Author name: Prasanna

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling InText Questions

NCERT In-text Question Page No. 59
Question 1.
Weigh (in kg) at least 20 children (girls and boys) of your class. Organise the date, and answer the following questions using this data.
(i) Who is the heaviest of all?
(ii) What is the most common weight?
(iii) What is the difference between your weight and that of your best friend?
Answer:
We assume that weights (in kg) of 20 children of the class are as follows:
27, 30, 28, 24, 25, 26, 21, 20, 18, 15,
28, 16, 27, 23, 24, 26, 22, 31, 18, 26
Organising the data in ascending order, we have.
15, 16, 18, 18, 20, 21, 22, 23, 24, 24,
25, 26, 26, 27, 27, 28,28, 30,31

(i) The child having weight 31 kg is the heaviest of all.
(ii) The most common weight is 26 Kg.
(iii) If my weight is 27, then required difference = 27 kg – 25 Kg = 2 Kg.

NCERT In-text Question Page No. 61
Question 1.
How would you find the average of your study hours for the whole week?
Answer:
In a week there are 7 days. The number of study hours everyday are different. Let the study hours for different days of the week are as follows:
Monday : 6 hours
Tuesday : 7 hours
Wednesday : 6 hours
Thursday : 8 hours
Friday : 8 hours
Saturday : 7 hours
Sunday : 7 hours
Total number of study hours.
= 6 + 7 + 6 + 8 + 8 + 7 + 7 = 40 hours
Number of total days = 7
∴ Arithmetic mean = \(\frac{\text { Total study hours }}{\text { Total Number of days }}\)
Thus, average study hours = 7 hours per day.

NCERT In-text Question Page No. 61
Question 1.
Find the mean of your sleeping hours daring one week?
Answer:
Let the sleeping hours be as follows:
Monday : 8 hours
Tuesday : 9 hours
Wednesday : 11 hours
Thursday : 8 hours
Friday : 10 hours
Saturday : 9 hours
Sunday : 8 hours
Sum of sleeping hours
= 8 + 9+11 + 8+10 + 9 + 8 = 63
Number of days = 7
∴ Arithmetic mean = \(\frac{\text { Sum of sleeping hours }}{\text { Number of days }}\)
= \(\frac{63}{7}\) hours = 9hours

Question 2.
Find at least 5 numbers between \(\frac{1}{2}\) and \(\frac{1}{3}\).
Answer:
The arithmetic mean of two given num¬bers is average between them

NCERT In-text Question Page No. 65
Question 1.
Find the mode of?
(i) 2, 6, 5, 3, 0, 3,4, 3, 2, 4, 5, 2, 4
(ii) 2, 14,16,12,14,14, 16, 14,10, 14,18,14
Answer:
(i) Given set of data is :
2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
Writing the numbers with same value together, we have:
0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6
∵ 2,3, and 4 occur in the given data highest times (3 times).
∴ Modes of the set of data are 2, 3 and 4.

(ii) Given set of data is :
2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Writing the numbers with same value together, we have:
2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18
∵ 14 occurs highest times (i.e. six times).
∴ Required mode is 14.

NCERT In-text Question Page No. 65-66
Question 1.
Find the mode of the following data.
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14
Answer:
Here the given set of data has large number of observations so, we put them in a table
Number Tally Marks Frequency

The highest frequency is 10, corresponding value is 15.
The required mode is 15.

Question 2.
Heights (in cm) of 25 children are given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 164, 163, 160, 165, 163, 162
What is the mode of their heights? What do we understand by mode here?
Answer:
Putting the data ina tabular form, we have

Since, 163 is occuring the maximum number of times.
∴ Mode is 163 cm.
By the mode, we mean that most of stu¬dents have height 163 cm.

Note :
Mode is modal value of a group which occurs most frequently in the data whereas mean gives us the average of all observations of the data.

NCERT In-text Question Page No. 66
Question 1.
Discuss with your friends and give:
(a) Two situations where mean would be an appropriate representative value to use, and
(b) Two situations where mode would be an appropriate representative value to use.
Answer:
(a) (i) For measures of heights of all students of a class, and
(ii) For weights of wheat stored in similar containers, the mean would give us a better picture of the data.

