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Comparing Quantities
Recalling Ratios and Percentages
We know, ratio means comparing two quantities.
A basket has two types of fruits, say, 20 apples and 5 oranges. Then, the ratio of the number of oranges to the number of apples = 5 : 20. The comparison can be done by using fractions as, \({5 \over 20}={1 \over 4}\)
The number of oranges is 1 4 th the number of apples. In terms of ratio, this is 1 : 4, read as, “1 is to 4”
(or)
Number of apples to number of oranges = \({20 \over 5}={4 \over 1}\) = which means, the number of apples is 4 times the number of oranges. This comparison can also be done using percentages
There are 5 oranges out of 25 fruits. So percentage of oranges is
\(\frac{5}{25}\times \frac{4}{4}=\frac{20}{100}\)=20%
[Denominator made 100].
By unitary method:
Out of 25 fruits, number of oranges are 5.
So out of 100 fruits, number of oranges
\(\frac{5}{25}\times 100=20\)
Since contains only apples and oranges,
So, percentage of apples + percentage of oranges = 100
or percentage of apples + 20 = 100
or percentage of apples = 100 – 20 = 80
Thus the basket has 20% oranges and 80% apples.
Example 1: A picnic is being planned in a school for Class VII. Girls are 60% of the total number of students and are 18 in number.
The picnic site is 55 km from the school and the transport company is charging at the rate of ₹ 12 per km. The total cost of refreshments will be₹4280.
Can you tell.
1. The ratio of the number of girls to the number of boys in the class?
2. The cost per head if two teachers are also going with the class?
3. If their first stop is at a place 22 km from the school, what per cent of the total distance of 55 km is this? What per cent of the distance is left to be covered?
Solution:
1. To find the ratio of girls to boys. Ashima and John came up with the following answers.
They needed to know the number of boys and also the total number of students.
So, the number of boys = 30 – 18 = 12.
Hence, ratio of the number of girls to the number of boys is 18 : 12 or 18 12 = 3 2 .
3 2 is written as 3 : 2 and read as 3 is to 2.
2. To find the cost per person.
Transportation charge = Distance both ways × Rate
= ₹ (55 × 2) × 12
= ₹110 × 12
= ₹ 1320
Total expenses = Refreshment charge + Transportation charge
= ₹4280 + ₹1320 = ₹ 5600
Total number of persons =18 girls + 12 boys + 2 teachers
= 32 persons
Ashima and John then used unitary method to find the cost per head. For 32 persons, amount spent would be ₹5600.
The amount spent for 1 person = ₹\(\frac{5600}{32}\)= ₹175.
3. The distance of the place where first stop was made = 22 km
To find the percentage of distance:
Ashima used this method:
\(\frac{22}{55}=\frac{22}{55}\times \frac{100}{100}=40%\)
She is multiplying the ratio by \(\frac{100}{100}=1\)
and converting to percentage.
(or)
John used the unitary method:
Out of 55 km, 22 km are travelled.
Out of 1 km,\(\frac{22}{55}\) km are travelled.
Out of 100 km \(\frac{22}{55}\times 100\)km are travelled.
That is 40% of the total distance is travelled.
Both came out with the same answer that the distance from their school of the place where they stopped at was 40% of the total distance they had to travel.
Therefore, the percent distance left to be travelled = 100% – 40% = 60%
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Comparing Quantities
Finding Discounts
Discount is a reduction given on the Marked Price (MP) of the article. This is generally given to attract customers to buy goods or to promote sales of the goods. You can find the discount by subtracting its sale price from its marked price.
So, Discount = Marked price – Sale price
Example : An item marked at ₹840 is sold for ₹ 714. What is the discount and discount %?
Solution: Discount = Marked Price – Sale Price
= ₹840 – ₹714 = ₹126
On marked price of ₹840, the discount is ₹ 126.
On MP of ₹ 100, how much will the discount be?
