Category: Class 6

MP Board Class 6th Maths Solutions Chapter 7 Fractions Ex 7.6

Question 1.
Solve:

Solution:

Question 2.
Sarita bought \(\frac{2}{5}\) metre of ribbon and Lalita \(\frac{3}{4}\) metre of ribbon. What is the total length of the
ribbon they bought?
Solution:
Ribbon bought by Sarita = \(\frac{2}{5}\) m
And Ribbon bought by Lalita = \(\frac{3}{4}\) m

Therefore, they bought 1\(\frac{3}{20}\) m of ribbon.

Question 3.
Naina was given 1\(\frac{1}{22}\) piece of cake and Najma was given 1\(\frac{1}{3}\) piece of cake. Find the total amount of cake was given to both of them.
Solution:
Cake taken by Naina = 1\(\frac{1}{2}\) piece
And cake taken by Najma = 1\(\frac{1}{3}\) piece
Total cake taken = \(1 \frac{1}{2}+1 \frac{1}{3}=\frac{3}{2}+\frac{4}{3}\)

Question 4.
Fill in the boxes:

Solution:

Question 5.
Complete the addition-subtraction box.

Solution:

Question 6.
A piece of wire \(\frac{7}{8}\) metre long broke into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?
Solution:
Total length of wire = \(\frac{7}{8}\) metre
Length of first part = \(\frac{1}{4}\) metre

Therefore, the length of remaining part is \(\frac{5}{8}\) metre

Question 7.
Nandini’s house is \(\frac{9}{10}\) metre km from her school. She 10
walked some distance and then took a bus for \(\frac{1}{2}\) metre km to reach the school. How far did she walk?
Solution:
Total distance between school and house = \(\frac{9}{10}\) km
Distance covered bv bus = \(\frac{1}{2}\) km

Therefore, distance covered by walking is \(\frac{2}{5}\) km.

Question 8.
Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is \(\frac{5}{6}\) th full and Samuel’s shelf is \(\frac{2}{5}\)th full. Whose bookshelf is more full? By what fraction?
Solution:

∴ Asha’s bookshelf is more covered than Samuel.
Difference \(=\frac{25}{30}-\frac{12}{30}=\frac{13}{30}\)

Question 9.
Jaidev takes 2\(\frac{1}{5}\) minutes to walk across the school ground. Rahul takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?
Solution:

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.5

Question 1.
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable
(a) 17 = x + 7
(b) (t – 7) > 5
(c) \(\frac{4}{2}\) = 2
(d) (7 × 3) – 19 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) – (12 × 4)
(j) 7 = (11 ×2) + p
(k) 20 = 5y
(l) \(\frac{3q}{2}\) < 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3× 5
(o) 7 – x = 5
Solution:
(a) 17 = x + 7 is an equation with variable as both the sides are equal. The variable is x.
(b) (t – 7) > 5 is not an equation as L.H.S. is greater than R.H.S.
(c) \(\frac{4}{2}\) = 2 is not an equation with variable.
(d) (7 × 3) – 19 = 8 is not an equation with variable.
(e) 5 × 4 – 8 = 2x is an equation with variable as both the sides are equal. The variable is x.
(f) x – 2 = 0 is an equation with variable as both the sides are equal. The variable is
(g) 2m < 30 is not an equation as L.H.S. is less than R.H.S.
(h) 2n + 1 = 11 is an equation with variable as both the sides are equal. The variable is n.
(i) 7 = (11 × 5) – (12 × 4) is not an equation with variable.
(j) 7 = (11 × 2) + p is an equation with variable as both the sides are equal. The variable is p.
(k) 20 = 5y is an equation with variable as both the sides are equal. The variable is
(l) \(\frac{3q}{2}\) < 5 is not an equation as L.H.S. is less than R.H.S.
(m) z + 12 > 24 is not an equation as L.H.S. is greater than R.H.S.
(n) 20 – (10 – 5) = 3 × 5 is not an equation with variable.
(o) 7 – x = 5 is an equation with variable as both the sides are equal. The variable is x.

Question 2.
Complete the entries in the third column of the table.


