Class 7

RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise.

Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise

Download and install Partial Fraction Calculator on Windows or MAC and enjoy Partial Fraction Calculator on big screen computers for free.

(Page 16)
Question
(i) You know that area of (RBSESolutions.com) a rectangle = length x breadth. But if length and breadth is given in fraction, then how will you find the area of rectangle?
(ii) Are you agree with that to find the area of a rectangle, we must know how to multiply two or more than fractional numbers?
Solution:
(i) If length and breadth are given in fractional numbers then the numerators are multiplied with numerators and denominators with denominators to find a fraction that is the area of a rectangle:
For example:
length = 4/5 unit and breadth = \(\frac { 3 }{ 7 }\)unit
Then area of rectangle = length x breadth
= \(\frac { 4 }{ 5 }\) x \(\frac { 3 }{ 7 }\) = \(\frac { 4\times 3 }{ 30 }\) = \(\frac { 5 }{ 7 }\)
= \(\frac { 12 }{ 35 }\) sq unit.
(ii) Yes. we are agree with that we must know, how to multiply two or more fractional numbers.

(Page 19)
Question
Complete (RBSESolutions.com) the table:

After completing the table, do you agree (RBSESolutions.com) that value of product of two proper fraction is less than the value of each fraction?
Solution:


yes, we are agree with that the product of two proper fraction is always less than the each fraction.

(Page 20)
Question
Complete (RBSESolutions.com) the table:

Solution:

On completing the table we can say that the product of two improper fractions is greater than each fraction.

(Page 24)
Question
Interchange the numerator (RBSESolutions.com) and denominotor with each of the function \(\frac { 1 }{ 5 }\) and \(\frac { 2 }{ 3 }\).

(ii) Fill in the blanks :

Solution:
(i) After changing the numerator and denominator of \(\frac { 1 }{ 5 }\) required fraction = \(\frac { 5 }{ 1 }\) and after changing the numerator and denominator of \(\frac { 2 }{ 3 }\) required fraction = \(\frac { 3 }{ 2 }\)

(Page 25)
Question
How can you (RBSESolutions.com) read those number
(1) 24.2 = Twenty four decimal two
(ii) 2.04 = Two decimal zero four
(iii) 325.52 = …………..
(iv) 56.32 = ……………
Solution:
(iii) 325.52 = Three hundred twenty five decimal five two.
(iv) 56.32 = Fifty six decimal three two.

How do we write 1/8 as a decimal? To convert 1/8 to a decimal, divide the denominator into the numerator.

(Page 26)
Question
See the following (RBSESolutions.com) table and fill up the blanks:

(ii) We can also write these numbers in their extended form
421.258 = 4 x 100 + 2 x 10 + 1 x 1 + 2 x \(\frac { 1 }{ 10 }\) + 5 x \(\frac { 1 }{ 100 }\) + 8 x \(\frac { 1 }{ 1000 }\)
Similarly write the (RBSESolutions.com) remaining numbers from above table.
120
(iii) Fill in the blanks : 120 m = \(\frac { 120 }{ 1000 }\)km = …… km.
Solution:

(ii) Extended form (RBSESolutions.com) of remaining numbers

(Page 30)
Put decimal in –
Question
1.52 x 1000 = ….
Solution:
1.52 x 1000 = 1520.00

(Page 34)
Question
(i) There are 8 black and 7 white (RBSESolutions.com) strips in a zebra crossing. So tell what part of total strips is the number of white strips?

(ii) On a day 100 people crossed the road by zebra crossing out of which 20 are men, 30 women, 10 children and 40 students. Show all these data’s In decimal.
Solution:
(i) Total strips 8 + 7 = 15
Part of white strips (RBSESolutions.com) related with total strips.

∴ White strips are \(\frac { 7 }{ 5 }\) part of the total strips.

(ii) Total number of people = 100

Do and Learn

(Page 14)

Let us write 5/8 as a decimal using the division method.

Question 1
Find five equivalent (RBSESolutions.com) fractions of \(\frac { 4 }{ 7 }\)
Solution:

∴ Equivalent fractions of \(\frac { 4 }{ 7 }\) are = \(\frac { 8 }{ 14 }\), \(\frac { 12 }{ 21 }\), \(\frac { 16 }{ 28 }\), \(\frac { 20 }{ 35 }\), and \(\frac { 24 }{ 42 }\)

Question 2
Compare and (RBSESolutions.com) write (<,= ,>)

Solution:

(Page 18)
Question
Solve

Solution:

(Page 18)
Question
Solve
(i) 5 x \(\frac { 1 }{ 2 }\) = ?
(ii) 1\(\frac { 4 }{ 9 }\) x 6 = ?
Solution:

(Page 18)
Question
Solve

Solution:

(Page 19)
Question
Find


Solution:

(Page 20)
Question
Find the product of one (RBSESolutions.com) proper and one improper fraction and prepare the table showing theresoult.

