Class 8

MP Board Class 8th Maths Solutions Chapter 15 Introduction to Graphs Ex 15.3

Question 1.
Draw the graphs for the following tables of values, with suitable scales on the axes.
(a) Cost of apples

(b) The horizontal line x-axis shows the time (in hours) and vertical line y-axis shows the distance (in km).

(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m.?
(ii) What was the time when the car had covered a distance of 100 km since it’s start?
(c) interest on deposits for a year.

(i) Does the graph pass through the origin?
(ii) Use the graph to find the interest on ₹ 2500 for a year.
(iii) To get an interest of ₹ 280 per year, how much money should be deposited?
Solution:
(a) The horizontal line x-axis shows the number of apples and vertical line y-axis shows the cost (in ₹) of apples.

(b) The horizontal line x-axis shows the time (in hours) and vertical line y-axis shows the distance (in km).

(i) The total distance covered by the car during the period 7.30 a.m. to 8 a.m. is (120 – 100) = 20 km.
(ii) At 7.30 a.m.car had covered a distance of 100 km since its start.
(c) The horizontal line x-axis shows deposit (in ₹) and vertical line y-axis shows simple interest (in ₹)

(i) Yes, graph passes through the origin.
(ii) Interest on ₹ 2500 is ₹ 200.
(iii) To get an interest of ₹ 280 per year, ₹ 3500 should be deposited.

Question 2.
Draw a graph for the following

Is it a linear graph?

Solution:
(i) The horizontal line x-axis shows a side of square (in cm) and the vertical line y-axis shows a perimeter (in cm).

Yes, it is a linear graph.
(ii) The horizontal line x-axis shows a side of a square (in cm) and the vertical line y-axis shows an area (in cm²).

No, it is not a linear graph.

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 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 4 Practical Geometry Ex 4.5

Question 1.
Draw the following.
(i) The square READ with RE = 5.1 cm.
(ii) A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
(iii) A rectangle with adjacent sides of lengths 5 cm and 4 cm.
(iv) A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm.
Solution:
(i) Since, all 4 sides of a square are equal and each of 4 angles is equal to 90°.
Steps of Construction:
Step-1: Draw RE = 5.1 cm.
Step-2 : Draw ∠REX = 90°.
Step-3 : Cut off EA = 5.1 cm on \(\overrightarrow{E X}\)
Step-4: With R as centre and radius equal to 5.1 cm, draw an arc.
Step-5 : With A as centre and radius equal to 5.1 cm, cut off another arc on the arc drawn in step-4 at point D.
Step-6 : Join DA and DR.
Hence, READ is the required square.

(ii) We know that diagonals of a rhombus bisect each other at right angles. Let AC = 5.2 cm and BD = 6.4 cm.
Steps of Construction:
Step-1: Draw AC = 5.2 cm.
Step-2 : Draw perpendicular bisector XY of AC which cut AC at point O.
Step-3 : Cut off OD = 3.2 cm on OX and OB = 3.2 cm on \(\overrightarrow{O Y}\).
Step-4 : Join AD, CD, AB and CB.
Hence, ABCD is the required rhombus.

(iii) In a rectangle, opposite sides are equal and each of 4 angles is equal to 90°.
Let AB = 5 cm and BC = 4 cm
∴ AB = DC = 5 cm and BC = AD = 4 cm.
Also, ∠A = ∠B = ∠C = ∠D = 90°.
Steps of Construction:
Step-1: Draw AB = 5 cm.
Step-2 : Draw ∠ABX = 90°.
Step-3 : Cut off BC = 4 cm on BX .
Step-4: With A as centre and radius equal to 4 cm, cut off an arc.
Step-5 : With C as centre and radius equal to 5 cm cut off another arc on the arc drawn in step-4 at point D.
Step-6 : Join AD and CD.
Hence, ABCD is the required rectangle.