Solution:

Question 3.
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
(a) 5m = 60        (10, 5, 12, 15)
(b) n + 12 = 20      (12, 8, 20, 0)
(c) p – 5 = 5       (0, 10, 5, -5)
(d) \(\frac{q}{2}\) = 7   (7, 2, 10, 14)
(e) r – 4 = 0     (4, -4, 8, 0)
(f) x + 4 = 2      (-2, 0, 2, 4)
Solution:
(a) 5m = 60
Putting the given values in L.H.S., we get
5 × 10 = 50
∵ L.H.S. ≠ R.H.S.
∴ m = 10 is not the solution.
5 × 5 = 25
∵ L.H.S. ≠ R.H.S.
∴ m = 5 is not the solution.
5 × 12 = 60
∵ L.H.S. = R.H.S.
∴ m = 12 is a solution.
5 x 15 = 75
∵ L.H.S. ≠ R.H.S.
∴ m = 15 is not the solution.

(b) n + 12 = 20
Putting the given values in L.H.S., we get
12 + 12 = 24
∵ L.H.S. ≠ R.H.S.
∴ n = 12 is not the solution.
8 + 12 = 20
∵ L.H.S. = R.H.S.
∴ n = 8 is a solution.
20 + 12 = 32
∵ L.H.S. ≠ R.H.S.
∴ n = 20 is not the solution.
0 + 12 = 12
∵ L.H.S. ≠ R.H.S.
∴ n = 0 is not the solution.

(c) p – 5 = 5
Putting the given values in L.H.S., we get
0 – 5 = -5
∵ L.H.S. ≠ R.H.S.
∴ p = 0 is not the solution.
10 – 5 = 5
∵ L.H.S. = R.H.S.
∴ p = 10 is a solution.
5 – 5 = 0
∵ L.H.S. ≠ R.H.S.
∴ p = 5 is not the solution.
-5 – 5 = -10
∵ L.H.S. ≠ R.H.S.
∴ p = -5 is not the solution.

(d) \(\frac{q}{2}\) = 7
Putting the given values in L.H.S., we get
\(\frac{7}{2}\)
∵ L.H.S. ≠ R.H.S.
∴ q = 7 is not the solution.
\(\frac{2}{2}\) = 1
∵ L.H.S. ≠ R.H.S.
∴ q = 2 is not the solution.
\(\frac{10}{2}\) = 5
∵ L.H.S. ≠ R.H.S.
∴ q = 10 is not the solution.
\(\frac{14}{2}\) = 7
∵ L.H.S. = R.H.S.
∴ q = 14 is a solution.

(e) r – 4 = 0
Putting the given values in L.H.S., we get
4 – 4 = 0
∵ L.H.S. = R.H.S.
∴ r = 4 is a solution.
-4 – 4 = -8
∵ L.H.S. ≠ R.H.S.
∴ r = -4 is not the solution.
8 – 4 = 4
∵ L.H.S. ≠ R.H.S.
∴ r = 8 is not the solution.
0 – 4 = -4
∵ L.H.S. ≠ R.H.S.
∴ r = 0 is not the solution.

(f) x + 4 = 2
Putting the given values in L.H.S., we get
-2 + 4 = 2
∵ L.H.S. = R.H.S.
∴ x = -2 is a solution.
0 + 4 = 4
∵ L.H.S. ≠ R.H.S.
∴ x = 0 is not the solution.
2 + 4 = 6
∵ L.H.S. ≠ R.H.S.
∴ x = 2 is not the solution.
4 + 4 = 8
∵ L.H.S. ≠ R.H.S.
∴ x = 4 is not the solution.

Question 4.
(a) Complete the table and by inspection of the table, find the solution to the equation m + 10 = 16.

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.

(c) Complete the table and find the solution of the equation z/3 = 4 using the table

(d) Complete the table and find the solution to the equation m – 7 = 3.

Solution:

Question 5.
Solve the following riddles, you may yourself construct such riddles.
Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!

(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!

(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!

