ncert-solutions-class-7-maths-ch-8-comparing-quantities

NCERT Solutions for Class 7Â Maths Chapter 8 Comparing Quantities

NCERT solutions for class 7 maths chapter 8 comparing quantities exercise 8.1

Question:1 Find the ratio of:

(a) Rs 5 to 50 paise
(b) 15 kg to 210 g

Answer:

(a) Rs 5 to 50 paise
First, convert the given quantities into the same units,

So, Rs 5 = $5 \times 100$ = 500 paise
Now, Ratio = $\frac{500}{50} = 10:1$

(b) 15 kg to 210 g
Converting the given quantities into the same unit.
So, 15 kg = $15 \times 1000$ = 15,000 g
Therefore, Required ratio
$\\=\frac{15000}{210}=500:7$

(c) 9 m to 27 cm
First, convert meter into centimetre
So, 9m = $9 \times 100 = 900 cm$
Therefore, the required ratio
$\frac{900}{27}=100:3$

(d) 30 days to 36 hours
Converting the given quantities into the same unit.
So, 30 days = $30 \times 24hr = 720 hrs$
Therefore, the ratio
$\frac{720}{36} = 20:1$

Question:2 In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?

Answer:

Given that,
6 student use 3 computers
So, 1 computer is used by $\frac{6}{3} =2$ students.

Therefore, for 24 students no. of computer required = $\frac{24}{2} =12$ computers

Hence the required number of computer is 12.

Question:3 Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = $3\: lakh \: km^{2}$ and area of UP = $2\: lakh \: km^{2}$ .

(i) How many people are there per $km^{2}$ in both these States?

(ii) Which State is less populated?

Answer:

Given that,
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs
Area of Rajasthan = $3\: lakh \: km^{2}$
and area of UP = $2\: lakh \: km^{2}$

Now,

(a) Number of peoples per
In Rajsthan, = $\frac{population}{area}$
$=\frac{570}{3}= 190$

In UP,
$\frac{population}{area} = \frac{1160}{2}$
= 580

(b) Rajasthan is less populated as we can see that the number of people in per is less

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: meaning of percentage

Question:1 Find the Percentage of children of different heights for the following data.

Height	Number of Children	In Fraction	In Percentage
110 cm	22
120 cm	25
128 cm	32
130 cm	21
Total	100

Answer:

Height	No. of children	In fraction	In percentage
110	22	22/100 =0.22	22%
120	25	25/100 = 0.25	25%
128	32	32/100 =0.32	32%
130	21	21/100 =0.21	21%
total	100	01	100%

Question:2 A shop has the following number of shoe pairs of different sizes.

Size 2 : 20 Size 3 : 30 Size 4 : 28 Size 5 : 14 Size 6 : 8

Write this information in tabular form as done earlier and find the Percentage of each shoe size available in the shop.

Answer:

Size	No. of shoes	Fraction	Percentage
2	20	$\frac{20}{100}=0.2$	$20 \%$
3	30	$\frac{30}{100}=0.3$	$30 \%$
4	28	$\frac{28}{100}=0.28$	$28 \%$
5	14	$\frac{14}{100}=0.14$	$14 \%$
6	8	$\frac{8}{100}=0.08$	$8 \%$
Total	100

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: when total is not hundred

Question:1 A collection of 10 chips with different colours is given.

Colour	Number	Fraction	Denominator Hundred	In Percentage
Green
Blue
Red
Total

Fill the table and find the percentage of chips of each colour.

Answer:

Colour	Number	Fraction	Denominator hundred	In percentage
Green	4	4/10=0.4	40/100	40%
Blue	3	3/10=0.3	30/100	30%
Red	3	3/10=0.3	30/100	30%
Total	10	1	100/100	100%

Question:2 Mala has a collection of bangles. She has 20 gold bangles and 10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as done in the above example?