(b) (i) If we are collecting the shoe size suitable for a particular age, and
(ii) A shopkeeper selling shirts needs to know suitable size to enrich the stock, then mode is more useful.

NCERT In-text Question Page No. 67
Question 1.
Your friend found the median and the mode of a given data. Describe and correct your friend’s error if any:
35, 32, 35, 42, 38, 32, 34
Median = 42, Mode = 32
Answer:
Set of data is: 35, 32, 35, 42, 38, 32, 34
Putting the given data in ascending order, we have:
32, 32, 34, 35, 35, 38, 42

(i)∵ The middle value of the data is 35.
∴ Median = 35

(ii)∵ The observations occuring maximum number of times are 32 and 35.
∴ The correct modes are 32 and 35.

NCERT In-text Question Page No. 71 & 72
Question 1.
The bar graph given below shows the result of a survey to test water resistant watches made by different companies.
Each of these companies claimed that their watches were water resistant. After a test the above results were revealed.
(a) Can you work out a fraction of the number of watches that leaked to the number tested for each company?
(b) Could you tell on this basis which company has better watches?

Answer:
(a) From the graph, we have :
Total number of watches tested for each company = 40
Fraction of ‘the numbers of watches that leaked’ to ‘the number of watches tested’ 20 1
For company A = \(\frac{20}{40}=\frac{1}{2}\)
For company B = \(\frac{10}{40}=\frac{1}{4}\)
For company C = \(\frac{15}{40}=\frac{3}{8}\)
For company D = \(\frac{25}{40}=\frac{5}{8}\)
(b) Obviously company B has better watches
∵ \(\frac { 1 }{ 4 }\) < \(\frac { 3 }{ 8 }\) > \(\frac { 1 }{ 2 }\) < \(\frac { 5 }{ 8 }\)

Question 2.
Sale of English and Hindi books in the years 1995, 1996, 1997 and 1998 are given below:

Draw a double bar graph and answer the following questions:
(a) In which year was the difference in the sale of the two language books least?
(b) Can you say that the demand for English books rose faster? Justify.
Answer:
(a) The difference in the sale of the two language books:
In 1995 is 500 – 350 = 150
In 1996 is 525 – 400 = 125
In 1997 is 600 – 450 = 150
In 1998 is 650 – 620 = 30
The difference in the scale of the two language books is least in the year 1998.

(b) Yes we can say that the demand of English books rose faster, during 1995 to 1998
(i) Sale of Hindi books rose from 500 to 650 i.e. it rose by 650 – 500 = 150
(ii) Sale of English books rose from 350 to 620 i.e. it rose by 620 – 350 = 270
270 is much greater than 150

NCERT In-text Question Page No. 74
Question 1.
Think of some situations, atleast 3 examples of each, that are certain to happen, some that are impossible and some that may or may not happen i.e., situations that have some chance of happening.
Answer:
(i) Situations that are certain to happen
(a) The sun rises in the east.
(b) Getting a number from 1 to 6 by throwing a dice.
(c) When a coin is tossed, then either a head or tail comes.

(ii) Situations that are impossible:
(a) Getting the number 8 by throwing a dice.
(b) To draw a five rupee coin from a bag containing one rupee coins.
(c) Drawing a red ball from a bag containing yellow and blue balls only.

(iii) Situations that have some chance of happening
(a) An ant rising to 5m height.
(b) To throw a dice and get an odd number.
(c) To toss a coin and get tail.

Note:
(i) A coin has two faces. The face with Ashoka Chakra is called ‘Head’ and other is called ‘Tail’.
(ii) A cube with six faces is called a dice. Its each face has making of dots numbering from 1 to 6.

NCERT In-text Question Page No. 75
Question 1.
Toss a coin 100 times and record the data. Find the number of times heads and tails occur in it.
Answer:
This is a group activity. Please do it yourself.

Question 2.
Aftaab threw a die 250 times and got the following table. Draw a bar graph for this data.
Answer:

NCERT In-text Question Page No. 76
Question 1.
Construct or think of five situations where outcomes do not have equal chances.
Answer:
In the following situations, the outcomes do not have equal chances:
(i) To draw red ball and to draw a yellow ball from a bag having 3 red balls and 5 yellow balls.
(ii) To throw a dice are getting 2 and getting an odd number.
(iii) To draw a flash card bearing ‘B’ and to draw a flash card bearing ‘C’ from a collection of 5 flash cards bearing A, B, C, D and E.
(iv) Choosing a boy and choosing a girl as class representative from a group of 10 boys and 15 girls.
(v) Choosing an even number and choosing an odd number from first five natural numbers.

