ব্যাখ্যা
Solution:
The average computer price three years ago was = 700 × 0.8
= 560 taka
Increase = 700 - 560 taka
= 140 taka
%increase = (140/560) × 100%
= 25%
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x = 80%
⇒ x = 80/100 = 4/5
⇒ x2 = 16/25 = 16/25 × 100% = 64%
x is larger than x2 (in percentage) = (80% - 64%) = 16%
x is larger than x2 = (80% - 64%) = 16%
x is larger than x2 (in percentage) = (16/64 ×100) = 25%
According to the question,
At 20% loss, 80 Tk selling price হলে cost = 100 Tk.
∴ 200 Tk. selling price হলে cost = (100 × 200)/80
= 250 Tk.
Question: A cuboidal room has its length, breadth, and height increased by 10%, 20%, and 50% respectively. Calculate the percentage change in the volume of the cuboid.
Solution:
Let,
Each side of the cuboid be 10 unit initially.
Initial Volume of the cuboid,
= length × breadth × height
= 10 × 10 × 10
= 1000 cubic unit.
After increment dimensions become,
Length = (10 + 10% of 10) = 11 unit.
Breadth = (10 + 20% of 10) = 12 unit.
Height = (10 + 50% of 10) = 15 unit.
Now, present volume = 11 × 12 × 15 = 1980 cubic unit.
Increase in volume = 1980 - 1000 = 980 cubic unit.
∴ percentage increase in volume = (980/1000) × 100 = 98%
New annual salary = 90,000
Salary increase = 15,000.
Original salary = 90,000 - 15,000.
= 75,000
% Increase = (15,000/ 75,000 )×100
=20%
Remaining books = 1 - 2/3 = 1/3
1/3 of the books = 41 books
So, 2/3 of the books = 41×(2/3) / (1/3) = 82
∴ Total amount received for the sold books = 82 × 3.20 = 262.4 TK
Question: The ratio of cost price and selling price is 4 : 5. The profit percentage is
Solution:
Given,
The ratio of cost price (C.P.) to selling price (S.P.) is 4:5.
Let,
The cost price be 4x and the selling price be 5x.
∴ Profit = S.P - C.P. = 5x - 4x = x
∴ Profit percent = (Profit/CP) × 100
= (x/4x) × 100
= (1/4) × 100
= 25%
Let the total expenditure = Tk.X
He spends for petrol = 8X/100
For his son's education = 15X/100
For his family = 52X/100
Therefore the remaining amount for savings = Tk. 2500 = X - (8X/100 + 15X/100 + 52X/100)
⇒ X - (8X/100 + 15X/100 + 52X/100) = Tk. 2500
⇒ X - 75X/100 = Tk. 2500
⇒ 25X/100 = Tk. 2500
⇒ 25X = Tk. 2500 × 100
⇒ X = (Tk. 2500 × 100)/25
⇒ X = Tk. 10000
Hence, the total expenditure = Tk. 10000.
Let the products actual price be x
ATQ, 76% of x = 1368
Or, 76x/100 = 1368
Or, x = (1368 × 100) / 76
= 1800
We know, I = pnr
= 750 × 5/100 × 4
= 150
Question: The population of a city grows by 10% every year. If the current population is 50,000, what will the population be after 2 years?
Solution:
Here we can use the compound interest based formula,
Population after n years
= P × [1 + (r/100)]n
∴ Population after 2 years = 50000 × [1 + (10/100)]²
= 50000 × (110/100)²
= 50000 × 1.21
= 60,500
Question: If a shopkeeper sells a bat for Tk. 600, he incurs a loss of 20%. At what price should he sell it to make a profit of 20%?
Solution:
Let the cost price (CP) of the bat be x.
Selling price = x - 20% of x
⇒ 600 = x - (20x/100)
⇒ 600 = 80x/100
⇒ 80x = 60000
⇒ x = 750 (This is the CP)
To gain,
20% = 750 + 20% of 750
= 750 + (20 × 750)/100
= 750 + 150
= Tk. 900
According to the question,
Cost Price : Marked Price
(100 - Discount) : (100 + Profit)
100 - 10 : 100 + 12
90 : 112
45 : 56
Question: A trader sells a bag for Tk. 2,400 at a loss of 20%. If his regular profit is 10%, find the regular selling price.
