TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.5 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.5

Question 1.
Construct ∠ABC = 60° without using protractor.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 1

Steps of construction:

  1. Draw a line l and choose a point B on it.
  2. Place the pointer of the compasses at B and draw an arc of convenient radius which cuts the line / at a point say C.
  3. Take C as centre and with the same radius (as in step 2) in draw an arc.
  4. Now take B as centre and with the same radius (as in step 2), draw another arc cutting the previous arc (drawn in step 2) at A.
  5. Join AB, we get ∠ABC whose measure is 60°.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5

Question 2.
Construct an angle of 120° with using protractor and compasses.
Answer:
With using compasses:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 2

Steps of construction:

  1.  Draw any ray OA.
  2. Place the pointer of the compasses at O with O as centre and any convenient radius draw an arc cutting OA at M.
  3. With M as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at P.
  4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
  5. Join OQ. Then ∠AOQ is the required angle.

With Using protractor:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 4

Steps of construction :

  1. Draw a ray QR of any length.
  2. Place the centre point of the protractor at
    Q and the line aligned with the \(\overline{\mathrm{QR}}\).
  3. Mark at a point P at 120°.
  4. Join QP. ∠PQR is the required angle.

Question 3.
Construct the following angles using ruler and compasses. Write the steps of construction in each case.
(i) 75°
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 3

Steps of construction:

  1. Draw any ray OA.
  2. Place the pointer of the compasses at O. With ‘O’ as centre and any convenient radius draw an arc cutting OA at M.
  3. With M as centre and without altering radius (as in step 2) draw an arc which, cuts the first arc at P.
  4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
  5. Join OQ. ∠AOQ = 120° (∵∠AOP + ∠POQ = 60°)
    ∠AOQ = ∠AOP + ∠POQ = 60° + 60° = 120°
  6. Draw OB the perpendicular bisector of ∠POQ.
    Now ∠POB = ∠QOB = 30°
    ∴∠AOB = 90°
  7. Draw OC the perpendicular bisector of ∠POB.
    Now ∠POC = ∠BOC = 15°
  8. ∠AOC = ∠AOP + ∠OC
    = 60° + 15°
    = 75°
    (or)
    ∠AOC = ∠AOB – ∠BOC
    = 90° – 15°
    = 75°
    ∠AOC is the required angle whose measure is 75°.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5

(ii) 15°
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 5

Steps of construction:

  1. Draw a line l and mark a point O on it.
  2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line / at a point say A.
  3. With the pointer at A (as centre) and the same radius as in the step – 2, now draw an arc that passes through O.
  4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.
  5. Draw the bisector OC of ∠AOB. Now ∠AOC = 30°.
    [∠AOC = ∠BOC = 30° ]
  6. Draw the bisector OD of ∠AOC. Now ∠AOD = 15°.
    [∠AOD = ∠COD = 15°]
  7. ∠AOD = 15° is the required angle.

(iii) 105°
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 6

Steps of construction:

  1. Draw any ray OM.
  2. Place the pointer of the compasses at O. With O as centre and any convenient radius draw an arc cutting OM at A.
  3. With A as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
  4. With B as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
  5. Join QC. ∠AOC = 120°.
  6. We know that ∠AOB = ∠BOC = \(\frac{1}{2}\) ∠AOC = \(\frac{1}{2}\) × 120° = 60°
  7. Draw OD the bisector of ∠BOC.
    ∴ ∠BOD = ∠DOC = \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) × 60° = 30°
  8. Draw OE the bisector of ∠DOC.
    ∴ ∠DOE = ∠EOC = \(\frac{1}{2}\) ∠DOC = \(\frac{1}{2}\) × 30° = 15°
  9. Now ∠AOE = ∠AOD + ∠DOE = 90° + 15° = 105°
  10. ∠AOE = 105° is the required angle.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5

Question 4.
Draw the angles given in Q. 3 using a protractor.
Answer:
The angles given in Q. 3 are 75°, 105° and 15°.

(i) 75°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 7

Steps of construction:

  1. Draw a ray QR of any length.
  2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
  3. Mark a point P at 75°.
  4. Join QP. ∠RQP is the required angle.

(ii) 105°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 8

Steps of construction:

  1. Draw a ray QR of any length.
  2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
  3. Mark a point P at 105°
  4. Join QP. ∠RQP is the required angle.

(iii) 15°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 9

Steps of construction:

  1. Draw a ray QR of any length.
  2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
  3. Mark a point P at 15°
  4. Join QP. ∠RQP is the required angle.