(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!
Solution:
(i) According to given information, we have
3(4) + x = 34 ⇒ 12 + x = 34
⇒ x = 34 – 12 ⇒ x = 22

(ii) According to given information, we have
x + 7 = 23
⇒ x = 23 – 7 ⇒ x = 16

(iii) According to given information, we have
x – 6 = 11
⇒ x = 11 + 6 ⇒ x = 17 .

(iv) According to given information, we have
x = 22-x ⇒ x + x = 22 ⇒ 2x = 22
⇒ x = \(\frac{22}{2}\) ⇒ x = 11

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.2

पाठ्य-पुस्तक पृष्ठ संख्या # 84-85

प्रश्न 1.
पाठ्य-पुस्तक में दी हुई वक्रों को
(i) खुली या
(ii) बंद वक्रों के रूप में वर्गीकृत कीजिए :
हल :
(i) खुली वक्र – (a) व (c)
(ii) बन्द वक्र – (b), (b) व (e)

प्रश्न 2.
निम्न को स्पष्ट करने के लिए रफ आकृतियाँ बनाइए:
(a) खुला वक्र
(b) बन्द वक्र
हल :

प्रश्न 3.
कोई भी बहुभुज खींचिए और उसके अभ्यन्तर को छायांकित (Shade) कीजिए।
हल :
ABCDE एक बहुभुज है जिसके अभ्यंतर को छायांकित किया गया है।

प्रश्न 4.
संलग्न आकृति को देखकर निम्न प्रश्नों के उत्तर दीजिए :
(a) क्या यह एक वक्र है ?
(b) क्या यह बन्द है?

हल :
(a) हाँ, यह एक वक्र है।
(b) हाँ, यह बन्द वक्र है।

प्रश्न 5.
रफ आकृतियाँ बनाकर, यदि सम्भव हो, तो निम्न को स्पष्ट कीजिए :
(a) एक बन्द आकृति जो बहुभुज नहीं है।
(b) केवल रेखाखण्डों से बनी हुई खुली वक्र
(c) दो भुजाओं वाला एक बहुभुज।
हल :
(a) बन्द आकृति जो बहुभुज नहीं है।

(b) रेखाखण्डों से बनी हुई खुली वक्र

(c) दो भुजाओं वाला बहुभुज असम्भव है।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.1

पाठ्य-पुस्तक पृष्ठ संख्या # 80-81

प्रश्न 1.
संलग्न आकृति का प्रयोग करके, निम्न के नाम लिखिए :
(a) पाँच बिन्दु
(b) एक रेखा
(c) चार किरणें
(d) पाँच रेखाखण्ड

हल :
(a) पाँच बिन्दु-O, B, C, D, E
(b) एक रेखा-इसमें कई रेखाएँ हो सकती हैं \(\stackrel{\leftrightarrow}{D E}, \stackrel{\leftrightarrow}{D O}, \stackrel{\leftrightarrow}{D B}, \overrightarrow{E O}\) आदि।
(c) चार किरणें- \(\overrightarrow{D B}, \overrightarrow{D E}, \overrightarrow{E B}, \overrightarrow{O E}\) आदि (अनेक किरणें हो सकती हैं)।
(d) पाँच रेखाखण्ड- \(\overline{D E}, \overline{D O}, \overline{E O}, \overline{O B}, \overline{E B}\) आदि (अनेक रेखाखण्ड हो सकते हैं)।

प्रश्न 2.
संलग्न आकृति में दी हुई रेखा के सभी सम्भव प्रकारों के नाम लिखिए। आप इन चार बिन्दुओं में से किसी भी बिन्दु का प्रयोग कर सकते हैं।

हल :
रेखाओं को निम्न नामों से व्यक्त कर सकते हैं
\(\stackrel{\leftrightarrow}{AB}, \stackrel{\leftrightarrow}{AC}, \stackrel{\leftrightarrow}{AD}, \stackrel{\leftrightarrow}{BA}, \stackrel{\leftrightarrow}{BC}, \stackrel{\leftrightarrow}{BD}, \stackrel{\leftrightarrow}{CA}, \stackrel{\leftrightarrow}{CB}, \stackrel{\leftrightarrow}{CD}, \stackrel{\leftrightarrow}{DA}, \stackrel{\leftrightarrow}{DB}, \stackrel{\leftrightarrow}{DC}\)

