Practice Set 3.2 Geometry 9th Standard Maths Part 2 Chapter 3 Triangles Solutions Maharashtra Board

9th Standard Maths 2 Practice Set 3.2 Chapter 3 Triangles Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

Class 9 Maths Part 2 Practice Set 3.2 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 1
By SSS test
∆ABC ≅ ∆PQR

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 2
By SAS test
∆ XYZ ≅ ∆LMN

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 3
By ASA test
∆PRQ ≅ ∆STU

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 4
By hypotenuse side test
∆LMN ≅ ∆PTR

Question 2.
Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 5
From the information shown in the figure,
In ∆ABC and ∆PQR,
∠ABC ≅ ∠PQR
seg BC ≅ seg QR
∠ACB ≅ ∠PRQ
∴ ∆ABC ≅ ∆PQR [ASA test]
∴ ∠BAC ≅ ∠QPR [Corresponding angles of congruent triangles]
seg AB ≅ segPQ and segAC ≅ seg PR [Corresponding sides of congruent triangles]

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 6
From the information shown in the figure,
In ∆PTQ and ∆STR,
seg PT ≅ seg ST
∠PTQ ≅ ∠STR [Vertically opposite angles]
seg TQ ≅ seg TR
∴ ∆PTQ ≅ ∆STR [SAS test]
∴ ∠TPQ ≅ ∠TSR and ∠TQP ≅ ∠TRS [Corresponding angles of congruent triangles]
seg PQ ≅ seg SR [Corresponding sides of congruent triangles]

Question 3.
From the information shown in the figure, state the test assuring the congruence of ∆ABC and ∆PQR. Write the remaining congruent parts of the triangles.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 7
Solution:
In ∆BAC and ∆PQR,
seg BA ≅ seg PQ
seg BC ≅ seg PR
∠BAC ≅ ∠PQR = 90° [Given]
∴ ∆BAC ≅ ∆PQR [Hypotenuse side test]
∴ seg AC ≅ seg QR [c.s.c.t.]
∠ABC ≅ ∠QPR and ∠ACB ≅ ∠QRP [c.a.c.t.]

Question 4.
As shown in the adjoining figure, in ∆LMN and ∆PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 8
Solution:
In ∆LMN and ∆PNM,
seg LM ≅ seg PN
seg LN ≅ seg PM [Given]
seg MN ≅ seg NM [Common side]
∴ ∆LMN ≅ ∆PNM [SSS test]
∴ ∠LMN ≅ ∠PNM,
∴ ∠MLN ≅ ∠NPM, and ∠LNM ≅ ∠PMN [c.a.c.t.]

Question 5.
In the adjoining figure, seg AB ≅ seg CB and seg AD ≅ seg CD. Prove that ∆ABD ≅ ∆CBD.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 9
Solution:
proof:
In ∆ABD and ∆CBD,
seg AB ≅ seg CB
seg AD ≅ seg CD [Given]
seg BD ≅ seg BD [Common side]
∴ ∆ABD ≅ ∆CBD [SSS test]

Question 6.
In the adjoining figure, ZP ≅ ZR, seg PQ ≅ seg RQ. Prove that APQT ≅ ARQS.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.2 10
Proof:
In ∆PQT and ∆RQS,
∠P ≅ ∠R
seg PQ ≅ seg RQ [Given]
∠Q ≅ ∠Q [Common angle]
∴ ∆PQT ≅ ∆RQS [ASA test]

Maharashtra Board Class 9 Maths Solutions

Practice Set 2.1 Geometry 9th Standard Maths Part 2 Chapter 2 Parallel Lines Solutions Maharashtra Board

9th Standard Maths 2 Practice Set 2.1 Chapter 2 Parallel Lines Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 2 Parallel Lines.

Class 9 Maths Part 2 Practice Set 2.1 Chapter 2 Parallel Lines Questions With Answers Maharashtra Board

Question :
In the given figure, line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 1
i. ∠RHD
ii. ∠PHG
iii. ∠HGS
iv. ∠MGK
Solution:
i. ∠DHP = 85° …..(i)
∠DHP + ∠RHD = 180° [Angles in a linear pair]
85° + ∠RHD = 180°
∴ ∠RHD = 180°- 85°
∴ ∠RHD = 95° …..(ii)

ii. ∠PHG = ∠RHD [Vertically opposite angles]
∴ ∠PHG = 95° [From (ii)]

iii. line RP || line MS and line DK is their transversal. [Corresponding angles]
∴ ∠HGS = ∠DHP …..(iii) [From (i)]

iv. ∠HGS = 85° [Vertically opposite angles]
∴ ∠MGK = ∠HGS ∠MGK = 85° [From (iii)]

Question 2.
In the given figure line p line q and line l and line m are tranversals.
Measures of some angles are shown. Hence find the measures of ∠a, ∠b, ∠c, ∠d.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 2
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 3
i. 110 + ∠a = 180° [Angles in a linear pair]
∴ ∠a = 180° – 110°
∴ ∠a = 70°

ii. consider ∠e as shown in the figure line p || line q, and line lis their transversal.
∠e + 110° = 180° [Interior angles]
∴ ∠e = 180° – 110°
∴ ∠e = 70°
But, ∠b = ∠e [Vertically opposite angles]
∴ ∠b = 70°

iii. line p || line q, and line m is their transversal.
∴ ∠c = 115° [Corresponding angles]

iv. 115° + ∠d = 180° [Angles in a linear pair]
∴ ∠d = 180° – 115°
∴ ∠d = 65°

Question 3.
In the given figure, line 11| line m and line n || line p. Find ∠a, ∠b, ∠c from the given measure of an angle.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 4
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 5
i. consider ∠d as shown in the figure
line l || line m, and line p is their transversal.
∴ ∠d = 45° [Corresponding angles]
Now, ∠d + ∠b = 180° [Angles in a linear pair]
∴ 45° +∠b = 180°
∴ ∠b = 180° – 45°
∴ ∠b = 135° …..(i)

ii. ∠a = ∠b [Vertically opposite angles]
∴ ∠a = 135° [From (i)]

iii. line n || line p, and line m is their transversal.
∴ ∠c = ∠b [Corresponding angles]
∴ ∠c = 135° [From (i)]

Question 4.
In the given figure, sides of ∠PQR and ∠XYZ are parallel to each other. Prove that, ∠PQR ≅ ∠XYZ.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 6
Given: Ray YZ || ray QRandray YX || ray QP
To prove: ∠PQR ≅ ∠XYZ
Construction: Extend ray YZ in the opposite direction. It intersects ray QP at point S.
Solution:
Proof:
Ray YX || ray QP [Given]
Ray YX || ray SP and seg SY is their transversal [P-S-Q]
∴ ∠XYZ ≅ ∠PSY ……(i) [Corresponding angles]
ray YZ || ray QR [Given]
ray SZ || ray QR and seg PQ is their transversal. [S-Y-Z]
∴ ∠PSY ≅ ∠SQR [Corresponding angles]
∴ ∠PSY ≅ ∠PQR …….. (ii) [P-S-Q]
∴ ∠PQR ≅ ∠XYZ [From (i) and (ii)]