Question 5.
Construct ∠ABC = 50° and then draw another angle ∠XYZ equal to ∠ABC without using a protactor.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 10

Steps of construction:

  1. Make an angle ∠ABC = 50° v
  2. Draw a line l and choose a point Y on it.
  3. Use the same compasses setting to draw an arc with Y as centre, cutting l in Z.
  4. Set your compasses to the length MN.
  5. With the same radius place the compasses pointer at Z and draw an arc to cut the arc drawn earlier at X.
  6. Join XY. This gives us ∠XYZ. It has the same measure as ∠ABC i.e,, 50°.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5

Question 6.
Construct ∠DEF = 60°. Bisect it, measure each half by using a protractor.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 11

Steps of construction :

  1. Draw a line P and mark a point ‘E’ on it.
  2. Place the pointer of the compasses at E and draw an arc of convenient radius which cuts the line \(\overleftrightarrow{\mathrm{PQ}}\) at a point say F.
  3. With the pointer at F (as centre), now draw an arc that passes through E.
  4. Let the two arcs intersect at D. Join ED. We get ∠DEF whose measure is 60°.
  5. Draw OC the bisector of ∠DEF.
  6. Now ∠FEC = ∠CED = 30°.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.4

Question 1.
Draw the following angles with the help of a protractor.
(i) ∠ABC = 65°
Answer:
∠ABC = 65°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 1

Steps of construction:

  1. Draw a ray BC of any length.
  2. Place the centre point of the protractor at B and the line aligned with the \(\overrightarrow{\mathrm{BC}}\).
  3. Mark a point A at 65°.
  4. Join BA. ∠ABC is the required angle.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4

(ii) ∠PQR = 136°
Answer:
∠ PQR = 136°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 2

Steps of construction:

  1. Draw a ray QR of any length.
  2. Place the centre point of the protractor at Q and the line aligned with the \(\overrightarrow{\mathrm{QR}}\).
  3. Mark a point P at 136°.
  4. Join QP. ∠PQR is the required angle.

(iii) ∠Y = 45°
Answer:
∠Y = 45°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 3

Steps of construction:

  1. Draw a ray YZ of any length.
  2. Place the centre point of the protractor
    at Y and the line aligned with the \(\overrightarrow{\mathrm{YZ}}\).
  3. Mark a point X at 45°.
  4. Join YX. ∠XYZ is the required angle.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4

(iv) ∠O =172°
Answer:
∠O = 172°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4

Steps of construction:

  1. Draw a ray OP of any length.
  2. Place the centre point of the protractor at 0 and the line aligned with the OP
  3. Mark a point N at 172°.
  4. Join ON. ∠NOP is the required angle.

Question 2.
Copy the following angles in your notebook and find their bisector.
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 5
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 6
Steps of construction :

  1. Draw a line l and choose a point P on it.
  2. Now place the compasses at A and draw an arc to cut the rays AC and AB.
  3. Use the same compasses setting to draw an arc with P as centre, cutting / at Q.
  4. Set your compasses with \(\overline{\mathrm{BC}}\) as the radius.
  5. Place the compasses pointer at Q and draw an arc to cut the existing arc at R.
  6. Join PR. This gives us ∠RPQ. It has the same measure as ∠CAB.
    This means ∠QPR has same measure as ∠BAC.
  7. Take Q as centre and with radius more than half of the length PQ, draw an arc. Now take R as centre and with the same radius, draw another arc intersecting the previous arc at S. Join PS. PS is the bisector of the given angle.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Try These

Question 1.
Can you now write the rule to form the following pattern with match-sticks ?
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 1
Answer:
The rule for the above figures is 3n.
[∵ 3 × 1, 3 × 2, 3 × 3, ………………]

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Question 2.
Find the rule for required number of matchsticks to form a pattern repeating ‘H’. How would the rule be for repeating the shape ‘L’ ?
Answer:
The required pattern of repeating ‘H’ is
‘3n’. (3 × 1, 3 × 2, ………………..)
The rule for ‘L’ is ‘2n’.

Question 3.
A line of shapes is constructed using matchsticks.
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 2
(i) Find the rule that shows how many sticks are needed to make a group of such shapes ?
(ii) How many matchsticks are needed to form a group of 12 shapes ?
Answer:
(i) Number of matchsticks that are used in each figure are
3 ; 5 ; 7 ; 9 ………………
Shape-1; Shape-2; Shape-3; Shape-4
2 × 1 + 1; 2 × 2 + 1 ; 2 × 3 + 1 ; 2 × 4 + 1 ………
In general (2n + 1) is the rule for the above shapes.

(ii) 2n + 1 = 2 × 12 + 1 = 25 [ ∵ n = 12]

(Try These)

Question 1.
Find the general rule for the perimeter of a rectangle. Use variables ‘l’ and ‘b’ for length and breadth of the rectangle respectively.
Answer:
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 3
Perimeter of a rectangle
= l + b + l + b
= 2l + 2b = 2 (l + b)

Question 2.
Find the general rule for the area of a square by using the variable ’s’ for the side of a square ?
Answer:
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 4
The rule for the area of a square is A = side × side
= s × s
A = s2

Question 3.
What would be the rule for perimeter of an Isosceles triangle ?
Answer:
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 5
The rule for the perimeter of an Isosceles triangle
= 3 × side
= 3 × a
P = 3a

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Do This

Question 1.
Find the nth term in the following sequences.
(i) 3, 6, 9, 12, …………………
(ii) 2, 5, 8, 11, …………………
(iii) 1, 8, 27, 64, 125, …………………
Answer:
(i) The nth> term of the above series is
3 × 1, 3 × 2, 3 × 3, ……………….. = 3n.

