Bank Math Master

Question: The price of cooking oil is reduced by 10%. A family buys 4 litres more for Tk. 720 after the reduction. What was the original price per litre?

Solution:
Let
Original price of cooking oil = x Tk/litre.
Original quantity = 720/x​ litre

New price = 0.90x Tk/litre
New quantity = 720/0.90x = 720/(9x/10) = (720 × 10)/9x = 800/x​ litre

ATQ,
(800/x​) - (720/x) = 4
⇒ (800 - 720)/x = 4
⇒ 80/x = 4
⇒ x = 80/4
∴ x = 20

 ∴ Original price of cooking oil = 20 Tk/litre.

Question: Find the simple interest on BDT 12000 at 8% per annum for 6 months.

Solution:
Principal, P = 12000 Taka
Time, n = 6 months = 6/12 = 1/2 years
Rate of interest, r = 8% = 8/100

Simple Interest, I = P × n × r
= 12000 × (1/2) × (8/100)
= (12000 × 1 × 8)/(2 × 100)
= 96000/200
= 480

∴ The simple interest is Tk. 480.

Question: A 42-liter mixture contains milk and water in a 3 : 4 ratio. How much milk should be added to the mixture to make the ratio equal?

Solution:
The ratio of milk to water is 3 : 4
Total portion = 3 + 4 = 7

Quantity of milk = 42 × (3/7) = 18 liters.
Quantity of water = 42 × (4/7) = 24 liters.

Let,
Quantity of milk to be added = x liters

According to the question,
(18 + x) : 24 = 1 : 1
⇒ (18 + x)/24 = 1/1
⇒ 18 + x = 24
⇒ x = 24 - 18
⇒ x = 6

∴ Quantity of milk to be added = 6 liters

Question: Aysha and Jara started a business by investing Tk. 85000 and Tk. 15000 each. What be will the ratio of profit earned after 2 years between Aysha and Jara respectively?

Solution:
Given,
Aysha's investment = Tk. 85,000
Jara's investment = Tk. 15,000
Time = 2 years for both

The ratio of profit earned after 2 years between Aysha and Jara respectively = (85000 × 2) : (15000 × 2)
= 170000 : 30000
= 17 : 3

Question: A train crosses a platform 300 meters long in 50 seconds and another platform 150 meters long in 35 seconds. What is the length of the train?

Solution:
Let the length of the train = x meters

Then,
For the first platform, the distance covered by the train = (x + 300) meters
And,
For the second platform, the distance covered by the train = (x + 150) meters

According to the question,
(x + 300)/50 = (x + 150)/35
⇒ 50(x + 150) = 35(x + 300)
⇒ 50x + 7500 = 35x + 10500
⇒ 50x - 35x = 10500 - 7500
⇒ 15x = 3000
⇒ x = 3000/15 
⇒ x = 200

∴ The length of the train is 200 meters

Question: A shopkeeper marks his goods 40% above the cost price and offers a discount of 20% on the marked price. What is his gain percent?

Solution:
Let,
the cost price (CP) be Tk. 100

Marked Price = 40% more than cost price
= 100 + 40
= 140 Tk.

Discount = 20% of 140
= (20/100) ​× 140
= 28 Tk.

Selling Price (SP)= 140 - 28 = 112 Tk.

∴ Profit = SP - CP = 112 - 100 = 12 Tk.

∴ Gain percent = 12%

Question: What is the compound amount of Tk. 4000 for 2 years at a rate of interest 5% per annum?

Solution:
Given,
Principal, P = 4000
Rate, r = 5% = 5/100 = 1/20
Time, n = 2 years

We know,
A = P(1 + r)ⁿ
= 4000 × (1 + 1/20)²
= 4000 × (21/20)²
= (4000 × 21 × 21)/(20 × 20)
= (4000 × 441)/400
= 4410

∴ The compound amount is Tk. 4410.