Product is greater than proper and smaller then improper fraction.
Solution:

Product is greater than proper and smaller than improper fraction.

Product is greater than proper and smaller than improper fraction.

Product is greater than (RBSESolutions.com) proper and smaller than improper fraction.

Product is greater than proper and smaller than improper fraction.

(Page 24)
Question
Solve

Solution:

(Page 24)
Question
Fill in the blanks

Solution:

(Page 25)
Question
Solve

Solution:

(Page 26)
Question
Which (RBSESolutions.com) number is smaller?
(i) 35.37 and 35.07
(ii) 262.327 and 262.372
Solution:
(j) 35.07
(ii) 262.327

(Page 29)
Question
Find the (RBSESolutions.com) value of
(1) 2.3 x 3.5
(ii) 37 x 5
(iii) 2.4 x 7.35
Solution:

(Page 32)
Question
Divide the given decimal (RBSESolutions.com) numbers by 10,100 and 1000?
(i) 132.4
(ii) 1.03
(iii) 40.033
(iv) 4.321
Solution:

(Page 32)
Question
(i) 6 ÷ 1.2
(ii) 9 ÷ 4.5
(iii) 48 ÷ 0.8
Solution:

(Page 33)
Question
Solve
(i) 7.75 ÷ 0.25
(ii) 5.6 ÷ 1.4
(iii) 42.8 ÷ 0.02
Solution:

We hope the RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise.

Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise

Do and Learn

(Page 159)
Question 1
Find the ratio of actual length and breadth (RBSESolutions.com) of mathematics book of class VII.
Solution:
Length of book = 27 cm.
Breadth = 15 cm

Question 2
Find the ratio of length and breadth of national flag by taking help from your teacher.
Solution:
Different size of national flag (RBSESolutions.com) length and breadth ratio as follows :

Question 3
Find the ratio of length and breadth (RBSESolutions.com) of your classroom by measuring it
Solution:
Let the length and breadth of room is 30 feet and 20 feet.
Ratio = 30 : 20 = \(\frac { 30 }{ 20 }\) = \(\frac { 3 }{ 2 }\) = 3 : 2

Question 4
Measure your height and measure the length by expanding both the hand completely. Now find out the ratio between these two quantities.
Solution:
Length of both hands = 60 cm
Self height = 120 cm
Ratio = 120 : 60
\(\frac { 120 }{ 60 }\) = \(\frac { 2 }{ 1 }\) = 2 : 1

(Page 162)

Percentage Difference calculator is a free online tool to find the percent difference between two numbers.

Question 1
25 students of class tell their (RBSESolutions.com) interests about games.
Kabbadi – 4 students
Cricket – 11 students
Chess – 6 students
Tennis – 3 students
Other games – 1 students
Express the number of students in percentage according to the interest in each game.
Solution:
Total students = 25

Percentage Difference calculator – online basic math function tool to calculate how much percent difference between two quantities or observations.

Question 2
Following results are obtained on taking (RBSESolutions.com) advice of 250 students of preference of menu of mid day meal is selected school of Jalore Panchayat.

Express the percentage of preference of each type of menu from the above results.
Solution:
Total number (RBSESolutions.com) of students = 250

(Page 163)
Question 1
Convert the following (RBSESolutions.com) fractions into per-centage.
(i) \(\frac { 5 }{ 8 }\)
(ii) \(\frac { 5 }{ 3 }\)
Solution:

Question 2
Convert the decimal(RBSESolutions.com) fractions into percentage.
(i) 0.5
(ii) 0.08
(iii) 0.225
(iv) 6.5
Solution:

Question 3
Convert the percentage (RBSESolutions.com) into simple fraction and decimal fraction
(i) 36%
(ii) 12\(\frac { 1 }{ 2 }\)%
(iii) 3.6%
Solution:

(Page 166)

convert CGPA into percentage? Read this blog to understand an easy process to calculate percentage from CGPA.