(ii) The nth term is 3 × 1 – 1, 3 × 2 – 1, 3 × 3 – 1, 3 × 4 – 1,
i.e., 3n – 1

(iii) The nth term is 1 × 1 × 1, 2 × 2 × 2, 3 × 3 × 3, ……. is
13, 23, 33 ………………….
i.e., n3

Question 2.
Write LHS and RHS of the following simple equations:
(i) 2x + 1 = 10
Answer:
2x + 1 = 10
LHS = 2x + 1, RHS = 10

(ii) 9 = y – 2
Answer:
9 = y – 2 ;
LHS = 9, RHS = y – 2

(iii) 3p + 5 = 2p + 10
Answer:
3p + 5 = 2p + 10
LHS = 3p + 5, RHS = 2p + 10

Question 3.
Write any two simple equations and give their LHS and RHS.
Answer:

Simple equation LHS RHS
1. 4x – 10 = 7x + 6 4x – 10 7x + 6
2. t – 2= 5 t – 2 5

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Question 4.
Find the solution of the equation ‘x – 4 = 2’ by Trail and Error method.
Answer:
Given equation is x – 4 = 2
If x = 1 ⇒ x – 4 = 2
⇒ 1 – 4 = 2
– 3 ≠ 2
If x = 3 ⇒ 3 – 4 = 2
– 1 ≠ 2
If x = 5 ⇒ 5 – 4 = 2 .
1 ≠ 2
If x = 6 ⇒ 6 – 4 = 2
2 = 2
x = 6 satisfies the given equation.
∴ x = 6 is the solution of the equation.

TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 12 Symmetry Ex 12.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 12 Symmetry Exercise 12.2

Question 1.
Write any five man made things which have two lines of symmetry.
Answer:
Blackboard, plain door of a house, the ruleif, writing pad, towel, sponze duster.

Question 2.
Write any five natural objects which have two or more than two lines of symmetry.
Answer:
Pumpkin, watermelon, apple, guava, boiled egg.

Question 3.
Find the number of lines of symmetry for the following shapes.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 1
Answer:
Fig (i) has 4 lines of symmetry.
Fig (ii) has 1 line of symmetry.
Fig (iii) has 2 lines of symmetry.
Fig (iv) has ‘O’ lines of symmetry.
Fig (v) has 4 lines of symmetry.
Fig (vi) has 2 lines of symmetry.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 2

Question 4.
Draw the possible number of lines of symmetry.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 3
Answer:
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 4

TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2

Question 5.
From the above problem complete the following table.

Shape Number of lines of symmetry
i) Equilateral triangle
ii) Isosceles triangle
iii) Scalene triangle
iv) Rhombus
v) Hexagon
vi) Circle

Answer:

Shape Number of lines of symmetry
i) Equilateral triangle 3
ii) Isosceles triangle 1
iii) Scalene triangle No line of symmetry
iv) Rhombus 2
v) Hexagon 6
vi) Circle Countless (Infinite)

Question 6.
A few folded sheets and designs drawn about the fold are given. In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 5
Answer:
TS 6th Class Maths Solutions Chapter 12 Symmetry Ex 12.2 6

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.3

Question 1.
Draw a line segment PQ = 5.8 cm and construct its perpendicular bisector using ruler and compasses.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3 1

Steps of construction:

  1. Draw a line segment \(\overline{\mathrm{PQ}}\) = 5.8 cm.
  2. Take P as centre and radius more than half of PQ draw two arcs above and below the line segment \(\overline{\mathrm{PQ}}\).
  3. Take Q as centre and with the same radius, draw two more arcs intersecting the previous arcs at A and B.
  4. Join A and B. This line intersects PQ at O.
  5. AB is the required perpendicular bisector of the line PQ.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3

Question 2.
Ravi made a line segment of length 8.6 Find the length of AC and BC.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3 2

Steps of construction :

  1. Draw a line segment AB = 8.6 cm.
  2. Take A as centre and radius more
    than half of the length AB, draw two arcs above and below the line segment AB. ,
  3. Take B as cfentre and with the same radius, draw two more arcs intersecting the previous- arcs at P and Q.
  4. Join PQ. This line intersects AB at C.
  5. Measure AC and BC. On measuring, it is noticed that AC = BC = 4.3 cm.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3

Question 3.
Using ruler and compasses, draw AB = 6.4 cm. Find its mid point.
Answer:
We can find the mid point of the line segment AB = 6.4 cm by drawing its perpendicular bisector.
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3 3

