Category: Class 8

MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 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:
Given that ABCD is a parallelogram.
(i) AD = BC [∵ In a parallelogram opposite sides are equal]
(ii) ∠DCB = ∠DAB [∵ In a parallelogram opposite angles are equal]
(iii) OC = OA [∵ In a parallelogram diagonals bisect each other]
(iv) m∠DAB + m∠CDA = 180° [∵ Adjacent angles in a parallelogram are supplementary]

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

Solution:
(i) ABCD is a parallelogram in which ∠B = 100° (given)
∴ ∠A + ∠B = 180° [∵ Sum of adjacent angles is 180°]
⇒ z +100° = 180°
⇒ z = 180° – 100° = 80°
Also ∠B = ∠D and ∠A = ∠C
[∵ Opposite angles are equal]
∴ ∠B = 100° = ∠D = y and ∠A = z = 80° = ∠C = x
∠ x = 80°, y = 100°, z = 80°.

(ii) y + 50° = 180° [∵ Sum of adjacent angles is 180°]
⇒ y = 180° – 50° = 130°
Also, y = x = 130° [∵ Opposite angles are equal in a parallelogram]
And 180°- z = 50° [Linear pair]
⇒ z = 180°- 50° ⇒ z = 130°

(iii) Clearly, x = 90°
[∵ Vertical opposite angles are equal] Also, x + y + 30° = 180°
[By using angle sum property of a triangle]
⇒ 90° + 30° + y = 180°
⇒ 120° + y = 180° ⇒ y = 180° – 120° = 60°
Since, alternate angles are equal in a parallelogram
∴ y = z = 60°
Thus x = 90°, y = 60° and z = 60°.

(iv) Since x + 80° = 180°
[Sum of adjacent angles is 180°] ⇒ x = 180° – 80° = 100°
Also 80° = y [∵ Opposite angles are equal in a parallelogram]
And x = 180° – z ⇒ 100° = 180° – 2° ⇒ 2 = 180° – 100° = 80°
Thus x = 100°, y = 80° and 2 = 80°.

(v) y = 112° [∵ Opposite angles are equal in a parallelogram]
y + x + 40° = 180°
[By angle sum property of a triangle]
⇒ 112°+ x+ 40° = 180°
⇒ x +152° = 180° ⇒ x = 180° – 152°
⇒ x = 28°
⇒ x = 2 = 28° [∵ Alternate angles are equal in a parallelogram]

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) Yes, but need not be true.
(ii) No, because in a parallelogram opposite sides are equal but here AD ≠ BC.
(iii) No, because in a parallelogram opposite angles are equal but here ∠A ≠ ∠C.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:

We can draw a figure of a kite in which exactly two opposite angles are equal.
Hence ∠D = ∠B.

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 given parallelogram.
Let ∠A = 3x and ∠B = 2x.
Since the sum of adjacent angles in a parallelogram is 180°
∴ m∠A + m∠B = 180°
⇒ 3x + 2x = 180°
⇒ 5x = 180° ⇒ x = 36°

∴ m∠A = 3 × 36° = 108° and m∠B = 2 × 36° = 72°
Also, m∠A = m∠C = 108°
[∵ In a parallelogram opposite angles are equal]
and m∠B = m∠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, in which, let m∠A = m∠B = x
Since sum of two adjacent angles is 180°
∴ m∠A + m∠B = 180°
⇒ x + x = 180° ⇒ 2x = 180°
⇒ x = 90°
Also m∠A = m∠C = 90°
[∵ In a parallelogram opposite angles are equal]
and m∠B = m∠D = 90°.

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

Solution:
y = 40°
[Since PO || HE ∴ alternate angles are equal]
70° = y + 2 [Exterior angle property of a triangle]
⇒ z = 70° – 40° = 30°
∠POH = 180° – 70° = 110° [Linear pair]
∠POH = x = 110° [Opposite angles are equal]
Thus x = 110°, y = 40°, 2 = 30°.

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

Solution:
(i) Since GUNS is a parallelogram.
∴ GS = NU and GU = SN
[∵ In a parallelogram opposite sides are equal]
⇒ 3x = 18 ⇒ x = 6
and 3y -1 = 26 ⇒ 3y = 1 + 26 = 27
⇒ y = 9
Thus x = 6 cm and y = 9 cm.

(ii) Since diagonals bisects each other in a parallelogram.
So, 20 = y + 7 ⇒ y = 13
Also, x + y= 16 ⇒ x = 16 – 13 = 3
Thus x = 3 cm and y = 13 cm.

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

Solution:
Since RISK and CLUE are parallelograms.
∴ ∠SKR = ∠RIS = 120° [∵ Opposite angles are equal]
Also, ∠ULC = ∠UEC = 70° [Opposite angles are equal]
∠RIS + ∠ISK = 180° [Adjacent angles are supplementary]
⇒ ∠ISK = 180° – 120° = 60°

In ∆OES, we have
70° + x + 60° = 180° [Angle sum property]
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°.

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

Solution:
In a trapezium, only one pair of opposite sides are parallel, whereas other pair of opposite sides are non-parallel.
∴ KLMN is a trapezium because MN || KL.
[∵ Sum of two adjacent interior angles is 180° = ( 80° + 100°)].

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

Solution:
We are given, \(\overline{A B} \| \overline{D C}\)
∴ Sum of two adjacent interior angles is 180°
i. e., ∠B + ∠C = 180° ⇒ 120° + ∠C= 180°
⇒ ∠C = 180° -120° = 60°.
Thus ∠C = 60°.

Question 12.
Find the measure of ∠P and ∠S if \(\overline{S P} \| \overline{R Q}\) in figure. (If you find m∠R, is there more than one method to find m∠P?)

