Class 6

MP Board Class 6th Maths Solutions Chapter 6 Integers Ex 6.2

Question 1.
Using the number line write the integer which is:
(a) 3 more than 5
(b) 5 more than -5
(c) 6 less than 2
(d) 3 less than -2
Solution:

Question 2.
Use number line and add the following integers:
(a) 9 +(-6)
(b) 5+ (-11)
(c) (- 1) + (- 7)
(d) (-5) +10
(e) (-1) + (- 2) + (- 3)
(f) (- 2) + 8 + (- 4)
Solution:

Question 3.
Add without using number line:
(a) 11 + (-7)
(b) (-13) + (+18)
(c) (- 10) + (+ 19)
(d) (- 250) + (+ 150)
(e) (-380) + (-270)
(f) (-217) + (-100)
Solution:
(a) 11 + (-7) = 11 – 7 = 4
(b) (-13) + (+18) = -13 + 18 = 5
(c) (-10) + (+19) = -10 + 19 = 9
(d) (-250) + (+150) = -250 + 150 = -100
(e) (-380) + (-270) = -380 – 270 = -650
(f) (-217) + (-100) = -217 – 100 = -317

Question 4.
Find the sum of:
(a) 137 and -354
(b) – 52 and 52
(c) – 312, 39 and 192
(d) – 50, – 200 and 300 .
Solution:
(a) 137 + (-354) = 137 – 354 = -217
(b) -52 + 52 = 0
(c) -312 + 39 + 192 = – 312 + 231 = -81
(d) -50 + (-200) + 300 = – 50 – 200 + 300
= – 250 + 300 = 50

Question 5.
Find the sum :
(a) (-7) + (-9) + 4 + 16
(b) (37) + (- 2) + (- 65) + (- 8)
Solution:
(a) (-7) + (-9) + 4 + 16 = -7 – 9 + 4 + 16
= -16 + 20 = 4

(b) (37) + (-2) + (-65) + (-8)
= 37 – 2 – 65 – 8
= 37 – 75 = – 38

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 3 संख्याओं के साथ खेलना Ex 3.2

पाठ्य-पुस्तक पृष्ठ संख्या # 56-57

प्रश्न 1.
बताइए कि किन्हीं दो संख्याओं का योग सम होता है या विषम होता है, यदि वे दोनों
(a) विषम संख्याएँ हों
(b) सम संख्याएँ हों।
हल :
(a) दो विषम संख्याओं का योग सम होता है।
(b) दो सम संख्याओं का योग सम होता है।

प्रश्न 2.
बताइए कि निम्नलिखित में कौन-सा कथन सत्य है और कौन-सा असत्य :
(a) तीन विषम संख्याओं का योग सम होता है।
(b) दो विषम संख्याओं और एक सम संख्या का योग सम होता है।
(c) तीन विषम संख्याओं का गुणनफल विषम होता है।
(d) यदि किसी सम संख्या को 2 से भाग दिया जाए तो भागफल सदैव विषम होता है।
(e) सभी अभाज्य संख्याएँ विषम हैं।
(f) अभाज्य संख्याओं के कोई गुणनखण्ड नहीं होते।
(g) दो अभाज्य संख्याओं का योग सदैव सम होता है।
(h) केवल 2 ही एक सम अभाज्य संख्या है।
(i) सभी सम संख्याएँ भाज्य संख्याएँ हैं।
(j) दो सम संख्याओं का गुणनफल सदैव सम होता है।
उत्तर-
(a) असत्य
(b) सत्य
(c) सत्य
(d) असत्य
(e) असत्य
(f) असत्य
(g) असत्य
(h) सत्य
(i) असत्य
(j) सत्य।

प्रश्न 3.
संख्या 13 और 31 अभाज्य संख्याएँ हैं। इन दोनों संख्याओं में दो अंक 1 और 3 हैं। 100 तक की संख्याओं में ऐसे अन्य सभी युग्म ज्ञात कीजिए।
हल :
100 तक की अभाज्य संख्याएँ हैं : 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
इनमें से समान इकाई वाली अभाज्य संख्याओं के युग्म हैं-17 और 71; 37 और 73; तथा 79 और 97

