CBSE Class 8

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

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals InText Questions

Try These (Page No. 43)

Take a regular hexagon Fig 3.10.

Question 1.
What is the sum of the measures of its exterior angles x, y, z, p, q, r?
Solution:
∠x + ∠y + ∠x + ∠z + ∠p + ∠q + ∠r = 360°
[∵ Sum of exterior angles of a polygon = 360°]

Question 2.
Is x = y = z = p = q = r? Why?
Solution:
Since, all the sides of the polygon are equal.
∴ It is a regular hexagon.
So, its interior angles are equal.
∴ x = (180° – a)
y = (180° – a)
z = (180° – a)
p = (180° – a)
q = (180° – a)
r = (180° – a)
∴ x = y = z = p = q = r

Question 3.
What is the measure of each?
(i) exterior angle
(ii) interior angle
Solution:
(i) x + y + z = p + q = r = 360° [sum of exterior angles = 360°]
and all these angles are equal
∴ Measure of each exterior angles = \(\frac{360^{\circ}}{6}\) = 60°

(ii) ∵ Exterior angle = 60°
∴ 180° – 60° = Interior angle
or 120° = Interior angle
or Measure of interior angle = 120°

Question 4.
Repeat this activity for the cases of
(i) a regular octagon
(ii) a regular 20-gon
Solution:
(i) In a regular octagon, number of sides (n) = 8
∴ Each exterior angle = \(\frac{360^{\circ}}{8}\) = 45°
∴ Each interior angle = 180° – 45° = 135°

(ii) For a regular 20-gon, the number of sides (n) = 20
∴ Each exterior angle = \(\frac{360^{\circ}}{20}\) = 18°
Thus, each interior angle = 180° – 18° = 162°

Try These (Page No. 47)

Question 5.
Take two identical set squares with angles 30° – 60° – 90° and place them adjacently to form a parallelogram as shown in Fig 3.20. Does this help you to verify the above property?

Solution:
The given figure helps us to verify that opposite sides of a parallelogram are of equal length.

Try These (Page No. 48)

Question 6.
Take two identical 30° – 60° – 90° set-squares and form a parallelogram as before. Does the figure obtain the help you to confirm the above property?
Solution:
The above figure also helps us to confirm that: opposite angles of a parallelogram are equal.

These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4

Question 1.
State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True

Question 2.
Identify all the quadrilaterals that have
(a) four sides of equal length
(b) four right angles.
Solution:
(a) square, rhombus
(b) square, rectangle

Question 3.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) A square is a 4-sided figure, so it is a quadrilateral.
(ii) The opposite sides of a square are parallel, so it is a parallelogram.
(iii) All the sides of a square are equal, so it is a rhombus.
(iv) Each angle of a square is a right-angle, so it is a rectangle.

Question 4.
Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) The diagonals of the following quadrilaterals bisect each other.
Square, rectangle, rhombus, and parallelogram
(ii) The diagonals are perpendicular bisectors of the following quadrilaterals.
Square and rhombus
(iii) The diagonals are equal in the case of square and rectangle.

Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution:
A rectangle is a convex quadrilateral as both of its diagonals lies in its interior.

Question 6.
ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B, and C (The dotted lines are drawn additionally to help you)

Solution:
Construction: Produce B to D such that BO = OD.
Join AD and CD
Proof: AO = OC (As O is a midpoint of AC)
BO = DO (By construction)
∴ ABCD is a parallelogram.
(As diagonals bisect each other)
Also, ∠B = 90°
∴ ABCD is a rectangle
∴ Diagonal AC = BD
\(\frac{1}{2}\) AC = \(\frac{1}{2}\) BD
AO = OB
Also, AO = OC
∴ AO = OB = OC
∴ O is equidistant from A, B and C.

These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition or property used.

(i) AD = __________
(ii) ∠DCB = __________
(iii) OC = __________
(iv) m∠DAB + m∠CDA = __________
Solution:
(i) AD = BC (opposite sides are equal)
(ii) ∠DCB = ∠DAB (opposite angles are equal)
(iii) OC = OA (Diagonals bisect each other)
(iv) m∠DAB + m∠CDA = 180° (Adjacent angles are supplementary)

Question 2.
Consider the following parallelograms. Find the values of the unknown x, y, z.

Solution:
(i) ∠y = 100° (opposite angles of a parallelogram are equal)

∠x + 100° = 180° (adjacent angles in a parallelogram)
∠x = 180° – 100° = 80°
∠z = ∠x = 80° (opposite angles of parallelogram are equal)
∴ ∠x = 80°, ∠y = 100°, ∠z = 80°

(ii) Opposite angles are equal.

