পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: 1.14 expressed as a percent of 1.9 is-

Solution:
Required Percentage = (1.14 ×100)/1.9
= (114 × 100 × 10)/(19 × 100)
= 60%

Question: The cost price of 40 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is-

Solution:
Let,
Cost price of each article be Tk. 1
Cost price of x articles = Tk. x
Selling price of x articles = Tk. 40

∴ Profit = Tk. (40 - x)

ATQ,
(40 - x)/x = 25%
Or, 100(40 - x)/x = 25
Or, 4000 - 100x = 25x
Or, 125x = 4000
∴ x = 32

Question: Find the compound interest after 2 years on Tk. 14000 at the rate of interest 5% per annum.

Solution:
Let the sum P = 14000.
Rate of interest = 5%
Period = 2 years

Hence, compound interest = 14000(1 + 5/100)² - 14000
= 14000(1.05)² - 14000
= 14000 × 1.05 × 1.05 - 14000
= 15435 - 14000
= 1435

Question: A bag contains 50p, 25p, and 10p coins in the ratio 2 : 5 : 3, amounting to Tk. 510. Find the number of 50p coins.

Solution:
Let the common ratio be 100k.
Number of 50p coins = 200k
Number of 25p coins = 500k
Number of 10p coins = 300k

Value of 50p coins = 0.5 × 200k = 100k
Value of 25p coins = 0.25 × 500k = 125k
Value of 10p coins = 0.1 × 300k = 30k

∴ Total value of all coins = 100k + 125k + 30k = 255k = 510 (given)
∴ k = 2

Therefore, Number of 50p coins = 200k = 200 × 2 = 400

Question: Two friends A and B started a business with an initial capital contribution of Tk. 1 lac and Tk. 2 lacs. At the end of the year, the business made a profit of Tk. 30,000. Find the share of A in the profit.

Solution:
We know that if the time period of investment is the same, profit/loss is divided by the ratio of the value of the investment.
Ratio of value of investment of A and B = 100000 : 200000 = 1 : 2
Ratio of share in profit = 1 : 2

Share of A in profit = (1/3) × 30,000 = Tk. 10000

Question: In what ratio, water must be mixed with fruit juice costing Tk. 24 per litre so that the juice would be worth of Tk. 20 per litre?

Solution:
Cost of 1 litre of water = Tk. 0 = cheaper quantity. 
Cost of 1 litre of juice = Tk. 24 = dearer quantity. 
And, the mean price = m = Tk. 20

Therefore, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c) = 4 : 20 = 1 : 5 
Hence, the required answer is 1 : 5. 

Question: In an examination 80% candidates passed in English and 85% candidates passed in Mathematics. If 73% candidates passed in both these subjects, then what per cent of candidates failed in both the subjects?

Solution:
Students passed in English = 80%
Students passed in Math's = 85%
Students passed in both subjects = 73%
Then, number of students passed in at least one subject
= (80 + 85) - 73
= 92%. 

Thus, students failed in both subjects = 100 - 92 = 8%.

Question: In a certain store, the profit is 80% of the cost. If the cost increases by 20% but the selling price remains constant, how much is the decrease in profit percentage?

Solution:
Let us assume CP = Tk. 100.
Then Profit = Tk. 80 and selling price = Tk. 180

The cost increases by 20% 
∴  New CP = Tk. 120,
SP = Tk. 180.

∴ Profit = 180 - 120 = Tk. 60
Profit % = (60/120) × 100% = 50%.

Therefore, Profit decreases by (80 - 50)% =  30%.

Question: Namita borrowed Tk. 50000 for 3 years at the rate of 3.5% per annum. Find the interest accumulated at the end of 3 years.​

Solution:
P = 50000
n = 3
r = 3.5%

∴ I = Pnr 
= (50000 × 3 × 3.5)/100
= 5250

Question: A mixture contains sugar solution and colored water in the ratio of 4 : 3. If 10 liters of colored water is added to the mixture, the ratio becomes 4: 5. Find the initial quantity of sugar solution in the given mixture.

