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Comparing Quantities - Class 8 : Notes

Ratio:
It is the quantitative relation between two quantities which reflects the relative size of both the quantities. It is simply the means to compare the quantities.

Example 1: Suppose in a classroom there are 20 boys and 40 girls. Find the ratio of boys to girls.
SolutionHere, the ratio of number of boys to the number of girls will be 20/40.
By using fraction, we can write 20/40 = 1/2.
Thus, the numbers of boys are half the number of girls in the classroom.
In terms of ratio, it is written as 1:2 and read as “1 is to 2”

Example 2: Find the ratio of 5 m to 10 km.
SolutionWe know that, 1 km = 1000 m
So, required ratio = 5 m / 10 km = 5m / (10 x 1000 m) = 1/2000.
Thus, ratio of 5 m to 10 km is 1:2000.

Percentage:
It is another means to compare the quantities.

Example 1: Suppose in a classroom there are 20 boys and 40 girls. Find the percentage of boys and girls.
SolutionWe know that, 20 students out of total 60 students are boys.
So, we can write 20/60 x 100 = 33.33 % boys are there in classroom.
And the percentage of girls will be 40/60 x 100 = 66.67 %
Thus, the classroom has 33.33 % of boys and 66.67% of girls in classroom.

Example 2: Convert 3:4 ratio into percentage.
Solution3:4 = 3/4 x 100 % = 75%

Example 3: A city has 50% people liking cricket, 40% liking football and the remaining liking other games. If the population of city is 50 lakh, then, find out how many per cent of people like other games. Also, calculate the number of people for each type of game.
SolutionHere, the percentage of people liking other games = (100 – 50 – 40)% = 10%
Given, total number of people = 50 lakh.
Now, number of people who like cricket = (50/100 x 50) = 25 lakh
Number of people who like football = (40/100 x 50) = 20 lakh
And number of people who like other games = (10/100 x 50) = 5 lakh

Finding the Increase or Decrease Percent:
To find the increase or decrease percent, multiply the percentage given by the actual quantity from which we need to increase or decrease. Then, to the actual quantity add or subtract the obtained quantity to get the final result.

Example 1: The price of certain item A was 20, 000 Rs. Now, some features were added to it and so the price was increased by 10%. Then find the new price.
SolutionFirstly, we will multiply the given percentage with the actual quantity. So, in this case we have,
10% x 20,000 Rs = 10/100 x 20,000
= 2,000 Rs
Now, we will add this amount to actual amount as the price has increased.
Therefore, 20,000 + 2,000 = 22,000 Rs will be the new price.

Example 2: The new salary of an employee is Rs 50,000 after he was given 15% increase in his salary. Find out his initial salary.
SolutionLet the intial salary be s.
Now, intial salary + increment = new salary.
Given, the increment is 15% of initial salary.
Thus, s + (15/100 x s) = 50,000
115s/100 = 50,000
s = 43,480 Rs.
Hence, his intitial salary was Rs 43,480.

Example 3: A zoo had  950 visitors on Sunday; while on Monday the number of visitors were just 170. Then calculate the per cent decrease in the people on Monday.
SolutionGiven, on Sunday 950 people went to zoo and on Monday 170 people went to zoo.
So, decrease in the number of people = 950 – 170 = 780.
Hence, %  decrease = (Decrease in no. of people/ no. of people who went to zoo on Sunday x 100)%
= (780/950 x 100) %
= 82 %

Estimating Percentages:
Follow the steps given below to estimate the percentage for finding n percent of x:
(i) Round off values of x and n to numbers whose simplification would be easy.
(ii) Multiply these rounded numbers.
(iii) Divide the multiplied answer by 100.

Example: Suppose price of an item is 565.90 Rs and a discount of 10% is given on this item. Calculate the amount needed to be paid.
SolutionWe will round off the price of item 565.90 Rs to its nearest tens which will be 570 Rs.
Now, we will multiply the two numbers 570 and 10, so we will have 5700 Rs.
At last, we will divide the answer by 100, so we will get 57 Rs.
Thus, the amount that needs to be paid will be 570 - 57 = 513 Rs.

Terms related to buying and selling:

1. Cost Price (CP):
The actual amount that a manufacturer spend to produce a product or to provide the service is known as the cost price.

Example: Suppose, it costs 30,000 Rs to a car manufacturer to produce a car, then this price will be known as its cost price.

2. Selling Price (SP):
The price at which a product is sold in the market is known as its selling price.

Example: Suppose, a car based company sell its car at 33,000 Rs then this price will be known as its selling price.

3. Profit:
The financial gain that is received on selling a product is known as its profit. In other words, it is the difference between the selling price and the cost price i.e.
Profit (P) = Selling Price (SP) – Cost Price (CP)
The profit in the percentage form can be expressed as P% = P/CP x 100.

Example 1: An item was manufactured at cost of 1000 Rs; but was sold at 1500 Rs. Find the profit gained on this item.
SolutionGiven, Cost Price (CP) = 1000 Rs; Selling Price (SP) = 1500 Rs
Therefore, Profit (P) = CP – SP = 1500 – 1000 = 500 Rs.
% of Profit = P/CP x 100 = 500/1500 x 100 = 33.33 %

Example 2: A shopkeeper bought 100 bulbs at the price of Rs 5 each. But, out of them 10 bulbs were not working. And the remaining were sold at the price of Rs 7 each. Find the gain%.
SolutionHere, cost price of 100 bulbs = 100 x 5 = 500 Rs.
Out of 100 bulbs, 10 bulbs were fused. So, number of bulbs left = 100 – 10 = 90.
Each bulb was sold at 7 Rs.
Thus, selling price of 90 bulbs = 90 x 7 = 630 Rs.
Since, SP > CP, Profit = 630 – 500 = 130 Rs.
%Profit = 130/500 x 100 = 26%.