प्रश्न 4.
20 से छोटी सभी अभाज्य और भाज्य संख्याएँ अलग-अलग लिखिए।
हल :
20 से छोटी अभाज्य संख्याएँ हैं-2, 3, 5, 7, 11, 13, 17 और 19
20 से छोटी भाज्य संख्याएँ हैं-4, 6, 8, 9, 10, 12, 14, 15, 16 और 18

प्रश्न 5.
1 और 10 के बीच में सबसे बड़ी अभाज्य संख्या लिखिए।
उत्तर-
1 और 10 के बीच में सबसे बड़ी अभाज्य संख्या = 7

प्रश्न 6.
निम्नलिखित को दो विषम अभाज्य संख्याओं के योग के रूप में व्यक्त कीजिए :
(a) 44
(b) 36
(c) 24
(d) 18
हल :
(a) 44 = 13 + 31
या 44 = 3 + 41 या 44 = 37 + 7

(b) 36 = 5 + 31
36 = 23 + 13 या 36 = 17 + 19

(c) 24 = 11 + 13 या
24 = 5 + 19 या 24 = 7 + 17

(d) 18 = 7 + 11 या 18 = 5 + 13

प्रश्न 7.
अभाज्य संख्याओं के ऐसे तीन युग्म लिखिए जिनका अंतर 2 हो।
हल :
अभाज्य संख्याओं के तीन युग्म जिनका अन्तर 2 हैं-
(i) 3 और 5,
(ii) 5 और 7
(iii) 11 और 13

प्रश्न 8.
निम्नलिखित में से कौन-सी संख्याएँ अभाज्य संख्याएँ हैं?
(i) 23
(ii) 51
(iii) 37
(d) 26
हल :
(a)∵23 = 1 x 23
∴23 अभाज्य संख्या है।

(b)∵51 = 1 x 51 = 3 x 17
∴51 के गुणनखण्ड = 1, 3, 17 और 51
(दो से अधिक गुणनखण्ड)
∴51 अभाज्य संख्या नहीं है।

(c)∵37 = 1 x 37
∴37 अभाज्य संख्या है।

(d)∵26 = 1 x 26 = 2 x 13
∴26 के गुणनखण्ड = 1, 2, 13 और 26
(दो से अधिक गुणनखण्ड)
∴26 अभाज्य संख्या नहीं है।

प्रश्न 9.
100 से छोटी सात क्रमागत भाज्य संख्याएँ लिखिए जिनके बीच में कोई अभाज्य संख्या नहीं है।
हल :
100 से छोटी सात क्रमागत भाज्य संख्याएँ जिनके बीच में कोई अभाज्य संख्या नहीं है
90, 91, 92, 93, 94, 95 और 96

प्रश्न 10.
निम्नलिखित संख्याओं में से प्रत्येक को तीन अभाज्य संख्याओं के योग के रूप में व्यक्त कीजिए :
(a) 21
(b) 31
(c) 53
(d) 61
हल :
(a) 21 = 3 + 5 + 13
(b) 31 = 3 + 5 + 23
(c) 53 = 13 + 17 + 23
(d) 61 = 7 + 13 + 41

प्रश्न 11.
20 से छोटी अभाज्य संख्याओं के ऐसे पाँच युग्म लिखिए जिनका योग 5 से विभाज्य (divisible) हो। (संकेत : 3 + 7 = 10)
हल :
20 से छोटी अभाज्य संख्याएँ हैं- 2, 3, 5, 7, 11, 13, 17 और 19.
2 + 3 = 5, 5, 5 से विभाज्य है।
3 + 7 = 10, 10, 5 से विभाज्य है।
2 + 13 = 15, 15, 5 से विभाज्य है।
7 + 13 = 20, 20, 5 से विभाज्य है।
3 + 17 = 20, 20, 5 से विभाज्य है।
अत: अभीष्ट अभाज्य संख्याओं के युग्म : 2 और 3; 3 और 7; 2 और 13; 7 और 13; तथा 3 और 17