Question: The speed of a boat in still water is 9 km/h. The time it takes to travel downstream is half the time it takes to travel upstream. What is the speed of the stream?

Solution:
Let the speed of the current be = x km/h

Then,
Downstream speed = (9 + x) km/h
Upstream speed = (9 - x) km/h

According to the question:
(9 + x) = 2(9 - x)
⇒ 9 + x = 18 - 2x
⇒ 2x + x = 18 - 9
⇒ 3x = 9
⇒ x = 9/3
⇒ x = 3

∴ Speed of the stream = 3 km/h

Question: 60% of a number is 30 less than three-fourth of the same number. Find the number.

Solution:
Let,
the number = x.

ATQ,
60% of x = (3/4) of x - 30
⇒ 60x/100 = (3x/4) - 30
⇒ 3x/5 = (3x/4) - 30
⇒ (3x/4) - (3x/5) = 30
⇒ (15x - 12x)/20 = 30
⇒ 3x/20 = 30
∴ x = 200

So, the number is 200.

Question: A certain principal amount, invested at simple interest, grows to Tk. 950 after 3 years and Tk. 1010 after 4 years. What is the original principal amount?

Solution:
Given,
Amount after 3 years = Tk. 950
Amount after 4 years = Tk. 1010
∴ Interest for 1 year = 1010 - 950 = Tk. 60

∴ Interest for 3 years = 60 × 3 = Tk. 180

∴ Principal = 950 - 180 = Tk. 770

Question: A biker was riding at 60 km/h, but if he had gone at 80 km/h, he could’ve gone 100 km more in the same time. How far did he actually travel?

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

Then,
a/60 = (a + 100)/80
⇒ a/6 = (a + 100)/8
⇒ 6(a + 100) = 8a
⇒ 6a + 600 = 8a
⇒ 8a - 6a = 600
⇒ 2a = 600
⇒ a = 600/2
⇒ a = 300 km

So the actual distance travelled = 300 km

Question: Find the rate of discount being given on a shirt whose selling price is Tk. 420 after deducting a discount of Tk. 80 on its marked price.

Solution:
দেওয়া আছে,
বিক্রয়মূল্য = 420 টাকা
কমিশন = 80 টাকা

∴ গায়ে লিখা দাম = (420 + 80) টাকা 
= 500 টাকা

500 টাকায় কমিশন দেয় 80 টাকা
∴ 1 টাকায় কমিশন দেয় 80/500 টাকা
∴ 100 টাকায় কমিশন দেয় (80 × 100)/500 টাকা
= 16 টাকা

Question: The ratio of the present ages of two friends is 3 : 5. After 7 years, the ratio becomes 2 : 3. What will be the ratio of their ages after 10 years?

Solution:
Let
The present age of the friend₁ = 3x
The present age of the friend₂ = 5x

After 7 years, their ages will be,
⇒ (3x + 7)/(5x + 7) = 2/3
 ⇒ 10x + 14 = 9x + 21
 ⇒ 10x - 9x = 21 - 14
∴ x = 7

Present age of friend₁ = 3x = 21 years
Present age of friend₂ = 5x = 35 years

Now, after 10 years,
friend₁ age = 21 + 10 = 31 years
friend₂ age = 35 + 10 = 45 years

∴ The ratio of their ages after 10 years,
= 31 : 45

Question: How much water should be added to 80 liters of pure milk to gain extra 20% profit when selling the mixture at the price of pure milk?

Solution:
Let’s assume,
Price of pure milk per liter = 100 Taka
So, the price of 50 liters of pure milk = 100 × 80 = 8000 Taka

Now assume,
Water added to the milk = x liters
Then the total quantity of the milk-water mixture = (80 + x) liters

Since the mixture is sold at the price of pure milk,
The selling price of (80 + x) liters = 100(80 + x) Taka

According to the question,
100(80 + x) = 8000 + 8000 of 20%
⇒ 8000 + 100x = 8000 + {8000 × (20/100)}
⇒ 8000 + 100x = 8000 + 1600
⇒ 8000 - 8000 + 100x = 1600
⇒ 100x = 1600
⇒ x = 1600/100
⇒ x = 16

∴ Amount of water to be added = 16 liters

Question: A fruit seller sells 20 oranges for Tk. 900 and suffers a loss equal to the cost price of 5 oranges. Find the cost price of one orange.

