CBSE Class 6

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2

Question 1.
Find the number of lines of symmetry for each of the following shapes :

Solution :

Question 2.
Copy the triangle in each of the following figures, on squared paper. In each case, draw the line(s) of symmetry if any and identify the type of triangle. (Some of you may like to trace the figures and try paper-folding first!)

Solution :

Question 3.
Complete the following table :

Solution :

Question 4.
Can you draw a triangle which has
(a) exactly one line of symmetry?
(b) exactly two lines of symmetry?
(c) exactly three lines of symmetry?
(d) no lines of symmetry?
Sketch a rough figure in each case.
Solution :
(a) Yes; an isosceles triangle

(b) No
(c) Yes; an equilateral triangle

(d) Yes, an equilateral triangle

Question 5.
On a squared paper, sketch the following:
(a) A triangle with a horizontal line of symmetry’ but no vertical line of symmetry.
(b) A quadrilateral with both horizontal and vertical lines of symmetry.
(c) A quadrilateral with a horizontal line of symmetry but no vertical line of symmetry.
(d) A hexagon with exactly two lines of symmetry.
(e) A hexagon with six lines of symmetry.
(Hint: It will be helpful if you first draw the lines of symmetry and then complete the figures.)
Solution :


Question 6.
Trace each figure and draw the lines of symmetry, if any :



Solution :

no line symmetry


Question 7.
Consider the letters of English alphabets A to Z. List among them the letters which have
(a) vertical lines of symmetry (like A)
(b) horizontal lines of symmetry (like B)
(c) no lines of symmetry (like Q)

Solution :
(a) A, H, I, M, O, T, U, V, W, X, Y
(b) B, C, D, E, H, I, K, O, X
(c) F, G, J, L, N, P, Q, R, S, Z

Question 8.
Given here are figures of a few folded sheets and designs drawn about the fold. In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.

Solution :

We hope the NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2.

• Ratio and Proportion Class 6 Ex 12.1
• Ratio and Proportion Class 6 Ex 12.3

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2

Question 1.
Determine if the following are in proportion:
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
(c) 24, 28, 36, 48
(d) 32, 48, 70, 210
(e) 4, 6, 8, 12
(f) 33, 44, 75, 100
Solution :





Question 2.
Write True (T) or False (F) against each of the following statements :
(a) 16: 24: : 20: 30
(b) 21: 6: 35: 10
(c) 12: 18: 28: 12
(d) 8: 9: : 24: 27
(e) 5.2: 3.9: : 3: 4
(f) 0.9: 0.36: : 10: 4
Solution :




Question 3.
Are the following statements true₹ I
(a) 40 Persons : 200 Persons =₹ 15: ₹75 ;
(b) 7.5 litre : 15 litre = 5 kg : 10 kg 1
(c) 99 kg : 45 kg = f 44 : ₹20
(d) 32 m : 64 m = 6 sec: 12 sec
(e) 45 km: 60 km = 12 hours : 15 hours.
Solution :


Since the two ratios are equal, therefore, the given statement is true.

Since the two ratios are equal, therefore, the given statement is true.

Since, the two ratios are equal, therefore, the given statement is true.


Since the two ratios are equal, therefore the given statement is true.

Since the two ratios are not equal, therefore the given statement is false.

Question 4.
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm: 1 m and 7 40 : 7160
(b) 39 litre : 65 litre and 6 bottle : 10 bottle
(c) 2 kg : 80 kg and 25 g : 625 g
(d) 200 ml: 2.5 litre and ₹4 ; ₹ 50.
Solution :
(a)
∴ 1 m= 100 cm

Since the two ratios are equal, therefore, the given ratios are in proportion. Middle terms are 1 m and ₹ 40. Extreme terms are 25 cm and ₹ 160.

Since the two ratios are equal, therefore, the given ratios are in proportion. Middle terms are 65 liters and 6 bottles. Extreme terms are 39 liters and 10 bottles.


Since the two ratios are not equal, therefore, the given ratios are not in proportion.

Since the two ratios are equal, therefore, the given ratios are in proportion. Middle terms are 2.5 litres and ₹ 4. Extreme terms are 200 ml and ₹ 50.

We hope the NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5.

• Algebra Class 6 Ex 11.1
• Algebra Class 6 Ex 11.2
• Algebra Class 6 Ex 11.3
• Algebra Class 6 Ex 11.4

NCERT Solutions for Class 6 Maths 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 x 3)-19 = 8
(e) 5 x 4 – 8 =2x
(f) x – 2 =0
(g) 2m<30
(h) 2n+1=11
(i) 7=(11 x 5) -(12x 4)
(j) 7=(11×2)+p
(k) 20=5y
(l) latex]\frac { 3q }{ 2 } [/latex]<5
(m) z+12>24
(n) 20-(10-5)=3×5
(o) 7-x =5
Solution.