प्रश्न 3.
संलग्न आकृति को देखकर नाम लिखिए :
(a) रेखाएँ जिसमें बिन्दु E सम्मिलित हैं
(b) A से होकर जाने वाली रेखा
(c) वह रेखा जिस पर O स्थित है
(d) प्रतिच्छेदी रेखाओं के दो युग्म

हल :
(a) ऐसी बहुत सी रेखाएँ है, जिनमें से एक है, \(\stackrel{\leftrightarrow}{EF}\) (या \(\stackrel{\leftrightarrow}{AE}\) )
(b) ऐसी बहुत सी रेखाएँ है, जिनमें से एक है, \(\stackrel{\leftrightarrow}{AE}\)
(c) \(\stackrel{\leftrightarrow}{CO}\) या \(\stackrel{\leftrightarrow}{CO}\)
(d) \(\stackrel{\leftrightarrow}{AE}\) और \(\stackrel{\leftrightarrow}{CO}\); \(\stackrel{\leftrightarrow}{AE}\) और \(\stackrel{\leftrightarrow}{EF}\)

प्रश्न 4.
निम्नलिखित से होकर कितनी रेखाएँ खींची जा सकती हैं ?
(a) एक बिन्दु
(b) दो बिन्दु
हल :
(a) एक बिन्दु से अनगिनत रेखाएँ खींची जा सकती हैं।
(b) दो बिन्दुओं से केवल एक रेखा खींची जा सकती है।

प्रश्न 5.
निम्नलिखित स्थितियों में से प्रत्येक के लिए एक रफ (Rough) आकृति बनाइए और चित रूप से उसे नामांकित कीजिए :
(a) बिन्दु P रेखाखण्ड \(\overline{A B}\) पर स्थित है।
(b) रेखाएँ XY और PQ बिन्दु M पर प्रतिच्छेद करती हैं।
(c) रेखा l पर E और F स्थित हैं, परन्तु D स्थित नहीं है।
(d) \(\stackrel{\leftrightarrow}{OP}\) और \(\stackrel{\leftrightarrow}{OQ}\) बिन्दु O पर मिलती हैं।
हल:

प्रश्न 6.
रेखा \(\stackrel{\leftrightarrow}{MN}\) की संलग्न आकृति को देखिए। इस आकृति के सन्दर्भ में बताइए कि निम्नलिखित कथन सत्य हैं या असत्य :

(a) Q, M, O और P रेखा \(\stackrel{\leftrightarrow}{MN}\) पर स्थित बिन्दु हैं।
(b) M, O और N रेखाखण्ड \(\overline{MN}\) पर स्थित बिन्दु हैं।
(c) M और N रेखाखण्ड \(\overline{MN}\) के अन्त बिन्दु हैं।
(d) O और N रेखाखण्ड \(\overline{OP}\) के अन्त बिन्दु हैं।
(e) M रेखाखण्ड \(\overline{QO}\) को दोनों अन्त बिन्दुओं में से एक बिन्दु है।
(f) M किरण \(\overrightarrow{OP}\) पर एक बिन्दु है।
(g) किरण \(\overrightarrow{OP}\) किरण \(\overrightarrow{QP}\) से भिन्न है।
(h) किरण \(\overrightarrow{O P}\) वही है जो किरण \(\overrightarrow{O M}\) है।
(i) किरण \(\overrightarrow{O M}\) किरण \(\overrightarrow{O P}\) के विपरीत (opposite) नहीं है।
(j) O किरण \(\overrightarrow{O P}\) का प्रारम्भिक बिन्दु नहीं है।
(k) N किरण \(\overrightarrow{N P}\) और \(\overrightarrow{N M}\) का प्रारम्भिक बिन्दु है।
उत्तर-
(a) सत्य,
(b) सत्य,
(c) सत्य
(d) असत्य,
(e) असत्य,
(f) भसत्य,
(g) सत्य,
(h) असत्य,
(i) असत्य,
(j) असत्य,
(k) सत्य।