Discount =\(\frac{126}{840}\times 100\)%=15%
You can also find discount when discount % is given.
Example : The list price of a frock is ₹ 220. A discount of 20% is announced on sales. What is the amount of discount on it and its sale price.
Solution: Marked price is same as the list price. 20% discount means that on ₹100 (MP), the discount is ₹20.
By unitary method, on ₹1 the discount will be ₹\(\frac{20}{100}\)
On ` 220, discount = ₹\(\frac{20}{100}\)×220 = ₹44
The sale price = (₹220 – ₹44) or ₹ 176
Rehana found the sale price like this — A discount of 20% means for a MP of ₹ 100, discount is ₹20. Hence the sale price is ₹80. Using unitary method, when MP is ₹100, sale price is ₹ 80;
When MP is ₹ 1, sale price is ₹\(\frac{80}{100}\).
Hence when MP is ₹220, sale price = \(\frac{80}{100}\)×220 = ₹ 176
Estimation in percentages
Your bill in a shop is ₹ 577.80 and the shopkeeper gives a discount of 15%. How would you estimate the amount to be paid?
(i) Round off the bill to the nearest tens of ₹577.80, i.e., to ₹ 580.
(ii) Find 10% of this, i.e., \(\frac{10}{100}\times 580=\)₹58.
(iii) Take half of this, i.e.,\(\frac{1}{2}\times 58\)= ₹29 .
(iv) Add the amounts in (ii) and (iii) to get ₹ 87. You could therefore reduce your bill amount by ₹ 87 or by about ₹85, which will be ₹495 approximately.
1. Try estimating 20% of the same bill amount.
2. Try finding 15% of ₹ 375.
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Comparing Quantities
Sales Tax/Value Added Tax/Goods and Services Tax
The teacher showed the class a bill in which the following heads were written.
Sales tax (ST) is charged by the government on the sale of an item. It is collected by the shopkeeper from the customer and given to the government. This is, therefore, always on the selling price of an item and is added to the value of the bill. There is another type of tax which is included in the prices known as Value Added Tax (VAT).
From July 1, 2017, Government of India introduced GST which stands for Goods and Services Tax which is levied on supply of goods or services or both
Example : (Finding Sales Tax) The cost of a pair of roller skates at a shop was ₹ 450. The sales tax charged was 5%. Find the bill amount.
Solution: On ₹100, the tax paid was ₹ 5.
On ₹450, the tax paid would be =\(\frac{5}{100}\times 450=\)₹ 22.50
Bill amount = Cost of item + Sales tax = ₹ 450 + ₹ 22.50 = ₹ 472.50.
Example : (Value Added Tax (VAT)) Waheeda bought an air cooler for ₹ 3300 including a tax of 10%. Find the price of the air cooler before VAT was added.
Solution: The price includes the VAT, i.e., the value added tax. Thus, a 10% VAT means if the price without VAT is ₹ 100 then price including VAT is ₹ 110. Now, when price including VAT is ₹ 110, original price is ₹ 100.
Hence when price including tax is ₹3300, the original price =\(\frac{100}{110}\times 3300=\)₹3000.
Example : Salim bought an article for ₹ 784 which included GST of 12% . What is the price of the article before GST was added?
Solution: Let original price of the article be ₹100. GST = 12%.
Price after GST is included = ₹ (100+12) = ₹112
When the selling price is ₹ 112 then original price = ₹ 100.
When the selling price is ₹ 784, then original price = \(\frac{100}{12}\times 784=\) ₹ 700
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Comparing Quantities
Compound Interest
You might have come across statements like “one year interest for FD (fixed deposit) in the bank @ 9% per annum” or ‘Savings account with interest @ 5% per annum’
Interest is the extra money paid by institutions like banks or post offices on money deposited (kept) with them. Interest is also paid by people when they borrow money. We already know how to calculate Simple Interest.