Solution:
Since \(\overline{S P} \| \overline{R Q}\)
Thus ∠R + ∠S = 180°
[∵ Sum of two adjacent interior angles is 180°]
⇒ 90° + ∠S = 180° ⇒ ∠S = 180° – 90° = 90°
∠P + ∠Q + ∠R + ∠S = 360° [∵ Sum of all angles of a quadrilateral is 360°]
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310°
⇒ ∠P = 50°
Thus ∠P = 50° and ∠S = 90°.
Also m∠P can be found as ∠P + ∠Q = 180°
[Adjacent angles are supplementary]
⇒ ∠P + 130° = 180°
⇒∠P = 180° -130° ⇒ ∠P = 50°

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 1.
Find x in the following figures.

Solution:
(a) Sum of the measures of the exterior angles of any polygon is 360°.
∴ 125° + 125° + x = 360°
⇒ 250° + x = 360°
⇒ x = 360° – 250° ⇒ x = 110°.
(b) Since, sum of the exterior angles of any polygon is 360°.
∴ 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) Let each exterior angle of a regular polygon who has 9 sides is equal to x.
Sum of exterior angles of any polygon is 360°.

Thus each exterior angle of a regular polygon of 9 sides is 40°.
(ii) Let each exterior angle of a regular polygon who has 15 sides is x.
Sum of all exterior angles of a polygon is 360°.

Thus each exterior angle of a regular polygon of 15 sides is 24°.

Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:
Total measure of all exterior angles = 360°
Measure of each exterior angle = 24°
Therefore, the number of sides = \(\frac{360^{\circ}}{24^{\circ}}\)
= 15
The polygon has 15 sides.

Question 4.
How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
Total measure of all exterior angles = 360°
Measure of each interior angle = 165°
Measure of each exterior angle = 180° – 165°
= 15°
Therefore, number of sides = \(\frac{360^{\circ}}{15^{\circ}}\) = 24

Question 5.
(a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why ?
Solution:
(a) No, because 22° is not a divisor of 360°.
(b) No, because each interior angle is 180° – 22° = 158°, which is not a divisor of 360°.

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) Since each angle of an equilateral triangle is 60°.
And equilateral triangle is a regular polygon.
∴ Minimum interior angle is 60° for a regular polygon.
(b) Since minimum interior angle of a regular polygon is 60°.
∴ Each exterior angle of a regular polygon = 180° – 60° = 120°.
∴ Possible maximum exterior angle of a regular polygon is 120°.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 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 curve : It is a curve that does not intersect itself. (1), (2), (5), (6) and (7) are simple curve.
(b) Simple closed curve : A closed curve if it does not pass through one point more than once (1), (2), (5), (6) and (7) are simple closed curve.
(c) Polygon : A simple closed curve made up of only line segments is called a polygon.
(1) and (2) are polygons.
(d) Convex polygon : A polygon that has all its interior angles less than 180°.
(2) is a convex polygon.
(e) Concave polygon – A polygon that has at least one interior angle greater than 180°. (1) is a concave polygon.

Question 2.
How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon
(c) A triangle
Solution:
A diagonal is a line segment joining two non consecutive vertices.
(a) Convex quadrilateral : Convex quadrilateral has 2 diagonals.
(b) A regular hexagon : A regular hexagon has 9 diagonals.
(c) A triangle : It has 0 diagonal, i.e., no diagonal.

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:
Sum of measures of four angles of a convex quadrilateral is 360°.
Example : Draw a figure given below and divide it into two triangles by joining AB, named CAB and DBA.
Now in ∆DBA, we have

Which shows that a quadrilateral which is not convex also have sum of measure of its angles is 360°.

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) If a convex polygon has 7 Sides, then angle sum = (7 – 2) × 180°
= 5 × 180° =900°.
(b) If a convex polygon has 8 sides, then angle sum = (8 – 2) × 180°
= 6 × 180° = 1080°.
(c) If a convex polygon has 10 sides, then angle sum = (10 – 2) × 180°
= 8 × 180° = 1440°.
(d) If a convex polygon has n sides, then angle sum = (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:
Regular polygon- A polygon, which has all sides of equal length and angles of equal measure is called a regular polygon.
(i) An equilateral triangle, as all 3 sides are equal and all 3 angles are also equal (= 60°).
(ii) A square, as it has all 4 sides equal and all 4 angles are also equal (= 90°).
(iii) A regular hexagon, as it has all 6 sides equal and all 6 angles equal (= 120°).

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


Solution:
(a) Let ABCD be a given quadrilateral.
∵ Sum of all angles of a quadrilateral is 360°
∴ m∠A + m∠B + m∠C + m∠D = 360°
⇒ 130° + 120° + x + 50° = 360°
⇒ 300° + x = 360°
⇒ x = 360° – 300°
⇒ x = 60°.

(b) Let ABCD be a given quadrilateral.
Sum of all angles of a quadrilateral is 360°.
∴ m∠A + m∠B + m∠C + m∠D = 360°
⇒ 90° + 60° + 70°’+ x = 360°
⇒ 220° + x = 360° D
⇒ x = 360° – 220°
⇒ x = 140°.

(c) Let ABCDE be a given polygon, which has 5 sides.

Now, sum of angles = (5 – 2) × 180°
= 3 × 180° = 540°.
Also, m∠1 = 180° – 70° = 110° [By linear pair] and m∠Z = 180° – 60° = 120° [By linear pair]
Thus m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 540°
⇒ 110° + 120° + x + x + 30° = 540°
⇒ 260° + 2x = 540°
⇒ 2x = 540° – 260°
⇒ 2x = 280° ∴ x = 140°.