Solution:
The initial ratio is 4 : 3.
Let ‘k’ be the common ratio.
Initial quantity of sugar solution = 4k liters
Initial quantity of colored water = 3k liters

Final quantity of sugar solution = 4k liters
Final quantity of colored water = 3k + 10 liters

Final ratio = 4k : (3k + 10) = 4 : 5
⇒ 20k = 12k + 40
⇒ 8k = 40
∴ k = 5

Therefore, the initial quantity of sugar solution in the given mixture = 4k = 4 × 5 = 20 liters

Question: On a 20% discount sale, an article costs Tk. 596. What was the original price of the article?

Solution:
If the selling price of the article is S, then
S - 20% of S = 596
⇒ S - S/5 = 596
⇒ 4S/5 = 596
⇒ S = (596 × 5)/4
∴ S = 745

Question: The angles in a triangle are in the ratio 3 : 4 : 5. Work out the size of each angle.

Solution:
The angles in a triangle add up to 180°.
Therefore 180° is the whole and we need to divide 180° in the ratio 3 : 4 : 5.

The total number of shares is 3 + 4 + 5 = 12
Each share is worth 180 ÷ 12 = 15°

3 shares is 3 × 15 = 45°
4 shares is 4 × 15 = 60°
5 shares is 5 × 15 = 75°

Question: If the price of the commodity is increased by 50% by what fraction must its consumption be reduced so as to keep the same expenditure on its consumption?

Solution:
Let, x kg of the commodity costs 100 taka 

Increased by 50%,
∴  x kg costs = 100 + 100 × .5 = 150 taka

In 100 taka, new quantity of commodity = (100 × x)/150 = 2x/3 kg 

Fraction must its consumption be reduced = {x - (2x/3)}/x
= (x/3)/x
= 1/3

Question: If the cost price of 18 pens is equal to the selling price of 12 pens, the gain percent is?

Solution:
Let the cost price of 1 pen is Tk. 1
Cost of 12 pens = Tk. 12

Selling price of 12 pens = Tk. 18
Gain = 18 - 12 = 6

Gain% = (gain/cost × 100)%
= (6/12 × 100)%
= 50%

Question: Parimal has two grandchildren, Jasmine, aged 2, and Holly, aged 4. Parimal divides Tk. 30 between them in the ratio of their ages. How much does Jasmine get?

Solution:
Tk. 30 is the whole amount.
Parimal divides Tk. 30 in the ratio 2 : 4.

The total number of shares is 2 + 4 = 6.
Each share is worth Tk. 30 ÷ 6 = Tk. 5.
∴ Jasmine gets 2 shares, 2 × 5 = Tk. 10

Question: Three friends A, B, and C started a business, each investing Tk. 10000. After 5 months A withdrew Tk. 3000, B withdrew Tk. 2000 and C invested Tk. 3000 more. At the end of the year, a total profit of Tk. 34600 was recorded. Find the share of C.

Solution:
We know that if the period of investment is not uniform, the gains/losses from the business are divided in the ratio of their inputs, where input is calculated as the product of an amount of investment and the time period of investment.
So, input = value of investment × period of investment, and here, the period of investment would be broken into parts as the investment is not uniform throughout the time period.
A’s input = (10000 × 5) + (7000 × 7) = 99000
B’s input = (10000 × 5) + (8000 × 7) = 106000
C’s input = (10000 × 5) + (13000 × 7) = 141000
∴ A : B : C = 99000 : 106000 : 141000
⇒ A : B : C = 99 : 106 : 141

C’s share = (141/346) × 34600 = Tk. 14100

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 
⇒ (24 - 3x)/8 + x = (40 - 5x)/8
⇒ 24 - 3x + 8x = 40 - 5x
⇒ 5x + 24 = 40 - 5x 
⇒ 10x = 16 
∴ x = 8/5. 

So, part of the mixture replaced = (8/5) × (1/8) = 1/5. 

Solution: Each side of a rectangular field diminished by 40%. By how much per cent is the area of the field diminished?

Solution:
Let,
The Original length of the rectangle be 20 unit and breadth be 10 unit.
Then Original Area = length × breadth = 20 × 10 = 200 Square unit.
40% decrease in each side, then
Length = (20 - 40% of 20) = 12 unit.
Breadth = (10 - 40% of 10) = 6 unit.
Now,
Area = 12 × 6 = 72 Square unit.
Decrease in area = 200 - 72 = 128 square unit.
% Decrease in Area = (128/200) × 100 = 64%

Question: A milkman purchases the milk at Tk. x per litre and sells it at Tk. 2x per litre still he mixes 2 litres water with every 6 litres of pure milk. What is the profit percentage?

