Bhagya

Students can download Maths Chapter 5 Statistics Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.1

Question 1.
Fill in the blanks.

1. The collected information is called ……..
2. An example of Primary data is ………
3. An example of Secondary Data is ……….
4. The tally marks for number 8 in standard form is ………

Solution:

1. Data
2. List of absentees in a class
3. Cricket scores gathered from a website
4. 

Question 2.
Viji threw a die 30 times and noted down the result each time as follows. Prepare a table on the numbers shown using Tally Marks.
1, 4, 3, 5, 5, 6, 6, 4, 3, 5, 4, 5, 6, 5, 2, 4, 2, 6, 5, 5, 6, 6, 4, 5, 6, 6, 5, 4, 1, 1
Solution:

Question 3.
The following list tells colours liked by 25 students. Prepare a table using Tally Marks.

Solution:

Question 4.
The following are the marks obtained by 30 students in a class test out of 20 in Mathematics subject.

Prepare a table using Tally Marks.

Question 5.
The table shows the number of calls recorded by a Fire Service Station in one year.

Complete the table and answer the following questions.
(i) Which type of call was recorded the most?
(ii) Which type of call was recorded the least?
(iii) How many calls were recorded in all?
(iv How many calls were recorded as False Alarms?
Solution:

(i) The call for “Other Fires” was recorded the most
(ii) The call for “Rescues” was recorded the least
(iii) The total of 35 calls was recorded
(iv) There are 7 calls were recorded as False alarm.

Objective Type Questions

Question 6.
The tally marks for the number 7 in standard form is ………

Solution:
(b)

Question 7.

(a) 5
(b) 8
(c) 9
(d) 10
Solution:
(c) 9

Question 8.
The plural form of ‘datum’ is ____
(a) datum
(b) datums
(c) data
(d) dates
Solution:
(c) data

Students can download Maths Chapter 4 Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions

I. Multiple choice questions

Question 1.
If a straight line intersects the sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, then \(\frac { AE }{ AC } \) = …………
(1) \(\frac { AD }{ DB } \)
(2) \(\frac { AD }{ DB } \)
(3) \(\frac { DE }{ BC } \)
(4) \(\frac { AD }{ EC } \)
Answer:
(2) \(\frac { AD }{ DB } \)

Hint:
By BPT theorem,
\(\frac { AE }{ AC } \) = \(\frac { AD }{ AB } \)

Question 2.
In ∆ABC, DE is || to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to ……..
(1) 6.5 cm
(2) 4.5 cm
(3) 3.5 cm
(4) 3.5 cm
Answer:
(2) 4.5 cm
Hint:
By BPT theorem,

\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 3 }{ 2 } \) = \(\frac { 2.7 }{ EC } \)
∴ EC = \(\frac{2.7 \times 2}{3}\) = 1.8 CM
AC = AE + EC
= 2.7 + 1.8 = 4.5 cm

Question 3.
In ∆PQR, RS is the bisector of ∠R. If PQ = 6 cm, QR = 8 cm, RP = 4 cm then PS is equal to …………..
(1) 2 cm
(2) 4 cm
(3) 3 cm
(4) 6 cm
Answer:
(2) 2 cm
Hint:

By ABT theorem,
\(\frac { PS }{ SQ } \) = \(\frac { PR }{ QR } \) ⇒ \(\frac { x }{ 6-x } \) = \(\frac { 4 }{ 8 } \)
24 – 4x = 8x ⇒ 24 = 12x
x = 2
PS = 2 cm

Question 4.
In figure, if \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \), ∠B = 40° and ∠C = 60°, then ∠BAD = ……………
(1) 30°
(2) 50°
(3) 80°
(4) 40°
Answer:
(4) 40°
Hint:
\(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)

AD is the internal bisector of ∠BAC
∠A + ∠B + ∠C = 180°
∠A + 40° + 60° = 180° ⇒ ∠A = 80°
∴ ∠BAC = \(\frac { 80 }{ 2 } \) = 40°

Question 5.
In the figure, the value x is equal to …………
(1) 4.2
(2) 3.2
(3) 0.8
(4) 0.4

Answer:
(2) 3.2
Hint:
By Thales theorem
(DE || BC)
\(\frac { AD }{ BD } \) = \(\frac { AE }{ EC } \)
\(\frac { x }{ 8 } \) = \(\frac { 4 }{ 10 } \) ⇒ x = \(\frac{8 \times 4}{10}\) = 3.2

Question 6.
In triangles ABC and DEF, ∠B = ∠E, ∠C = ∠F, then ………….
(1) \(\frac { AB }{ DE } \) = \(\frac { CA }{ EF } \)
(2) \(\frac { BC }{ EF } \) = \(\frac { AB }{ FD } \)
(3) \(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
(4) \(\frac { CA }{ FD } \) = \(\frac { AB }{ EF } \)
Answer:
(3) \(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
Hint:

Question 7.
From the given figure, identify the wrong statement.