Example : A sum of ₹ 10,000 is borrowed at a rate of interest 15% per annum for 2 years. Find the simple interest on this sum and the amount to be paid at the end of 2 years.
Solution: On ₹ 100, interest charged for 1 year is ₹ 15.
So, on ₹ 10,000, interest charged \(\frac{15}{100}\times 1000\)= ₹ 1500
Interest for 2 years = ₹ 1500 × 2 = ₹ 3000
Amount to be paid at the end of 2 years = Principal + Interest = ₹ 10000 + ₹ 3000 =₹13000
Let us take an example and find the interest year by year. Each year our sum or principal changes.
Calculating Compound Interest
A sum of ₹20,000 is borrowed by Heena for 2 years at an interest of 8% compounded annually. Find the Compound Interest (C.I.) and the amount she has to pay at the end of 2 years.
Aslam asked the teacher whether this means that they should find the interest year by year. The teacher said ‘yes’, and asked him to use the following
steps : 1. Find the Simple Interest (S.I.) for one year.
Let the principal for the first year be P1 .
Here, P1 = ₹20,000 SI1 = SI at 8% p.a. for 1st year = ₹\(\frac{20000}{8}\times 100\)= ₹ 1600
2. Then find the amount which will be paid or received. This becomes principal for the next year. Amount at the end of 1st year = P1+ SI1= ₹20000 + ₹1600 = ₹ 21600 = P2 (Principal for 2nd year)
3. Again find the interest on this sum for another year.
SI2 = SI at 8% p.a.for 2nd year = ₹ 21600 8 100 × = ₹1728
4. Find the amount which has to be paid or received at the end of second year.
Amount at the end of 2nd year = P2 + SI2 = ₹21600 +₹1728 =₹23328
Total interest given = ₹ 1600 + ₹1728 =₹3328
Reeta asked whether the amount would be different for simple interest. The teacher told her to find the interest for two years and see for herself.
SI for 2 years =₹\(\frac{20000\times 8\times 2}{100}\)= ₹3200
Reeta said that when compound interest was used Heena would pay ₹128 more. Let us look at the difference between simple interest and compound interest. We start with ₹100. Try completing the chart.
Note that in 3 years,
Interest earned by Simple Interest = ₹(130 – 100) =₹30, whereas,
Interest earned by Compound Interest = ₹ (133.10 – 100) = ₹ 33.10
Note also that the Principal remains the same under Simple Interest, while it changes year after year under compound interest.
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Comparing Quantities
Deducing a Formula for Compound Interest
Zubeda asked her teacher, ‘Is there an easier way to find compound interest?’ The teacher said ‘There is a shorter way of finding compound interest. Let us try to find it.’
Suppose P1 is the sum on which interest is compounded annually at a rate of R% per annum
Let P1 =₹5000 and R = 5. Then by the steps mentioned above
1.SI1 = \(\frac{\text{5}000\times \text{ 5 }\times \text{1}}{100}\text{ }\)
So,A1=₹5000+\(\frac{\text{5}000\times \text{ 5 }\times \text{1}}{100}\text{ }\)
=₹5000+\(\left( 1+\frac{5}{100} \right)\)
(or)
\(\begin{align} & \text{S}{{\text{I}}_{1}}\text{ }=\frac{{{P}_{1}}\times \text{ R }\times \text{1}}{100}\text{ } \\ & (or){{A}_{1}}={{P}_{1}}+\text{S}{{\text{I}}_{1}}={{P}_{1}}+\frac{{{P}_{1}}R}{100} \\ & {{P}_{1}}\left( 1+\frac{R}{100} \right)={{P}_{2}} \\ \end{align} \)