Question 1
The population of a village is increased (RBSESolutions.com) to 15000 from 12000 in last 10 years. Then what is the percentage growth of population ?
Solution:
Population of village = 12000
Increased population of village in last 10 year = 15000
Increase in population = (15000 – 12000) = 3000

Question 2
Express rate of growth or loss in the (RBSESolutions.com) form of percentage in the following :
(1) The cost of electricity per unit has increased to ₹6.00 from ₹3.50
(2) The cost of 100 envelopes has decreased to ₹80 from ₹100.
Solution:
(1) Original rate per unit = ₹3.50
Increased rate per unit = ₹6
increase = 6 – 3.50 = ₹2.50

(2)100 envelopes cost = ₹100
New price = ₹80
Decrease in price = 100 – 80 = ₹20
Decrease % = \(\frac { 20 }{ 100 }\) x 100% = 20%

(Page 168)
Question 1
Mahaveer bought 5 bags of sugar for ₹16000. He (RBSESolutions.com) spent ₹200 for taxi fare, ₹120 for labour charges and ₹200 for transporttation charges. Find the selling price of per kg of sugar so as to earn a profit of ₹3 on each kg.
Solution:
Cost of 5 bags of sugar = ₹16000
Other expenses = ₹200 + ₹120 + ₹200
= ₹520
∴ Total C.P. of 5 bags of sugar
= ₹16000 + ₹520 = ₹16520
∵ Profit on 1 kg = ₹3
Profit on 5 bags or 500 kg sugar
= 500 x 3 = ₹1500
So, Profit = S.P – C.P
or S.P = C.P + Profit
= 16520+ 1500 = ₹18020
So, S.P. of 5 bags = ₹18020
S.P. of 1 bag = \(\frac { 1820 }{ 5 }\) = ₹3604

Question 2
Manoj bought a second hand car for ₹1,50,000. He spent ₹60,000 on Its (RBSESolutions.com) engine repair and ₹15000 on replacing the tyre tubes. Manoj sold this car to Jitendra for ₹2,10,000 Calculate loss or gain in this business.
Solution:
Cost of car = ₹1,50,000
Expense on engine = ₹60,000
Expense on tyre & tube = ₹15,000
Total cost price = ₹2,25,000
Selling price = ₹2,10,000
Manoj has a loss
So, Loss = C.P – S.P
= 2,25,000 – 2,10,000
Loss = ₹15,000
Manoj has a loss of ₹ 15,000

Percentage Difference calculator – online basic math function tool to calculate how much percent difference between two quantities or observations.

(Page 170)
Ashok has borrowed ₹50000/- for the construction (RBSESolutions.com) of his house. This borrowed amount is known as Principal. He repays ₹55000 to that institute after one year.
Ashok paid ₹5,000/- extra on ₹50,000. This extra amount is known as interest.
The amount of interest depends on following things:
1. Borrowed money (Principal)
2. Time (Period for which the amount is borrowed).
3. Rate (The extra amount paid on per hundred) which is determined on the basis of per mounth/per annum.
You can find the amount you have to pay at the end of the year by adding sum borrowed and the interest
i.e., Amount = Principal + Interest

Question 1
How much interest would Ashok have pay after 2 years if he (RBSESolutions.com) was not able to return the money to the instutute after 1 year ?
Solution:
Because after one year he pays ₹5000/-
as interest so in 2 years he has to pay ₹10000/- as interest.

Question 2
How much money in (RBSESolutions.com) total has to be paid including interest?
Solution:
Amount = Principal + Interest
= 50,000 + 10,000
= ₹60,000/-

We hope the RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities Exercise Additional Questions.

Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities Additional Questions

Multiple Choice Questions

Question 1
Value of 5% is :
(A) 0.05
(B) 0.5
(C) 5
(D) 0.005

the percentage difference calculated dividing the absolute values.

Question 2
Value of \(\frac { 1 }{ 2 }\) in (RBSESolutions.com) percentage is :
(A) 80 %
(B) 50 %
(C) 25 %
(D) 15 %

Question 3
Express \(\frac { 4 }{ 5 }\) in ratio is :
(A) 4 : 5
(B) 5 : 4
(C) 1 : 2
(D) 2 : 3

Question 4
\(\frac { 2 }{ 3 }\) is (RBSESolutions.com) equivalent to
(A) 66\(\frac { 2 }{ 3 }\)
(B) 65\(\frac { 2 }{ 3 }\)
(C) 62\(\frac { 2 }{ 3 }\)
(D) 60\(\frac { 2 }{ 3 }\)

Our percentage off calculator helps to calculate percent off by given number.