प्रश्न 12.
निम्न में रिक्त स्थानों को भरिए:
(a) वह संख्या जिसके केवल दो गुणनखण्ड हों ……… कहलाती है।
(b) वह संख्या जिसके दो से अधिक गुणनखण्ड हों एक …… कहलाती है।
(c) 1 न तो …… है और न ही ……. ।
(d) सबसे छोटी अभाज्य संख्या ……… है।
(e) सबसे छोटी भाज्य संख्या …….. है।
(f) सबसे छोटी सम संख्या …. है।
उत्तर-
(a) अभाज्य संख्या,
(b) भाज्य संख्या,
(c) अभाज्य संख्या, भाज्य संख्या
(d) 2,
(e) 4
(f) 2.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 6 Integers Ex 6.1

Question 1.
Write opposites of the following :
(a) Increase in weight
(b) 30 km north
(c) 326 BC
(d) Loss of ₹ 700
(e) 100 m above sea level
Solution:
(a) Decrease in weight
(b) 30 km south
(c) 326 AD
(d) Profit of ₹ 700
(e) 100 m below sea level

Question 2.
Represent the following numbers as integers with appropriate signs.
(a) An aeroplane is flying at a height, two thousand metre above the ground.
(b) A submarine is moving at a depth, eight hundred metre below the sea level.
(c) A deposit of rupees two hundred.
(d) Withdrawal of rupees seven hundred.
Solution:
(a) Two thousand metre above the ground = + 2000 metres
(b) Eight hundred metre below the sea level = – 800 metres
(c) Deposit of two hundred rupees = + 200 Rupees
(d) Withdrawal of seven hundred rupees = -700 Rupees

Question 3.
Represent the following numbers on a number line:
(a) +5
(b) -10
(c) + 8
(d) -1
(e) -6
Solution:

Question 4.
Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points :
(a) If point D is + 8, then which point is-8?
(b) Is point G a negative integer or a positive integer?
(c) Write integers for points B and E.
(d) Which point marked on this number line has the least value?
(e) Arrange all the points in decreasing order of value.

Solution:
(a) We have, point D is +8. Therefore, 16 steps to the down from D is -8 i.e., the point F.
(b) Point G is a negative integer.
(c) Point B is four steps down from point D.
∴ Value of point B = +8 – 4 = +4
Point E is eighteen steps down from point D.
∴ Value of point E = +8 – 18 = -10
(d) Since, point E is located in the bottom. So, point E has the least value.
(e) Decreasing order of all the points is, D, C, B, A, O, H, G, F, E

Question 5.
Following is the list of temperatures of five places in India on a particular day of the year.

(a) Write the temperatures of these places in the form of integers in the blank column.
(b) Following is the number line representing the temperature in degree Celsius.

Plot the name of the city against its temperature.
(c) Which is the coolest place?
(d) Write the names of the places where temperatures are above 10°C.
Solution:

(c) Siachin is the coolest place.
(d) Ahmedabad and Delhi have temperature above 10°C.

Question 6.
In each of the following pairs, which number is to the right of the other on the number line?
(a) 2, 9
(b) -3, -8
(c) 0, -1
(d) -11, 10
(e) – 6, 6
(f) 1,-100
Solution:
(a) 9 is right to 2
(b) -3 is right to -8
(c) 0 is right to -1
(d) 10 is right to -11
(e) 6 is right to -6
(f) 1 is right to-100

Question 7.
Write all the integers between the given pairs (write them in the increasing order.)
(a) 0 and -7
(b) -4 and 4
(c) – 8 and – 15
(d) -30 and -23
(a) The integers between 0 and -7 are -6, -5, -4, -3, -2, -1
(b) The integers between -4 and 4 are -3, -2, -1, 0, 1, 2, 3
(c) The integers between -8 and -15 are -14, -13, -12,-11,-10, -9
(d) The integers between -30 and -23 are -29, -28, -27, -26, -25, -24

Question 8.
(a) Write four negative integers greater than -20.
(b) Write four negative integers less than -10.
Solution:
(a) There are 19 negative integers which are greater than -20. Four of them are -19, -18, -17, -16
(b) There are infinite negative integers which are less than -10. Four of them are -11, -12, -13, -14

Question 9.
For the following statements, write True (T) or False (F). If the statement is false, correct the statement.
(a) – 8 is to the right of – 10 on a number line.
(b) – 100 is to the right of – 50 on a number line.
(c) Smallest negative integer is – 1.
(d) – 26 is greater than – 25.
Solution:
(a) True
(b) False
Since -100 is to the left of -50 on the number line.
(c) False
Since -1 is the greatest negative integer.
(d) False
Since -26 is less than -25.