Question 5.
In the given figure, line AB || line CD and line PQ is transversal. Measure of one of the angles is given. Hence find the measures of the following angles.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 7
i. ∠ART
ii. ∠CTQ
iii. ∠DTQ
iv. ∠PRB
Solution:
i. ∠BRT = 105° ….(i)
∠ART + ∠BRT = 180° [Angles in a linear pair]
∴ ∠ART + 105° = 180°
∴ ∠ART = 180° – 105°
∴ ∠ART = 75° …(ii)

ii. line AB || line CD and line PQ is their transversal.
∴ ∠CTQ = ∠ART [Corresponding angles]
∴ ∠CTQ = 75° [From (ii)]

iii. line AB || line CD and line PQ is their transversal.
∴ ∠DTQ = ∠BRT [Corresponding angles]
∴ ∠DTQ = 105° [From (i)]

iv. ∠PRB = ∠ART [Vertically opposite angles]
∴ ∠PRB = 75° [From (ii)]

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Practice Set 2.1 Intext Questions and Activities

Question 1.
Angles formed by two lines and their transversal. (Textbook pg, no. 13)
When a transversal (line n) intersects two lines (line l and m) in two distinct points, 8 angles are formed as shown in the figure. Pairs of angles formed out of these angles are as follows:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 8
Pairs of corresponding angles
i. ∠d, ∠h
ii. ∠a, ∠e
iii. ∠c, ∠g
iv. ∠b, ∠f

Pairs of alternate interior angles
i. ∠c, ∠e
ii. ∠b, ∠h

Pairs of alternate exterior angles
i. ∠d, ∠f
ii. ∠a, ∠g

Pairs of interior angles on the same side of the transversal
i. ∠c, ∠h
ii. ∠b, ∠e

Some important properties:
1. When two lines intersect, the pairs of vertically opposite angles formed are congruent.
Example:
In the given diagram,
line l and m intersect at point P.
The pairs of vertically opposite angles that are congruent are:
i. ∠a ≅ ∠b
ii. ∠c ≅ ∠d

2. The angles in a linear pair are supplementary.
Example:
For the given diagram,
∠a and ∠c are in linear pair
∴ ∠a + ∠c = 180°
Also, ∠d and ∠b are in linear pair
∴ ∠d + ∠b = 180°
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 9
3. When one pair of corresponding angles is congruent, then all the remaining pairs of corresponding angles are congruent.
Example:
In the given diagram,
If ∠a ≅ ∠b
then ∠e ≅ ∠f, ∠c ≅ ∠d and ∠g ≅ ∠h

4. When one pair of alternate angles is congruent, then all the remaining pairs of alternate angles are congruent.
Example:
For the given diagram,
If ∠e ≅ ∠d, then ∠g ≅ ∠b
Also, ∠a ≅ ∠h and ∠c ≅ ∠f
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 10

5. When one pair of interior angles on one side of the transversal is supplementary, then the other pair of interior angles is also supplementary.
Example:
For the given diagram,
If ∠e + ∠b = 180°, then ∠g + ∠d = 180°.

Maharashtra Board Class 9 Maths Solutions

Problem Set 1 Geometry 9th Standard Maths Part 2 Chapter 1 Basic Concepts in Geometry Solutions Maharashtra Board

9th Standard Maths 2 Problem Set 1 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

Class 9 Maths Part 2 Problem Set 1 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
Select the correct alternative answer for the questions given below.

i. How many midpoints does a segment have ?
(A) only one
(B) two
(C) three
(D) many
Answer:
(A) only one

ii. How many points are there in the intersection of two distinct lines ?
(A) infinite
(B) two
(C) one
(D) not a single
Answer:
(C) one

iii. How many lines are determined by three distinct points?
(A) two
(B) three
(C) one or three
(D) six
Answer:
(C) one or three

iv. Find d(A, B), if co-ordinates of A and B are – 2 and 5 respectively.
(A) -2
(B) 5
(C) 7
(D) 3
Answer:
Since, 5 > -2
∴ d(A, B) = 5 – (-2) = 5+2 = 7
(C) 7

v. If P – Q – R and d(P, Q) = 2, d(P, R) = 10, then find d(Q, R).
(A) 12
(B) 8
(C) √96
(D) 20
Answer:
d(P, R) = d(P, Q) + d(Q, R)
∴ 10 = 2 + d(Q, R)
∴ d(Q, R) = 8
(B) 8

Question 2.
On a number line, co-ordinates of P, Q, R are 3,-5 and 6 respectively. State with reason whether the following statements are true or false.
i. d(p, Q) + d(Q, R) = d(P, R)
ii. d(P, R) + d(R, Q) = d(P, Q)
iii. d(R, P) + d(P, Q) = d(R, Q)
iv. d(P, Q) – d(P, R) = d(Q, R)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 1
Co-ordinate of the point P is 3.
Co-ordinate of the point Q is -5.
Since, 3 > -5
d(P, Q) = 3 – (-5) = 3 + 5
∴ d(P,Q) = 8
Co-ordinate of the point Q is -5.
Co-ordinate of the point R is 6.
Since, 6 > -5
d(Q, R) = 6 – (-5) = 6 + 5
∴ d(Q, R) = 11
Co-ordinate of the point P is 3.
Co-ordinate of the point R is 6.
Since, 6 > 3
d(P, R) = 6 – 3
∴ d(P, R) = 3

i. d(P, Q) + d(Q, R) = 8 + 11
= 19 …(i)
d(P, R) = 3 …(ii)
∴ d(P, Q) + d(Q, R) ≠ d(P, R) … [From (i) and (ii)]
∴ The given statement is false.

ii. d(P, R) + d(R, Q) = 3 + 11
d(P,Q) = 8 …(ii)
∴ d(P, R) + d(R, Q) + d(P, Q) …[From (i) and (ii)]
∴ The given statement is false.

iii. d(R, P) + d(P, Q) = 3 + 8
= 11 …(i)
d(R, Q) =11 . -(ii)
∴ d(R,P) + d(P,Q) = d(R,Q) ….[From (i) and (ii)]
∴ The given statement is true.

iv. d(P, Q) – d(P, R) = 8 – 3
= 5 …(i)
d(Q,R) = 11 ..(h)
∴ d(P, Q) – d(P, R) ≠ d(Q, R) …[From (i) and (ii)]
∴ The given statement is false.

Question 3.
Co-ordinates of some pairs of points are given below. Hence find the distance between each pair.
i. 3,6
ii. -9, -1
iii. A, 5
iv. 0,-2
v. x + 3, x – 3
vi. -25, -47
vii. 80, -85
Solution:
i. Co-ordinate of first point is 3.
Co-ordinate of second point is 6.
Since, 6 > 3
∴ Distance between the points = 6 – 3 = 3

ii. Co-ordinate of first point is -9.
Co-ordinate of second point is -1.
Since, -1 > -9
∴ Distance between the points = -1 – (-9) = -1+9 = 8

iii. Co-ordinate of first point is -4.
Co-ordinate of second point is 5.
Since, 5 > -4
∴ Distance between the points = 5 – (-4)
= 5 + 4 = 9

iv. Co-ordinate of first point is 0.
Co-ordinate of second point is -2. Since,
0 > – 2
∴ Distance between the points = 0 – (-2)
= 0 + 2
= 2

v. Co-ordinate of first point is x + 3.
Co-ordinate of second point is x – 3.
Since, x + 3 > x – 3
∴ Distance between the points = x + 3 – (x – 3)
= x + 3 – x + 3 = 3 + 3
= 6

vi. Co-ordinate of first point is -25.
Co-ordinate of second point is -47.
Since, -25 > -47
∴ Distance between the points = -25 – (-47)
= -25 + 47
= 22

vii. Co-ordinate of first point is 80.
Co-ordinate of second point is -85.
Since, 80 > -85
∴ Distance between the points = 80 – (-85)
= 80 + 85
= 165