पाठ्य-पुस्तक पृष्ठ संख्या # 83

इन्हें कीजिए

प्रश्न 1.
निम्न की सहायता से बहुभुज बनाने का प्रयत्न कीजिए :
1. माचिस की पाँच तीलियाँ
2. माचिस की चार तीलियाँ
3. माचिस की तीन तीलियाँ
4. माचिस की दो तीलियाँ
उपर्युक्त में से किस स्थिति में यह सम्भव नहीं हुआ ? क्यों?
हल :

4. माचिस की दो तीलियाँ-कोई बहुभुज सम्भव नहीं। चूँकि बहुभुज रेखाखण्डों से घिरी बन्द आकृति है। यह सम्भव नहीं कि दो रेखाखण्डों से एक बन्द घिरी हुई आकृति बनाई जाए। इसलिए दो माचिस की तीलियों से बहुभुज बनाना सम्भव नहीं है।

पाठ्य-पुस्तक पृष्ठ संख्या # 84

प्रश्न 1.
B और C इसके अन्य दो शीर्ष हैं। क्या आप इन बिन्दुओं पर मिलने वाली भुजाओं के नाम लिख सकते हैं?
हल :
इन भुजाओं के नाम हैं जिनके शीर्ष B और C हैं :
AB और BC, BC और CD

प्रश्न 2.
क्या AB और BC आसन्न भुजाएँ हैं? AE और DC के बारे में आप क्या कह सकते हैं?
हल :
चूँकि AB और BC में एक उभयनिष्ठ अंत बिन्दु है, इसलिए ये बहुभुज की आसन्न भुजाएँ हैं। AE और DC आसन्न भुजाएँ नहीं हैं। क्योंकि इनका उभयनिष्ठ अन्त बिन्दु नहीं है।

प्रश्न 3.
क्या रेखाखण्ड \(\overline{B C}\) एक विकर्ण है? क्यों या क्यों नहीं?
हल :
नहीं, रेखाखण्ड \(\overline{B C}\) विकर्ण नहीं है क्योंकि BC आसन्न शीर्षों को मिलाने वाला रेखाखण्ड है।

प्रश्न 4.
क्या आप आसन्न शीर्षों को जोड़कर विकर्ण प्राप्त कर सकते हैं?
हल :
नहीं, हम आसन्न शीर्षों को जोड़कर विकर्ण प्राप्त नहीं कर सकते हैं।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 7 Fractions Ex 7.5

Question 1.
Write these fractions appropriately as additions or subtractions:

Solution:

Question 2.
Solve:

Solution:

Question 3.
Shubham painted \(\frac{2}{3}\) of the wall space in his room. His sister Madhavi helped and painted \(\frac{1}{3}\) of the wall space. How much did they paint together?
Solution:
Fraction of wall painted by Shubham = \(\frac{2}{3}\)
Fraction of wall painted by Madhavi = \(\frac{1}{3}\)
Total painting by both of them \(=\frac{2}{3}+\frac{1}{3}=\frac{2+1}{3}=\frac{3}{3}=1\)
Therefore, they painted complete wall.

Question 4.
Fill in the missing fractions.

Solution:

Question 5.
Javed was given \(\frac{5}{7}\) of a basket of oranges. What fraction of oranges was left in the basket?
Solution:
Consider the total number of oranges to be the whole portion or 1.
Fraction of oranges left = \(1-\frac{5}{7}\)

Thus, \(\frac{2}{7}\) oranges was left in the basket.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.4

Question 1.
Answer the following:
(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
(b) The length of a rectangular hall is 4 metres less than 3 times the breadth of the hall. What is the length, if the breadth is b metres?
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Solution:
(a) Sarita’s present age = y years
(i) After 5 years, her age will be (y + 5) years.
(ii) 3 years ago, her age was (y – 3) years.
(iii) Age of her grandfather = 6 × (Sarita’s age) = (6 × y) years = 6y years
(iv) Age of her grandmother = age of her grandfather – 2 years = (6y – 2) years
(v) Age of her father = 3 (Sarita’s age) + 5 years = (3y + 5) years