Answer:

We have,
Number of gold bangles = 20
Number of silver bangles = 10
Total bangles = 30

Therefore, the percentage of gold bangles
$\frac{20}{30}\times 100 = 66.66\%$

and the percentage of the silver bangles
$\frac{10}{30}\times 100 = 33.33\%$

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: converting decimals to percentage

Question:1(a) Convert the following to per cents:

$(a)\frac{12}{16}$

Answer:

$(a)\frac{12}{16}$
To convert into a percentage,
$\frac{12\times 100}{16}=\frac{1200}{16}\%=75\%$

Question:(b) Convert the following to per cents:

3.5

Answer:

(b) 3.5
To convert into a percentage,
$\frac{35\times 100}{10\times 10}=\frac{3500}{10}\%=350\%$

Question: Convert the following to per cents:

$(c)\frac{49}{50}$

Answer:

$(c)\frac{49}{50}$

To convert into a percentage,
$\frac{49\times 100}{50}=\frac{4900}{50}\%=98\%$

Question:1(d) Convert the following to per cents:

$(d)\frac{2}{2}$

Answer:

$(d)\frac{2}{2}$
To convert into a percentage,
$\frac{2\times 100}{2}=\frac{200}{2}\%=100\%$

Question:(e) Convert the following to per cents:

0.05

Answer:

(e) 0.05
To convert into a percentage,
$\frac{5\times 100}{100}=\frac{500}{100}\%=5\%$

Question:2(i) Out of 32 students, 8 are absent. What per cent of the students are absent?

Answer:

We have a total number of student = 32
and the absent student = 8

Therefore, the percentage of absent students
$=\frac{8}{32}\times 100 =25\%$

Question:2(ii) There are 25 radios, 16 of them are out of order. What per cent of radios are out of order?

Answer:

We have a total radio = 25
and the radios that are out of order = 16

Therefore, the percentage of out of order radios
$=\frac{16}{25}\times 100 =64\%$

Question:2(iii) A shop has 500 items, out of which 5 are defective. What per cent are defective?

Answer:

Given that,
The shop has total items =500
Defective item = 5

Therefore, the percentage of the defective item
$\Rightarrow \frac{5}{500}\times 100 = \frac{1}{100}\times 100 =1\%$

Question:2(iv) There are 120 voters, 90 of them voted yes. What per cent voted yes?

Answer:

Given that,
The total no. of voters =120
Voted yes = 90

Therefore, the percentage of voted yes
$\Rightarrow \frac{90}{120}\times 100 = \frac{9}{12}\times 100= \frac{3}{4}\times 100=75\%$

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: converting percentage to fractions or decimals

Question:1(i) 35% + _______ = 100%

Answer:

Let the blank space be X
therefore, 35% + X = 100%
So, X = 100% - 35%
= 65%

Question:1(ii) 64% + 20% +________ % = 100%

Answer:

Let the blank space be X
therefore,
64% + 20% + X = 100%
So, X = 100% - 64% - 20%
= 16%

Question:1(iii) 45% = 100% – _________ %

Answer:

Let the blank space be X
therefore,
45% = 100% - X
So, X = 100% - 45%
= 55%

Question:1(iv) 70% = ______% – 30%

Answer:

Let the blank space be X
therefore,
70% = X - 30 %
So, X = 70%- 30%
= 100%

Question:2 If 65% of students in a class have a bicycle, what per cent of the student do not have bicycles?

Answer:

Given that,
65 % of students in a class have a bicycle.
So, the remaining percent of students have no bicycle
= 100% - 65%
= 35%

Hence 35% of students have no bicycle.

Question:3 We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what per cent are mangoes?

Answer:

We have,

A basket is full of apples, oranges and mangoes.
50% are apples, 30% are oranges.

So, the remaining percent are mangoes = 100% - 50% - 30%
= 20%

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: fun with estimation

Question:1(a) What per cent of this figure is shaded?

You can make some more figures yourself and ask your friends to estimate the shaded parts.

Answer:

In the above figure, there are a total of 4 parts and out of which 3 parts are shaded
therefore, the percentage of the shaded part
<img alt="\Rightarrow \frac{3}{4}\times 100 = 75\%" height="38"

src="https://lh3.googleusercontent.com/YexaMs3GB9QgBL2qTxbYdiwFBd3iqBDysOc5PYBl44s98G9amKYrDVTl6DVtQwC2Bb1F9bZcRurmKu1VTDTlV2xmd5k1rbgEANyrhm78bPHBfLzfAL-gQw80H8ZBeSfGVA5d1aQ" style="margin-left: 0px; margin-top: 0px;" width="140" />

Question:1(b) What per cent of this figure is shaded?

You can make some more figures yourself and ask your friends to estimate the shaded parts.