Question 4.
Co-ordinate of point P on a number line is – 7. Find the co-ordinates of points on the number line which are at a distance of 8 units from point P.
Solution:
Let point Q be at a distance of 8 units from P and on left side of P
Let point R be at a distance of 8 units from P and on right side of P.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 2

i. Let the co-ordinate of point Q be x.
Co-ordinate of point P is -7.
Since, point Q is to the left of point P.
∴ -7 > x
∴ d(P, Q) = -7 -x
∴8 = -7 – x
∴ x = – 7 – 8
∴x = -15

ii. Let the co-ordinate of point R be y.
Co-ordinate of point P is -7.
Since, point R is to the right of point P.
∴ y > -7
∴ d(P, R) = 7- (-7)
∴ 8 = y + 7
∴ 8 – 7 = 7
∴ y = 1
∴ The co-ordinates of the points at a distance of 8 units from P are -15 and 1.

Question 5.
Answer the following questions.
i. If A – B – C and d(A, C) = 17, d(B, C) = 6.5, then d (A, B) = ?
ii. If P – Q – R and d(P, Q) = 3.4, d(Q, R) = 5.7, then d(P, R) = ?
Solution:
i. Given, (A, C) = 17, d(B, C) = 6.5
d(A, C) = d(A, B) + d(B, C) …[A – B – C]
∴ 17 = d(A, B) + 6.5
∴ d(A,B)= 17 – 6.5
∴ d(A, B) = 10.5

ii. Given, d(P, Q) = 3.4, d(Q, R) = 5.7
d(P,R) = d(P,Q) + d(Q,R) …[P – Q – R]
= 34 + 5.7
∴ d(P, R) = 9.1

Question 6.
Co-ordinate of point A on a number line is 1. What are the co-ordinates of points on the number line which are at a distance of 7 units from A ?
Solution:
Let point C be at a distance of 7 units from A and on left side of A
Let point B be at a distance of 7 units from A and on right side of A.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 3
i. Let the co-ordinate of point C be x.
Co-ordinate of point A is 1.
Since, point C is to the left of point A.
∴ 1 > x
∴ d(A, C) = 1 – x
∴ 7 = 1 -x
∴x = 1 – 7
∴ x = – 6

ii. Let the co-ordinate of point B be y.
Co-ordinate of point A is 1.
Since, point B is to the right of point A.
∴y > 1
∴ d(A, B) = 7 – 1
∴ 7 = y – 1
∴ 7 + 1 = 7
∴ 7 = 8
∴ The co-ordinates of the points at a distance of 7 units from A are -6 and 8.

Question 7.
Write the following statements in conditional form.
i. Every rhombus is a square.
ii. Annies in a linear pair are supplementary.
iii. A triangle is a figure formed by three segments
iv. A number having only two divisors is called a prime number.
Solution:
i If a quadrilateral is a rhombus, then it is a square.
ii. If iwo angles are in a linear pair, then they are supplementary.
iii. If a figure is a triangle, then it is formed by three segments.
iv. If a number has only two divisors, then it is a prime number.

Question 8.
Write the converse of each of the following statements.
i. If the sum of measures of angles in a figure is 180°, then the figure is a triangle.
ii If the sum of measures of two angles is 90°, thfcn they are eomplement of each other.
iii. If the corresponding angles formed by a transversal of two lines are congruent, then the two lines are parallel.
iv. If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
Answer:
i. If a figure is a triangle, then the sum of the measures of its angles is 180°.
ii. if two angles are eomplement of each other, then sum of their measures is 90°,
iii. If two lines are parallel, then the corresponding angles formed by a transversal of two lines are congruent.
iv. If a number is divisible by 3, then the sum of its digits is also divisible by 3.

Question 9.
Write the antecedent (given part) and the consequent (part to be proved) in the following statements.
i. If all sides of a triangle are congruent, then its all angles are congruent.
ii. The diagonals of a parallelogram bisect each other.
Answer:
i. If all sides of a triangle are congruent, then its all angles are congruent.
Antecedent (Given): All the sides of the triangle are congruent.
Consequent (To prove): All the angles are congruent.

ii. The diagonals of a parallelogram bisect each other.
Conditional statement: “If a quadrilateral is a parallelogram then its diagonals bisect each other.
Antecedent (Given): Quadrilateral is a parallelogram.
Consequent (To prove): Its diagonals bisect each other.

Question 10.
Draw a labelled figure showing information in each of the following statements and write the antecedent and the consequent.
i. Two equilateral triangles are similar.
ii. If angles in a linear pair are congruent, then each of them is a right angle.
iii. If the altitudes drawn on two sides of a triangle are congruent, then these two sides are congruent.
Answer:
i. Two equilateral triangles are similar.
Conditional statement: “If two triangles are equilateral, then they are similar.
Antecedent (Given): Two triangles are equailateral.
i.e. ∆ABC and ∆PQR are equilatral triangle.
Consequent (To prove): Triangles are similar
i.e. ∆ABC ∼ ∆PQR
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 4

ii. If angles in a linear pair are congruent, then each of them is a right angle.
Antecedent (Given): Angles in a linear pair are congrunent.
∠ABC and ∠ABD are angles in a linear pair i.e. ∠ABC = ∠ABD
Consequent (To prove): Each angle is a right angle.
i.e. ∠ABC – ∠ABD = 90°
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 5

iii. If the altitudes drawn on two sides of a triangle are congruent, then these two sides are congruent.
Antecedent (Given): Altitude drawn on two sides of triangle are congrunent.
In ∆ABC, AD ⊥ BC . and BE ⊥ AC. seg AD ≅ seg BE

Consequent (To prove): Two sides are congruent.
side BC ≅ side AC A
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 6

Maharashtra Board Class 9 Maths Chapter 1 Basic Concepts in Geometry Problem Set 1 Intext Questions and Activities

Question 1.
Points A, B, C are given below. Check, with a stretched thread, whether the three points are collinear or not. If they are collinear, write which one of them is between the other two. (Textbook pg. no. 4)
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 7
Answer:
Point B is between the points A and C.

Question 2.
Given below are four points P, Q, R, and S. Check which three of them are collinear and which three are non collinear. In the case of three collinear points, state which of them is between the other two. (Textbook pg. no. 4)
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 8
Answer:
Points P, R and S are collinear.
Point R is between the points P and S.

Question 3.
Students are asked to stand in a line for mass drill. How will you check whether the students standing are in a line or not ? (Textbook pg. no. 4)
Answer:
If one stands in front of the line and observes only the first student standing in the line, then all the students standing in that line are collinear i.e., standing in the same line. We can use this property of collinearity to check whether the students are standing in the same line or not.

Question 4.
How had you verified that light rays travel in a straight line? Recall an experiment in science which you have done in a previous standard. (Textbook pg. no. 4)
Answer:
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Problem Set 1 9

The flame of the candle can be seen only when the pin holes in all cardboards are in the same straight line. We can use the set up shown in the figure above to verify that light rays travels in a straight line.