Solution:
Let,
cost price of 1 orange is = Tk. x
∴ cost price of 20 oranges is = Tk. 20x
∴ cost price of 5 oranges is = Tk. 5x

We know,
Cost price - Selling price = Loss
20x - 900 = 5x
⇒ 20x - 5x = 900
⇒ 15x = 900
⇒ x = 900/15
∴ x = 60

∴ Cost price of 1 orange is Tk. 60

Question: Two numbers are in the ratio 2 : 3. If 4 is subtracted from the first number, the ratio becomes 1 : 2. What are the numbers?

Solution:
Let the two numbers be: 2x and 3x

According to the question,
(2x - 4)/3x = 1/2
⇒ 2(2x - 4) = 3x
⇒ 4x - 8 = 3x
⇒ x = 8

∴ First number = 2 × 8 = 16
∴ Second number = 3 × 8 = 24

Question: A fighter jet covers a certain distance at a speed of 1200 km/h in 5 hours. What speed must it maintain to cover the same distance in 250 minutes?

Solution:
Total distance = Speed × Time
= (1200 × 5) km
= 6000 km

Given time = 250 minutes = (250/60) hours= 25/6 hours

∴ Required speed = Distance/Time
= {6000/(25/6)} km/h
= {6000 × (6/25)} km/h
= (240 × 6) km/h
= 1440 km/h

প্রশ্ন: Sumon sells a mobile phone to Tania at a profit of 20%. Tania then sells that phone to Rafi for Tk. 1800, making a profit of 25%. At what price did Sumon purchase the mobile phone?

সমাধান:
২৫% লাভে,
তানিয়ার বিক্রয়মূল্য ১২৫ টাকা হলে ক্রয়মূল্য ১০০ টাকা
∴ তানিয়ার বিক্রয়মূল্য ১ টাকা হলে ক্রয়মূল্য ১০০/১২৫ টাকা
∴ তানিয়ার বিক্রয়মূল্য ১৮০০ টাকা হলে ক্রয়মূল্য (১০০ × ১৮০০)/১২৫ টাকা
= ১৪৪০ টাকা

এখানে,
তানিয়ার ক্রয়মূল্য = সুমনের বিক্রয়মূল্য

২০% লাভে,
সুমনের বিক্রয়মূল্য ১২০ টাকা হলে ক্রয়মূল্য = ১০০ টাকা
∴ সুমনের বিক্রয়মূল্য ১ টাকা হলে ক্রয়মূল্য = ১০০/১২০ টাকা
∴ সুমনের বিক্রয়মূল্য ১৪৪০ টাকা হলে ক্রয়মূল্য = (১০০ × ১৪৪০)/১২০ = ১২০০ টাকা

 ∴ সুমনের-এর ক্রয়মূল্য = ১২০০ টাকা

Question: What is the difference between simple and compound interest at 10% per annum on a sum of Tk. 6000 at the end of 2 years?

Solution:
Principal (P) = Tk. 6000
Rate (r) = 10% per annum
Time (n) = 2 years

Simple Interest (SI):
SI = (P × R × T)/100
= (6000 × 10 × 2)/100
= 120000/100
= Tk. 1200

Compound Interest (CI):
Amount (A) = P × (1 + r/100)ⁿ
= 6000 × (1 + 10/100)²
= 6000 × (1.1)²
= 6000 × 1.21
= Tk. 7260

∴ CI = A - P = 7260 - 6000
= Tk. 1260

∴ Difference between CI and SI = 1260 - 1200
= Tk. 60