(d) Let ABCDE be a given pentagon.

∴ Sum of angles of a given pentagon
ABODE = (5 – 2) × 180° = 540°.
∴ m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 540°
⇒ x + x + x + x + x = 540°
⇒ 5x = 540° ⇒ x = 108°.

Question 7.
(a) Find x + y + z.
(b) Find x + y + z + w.

Solution:
(a) Let ABC be a given triangle.

Sum of angles of a triangle is 180°.
m∠A + m∠B + m∠C = 180°
⇒ 30° + m∠B + 90° = 180°
⇒ m∠B + 120° = 180°
⇒ m∠B = 180° – 120° = 60°
⇒ m∠B = 60° …(i)
Clearly, x + 90° = 180° [By linear pair]
⇒ x = 180° – 90° ⇒ x = 90° …… (ii)
Also z + 30° = 180° [By linear pair]
⇒ z = 180° – 30° ⇒ z = 150° …….. (iii)
and y + 60° = 180° [By linear pair]
⇒ y = 180° – 60° => y = 120° ……. (iv)
∴ By using (ii), (iii) and (iv), we get
x + y + z = 90° + 120° + 150°
⇒ x + y + z = 360°.

(b) Let ABCD be a given quadrilateral.

Sum of angles of a quadrilateral is 360°.
∴ m∠1 + m∠2 + m∠3 + m∠4 = 360°
⇒ m∠1 + 120° + 80° + 60° = 360° ⇒ m∠1 + 260° = 360°
⇒ m∠1 = 360° – 260° ⇒ m∠l = 100°
Clearly, w + 100° = 180° [By linear pair] ⇒ w = 180° – 100° = 80° ….(i)
x + 120° = 180° [By linear pair]
⇒ x = 180° – 120° = 60° y + 80° = 180°
⇒ y = 180° – 80° = 100° Also, z + 60° = 180°
⇒ z = 180° – 60° ⇒ z = 120° ….(iv)
Thus by (i), (ii), (iii) & (iv), we get
w + x + y + z = 80° + 60° + 100° + 120° = 360°

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.6

Solve the following equations.

Question 1.
\(\frac{8 x-3}{3 x}=2\)
Solution:
We have, \(\frac{8 x-3}{3 x}\) = 2
⇒ 8x – 3 = 2 × 3
⇒ 8x – 3 = 6x ⇒ 8x – 6x = 3 ⇒ 2x = 3
⇒ x = \(\frac{3}{2}\), which is the required solution.

Question 2.
\(\frac{9 x}{7-6 x}=15\)
Solution:

Question 3.
\(\frac{z}{z+15}=\frac{4}{9}\)
Solution:

Question 4.
\(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)
Solution:

Question 5.
\(\frac{7 y+4}{y+2}=\frac{-4}{3}\)
Solution:

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 ages of Hari and Harry be 5x and 7x respectively.
After 4 years, Hari’s age = (5x + 4) years and Harry’s age = (7x + 4) years

∴ Hari’s present age = 5x = 5 × 4 = 20 years and Harry’s present age = 7x = 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 of the rational number be x.
∴ The denominator is x + 8.
According to question, we have
New numerator = x + 17 and
New denominator = (x + 8) – 1 = x + 7

⇒ 2(x + 17) = 3(x + 7) ⇒ 2x + 34 = 3x + 21
⇒ 34 – 21 = 3x – 2x ⇒ 13 = x
Thus, numerator = x = 13
and denominator = x + 8 = 13 + 8 = 21
Hence, the rational number is \(\frac{13}{21}\).

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5

Solve the following linear equations.

Question 1.
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:

which is the required solution.

Question 2.
\(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)
Solution:

⇒ n = 6, which is the required solution.

Question 3.
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:

⇒ x = -5, which is the required solution.

Question 4.
\(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:

⇒ x = 8, which is the required solution.

Question 5.
\(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
Solution:

⇒ t = 2, which is the required solution.

Question 6.
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Solution:

Simplify and solve the following linear equations.

Question 7.
3(t – 3) = 5(2t + 1)
Solution:
We have, 3(t – 3) = 5(2t + 1)
⇒ 31 – 9 = 101 + 5
⇒ – 9 – 5 = 101 – 31 ⇒ -14 = 71
⇒ 1 = – 2, which is the required solution.

Question 8.
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Solution:
We have,
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
⇒ 15 y – 60 – 2y + 18 + 5y + 30 = 0
⇒ 18y – 12 = 0 ⇒ 18y = 12

⇒ y = \(\frac{2}{3}\), which is the required solution.

Question 9.
3(5z – 7) – 2(9z -11) = 4(8z – 13) – 17
Solution:
We have,
3(5z – 7) – 2(9z -11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17
⇒ 15z – 18z – 32z = 21 – 22 – 52 – 17 – 70
⇒ -35z = -70 ⇒ z = \(\frac{-70}{-35}\)
⇒ z = 2, which is the required solution.