Std 9 Maths Solutions Maharashtra State Board 

Practice Set 2.2 Geometry 9th Standard Maths Part 2 Chapter 2 Parallel Lines Solutions Maharashtra Board

9th Standard Maths 2 Practice Set 2.2 Chapter 2 Parallel Lines Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 2 Parallel Lines.

Class 9 Maths Part 2 Practice Set 2.2 Chapter 2 Parallel Lines Questions With Answers Maharashtra Board

Question 1.
In the given figure, y = 108° and x = 71°. Are the lines m and n parallel? Justify?
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 1
Solution:
y = 108°, x = 71° …[Given]
x + y = 71° + 108°
= 179°
∴ x + y = 180°
∴ The angles x andy are not supplementary.
∴ The angles do not satisfy the interior angles test for parallel lines
∴ line m and line n are not parallel lines.

Question 2.
In the given figure, if ∠a = ∠b then prove that line l || line m.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 2
Given: ∠a ≅ ∠b
To prove: line l| line m
Solution:
Proof:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 3
consider ∠c as shown in the figure ∠a ≅ ∠c …….. (i) [Vertically opposite angles]
But, ∠a ≅ ∠b I (ii) [Given]
∴ ∠b ≅ ∠c [From (i) and (ii)]
But, ∠b and ∠c are corresponding angles on lines l and m when line n is the transversal.
∴ line l || line m. [Corresponding angles test]

Question 3.
In the given figure, if ∠a ≅ ∠b and ∠x ≅ ∠y, then prove that line l | line n.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 4
Given: ∠a ≅ ∠b and ∠x ≅ ∠y
To prove: line l | line n
Solution:
Proof:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 5
∠a = ∠b [Given]
But, ∠a and ∠b are corresponding angles on lines l and m when line k is the transversal.
∴ line l || line m ….(i) [Corresponding angles test]
∠x ≅ ∠y [Given]
But, ∠x and ∠y are alternate angles on lines m and n when seg PQ is the transversal,
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 6
∴ line m || line n ……(ii) [Alternate angles test]
∴ From (i) and (ii),
line l || line m || line n
i.e., line l || line n

Question 4.
In the given figure, if ray BA || ray DE, ∠C = 50° and ∠D = 100°. Find the measure of ∠ABC.
(Hint: Draw a line passing through point C and parallel to line AB.)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 7
Solution:
Draw a line FG passing through point
C and parallel to line AB
line FG || ray BA …….(i) [Construction]
Ray BA || ray DE ….(ii) [Given]
line FG || ray BA || ray DE …(iii) [From (i) and (ii)]
line FG||rayDE [From (iii)]
and seg DC is their transvensal
∴ ∠ DCF = ∠ EDC [Alternate angles]
∴ ∠ DCF = 100° [∵ ∠D = 100°]
Now, ∠ DCF = ∠ BCF + ∠ BCD [Angle addition property]
∴ 100° = ∠BCF + 50°
∴ 100° – 50° = ∠BCF
∴ ∠BCF = 50° ….(iv)
Now, line FG || ray BA and seg BC is their transversal.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 8
∴ ∠ABC + ∠BCF = 180° [Interior angles]
∴ ∠ABC + 50° = 180° [From (iv)]
∴ ∠ABC = 180°- 50°
∴ ∠ABC = 130°

Question 5.
In the given figure, ray AE || ray BD, ray AF is the bisector of ∠EAB and ray BC is the bisector of ∠ABD. Prove that line AF || line BC.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 9
Given: Ray AE || ray BD, and
ray AF and ray BC are the bisectors of ∠EAB and ∠ABD respectively.
To prove: line AF || line BC
Solution:
Proof:
Ray AE || ray BD and seg AB is their transversal.
∴ ∠EAB = ∠ABD ….(i) [Alternate angles]
∠FAB = \(\frac { 1 }{ 2 }\)∠EAB [Ray AF bisects ∠EAB]
∴ 2∠FAB = ∠EAB …..(ii)
∠CBA = \(\frac { 1 }{ 2 }\)∠ABD [Ray BC bisects ∠ABD]
∴ 2∠CBA = ∠ABD …(iii)
∴ 2∠FAB = 2∠CBA [From (i), (ii) and (iii)]
∴ ∠FAB = ∠CBA
But, ∠FAB and ∠ABC are alternate angles on lines AF and BC when seg AB is the transversal.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 10
∴ line AF || line BC [Alternate angles test]

Question 6.
A transversal EF of line AB and line CD intersects the lines at points P and Q respectively. Ray PR and ray QS are parallel and bisectors of ∠BPQ and ∠PQC respectively. Prove that line AB || line CD.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 11
Given: Ray PR || ray QS
Ray PR and ray QS are the bisectors of ∠BPQ and ∠PQC respectively.
To prove: line AB || line CD
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 12
Proof:
Ray PR || ray QS and seg PQ is their transversal.
∠RPQ = ∠SQP ….(i) [Alternate angles]
∠RPQ = \(\frac { 1 }{ 2 }\)∠BPQ …. (ii) [Ray PR bisects ∠BPQ]
∠SQP = \(\frac { 1 }{ 2 }\)∠PQC [Ray QS bisects ∠PQC]
∴ \(\frac { 1 }{ 2 }\)∠BPQ = \(\frac { 1 }{ 2 }\)∠PQC
∴ ∠BPQ = ∠PQC
But, ∠BPQ and ∠PQC are alternate angles on lines AB and CD when line EF is the transversal.
∴ line AB || line CD [Alternate angles test]

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Practice Set 2.2 Intext Questions and Activities

Question 1.
In the given figure, how will you decide whether line ¡ and line m are parallel or not? (Textbook pg. no. 19)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 13
Answer:
In the figure, we observe that line I and line m are coplanar and do not intersect each other.
∴ Line l and line m are parallel lines.

Maharashtra Board Class 9 Maths Solutions

Practice Set 1.3 Geometry 9th Standard Maths Part 2 Chapter 1 Basic Concepts in Geometry Solutions Maharashtra Board

9th Standard Maths 2 Practice Set 1.3 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

Class 9 Maths Part 2 Practice Set 1.3 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
Write the following statements in ‘if-then’ form.
i. The opposite angles of a parallelogram are congruent.
ii. The diagonals of a rectangle are congruent.
iii. In an isosceles triangle, the segment joining the vertex and the midpoint of the base is perpendicular to the base.
Answer:
i. If a quadrilateral is a parallelogram, then its opposite angles are congruent.
ii. If a quadrilateral is a rectangle, then its diagonals are congruent.
iii. If a triangle is isosceles triangle, then segment joining the vertex of a triangle and midpoint of the base is perpendicular to the base.

Question 2.
Write converses of the following statements.
i. The alternate angles formed by two parallel lines and their transversal are congruent.
ii. If a pair of the interior angles made by a transversal of two lines are supplementary, then the lines are parallel.
iii. The diagonals of a rectangle are congruent.
Answer:
i. If the alternate angles made by two lines and their transversal are congruent, then the two lines are parallel.
ii. If two parallel lines are intersected by a transversal, then the interior angles formed bv the transversal are supplementary.
iii. If the diagonals of a quadrilateral are congruent, then that quadrilateral is a rectangle.