(1) ∆ADB ~ ∆ABC
(2) ∆ABD ~ ∆ABC
(3) ∆BDC ~ ∆ABC
(4) ∆ADB ~ ∆BDC
Answer:
(2) ∆ ABD ~ ∆ ABC

Question 8.
If a vertical stick 12 m long casts a shadow 8 m long on the ground and at the same time a tower casts a shadow 40 m long on the ground, then the height of the tower is …………..
(1) 40 m
(2) 50 m
(3) 75 m
(4) 60 m
Answer:
(4) 60 m
Hint:

\(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
\(\frac { 12 }{ h } \) = \(\frac { 8 }{ 40 } \); h = \(\frac{40 \times 12}{8}\) = 60

Question 9.
The sides of two similar triangles are in the ratio 2:3, then their areas are in the ratio ………….
(1) 9 : 4
(2) 4 : 9
(3) 2 : 3
(4) 3 : 2
Answer:
(2) 4 : 9
Hint:
Ratio of the Area of two similar triangle
= 2² : 3² = 4 : 9
[squares of their corresponding sides]

Question 10.
Triangles ABC and DEF are similar. If their areas are 100 cm² and 49 cm² respectively and BC is 8.2 cm then EF = …………….
(1) 5.47 cm
(2) 5.74 cm
(3) 6.47 cm
(4) 6.74 cm
Answer:
(2) 5.74 cm
Hint:

Question 11.
The perimeters of two similar triangles are 24 cm and 18 cm respectively.
If one side of the first triangle is 8 cm, then the corresponding side of the other triangle is ………
(1) 4 cm
(2) 3 cm
(3) 9 cm
(4) 6 cm
Answer:
(4) 6 cm
Hint:

Question 12.
A point P is 26 cm away from the centre O of a circle and PT is the tangent drawn from P to the circle 10 cm, then OT is equal to …………
(1) 36 cm
(2) 20 cm
(3) 18 cm
(4) 24 cm
Answer:
(4) 24 cm
Hint:

OT² = OP² – PT²
= 26² – 10²
= (26 + 10) (26 – 10)
OT² = 36 × 16
OT = 6 × 4 = 24 cm

Question 13.
In the figure, if ∠PAB = 120° then ∠BPT = ……….

(1) 120°
(2) 30°
(3) 40°
(4) 60°
Answer:
(4) 60°
Hint:
∠BCP + ∠PAB = 180°
(sum of the opposite angles of a cyclic quadrilateral)
∠BCP = 180° – 120° = 60°
∠BPT = 60°
(By tangent chord theorem)

Question 14.
If the tangents PA and PB from an external point P to circle with centre O are inclined to each other at an angle of 40°, then ∠POA = ………………
(1) 70°
(2) 80°
(3) 50°
(4) 60°
Answer:
(1) 70°
Hint:
∠OPA = \(\frac { 40 }{ 2 } \) = 20°
In ∆OAP.
∠POA + ∠OAP + ∠APO = 180°
(sum of the angles of a triangle)

∠POA + 90° + 20° = 180°
∠POA = 180° – 110° = 70°

Question 15.
In the figure, PA and PB are tangents to the circle drawn from an external point P. Also CD is a tangent to the circle at Q. If PA = 8 cm and CQ = 3 cm, then PC is equal to ……….

(1) 11 cm
(2) 5 cm
(3) 24 cm
(4) 38 cm
Answer:
(2) 5 cm
Hint:
PA = PB (tangent of a circle)
PB = 8 cm
PC + BC = 8
PA + QC = (BC = QC tangent)
PC + 3 = 8
∴ PC = 8 cm – 3 cm = 5 cm

Question 16.
∆ABC is a right angled triangle where ∠B = 90° and BD ⊥ AC. If BD = 8 cm, AD = 4 cm, then CD is …………
(1) 24 cm
(2) 16 cm
(3) 32 cm
(4) 8 cm
Answer:
(2) 16 cm
Hint:
∆DCB ~ ∆DBA
\(\frac { DC }{ DB } \) = \(\frac { DB }{ DA } \)