Answer:

In the above figure,
The total shaded part =
$\Rightarrow \frac{1}{4} +\frac{1}{8}+\frac{1}{8} = \frac{1}{2}$

Therefore, the percentage of the shaded part
$\Rightarrow \frac{1}{2}\times 100 = 50\%$

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: converting percentage to how many

Question:1(a) Find: 50% of 164

Answer:

50% of 164
$\\\Rightarrow \frac{50}{100}\times 164\\ \Rightarrow\frac{1}{2}\times 164 = 82$

So 50% of 164 is 82, that is half of 164

Question:1(b) Find: 75% of 12

Answer:

75% of 12
$\\\Rightarrow \frac{75}{100}\times 12\\ \Rightarrow\frac{3}{4}\times 12 = 9$

Question:1(c) Find: $(c) 12\tfrac{1}{2}\: ^{o}/_{o}\ of\ 64$

Answer:

$(c) 12\tfrac{1}{2}\: ^{o}/_{o}\ of\ 64$

it means 12.5% of 64
Therefore,
$\\\Rightarrow \frac{12.5}{100}\times 64\\ \Rightarrow 0.125\times 64 = 8$

Question:2 8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain.

Answer:

We have,
8% children of a class of 25 like getting wet in the rain

therefore, the number of children get wet
8 % of 25
$\frac{8}{100}\times 25 = 2$

Hence out of 25 only 2 children get wet in the rain

Question:1 9 is 25% of what number?

Answer:

Let the number be X
therefore, 25% of X = 9
$\\\frac{25}{100}\times X = 9\\ X = \frac{900}{25}=36$

Question:2 75% of what number is 15?

Answer:

Let the number be X
therefore, 75% of X = 15
$\\\Rightarrow \frac{75}{100}\times X = 15\\ \Rightarrow X = \frac{1500}{75}=20$

NCERT solutions for class 7 maths chapter 8 comparing quantities exercise 8.2

Question:1 Convert the given fractional numbers to per cents.

$(a)\frac{1}{8}$
$(b)\frac{5}{4}$
$(c)\frac{3}{40}$
$(d)\frac{2}{7}$

Answer:

To convert the given fractional into the per cent we have to do multiplication in numerator and denominator by 100
Now,
$(a)\frac{1}{8}$
$\Rightarrow \frac{1}{8}\times100 = \frac{100}{8}\%=12.5\%$

$(b)\frac{5}{4}$
$\Rightarrow \frac{5}{4}\times100 = \frac{500}{4}\%=125\%$

$(c)\frac{3}{40}$
$\Rightarrow \frac{3}{40}\times 100 = \frac{300}{40}\%=7.5\%$

$(d)\frac{2}{7}$
$\Rightarrow \frac{2}{7}\times100 = \frac{200}{7}\%=28.57\%$

Question:2 Convert the given decimal fractions to per cents.

(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35

Answer:

To convert the given fractional decimal into a per cent, multiply the denominator and numerator by 100.
So,
(a) 0.65
$0.65\times 100=65\%$

(b) 2.1
$2.1\times 100=210\%$

(d) 12.35
$12.35\times100=1235\%$

Question:3(i) Estimate what part of the figure is coloured and hence find the per cent which is coloured.

Answer:

We have,
The fraction of the coloured part = $\frac{1}{4}$
Therefore,
$\frac{1}{4}\times \frac{100}{100}=\frac{100}{4}\%=25\%$

Question:3(ii) Estimate what part of the figure is coloured and hence find the per cent which is coloured.

Answer:

In the given figure, the fraction of the coloured part is
Therefore, Percentage of the coloured part
$=\frac{3\times 100}{5\times 100} = \frac{300}{5}\% = 60\%$

Question:3(iii) Estimate what part of the figures is coloured and hence find the per cent which is coloured.

Answer:

In the given figure, the fraction of the coloured part is
Therefore, Percentage of the coloured part
$=\frac{3\times 100}{8\times 100}=\frac{300}{8}\% = 37.5\%$

Question:4 Find: (a) 15 $\%$ of 250
(b) 1 $\%$ of 1 hour
(c) 20 $\%$ of Rs 2500
(d) 75 $\%$ of 1 kg

Answer:

Here per cent implies for ( $\% \rightarrow \frac{1}{100}$ )

Therefore,
(a) 15% of 250
$15\times \frac{1}{100}\times 250 =\frac{75}{2}$
= 37.5

(b) 1 $\%$ of 1 hour
= 1 $\%$ of 60 minutes
$=\frac{1}{100}\times 60 =\frac{3}{5}\ min$
$=\frac{3}{5}\times 60 = 36\ sec$

(d) 75 $\%$ of 1 kg
We know that 1kg = 1000 gm, therefore,
$\frac{75}{100}\times 1000g=750g$
=0.75 kg

Question:5 Find the whole quantity if

(a) 5% of it is 600.
(b) 12% of it is Rs 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 litres.