(iv) Here, data given is incomplete. Since we know that to draw a quadrilateral at least five parts are necessary. In the present case, OK = 5.5 cm & KA = 4.2 cm is given.
We know that opposite sides of a parallelogram are equal.
∴ OK = YA = 5.5 cm and KA = OY = 4.2 cm
Here, only four parts are given. This means that one more part is necessary.
So, either one angle or diagonal of a parallelogram is required to construct it.
Hence, parallelogram OKAY cannot be drawn.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.4

Question 1.
Construct the following quadrilaterals.
(i) Quadrilateral
DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

(ii) Quadrilateral
TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Solution:
(i) Steps of Construction:
Step-1: Draw EA = 5 cm.
Step-2: Make ∠AEX = 60° and ∠EAY = 90°.
Step-3 : Cut off arcs AR = 4.5 cm on \(\overrightarrow{A Y}\) and ED = 4 cm on \(\overrightarrow{E X}\).
Step-4: Join DR.
Hence, DEAR is the required quadrilateral.

(ii) Steps of Construction :
Step-1: Draw RU = 3 cm.
Step-2 : Make ∠URX = 75° and ∠RUY = 120°
Step-3 : Cut off RT = 3.5 cm on \(\overrightarrow{R X}\) and UE = 4 cm on \(\overrightarrow{U Y}\).
Step-4: Join TE.
Hence, TRUE is the required quadrilateral.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question 1.
Construct the following quadrilaterals.
(i) Quadrilateral
MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 1050°
∠R = 105°

(ii) Quadrilateral
PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 6.5 cm
∠N = 85°

(iii) Parallelogram
HEAR
HE =5 cm
EA = 6 cm
∠B = 85°

(iv) Rectangle
OKAY
OK = 7 cm
KA = 5 cm
Solution:
(i) Steps of Construction:
Step-1: Draw MO = 6 cm
Step-2 : Make ∠MOX = 105° and ∠OMY = 60°.
Step-3 : Cut off OR = 4.5 cm on OX.
Step-4 : At point R, draw ∠ORZ = 105° which cuts MY at E.
Hence, MORE is the required quadrilateral.

(ii) Here, ∠P = 90°, ∠A = 110°, ∠N = 85°
∴ ∠L = 360° – (∠P + ∠A + ∠N)
= 360° – (90° + 110° + 85°)
= 360°- 285° = 75° ….. (A)
Steps of Construction:
Step-1: Draw PL = 4 cm.
Step-2 : Make ∠LPX = 90° and ∠PLY = 75°. [From (A)]
Step-3 : Cut off LA = 6.5 cm on \(\overrightarrow{L Y}\).
Step-4: Draw ∠LAZ = 110° which cut \(\overrightarrow{P X}\) at point N.
Hence, PLAN is the required quadrilateral.

(iii) Since, opposite sides and angles of a parallelogram are equal i.e., ∠R = ∠E = 85°, HE = RA = 5 cm and EA = HR = 6 cm.
Steps of Construction:
Step-1: Draw HE = 5 cm.
Step-2 : Draw ∠HEX = 85°.
Step-3 : Cut off EA = 6 cm on \(\overrightarrow{E X}\)
Step-4: With H as centre and radius equal to 6 cm, draw an arc.
Step-5: With A as centre and radius equal to 5 cm, cut another arc on the arc drawn in step-4 at point R.
Step-6 : Join HR and RA.
Hence HEAR is the required parallelogram.

(iv) We know that each of the four angles of a rectangle is equal to 90° and opposite sides are also equal.
OK = YA and KA = OY
Steps of Construction:
Step-1: Draw OK = 7 cm.
Step-2 : Make ∠OKX = 90°.
Step-3 : Cut off KA = 5 cm on \(\overrightarrow{K X}\).
Step-4: With O as centre and radius equal to 5 cm, cut an arc.
Step-5: With A as centre and radius equal to 7 cm cut another arc on the arc drawn in step-4 at point Y.
Step-6 : Join OY and YA.
Hence, OKAY is the required rectangle.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.2

Question 1.
Construct the following quadrilaterals
(i) Quadrilateral
LIFT
LI = 4 cm
IF = 3 cm
TL = 2.5 cm
LF= 4.5 cm
IT =4 cm

(ii) Quadrilateral
GOLD
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm

(iii) Rhombus BEND
BN = 5.6 cm,
DE = 6.5 cm
Solution:
(i) Steps of Construction:
Step-1: Draw LI = 4 cm.
Step-2: With L as centre and radius equal to 2.5 cm, cut an arc.
Step-3: With I as centre and radius equal to 4 cm, cut another arc on the arc drawn in step-2 at point T.
Step-4 : With I as centre and radius equal to 3 cm, cut an arc.
Step-5: With L as centre and radius equal to 4.5 cm, cut another arc on the arc drawn in step-4 at point F.
Step-6 : Join LT, IT, LF, IF and TF.
Hence, LIFT is the required quadrilateral.

