,
2(7x + 5x) = 72
⇒ 12x = 36
∴ x = 3

Now,
Length = 7x = 7 × 3 = 21 cm
Breadth = 5x = 5 × 3 = 15 cm

∴ sides will be of 21 cm and 15 cm.

Question: A trader mixes 6ltr of milk costing 500 TK. with 7ltr of milk costing 600 TK. per litre. The trader also mixes some quantity of water to the mixture so as to bring the price to 480 TK. per litre. How many litres of water is added?

Solution:
Let us consider, the water be 'W' litre

Here,
(6 × 500 + 7 × 600)/(13 + W) = 480 
⇒ 3000 + 4200 = 6240 + 480W
⇒ 480W = 7200 - 6240
⇒ W = 960/480
∴ W = 2

W is the amount of water added, W = 2 litre

Question: The ratio of two numbers is 2 : 3 and their product is 726. The smallest number between the two numbers is-

Solution:
Given that,
Ratio of two numbers is 2 : 3 and Their product is 726.

Let the numbers be 2n and 3n.

So according to the question :
⇒ 2n × 3n = 726
⇒ 6n² = 726
⇒ n² = 726/6
⇒ n² = 121
⇒ n = √121 [Ignoring the negative value]
∴ n = 11
 
Now,
Smaller number = 2n = 2 × 11 = 22
Larger number = 3n = 3 × 11 = 33

Therefore, Smaller number is 22.

Question: A 10% stock yields 8%. What is the market value of the stock?

Solution:
Earn Tk. 8 when market value Tk. 100
Earn Tk. 1 when market value Tk. 100/8
Earn Tk. 10 when market value Tk. (100 × 10)/8
= Tk. 125

Question: A dairy farmer's can contains 6 litres of milk. His wife adds some water to it such that milk and water are in the ratio 4 ∶ 1. How many litres of milk should the farmer add so that the milk and water are in the ratio 5 ∶ 1?

Solution:
Given that,
Initially, 6 litres of milk (no water yet)
Wife adds water so that milk : water = 4 : 1
After that, farmer adds some milk so that the new ratio is 5 : 1

Now, find the amount of water added by the wife,
Milk : Water = 4 : 1
Milk = 6 litres
Let water = w litres
From the ratio we get,
⇒ 6/w = 4/1
⇒ 4w = 6
⇒ w = 6/4
⇒ w = 1.5
So, after wife adds water, mixture = 6 litres milk + 1.5 litres water.

And,
Let the farmer add x litres of milk
Now, Milk = 6 + x litres
Water = 1.5 litres
After farmer adds milk, new ratio is,
⇒ (6 + x)/1.5 = 5/1
⇒ 6 + x = 7.5
⇒ x = 7.5 - 6
∴ x = 1.5

The farmer should add 1.5 litres of milk.

Question: X, Y, and Z started a business by investing Tk. 150000, Tk. 250000 and Tk. 100,000 respectively. Find the share of Y, out of an annual profit of Tk. 100,000.

Solution:
Ratio of shares of A, B and C = Ratio of their investment
X : Y : Z = 150000 : 250000 : 100000
= 3 : 5 : 2

∴ Total parts = 3 + 5 + 2 = 10

∴ Share of Y = Tk. [(100000 × 5)/ 10] = 50000

Question: There are 174 students in the first three standards. The ratio of number of students in the first and the second standards is 2 : 3, while that of students in standards second and third is 4 : 3. Find the number of students in 2nd standard.

Solution:
Total students = 174.
Ratio of students in 1st and 2nd standards = 2 : 3 = (2 × 4) : (3 × 4) = 8 : 12

Ratio of students in 2nd and 3rd standards = 4 : 3 = (4 × 3) : (3 × 3) = 12 : 9
Hence combined ratio i.e. 1st : 2nd: 3rd is = 8 : 12 : 9.

∴ Number of students in 2nd standard = (174 × 12)/29 = 72

Question: A and B together have Tk 2250. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does A have?

Solution:
ATQ,
4A/15 = 2B/5
⇒ A = (2B × 15)/(5 × 4)
⇒ A = 3B/2
⇒ A/B = 3/2
⇒ A/B = 3/2

∴ A's share = 2250(3/5) = Tk. 1350

Question: Which number when added to each of the numbers 7, 14 and 28 would make the sums to be in continued proportion?

Solution:
Let the number to be added is x.