DB² = DC × DA
8² = DC × 4
DC = \(\frac { 64 }{ 4 } \) = 16 cm

Question 17.
The areas of two similar triangles are 16 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 3 cm, then the corresponding altitude of the other triangle is
(1) 6.5 cm
(2) 6 cm
(3) 4 cm
(4) 4.5 cm
Answer:
(4) 4.5 cm
Hint:

Question 18.
The perimeter of two similar triangles ∆ABC and ∆DEF are 36 cm and 24 cm respectively. If DE =10 cm, then AB is …………
(1) 12 cm
(2) 20 cm
(3) 15 cm
(4) 18 cm
Answer:
(3) 15 cm
Hint:
Perimeter of

Question 19.
In the given diagram θ is ………….
(1) 15°
(2) 30°
(3) 45°
(4) 60°

Answer:
(2) 30°
Hint:
∠BAD = 180° – 150° = 30°
= 180° – 150° = 30°
∠DAC = θ = 30°

Question 20.
If AD is the bisector of ∠A then AC is …………

(1) 12
(2) 16
(3) 18
(4) 20
Answer:
(4) 20
Hint. In ∆ABC, AD is the internal bisector of ∠A
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
\(\frac { 4 }{ 10 } \) = \(\frac { 8 }{ x } \)
4x = 10 × 8
x = \(\frac{10 \times 8}{4}\) = 20 cm

Question 21.
In ∆ABC and ∆DEF, ∠A = ∠E and ∠B = ∠F. Then AB : AC is ………….
(1) DE : DF
(2) DE : EF
(3) EF : ED
(4) DF : EF
Answer:
(3) EF : ED
Hint:

\(\frac { AB }{ EF } \) = \(\frac { BC }{ FD } \) = \(\frac { AC }{ ED } \)

Question 22.
Two circles of radius 8.2 cm and 3.6 cm touch each other externally, the distance between their centres is ………….
(1) 1.8 cm
(2) 4.1 cm
(3) 4.6 cm
(4) 11.8 cm
Answer:
(4) 11.8 cm
Hint:
Distance between the two centres

= r₁ + r₂
= 8.2 + 3.6
= 11.8 cm

Question 23.
In the given diagram PA and PB are tangents drawn from P to a circle with centre O. ∠OPA = 35° then a and b is …………

(1) a = 30°, b = 60°
(2) a = 35°, b = 55°
(3) a = 40°, b = 50°
(4) a = 45°, b = 45°
Answer:
(2) a = 35°, b = 55°
Hint:
∠OAP = 90° (tangent of the circle)
∠AOP + ∠OPA + ∠PAO = 180°
b + 35° + 90° = 180°
b = 180° – 125°
= 55°
OP is the angle bisector of ∠P
∴ a = 35°

II. Answer the following questions

Question 1.
The image of a man of height 1.8 m, is of length 1.5 cm on the film of a camera. If the film is 3 cm from the lens of the camera, how far is the man from the camera?
Solution:
Let AB be the height of the man, CD be the height of the image of the man of height 1.8 m (180 cm). LM be the distance between man and lens. LN be the distance between Lens and the film.

Given, AB = 1.8 m (180 cm)
CD = 1.5 cm and LN = 3 cm
Consider ∆ LAB and ∆ LCD
∠ALB = ∠DLC (vertically opposite angles)
∠LAB = ∠LDC (alternate angles) (AB || CD)
∴ ∆ LAB ~ ∆ LDC (AA similarity)
∴ \(\frac { AB }{ CD } \) = \(\frac { LM }{ LN } \) ⇒ \(\frac { 180 }{ 1.5 } \) = \(\frac { x }{ 3 } \) ⇒ 1.5x = 180 × 3
x = \(\frac{180 \times 3}{1.5}\) = \(\frac{180 \times 3 \times 10}{15}\)
x = 360 cm (or) 3.6 m
∴ The distance between the man and the camera = 3.6 cm

Question 2.
A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 0. 6 m/sec. If the lamp is 3.6 m above the ground level, then find the length of her shadow after 4 seconds.
Solution:
Let AB be the height of the lamp-post above the ground level.
AB = 3.6 m = 360 cm
Let CD be the height of the girl.
CD = 1.2 m = 120 cm
The distance travelled by the girl in 4 seconds (AC)
= 4 × 0.6 = 2.4m = 240 cm
Consider ∆ECD and ∆EAB
Given (CD || AB)
∠EAB = ∠ECD = 90°
∠E is common