(b) Breadth of the hall = b m
Length of the hall = 3 (breadth) – 4m = 3b m – 4 m = (3b – 4) m

(c) Height of the box = h cm
Length of the box = 5 × height = 5 h cm
Breadth of the box = Length – 10 cm = (5h – 10) cm

(d) Meena’s position = s
Beena’s position = 8 steps ahead = s + 8
Leena’s position = 7 steps behind = s – 7
∴ Total number of steps = 4s – 10

(e) Speed of the bus = v km/h
Distance travelled in 5 hours = 5v km
Remaining distance = 20 km
Therefore, total distance = (5v + 20) km

Question 2.
Change the following statements using expressions into statements in ordinary language.
(For example. Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.)
(a) A notebook costs ₹ p. A book costs ₹ 3 p.
(b) Tony puts q marbles on the table. He has 8 q marbles in his box.
(c) Our class has n students. The school has 20n students.
(d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Solution:
(a) A book costs 3 times the cost of a notebook.
(b) The number of marbles in Tony’s box is 8 times the marbles on the table.
(c) Total number of students in the school is 20 times the number of students in our class.
(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.

Question 3.
(a) Given Munnu’s age to be x years, can you guess what (x – 2) may show?
{Hint: Think of Munnu’s younger brother.)
Can you guess what (x + 4) may show? What (3x+ 7) may show?
(b) Given Sara’s age today to be y years. Think of her age in the future or in the past.
What will the following expression indicate? y + 7, y – 3, y + \(4 \frac{1}{2}\), y – \(2 \frac{1}{2}\).
(c) Given n students in the class like football, what may 2n show? What may \(\frac{n}{2}\) show?
(Hint: Think of games other than football).
Solution:
(a) Munnu’s age = x years Age of his. younger brother who is 2 years younger than him = (x – 2) years
Age of his elder brother who is 4 years elder than him = (x + 4) years
Age of his father whose age is 7 years more than thrice of his age = (3x + 7) years

(b) Sara’s present age = y years
Her age in past = (y – 3), (y – \(2 \frac{1}{2}\))
Her age in future = (y + 7), (y + \(4 \frac{1}{2}\))

(c) Number of students like football = n
Number of students like hockey is twice the number of students liking football,
i. e., 2n
Number of students like tennis is half the number of students liking football,
i.e., \(\frac{n}{2}\)

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Intext Questions

पाठ्य-पुस्तक पृष्ठ संख्या # 75

प्रयास कीजिए

प्रश्न 1.
अपनी पेंसिल के नुकीले सिरे से एक कागज पर चार बिन्दु अंकित कीजिए तथा उन्हें नाम A, C, P और H दीजिए। इन बिन्दुओं को विभिन्न प्रकार से नाम दीजिए। नाम देने का एक प्रकार संलग्न आकृति के अनुसार हो सकता है।

हल :
हम इन बिन्दुओं को निम्न प्रकार से नाम दे सकते हैं :

प्रश्न 2.
आसमान में एक तारा हमें एक बिन्दु के अवधारण का आभास कराता है। अपने दैनिक जीवन से इसी प्रकार की पाँच स्थितियाँ चुनकर दीजिए।
हल :
पेंसिल की नोंक, सुई की नोंक, कागज का कोना, डेस्क का कोना, वर्ग का कोना।

पाठ्य-पुस्तक पृष्ठ संख्या # 76-77

रेखाखण्ड के कुछ उदाहरण – मेज का किनारा, किसी आयत/वर्ग की भुजा, घन/ घनाभ का किनारा, दोनों सिरों पर कसा हुआ धागा, A, B को जोड़ने वाला सबसे छोटा रास्ता।

प्रयास कीजिए

प्रश्न 1.
संलग्न आकृति में दिए रेखाखण्डों को नाम दीजिए। क्या A प्रत्येक रेखाखण्ड का एक अन्त बिन्दु है ?