Solution:
Selling price = 2,400 Taka
Loss % = 20%
Original cost price = 2400 / 0.8 = 3000 Taka
Regular profit = 10%
Regular Selling price = 3000 × 1.1 = 3300 Taka
∴ Regular Selling price = 3300 Taka
Question: What would be the value of 20% of m as a percentage of p, if 8% of m = 4% of p?
Solution:
8% of m = 4% of p
⇒ 8m/100 = 4p/100
⇒ m = p/2
⇒ 20% of m = (p/2) × 20%
⇒ 20% of m = 10% of p
Question: P is 6 times greater than Q then by what percent is Q smaller than P?
Solution:
Let Q = 10.
Then, P = 60.
Q is 50 less than P.
Q, % less than P = (50/60) × 100
= 83.33%
Let,
The C.P (cost price) of 1 kg goods be 1 tk
Then, S.P. of {100 - (25/2)% of 1 kg}
Or, 875 gm goods = .9375 tk
S.P of 1 kg goods = (.9375/875 × 1000) tk
= 1 (1/14) tk
∴ Profit % = (1/14 × 100)% = 7(1/7)%
Answer: 7(1/7)%
Question: 605 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20% of the total number of children. How many sweets did each child receive?
Solution:
Let Children = X
A/Q,
605/X = 20% of X
⇒ 605/X = X/5
⇒, X2 =5 × 605
⇒ X2 = 52 × 112
X = 55
So each children receive = 605/55
= 11
Question: An article has a marked price of Tk. 500. If two successive discounts of x% and 5% reduce the selling price to Tk. 427.50, determine the value of x.
Solution:
Marked Price = Tk. 500
Final Selling Price after two successive discounts = Tk. 427.50
Let the first discount = x%
Second discount = 5%
First discount be:
500 × (1 - x/100) × (1 - 0.05) = 427.50
⇒ 500 × (1 - x/100) × 0.95 = 427.50
⇒ 475 × (1 - x/100) = 427.50
⇒ (1 - x/100) = 427.50 / 475
⇒ 1 - x/100 = 0.9
⇒ - x/100 = 0.9 - 1
⇒ - x/100 = - 0.1
⇒ x = 0.1 × 100
⇒ x = 10%
∴ First discount = 10%
Let s=shirt and t=tie
Need to solve:
3s + 5t = 23 and 5s + 1t = 20
Look for a multiple of one (or both) formula that will match the quantity of the second. One option is to multiply the second equation by 5:
5x (5s + 1t) = 5x 20
Or, 25s + 5t = 100
Subtract the first formula :
25s + 5t - 3s - 5t = 100 - 23
Or, 22s = 77
So shirts are 77 / 22 = Tk. 3.50 each
Let,
He purchases 6 liters milk.
So, cost of 6 liters = 6x tk.
After mixing 2 liters waters, he sells , (6 + 2) = 8 liters
Now, selling price of 8 liters = 8 × 2x = 16x tk
Profit = 16x – 6x = 10x tk.
∴ Profit percentage = (10x/6x) × 100 = 166.66%
Let the cost price x.
Therefore, 90% of 15000 = 108% of x .
Or, x = 90 × 15000108
Or, x = Tk. 12500
Total marks obtained by the student = 55% of 800
= {(55/100) × 800}
= 440
∴ Marks scored in English
= 15% of 440
= {(15/100) × 440}
= 66
• Question: In an examination, 36% are pass marks. If an examine gets 17 marks and fails by 10 marks, what is the maximum mark?