.\(\begin{align} & 2.\text{ S}{{\text{I}}_{2}}\text{ }=₹5000\left( 1+\frac{5}{100} \right)\times \frac{5\times 1}{100} \\ & =\frac{₹5000\times 5}{100}\left( 1+\frac{5}{100} \right) \\ & {{A}_{2}}=₹5000\left( 1+\frac{5}{100} \right)+\frac{5000\times 5}{100}\left( 1+\frac{5}{100} \right) \\ & =₹5000\left( 1+\frac{5}{100} \right)\left( 1+\frac{5}{100} \right) \\ & =₹5000{{\left( 1+\frac{5}{100} \right)}^{2}} \\ \end{align} \) (or)\(\begin{align} & {{A}_{2}}={{P}_{2}}+\text{S}{{\text{I}}_{2}}={{P}_{1}}+\frac{{{P}_{1}}R}{100} \\ & ={{P}_{1}}\left( 1+\frac{R}{100} \right)+{{P}_{1}}\frac{R}{100}\left( 1+\frac{R}{100} \right) \\ & ={{P}_{1}}\left( 1+\frac{R}{100} \right)\left( 1+\frac{R}{100} \right) \\ & ={{P}_{1}}{{\left( 1+\frac{R}{100} \right)}^{2}}={{p}_{3}} \\ \end{align} \)
Proceeding in this way the amount at the end of n years will be
\({{A}_{n}}={{P}_{1}}{{\left( 1+\frac{R}{100} \right)}^{n}}\)
Or, we can say \(A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\)So, Zubeda said, but using this we get only the formula for the amount to be paid at the end of n years, and not the formula for compound interest. Aruna at once said that we know CI = A – P, so we can easily find the compound interest too.
Example : Find CI on ₹ 12600 for 2 years at 10% per annum compounded annually
Solution: We have, \(A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\), where Principal (P) =₹12600, Rate (R) = 10,
Number of years (n) = 2
= ₹\({{\left( 1+\frac{10}{100} \right)}^{n}}=₹12600{{\left( \frac{11}{10} \right)}^{2}}\)
= ₹12600× \(\frac{11}{10}×\frac{11}{10}\)=₹15246
CI = A – P = ₹15246 – ₹12600 =₹2646
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Comparing Quantities
Applications of Compound Interest Formula
There are some situations where we could use the formula for calculation of amount in CI. Here are a few.
(i) Increase (or decrease) in population.
(ii) The growth of a bacteria if the rate of growth is known.
(iii) The value of an item, if its price increases or decreases in the intermediate years
Example : The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.
Solution: There is 5% increase in population every year, so every new year has new population. Thus, we can say it is increasing in compounded form. Population in the beginning of 1998 = 20000 (we treat this as the principal for the 1st year)
Increase at 5% = \(\frac{5}{100}\times 20000=1000\)= 1000
Population in 1999 = 20000 + 1000 = 21000
Increase at 5% = \(\frac{5}{100}\times 21000=1050\)
Population in 2000 = 21000 + 1050 = 22050
Increase at 5% = \(\frac{5}{100}\times 22050=1102.5\)
At the end of 2000 the population = 22050 + 1102.5 = 23152.5
or, Population at the end of 2000 = 20000\({{\left( 1+\frac{5}{100} \right)}^{3}}\)
\(\begin{align} & 20000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20} \\ & =23152.5 \\ \end{align}\)
So, the estimated population = 23153.
Aruna asked what is to be done if there is a decrease. The teacher then considered the following example.
Example: A TV was bought at a price of ` 21,000. After one year the value of the TV was depreciated by 5% (Depreciation means reduction of value due to use and age of the item). Find the value of the TV after one year.
Solution:
Principal = ₹ 21,000
Reduction = 5% of ₹21000 per year
\(=\frac{21000\times 5\times 1}{100}=₹1050\)
value at the end of 1 year = ₹ 21000 –₹1050 =₹19,950
Alternately, We may directly get this as follows:
value at the end of 1 year = ₹ 21000\(\left( 1-\frac{5}{100} \right)\)
₹21000 × \(19 \over 20\) =₹ 19,950