(ii) Steps of Construction:
Step-1: Draw OL = 7.5 cm.
Step-2: With L as centre and radius equal to 5 cm cut an arc.
Step-3 : With O as centre and radius equal to 10 cm, cut another arc on the arc drawn in step-2 at point D.
Step-4: With L as centre and radius equal to 6 cm, cut another arc.
Step-5: With D as centre and radius equal to 6 cm cut an arc on arc drawn in step-4 at point G.
Step-6 : Join LD, LG, OG, OD and DG. Hence, GOLD is the required quadrilateral.

(iii) Here, BN = 5.6 cm and DE = 6.5 cm are given. These two sides are diagonals of a rhombus BEND. We know that diagonals of a rhombus bisect each other at right angles.

Steps of Construction:
Step-1: Draw BN = 5.6 cm.
Step-2 : Draw perpendicular bisector XY of BN which intersect BN at O.
Step-3 : Cut off OD = OF = 3.25 cm on \(\overrightarrow{\mathrm{OX}}\) and \(\overrightarrow{\mathrm{OY}}\) respectively.
Step-4 : Join BD, ND, BE and NE.
Hence, BEND is the required rhombus.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 12 घातांक और घात Intext Questions

MP Board Class 8th Maths Chapter 12 पाठान्तर्गत प्रश्नोत्तर

पाठ्य-पुस्तक पृष्ठ संख्या # 201

भूमिका

प्रश्न 1.
2^-2 किसके बराबर है?
हल:
2^-2 = \(\frac { 1 }{ 2^{ 2 } } \) = \(\frac{1}{4}\)

ऋणात्मक घातांकों की घात

प्रश्न 1.
10^-10 किसके बराबर है?
हल:
∴ a^-m = \(\frac { 1 }{ a^{ m } } \)
10^-10 = \(\frac { 1 }{ 10^{ 10 } } \)

पाठ्य-पुस्तक पृष्ठ संख्या # 202

प्रयास कीजिए (क्रमांक 12.1)

प्रश्न 1.
गुणात्मक प्रतिलोम लिखिए –

1. 2^-4
2. 10^-5
3. 7^-2
4. 5^-3
5. 10^-100

हल:

1. 2^-4 का गुणात्मक प्रतिलोम = 2⁴
2. 10^-5 का गुणात्मक प्रतिलोम = 10⁵
3. 7^-2 का गुणात्मक प्रतिलोम = 7²
4. 5^-3 का गुणात्मक प्रतिलोम = 5³
5. 10^-100 का गुणात्मक प्रतिलोम = 10¹⁰⁰

प्रयास कीजिए (क्रमांक 12.2)

प्रश्न 1.
घातांकों का उपयोग करते हुए निम्न को विस्तारित रूप में लिखिए –

1. 1025.63
2. 1256.249

हल:
1. 1025.63 = 1 x 1000 + 2 x 10 + 5 x 1 + \(\frac{6}{10}\) + \(\frac{3}{100}\)
= 1 x 10³ + 2 x 10¹ + 5 x 10⁰ + 6 x 10^-1 + 3 x 10^-2

2. 1256.249 = 1 x 1000 + 2 x 100 + 5 x 10 + 6 x 1 + \(\frac{2}{10}\) + \(\frac{4}{100}\) + \(\frac{9}{1000}\)
= 1 x 10³ + 2 x 10² + 5 x 10¹ + 6 x 10⁰ + 2 x 10^-1 + 4 x 10^-2 + 9 x 10^-3

पाठ्य-पुस्तक पृष्ठ संख्या # 203

प्रयास कीजिए (क्रमांक 12.3)

प्रश्न 1.
घातांक रूप से सरल कीजिए और लिखिए –

1. (-2)^-3 x (-2)^-4
2. p³ x p^-10
3. 3² x 3^-5 x 3⁶

हल:
1. (-2)^-3 x (-2)^-4 = (-2) ^{(-3) + (-4)} = (-2)^-7
(∵ a^m x aⁿ = a^{m + 1})

2. p³ x p^-10 = p^3-10 = p^-7 = \(\frac { 1 }{ p^{ 7 } } \)

3. 3² x 3^-5 x 3⁶ = 3^2-5+6 = 3³

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