Question 10.
Draw a number line and answer the following :
(a) Which number will we reach if we move 4 numbers to the right of – 2.
(b) Which number will we reach if we move 5 numbers to the left of 1.
(c) If we are at – 8 on the number line, in which direction should we move to reach – 13?
(d) If we are at – 6 on the number line, in which direction should we move to reach – 1 ?
Solution:

Thus, we will reach 2 if we move 4 numbers to the right of -2

Thus, we will reach -4 if we move 5 numbers to the left of 1.

Thus, we should move 5 numbers to the left of -8 to reach -13.

Thus, we should move 5 numbers to the right of -6 to reach -1.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 3 संख्याओं के साथ खेलना Ex 3.1

प्रश्न 1.
निम्नलिखित संख्याओं के सभी गुणनखण्ड लिखिए :
(a) 24
(b) 15
(c) 21
(d) 27
(e) 12
(f) 20
(g) 18
(h) 23
(i) 36
हल :
(a)∵ 24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6
∴ 24 के सभी गुणनखण्ड = 1, 2, 3, 4, 6, 8, 12 और 24

(b)∵ 15 = 1 x 15
15 = 3 x 5
∴ 15 के सभी गुणनखण्ड = 1, 3, 5 और 15

(c)∵ 21 = 1 x 21
21 = 3 x 7
∴ 21 के सभी गुणनखण्ड = 1, 3, 7 और 21

(d)∵ 27 = 1 x 27
27 = 3 x 9
∴ 27 के सभी गुणनखण्ड = 1, 3, 9 और 27

(e)∵ 12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
∴ 12 के सभी गुणनखण्ड = 1, 2, 3, 4, 6 और 12

(f)∵ 20 = 1 x 20
20 = 2 x 10
20 = 4 x 5
∴ 20 के सभी गुणनखण्ड = 1, 2, 4, 5, 10 और 20

(g)∵ 8 = 1 x 18
18 = 2 x 9
18 = 3 x 6
∴ 18 के सभी गुणनखण्ड = 1, 2, 3, 6, 9, और 18

(h)∵ 23 = 1 x 23
∴ 23 के सभी गुणनखण्ड = 1 और 23

(i)∵ 36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 6 x 6
∴ 36 के सभी गुणनखण्ड = 1, 2, 3, 4, 6, 9, 12, 18 और 36

प्रश्न 2.
निम्न संख्याओं के प्रथम पाँच गुणज लिखिए :
(a)5
(b)8
(c)9
हल :
(a) 5 x 1 = 5,
5 x 2 = 10,
5 x 3 = 15,
5 x 4 = 20,
5 x 5 = 25
अत: 5 के गुणज = 5, 10, 15, 20 और 25

(b) 8 x 1 = 8,
8 x 2 = 16,
8 x 3 = 24,
8 x 4 = 32,
8 x 5 = 40
अत: 8 के गुणज = 8, 16, 24, 32 और 40

(c) 9 x 1 = 9,
9 x 2 = 18,
9 x 3 = 27
9 x 4 = 36,
9 x 5 = 45
अत : 9 के गुणन = 9, 18, 27, 36 और 45

प्रश्न 3.
स्तम्भ 1 की संख्याओं का स्तम्भ 2 के साथ मिलान कीजिए:

उत्तर-
(i) → (b),
(ii) → (d),
(iii) → (a),
(iv) → (f),
(v) → (e).