Std 9 Maths Solutions Maharashtra State Board 

Practice Set 1.2 Geometry 9th Standard Maths Part 2 Chapter 1 Basic Concepts in Geometry Solutions Maharashtra Board

9th Standard Maths 2 Practice Set 1.2 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

Class 9 Maths Part 2 Practice Set 1.2 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
The following table shows points on a number line and their co-ordinates. Decide whether the pair of segments given below the table are congruent or not.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.2 1
i. seg DE and seg AB
ii. seg BC and seg AD
iii. seg BE and seg AD
Solution:
i. Co-ordinate of the point E is 9.
Co-ordinate of the point D is -7.
Since, 9 > -7
∴ d(D, E) = 9 – (-7) = 9 + 7 = 16
∴ l(DE) = 16 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point B is 5.
Since, 5 > -3
∴ d(A, B) = 5 – (-3) = 5 + 3 = 8
∴ l(AB) = 8 …(ii)
∴ l(DE) ≠ l(AB) …[From (i) and (ii)]
∴ seg DE and seg AB are not congruent.

ii. Co-ordinate of the point B is 5.
Co-ordinate of the point C is 2.
Since, 5 > 2
∴ d(B, C) = 5 – 2 = 3
∴ l(BC) = 3 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = -3 + 7 = 4
∴ l(AD) = 4 . ..(ii)
∴ l(BC) ≠ l(AD) … [From (i) and (ii)]
∴ seg BC and seg AD are not congruent.

iii. Co-ordinate of the point E is 9.
Co-ordinate of the point B is 5.
Since, 9 > 5
∴ d(B, E) = 9 – 5 = 4
∴ l(BE) = 4 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = 4
∴ l(AD) = 4 …(ii)
∴ l(BE) =l(AD) …[From (i) and (ii)]
∴ seg BE and seg AD are congruent.
i.e, seg BE ≅ seg AD

Question 2.
Point M is the midpoint of seg AB. If AB = 8, then find the length of AM.
Solution:
Point M is the midpoint of seg AB and l(AB) = 8. …[Given]
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.2 2

Question 3.
Point P is the midpoint of seg CD. If CP = 2.5, find l(CD).
Solution:
Point P is the midpoint of seg CD and l(CP) = 2.5 …[Given]
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.2 3
∴ l(CD) = 2.5 x 2
∴ l(CD) = 5

Question 4.
If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments.
Solution:
Given, l(AB) = 5 cm, l(BP) = 2 cm,
l(AP) = 3.4 cm … [Given]
r Since, 2 < 3.4 < 5
∴ l(BP) < l(AP) < l(AB)
i.e., seg BP < seg AP < seg AB

Question 5.
Write the answers to the following questions with reference to the figure given below:
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.2 4
i. Write the name of the opposite ray of ray RP
ii. Write the intersection set of ray PQ and ray RP.
iii. Write the union set of ray PQ and ray QR.
iv. State the rays of which seg QR is a subset.
v. Write the pair of opposite rays with common end point R.
vi. Write any two rays with common end point S.
vii. Write the intersection set of ray SP and ray ST.
Answer:
i. Ray RS or ray RT
ii. Ray PQ
iii. Line QR
iv. Ray QR, ray QS, ray QT, ray RQ, ray SQ, ray TQ
v. Ray RP and ray RS, ray RQ and ray RT vi. Ray ST, ray SR
vii. Point S

Question 6.
Answer the questions with the help of figure given below.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.2 5
i. State the points which are equidistant from point B.
ii. Write a pair of points equidistant from point iii. Find d(U,V), d(P,C), d(V,B), d(U, L).
Answer:
i. Points equidistant from point B are a. A and C, because d(B, A) = d(B, C) = 2 b. D and P, because d(B, D) = d(B, P) = 4
ii. Points equidistant from point Q are a. L and U, because d(Q, L) = d(Q, U) = 1 b. P and R, because d(P, Q) = d(Q, R) = 2
iii. a. Co-ordinate of the point U is -5. Co-ordinate of the point V is 5. Since, 5 > -5
∴ d(U, V) = 5 – (-5)
= 5 + 5
∴ d(U, V) = 10

b. Co-ordinate of the point P is -2.
Co-ordinate of the point C is 4.
Since, 4 > -2
∴ d(P, C) = 4 – (-2)
= 4 + 2
∴ d(P, C) = 6

c. Co-ordinate of the point V is 5.
Co-ordinate of the point B is 2.
Since, 5 > 2
∴ d(V, B) = 5 – 2
∴ d(V, B) = 3

d. Co-ordinate of the point U is -5.
Co-ordinate of the point L is -3.
Since, -3 > -5
∴ d(U, L) = -3 – (-5)
= -3 + 5
∴ d(U, L) = 2

Std 9 Maths Solutions Maharashtra State Board 

Practice Set 1.1 Geometry 9th Standard Maths Part 2 Chapter 1 Basic Concepts in Geometry Solutions Maharashtra Board

9th Standard Maths 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

Class 9 Maths Part 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
Find the distances with the help of the number line given below.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 1
i. d(B, E)
ii. d (J, J)
iii. d(P, C)
iv. d(J, H)
v. d(K, O)
vi. d(O, E)
vii. d(P, J)
viii. d(Q, B)
Solution:
i. Co-ordinate of the point B is 2.
Co-ordinate of the point E is 5.
Since, 5 > 2
∴ d(B, E) = 5 – 2
∴ d(B, E) = 3

ii. Co-ordinate of the point J is -2.
Co-ordinate of the point A is 1.
Since, 1 > -2
∴ d(J, A) = 1 – (-2)
= 1 + 2
∴ d(J, A) = 3

iii. Co-ordinate of the point P is -4.
Co-ordinate of the point C is 3.
Since, 3 > -4
∴ d(P,C) = 3 – (-4)
= 3 + 4
∴ d(P,C) = 7

iv. Co-ordinate of the point J is -2.
Co-ordinate of the point H is -1.
Since, -1 > -2
∴ d(J,H) = – 1 – (-2)
= -1 + 2
∴ d(J,H) = 1

v. Co-ordinate of the point K is -3.
Co-ordinate of the point O is 0.
Since,0 > -3
∴ d(K, O) = 0 – (-3)
= 0 + 3
∴ d(K, O) = 3

vi. Co-ordinate of the point O is 0.
∴ Co-ordinate of the point E is 5.
Since, 5 > 0
∴ d(O, E) = 5 – 0
∴ d(O, E) = 5

vii. Co-ordinate of the point P is -4.
Co-ordinate of the point J is -2.
Since -2 > -4
∴ d(P, J) = -2 – (-4)
= – 2+ 4
∴ d(P, J) = 2

viii. Co-ordinate of the point Q is -5.
Co-ordinate of the point B is 2.
Since,2 > -5
∴ d(Q,B) = 2 – (-5)
= 2 + 5
∴ d(Q, B) = 7