If three numbers a, b, and c are in continued proportion, then,
a/b = b/c
⇒ b² =  ac ..... (1)

Now,
7 + x, 14 + x, 28 + x

From (1) we get,
∴ (14 + x)² = (7 + x)(28 + x)
⇒ 14² + 2 × 14 × x + x² = 7 × 28 + 7x + 28x + x²
⇒ 196 + 35x + x² = 196 + 28x + x²
⇒ 35x = 28x
⇒ 35x - 28x = 0
⇒ 7x = 0
∴ x = 0

Question: The capital stock of a company is Tk. 300000 and is divided into 3000 shares. If the company declares a total dividend of Tk. 45000, how much will Rina receive for her 40 shares?

Solution:
3000 shares income Tk. 45000
1 share incomes Tk. 45000/3000
40 shares income Tk. (45000 × 40)/3000
= Tk. 600

∴ Rina will receive Tk. 600 as her share of the dividend.

Question: In a 60-liter mixture, the ratio of juice and water is 5:1. How much water must be added to make the ratio 2:1?

সমাধান:
মোট মিশ্রণের পরিমাণ = ৬০ লিটার
জুস : পানি = ৫ : ১
∴ জুসের পরিমাণ = (৫/৬) × ৬০ = ৫০ লিটার
∴ পানির পরিমাণ = (১/৬) × ৬০ = ১০ লিটার

মনে করি,
ক লিটার পানি যোগ করতে হবে।
⇒ নতুন পানির পরিমাণ = ১০ + ক লিটার
জুস অপরিবর্তিত = ৫০ লিটার

প্রশ্নমতে,
৫০/(১০ + ক) = ২/১
⇒ ৫০ = ২ × (১০ + ক)
⇒ ৫০ = ২০ + ২ক
⇒ ২ক = ৩০
⇒ ক = ১৫

∴ ১৫ লিটার পানি যোগ করতে হবে।

Question: Seats for English, History, and Geography in a school are in the ratio 3 : 4 : 6. There is a proposal to increase these seats by 20%, 25%, and 50% respectively. What will be the new ratio of increased seats?

Solution:
Originally, let the number of seats for English, History and Geography 3x, 4x, 6x respectively.

Number of increased seats are,
⇒ (120% of 3x), (125% of 4x) and (150% of 6x)
⇒ (120/100) × 3x, (125/100) × 4x and (150/100)  × 6x
⇒ 18x/5, 5x, and 9x

∴ New ratio of, English : History : Geography = 18x/5 : 5x : 9x
= 18x : 25x : 45x
= 18 : 25 : 45

Question: If 4(Nahid's Capital ) = 6(Safin's Capital ) = 10(Robin's Capital ), then out of the total profit of Tk. 4650 , Robin will receive-

Solution:
Let
Nahid's capital = a
Safin's capital = b
Robin's capital = c

ATQ,
4a = 6b = 10c
⇒ 2a = 3b = 5c
∴ b = 2a/3

∴ c = 2a/5

Nahid : Safin : Robin = a : 2a/3 : 2a/5
= 15 : 10 : 6 [Multiply by 15]

Robin's share = 4650 × (6/31)
= 150 × 6
= Tk. 900

Question: A man purchased 400 shares of the face value of Tk. 100 each from the market at Tk. 125 per share. If a dividend of 20% is declared, find his earning percent on the investment.

Solution:
Given,
Price of 1 share = Tk. 125.
∴ Price of 400 share = Tk. (125 × 400)
= Tk. 50000

Dividend per share=20% of 100 = Tk. 20
∴ Total dividend income = 400 × 20 = Tk. 8000

∴ Earning % = (Dividend/Market Price​) × 100
=(8000/50000) × 100
= 16%.

Question: If the annual income from a 5% stock at 90 is Tk. 40 more than the income from a 6% stock at 120, then what is the total investment?

Solution:
Let,
The investment = Tk. x

5% stock at 90,
income from Tk. 90 = Tk. 5
∴ income from Tk. x = Tk. 5x/90
= Tk. x/18 

6% stock at 120,
income from Tk. 120 = Tk. 6
∴ income from Tk. x = Tk. 6x/120
= Tk. x/20

ATQ,
x/18 - x/20 = 40
⇒ (10x - 9x)/180 = 40
⇒ x = 40 × 180
∴ x = 7200

Question: A mixture of 150 liters of milk and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

Solution:
Amount of water in the 150 liters mixture = 20% of 150
= 1/5 of 150
= 30 liters

Let,
P liters of water be added.
So, new amount of water = (30 + P)
and new total mixture = (150 + P)

ATQ,
(30 + P) = 25% of (150 + P)
⇒ 30 + P = (25/100) × (150 + P)
⇒ 30 + P = (1/4) × (150 + P)
⇒ 120 + 4P = 150 + P
⇒ 4P - P = 150 - 120
⇒ 3P = 30
∴ P = 10

∴ 10 liters more water should be added.