प्रश्न 4.
9 के सभी गुणज ज्ञात कीजिए जो 100 से कम हों।
हल:
∵ 9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
9 x 11 = 99
∴ 9 के 100 से कम गुणज = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 और 99

पाठ्य-पुस्तक पृष्ठ संख्या # 55

प्रश्न 1.
क्या 15 एक भाज्य संख्या है? 18 तथा 25 के बारे में आप क्या सोचते हैं ?
हल :
(i) 15 के सम्भावित गुणनखण्ड = 1, 3, 5 और 15
∵ 15 के दो से अधिक गुणनखण्ड हैं।
∴ 15 भाज्य संख्या है।

(ii) 18 के सम्भावित गुणनखण्ड = 1, 2, 3, 6, 9 और 18
∵ 18 के दो से अधिक गुणनखण्ड हैं।
∴ 18 भाज्य संख्या है।

(iii) 25 के सम्भावित गुणनखण्ड = 1, 5, 25
∵ 25 के दो से अधिक गुणनखण्ड हैं।
∴ 25 भाज्य संख्या है।

प्रयास कीजिए

प्रश्न 1.
ध्यान दीजिए कि 2 x 3 + 1 = 7 एक अभाज्य संख्या है। यहाँ 2 के एक गुणज में 1 जोड़कर एक अभाज्य संख्या प्राप्त की गई है। क्या आप इस प्रकार से कुछ और अभाज्य संख्याएँ ज्ञात कर सकते हैं ?
हल :
2 x 2 + 1 = 5, जो कि एक अभाज्य संख्या है।
2 x 5 + 1 = 11, जो कि एक अभाज्य संख्या है।
2 x 6 + 1 = 13, जो कि एक अभाज्य संख्या है।
2 x 8 + 1 = 17, जो कि एक अभाज्य संख्या है।
2 x 9 + 1 = 19, जो कि एक अभाज्य संख्या है।
2 x 11 + 1 = 23, जो कि एक अभाज्य संख्या है।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 3 संख्याओं के साथ खेलना Intext Questions

पाठ्य-पुस्तक पृष्ठ संख्या # 51

प्रयास कीजिए

प्रश्न 1.
45, 30 और 36 के सम्भावित गुणनखण्ड कीजिए।
हल :
45 के गुणनखण्ड = 1, 3, 5, 9, 15 और 45
30 के गुणनखण्ड = 1, 2, 3, 5, 6, 10, 15 और 30
36 के गुणनखण्ड = 1, 2, 3, 4, 6, 9, 12, 18 और 36

पाठ्य-पुस्तक पृष्ठ संख्या # 53

प्रश्न 1.
क्या 10 एक सम्पूर्ण संख्या है ?
हल :
10 के सम्भावित गुणनखण्ड = 1, 2, 5 और 10
गुणनखण्डों का योग = 1 + 2 + 5 + 10 = 18
संख्या का दो गुना = 2 x 10 = 20
स्पष्टतः 18 ≠ 20
∴10 एक सम्पूर्ण संख्या नहीं है।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.9

Question 1.
Match the following :

Give two new examples of each shape.
Solution:
(a) ➝ (ii)
(b) ➝ (iv)
(c) ➝ (v)
(d) ➝ (iii)
(e) ➝ (i)


Two examples of cone : Ice-cream, Birthday cap
Two examples of sphere : Ball, Rasgulla
Two example of Cylinder : Pipe, Can
Two examples of Cuboid : Box, Brick
Two examples of a Pyramid : Roof of the house, Cheese grater

Question 2.
What shape is
(a) Your instrument box?
(b) A brick?
(c) A match box?
(d) A road-roller?
(e) A sweet laddu?
Solution:
(a) Cuboid
(b) Cuboid
(c) Cuboid
(d) Cylinder
(e) Sphere

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.8

Question 1.
Examine whether the following are polygons. If any one among them is not, say why?

Solution:
(a) As it is not a closed figure, therefore, it is not a polygon.
(b) It is a polygon because it is closed by line segments.
(c) It is not a polygon because it is not made by line segments.
(d) It is not a polygon because it is not made by only line segments and also it has curved surface.

Question 2.
Name each polygon.