Steps of construction:

  1.  Draw a line segment AB = 6.4 cm.
  2. Take A as centre and radius more than half of the length AB draw two arcs above and below the line segment AB.
  3. Take B as centre and with the same radius draw two more arcs intersecting the previous arcs at M and N.
  4. Join MN. This line intersects AB at O. ‘O’ is the mid point of AB.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.2

Question 1.
Construct a circle with centre M and radius 4 cm.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 1

Steps of construction:

  1. Open the compasses for 4 cm radius.
  2. Mark a point with a sharp pencil. This is the centre. Mark it as ‘O’.
  3. Place the pointer of the compasses firmly at ‘O’.
  4. Without moving its metal point, now slowly rotate the pencil till it comes back to the starting point.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2

Question 2.
Construct a circle with centre X and diameter 10 cm.
Answer:
Diameter of the required circle = 10 cm
Radius of the circle = \(\frac{10}{2}\) = 5 cm
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 2

Steps of construction:

  1. Open the compasses for 5 cm radius.
  2. Mark ’a point with a sharp pencil. This is the centre. Mark it as ‘X’. ‘
  3. Place the pointer of the compasses firmly at X.
  4. Without moving its metal point, now slowly rotate the pencil till it comes back to the starting point.

Question 3.
Draw four circles of radii 2 cm, 3 cm, 4 cm and 5 cm with the same centre P.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 3

Steps of construction:

  1. Mark a point with sharp pencil. This is the centre mark it as ‘P’.
  2. Place the pointer of the compasses firmly at P.
  3. Without moving its metal point, slowly rotate the pencil till it comes back to the, starting point.
  4. With ‘P’ as centre and radii 3, 4 and 5 cm, repeat the procedure given in the steps 2 and 3.
  5. Now we have four circles with radii 2 cm, 3 cm, 4 cm and 5 cm having centre ‘P’. These circles are called concentric circles.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2

Question 4.
Draw any circle and mark three points A, B and C such that
(i) A Is on the circle.
(ii) B is in the interior of the circle.
(iii) C is in the exterior of the circle.
Answer:
‘O’ is the centre of the circle.
(i) A is on the circle.
(ii) B is in the interior of the circle.
(iii) C is in the exterior of the circle.
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 4

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.1

Question 1.
Construct a line segment of length 6.9 cm using a ruler and compasses.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 1

Steps of construction:

  1. Draw a line l. Mark a point A on the line l.
  2. Place the metal pointer of the compasses on the zero mark of the ruler. Open the compasses so that pencil point touches 6.9 cm mark on the ruler.
  3. Place the pointer on A on the line l and draw an arc to cut the line. Mark the point where the arc cuts the line as B.
  4. On the line l, we got the line segment AB of required length.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1

Question 2.
Construct a line segment of length 4.3 cm using the ruler.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 2

Steps of construction:

  1. Place the ruler on paper and hold it firmly. Mark a point with a sharp edged pencil against 0 cm mark of the ruler. Name the point as A.
  2. Mark another point against 3 small divisions just after the 4 cm mark. Name this point as B.
  3. Join points A and B along the edge of the ruler.
  4. AB is the required line segment of length 4.3 cm.

Question 3.
Construct a line segment MN of length 6 cm. Mark any point O on it. Measure MO, ON and MN. What do you observe ?
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 3

Steps of construction :

  1. Draw a line l. Mark a point M on the line l.
  2. Place the metal pointer of the compasses on the zero mark of the ruler. Open it to
    place the pencil point upto the 6 cm mark.
  3. Taking caution that the opening of the compasses has not changed, place the pointer on M and swing an arc to cut / at N.
  4. MN is a line segment of required length.
  5. Mark any point ‘O’ on MN.
  6. Measure the length of MO and ON. It is found that MO = 3.8 cm and ON = 2.2 cm MO + ON = 3.8 +2.2 = 6 cm
    It is noticed that \(\overline{\mathrm{MO}}+\overline{\mathrm{ON}}\) = \(\overline{\mathrm{MN}}\)

Question 4.
Draw a line segment \(\overline{\mathbf{A B}}\) of length 12 cm. Mark a point C on the line segment \(\overline{\mathbf{A B}}\), such that \(\overline{\mathbf{A C}}\) = 5.6 cm. What should be the length of \(\overline{\mathbf{C B}}\)? Measure the length of \(\overline{\mathbf{C B}}\).
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 4

Steps of construction:

  1. Draw a line l. Mark a point A on the line l.
  2. Place the metal pointer of the compasses on the zero mark of the ruler. Open it to place the pencil point upto the 12 cm mark.
  3. Taking caution that the opening of the compasses has not changed, place the pointer on A and swing an arc to cut / at B.
  4. \(\overline{\mathbf{A B}}\) is a line segment of required length.
  5. Similarly mark the point C on l such that \(\overline{\mathbf{A C}}\) = 5.6 cm.
  6. The length of \(\overline{\mathbf{C B}}\) should be 6.4 cm = (12 – 5.6) on measuring, the length of \(\overline{\mathbf{C B}}\) = 6.4 cm.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1

