Comparing Quantities - Class 8 - Mathematics
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Examples - Comparing Quantities | NCERT | Mathematics | Class 8
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.
To find the total number of students in Class VII based on the given information, we can set up an equation. If girls make up $60%$ of the total number of students and there are $18$ girls, we can express this as:
$$ 0.60 \times \text{Total Number of Students} = 18 $$
Solving for the total number of students, we have:
$$ \text{Total Number of Students} = \frac{18}{0.60} $$
By calculating this, we get:
$$ \text{Total Number of Students} = 30 $$
Hence, there are $30$ students in Class VII.
An item marked at ₹ 840 is sold for ₹ 714 . What is the discount and discount $\%$ ?
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Sign up nowThe 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.
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Sign up now(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.
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Sign up now(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.
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Sign up nowSalim bought an article for ₹ 784 which included GST of $12 \%$. What is the price of the article before GST was added?
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Sign up nowA 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.
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Sign up nowFind CI on ₹ 12600 for 2 years at $10 \%$ per annum compounded annually.
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Sign up nowThe 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.
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Sign up nowA 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.
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Sign up nowExercise 7.1 - Comparing Quantities | NCERT | Mathematics | Class 8
Find the ratio of the following.
(a) Speed of a cycle $15 \mathrm{~km}$ per hour to the speed of scooter $30 \mathrm{~km}$ per hour.
(b) $5 \mathrm{~m}$ to $10 \mathrm{~km}$
(c) 50 paise to ₹5
To find the ratios, we need to make sure the units are consistent across each item being compared, and then simply divide the first quantity by the second.
(a) Speed of a cycle ($15 , \mathrm{km/h}$) to the speed of a scooter ($30 , \mathrm{km/h}$):
The units are already consistent (both in $\mathrm{km/h}$), so we can directly find the ratio by dividing:
$$\frac{15 , \mathrm{km/h}}{30 , \mathrm{km/h}} = 0.5$$
Thus, the ratio is $1:2$.
(b) $5 , \mathrm{m}$ to $10 , \mathrm{km}$:
To make the units consistent, we'll convert $10 , \mathrm{km}$ to meters (since $1 , \mathrm{km} = 1000 , \mathrm{m}$):
$$10 , \mathrm{km} = 10 \times 1000 , \mathrm{m} = 10000 , \mathrm{m} $$
Now, we can find the ratio:
$$\frac{5 , \mathrm{m}}{10000 , \mathrm{m}} = 0.0005$$
Thus, the ratio is $1:2000$.
(c) 50 paise to ₹5:
We need to express both quantities in the same unit. Since $1 , \mathrm{₹} = 100 , \mathrm{paise}$:
$$₹5 = 5 \times 100 , \mathrm{paise} = 500 , \mathrm{paise}$$
Now, we can find the ratio:
$$\frac{50 , \mathrm{paise}}{500 , \mathrm{paise}} = 0.1$$
Thus, the ratio is $1:10$.
Convert the following ratios to percentages.
(a) $3: 4$
(b) $2: 3$
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Sign up now$72 \%$ of 25 students are interested in mathematics. How many are not interested in mathematics?
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Sign up nowA football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
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Sign up nowIf Chameli had ₹600 left after spending $75 \%$ of her money, how much did she have in the beginning?
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Sign up nowIf $60 \%$ people in a city like cricket, $30 \%$ like football and the remaining like other games, then what per cent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.
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Sign up nowExercise 7.2 - Comparing Quantities | NCERT | Mathematics | Class 8
During a sale, a shop offered a discount of $10 \%$ on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?
To find the total amount a customer has to pay for a pair of jeans and two shirts during a sale with a 10% discount, we perform the following steps:
Calculate the original total price of the items without the discount.
Calculate the discount amount based on the original total price.
Subtract the discount amount from the original total price to get the final price the customer pays.
Let's do the calculations:
Original Total Price
Price of jeans: ₹ 1450
Price of one shirt: ₹ 850
Total price for two shirts: ₹ 850 * 2 = ₹ 1700
Total original price of jeans and shirts together: ₹ 1450 + ₹ 1700 = ₹ 3150
Discount Amount
Discount: 10% of total original price = 0.10 * ₹ 3150 = ₹ 315
Final Price After Discount
Final price = Original total price - Discount = ₹ 3150 - ₹ 315 = ₹ 2835
So, a customer would have to pay ₹ 2835 for a pair of jeans and two shirts during the sale.
The price of a TV is $₹13,000$. The sales tax charged on it is at the rate of $12 \%$. Find the amount that Vinod will have to pay if he buys it.