Make two more examples of each of these.
Solution:
(a) Quadrilateral
(b) Triangle
(c) Pentagon
(d) Octagon
Two more examples of each :

Question 3.
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.

Solution:
ABCDEF is a regular hexagon and ∆AEF is a triangle formed by joining AE.
Hence, ∆AEF is an isosceles triangle.

Question 4.
Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution:

ABCDEFGH is a regular octagon and CDGH is a rectangle formed by joining C and H; D and G.

Question 5.
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution:

ABCDE is the required pentagon and its diagonals are AD, AC, BE, BD and CE.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 7 Fractions Ex 7.4

Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<‘ ‘=’ ‘>’ between the fractions:

(c) Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}\) and \(\frac{6}{6}\) on the number line. Put appropriate signs between the fractions given.

Solution:

Question 2.
Compare the fractions and put an appropriate sign.

Solution:
(a)
\(\frac{3}{6}\) and \(\frac{5}{6}\) are like fractions.
Also, denominator of \(\frac{5}{6}\) is greater than denominator of \(\frac{3}{6}\)

(b)
\(\frac{1}{7}\) and \(\frac{1}{4}\) are unlike fractions with same numerator. Also, denominator of \(\frac{1}{7}\) is greater than denominator of \(\frac{1}{4}\)

(c)
\(\frac{4}{5}\) and \(\frac{5}{5}\) are like fractions.
Also, numerator of \(\frac{5}{5}\) is greater than numerator of \(\frac{4}{5}\)

(d) \(\frac{3}{5}\) and \(\frac{3}{7}\) are unlike fractions with same numerator. Also, denominator of \(\frac{3}{7}\) is greater than denominator of \(\frac{3}{5}\).

Question 3.
Make five more such pairs and put appropriate signs.
Solution:

Question 4.
Look at the figures and write ‘<‘ or ‘>’, ‘=’ between the given pairs of fractions.

make five more such problems and solve them with your friends.
Solution:

Question 5.
How quickly can you do this ? Fill appropriate sign (‘<‘, ‘=’, ‘>’)

Solution:
(a) \(\frac{1}{2}\) and \(\frac{1}{5}\) are unlike fractions with same numerator. Also denominator of \(\frac{1}{5}\) is greater than denominator of \(\frac{1}{2}\).

(b) \(\frac{2}{3}\) and \(\frac{3}{6}\) are unlike fractions with different numerator.

(c) \(\frac{3}{5}\) and \(\frac{2}{3}\) are unlike fractions with different numerator.

(d) \(\frac{3}{4}\) and \(\frac{2}{8}\) are unlike fractions with different numerators.

(e) \(\frac{3}{5}\) and \(\frac{6}{5}\) are like fractions. Also, numerator of \(\frac{6}{5}\) is greater than numerator of \(\frac{3}{9}\).

(f) \(\frac{7}{9}\) and \(\frac{3}{9}\) are like fractions. Also, numerator of \(\frac{7}{9}\) is greater than numerator of \(\frac{3}{9}\).

(g) \(\frac{1}{4}\) and \(\frac{1}{2}\) are unlike fractions with different numerators.

Question 6.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.


Solution:

Question 7.
Find answers to the following. Write and indicate how you solved them.


Solution:

Question 8.
Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac{2}{5}\) of the same book. Who read less ?
Solution:
Ila read 25 pages out of 100 pages.
Fraction of reading the pages

Question 9.
Rafiq \(\frac{3}{6}\) of an hour, while Rohit exercised for \(\frac{3}{4}\) of an hour. Who exercised for a longer time?
Solution:
Rafiq exercised \(\frac{3}{6}\) of an hour.
Rafiq exercised \(\frac{3}{4}\) of an hour.
Since, \(\frac{3}{4}>\frac{3}{6}\)
Therefore, Rohit exercised for a longer time.

Question 10.
In a Class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Solution:
In class A, 20 passed in first class out of 25.
∴ Fraction of first class passed students \(=\frac{20}{25}=\frac{4}{5}\)
In class B, 24 passed in first class out of 30.
∴ Fraction of first class passed students \(=\frac{24}{30}=\frac{4}{5}\)
Hence, both classes have same fraction of student getting first class.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.2

Question 1.
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Solution:
Side of equilateral triangle = l
Therefore, perimeter of equilateral triangle = 3 × side = 3l

Question 2.
The side of a regular hexagon (see fig.) is denoted by l.