Question 2.
If the co-ordinate of A is x and that of B is . y, find d(A, B).
i. x = 1, y = 7
ii. x = 6, y = -2
iii. x = -3, y = 7
iv. x = -4, y = -5
v. x = -3, y = -6
vi. x = 4, y = -8
Solution:
i. Co-ordinate of point A is x = 1.
Co-ordinate of point B is y = 7
Since, 7 > 1
∴ d(A, B) = 7 – 1
∴ d(A, B) = 6

ii. Co-ordinate of point A is x = 6.
Co-ordinate of point B is y = -2.
Since, 6 > -2
∴ d(A, B) = 6 – ( -2) = 6 + 2
∴ d(A, B) = 8

iii. Co-ordinate of point A is x = -3.
Co-ordinate of point B is y = 7.
Since, 7 > -3
∴ d(A, B) = 7 – (-3) = 7 + 3
∴ d(A, B) = 10

iv. Co-ordinate of point A is x = -4.
Co-ordinate of point B is y = -5.
Since, -4 > -5
∴ d(A, B) = -4 – (-5)
= -4 + 5
∴ d(A, B) = 1

v. Co-ordinate of point A is x =-3.
Co-ordinate of point B is y = -6.
Since, -3 > -6
∴ d(A, B) = -3 – (-6)
= -3 + 6
∴ d(A, B) = 3

vi. Co-ordinate of point A is x = 4.
Co-ordinate of point B is y = -8.
Since, 4 > -8
∴ d(A, B) = 4 – (-8)
= 4 + 8
∴d(A, B) = 12

Question 3.
From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
i. d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
ii. d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
iii. d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
iv. d(L, M) =11, d(M, N) = 12, d(N, L) = 8
v. d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
vi. d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Solution:
i. Given, d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
d(P, Q) = 10 …(i)
d(P, R) + d(Q, R) = 7 + 3 = 10 .. .(ii)
∴ d(P, Q) = d(P, R) + d(Q, R) …[From (i) and (ii)]
∴ Point R is between the points P and Q
i. e., P – R – Q or Q – R – P.
∴ Points P, R, Q are collinear.

ii. Given, d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
d(R, S) = 8 …(i)
d(S, T) + d(R, T) = 6 + 4 = 10 …(h)
∴ d(R, S) ≠ d(S, T) + d(R, T) … [From (i) and (ii)]
∴ The given points are not collinear.

iii. Given, d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
d(A, B) = 16 …(i)
d(C, A) + d(B, C) = 9 + 7 = 16 …(ii)
∴ d(A, B) = d(C, A) + d(B, C) …[From(i) and (ii)]
∴ Point C is between the points A and B.
i. e., A – C – B or B – C – A.
∴ Points A, C, B are collinear

iv. Given, d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
d(M, N) = 12 …(i)
d(L, M) + d(N, L) = 11 + 8 = 19 …(ii)
∴d(M, N) + d(L, M) + d(N, L) … [From (i) and (ii)]
∴ The given points are not collinear.

v. Given, d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
d(X, Y) = 15 …(i)
d(X,Z) + d(Y, Z) = 8 + 7= 15 …(ii)
∴ d(X, Y) = d(X, Z) + d(Y, Z) …[From (i) and (ii)]
∴ Point Z is between the points X and Y
i. e.,X – Z – Y or Y – Z – X.
∴ Points X, Z, Y are collinear.

vi. Given, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
d(E, F) = 8 …(i)
d(D, E) + d(D, F) = 5 + 6 = 11 …(ii)
∴ d(E, F) ≠ d(D, E) + d(D, F) … [From (i) and (ii)]
∴ The given points are not collinear.

Question 4.
On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.
Solution:
Given, d(A, C) = 10, d(C, B) = 8.

Case I: Points A, B, C are such that, A – B – C.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 2
∴ d(A, C) = d(A, B) + d(B, C)
∴ 10 = d(A, B) + 8
∴ d(A, B) = 10 – 8
∴ d(A, B) = 2

Case II: Points A, B, C are such that, A – C – B.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 3
∴ d(A, B) = d(A, C) + d(C, B)
= 10 + 8
∴ d(A, B) = 18

Case III: Points A, B, C are such that, B – A – C.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 4
From the diagram,
d (A, C) > d(B, C)
Which is not possible
∴ Point A is not between B and C.
∴ d(A, B) = 2 or d(A, B) = 18.

Question 5.
Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).
Solution:
Given,d(X, Y) = 17, d(Y, Z) = 8
Case I: Points X, Y, Z are such that, X – Y – Z.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 5
∴ d(X, Z) = d(X, Y) + d(Y, Z)
= 17 + 8
∴ d(X, Z) = 25

Case II: Points X, Y, Z are such that, X – Z – Y.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 6
∴ d(X,Y) = d(X,Z) + d(Z,Y)
∴ 17 = d(X, Z) + 8
∴ d(X, Z) = 17 – 8
∴ d(X, Z) = 9

Case III: Points X, Y, Z are such that, Z – X – Y.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 7
From the diagram,
d(X, Y) > d (Y, Z)
Which is not possible
∴ Point X is not between Z and Y.
∴ d(X, Z) = 25 or d(X, Z) = 9.

Question 6.
Sketch proper figure and write the answers of the following questions. [2 Marks each]
i. If A – B – C and l(AC) = 11,
l(BC) = 6.5, then l(AB) = ?
ii. If R – S – T and l(ST) = 3.7,
l(RS) = 2.5, then l(RT) = ?
iii. If X – Y – Z and l(XZ) = 3√7,
l(XY) = √7, then l(YZ) = ?
Solution:
i. Given, l(AC) =11, l(BC) = 6.5
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 8
l(AC) = l(AB) + l(BC) … [A – B – C]
∴ 11= l(AB) + 6.5
∴ l(AB) = 11 – 6.5
∴ l(AB) = 4.5

ii. Given, l(ST) = 3.7, l(RS) = 2.5
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 9
l(RT) = l(RS) + l(ST) … [R – S – T]
= 2.5 + 3.7
∴ (RT) = 6.2

iii. l(XZ) = 3√7 , l(XY) = √7,
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 10
l(XZ) = l(X Y) + l(YZ) … [X – Y – Z]
∴ 3 √7 ⇒ √7 + l(YZ)
∴ l(YZ)= 3√7 – √7
∴ l(YZ) = 2 √7

Question 7.
Which figure is formed by three non-collinear points?
Solution:
Three non-collinear points form a triangle.

Std 9 Maths Solutions Maharashtra State Board 

Problem Set 7 Algebra 9th Standard Maths Part 1 Chapter 7 Statistics Solutions Maharashtra Board

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Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

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Question 1.
Write the correct alternative answer for each of the following questions.