हल :
दी हुई आकृति में \(\overline{A B}\) (या \(\overline{B A}\) ) तथा \(\overline{A C}\) (या \(\overline{C A}\) ) रेखाखण्ड हैं।
हाँ, A प्रत्येक रेखाखण्ड का अन्त बिन्दु है।

प्रतिच्छेदी रेखा युग्मों के उदाहरण – घन के आसन्न किनारे, श्यामपट की आसन्न भुजाएँ, अंग्रेजी वर्णमाला के अक्षर V तथा L

पाठ्य-पुस्तक पृष्ठ संख्या # 78

इन्हें कीजिए

प्रश्न 1.
एक कागज लीजिए। इसे दो बार मोडिए (और मोड़ के निशान बनाइए) ताकि दो प्रतिच्छेदी रेखाएँ प्राप्त हो जाएँ और चर्चा कीजिए :
(a) क्या दो रेखाएँ एक से अधिक बिन्दुओं पर प्रतिच्छेद कर सकती हैं?
(b) क्या दो से अधिक रेखाएँ एक ही बिन्दु पर प्रतिच्छेद कर सकती हैं?
हल :
(a) नहीं, दो रेखाएँ एक से अधिक बिन्दुओं पर प्रतिच्छेद नहीं कर सकती हैं।
(b) हाँ, दो से अधिक रेखाएँ एक ही बिन्दु पर प्रतिच्छेद कर सकती हैं।

सोचिए, चर्चा कीजिए एवं लिखिए

आप समान्तर रेखाओं को और कहाँ देखते हैं ? इनके 10 उदाहरण ज्ञात करने का प्रयत्न कीजिए।
हल :
हम समान्तर रेखाओं को निम्नांकित में देख सकते हैं
पैमाने के सम्मुख किनारे, आयत के सम्मुख किनारे, श्यामपट के सम्मुख किनारे, खिड़की की सलाखें, रेल की पटरी, मेज के किनारे, घनाभ के किनारे, घन के किनारे, पेज के किनारे, अभ्यास-पुस्तिका/किताब के सम्मुख किनारे आदि।

पाठ्य-पुस्तक पृष्ठ संख्या # 79

सोचिए, चर्चा कीजिए एवं लिखिए

प्रश्न 1.
यदि \(\overrightarrow{P Q}\) एक किरण है, तो
(a) इसका प्रारम्भिक बिन्दु क्या है ?
(b) बिन्दु ए किरण पर कहाँ स्थित है ?
(c) क्या हम कह सकते हैं कि ए इस किरण का Q प्रारम्भिक बिन्दु है ?

हल :
\(\overrightarrow{P Q}\) एक किरण है :
(a) इसका प्रारम्भिक बिन्दु P है।
(b) बिन्दु Q, किरण \(\overrightarrow{P Q}\) पर स्थित है।
(c) नहीं, Q इस किरण का प्रारम्भिक बिन्दु नहीं है।
(Q, \(\overrightarrow{Q P}\) का प्रारम्भिक बिन्दु हो सकता है।)

पाठ्य-पुस्तक पृष्ठ संख्या # 80

प्रयास कीजिए

प्रश्न 1.
संलग्न दी आकृति में दर्शाई गई किरणों के नाम लिखिए।

हल :
संलग्न चित्र में \(\overrightarrow{T A}, \overrightarrow{T N}, \overrightarrow{T B}\) और \(\overrightarrow{N B}\) किरणें हैं।

प्रश्न 2.
क्या T इन सभी किरणों का प्रारम्भिक बिन्दु है ?
हल :
नहीं, T किरण \(\overrightarrow{N B}\) का प्रारम्भिक बिन्दु नहीं है।

संलग्न आकृति में एक किरण OA दी है। यह O से प्रारम्भ होती है और A से होकर जाती है। यह किरण बिन्दु B से होकर भी जाती है।