Express the perimeter of the hexagon using l. (Hint: A regular hexagon has all its six sides equal in length.)
Solution:
Side of regular hexagon = l
Therefore, perimeter of regular hexagon = 6 × side = 6l

Question 3.
A cube is a three-dimensional figure as shown in the given figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.

Solution:
Length of one edge of a cube = l
Number of edges in a cube = 12
Therefore, total length of the edges of a cube = 12 × l = 12l

Question 4.
The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure, AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).

Solution:
Since, diameter of the circle is always twice the radius of the circle.
Therefore, d = 2r

Question 5.
To find sum of three numbers 14, 27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54.
Thus, (14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.
Solution:
(a + b) + c = a + (b + c), where a, b and c are any three numbers.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.
(a) A pattern of letter T as
(b) A pattern of letter Z as
(c) A pattern of letter U as
(d) A pattern of letter V as
(e) A pattern of letter E as
(f) A pattern of letter S as
(g) A pattern of letter A as
Solution:
(a) Number of matchsticks required to make one
∴ Number of matchsticks required to
make a pattern of letter T as

(b) Number of matchsticks required to make one
∴ Number of matchsticks required to make a pattern of letter Z as

(c) Number of matchsticks required to make one
∴Number of matchsticks required to make a pattern of letter U as

(d) Number of matchsticks required to make one
Number of matchsticks required to make a pattern of letter V as

(e) Number of matchsticks required to make one
∴ Number of matchsticks required to make a pattern of letter E as

(f) Number of matchsticks required to make one
∴ Number of matchsticks required to make a pattern of letter S as

(g) Number of matchsticks required to make one
∴ Number of matchsticks required to make a pattern of letter A as

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Solution:
Part (a) & (d) i.e., letter T and V have same rule as that given by L.
Because the number of matchsticks required in each of them is 2.

Question 3.
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution:
Let number of rows = n
Number of cadets in each row = 5
Therefore, total number of cadets = 5M

Question 4.
If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution:
Let number of boxes = b
Number of mangoes in each box = 50
Therefore, total number of mangoes = 50b

Question 5.
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Solution:
Let number of students = s
Number of pencils distributed to each student = 5
Therefore, total number of pencils needed = 5s

Question 6.
A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Solution:
Let flying time of the bird be t minutes.
Distance covered by the bird in 1 minute = 1 km
∴ Distance covered by the bird in t minutes = 1 × t km = t km

Question 7.
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
Solution:
Number of dots in each row = 9
Number of rows = r
Therefore, total number of dots = 9r
When there are 8 rows, then number of dots = 9 × 8 = 72
When there are 10 rows, then number of dots = 9 × 10 = 90

Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Solution:
Radha’s age = x years
Since, Leela is 4 years younger than Radha.
Therefore, Leela’s age = (x – 4) years.

Question 9.
Mother has made laddus. She gives Vome laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution:
Number of laddus gave away = l
Number of remaining laddus = 5
∴ Total number of laddus = (l + 5)

Question 10.
Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Solution:
Number of oranges in a small box = x
Number of smaller boxes = 2
Therefore, total number of oranges in smaller boxes = 2x
Number of remaining oranges = 10
Thus, number of oranges in the larger box = 2x + 10

Question 11.
(a) Look at the following matchstick pattern of squares (see fig.).The squares are not separate. Two neighbouring squares have a common matchstick.
Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)

(b) The given figure gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.

Solution:
(a)

If we remove 1 matchstick from each figure, then they make multiples of 3, i.e., 3, 6, 9, 12, …………
So the required equation = 3x + 1, where x is number of squares.

(b)

If we remove 1 matchstick from each figure, then they make multiples of 2, i.e., 2, 4, 6, 8, ………..
So the required equation = 2x + 1, where x is number of triangles.

MP Board Class 6th Maths Solutions