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of number m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Answer:
(i) y – z
(ii) \(\frac { 1 }{ 2 }\)(x + y)
(iii) z²
(iv) \(\frac { 1 }{ 4 }\)pq
(v) x² + y²
(vi) 5 + 3mn
(vii) 10 – yz
(viii) ab – (a + b)

Question 2.
(i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x²
(c) y – y³
(d) 5xy² + 7x²y
(e) – ab + 2b² – 3a²

(ii) Identify terms and factors in the expressions given below:
(a) – 4x + 5
(b) – 4x + 5y
(c) 5y + 3y²
(d) xy + 2x²y²
(e) pq + q
(f) 1.2ab- 2.4 b + 3.6a
(g) \(\frac { 3 }{ 4 }\)x + \(\frac { 1 }{ 4 }\)
(h) 0.1p² + 0.2p²
Answer:
(i) (a) x – 3
Terms : x; – 3 Factors x; -3
1 + x + x²

(b) 1 + x + x²
Terms = 1; x; x²
Factors = 1; x; x, x (x² )

(c) y-y³
Terms: y; -y³
Factors y;-1,y,y,y

(d) 5xy² + 7x² y
Terms: 5xy²; 7x² y
Factors: 5, x, y, y; 7, x, x, y

(e) -ab + 2b² – 3a²
Terms: -ab; 2b², -3a²
Factors: -1, a, b; 2; b; b; 3, a, a

(ii)

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 – 3t²
(ii) 1 + t + t² + t³
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) -p²q² + 7pq
(vi) 1.2a + 0.8b
(vii) 3.14r²
(viii) 2(l + b)
(ix) 0.1y + 0.01y²
Answer:

Question 4.
(a) Identify terms which contain x and give the coefficient of x.
(i) y²x + y
(ii) 13y² – 8yx
(iii) x + y + 2
(iv) 5 + z + zx
(iv) 1+x + xy
(vi) 12xy² + 25
(vii) 7x + xy²

(b) Identify terms which contain y² and give the coefficient of y².
(i) 8-xy²
(ii) 5y² + 7x
(iii) 2x²y – 15xy² + 7y²
Answer:

(b)

Question 5.
Classify into monomials, binomials and trinomials.
(i) 4y – 7z
(ii) y²
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p²q – 4pq²
(viii) 7mn
(ix) z² – 3z + 8
(x) a² + b²
(xi) z² + z
(xii) 1 + x + x²
Answer:
(i) 4y – 7z
The expression 4y – 7z is having two unlike terms (4y and – 7z)
∴ It is a binomial.

(ii) y²
The expression y² is having only one term (y²)
∴ It is a monomial.

(iii) x + y – xy
The expression x + y – xy is having three terms (x, y and – xy)
∴ It is a trinomial.

(iv) 100
The expression 100 is having only one term (100)
∴ It is a monomial.

(v) ab – a – b.
The expression ab – a – b is having three terms (ab, -a, and -b)
∴ The expression is a trinomial.

(vi) 5 – 3t
The expression 5 – 3t is having two terms (5 and -3t)
∴ It is a binomial expression.

(vii) 4p²q – 4pq²
The expression 4p²q – 4pq² is having
two unlike terms (4p²q and – 4pq²)
∴ The expression is a binomial.

(viii) 7mn
The expression 7mn is having only one term (ie 7mn)
∴ The expression is a monomial.

(ix) z²– 3z + 8
The expression z² – 3z + 8 is having three terms (ie z², – 3z and 8)
∴ The expression is a trinomial.

(x) a² + b²
The expression (a² + b²) is having two unlike terms (a² and b²)
∴ It is a binomial expression.

(xi) z² + z
The expression z² + z is having two unlike terms (z2 and z)
∴ The expression in binomial.

(xii) 1 + x + x²
The expression 1 + x + x² is having three terms (1, x and x²)
∴ The expression is a trinomial.