i. Which of the following data is not primary ?
(A) By visiting a certain class, gathering information about attendance of students.
(B) By actual visit to homes, to find number of family members.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
(D) Review the cleanliness status of canals by actually visiting them.
Answer:
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

ii. What is the upper class limit for the class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(B) 35

iii. What is the class-mark of class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(D) 30

iv. If the classes are 0 – 10, 10 – 20, 20 – 30, …, then in which class should the observation 10 be included?
(A) 0 – 10
(B) 10 – 20
(C) 0 – 10 and 10-20 in these 2 classes
(D) 20 – 30
Answer:
(B) 10 – 20

v. If \(\overline { x }\) is the mean of x1, x2, ……. , xn and \(\overline { y }\) is the mean of y1, y2, ….. yn and \(\overline { z }\) is the mean of x1,x2, …… , xn , y1, y2, …. yn , then z = ?
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 1
Answer:
x1, x2, x3, ……. , xn
∴ \(\overline{x}=\frac{\sum x}{\mathrm{n}}\)
∴ n\(\overline{x}\) = ∑x
Similarly, n\(\overline{y}\) = ∑y
Now,
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 2
\(\text { (A) } \frac{\overline{x}+\overline{y}}{2}\)

vi. The mean of five numbers is 50, out of which mean of 4 numbers is 46, find the 5th number.
(A) 4
(B) 20
(C) 434
(D) 66
Answer:
5th number = Sum of five numbers – Sum of four numbers
= (5 x 50) – (4 x 46)
= 250 – 184
= 66
(D) 66

vii. Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
Answer:
New mean = \(\frac { 4000-30+70 }{ 100 }\)
= 40.4
(B) 40.4

viii. What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?
(A) 15
(B) 20
(C) 19
(D) 25
Answer:
(A) 15

ix. What is the median of 7, 10, 7, 5, 9, 10 ?
(A) 7
(B) 9
(C) 8
(D) 10
Answer:
(C) 8

x. From following table, what is the cumulative frequency of less than type for the class 30 – 40?
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 3
(A) 13
(B) 15
(C) 35
(D) 22
Answer:
Cumulative frequency of less than type for the class 30 – 40 = 7 + 3 + 12 + 13 = 35
(C) 35

Question 2.
The mean salary of 20 workers is ₹10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
The mean salary of 20 workers is ₹ 10,250.
∴ Sum of the salaries of 20 workers
= 20 x 10,250
= ₹ 2,05,000 …(i)
If the superintendent’s salary is added, then mean increases by 750
new mean = 10, 250 + 750 = 11,000
Total number of people after adding superintendent = 20 + 1 = 21
∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = ₹ 2,31,000 …(ii)
∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers
= 2, 31,00 – 2,05,000 …[From (i) and (ii)]
= 26,000
∴ The salary of the office superintendent is ₹ 26,000.

Question 3.
The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.
Solution:
∴ \( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77
∴ sum of 9 numbers = 11 x 9 = 693 …(i)
If one more number is added, then mean increases by 5
mean of 10 numbers = 77 + 5 = 82
∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)
∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)]
= 127
∴ The number added in the data is 127.

Question 4.
The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 4
i. Number of days for which the maximum temperature was less than 34°C
= 8 + 8 + 8 = 24
ii. Number of days for which the maximum temperature was 34°C or more than 34°C
= 5 + 1 = 6

Question 5.
If the mean of the following data is 20.2, then find the value of p.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 5
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 6
∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
∴ p = \(\frac { 4 }{ 0.2 }\) = \(\frac { 40 }{ 2 }\) = 20
∴ p = 20

Question 6.
There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 – 40, 40 – 50, …. prepare the less than type cumulative frequency table. Using the table, answer the following questions:

i. How many students have scored marks less than 80?
ii. How many students have scored marks less than 40?
iii. How many students have scored marks less than 60?
Solution:
Class
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 7
i. 66 students have scored marks less than 80.
ii. 14 students have scored marks less than 40.
iii. 45 students have scored marks less than 60.

Question 7.
By using data in example (6), and taking classes 30 – 40, 40 – 50,… prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 8
i. 11 students have scored marks 70 or more than 70.
ii. 68 students have scored marks 30 or more than 30.

Question 8.
There are 10 observations arranged in ascending order as given below.
45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74. The median of these observations is 53.
Find the value of JC. Also find the mean and the mode of the data.
Solution:
i. Given data in ascending order:
45,47, 50, 52, x, JC+2, 60, 62, 63, 74.
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 9
∴ 106 = 2x + 2
∴ 106 – 2 = 2x
∴ 104 = 2x
∴ x = 52
∴ The given data becomes:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.

Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Problem Set 7 10
∴ The mean of the given data is 55.9.

iii. Given data in ascending order:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The observation repeated maximum number of times = 52
∴ The mode of the given data is 52.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Problem Set 7 Intext Questions and Activities

Question 1.
To show following information diagrammatically, which type of bar- diagram is suitable?
i. Literacy percentage of four villages.
ii. The expenses of a family on various items.
iii. The numbers of girls and boys in each of five divisions.
iv. The number of people visiting a science exhibition on each of three days.
v. The maximum and minimum temperature of your town during the months from January to June.
vi. While driving a two-wheeler, number of people wearing helmets and not wearing helmet in 100 families.
(Textbook pg. no. 112)
Solution:
i. Percentage bar diagram
ii. Sub-divided bar diagram
iii. Sub-divided bar diagram
iv. Sub-divided bar diagram
v. Sub-divided bar diagram
vi. Sub-divided bar diagram

Question 2.
You gather information for several reasons. Take a few examples and discuss whether the data is primary or secondary.
(Textbook pg. no, 113)
[Students should attempt the above activity on their own.]

Maharashtra Board Class 9 Maths Solutions

Practice Set 7.5 Algebra 9th Standard Maths Part 1 Chapter 7 Statistics Solutions Maharashtra Board

9th Standard Maths 1 Practice Set 7.5 Chapter 7 Statistics Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Class 9 Maths Part 1 Practice Set 7.5 Chapter 7 Statistics Questions With Answers Maharashtra Board

Question 1.
Yield of soyabean per acre in quintal in Mukund’s field for 7 years was 10, 7, 5,3, 9, 6, 9. Find the mean of yield per acre.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 1
Mean = 7
The mean of yield per acre is 7 quintals.

Question 2.
Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.
Solution:
Given data in ascending order:
59, 68, 70, 74, 75, 75, 80
∴ Number of observations(n) = 7 (i.e., odd)
∴ Median is the middle most observation
Here, 4th number is at the middle position, which is = 74
∴ The median of the given data is 74.

Question 3.
The marks (out of 100) obtained by 7 students in Mathematics examination are given below. Find the mode for these marks.
99, 100, 95, 100, 100, 60, 90
Solution:
Given data in ascending order:
60, 90, 95, 99, 100, 100, 100
Here, the observation repeated maximum number of times = 100
∴ The mode of the given data is 100.

Question 4.
The monthly salaries in rupees of 30 workers in a factory are given below.
5000, 7000, 3000, 4000, 4000, 3000, 3000,
3000, 8000, 4000, 4000, 9000, 3000, 5000,
5000, 4000, 4000, 3000, 5000, 5000, 6000,
8000, 3000, 3000, 6000, 7000, 7000, 6000,
6000, 4000
From the above data find the mean of monthly salary.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 2
∴ The mean of monthly salary is ₹ 4900.

Question 5.
In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows:
60, 70, 90, 95, 50, 65, 70, 80, 85, 95.
Find the median of the weights of tomatoes.
Solution:
Given data in ascending order:
50, 60, 65, 70, 70, 80 85, 90, 95, 95
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, 5th and 6th numbers are in the middle position
∴ Median = \(\frac { 70+80 }{ 2 }\)
∴ Median = \(\frac { 150 }{ 2 }\)
∴ The median of the weights of tomatoes is 75 grams.

Question 6.
A hockey player has scored following number of goals in 9 matches: 5, 4, 0, 2, 2, 4, 4, 3,3.
Find the mean, median and mode of the data.
Solution:
i. Given data: 5, 4, 0, 2, 2, 4, 4, 3, 3.
Total number of observations = 9
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 3
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 4
∴ The mean of the given data is 3.

ii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
∴ Number of observations(n) = 9 (i.e., odd)
∴ Median is the middle most observation
Here, the 5th number is at the middle position, which is 3.
∴ The median of the given data is 3.

iii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
Here, the observation repeated maximum number of times = 4
∴ The mode of the given data is 4.