Solution: 
the buying price of 6 litres of milk is = 6x Tk.

after mixing 2 litres of water the total amount of mixture is = 8 litres

selling 2x per litres the total selling price is = 16x
∴ profit = 16x - 6x = 10x

profit in percentage = (10x/6x)100%
= 166.66%

Question: A machine is purchased for Tk. 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?

Solution:
To find the depreciated value of the machine after 2 years, we can use the formula for depreciation:
A = P(1 - r/100)^t
P is the initial value of the machine = 625,000 Tk,
r is the rate of depreciation per annum = 8%,
t is the time in years = 2.

∴ A = 625000(1 - 8/100)²
= 625000 × 0.92 × 0.92
= 529000

Question: The ratio of men : women working in a company is 3 : 5. What proportion of the employees are women?

Solution:
In this company, the ratio of men : women is 3 : 5
so for every 3 men there are 5 women.
This means that for every 8 employees, 5 of them are women.
Therefore 5/8 of the employees are women.

Question: A tradesman marks his goods 10% above his cost price. If he allows his customers 10% discount on the marked price, how much profit or loss does he make, if any?

Solution:
Let the cost price be Tk. 100.
Marked price = Tk. 110 (10% above CP)
Discount = 10% on the marked price = 10% of 110 = Tk. 11
∴ Selling price = 110 - 11 = Tk. 99
∴ Loss = CP - SP = 100 - 99 = Tk. 1 
∴ % of loss = 1%

Question: A container contains 40 liters of milk. From this container 4 liters of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Solution:
Milk contained by the container now 
= 40(1 - 4/40)³ 
= 40(1 - 1/10)³  
= 40 × 0.9 × 0.9 × 0.9 
= 29.16 

Question: 220% of a number X is 44. What is 44% of X.

Solution:
Given,
220% of X = 44
⇒ (220X)/100 = 44
⇒ X = (44 × 100)/220
Or, X = 20

Thus,
44% of 20
= (44 × 20)/100
= 8.8

Question: An article is sold for Tk. 3600 at a profit of 25%. What would have been the actual profit or loss if it had been sold at Tk. 2700?

Solution: 
The cost price = (3600 × 100)/125 = 2880
New selling price = Tk. 2700
∴  Loss = 2880 - 2700 = 180
∴ Loss percentage = (100 × 180)/2880 = 6.25%.

Question: If Tk. 3000 is loaned for 4 months at a 4.5% annual rate, how much interest is earned?

Solution:
P = 3000
n = 4 months = 4/12 years = 1/3 year
r = 4.5%

I = Pnr
= {3000 × (1/3) × (4.5)}/100
= 45

Question: The ratio of cups of flour : cups of water in a pizza dough recipe is 9 : 4. A pizza restaurant makes a large quantity of dough, using 36 cups of flour. How much water should they use?

Solution:
Let,
cups of flour 9x
cups of water 4x

∴ 9x = 36
⇒ x = 4

∴ cups of water 4x = 4 × 4 = 16 cups

Question: A invested Tk. 70000 in a business. After a few months, B joined him with Tk. 60000. At the end of the year, the total profit was divided between them in the ratio of 2 : 1. After how many months did B join?

Solution:
Let A work alone for ‘n’ months.
A’s input = 70000 × 12
B’s input = 60000 × (12 - n)

So, (70000 × 12)/[60000 × (12 - n] = 2 / 1
⇒(7 × 12) / [6 × (12 - n)] = 2 / 1
⇒ 12 - n = 7
∴ n = 5
Therefore, B joined after 5 months.

Question: One type of liquid contains 25% of benzene, the other contains 30% of benzene. A can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of benzene in the new mixture.

Solution:
Let the percentage of benzene in new mixure = X
(30 - X)/(X - 25) = 6/4 = 3/2 
⇒ 60 - 2X = 3x - 75
⇒ 5X = 135
∴ X = 27

required percentage of benzene = 27%