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Sign up nowArun bought a pair of skates at a sale where the discount given was $20 \%$. If the amount he pays is ₹ 1,600 , find the marked price.
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Sign up nowI purchased a hair-dryer for ₹ 5,400 including $8 \%$ VAT. Find the price before VAT was added.
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Sign up nowAn article was purchased for ₹ 1239 including GST of $18 \%$. Find the price of the article before GST was added?
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Sign up nowExercise 7.3 - Comparing Quantities | NCERT | Mathematics | Class 8
The population of a place increased to 54,000 in 2003 at a rate of $5 \%$ per annum
(i) Find the population in 2001.
(ii) what would be its population in 2005?
To determine the population at a certain year given an annual growth rate, we can use the formula related to exponential growth:
$$ P = P_0 (1 + r)^n $$
Where:
- $P$ is the future population
- $P_0$ is the initial population
- $r$ is the annual growth rate (expressed as a decimal)
- $n$ is the number of years
For 2001 (2 years earlier than 2003)
Given the population in 2003 $(P)$ is 54,000 and the growth rate is $5% = 0.05$ per annum, we need to find the population in 2001, which is $n = -2$ years from 2003. Rearranging the formula to solve for $P_0$:
$$ P_0 = \frac{P}{(1 + r)^n} $$
Substituting in the given values:
$$ P_0 = \frac{54000}{(1 + 0.05)^{-2}} $$
$$ P_0 = \frac{54000}{(1.05)^{-2}} $$
$$ P_0 = 54000 \times (1.05)^2 $$
$$ P_0 = 54000 \times 1.1025 $$
$$ P_0 = 59535 $$
Therefore, the population in 2001 was approximately 59,535.
For 2005 (2 years after 2003)
Now, we compute the population in 2005, which is $n = 2$ years after 2003 using the formula given by $P = P_0 (1 + r)^n$
Given:
- $P_0 = 54000$
- $n = 2$
- $r = 0.05$
Substituting these values into the formula:
$$ P = 54000 \times (1 + 0.05)^2 $$
$$ P = 54000 \times (1.05)^2 $$
$$ P = 54000 \times 1.1025 $$
$$ P = 59535 $$
So, the population in 2005 would also be approximately 59,535, which is clearly an error in my calculation. Let's correct the population calculation for 2005.
For 2005, the correct computation should be:
$$ P = 54000 \times (1.05)^2 $$
Let's correctly compute the population for 2005.
Correct Calculation for 2005
$$ P = 54000 \times (1.05)^2 $$
$$ P = 54000 \times 1.1025 $$
$$ P = 59670 $$
Thus, the corrected population in 2005 would be approximately 59,670, not 59,535 as incorrectly repeated earlier.
Apologies for the confusion. The correct population for 2005 based on a $5%$ growth rate would indeed be higher than that for 2003, calculated correctly as approximately 59,670.
In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of $2.5 \%$ per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
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Sign up nowA scooter was bought at $₹ 42,000$. Its value depreciated at the rate of $8 \%$ per annum. Find its value after one year.
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Sign up nowExtra Questions - Comparing Quantities | NCERT | Mathematics | Class 8
Which of the following is equal to $200\%$ of 12?
A. 24
B. $40%$ of 60
C. $80%$ of 32
D. $\mathbf{16}$
To find $200\%$ of 12, we calculate:
$$ 200% \text{ of } 12 = \frac{200}{100} \times 12 = 24 $$
To check which options are equal to this value:
Option A: 24
This is straightforwardly equal to $24$.Option B: $40\%$ of 60
We need to check this by calculation:$$ 40% \text{ of } 60 = \frac{40}{100} \times 60 = 24 $$
Option C: $80\%$ of 32
Let's check this too:$$ 80% \text{ of } 32 = \frac{80}{100} \times 32 = 25.6 $$
From the computations:
Option A: $24$ matches, corresponding to $200%$ of $12$.
Option B: Also computes to $24$, thus it matches too.
Option C: Computes to $25.6$, which does not match.
Thus, Options A and B are correct as both equal $200%$ of $12$.
Two cylinders $A$ and $B$ are formed by folding a rectangular sheet of dimensions $20 \mathrm{~cm} \times 10 \mathrm{~cm}$ along its length and also along its breadth, respectively. Then, the volume of $A$ is _________ of the volume of $B$.
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Sign up nowDavid cuts a cake into two equal halves. He cuts one half into smaller pieces of equal size. Each of these small pieces weighs twenty grams. If he has seven pieces in all, how heavy is the original cake?
A) 120 grams
B) 180 grams
C) 300 grams
D) 240 grams
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