Question 10.
0.25 (4f- 3) = 0.05(10f – 9)
Solution:
We have, 0.25(4f – 3) = 0.05(10f – 9)
⇒ f – 0.75 = 0.5f – 0.45

⇒ f = 0.6, which is the required solution.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 13 सीधा और प्रतिलोम समानुपात Intext Questions

MP Board Class 8th Maths Chapter 13 पाठान्तर्गत प्रश्नोत्तर

पाठ्य-पुस्तक पृष्ठ संख्या # 209

भूमिका

प्रश्न 1.
मोहन स्वयं अपने और अपनी बहन के लिए चाय बनाता है। वह 300 mL पानी, 2 चम्मच चीनी, 1 चम्मच चाय पत्ती और 50 mL दूध का उपयोग करता है। यदि वह पाँच व्यक्तियों के लिए चाय बनाए, तो उसे प्रत्येक वस्तु की कितनी मात्रा की आवश्यकता होगी?
हल:
यहाँ, दो व्यक्तियों के लिए पानी = 300 mL, चीनी = 2 चम्मच, चायपत्ती = 1 चम्मच, दूध = 50 mL.
∴ 2 व्यक्तियों के लिये पानी की मात्रा = 300 mL
∴ 5 व्यक्तियों के लिए पानी की मात्रा = 5 x \(\frac{300mL}{2}\)
= 750 mL
∴ 2 व्यक्तियों के लिए चीनी की मात्रा = 2 चम्मच
∴ 5 व्यक्तियों के लिए चीनी की मात्रा = \(\frac{2×5}{2}\) चम्मच
= 5 चम्मच
∴ 2 व्यक्तियों के लिए चायपत्ती = 1 चम्मच
∴ 5 व्यक्तियों के लिए चायपत्ती = \(\frac{1}{2}\) x 5
= 2\(\frac{1}{2}\) चम्मच
∴ 2 व्यक्तियों के लिए दूध की मात्रा = 50 mL
∴ 5 व्यक्तियों के लिए दूध की मात्रा = \(\frac{50}{2}\) x 5 mL
= 125 mL
अतः मोहन को 5 व्यक्तियों के लिए चाय बनाने के लिए 750 mL पानी, 5 चम्मच चीनी, 29 चम्मच चायपत्ती और 125 mL दूध की आवश्यकता होगी।

प्रश्न 2.
यदि दो विद्यार्थी किसी सभा के लिए कुर्सियाँ व्यवस्थित करने में 20 मिनट का समय लगाते हैं तो इसी कार्य को करने में 5 विद्यार्थी कितना समय लेंगे?
हल:
∴ 2 विद्यार्थियों को कुर्सियाँ व्यवस्थित करने में लगा समय = 20 मिनट
∴ 5 विद्यार्थियों को कुर्सियाँ व्यवस्थित करने में लगा समय = 2072 मिनट
= 8 मिनट
अत: 5 विद्यार्थियों को कुर्सियाँ व्यवस्थित करने में 8 मिनट लगेंगे।

प्रश्न 3.
ऐसी पाँच और स्थितियाँ लिखिए, जहाँ एक राशि में परिवर्तन होने से दूसरी राशि में परिवर्तन होता है।
उत्तर:
इस प्रकार की पाँच स्थितियाँ निम्नलिखित हैं –

1. यदि हम बैंक से अधिक धन उधार लेंगे तो हमें अधिक ब्याज देना होगा।
2. किसी गैस पर दबाब बढ़ाने से गैस का आयतन कम हो जाएगा।
3. किसी कार्य को करने के लिए मजदूरों की संख्या बढ़ाने पर पहले की अपेक्षा कम दिन लगेंगे।
4. अधिक दूरी तय करने के लिए किसी वाहन को अधिक पेट्रोल/डीजल की आवश्यकता होगी।
5. किसी मैस में विद्यार्थियों की संख्या बढ़ जाने पर पहले की अपेक्षा अधिक भोजन की आवश्यकता होगी।

पाठ्यपुस्तक पृष्ठ संख्या # 210 सीधा समानुपात

प्रश्न 1.
निम्नलिखित सारणी का अध्ययन कीजिए –
हल:

प्रश्न 2.
अब निम्नलिखित सारणी को पूरा कीजिए –
हल:

पाठ्य-पुस्तक पृष्ठ संख्या # 211-212

इन्हें कीजिए (क्रमांक 13.1)

प्रश्न 1.
एक घड़ी लीजिए और उसकी मिनट वाली (बड़ी) सुई को 12 पर स्थिर कीजिए।
मिनट की सुई द्वारा अपनी प्रारम्भिक स्थिति में घूमे गए कोणों एवं बीते हुए समय को निम्नलिखित सारणी के रूप में लिखिए –

आप T और A के बारे में क्या देखते हैं? क्या इनमें साथ-साथ वृद्धि होती है? क्या – प्रत्येक समय वही रहता है?
क्या मिनट की सुई द्वारा घूमा गया कोण व्यतीत हुए समय के अनुक्रमानुपाती (directly proportional) है? हाँ! उपर्युक्त सारणी से, आप यह भी देख सकते हैं कि –
T₁ : T₂ = A₁ : A₂, क्योंकि
T₁ : T₂ = 15 : 30 = 1 : 2
A₁ : A₂ = 90 : 180 = 1 : 2
जाँच कीजिए कि क्या T₂ : T₃ = A₂ : A₃ तथा T₃ : T₄ = A₃ : A₄ है।
आपस्वयं अपने समय अन्तराल लेकर, इस क्रियाकलाप को दोहरा सकते हैं।
हल:
घूमा गया कोण –
A₂ → 180°
A₃ → 270°
A₄ → 360°

यहाँ, हम देखते हैं कि T में वृद्धि होने पर A में वृद्धि होती है।
हाँ, इनमें साथ-साथ वृद्धि होती है।
हाँ, प्रत्येक समय \(\frac{T}{A}\) = \(\frac{1}{6}\) रहता है।
“हाँ, मिनट की सुई द्वारा घूमा गया कोण व्यतीत हुए समय के अनुक्रमानुपाती है।

यहाँ, सत्यापन होता है।

प्रश्न 2.
अपने मित्र से निम्नलिखित सारणी के भरने के लिए कहिए तथा उसकी आयु और उसकी माँ गत आयु का अनुपात ज्ञात करने के लिए भी कहिए –