Answer:

(a) Let the whole quantity be X
Therefore, 5 % of X = 600
$\\\Rightarrow \frac{5}{100}\times X=600\\ \Rightarrow X = \frac{600\time 100}{5}=12000$
Thus the required whole quantity is 12000

(b) 12 % of X = Rs 1080
$\\\Rightarrow \frac{12}{100}\times X=1080\\ \Rightarrow X = \frac{1080\time 100}{12}=9000$

(d) 70% of X = 14 minutes
$\\\Rightarrow \frac{70}{100}\times X=14\ min\\ \Rightarrow X = \frac{1400\time 100}{70}=20\ min$

(e) 8 % of X = 40 litre
$\\\Rightarrow \frac{8}{100}\times X=40\ L\\ \Rightarrow X = \frac{4000}{8}=500\ L$

Question:6 Convert given per cent to decimal fraction and also to fraction in simplest forms:

(a) 25%
(b) 150%
(c) 20%
(d) 5%

Answer:

(a) 25 %
$\Rightarrow \frac{25}{100} =0.25 = \frac{1}{4}$

(b) 150%
$\Rightarrow \frac{150}{100} =1.5 = \frac{3}{2}$

(d) 5%
$\Rightarrow \frac{5}{100} =0.05 = \frac{1}{20}$

Question:7 In a city, 30% are females, 40% are males and remaining are children. What per cent are children?

Answer:

Given that,
30 % are female and
40 % are males
Total percentage of females and males
= 30 % + 40% = 70%
Therefore, Percentage of children
= (100% -70%) = 30 %

Question:8 Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

Answer:

Given that,
Total number of voters = 15000
Percentage of voters who voted = 60%
Therefore, remaining 40% voters didn't vote

Thus, the number of people who did not vote
= $40\%\ of\ 15000$
$=\frac{40}{100}\times 15000$

Question:9 Meeta saves Rs 4000 from her salary. If this is 10% of her salary. What is her salary?

Answer:

Given that,
10 % of her salary = 4000
Let her total salary be X
Therefore,
10% of X = 4000
$\Rightarrow \frac{10}{100}\times X = 4000$
$X = \frac{4000}{10}\times 100$
X = Rs 40,000

Question:10 A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?

Answer:

Given that,
Total match played by the cricket team = 20
and, a match won = 25%

According to question,
Number of winning match = 25 % of 20
$\\\Rightarrow \frac{25}{100}\times 20=5\\$

hence they won total 5 matches out of 20 match

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: ratio of percents

Question:1 Divide 15 sweets between Manu and Sonu so that they get 20 % and 80 % of them respectively.

Answer:

We have 15 sweets.
Manu wants 20% and Sonu wants 80% of the sweets respectively.

Therefore, 20% of 15
= $\frac{20}{100}\times 15 =3$

and 80% of 15 sweets
$\frac{80}{100}\times 15 =12$

Hence give 3 sweets to Manu and 12 sweets to Sonu.

Question:2 If the angles of a triangle are in the ratio 2 : 3 : 4. Find the value of each angle.

Answer:

We have,
angles of a triangle are in the ratio 2 : 3: 4.
Let the angle be $2x,3x\and\ 4x$ respectively

And the sum of all the angle
[by angle sum property]
so,
$x =\frac{180}{9}=20$

now, the angles are
40, 60 and 80

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: increase or decrease as per cent

Question:1 Find Percentage of increase or decrease:

– Price of shirt decreased from Rs 280 to Rs 210.

– Marks in a test increased from 20 to 30.

Answer:

(i) The decrease in price is 280-210=70

Percentage of decrease
$=\frac{70}{280}\times 100$
= 25%

(ii) increase of mark is 30-20=10

Percentage of increase
$=\frac{10}{30}\times 100$
= 33.33%

Question:2 My mother says, in her childhood petrol was Rs 1 a litre. It is Rs 52 per litre today. By what Percentage has the price gone up?

Answer:

Total increase in petrol = 52 -1 = Rs51

Therefore, the percentage of price increase
$\Rightarrow \frac{51}{52}\times 100 = 98.07\%$

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: profit or loss as a percentage

Question:1 A shopkeeper bought a chair for Rs 375 and sold it for Rs 400. Find the gain Percentage.