∠1 = 50° (opposite angles are equal)
∠1 + ∠z = 180° (Linear pair)
50 + ∠z = 180
∠z = 180° – 50°= 130°
x + ∠1 + y + 50° = 360°
⇒ x + 50° + y + 50° = 360°
⇒ x + y + 100° = 360°
⇒ x + y = 360° – 100°
⇒ x + y = 260°
∴ x = y (opposite angles of a parallelogram)
x = \(\frac{260^{\circ}}{2}\) = 130°
Thus x = 130°, y = 130° and z = 130°.

(iii) x = 90° (vertically opposite angles are equal)

x + y + 30° = 180°
(sum of the angles of a triangle = 180°)
90° + y + 30° = 180°
y + 120° = 180°
∠y = 180° – 120° = 60°
In the parallelogram ABCD
AD || BC and BD is a transversal.
∴ y = z (alternate angles are equal)
z = y = 60°
Thus x = 90°, y = 60° and z = 60°

(iv) In the parallelogram ABCD
y = 80° (opposite angles are equal)
AD || BC and CD is a transversal equal)

∴ z = 80° (corresponding angles are equal)
x + 80° = 180° (sum of the adjacent angle is 180°)
x = 180° – 80° = 100°
Thus, x = 100°, y = 80° and z = 80°

(v) y = 112° (in a parallelogram, opposite angles are equal)

In ΔACD,
x + y + 40° = 180° (sum of the angles of a triangle is 180°)
⇒ x + 112° + 40° = 180°
⇒ x + 152°= 180°
⇒ x = 180° – 152° = 28°
∠C + ∠B = 180° (sum of the adjacent angles of a parallelogram is 180°)
⇒ 40° + z + 112° = 180°
⇒ z + 152° = 180°
⇒ z = 180° – 152° = 28°
Thus, x = 28°, y = 112° and z = 28°.

Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Solution:
(i) In a quadrilateral ABCD
∠D + ∠B = 180° can be, but need not be
∴ The quadrilateral may be a parallelogram but not always

(ii) In a quadrilateral ABCD
AB = DC = 8 cm
AD = 4 cm
BC = 4.4 cm
∴ Opposite sides AD and BC are not equal.
∴ It cannot be a parallelogram.

(iii) In a quadrilateral ABCD
∠A = 70° and ∠C = 65°
∵ Opposite angles ∠A ≠ ∠C
∴ It cannot be a parallelogram.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
In the adjoining figure, ABCD is not a parallelogram such that opposite angles ∠B and ∠D are equal. It is a kite.

Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively.
Since adjacent angles are supplementary.
∴ ∠A + ∠B = 180°
⇒ 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = \(\frac{180^{\circ}}{5}\) = 36°
∠A = 3 × 36° = 108°
and ∠B = 2 × 36° = 72°
Since opposite angles are equal.
∴ ∠D = ∠B = 72° and ∠C = ∠A = 108°
∴ ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B
Since ∠A + ∠B = 180°
∠A = ∠B = \(\frac{180^{\circ}}{2}\) = 90°
Since opposite angles of a parallelogram are equal
∴ ∠A = ∠C = 90° and ∠B = ∠D = 90°
Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°

Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Solution:
y + z = 70°
(An exterior angles of a triangle is equal to the sum of the opposite interior angles)
∵ In ∆HOP
∠HOP = 180° – 70° = 110° (sum of the adjacent angle is 180°)
Now, x = ∠HOP = 110° (opposite angles of a parallelogram are equal)
EH || OP and PH is a transversal
∴ y = 40° (alternate angles are equal)
From (1),
y + z = 70°
⇒ 40° + z = 70°
⇒ z = 70° – 40° = 30°
Thus x = 110°, y = 40° and z = 30°

Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:
(i) GUNS is a parallelogram.
GS = NU (opposite sides are equal)
3x = 18
⇒ x = \(\frac{18}{3}\) = 6
In the parallelogram GUNS
GU = SN (opposite sides are equal)
⇒ 3y – 1 = 26
⇒ 3y = 26 + 1
⇒ 3y = 27
⇒ y = \(\frac{27}{3}\) = 9
Thus, x = 6 cm and y = 9 cm.

(ii) RUNS is a parallelogram and the di-agonal RN and US bisect each other.
∴ y + 7 = 20
⇒ y = 20 – 7 = 13
x + y = 16
⇒ x + 13 = 16
⇒ x = 16 – 13 = 3
Thus, x = 3 cm and y = 13 cm

Question 9.
In the figure given below both RISK and CLUE are parallelograms. Find the value of x.