आप क्या देखते हैं? क्या F और M में साथ-साथ वृद्धि (या कमी) होती है? क्या \(\frac{F}{M}\) प्रत्येक बार वही रहता है? नहीं। आप इस क्रियाकलाप को अपने अन्य मित्रों के साथ दोहरा सकते हैं तथा अपने प्रेक्षणों को लिख सकते हैं।
हल:
सारणी को भरने पर,

यहाँ, हम देखते हैं कि F और M में साथ-साथ वृद्धि (या कमी) होती है।
नहीं, \(\frac{F}{M}\) प्रत्येक बार वही नहीं है।
इस क्रियाकलाप को हम अपने अन्य मित्रों के साथ दोहरा सकते हैं। हम यही स्थिति पाएँगे।

पाठ्य-पुस्तक पृष्ठ संख्या # 212

प्रयास कीजिए (क्रमांक 13.1)

प्रश्न 1.
निम्नलिखित सारणियों को देखिए तथा ज्ञात कीजिए कि क्या x और y अनुक्रमानुपाती हैं –
1.

2.

3.

हल:
1.

अतः x और ‘ के संगत मानों का अनुपात , ही रहता है। इसलिए, x और y अनुक्रमानुपाती हैं जिनका अनुपात \(\frac{1}{2}\) अचर है।

2.
यहाँ,

यहाँ x और y का अनुपात अचर नहीं हैं। इसलिए, x और y अनुक्रमानुपाती नहीं हैं।

3.

यहाँ, हम देखते हैं कि x और y के संगत अनुपात अचर नहीं हैं।
अतः x और y अनुक्रमानुपाती नहीं हैं। उत्तर

प्रश्न 2.
मूलधन = ₹ 1,000 ब्याज दर = 8% वार्षिक निम्नलिखित सारणी को भरिए तथा ज्ञात कीजिए कि किस प्रकार ब्याज (साधारण या चक्रवृद्धि) समय अवधि के साथ प्रत्यक्ष अनुपात में बदलता या परिवर्तित होता है।

हल:
यहाँ, मूलधन = ₹ 1,000, ब्याज दर = 8% वार्षिक
साधारण ब्याज = \(\frac{pxrxt}{100}\)

= ₹ 1259.712 – ₹ 1000
= ₹ 259.712
अब, सारणी भरने पर

अतः साधारण ब्याज समय अवधि के साथ प्रत्यक्ष अनुपात में बदलता है।
लेकिन चक्रवृद्धि ब्याज समय अवधि के साथ प्रत्यक्ष अनुपात में नहीं बदलता है।

सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 13.1)

प्रश्न 1.
यदि हम समय अवधि और ब्याज की दर स्थिर रखें, तो साधारण ब्याज मूलधन के साथ प्रत्यक्ष अनुपात में परिवर्तित होता है। क्या ऐसा ही सम्बन्ध चक्रवृद्धि ब्याज के लिए भी होगा? क्यों?
हल:
नहीं, ऐसा सम्बन्ध चक्रवृद्धि ब्याज के लिए नहीं होगा। क्योंकि चक्रवृद्धि ब्याज में मूलधन समय अवधि के साथ बदलता रहता है।

पाठ्य-पुस्तक पृष्ठ संख्या # 215

इन्हें कीजिए (क्रमांक 13.2)

प्रश्न 1.
अपने राज्य का एक मानचित्र लीजिए। वहाँ पर प्रयुक्त पैमाने को लिख लीजिए। पैमाने (तनसमत) का प्रयोग करते हुए, मानचित्र पर किन्हीं दो नगरों की दूरी मापिए। इन दोनों नगरों के बीच की वास्तविक दूरी परिकलित कीजिए।
हल:
माना कि पैमाना 1 सेमी. = 200 किमी
माना कि दो नगरों के बीच की दूरी = 4 सेमी
तब, दो नगरों के बीच वास्तविक दूरी = 4 x 200 किमी
= 800 किमी

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4

Question 1.
Amina thinks of a number and subtracts \(\frac{5}{2}\) from it. She multiplies the result by 8.
The result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let the number be x.
After subtracting \(\frac{5}{2}\) from x and multiplying the result by 8, we get \(\left(x-\frac{5}{2}\right) \times 8\)

⇒ (2x – 5) × 4 = 3x ⇒ 8x – 20 = 3x
⇒ 8x – 3x = 20 ⇒ 5x = 20 ⇒ x = 4
∴ The number that Amina thought of is 4.

Question 2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let one number be x.
∴ Another number be 5x.
Now, according to question,
2(x + 21) = 5x + 21 ⇒ 2x + 42 = 5x + 21
⇒ 42 – 21 = 5x – 2x ⇒ 21 = 3x

∴ One number is 7 and another number is 5 × 7 = 35.

Question 3.
Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Solution:
Let the digit at the units place be x.
∴The digit at the tens place be 9 – x.
⇒ The given two-digit number is
10 × (9 – x) + x = 90 – 10x + x = 90 – 9x
After interchanging the digits, we get a new number i.e., 10 × x + (9 – x) = 10x + 9 – x = 9x + 9
Now, according to question,
9x + 9 – (90 – 9x) = 27 ⇒ 9x + 9 – 90 + 9x = 27
⇒ 18x – 81 = 27 ⇒ 18x = 27 + 81

The units digit is 6 and tens digit is 9 – 6 = 3
∴ The required two-digit number is 36.