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


Solution.
(a) Yes
(b) Yes
(c) No
(d) No
(e) No
(f) Yes
(g) No
(h) No
(i)  Yes
(j) Yes
(k) No
(l) No
(m) No
(n) No
(o) No
(p) No
(q) Yes

Question 3.
Pick out the solution from the values given bracket next to each equation. Show that the 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 – . y (-2, 0, 2. 4)
Solution.
(a) Solution is 12
For m = 10
L.H.S. = 5 x 10 = 50 which is not equal to R.H.S.
For m = 5
L.H.S. = 5 x 5 = 25 which is not equal to R.H.S.
For m = 15
L.H.S. = 5 x 15 = 75 which is not equal to R.H.S.

(b) Solution is 8
For n = 12
L.H.S. = 12+12
= 24 which is not equal to R.H.S.
For n = 20
L.H.S. = 20+12
= 32 which is not equal to R.H.S.
For n =0
L.H.S. = 0+12
= 12 which is not equal to R.H.S.

(c) Solution is 10
For p = 0
L.H.S. = 0 – 5 = – 5 which is not equal to R.H.S.
For p = 5
L.H.S. = 5-5 = 0 which is not equal to R.H.S.
For p = – 5
L.H.S. = – 5 – 5 = – 10 which is not equal to R.H.S.

(d) Solution is 14
For g = 7
L.H.S. = \(\frac { q }{ 2 } \) which is not equal to R.H.S.
For q = 2
L.H.S. = \(\frac { 2 }{ 2 } \) = 1 which is not equal to R.H.S.
For q = 10
L.H.S. = \(\frac { 10 }{ 2 } \) = 5 which is not equal to R.H.S.

(e) Solution is 4
For r = – 4
L.H.S. = -4-4
= – 8 which is not equal to R.H.S.
For r = 8
L.H.S. = 8-4 = 4 which is not equal to R.H.S.
For r = 0
L.H.S. = 0 – 4 = – 4 which is not equal to R.H.S.

(f) Solution is – 2
For x = 0
L.H.S. = 0 + 4 = 4 which is not equal to R.H.S. For x = 2
L.H.S. = 2 + 4 = 6 which is not equal to R.H.S.
For x : = 4
L.H.S. = 4 + 4 = 8 which is not equal to R.H.S.

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 using the table the solution of the equation z/3 = 4

(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 account 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) Let the required number be x.
Number of corners of the square = 4
Number obtained on counting every comer thrice = 4 x 3=12
According to the question, x + 12 = 34 ⇒ x = 34 -12 = 22
Hence, I am 22.

(ii) Let the required number be x.
Number of players in a cricket team = 11
According to the question.
x – 6 = 11  ⇒ x =11+6=17
Hence, lam 17.

(iii) Let the required number be v.
Number of days of a week = 7
According to the question.
x + 7 = 23    ⇒ x = 23-7 =16
Hence, lam 16.

(iv) Let the required number be x.
According to the question.
22 – x = x ⇒ x + x = 22
2x = 22    ⇒ x = \(\frac { 10 }{ 2 } \) = 11
Hence, lam 11.

We hope the NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4.

• Algebra Class 6 Ex 11.1
• Algebra Class 6 Ex 11.2
• Algebra Class 6 Ex 11.3
• Algebra Class 6 Ex 11.5

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4

Question 1.
Answer the following :
(a) Take Sarita’s present age to he 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’sfather’s age is 5years more than 3 times Sarita’s age. What is her father’s age?
(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?
(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 hilltop. Meena is at step s, Beena is 8 steps ahead and Leena7 steps behind. Where are Beena and Leena? The total number of steps to the hilltop is 10 less than 4 times what Meena has reached. Express the total number of steps using.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has traveled 5 hours,
Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using.

Solution.
(a) (y + 5) years
(ii) (y – 3) years
(iii) 6y years
(iv) (6y – 2) years
(v) (3y + 5) years
(b) Length (l) = (3b – 4) meters
(c) Length of the box = 5h cm
Breadth of the box = (5h – 10) cm
(d) Beena is at step (s + 8)
Leena is at step (s – 7)
Total number of steps = 4s – 10.
(e) Speed of the bus = v km/hr
Distance travelled in 5 hours = 5v km.
∴Total distance = (5v + 20) km

Question 2.
Change the following statements using expressions into statements in the 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 Rs p. A book costs Rs 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 three times the cost of a notebook.
(b) Tony’s box contains 8 times the marbles on the table.
(c) The total number of students in the school in 20 times that of 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 in an arrangement 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 — show?(Hint: Think of games other than football).
Solution.
(a) (x – 2) may show the age is Munnu’s younger brother. (a + 4) may show the age of Munnu’s elder brother. (3a+ 7) may show the age of Munnu’s grand mother.
(b) y + l indicates her age 7 years from now.
y – 3 indicates her age 3 years back.
y+ 4\(\frac { 1 }{ 2 } \) indicates her age 4\(\frac { 1 }{ 2 } \) years from now
y-2\(\frac { 1 }{ 2 } \) indicates her age 2\(\frac { 1 }{ 2 } \) years back.
(c) 2n may show the number of students who like cricket.
\(\frac { n }{ 2 } \)may show the number of students who like hockey.

We hope the NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4, drop a comment below and we will get back to you at the earliest.