Question 5.
Given that AB = 12 cm
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 5
(i) From the figure measure the lengths of the following line segments.
(a) \(\overline{\mathrm{CD}}\)
(b) \(\overline{\mathrm{DB}}\)
(c) \(\overline{\mathrm{EA}}\)
(d) \(\overline{\mathrm{AD}}\)
Answer:
(a) \(\overline{\mathrm{CD}}\) = 2.8 cm
(b) \(\overline{\mathrm{DB}}\) = 4.3 cm
(c) \(\overline{\mathrm{EA}}\) = 17.3 cm
(d) \(\overline{\mathrm{AD}}\) = 16.3 cm

(ii) Verify \(\overline{\mathbf{A E}}-\overline{\mathbf{C E}}\) = \(\overline{\mathbf{A C}}\)
Answer:
\(\overline{\mathbf{A E}}-\overline{\mathbf{C E}}\) = 17.3 cm – 3.8 cm = 13.5 cm
\(\overline{\mathrm{BC}}\) = 1.5 cm
\(\overline{\mathrm{AC}}\) = \(\overline{\mathbf{A B}}+\overline{\mathbf{B C}}\)
= 12 cm + 1.5 cm
= 13.5 cm
∴ \(\overline{\mathrm{AE}}-\overline{\mathrm{CE}}\) = \(\overline{\mathrm{AC}}\)

Question 6.
\(\overline{\mathrm{AB}}\) = 3.8 cm. Construct \(\overline{\mathrm{MN}}\) by compasses such that the length of \(\overline{\mathrm{MN}}\) = 3AB’is thrice that of \(\overline{\mathrm{AB}}\). Verify this with the help of a ruler.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 6

Steps of construction:

  1. Draw a line l. Mark a point M on it.
  2. Place the compasses pointer on the zero mark of the ruler.-Open it to place the pencil upto the 3.8 cm mark.
  3. Taking caution that the opening of the compasses has not changed, place the pointer on M and swing an arc to cut l at A.
  4. \(\overline{\mathrm{MA}}\) is a line segment of 3.8 cm.
  5. Similarly place the pointer on A and swing an arc to cut l at B such that MA = AB.
  6. Again place the pointer on B and swing an arc to cut l at N such that AB = BN.
  7. Now \(\overline{\mathrm{MN}}\) is a line segment of 11.4 cm.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 11 Ratio and Proportion InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

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Question 1.
Observe the example and fill in the blanks
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 1
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 2
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 3

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

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Question 1.
In the given figure, find the ratio of
i) Shaded parts to unshaded parts.
ii) Shaded parts to total parts.
iii) Unshaded parts to total parts.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 4
Answer:
i) Shaded parts to unshaded parts = 4: 1
ii) Shaded parts to total parts = 4 : 5
iii) Unshaded parts to total parts = 1: 5

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Question 1.
Complete the following table.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 5
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 6

Question 2.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 7
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 8

Question 3.
In the following figures express the ratio of shaded parts to unshaded parts in the simplest terms.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 9
Answer:
Number of shaded parts = 4
Number of unshaded parts = 12
Ratio of shaded parts to unshaded parts = 4:12
= 1:3

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 10
Answer:
Number of shaded parts = 3
Number of unshaded parts = 6
Ratio of shaded parts to unshaded parts = 3:6
= 1:2

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 11
Answer:
Number of shaded parts = 4
Number of unshaded parts = 4
Ratio of shaded parts to unshaded parts = 4 : 4
= 1:1

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 12
Answer:
Number of shaded parts = 6
Number of unshaded parts = 6
Ratio of shaded parts to unshaded parts = 6:6

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 13
Answer:
Number of shaded parts = 8
Number of unshaded parts = 12
Ratio of shaded parts to unshaded parts = 8 : 12
= 2:3

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 14
Answer:
Number of shaded parts = 3
Number of unshaded parts = 6
Ratio of shaded parts to unshaded parts = 3:6
= 1:2

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions

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Question 1.
Make a pattern with squared tiles using black and white tiles In the ratIo 2:5. There are many possible ways.
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 15
One of the way is 2:5
2 × 5 : 5 × 5
10 : 25

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Question 1.
i) In the given square rule paper 5 squares, colour 3 squares red and 2 squares green.
ii) If 10 squares are given, find how many are to be red and how maiy of them are to be green so that it becomes proportionate to the figure.
iii) If there are 15 squares then colour them accordingly.
TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion InText Questions 16
Answer:
Student Activity.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.4

Students can practice TS SCERT Class 6 Maths Solutions Chapter 11 Ratio and Proportion Ex 11.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Exercise 11.4