Question 4.
One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this number and add the resulting number to the original number, you get 88. What is the original number?
Solution:
Let the digit at units place be x and the digit at tens place be 3x.
∴ The given number is 10 × 3x + x = 30x + x
= 31 x
After interchanging the digits, we get a new number i.e., 10 × x + 3x = 10x + 3x = 13x
Now, according to question,
31x + 13x = 88
⇒ 44x = 88 ⇒ x = 2
∴ The digit at units place is 2 and the digit at tens place is 3 × 2 = 6.
Thus, the required number is 62 or 26.

Question 5.
Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Solution:
Let Shobo’s present age be x years
∴ Shobo’s mother’s present age = 6x years
After 5 years,
Shobo’s age = (x + 5) years
According to question,

⇒ x + 5 = 2x
⇒ 5 = 2x – x ⇒ 5 = x
∴ Shobo’s present age is 5 years and Shobo’s mother’s present age is 6 × 5 = 30 years.

Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11 : 4. At the rate ₹ 100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Since, length and breadth are in the ratio 11 : 4. So, let length is 11x and breadth is4x.
Now, perimeter of the plot

Perimeter of rectangular plot = 2 (length + breadth) = 2 (11x + 4x) = 2 × 15x = 30x
⇒ 30x = 750 ⇒ x = \(\frac{750}{30}\) = 25
∴ Length of the plot = 11x = 11 × 25 = 275 m and breadth of the plot = 4x = 4 × 25 = 100 m

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 2 metres of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36,660. How much trouser material did he buy?
Solution:
Let material bought for trousers = 2x metres
and material bought for shirts = 3x metres
Cost of material bought for shirts = ₹ 50 × 3x
= ₹ 150x
and cost of material bought for trousers
= ₹ 90 × 2x = ₹ 180x

Thus, material bought for trousers = 2x m
= 2 × 100.16 m
= 200.32 m

Question 8.
Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Solution:
Let number of deer in the herd be x.

Therefore, number of deer in the herd is 72.

Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution:
Let granddaughter’s present age be x years.
Grandfather’s present age = 10x years …(i)
Also, grandfather’s age = (x + 54) years …(ii)
From (i) and (ii), we get x + 54 = 10x
⇒ 54 = 10x – x ⇒ 54 = 9x ⇒ 6 = x
∴ Granddaughter’s present age = 6 years and grandfather’s present age = 10 × 6
= 60 years

Question 10.
Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution:
Let the present age of Aman’s son be x years.
∴ The present age of Aman = 3x years
10 years ago,
Aman’s son’s age = (x -10) years and Aman’s age = (3x – 10) years
Now, according to question,
3x – 10 = 5(x -10) ⇒ 3x – 10 = 5x – 50
⇒ 50 – 10 = 5x – 3x ⇒ 40 = 2x
⇒ 20 = x
Thus, Aman’s present age = 3 × 20 = 60 years and his son’s age = 20 years

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3

Solve the following equations and check your results.

Question 1.
3x = 2x + 18
Solution:
We have, 3x = 2x + 18
Transposing 2x to L.H.S., we get
3x – 2x = 18 ⇒ x = 18
Checking: L.H.S. = 3 × 18 = 54
R.H.S. = 2x + 18 = 2 × 18 + 18 = 36 + 18 = 54
Thus, L.H.S. = R.H.S.

Question 2.
5t – 3 = 3t – 5
Solution:
We have, 5t – 3 = 3t – 5
Transposing 3t to L.H.S., we get
5t – 3 – 3t = – 5
⇒ 2t – 3 = – 5
Transposing – 3 to R.H.S., we get
2t = -5 + 3
⇒ 2t = – 2
Dividing both sides by 2, we get

Checking : L.H.S. = 5t – 3 = 5 × (-1) – 3 = -5 – 3 = -8
R.H.S. = 3t – 5 = 3 × (-1) – 5 = -3 – 5 = -8
Thus, L.H.S. = R.H.S.

Question 3.
5x + 9 = 5 + 3x
Solution:
We have, 5x + 9 = 5 + 3x
Transposing 3x to L.H.S., we get
5x + 9 – 3x = 5
⇒ 2x + 9 = 5
Now, transposing 9 to R.H.S., we get
2x = 5 – 9 = 4 ⇒ 2x = – 4
Dividing both sides by 2, we get \(\frac{2 x}{2}=\frac{-4}{2}\)
Checking: L.H.S. = 5x + 9 = 5 × (-2) + 9 = -10 + 9 = -1
R.H.S. = 5 + 3x = 5 + 3 (-2) = 5 – 6 = – 1.
Thus, L.H.S. = R.H.S.

Question 4.
4z + 3 = 6 + 2z
Solution:
We have, 4z + 3 = 6 + 2z
Transposing 2z to L.H.S., we get
4z + 3-2z = 6 ⇒ 2z + 3 = 6
Transposing 3 to R.H.S., we get
2z = 6 – 3 ⇒ 2z = 3

Question 5.
2x – 1 = 14 – x
Solution:
We have, 2x -1 = 14 – x
Transposing – x to L.H.S., we get
2x – 1 + x = 14 ⇒ 3x – 1 = 14
Transposing – 1 to R.H.S., we get
3x = 14 + 1 ⇒ 3x = 15
Dividing both sides by 3, we get \(\frac{3 x}{3}=\frac{15}{3}\)
⇒ x = 15
Checking: L.H.S. = 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9
R.H.S. = 14 – x = 14 – 5 = 9
Thus, L.H.S. = R.H.S.