Question 6.
State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
(ii) -7x, \(\frac { 5 }{ 2 }\)x
(iii) -29x, -29y
(iv) 14xy, 42yx
(v) 4m²p, 4mp²
(vi) 12xz, 12x²z²
Answer:
(i) 1, 100 is a pair of like terms.
(ii) \(\frac { 5 }{ 2 }\)x is a pair of like terms.
(iii) – 29x, – 29y is a pair of unlike terms.
(iv) 14xy, 4²yx is a pair of like terms.
(v) 4m²p, 4mp² is a pair of unlike terms.
(vi) 12xz; 12x²z² is a pair of unlike terms.

Question 7.
Identify like terms in the following:
(a) -xy², – 4yx², 8x², 2xy², 7y, -11x², 100x, -11yx, 20x²y, – 6x², y, 2xy, 3x
(b) 10pq, 7p, 8q, – p²q², -7pq, -100q, -23, 12q²p², -5p², 41, 2405p, 78qp,
13p²q, qp²,701p²
Answer:
(a) -xy² and 2xy², – 4yx² and 20x² y, 8x², -11x² and – 6x²; 7y and y; lOOx and 3x; -1 lyx and 2xy are like terms.
(b) 10pq, – 7pq and 78qp.
7p and 2405p; 8q and 100q; – p²q² and 12q²p²; – 23 and 41; – 5p² and 701p²; 13p²q and qp² are like terms.

These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1

Question 1.
Given here are some figures.

Classify each of them on the basis of the following.
(a) Simple curve
(b) Simple closed curve
(c) Polygon
(d) Convex polygon
(e) Concave polygon
Solution:
(a) Simple curves are: (1), (2), (5) (6) and (7)
(b) Simple closed curves are: (1), (2), (5), (6) and (7) and (4)
(c) Polygons are: (1) and (2)
(d) Convex polygon is (2)
(e) Concave polygon is (1) and (4)

Question 2.
How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon
(c) A triangle
Solution:
Note: Number of diagonals in a polygon having n sides
n-side = \(\left[\frac{n(n-1)}{2}-n\right]\)
(a) In quadrilateral, number of sides (n) = 4
Number of diagonals = 2
\(\left[\frac{4(4-1)}{2}-4=\frac{4 \times 3}{2}-4=6-4=2\right]\)

(b) In a regular hexagon number of sides, (n) = 6
\(\left[\frac{6(6-1)}{2}-6=\frac{6 \times 5}{2}-6=15-6=9\right]\)

(c) In a triangle, number of sides (n) = 3
∴ Number of diagonals = \(\frac{n(n-1)}{2}-n\)
= \(\frac{3(3-1)}{2}\) – 3
= \(\frac{3 \times 2}{2}\) – 3
= 3 – 3
= 0

Question 3.
What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)
Solution:
The sum of the measures of the angles of a convex quadrilateral is 360°
Yes, this property holds, even if the quad-rilateral is not convex.

Question 4.
Examine the table.
Each (figure is divided into triangles and the sum of the angles deduced from that).

What can you say about the angle sum of a convex polygon with number of sides?
(a) 7
(b) 8
(c) 10
(d) n
Solution:
(a) When n = 7
Sum of interior angles of a polygon = (7 – 2) × 180°
= 5 × 180°
= 900°

(b) When n = 8
Sum of interior angles of a polygon = (8 – 2) × 180°
= 6 × 180°
= 1080°

(c) When n = 10
Sum of interior angles of a polygon having 10 sides = (10 – 2) × 180°
= 8 × 180°
= 1440°

(d) When n = n
Sum of interior angles of a polygon = (n – 2) × 180°

Question 5.
What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides
(ii) 4 sides
(iii) 6 sides
Solution:
A polygon is said to be a regular polygon if
(a) The measures of its interior angles are equal
(b) The length of its sides are equal
The name of a regular polygon having
(i) 3 sides is ‘equilateral triangle’
(ii) 4 sides is ‘square’.
(iii) 6 sides is ‘regular hexagon’.

Question 6.
Find the angle measure x in the following figures.