Answer:

We have,
SP = 400
CP =375
So, SP - CP = 25

SInce SP > CP
Therefore, Gain(%)
$\Rightarrow \frac{25}{375}\times 100 = 6.6\%$

Question:2 Cost of an item is Rs 50. It was sold with a profit of 12%. Find the selling price.

Answer:

We have,
CP = Rs 50
Profit = 12 %
SP = ?

Therefore,
$SP = CP(1+\frac{profit\%}{100})$
Put the values in the above equation, we get

$\\= 50(1+\frac{12}{100})\\ =50 \times 1.12\\ = Rs\ 56$

Question:3 An article was sold for Rs 250 with a profit of 5%. What was its cost price?

Answer:

We have,
SP = Rs 250
Profit = 5%
CP =?

Therefore,
$SP = CP (1+\frac{profit\%}{100})$
put the values in the above equation, we get

$\\\Rightarrow 250 = CP (1+\frac{5}{100})\\ \Rightarrow 250 = CP\times \frac{21}{20}\\$
$CP = \frac{250\times 20}{21} = 238.095$

Question:4 An item was sold for Rs 540 at a loss of 5%. What was its cost price?

Answer:

We have,
SP = Rs 540
Loss = 5%
CP =?

Therefore,
$SP = CP(1-\frac{loss\%}{100})$
put the values in the above equation, we get

$\\\Rightarrow 540 = CP(1-\frac{5}{100})\\ \Rightarrow540 = CP\times \frac{19}{20}$

$\Rightarrow CP = \frac{540 \times 20}{19} = Rs\ 568.42$

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: interest for multiple years

Question:1 Rs 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one year.

Answer:

Given that,
Principal = 10,000
Rate= 5% p.a
T = 1 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{10,000\times 5\times 1}{100}$
$= Rs\ 500$

Question:2 Rs 3,500 is given at 7% p.a. rate of interest. Find the interest which will be received at the end of two years.

Answer:

Given that,
Principal = 3,500
Rate= 7% p.a
T = 2 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{3,500\times 7\times 2}{100}$
$= Rs\ 490$

Question:3 Rs 6,050 is borrowed at 6.5% rate of interest p.a.. Find the interest and the amount to be paid at the end of 3 years.

Answer:

Given that,
Principal = 6050
Rate= 6.5% p.a
T = 3 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{6050\times 6.5\times 3}{100}$
$= Rs\ 1179.75$

So, the total amount to be paid = Rs 1179.75 + Rs 6050 =Rs 7229.75

Question:4 Rs 7,000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the amount to be paid at the end of the second year.

Answer:

Given that,
Principal = 7,000
Rate= 3.5% p.a
T = 2 year

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$= \frac{7000\times 3.5\times 2}{100}$
$= Rs\ 490$

So, the total amount to be paid = Rs 490 + Rs 7000 =Rs 7490

NCERT solutions for class 7 maths chapter 8 comparing quantities topic: interest for multiple years

Question:1 You have Rs 2,400 in your account and the interest rate is 5%. After how many years would you earn Rs 240 as interest.

Answer:

Given that,
Principal = 2,400
Rate= 5% p.a
T = ?
Interest = Rs 240

Therefore,
$Interest = \frac{P\times R\times T}{100}$
Substituting the values in the above formula, we get
$240= \frac{2,400\times 7\times T}{100}$
$T = \frac{10}{5} = 2\ year$

Question:2 On a certain sum the interest paid after 3 years is Rs 450 at 5% rate of interest per annum. Find the sum.

Answer:

Given,
Simple Interest, S.I.= <img alt="Rs. \ 450" height="14"

src="https://lh5.googleusercontent.com/BeVpGpNJBp7OYUxeVhehmmVwXvCAwOeukrLz4hvQlg0ISxIafjzRFPD5xlYQ01rz5bQBQhLEnEe8PlzFZjhWKDZ291Uq-5MvMk4Jszf1zUx3tOSKSQAHqTOiWVdFO-LC-Lt0mpg" style="margin-left: 0px; margin-top: 0px;" width="60" />
Time, $T = 3\ yrs$
Rate, $R= 5\%$
We know,
$P= S.I \times \frac{100}{r} \times t$
$\\ =450 \times \frac{100}{3} \times 5 \\ = 3000$

Therefore, the sum is $Rs.\ 3000$

NCERT solutions for class 7 maths chapter 8 comparing quantities exercise 8.3

Question:1(a) Tell what is the profit or loss in the following transaction. Also find profit per cent or loss per cent in each case.