Question 6.
8x + 4 = 3(x – 1) + 7
Solution:
We have, 8x + 4 = 3(x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7 ⇒ 8x + 4 = 3x + 4
Transposing 3x to L.H.S., we get
8x + 4 – 3x = 4 ⇒ 5x + 4 = 4
Transposing 4 to R.H.S., we get
⇒ 5x = 4 – 4 ⇒ 5x = 0
Dividing both sides by 5, we get

Checking: L.H.S. = 8x + 4 = 8 × 0 + 4 = 4
R.H.S. = 3(x – 1) + 7 = 3(0 – 1) + 7 = -3 + 7 = 4
Thus, L.H.S. = R.H.S.

Question 7.
x = \(\frac{4}{5}\)(x + 10)
Solution:
We have, x = \(\frac{4}{5}\)(x + 10)
Multiplying both sides by 5, we get
5x = 4(x + 10) ⇒ 5x = 4x + 40
Transposing 4x to L.H.S., we get 5x – 4x = 40 ⇒ x = 40
Checking : L.H.S. = x = 40

Thus, L.H.S. = R.H.S.

Question 8.
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
Solution:

Question 9.
\(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Solution:

Question 10.
3m = 5m – \(\frac{8}{5}\)
Solution:

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2

Question 1.
If you subtract \(\frac{1}{2}\) from a number and multiply the result by \(\frac{1}{2}\), you get \(\frac{1}{8}\).What is the number?
Solution:
Let the number be x.

Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What is the length and the breadth of the pool?
Solution:
Assume that the breadth of the rectangular pool be x m.
∴ Length = 2 + 2x
Perimeter of the rectangular pool = 2(Length + Breadth)
⇒ 154 = 2(2 + 2x + x)
⇒ 154 = 2(2 + 3x)
⇒ 154 = 4 + 6x
Transposing 4 to L.H.S., we get
154 – 4 = 6x ⇒ 150 = 6x
Now, dividing both sides by 6, we get

∴ The breadth of the pool is 25 m and length of the pool = (2 + 2 × 25) m = (2 + 50) m = 52 m.

Question 3.
The base of an isosceles triangle is \(\frac{4}{3}\) cm. The perimeter of the triangle is 4\frac{2}{15} cm. What is the length of either of the remaining equal sides?
Solution:
The base of an isosceles triangle = \(\frac{4}{3}\) cm
Let x cm be the length of both of the remaining equal sides of the isosceles triangle.
Perimeter of the isosceles ∆ABC = AB + BC + CA

Transposing \(\frac{4}{3}\) to L.H.S., we get

∴ The length of either of the remaining equal sides is 1\(\frac{2}{5}\) cm.

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let one number be x and other number is 15 + x.
Now, x + 15 + x = 95 ⇒ 2x + 15 = 95
Transposing 15 to R.H.S., we get
2x = 95 – 15 ⇒ 2x = 80
Now, dividing both sides by 2, we get

∴ One number is 40 and other number is 15 + 40 = 55.

Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Since, two numbers are in the ratio 5 : 3.
Let the two numbers be 5x and 3x.
Now, 5x – 3x = 18 ⇒ 2x = 18
Dividing both sides by 2, we get

⇒ x = 9
∴ The required numbers are 5x = 5 × 9 = 45 and 3x = 3 × 9 = 27.

Question 6.
Three consecutive integers add up to 51. What are these integers?
Solution:
Let three consecutive integers be x, (x + 1) and (x + 2).
Now, x+x + 1+ x + 2 = 51 ⇒ 3x + 3 = 51
Transposing 3 to R.H.S., we get
3x = 51 – 3
⇒ 3x = 48
Dividing both sides by 3, we get

Thus, the required three consecutive integers are 16,17 and 18.

Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be 8x, 8(x + 1) and 8(x + 2).
Now, 8x + 8x + 8 + 8x + 16 = 888
⇒ 24x + 24 = 888
Transposing 24 to R.H.S., we get
24x = 888 – 24 ⇒ 24x = 864
Dividing both sides by 24, we get

Thus, the required three consecutive multiples of 8 are 8 × 36 = 288, 8 × 37 = 296 and 8 × 38 = 304.

Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let three consecutive integers be x, (x + 1) and (x + 2).
When they are multiplied by 2, 3 and 4, we get 2x, 3(x +1) and 4(x + 2) i.e., 2x, (3x + 3) and (4x + 8) respectively.
Now, 2x + 3x + 3 + 4x + 8 = 74 × 9x + 11 = 74
Transposing 11 to R.H.S., we get
9x = 74 – 11 ⇒ 9x = 63
Now, dividing both sides by 9, we get

Thus, the required three consecutive integers are 7, 8 and 9.

Question 9.
The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Solution:
Since, the ages of Rahul and Haroon are in the ratio 5 : 7.
Let Rahul’s present age be 5x years and Haroon’s present age be 7x years.
After 4 years, we have
Rahul’s age = 5x + 4 and Haroon’s age = 7x + 4
Now, 5x + 4 + 7x + 4 = 56 ⇒ 12x + 8 = 56
Transposing 8 to R.H.S., we get
12x = 56 – 8 ⇒ 12x = 48
Now, dividing both sides by 12, we get

Thus, Rahul’s present age = 5 × 4 = 20 years and Haroon’s present age = 7 × 4 = 28 years

Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Solution:
We know that the number of boys and girls are in ratio 7 : 5.
Let the number of boys be 7x and the number of girls be 5x.
Now, 7x = 5x + 8
Transposing 5x to L.H.S., we get
7x – 5x = 8 ⇒ 2x = 8
Dividing both sides by 2, we get

Thus, the required number of boys = 7 × 4 = 28 and the required number of girls = 5 × 4 = 20
Hence, total number of students = 28 + 20 = 48

Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let Baichung’s grandfather’s age be x years.
∴ Baichung’s father’s age = (x – 26) years
Baichung’s age = [(x – 26) – 29] years = (x – 55) years
Since, sum of the ages of all the three is 135 years. So, x + x – 26 + x – 55 = 135
⇒ 3x – 81 = 135
Transposing – 81 to R.H.S., we get
3x = 135 + 81 ⇒ 3x = 216
Dividing both sides by 3, we get

Thus, Baichung’s grandfather’s age = 72 years
Baichung’s father’s age = (72 – 26) years = 46 years
Baichung’s age = (72 – 55) years = 17 years

Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let the present age of Ravi be x years.
After 15 years, Ravi’s age = (15 + x) years
Now, 15 + x = 4x
Transposing x to R.H.S., we get
15 = 4x – x
⇒ 15 = 3x
Dividing both sides by 3, we get

Thus, the present age of Ravi is 5 years.