Solution:
(a) The sum of interior angles of a quadrilateral is 360°
∴ x + 120° + 130° + 50° = 360°
⇒ x + 300° = 360°
⇒ x = 360° – 300°
⇒ x = 60°

(b) The sum of the interior angles of a quadrilateral is 360°
∴ x + 70° + 60° + 90° = 360°
⇒ x + 220° = 360°
⇒ x = 360° – 220°
⇒ x = 140°

(c) Interior angles are 30°, x°, (180 – 60°), (180° – 70°) and x°
i.e., 30°, x°, 120°, 110° and x°
The given figure is a pentagon.
Sum of interior angles of a pentagon = 540°
∴ 30° + x ° + 120° + 110° + x = 540°
⇒ 2x° + 260° = 540°
⇒ 2x° = 540° – 260°
⇒ 2x°= 280°
⇒ x = \(\frac{280^{\circ}}{2}\)
⇒ x = 140°
The measure of x is 140°.

(d) It is a regular pentagon.
Sum of all the interior angles of regular pentagon = 540°
It’s each angle is equal to x°.
x° + x° + x° + x° + x° = 540°
⇒ 5x° = 540°
⇒ x = \(\frac{540^{\circ}}{5}\)
⇒ x = 108°
The measure of x is 108°.

Question 7.

(a) Find x + y + z

(b) Find x + y + z + w
Solution:
(a) x + 90° = 180° (Linear Pair)
x = 180° – 90° = 90°
y = 30° + 90° [∵ exterior angle of a triangle is equal to the sum of interior opposite angles]
⇒ y = 120°
z + 30° = 180° (Linear pair)
⇒ z = 180° – 30° = 150°
∴ x + y + z = 90° + 120° + 150° = 360°

(b) The sum of interior angles of a quadrilateral = 360°
⇒ ∠1 + 120° + 80° + 60° = 360°
⇒ ∠1 + 260° = 360°
⇒ ∠1 = 360° – 260° = 100°
Now, x + 120° = 180° (linear pair)
x = 180° – 120° = 60°
y + 80° = 180° (Linear pair)
⇒ y = 180° – 80° = 100°
z + 60° = 180° (Linear pair)
⇒ z = 180° – 60°
⇒ z = 120°
w + ∠1 = 180° (Linear Pair)
⇒ w + 100°= 180°
⇒ w = 180° – 100° = 80°
Thus x + y + z + w = 60° + 100° + 120° + 80° = 360°

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6

Solve the following equations.

Question 1.
\(\frac{8 x-3}{3 x}=2\)
Solution:
\(\frac{8 x-3}{3 x}=2\)
By cross-multiplication we get,
8x – 3 = 2(3x)
8x – 3 = 6x
By transposing -3 to the R.H.S. and 6x to L.H.S.
8x – 6x = 3
2x = 3
Dividing both sides by 2
\(\frac{2 x}{2}=\frac{3}{2}\)
∴ x = \(\frac{3}{2}\)

Question 2.
\(\frac{9 x}{7-6 x}=15\)
Solution:
\(\frac{9 x}{7-6 x}=15\)
By cross-multiplication, we get
9x = 15(7 – 6x)
9x = 105 – 90x
By transposing -90x to L.H.S., we get
9x + 90x = 105
99x = 105
Dividing both sides by 99, we get
\(\frac{99 \mathrm{x}}{99}=\frac{105}{99}\)
∴ x = \(\frac{35}{33}\)

Question 3.
\(\frac{z}{z+15}=\frac{4}{9}\)
Solution:
\(\frac{z}{z+15}=\frac{4}{9}\)
By cross-multiplication, we get
9z = 4(z + 15)
9z = 4z + 60
Transposing 4z to L.H.S, we get
9z – 4z = 60
5z = 60
Dividing both sides by 5, we get
\(\frac{5 z}{5}=\frac{60}{5}\)
∴ z = 12

Question 4.
\(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)
Solution:
\(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)
By cross-multiplication, we get
5(3y + 4) = -2(2 – 6y)
15y + 20 = -4 + 12y
Transposing 12y to L.H.S. and 20 to R.H.S.
15y – 12y = -4 – 20
3y = -24
Dividing both sides by 3, we get
\(\frac{3 y}{3}=-\frac{24}{3}\)
∴ y = -8