Gardening shears bought for Rs 250 and sold for Rs 325.

Answer:

Given that,
Selling price = Rs 325
Cost price = Rs 250
Since SP > CP
Therefore, profit = SP-CP = Rs 75

And, Profit %
$\Rightarrow \frac{75}{250}\times 100 = 30\%$

Question:1(b) Tell what is the profit or loss in the following transaction. Also find profit per cent or loss per cent in each case.

A refrigerator bought for Rs 12,000 and sold at Rs 13,500.

Answer:

Given that,
CP = Rs 12000
SP = Rs 13500

SInce SP > CP
Therefore, Profit = SP - CP = 1500

and, Profit %
$\Rightarrow \frac{1500}{12000}\times 100 = 12.5\%$

Question:1(c) Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.

A cupboard bought for Rs 2,500 and sold at Rs 3,000.

Answer:

We have,
CP = Rs 2500
SP = Rs 3000

Since SP > CP
Therefore, Profit = SP - CP = 500

and, Profit %
$\Rightarrow \frac{500}{2500}\times 100 = 20\%$

Question:1(d) Tell what is the profit or loss in the following transaction. Also find profit per cent or loss per cent in each case.

A skirt bought for Rs 250 and sold at Rs 150.

Answer:

We have,
SP = Rs 150
CP = Rs 250

Since CP > SP
Therefore, Loss = CP-SP = Rs 100

and, Loss%
$\Rightarrow \frac{100}{250}\times 100= 40\%$

Question:2(a) Convert each part of the ratio to percentage:

3 : 1

Answer:

(a) 3: 1
The total sum of the ratio is 3 + 1 = 4
Therefore, the percentage of the first part
$\frac{3}{4}\times 100 = 75\%$

Percentage of the second part
$=\frac{1}{4}\times 100 = 25\%$

Question:2(b) Convert each part of the ratio to percentage:

2 : 3: 5

Answer:

(b) 2 : 3: 5
Sum of the ratio part = 2 + 3 + 5 = 10
Therefore, the percentage of the first part
$=\frac{2}{10}\times 100 = 20\%$

the percentage of the second part
$=\frac{3}{10}\times 100 = 30\%$

the percentage of the third part
$=\frac{5}{10}\times 100 = 50\%$

Question:2(c) Convert each part of the ratio to percentage:

1:4

Answer:

the percentage of the second part
$=\frac{4}{5}\times 100 = 80\%$

Question:2(d) Convert each part of the ratio to percentage:

1 : 2 : 5

Answer:

(d) 1 : 2 : 5

Sum of the ratio part = 1 + 2 + 5 = 8
therefore, the percentage of the first part
$=\frac{1}{8}\times 100 = 12.5\%$

the percentage of the second part
$=\frac{2}{8}\times 100 = 25\%$

the percentage of the third part
$=\frac{5}{8}\times 100 = 62.5\%$

Question:3 The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

Answer:

Given that,
Initial population = 25,000
Final population = 24,500

Total decrement = 1000 (25000 - 24500)
Therefore, percentage in decrease
$\Rightarrow \frac{500}{25000}\times100 = 2\%$

Question:4 Arun bought a car for Rs 3,50,000. The next year, the price went upto Rs 3,70,000. What was the Percentage of price increase?

Answer:

Given that,
Original price (OP) = 3,50,000
Increased price (IP) = 3,70,000

Therefore, increase in price = OP - IP = 20,000

Thus, percentage of the increase price
$=\frac{20,000}{3,50,000}\times 100 = 5.71\%$

Question:5 I buy a T.V. for Rs 10,000 and sell it at a profit of 20%. How much money do I get for it?

Answer:

Here we have,
CP = Rs 10,000
Profit = $20\%$
SP =?

We know that,
$SP =CP(1+\frac{profit}{100})$ ...........(i)

By substituting the values in eq (i), we get

$SP =10,000(1+\frac{20}{100})$
$=10,000\times (\frac{6}{5})$
= Rs 12,000

Question:6 Juhi sells a washing machine for Rs 13,500. She loses 20% in the bargain. What was the price at which she bought it?