Question 5
Formula to find out profit percentage is :

Question 6
Formula to (RBSESolutions.com) find out % increase

Question 7
10 % of 50 will be :
(A)10
(B) 50
(C) 40
(D) 5

Question 8
What will be value of 75% in (RBSESolutions.com) decimal fractions?
(A) 73
(B) 7.3
(C) 0.73
(D) 0.073

Question 9
Mohan purchase a pen ₹10 and sold it in ₹12. Then profit will be :
(A) ₹2
(B) ₹3
(C) ₹4
(D) ₹6

Question 10
Suresh purchased a book in ₹120 and (RBSESolutions.com) sold it on 10% loss then its selling price is :
(A) ₹100
(B) ₹110
(C) ₹108
(D) ₹115

Answer:
(1) (A), (2) (B), (3) (A), (4) (A), (5) (B), (6) (A), (7) (D), (8) (C), (9) (A), (10) (C)

You can convert CGPA to percentage with the simple formula of mutiplying CGPA with 9.5.

Fill in the blanks
(i) 10% of 100 will be …………..
(ii) Ratio of 2 km to 400 km will be ……………
(iii) Fraction \(\frac { 12 }{ 16 }\) is in percentage (RBSESolutions.com) form will be …………
Answer:
(i) 10,
(ii) 1 : 200,
(iii) 75%

True/False
(i) The ratio of ₹5 and 50 paise is 10 : 1.
(ii) 75% of 12 is 12\(\frac { 3 }{ 4 }\)
(iii) 35 % + 65 % = 100 %.
(iv) Fraction \(\frac { 1 }{ 8 }\) is (RBSESolutions.com) equivalent 12.5%.
Answer:
(i) True
(ii) False
(iii) True
(iv) True

Very Short Answer Type Questions

Question 1
Collection of 10 chips of various (RBSESolutions.com) colours is as follows :

Fill the table and find out the percentage of every colour chips?
Solution:
The required percentage (RBSESolutions.com) of every colour chips is as follows :

Question 2
Find out following :
(a) 50% of 164
(b) 75% of 12
(c) 12\(\frac { 1 }{ 2 }\)% of 64
Solution:

Question 3
In the class of 25 children 8% children (RBSESolutions.com) like to wet in rain. Find out the number of children who like to wet in rain.
Solution:
Number of children who like to wet in rain = 8% of 25
= \(\frac { 8 }{ 100 }\) x 25 = 2

Question 4
9 is which number’s 25%?
Solution:
Let the number is x
25% of x = 9
⇒ \(\frac { 25 }{ 100 }\) × x = 9 ⇒\(\frac { 1 }{ 4 }\) × x = 9
⇒ x = 9 x 4 = 36
∴ Required number is 36.

Question 5
15 is which (RBSESolutions.com) number’s 75%?
Solution:
Let the number is x
75% of x = 15
⇒ \(\frac { 75 }{ 100 }\) x 1 = 15
\(\frac { 3 }{ 4 }\) x a = 15
⇒ 44 = 15 x \(\frac { 4 }{ 3 }\)
= 5 x 4 = 20
∴ Required number is 20.

Short Answer Type Questions

Question 1
In a computer lab 3 computers are required for 6 students. Find out (RBSESolutions.com) how many computers are required for 24 students?
Solution:
In a computer lab,
∵ 6 students required computers = 3
∴1 student required computers = \(\frac { 3 }{ 6 }\)
∴ 24 students required computers = \(\frac { 3 }{ 6 }\) x 24 = 3 x 4 = 12

Question 2
Population of Rajasthan is 570 lac and population (RBSESolutions.com) of Uttar Pradesh is 1660 lac. Area of Rajasthan is 3 Lac square km and area of Uttar Pradesh is 2 lac square km.
(i) In both states how many people are per square km.
(ii) Which state is having less population density?
Solution:
(i) Population of Rajasthan = 570 lac
Area of Rajasthan = 3 lac sq. km
∴ Number of people per square km = \(\frac { 570 }{ 3 }\) = 190
And population (RBSESolutions.com) of Uttar Pradesh = 1660 lac
Area of Uttar Pradesh = 2 lac sq. km
∴ Number of people per square km
= \(\frac { 1660 }{ 2 }\)
= 830
(ii) Because per square km less people are living in Rajasthan. So, Rajasthan is having less population density.