Question 5.
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
Solution:
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
By cross-multiplication, we get
3(7y + 4) = -4(y + 2)
21y + 12 = -4y – 8
Transposing +12 to R.H.S. and -4y to L.H.S.
21y + 4y = -8 – 12
25y = -20
Dividing both sides by 25, we get
\(\frac{25 y}{25}=\frac{-20}{25}\)
∴ y = \(-\frac{4}{5}\)

Question 6.
The ages of Hari and Harry are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. Find their present ages.
Solution:
Let the present age of Hari = 5x years
and the present age of Harry = 7x years
After four years
Age of Hari = (5x + 4) years
Age of Harry = (7x + 4) years
According to the question
(5x + 4) : (7x + 4) = 3 : 4
4(5x + 4) = 3(7x + 4)
[Product of the extremes is equal to the product of the means]
20x + 16 = 21x + 12
Transposing 16 to R.H.S. and 21x to L.H.S., we get
20x – 21x = 12 – 16
-x = -4
∴ x = 4
Present age of Hari = 5 × 4 = 20 years
Present age of Harry = 7 × 4 = 28 years

Question 7.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \(\frac{3}{2}\). Find the rational number.
Solution:
Let the numerator be ‘x’
then the denominator is x + 8
∴ The fraction is \(\frac{\mathrm{x}}{\mathrm{x}+8}\)
According to the question, we get
\(\frac{x+17}{(x+8)-1}=\frac{3}{2}\)
\(\frac{x+17}{x+8-1}=\frac{3}{2}\)
\(\frac{x+17}{x+7}=\frac{3}{2}\)
By cross-multiplication, we get
3(x + 7) = 2(x + 17)
3x + 21 = 2x + 34
By transposing 21 to R.H.S. and 2x to L.H.S., we get
3x – 2x = 34 – 21
x = 13
∴ The fraction is \(\frac{13}{13+8}\) = \(\frac{13}{21}\)
or The rational number = \(\frac{13}{21}\)

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5

Solve the following linear equations.

Question 1.
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Transposing \(-\frac{1}{5}\) to R.H.S. and \(\frac{\mathrm{x}}{3}\) to L.H.S
\(\frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5}\)
\(\frac{3 x-2 x}{6}=\frac{5+4}{20}\)
\(\frac{x}{6}=\frac{9}{20}\)
By cross-multiplication, we get
20(x) = 9 × 6
20x = 54
Dividing both sides by 20
\(\frac{20 \mathrm{x}}{20}=\frac{54}{20}\)
x = \(\frac{27}{10}\)

Question 2.
\(\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21\)
Solution:
\(\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21\)
L.C.M of 2, 4 and 6 = 12
Multiplying each term by 12, we get
\(12 \times \frac{\mathrm{n}}{2}-12 \times \frac{3 \mathrm{n}}{4}+12 \times \frac{5 \mathrm{n}}{6}=21 \times 12\)
6n – 9n + 10n = 252
7n = 252
Dividing both sides by 7
\(\frac{7 \mathrm{n}}{7}=\frac{252}{7}\)
n = 36

Question 3.
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Transposing 7 to R.H.S. and \(\frac{-5 x}{2}\) to L.H.S
\(x-\frac{8 x}{3}+\frac{5 x}{2}=\frac{17}{6}-7\)
L.C.M of 3, 2 and 6 = 6
Multiplying each term by 6, we get
\(6 \times x-6 \times \frac{8 x}{3}+6 \times \frac{5 x}{2}=6 \times \frac{17}{6}-6 \times 7\)
6x – 16x + 15x = 17 – 42
5x = -25
Dividing both sides by 5
\(\frac{5 x}{5}=-\frac{25}{5}\)
x = -5

Question 4.
\(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:
\(\frac{x-5}{3}=\frac{x-3}{5}\)
By cross-multiplication we get
5(x – 5) = 3(x – 3)
5x – 25 = 3x – 9
Transposing -25 to R.H.S and 3x to L.H.S.
5x – 3x = 25 – 9
2x = 16
Dividing both sides by 2
\(\frac{2 \mathrm{x}}{2}\) = \(\frac{16}{2}\)
x = 8