Question 1.
If three apples cost ₹ 45, how much would five apples cost ₹ *
Answer:
Cost of 3 apples = ₹ 45
Cost of each apple = ₹45 ÷ 3
= ₹ 15
∴ Cost of 5 apples = ₹ 15 × 5
= ₹ 75

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.4

Question 2.
Laxmi bought 7 books for a total of ₹ 56. How much would she pay for just 3 books ?
Answer:
Cost of 7 books = ₹ 56
Cost of one book = ₹ 56 ÷ 7 = ₹ 8
∴ Cost of 3 books = ₹ 8 × 3 = ₹ 24

Question 3.
Reena wants to prepare vegetable pulao, she needs 300 grams of rice, if she has to feed 4 people. How much of rice is needed if the same pulao is prepared for 7 people ?
Answer:
Quantity of rice required for the preparation of vegetable pulao for 4 people is 300 grams.
Quantity of rice needed to prepare pulao for 1 person = \(\frac{300}{4}\) = 75 grams
Quantity of rice needed to prepare pulao for 7 people = 75 × 7 grams = 525 grams

Question 4.
The cost of 16 chairs is ₹ 3600. Find the number of chairs that can be purchased for ₹ 4500.
Answer:
Cost of 16 chairs = ₹ 3600
Cost of one chair = \(\frac{3600}{16}\) = ₹ 225
Number of chairs that can purchased for ₹ 4500 = \(\frac{4500}{225}\)
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 1

Question 5.
A train moving at a constant speed covers a distance of 90 km in 2 hours. Find the time taken by the train to cover a distance of 540 km at the same speed.
Answer:
Time taken by the train to cover a distance of 90 km in 2 hours.
It covers 45 km in = \(\left(=\frac{90}{2}\right)\) 1 hour.
Time taken by the train to cover 540 km 540
= – \(\frac{540}{45}\) hours
= 12 hours

Question 6.
The income of Kumar for 3 months is ₹ 15,000. If his monthly income remains the same then,
(i) How much will he earn in 5 months ?
(ii) In how many months will he earn ₹ 95,000 ?
Answer:
The income of Kumar for 3 months is ₹ 15,000.
The income of Kumar for 1 month is = ₹ \(\frac{15,000}{3}\) = ₹ 5,000
Kumar earns ₹ 5000 in one month.

(i) The income of Kumar for 5 months = ₹ 5000 × 5 = ₹ 25,000

(ii) The time needed to earn ₹ 95,000
= \(\frac{95,000}{5,000}\) = 19 months
∴ In 1 year and 7 months, he will earn ₹ 95,000.

Question 7.
If the cost of 7 meters of cloth is ₹ 294, find the cost of 5 m of cloth.
Answer:
Cost of 7 meters of cloth = ₹ 294
Cost of 1 meter of cloth = ₹ \(\frac{294}{7}\) = ₹ 42
The cost of 5 meters of cloth = ₹ 42 × 5 = ₹ 210

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.4

Question 8.
A farmer has sheep and cows in the ratio 8:3.
(i) How many sheep has the farmer, if he has 180 cows ?
(ii) Find the ratio of the number of sheep to the total number of animals the farmer has.
(iii) Find the ratio of the total number of animals with the farmer to the number of cows with him.
Answer:
The ratio of sheep and cows is 8 : 3.
(i) Total number of parts = 8 + 3 = 11
Value of 3 parts = 180
Value of 1 part = \(\frac{180}{3}\) = 60
The number of sheep that the farmer has = 60 × 8 = 480
Total number of animals that the farmer has = 180 + 480 = 660

(ii) The ratio of the number of sheep to that of total animals = 480 : 660
= 8 : 11

(iii) The ratio of the number of animals to that of cows = 660 : 180 = 11:3

Question 9.
Are 3, 5, 15, 9 in proportion ? If we change their order, can we think of proportional pairs ? Write as many proportionality statements as you can for the above example ?
Answer:
3, 5, 15, 9 are the given numbers.
Ratio of 3 to 5 = 3 : 5
Ratio of 15 to 9 = 15 : 9 (i.e) 5 : 3
Since 3 : 5 5 : 3
3, 5, 15, 9 are not in proportion. .
We can think of proportional pairs if the order is changed.
3, 9, 5, 15 are in proportion.
5, 3, 15, 9 are in proportion.
3, 5, 9,15 are in proportion.
9, 3, 15, 5 are also in proportion.