Question 7.
The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What Was the correct mean?
Solution:
Here, mean = 80, number of observations = 50
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
∴ The sum of 50 observations = 80 x 50
= 4000
One of the observation was 19. However, by mistake it was recorded as 91.
Sum of observations after correction = sum of 50 observation + correct observation – incorrect observation
= 4000 + 19 – 91
= 3928
∴ Corrected mean
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 5
= 78.56
∴ The corrected mean is 78.56.

Question 8.
Following 10 observations are arranged in ascending order as follows. 2, 3 , 5 , 9, x + 1, x + 3, 14, 16, 19, 20. If the median of the data is 11, find the value of x.
Solution:
Given data in ascending order :
2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20.
∴ Number if observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴ \( \text { Median }=\frac{(x+1)+(x+3)}{2}\)
∴ 11 = \(\frac { 2x+4 }{ 2 }\)
∴ 22 = 2x + 4
∴ 22 – 4 = 2x
∴ 18 = 2x
∴ x = 9

Question 9.
The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observations is 25. Find the 18th observation.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations
= Mean x Total number of observations
The mean of 35 observations is 20
∴ Sum of 35 observations = 20 x 35 = 700 ,..(i)
The mean of first 18 observations is 15
Sum of first 18 observations =15 x 18
= 270 …(ii)
The mean of last 18 observations is 25 Sum of last 18 observations = 25 x 18
= 450 …(iii)
∴ 18th observation = (Sum of first 18 observations + Sum of last 18 observations) – (Sum of 35 observations)
= (270 + 450) – (700) … [From (i), (ii) and (iii)]
= 720 – 700 = 20
The 18th observation is 20.

Question 10.
The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
The mean of 5 observations is 50
Sum of 5 observations = 50 x 5 = 250 …(i)
One observation was removed and mean of remaining data is 45.
Total number of observations after removing one observation = 5 – 1 = 4
Now, mean of 4 observations is 45.
∴ Sum of 4 observations = 45 x 4 = 180 …(ii)
∴ Observation which was removed
= Sum of 5 observations – Sum of 4 observations = 250 – 180 … [From (i) and (ii)]
= 70
∴ The observation which was removed is 70.

Question 11.
There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.
Solution:
Total number of students = 40
Number of boys =15
∴ Number of girls = 40 – 15 = 25
The mean of marks obtained by 15 boys is 33
Here, sum of the marks obtained by boys
= 33 x 15
= 495 …(i)
The mean of marks obtained by 25 girls is 35 Sum of the marks obtained by girls = 35 x 25
= 875 …(ii)
Sum of the marks obtained by boys and girls = 495 + 875 … [From (i) and (ii)]
= 1370
∴ Mean of all the students
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 6
= 34.25
∴ The mean of all the students in the class is 34.25.

Question 12.
The weights of 10 students (in kg) are given below:
40, 35, 42, 43, 37, 35, 37, 37, 42, 37. Find the mode of the data.
Solution:
Given data in ascending order:
35, 35, 37, 37, 37, 37, 40, 42, 42, 43
∴ The observation repeated maximum number of times = 37
∴ Mode of the given data is 37 kg

Question 13.
In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 7
Solution:
Here, the maximum frequency is 25.
Since, Mode = observations having maximum frequency
∴ The mode of the given data is 2.

Question 14.
Find the mode of the following data.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 8
Solution:
Here, the maximum frequency is 9.
Since, Mode = observations having maximum frequency
But, this is the frequency of two observations.
∴ Mode = 35 and 37

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.5 Intext Questions and Activities

Question 1.
The first unit test of 40 marks was conducted for a class of 35 students. The marks obtained by the students were as follows. Find the mean of the marks.
40, 35, 30, 25, 23, 20, 14, 15, 16, 20, 17, 37, 37, 20, 36, 16, 30, 25, 25, 36, 37, 39, 39, 40, 15, 16, 17, 30, 16, 39, 40, 35, 37, 23, 16.
(Textbook pg, no. 123)
Solution:
Here, we can add all observations, but it will be a tedious job. It is easy to make frequency distribution table to calculate mean.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 9
= 27.31 marks (approximately)
∴ The mean of the mark is 27.31.

Maharashtra Board Class 9 Maths Solutions

Practice Set 7.4 Algebra 9th Standard Maths Part 1 Chapter 7 Statistics Solutions Maharashtra Board

9th Standard Maths 1 Practice Set 7.4 Chapter 7 Statistics Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Class 9 Maths Part 1 Practice Set 7.4 Chapter 7 Statistics Questions With Answers Maharashtra Board

Question 1.
Complete the following cumulative frequency table:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 1
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 2

Question 2.
Complete the following Cumulative Frequency Table:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 3
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 4

Question 3.
The data is given for 62 students in a certain class regarding their mathematics marks out of 100. Take the classes 0 – 10, 10 – 20,… and prepare frequency distribution
table and cumulative frequency table more than or equal to type.
55. 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10,
75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64,
42. 58. 31, 82, 27, 11, 78, 97, 07, 22, 27, 36,
35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17,
77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45.
47,49
From the prcparcd table, answer the following questions :
i. How many students obtained marks 40 or above 40?
ii. How many students obtained marks 90 or above 90?
iii. How many students obtained marks 60 or above 60?
iv. What is the cumulative frequency of equal to or more than type of the class 0 – 10?
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 5
i. 38 students obtained marks 40 or above 40.
ii. 3 students obtained marks 90 or above 90.
iii. 19 students obtained marks 60 or above 60.
iv. Cumulative frequency of equal to or more than type of the class 0 – 10 is 62.

Question 4.
Using the data In example (3) above, prepare less than type cumulative frequency table and answer the following questions.
i. How many students obtained less than 40 marks?
ii. How many students obtained less than 10 marks?
iii. How many students obtained less than 60 marks?
iv. Find the cumulative frequency of the class 50 – 60.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 6
i. 24 students obtained less than 40 marks.
ii. 3 students obtained less than 10 marks.
iii. 43 students obtained less than 60 marks.
iv. Cumulative frequency of the class 50 – 60 is 43.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.4 Intext Questions and Activities

Question 1.
The following information is regarding marks in mathematics, obtained out of 40, scored by 50 students of 9th std. ¡n the first unit test. (Textbook pg. no. 120)
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 7
From the table, fill in the blanks in the following statements.
i. For class interval 10 – 20 the lower class limit is _____ and upper class limit is _____
ii. How many students obtained marks less than 10? 2
iii. How many students obtained marks less than 20? 2 + ____ = 14
iv. How many students obtained marks less than 30? ______ + _____ = 34
v. How many students obtained marks less than 40? ______ + ______ =50
Solution:
i. 10, 20
iii. 12
iv. 14 + 20
v. 34 + 16

Question 2.
A sports club has organised a table-tennis tournaments. The following table gives the distribution of players ages. Find the cumulative frequencies equal to or more than the lower class limit and complete the table (Textbook pg. no. 121)
Solution:
Equal to lower limit or more than lower limit type of cumulative table.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.4 8

Maharashtra Board Class 9 Maths Solutions