प्रश्न a.
(i) क्या आप इसे \(\overrightarrow{O B}\) भी कह सकते हैं ? क्यों? यहाँ \(\overrightarrow{O A}\) और \(\overrightarrow{O B}\) एक ही किरण को दर्शाते हैं।
(ii) क्या हम किरण \(\overrightarrow{O A}\) को किरण \(\overrightarrow{A O}\) लिख सकते हैं ? क्यों ? या क्यों नहीं ?
(iii) पाँच किरणें खींचिए और उनके उचित नाम लिखिए।
इन किरणों के सिरे पर लगे तीर क्या दर्शाते हैं ?
हल :
(i) हाँ, हम इसे \(\overrightarrow{O B}\) भी कह सकते हैं। क्योंकि एक किरण की कोई निश्चित लम्बाई नहीं होती है। किरण को अनिश्चित रूप से बढ़ाया जा सकता है। इसलिए \(\overrightarrow{O A}\) और \(\overrightarrow{O B}\) समान किरणें हैं।

(ii) यहाँ, \(\overrightarrow{O A}\) एक किरण है, जिसका प्रारम्भिक बिन्दु O हैं। इसे O से A की दिशा में अनिश्चित रूप से बढ़ाया गया है। दूसरी किरण \(\overrightarrow{A O}\) है, जिसका प्रारम्भिक बिन्दु A है। इसे A से O की दिशा में अनिश्चित रूप से बढ़ाया गया है।
अतः \(\overrightarrow{A O}\) और \(\overrightarrow{O A}\) अलग-अलग किरणें हैं। इस प्रकार \(\overrightarrow{O A}\) को \(\overrightarrow{A O}\) नहीं लिखा जा सकता है।

(iii) पाँच किरणें निम्नलिखित हैं

किरणों के सिरों पर लगे तीर दर्शाते हैं कि इन किरणों को तीर की दिशा में अनिश्चित रूप से बढ़ाया जा सकता है।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.3

Question 1.
Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
(Hint : Three possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 × 8) + 7; make the other expressions.)
Solution:
(a) (8 × 5) – 7
(b) (8 + 5) – 7
(c) (8 × 7) – 5
(d) (8 + 7) – 5
(e) 5 × (7 + 8)
(f) 5 + (7 × 8)
(g) 5 + (8 – 7)
(h) 5 – (7 + 8)

Question 2.
Which out of the following are expressions with numbers only?
(a) y + 3
(b) (7 × 20) – 8z
(c) 5 (21 – 7) + 7 × 2
(d) 5
(e) 3 x
(f) 5 – 5n
(g) (7 × 20) – (5 × 10) – 45 + p
Solution:
(c) and (d) have expressions with numbers only.

Question 3.
Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
(a) z + 1, z – 1, y + 17, y – 17
(b) 17y, \(\frac{y}{17}\), 5z
(c) 2y + 17, 2y – 17
(d) 7m, – 7m + 3, – 7m – 3
Solution:
(a) z + 1 = 1 added to z
z – 1 = 1 subtracted from z
y + 17 = 17 added to y
y – 17 = 17 subtracted from y

(b) 17y = y multiplied by 17
\(\frac{y}{17}\) = y divided by 17
5z = z multiplied by 5 .

(c) 2y + 17 = First y multiplied by 2, then 17 added to the product
2y – 17 = First y multiplied by 2, then 17 subtracted from the product

(d) 7m = m multiplied by 7
-7m + 3 = First m multiplied by -7, then 3 added to the product
-7m – 3 = First m multiplied by -7, then 3 subtracted from the product

Question 4.
Give expressions for the following cases.
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from – m
(f) -p multiplied by 5
(g) -p divided by 5
(h) p multiplied by – 5
Solution:
(a) p + 7
(b) p – 7
(c) 7p
(d) \(\frac{p}{7}\)
(e) -m – 7
(f) -5p
(g) \(\frac{-p}{5}\)
(h) -5p

Question 5.
Give expressions in the following cases.
(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by – 5 and the result is added to 16.
Solution:
(a) 2m + 11
(b) 2m – 11
(c) 5y + 3
(d) 5y – 3
(e) -8y
(f) -8y + 5
(g) 16 – 5 y
(h) -5y + 16

Question 6.
(a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution:
(a) t + 4, t – 4, 4 – t, 4t, \(\frac{t}{4}\), \(\frac{4}{t}\)
(b) 2 y + 7, 2y – 7, 7y + 2,7y – 2 and so on.

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