Question 3
Find out
(a) 15% of 250
(b) 1% of an hour
(c) 20% of 2500
(d) 75% of 1kg
Solution:
(a) 15% of 250 = \(\frac { 15 }{ 100 }\) x 250
= 15 x 2.5 = 37.5

(b) 1% of 1 hour = 1% of 60 minutes
= \(\frac { 1 }{ 100 }\) x 60 = \(\frac { 6 }{ 100 }\) minutes
= \(\frac { 6 }{ 10 }\) x 60 second = 36 second

(c) 20% of 2500 = \(\frac { 20 }{ 100 }\) x 2500
= 20 x 25 = 500

(d) 75% of 1 kg = 75% of 1000 gm
= \(\frac { 75 }{ 100 }\) x 1000 = 750 gm = 0.75 kg.

Long Answer Type Questions

Question 1
In chalk powder calcium, carbon and oxygen (RBSESolutions.com) ratio are 10 : 3 : 12. Find out the carbon percentage.
Solution:
Calcium : Carbon : Oxygen
10 : 3 : 12
Total of ratio = 10 + 3 + 12 = 25
Parts of carbon in chalk = \(\frac { 3 }{ 25 }\)
∴ Percentage of carbon in chalk = ( \(\frac { 3 }{ 25 }\) x 100)%
= (3 x 40)%
= 12%

Question 2
Amina purchase a book in ₹275 and sold it on 15 percent loss, then (RBSESolutions.com) find out the selling price of the book.
Solution:
C.P. of book = ₹275
Loss = 15%

Question 3
In every condition after 3 years how much amount (RBSESolutions.com) has to be paid if principal = ₹1200, rate = 12% P.A.
Solution:
Principal = ₹1200, Rate = 12%. Time = 3 years

Question 4
Arun purchased a car in ₹3,50,000 and next year its price (RBSESolutions.com) has increased up to ₹3,70,000. Find the percentage increase in the price of car.
Solution:
Original Price = ₹3,50,000
Increased Price = ₹3,70,000
Increment in Price = ₹(3,70,000 – 3,50,000)
= ₹20,000

Question 5
I have purchased a T.V. in ₹10,000 and taking 20% profit it has (RBSESolutions.com) been sold out. How much money I received on selling of this ?
Solution:
C.P. = ₹10,000, Profit = 20%

Question 6
On selling a washing machine in ₹13500, Juhi bears a loss of 20%. What was (RBSESolutions.com) the cost price of washing machine?
Solution:
S.P. = ₹13,500, loss = 20%

∴ C.P. of washing machine = ₹16,875

We hope the RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 7 Maths Chapter 4 Rational Numbers Ex 4.1 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 4 Rational Numbers Exercise 4.1.

Rajasthan Board RBSE Class 7 Maths Chapter 4 Rational Numbers Ex 4.1

Welcome to our step-by-step math rational expressions calculator.

Question 1
Write five rational numbers equivalent to (RBSESolutions.com) the following ratinal numbers

Solution:
(i) Five equivalent rational numbers of –\(\frac { 2 }{ 3 }\) are

(ii) Five equivalent rational (RBSESolutions.com) numbers of \(\frac { 1 }{ 5 }\) are

(iii) Five equivalent rational (RBSESolutions.com) numbers of \(\frac { -5 }{ 3 }\) are

(iv) Five equivalent rational (RBSESolutions.com) numbers of \(\frac { 4 }{ -9 }\) are

.

Question 2
Write three such (RBSESolutions.com) rational numbers equivalent to \(\frac { -5 }{ 12 }\) in which the denominator is 60, – 96 and 108.
Solution:

Question 3
Write three such rational (RBSESolutions.com) numbers equivalent to \(\frac { -3 }{ 7 }\) in which the numerator is 24, – 60 and 75.
Solution:

Question 4
Write the following rational numbers in the (RBSESolutions.com) simplest from (standard forms) :

Solution:

Question 5
Represent the following rational numbers (RBSESolutions.com) on number line :

Solution:
(i) We know that \(\frac { 3 }{ 5 }\) is greater than 0, so \(\frac { 3 }{ 5 }\) will be in between 0 and 1.
Now to represent \(\frac { 3 }{ 5 }\) on number line, the distance between 0 and S is divided in 5 equal parts and take three parts from it to mark point P. This way point P represents point \(\frac { 3 }{ 5 }\) .