Answer:

We have,
Sp of the washing m/c = Rs 13,500
Loss(%) = 20%
CP = ?

We know that,
$SP = CP (1-\frac{Loss}{100})$
Putting the values in the above equation we get;

$\\13,500= CP (1-\frac{20}{100})\\ 13500 = CP (1-\frac{1}{5}) = CP \times (\frac{4}{5})$
Therefore,
$CP = 13,500\times \frac{5}{4} = Rs\ 16875$

Question:7(i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.

Answer:

We have,
The ratio of calcium, carbon and oxygen in the chalk = 10 : 3: 12
Now, Sum of the ratio = 10 + 3 + 12 = 25

Therefore, the percentage of carbon in the chalk
$\Rightarrow \frac{3}{25}\times 100 =12\%$

Question:7(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?

Answer:

We have the weight of the carbon = 3g
Therefore, the weight of the chalk
$\Rightarrow \frac{3}{3}\times 25g = 25g$

Hence the weight of the chalk is 25g

Question:8 Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sell it for?

Answer:

Here we have,
CP of the book = Rs 275
Loss = 15%

We know that,
$SP = CP (1-\frac{Loss}{100}) = 275(1-\frac{15}{100})$
$= 275\times (\frac{85}{100})$
$= Rs\ 233.75$

Hence the required selling price is Rs 233.75

Question:9(a) Find the amount to be paid at the end of 3 years in this case:

Principal = Rs 1,200 at 12% p.a.

Answer:

Given that,
Principal (PA)= Rs 1,200 at 12% p.a. (Rate of interest)
and T = 3 year
We know that,
$Interest = \frac{P\times R\times T}{100}$
Now, putting the values in the above equation, we get
$\Rightarrow \frac{1200\times 12\times 3}{100} = Rs\ 432$

So, Amount = PA + PI = 1200 + 432 = Rs 1632

Question:9(b) Find the amount to be paid at the end of 3 years in this case:

(b) Principal = Rs 7,500 at 5% p.a.

Answer:

Given that,
Principal (PA)= Rs 7,500 at 5% p.a. (Rate of interest)
and T = 3 year
We know that,
$Interest = \frac{P\times R\times T}{100}$
Now, putting the values in the above equation, we get
$\Rightarrow \frac{7,500\times 5\times 3}{100} = Rs\ 1125$

So, Amount = PA + PI = 7,500 + 1125 = Rs 8625

Question:10 What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years?

Answer:

Given that,
Principal = Rs 56,000
Interest = Rs 280
Time = 2 year

Rate= ?

Therefore,
$Rate = \frac{100\times I}{P\times T}$
$= \frac{100\times 280}{56,000\times 2} = 0.25\%$

Question:11 If Meena gives an interest of Rs 45 for one year at 9% rate p.a. What is the sum she has borrowed?

Answer:

Given that,
Principal =?
Interest = Rs 45
Time = 1 year

Rate= 9% p.a

Therefore,
$Principal = \frac{100\times I}{R\times T}$
$= \frac{100\times 45}{9\times 1} = Rs\ 500$

Hence she borrowed total Rs 500

Responsive Table

Chapter No.Chapter Name

Chapter 1	NCERT solutions for class 7 maths chapter 1 Integers
Chapter 2	NCERT solutions for class 7 maths chapter 2 Fractions and Decimals
Chapter 3	NCERT solutions for class 7 maths chapter 3 Data Handling
Chapter 4	NCERT solutions for class 7 maths chapter 4 Simple Equations
Chapter 5	NCERT solutions for class 7 maths chapter 5 Lines and Angles
Chapter 6	NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties
Chapter 7	NCERT solutions for class 7 maths chapter 7 Congruence of Triangles
Chapter 8	NCERT solutions for class 7 maths chapter 8 comparing quantities
Chapter 9	NCERT solutions for class 7 maths chapter 9 Rational Numbers
Chapter 10	NCERT solutions for class 7 maths chapter 10 Practical Geometry
Chapter 11	NCERT solutions for class 7 maths chapter 11 Perimeter and Area
Chapter 12	NCERT solutions for class 7 maths chapter 12 Algebraic Expressions
Chapter 13	NCERT solutions for class 7 maths chapter 13 Exponents and Powers
Chapter 14	NCERT solutions for class 7 maths chapter 14 Symmetry
Chapter 15	NCERT solutions for class 7 maths chapter 15 visualising-solid-shapes

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