Solution:
RISK is a parallelogram
∠R + ∠K = 180° (adjacent angles of a parallelogram are supplementary)
∴ ∠R + 120° = 180°
∠R = 180° – 120° = 60°
In the parallelogram RISK
∠R = ∠S (opposite angles are equal)
∠S = 60°
CLUE is also a parallelogram.
∠E = ∠L = 70° (opposite angle of a parallelogram)
∠E = 70°
Now, in triangle ESO
∠E + ∠S + x = 180°
⇒ 70° + 60° + x = 180°
⇒ 130° + x = 180°
⇒ x = 180° – 130°
⇒ x = 50°

Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel?

Solution:
∠KLM + ∠NML = 80° + 100° = 180°
∴ KL || NM (The sum of consecutive interior angles is 180°)
∴ The given figure KLMN is a trapezium.

Question 11.
Find m∠C in the figure given below if AB || DC.

Solution:
AB || DC and BC is a transversal,
∵ m∠B + m∠C = 180°
sum of interior angles is 180°
m∠C = 180° – m∠B
∴ m∠C = 180° – 120° = 60°

Question 12.
Find the measure of ∠P and ∠S if SP || QR in figure (if you find m∠R, is there more than one method to find m∠P)

Solution:
PQRS is a trapezium such that SP || RQ and PQ is a transversal.
∴ m∠P + m∠Q = 180° (Interior angles are supplementary)
m∠P + 130° = 180°
m∠P = 180° – 130° = 50°
Also, m∠S + m∠R = 180°
⇒ m∠S + 90° = 180°
⇒ m∠S = 180° – 90°
⇒ m∠S = 90°
m∠P + m∠Q + m∠R + m∠S = 360° (sum of the angles of a quadrilateral is 360°)
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 310° = 360°
⇒ m∠P = 360° – 310°
⇒ m∠P = 50°
Hence, m∠P = 50° and m∠S = 90°

These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2

Question 1.
Find x in the following figures.

Solution:
(a) Sum of all the exterior angles of a polygon = 360°
⇒ 125° + 125° + x° = 360°
⇒ 250° + x = 360°
⇒ x = 360° – 250°
⇒ x = 110°

(b) x + 90° + 60° + 90° + 70° = 360°
⇒ x + 310° = 360°
⇒ x = 360° – 310°
⇒ x = 50°

Question 2.
Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution:
(i) Number of sides (n) = 9
Number of exterior angles = 9
The given polygon is a regular polygon
All the exterior angles are equal
Measure of an exterior angle = \(\frac{360^{\circ}}{9}\) = 40°

(ii) Number of sides of regular polygon = 15
Number of equal exterior angles =15
The sum of all the exterior angles = 360°
The measure of each exterior angle = \(\frac{360^{\circ}}{15}\) = 24°

Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:
For a regular polygon, measure of each angle is equal
Sum of all the exterior angles = 360°
Measure of an exterior angle = 24°
Number of sides = \(\frac{360^{\circ}}{24^{\circ}}\) = 15
Thus, there are 15 sides of the regular polygon.

Question 4.
How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
The given polygon is a regular polygon.
Each interior angle = 165°
Each exterior angle =180° – 165° = 15°
Number of sides = \(\frac{360^{\circ}}{15^{\circ}}\) = 24
Thus, there are 24 sides of the polygon.

Question 5.
(a) Is it possible to have a regular polygon with measure of each exterior angle 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Solution:
(a) Each exterior angle = 22°
∴ Number of sides = \(\frac{360^{\circ}}{22^{\circ}}=\frac{180^{\circ}}{11^{\circ}}\)
If it is a regular polygon, then its number of sides must be a whole number.
Here \(\frac{180^{\circ}}{11^{\circ}}\) is not a whole number.
∴ 22° cannot be an exterior angle of a regular polygon.

(b) If 22° is an interior angle, then
(180° – 22°) = 158° is an exterior angle.
∴ Number of sides = \(\frac{360^{\circ}}{158^{\circ}}=\frac{180^{\circ}}{79^{\circ}}\)
\(\frac{180^{\circ}}{79^{\circ}}\) is not a whole number
∴ 22° cannot be an interior angle of a regular polygon.

Question 6.
(a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Solution:
(a) The minimum number of sides of a polygon = 3
The regular polygon of 3 sides is an equilateral triangle.
∴ Each interior angle of an equilateral triangle is 60°.

(b) The sum of an exterior angle and its corresponding interior angle is 180°.
Minimum interior angle of a regular polygon is 60°.
∴ The maximum exterior angle of a regular polygon = 180° – 60° = 120°

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 = \(\frac{13}{21}\)