Question 10.
The temperature has dropped by 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how much more will the temperature drop in the next 10 days ?
Answer:
In 30 days the temperature dropped by 15°, Celsius
In 1 day the temperature dropped by 150 \(\frac{15^{\circ}}{30}\) Celsius
In 10 days the temperature will drop 15 by \(\frac{15}{30}\) × 10 = 5° Celsius

Question 11.
Fill in the following blanks.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 2
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 3

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.4

Question 12.
(i) Ratio of breadth and length of a hall is 2:5. Complete the following table that shows some possible breadths and lengths of the hall.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 4
Add 3 more of your choice.
(ii) Find the ratio of length to breadth of your classroom.
Answer:
(i) Ratio of breadth and length of a hall is 2 :5.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 5

(ii) Length of our classroom = 5 m
Breadth of our classroom = 4 m
The ratio of length to breadth of our classroom = 5m : 4m = 5 : 4

Question 13.
Geetha earns ₹ 12,000 a month, out of which she saves ₹ 3,000. Find the ratio of her
(i) Expenditure to savings
(ii) Savings to her income
(iii) Expenditure to her income
Answer:
Earning of Geetha = ₹ 12,000 per month
Geetha’s saving = ₹ 3,000
Geetha’s expenditure
= ₹ 12,000 – ₹ 3000 = ₹ 9,000
(i) The ratio of Geetha’s expenditure to savings is 9000 : 3000 = 3:1
(ii) The ratio of Geetha’s savings to her income is 3000 :12000 = 1:4
(iii) The ratio of Geetha’s expenditure to her income is 9000 : 12000 = 3:4

Question 14.
There are 45 persons working in an office. The number of females is 25 and the remaining are males. Find the ratio of .
(i) The number of females to number of males.
(ii) The number of males to the number of females.
Answer:
The number of persons in the office . =45
The number of females = 25
The number of males = 45 – 25 = 20
(i) The ratio of the number of females to the number of males = 25 : 20 = 5 : 4
(ii) The ratio of number of males to the number of females = 20 : 25 = 4 : 5

Question 15.
A bag of sweets contain yellow and green sweets. For every 2 yellow sweets, there are 6 green sweets. Complete this table based on the above information.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 6
Answer the following questions.
(i) What is the ratio of green to yellow sweets ?
(ii) If you have 8 yellow sweets, how many green sweets will you have ?
(iii) If there are 32 sweets in the medium sized bag. How many will be yellow ?
(iv) In the super fat size bag there are 40 sweets. How many will be green ?
(v) In a bag if there are 16 yellow sweets. How many total sweets are in the bag?
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 7

(i) The ratio of green to yellow sweets is 6 : 2 (i.e.,) 3:1
(ii) For 2 yellow sweets, there are 6 green sweets. For 8 yellow sweets, there are \(\frac{6}{2}\) × 8 = 24
(iii) In a bag of 32 sweets, the number of yellow sweets = \(\frac{2}{8}\) × 32 = 8
(iv) In a bag of 40 sweets, the number of green sweets = \(\frac{6}{8}\) × 40 = 30
(v) For 16 yellow sweets, total sweets in the bag = \(\frac{8}{2}\) × 16 = 64

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.4

Question 16.
In a school survey it was found that for every 4 girls there were 5 boys. Fill In the following table.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 8
Now answer these questions:
(i) What Is the ratio of girls to boys ?
(ii) In a class of 27 children, how many would be girls ?
(iii) There are 54 children In a class. How many are boys ?
(iv) If 20 girls join in a year. How many boys would join ?
Answer:
For every 4 girls, there were 5 boys.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.4 9

(i) The ratio of girls to boys is 4 : 5
(ii) The number of girls in a class of 27 children is \(\frac{4}{9}\) × 27 = 12
(iii) The number of boys in a class of 54 students is \(\frac{5}{9}\) × 54 = 30
(iv) If 20 girls join in a year, the number of boys that would join is \(\frac{5}{4}\) × 20 = 25

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.3

Students can practice TS SCERT Class 6 Maths Solutions Chapter 11 Ratio and Proportion Ex 11.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Exercise 11.3

Question 1.
A bag of 25 marbles is shared between Rahul and Kiran in the ratio 2:3.
(i) How many marbles does Kiran receive?
(ii) How many marbles does Rahul receive?
Answer:
Ratio of Rahuls marbles to Kiran’s marbles is 2 : 3.
Total parts = 2 + 3 = 5
Total number of marbles = 25
(i) Kirans share of marbles = \(\frac{25 \times 3}{5}\) = 15
(ii) Rahuls share of marbles = \(\frac{25 \times 2}{5}\) = 10

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.3

Question 2.
A point X on AB = 14 cm divides it in the ratio 3 : 4. Find the length of AX and XB.
Answer:
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.3 1
Length of line segment AB = 14 cm.
Ratio of line segments \(\overline{\mathrm{AX}}\) and \(\overline{\mathrm{XB}}\) is 3:4
Total parts = 3 + 4 = 7
Length of 7 parts = 14 cm
Length of each part = \(\frac{14}{7}\) = 2 cm
Length of the line segment \(\overline{\mathrm{AX}}\) = 3 × 2 = 6 cm
Length of the line segment \(\overline{\mathrm{XB}}\) = 4 × 2 = 8 cm