Solution:
Pass mark = (17 + 10) = 27
Let maximum marks be x
Then 36% of x = 27
Or, 36x/100 = 27
Or, 36x = 2700
Hence, x = 75
∴ The maximum mark is 75
Question: Mr. Ali is a trader. He mixes 26 kg of rice at Tk. 20 per kg with 30 kg of rice of other variety at Tk. 36 per kg and sells the mixture at Tk. 32 per kg. His profit percent is-
Solution:
Cost Price of 56 kg rice = {(26 × 20) + (30 × 36)}
= (520 + 1080)
= 1600 taka
Selling Price of 56 kg rice = (56 × 32)
= 1792 taka
∴ Profit = 1792 - 1600
= 192 taka
∴ Profit percentage = (192/1600) × 100%
= 12%
At 25% discount, 75% = 180
So, 100% = (180×100)/75 = 240
At 10% loss, He got = (52000×90) / 100 = 46800 tk
His loss is = 52000 - 46800 = 5200
At 20% profit, He receives = (46800×120)/100 = 56160
Profit, 56160 - 46800 = 9360
∴ His net profit = 9360 - 5200 = 4160
15% discount - এ T - shirt টির বিক্রয়মূল্য 85 টাকা হলে প্রকৃত মূল্য 100 টাকা
∴ বিক্রয়মূল্য 272 টাকা হলে প্রকৃত মূল্য = (100 × 272) / 85 = 320 টাকা
Question: A worker union contract specifies a 6% salary increase plus a Tk. 450 bonus for each worker. For a worker, this is equivalent to an 8% salary increase. What was this worker's salary before the new contract?
Solution:
ধরি, কর্মীর পূর্বের বেতন = x টাকা।
6% বৃদ্ধিতে বেতন = x + x এর 6%
= x + (6x/100) = 106x/100
বোনাস হিসেবে 450 টাকা যোগ করলে মোট বেতন = (106x/100) + 450
8% বৃদ্ধিতে বেতন = x + x এর 8%
= x + (8x/100) = (108x/100)
প্রশ্নমতে,
(106x/100) + 450 = (108x/100)
⇒ 450 = (108x/100) - (106x/100)
⇒ 450 = (2x/100)
⇒ x = (450 × 100)/2
∴ x = 22500
অর্থাৎ, কর্মীর পূর্ববর্তী বেতন ছিল 22500 টাকা।
Now, in this numerical, there is no common loss and gain %.
Hence, solve it by making equations.
Let the cost price of the camel be x.
As the cost of camel and carriage = Tk 5000
Cost of carriage = Tk (5000 – x)
After selling the camel he gains 20% and on carriage a loss of 10%. But on the whole, he gains 3%.
Therefore,
20% of x – 10% of (5000 – x) = 3% of 5000
⇒ (20/100) × x - (10/100) × (5000 – x) = (3/100) × 5000
⇒ x/5 - (5000 – x)/10 = 150
⇒ 10x/5 - {(5000 – x) × 10}/10 = 150 × 10
⇒ 10x/5 - (5000 – x) = 1500
⇒ 2x - 5000 + x = 1500
⇒ 3x=1500 + 5000
⇒ x = 2166.67.
The cost of camel = Tk. 2166.67
Let cost price per item be 1.
cp of 15 items = 15
sp of 15 items = cost price of 20 items= 20
So, profit = 20 - 15 = 5.
∴ required profit % = (5/15) × 100 = 33.3%
Question: If b is equal to 20% of a, then what is b% of 20 equal to?
Solution:
20% of a = b
⇒ 20a/100 = b
∴ b% of 20
= (b/100) × 20
= (20a/100) × (1/100) × 20
= 4a × (1/100)
= 4% of a
Question: Population of a town increase 2.5% annually but is decreased by 0.5% every year due to migration. What will be the percentage increase in 2 years?
Solution:
Net percentage increase in Population = (2.5 - 0.5) = 2% each year.
Assuming initial population = 100
Population of Town after 1st year = (100 + 2% of 100)
= 100 + 2 = 102
Population of Town after 2nd year = (102 + 2% of 102)
= 102 + 2.04 = 104.04
Increase = 104.04 - 100 = 4.04
∴ Percentage increase = (4.04/100) × 100% = 4.04%
Selling price of egg Tk (M/6) × 12 per dozen
= 2M
profit = 2M - M = M
profit percentage = (M/M) × 100
= 100%.