Question 13.
A rational number is such that when you multiply it by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, you get \(-\frac{7}{12}\).What is the number?
Solution:

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?
Solution:
Let the number of notes of ₹ 100, ₹ 50 and ₹ 10 be 2x, 3x and 5x, respectively.
∴ The amount Lakshmi has from ₹ 100 notes : ₹ (2x × 100) = ₹ 200x from ₹ 50 notes : ₹ (3x × 50) = ₹ 150x from ₹ 10 notes : ₹ (5x × 10) = ₹ 50x Now, 200x + 150x + 50x = 4,00,000
⇒ 400x = 4,00,000 ⇒ x = 1000
∴ Number of notes of ₹ 100 = 2x
= 2 × 1000 = 2000
Number of notes of ₹ 50 = 3x = 3 × 1000 = 3000
and number of notes of ₹ 10 = 5x = 5 × 1000
= 5000

Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins be x.
Number of ₹ 2 coins = 3x and number of ₹ 1 coins = 160 – x – 3x
The total amount
from ₹ 5 coins : ₹ 5 × x = × 5x
from ₹ 2 coins : ₹ 2 × 3x = ₹ 6x
and from ₹ 1 coins : ₹ 1[160 – x – 3x]
= ₹ [160 – 4x]
Now, 5x + 6x + 160 – 4x = 300
⇒ 7x + 160 = 300
Transposing 160 to R.H.S., we get
7x = 300 – 160
⇒ 7x = 140
Dividing both sides by 7, we get

⇒ x = 20
The number of ₹ 1 coins = 160 – 20 – 3 × 20 = 160 – 20 – 60 = 80
Number of ₹ 2 coins = 3 × 20 = 60
Number of ₹ 5 coins = 20

Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners be x.
Then, the number of losers will be 63 – x.
Since, the winner gets a prize of ₹ 100 and a loser gets a prize of ₹ 25.
Amount got by winners = ₹ 100x
And amount got by losers = ₹ (63 – x) × 25 But total prize money is ₹ 3000.
Therefore, 100x + (63 – x) × 25 = 3000
⇒ 100x + 63 × 25 – 25x = 3000
⇒ 75x + 1575 = 3000

So, the number of winners is 19.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1

Solve the following equations.

Question 1.
x – 2 = 7.
Solution:
We have, x – 2 = 7
Transposing -2 to R.H.S., we get
x = 7 + 2 ⇒ x = 9, which is the required solution.

Question 2.
y + 3 = 10.
Solution:
We have, y + 3 = 10
Transposing 3 to R.H.S., we get y = 10 – 3
⇒ y = 7, which is the required solution.

Question 3.
6 = z + 2.
Solution:
We have, 6 = z + 2
Transposing 2 to L.H.S., we get
6 – 2 = z
⇒ z = 4, which is the required solution.

Question 4.
\(\frac{3}{7}\) + x = \(\frac{17}{7}\).
Solution:

⇒ x = 2, which is the required solution.

Question 5.
6x = 12.
Solution:
We have, 6x = 12
Dividing both sides by 6, we get
\(\frac{6 x}{6}=\frac{12}{6}\)
⇒ x = 2, which is the required solution.

Question 6.
\(\frac{t}{5}=10\)
Solution:
We have, \(\frac{t}{5}\) = 10
Multiplying both sides by 5, we get

⇒ t = 50, which is the required solution.

Question 7.
\(\frac{2 x}{3}\) = 18
Solution:
We have,
\(\frac{2 x}{3}\) = 18
Dividing both sides by 2, we get

Now, multiplying both sides by 3, we get

⇒ x = 27, which is the required solution.

Question 8.
\(1.6=\frac{y}{1.5}\)
Solution:
We have, \(1.6=\frac{y}{1.5}\)
Multiplying both sides by 1.5, we get

⇒ y = 2.4, which is the required solution.

Question 9.
7x – 9 = 16.
Solution:
We have, 7x – 9 = 16
Transposing -9 to R.H.S., we get
7x = 16 + 9
⇒ 7x = 25
Now, dividing both sides by 7, we get

⇒ x = \(\frac{25}{7}\), which is the required solution.

Question 10.
14y – 8 = 13.
Solution:
We have, 14y -8 = 13
Transposing -8 to R.H.S., we get
14y = 13 + 8 ⇒ 14y = 21
Now, dividing both sides by 14, we get

Question 11.
17 + 6p = 9.
Solution:
We have, 17 + 6p = 9
Transposing 17 to R.H.S., we get 6p = 9 – 17
⇒ 6p = – 8
Now, dividing both sides by 6, we get

Question 12.
\(\frac{x}{3}+1=\frac{7}{15}\)
Solution:

Now, multiplying both sides by 3, we get

⇒ x = \(-\frac{8}{5}\), which is the required solution.

MP Board Class 8th Maths Solutions