Question 5.
\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}\)
Solution:
\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}\)
Transposing -t to L.H.S., we get
\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}+\mathrm{t}=\frac{2}{3}\)
L.C.M of 4 and 3 is 12
Multiplying each term by 12
\(12 \frac{(3 \mathrm{t}-2)}{4}-12 \frac{(2 \mathrm{t}+3)}{3}+12 \mathrm{t}=12 \times \frac{2}{3}\)
3(3t – 2) – 4(2t + 3) + 12t = 8
9t – 6 – 8t – 12 + 12t = 8
13t – 18 = 8
Transposing -18 to R.H.S., we get
13t = 8 + 18
13t = 26
Dividing both sides by 13
\(\frac{13 t}{13}=\frac{26}{13}\)
t = 2

Question 6.
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Solution:
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Transposing \(\left(\frac{m-2}{3}\right)\) to L.H.S. we get
\(\mathrm{m}-\frac{\mathrm{m}-1}{2}+\frac{\mathrm{m}-2}{3}=1\)
L.C.M of 2 and 3 is 6
Multiplying each term by 6, we get
6m – \(\frac{6(m-1)}{2}+\frac{6(m-2)}{3}\) = 1 × 6
6m – 3(m – 1) + 2(m – 2) = 6
6m – 3m + 3 + 2m – 4 = 6
5m – 1 = 6
Transposing -1 to the R.H.S., we get
5m = 6 + 1
5m = 7
Dividing both sides by 5, we get
\(\frac{5 m}{5}=\frac{7}{5}\)
m = \(\frac{7}{5}\)

Simplify and solve the following linear equations.

Question 7.
3(t – 3) = 5 (2t + 1)
Solution:
3(t – 3) = 5 (2t + 1)
Opening the brackets, we get
3t – 9 = 10t + 5
Transposing -9 to R.H.S. and 10t to L.H.S.
3t – 10t = 5 + 9
-7t = 14
Dividing both sides by -7, we get
\(\frac{-7 t}{-7}=\frac{14}{-7}\)
t = -2

Question 8.
15(y – 4) – 2(y – 9) + 5 (y + 6) = 0
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Opening the brackets, we get
15y – 60 – 2y + 18 + 5y + 30 = 0
15y – 2y + 5y – 60 + 18 + 30 = 0
18y – 12 = 0
Transposing -12 to R.H.S.
18y = 12
Dividing both sides by 18, we get.
\(\frac{18 y}{18}=\frac{12}{18}\)
y = \(\frac{2}{3}\)

Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Opening the brackets, we get
15z – 21 – 18z + 22 = 32z – 52 – 17
15z – 18z + 22 – 21 = 32z – 69
-3z + 1 = 32z – 69
Transposing 1 to R.H.S. and 32z to L.H.S.
-3z – 32z = – 69 – 1
-35z = -70
Dividing both sides by -35, we get
\(\frac{-35 z}{-35}=\frac{-70}{-35}\)
z = 2

Question 10.
0.25(4f – 3) = 0.05(10f – 9)
Solution:
0.25(4f – 3) = 0.05(10f – 9)
Opening the brackets, we get
0.25(4f) – 0.25 × 3 = 0.05 × 10f – 0.05 × 9
f – 0.75 = 0.5f – 0.45
f – 0.5f = 0.75 – 0.45
0.5f = 0.3
Dividing both sides by 0.5, we get
\(\frac{0.5 \mathrm{f}}{0.5}=\frac{0.3}{0.5}\)
f = \(\frac{3}{5}\) = 0.6
f = 0.6

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.4

Question 1.
Tell whether the following is certain to happen, impossible, can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
Answer:
The chances are
(i) It is certain to happen.
(ii) It can happen but not certain.
(iii) It is impossible.
(iv) It can happen but not certain.
(v) It can happen but not certain.

Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with number 5?
Answer:
Total number of marbles = 6
Total number of possible outcomes = 6
(i) Probability of drawing marble with number 2
p (2) = \(\frac{1}{6}\)
(ii) Probability of drawing marble with number 5
p (5) = \(\frac{1}{6}\)

Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Answer:
Since there are two faces of a coin head (H) and tail (T), when a coin is tossed, then either a ‘H’ or a ‘T’ comes.
successfully achieved outcome = 1
Total number of possible outcomes= 2
[ Head + Tail = 1 + 1 = 2]
Required probability = \(\frac{1}{2}\)