4. Loss:
The financial negative revenue that is received on selling a product is known as its loss. In other words, it is the opposite of the profit.

Example 1: An item was purchased at cost of 1000 Rs; but was sold at 500 Rs. Find the loss of this item.
SolutionGiven, Cost Price (CP) = 1000 Rs; Selling Price (SP) = 500 Rs
Therefore, Loss (L) = CP – SP = 1000 – 500 = 500 Rs.
% of Loss = L/CP x 100 = 500/1000 x 100 = 50 %

Example 2: A fan was purchased for Rs 1500 and was sold at loss of 10%. Find the selling price of fan.
SolutionGiven, Cost Price of fan = 1500 Rs.
Now, it is sold at 10% loss. This means if CP is 100 Rs, SP is 90 Rs.
Hence, Selling Price of fan = 90/100 x 1500 = 1350 Rs.
Thus, there is loss of 1350 Rs on selling fan.

5. Sales Tax (ST):
It is charge we pay to the government when we purchase items. The price of this tax is added to the selling price of an item.
Hence, the amount we pay to shopkeeper is then given to the government. Nowadays, the prices include the tax known as Value Added Tax (VAT). Example 1: A sales tax of 5% is added to an item whose selling price is 500 Rs. Find the bill amount.
SolutionThe tax required to be paid for the item will be = 5/100 x 500 = 25 Rs
Hence, the bill amount will be selling price + tax = 500 Rs + 25 Rs = 525 Rs.

Example 2: A machine was purchased at Rs 6000 including 5% VAT. Find the price before VAT was added.
SolutionGiven, 5% VAT is included in the price of Rs 6000. Hence, 5% VAT means that if price without VAT is Rs 100, then price including VAT will be Rs 105.
So, original price = (100/105 x 6000) = 5714 Rs

6. Compound Interest:
It is the interest calculated on both the basic principal as well as the interest earned till date. Example
: Ajay borrowed Rs 50,000 for 2 years at an interest of 10% compounded annually. Find the Compound Interest and the amount to be paid at the end of two years.
SolutionLet us denote the principal for the first year by P1. Given, P1 = 50,000 Rs.
The simple interest for first year will be = 50,000 x 10/1000 = 500 Rs
Now, the total amount at the end of first year will be = 50,000 Rs + 500 Rs = 50,500 Rs
The simple interest for second year will be = 50,500 x 10/1000 = 505 Rs
Now, the total amount to be paid at the end of second year will be = 50,500 Rs + 505 Rs = 51,005 Rs
Total interest paid = 500 Rs + 505 Rs = 1005 Rs

Formula for computing Compound Interest:
A = P (1 + R/100) n
Where, A is the amount
P is the principal amount
R is the rate of interest
n is the number of years.

Example 1: Calculate the compound interest for an amount of 10,000 Rs taken at 10% for 2 years.
SolutionGiven, P = 10,000 Rs, R = 10, n = 2.
We know that, A = P (1 + R/100) n
Substituting the values, we get, A = 10,000 x (1 + 10/100) 2
A = 10,000 x 1.21 = 12,100 Rs
Thus, CI = A – P = 12,100 Rs – 10,000 Rs = 2100 Rs

Example 2: A sum of Rs 15,000 was borrowed for 2 ½  years at an interest of 10% compounded annually. Find the Compound Interest and the amount to be paid.
SolutionGiven, Principal = 15,000 Rs , Rate = 10% annually, No of years = 2 ½ years
Here, the amount for 2 years will be A = 15,000 x (1 + 10/100)2 = 18,150 Rs.
Now, for next six months, principal amount will be 18,150 Rs. So, SI will be
SI = (15000 x ½ x 10)/100 = 750 Rs.
Hence, Interest for first 2 years = (18150 – 15000) = 3150 Rs.
And interest for next ½ year = 750 Rs.
Therefore, total CI = 3150 + 750 = 3900 Rs.
A = P + CI = 15000 + 3900 = 18,900 Rs.

Rate Compounded Annually or Half Yearly:
When the interest is compounded half yearly rather than yearly, then we say that the rate is compounded annually or half yearly.
If the interest is compounded half yearly, then we need to calculate the interest twice. And in this case, time period will get double and the rate will get half.

Example 1: Find the compound interest on 10,000 Rs for 2 years at 10% per annum when compounded half-yearly.
SolutionHere, P = 10,000 Rs; R = 10% per annum = 5% per half - yearly; n = 2 years = 4 half years.
Amount A = 10,000 x ( 1 + 5/100) 4 = 12,155 Rs.
Thus, CI = 12,155 Rs - 10,000 Rs = 2155 Rs.

Example 2: Find the amount which Ram will get on Rs 5000, if he gave it for 18 months at 12½ %  per annum, interest being compounded half yearly.
SolutionGiven, Principal = 5000 Rs, Rate = 12½ % per annum = 25/4 % per half yearly, n = 18 months
Here, there will be 3 half years in 8 months.
Hence, A = 5000 (1 + (25/4)/100)3 = 6000 Rs. Thus, the required amount is 6,000 Rs.

Applications of Compound Interest Formula:
There are many practical situations where we are in need to calculate CI. Some of the examples are given below:
(i) The rate of population growth. (ii) The rate of growth of bacteria. (iii) The increment or decrement in price of certain item.

• Anonymous

nice

• Anonymous

Nice notes

• Anonymous

Good notes