Question 3.
Geetha and Laxmi won ₹ 1,050 in a game. They agreed to share the amount in the ratio of 3 : 4. How much does each person receive ?
Answer:
Amount won in the game and shared between Geetha and Laxmi is Rs. 1,050.
Ratio of Geetha’s amount to Laxmi’s amount is 3 : 4.
Total number of parts = 3 + 4 = 7
Value of 7 parts = 1,050
Therefore, value of each part = \(\frac{1050}{7}\) = 150
Geetha’s share = Rs. 150 × 3
Laxmi’s share = Rs. 150 × 4

Question 4.
Divide ₹ 3600 between Satya and Vishnu in the ratio 3:5.
Answer:
Amount that is divided between Satya and Vishnu = ₹ 3,600
Ratio of Satya’s amount to Vishnu’s amount is 3 : 5
Total parts =3 + 5 = 8
Value of 8 parts = ₹ 3,600
Value of each part = ₹ \(\frac{3600}{8}\) = Rs. 450
Satya’s share = ₹ 450 × 3 = ₹ 1350
Vishnu’s share = ₹ 450 × 5 = ₹ 2250

Question 5.
Two numbers are in the ratio 5 : 6. If the sum of the numbers is 132, find the two numbers.
Answer:
Sum of the two numbers = 132
Ratio of two numbers is 5 : 6.
Total parts = 5 + 6 = 11
Value of 11 parts = 132
∴ Value of each part = \(\frac{132}{11}\) = 12
The first number = 12 × 5 = 60
The second number = 12 × 6 = 72
The two numbers are 60 and 72

Question 6.
Estimate the ratio in which X divides AB and then check your estimation by measuring it.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.3 2
Answer:
The ratio of \(\overline{\mathrm{AX}}\) to \(\overline{\mathrm{XB}}\) is 1 : 1
Verification : \(\overline{\mathrm{AX}}\) = 3.65 ; \(\overline{\mathrm{XB}}\) = 3.65

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.3

Question 7.
The income and savings of an employee are in the ratio 11: 2. If his expenditure is ₹ 5346. Then find his income and savings.
Answer:
The ratio of the income and savings of an employee is 11 : 2.
If income is 11 parts, savings 2 parts, then the expenditure will be equal to 9(= 11-2) parts.
Value of 9 parts = ₹ 5346
∴ Value of each part = \(\frac{5346}{9}\) = ₹594
Employee’s income = ₹ 594 × 11 = ₹ 6534
Employee’s savings = ₹ 594 × 2 = ₹ 1188

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.1

Students can practice TS SCERT Class 6 Maths Solutions Chapter 11 Ratio and Proportion Ex 11.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Exercise 11.1

Question 1.
Complete the following table.
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.1 1
Answer:

First quantity Second Quantity Ratio
(i) 3 5 3 : 5
(ii) 7 11 7 : 11
(iii) 2 3 2 : 3
(iv) 5 8 5 : 8
(v) 3 5 3 : 5

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.1

Question 2.
Compare:
TS 6th Class Maths Solutions Chapter 11 Ratio and proportion Ex 11.1 2

(i) Number of blue coloured squares is ………………. times the number of blue red colour squares.
Answer:
Red coloured squares are 4 and blue coloured squares are 6.
So, blue coloured squares is \(\frac{3}{2}\) times the number of red colour squares.

(ii) Number of red coloured squares is ………………… times of the number of blue red coloured squares.
Answer:
\(\frac{2}{3}\)

(iii) Find the ratio of number of blue squares to the number of red squares.
Answer:
6 : 4 (or) 3 : 2

Question 3.
Solve the following.
(i) A milk man adds 250 ml of water to 1 litre of milk. Find the ratio of water to milk.
Answer:
Quantity of water = 250 ml
Quantity of milk = 1 litre = 1000 ml (∵ 1 litre = 1000 ml)
∴ Ratio of water to milk = 250 : 1000 = 1:4

(ii) Satya’s mother bought 4 kg pulses and 50g chilli powder. Find the ratio of weights of chilli powder to pulses. What is the ratio of weights of the pulses to chilli powder ?
Answer:
Weight of pulses = 4 kg
= 4 × 1000 gms (∵ 1 kg = 1000 grams)
= 4000 gms
Weight of chilli powder = 50 gms
Ratio of weights of chilli powder to pulses = 50 : 4000 = 1 : 80
Ratio of weights of the pulses to chilli powder = 4000 : 50 = 80 : 1

TS 6th Class Maths Solutions Chapter 11 Ratio and Proportion Ex 11.1

(iii) Rani takes 30 minutes to reach school from home. Ismail takes √2 an hour to cover the same distance. Find the ratio of time taken by Rani to the time taken by Ismail.
Answer:
Time taken by Rani to reach school from home = 30 minutes
Time taken by Ismail to cover the same distance = ½ hour = 30 minutes (∵ 1 hour = 60 minutes)
Ratio of time taken by Rani to the time taken by Ismail = 30 : 30 = 1 : 1