(ii) We know (RBSESolutions.com) that \(\frac { 7 }{ 8 }\) is greater than 0 and less than 1.
∴ \(\frac { 7 }{ 8 }\) will be in between 0 and 1.
Now to represent \(\frac { 7 }{ 8 }\) on number line, distance between 0 and 1 is divided in 8 equal parts and take 7 parts from it to mark point P. Thus point P represent point \(\frac { 7 }{ 8 }\).

(iii) We know that \(\frac { -8 }{ 3 }\) or -2\(\frac { 2 }{ 3 }\), is greater than -3 and less than – 2 so \(\frac { -8 }{ 3 }\), will be in between -2 and – 3.
Now on number line, the (RBSESolutions.com) distance between – 2 and – 3 divided in 3 equal parts and take 2 parts from it to mark point P. This way point P represents point \(\frac { -8 }{ 3 }\).

(iv) We know that -2\(\frac { 1 }{ 2 }\) or \(\frac { -5 }{ 2 }\), is greater from – 3 and less than – 2 so \(\frac { -5 }{ 2 }\) will be in between – 2 and -3.
Now to represent \(\frac { -5 }{ 2 }\) on number line, the distance between -2 and -3 is divided in 2 equal parts and take one part from it to mark point P.
This way point P represent (RBSESolutions.com) the point –\(\frac { 5 }{ 2 }\) ( -2\(\frac { 1 }{ 2 }\)) or number line.

(v) We know that \(\frac { 5 }{ 7 }\) is greater than 0 and less than 1.
\(\frac { 5 }{ 7 }\), will be in between 1 and 1.
Now to represent \(\frac { 5 }{ 7 }\) on number line, the distance between 0 and 1 is divided in 7 equal parts and take 5 parts to make point P. This way point P

represents \(\frac { 5 }{ 7 }\) on number line.

Question 6
Choose the correct (RBSESolutions.com) sign from <, >, = and fill in the blanks :

Solution:
(i) We know that positive rational numbers are greater than negative rational numbers

(ii) L.C.M. of (RBSESolutions.com) denominator 4 and 3 is 12

(iii) L.C.M. of (RBSESolutions.com) denominator – 5 and 3 is 15

(iv) L.C.M. of 7 and 2 is 14

(vi) We know that negative rational (RBSESolutions.com) numbers are smaller than positive rational number.

Question 7
Write five rational numbers between fol-lowing rational numbers :

Solution:

We know that integers (RBSESolutions.com) between -9 and -3 are
⇒ -8 < -7 < -6 < -5 < -4

∴ Five rational numbers between – 9 and – 3 will be:

We know that integers (RBSESolutions.com) between 0 and -6 are

∴ Five rational numbers between 0 and -1

∴ Five rational numbers (RBSESolutions.com) between -56 and – 50 are

We know that integers between 12 and 6 are :

∴ Five rational (RBSESolutions.com) numbers between \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 4 }\) are

(v) We know that integers between 2 and -4 are:
-4 < -3 < -2 < -1 < 0 < 1 < 2
∴ Five rational numbers between \(\frac { 2 }{ 5 }\) and \(\frac { -4 }{ 5 }\) will be

Question 8
Write three more rational (RBSESolutions.com) numbers in each of the following :

Solution:

Question 9
Write the (RBSESolutions.com) following rational number in increasing order :

Solution:
(i) ∵ Denominators of given rational numbers are positive.
∴ L.C.M of 2, 2, 4 and 4 = 4 on making denominator of each = 4

∵ Denominators of (RBSESolutions.com) rational numbers are positive
∴ L.C.M. of 2, 4 and 7 = 28 on making the denominator of each = 28

∴ Increasing order is

∵ Denominator of rational (RBSESolutions.com) numbers are positive.
∴ L.C.M of 11, 15, 1, 1 and 15 = 165
on making denominator of each = 165

∵ Denominator of rational number are positive
∴ L.C.M of 5, 7, 6, 9 = 630 on making (RBSESolutions.com) denominator of each = 165

Question 10
Write the following rational (RBSESolutions.com) numbers is decreasing order :

Solution:

on making (RBSESolutions.com) the denominator 48 of each

∵ Denominator of rational number (RBSESolutions.com) are positive
∴ L.C.M of 6, 6, 9, 12 = 36
on making denominator 36 of each


on making (RBSESolutions.com) denominator 6 of each

on making (RBSESolutions.com) denominator 30 of each

We hope the RBSE Solutions for Class 7 Maths Chapter 4 Rational Numbers Ex 4.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 4 Rational Numbers Exercise 4.1, drop a comment below and we will get back to you at the earliest.