∴ ∆ EAB = ∆ ECD = 90°
\(\frac { AB }{ CD } \) = \(\frac { AE }{ CE } \)
= \(\frac { x+240 }{ x } \) ⇒ 3 = \(\frac { x+240 }{ x } \)
3x = x + 240
2x = 240 ⇒ x = \(\frac { 240 }{ 2 } \) = 120 cm
∴ Lenght of girls shadow after 4 seconds = 120 cm

Question 3.
In ∆ ABC, AB = AC and BC = 6 cm. D is a point on the side AC such that AD = 5 cm and CD = 4 cm. Show that ∆BCD ~ ∆ACB and hence find BD.
Solution:
Given, In ∆ ABC, AB = AC = 9 cm and BC = 6 cm, AD = 5 cm and CD = 4 cm
\(\frac { BC }{ AC } \) = \(\frac { 6 }{ 9 } \) = \(\frac { 2 }{ 3 } \) …..(1)
\(\frac { CD }{ CB } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \) ……..(2)

From (1) and (2) we get,
\(\frac { BC }{ AC } \) = \(\frac { CD }{ CB } \)
In ∆ BCD and ∆ ACB
∠C = ∠C (common angle)
\(\frac { BC }{ AC } \) = \(\frac { CD }{ CB } \)
∴ ∆ BCD ~ ∆ ACB
\(\frac { BD }{ AB } \) = \(\frac { BC }{ AC } \) ⇒ \(\frac { BD }{ 9 } \) = \(\frac { 6 }{ 9 } \) ⇒ ∴ BD = 6 cm

Question 4.
The lengths of three sides of a triangle ABC are 6 cm, 4 cm and 9 cm. ∆PQR ~ ∆ABC. One of the lengths of sides of APQR is 35 cm. What is the greatest perimeter possible for ∆PQR?
Solution:
Given, ∆ PQR ~ ∆ ABC

Perimeter of ∆ ABC = 6 + 4 + 9 = 19 cm
When the perimeter of ∆ PQR is the greatest only the corresponding side QR must be equal to 35 cm

Question 5.
A man sees the top of a tower in a mirror which is at a distance of 87.6 m from the tower. The mirror is on the ground, facing upward. The man is 0.4 m away from the mirror, and the distance of his eye level from the ground is 1.5 m. How tall is the tower? (The foot of man, the mirror and the foot of the tower lie along a straight line).
Solution:
Let AB and ED be the heights of the man and the tower respectively. Let C be the point of incidence of the tower in the mirror.
In ∆ ABC and ∆ EDC, we have

∠ABC = ∠EDC = 90°
∠BCA = ∠DCE
(angular elevation is same at the same instant, i.e., the angle of incidence and the angle of reflection are same.)
∴ ∆ ABC ~ ∆ EDC (AA similarity criterion)
Thus,
\(\frac { ED }{ AB } \) = \(\frac { DC }{ BC } \) (corresponding sides are proportional)
ED = \(\frac { DC }{ BC } \) × AB = \(\frac { 87.6 }{ 0.4 } \) × 1.5 = 328.5
Hence, the height of the tower is 328.5 m.

Question 6.
In ∆ PQR, given that S is a point on PQ such that ST || QR and \(\frac { PS }{ SQ } \) = \(\frac { 3 }{ 5 } \). If PR = 5.6 cm, then find PT.
Solution:
In ∆ PQR, we have ST || QR and by Thales theorem,
\(\frac { PS }{ SQ } \) = \(\frac { PT }{ TR } \) …….(1)
Let PT = x
Thus, TR = PR – PT = 5.6 – x

From (1), we get PT = TR (\(\frac { PS }{ SQ } \))
x = (5.6 – x) (\(\frac { 3 }{ 5 } \))
5x = 16.8 – 3x
8x = 16.8
x = \(\frac { 16.8 }{ 8 } \) = 2.1
That is PT = 2.1 cm

Question 7.
In a ∆ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3, then find the value of x.
Solution:
In ∆ ABC, DE || BC. By BPT theorem. (Thales theorem)
We get \(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \) ⇒ \(\frac { 4x-3 }{ 3x-1 } \) = \(\frac { 8x-7 }{ 5x-3 } \)

(8x – 7) (3x – 1) = (4x – 3) (5x – 3)
24x² – 8x – 21x + 7 = 20x² – 12x – 15x + 9
24x² – 8x – 21x + 7 = 20x² – 27x + 9
24x² – 29x + 7 – 20x² + 27x – 9 = 0
∴ 4x² – 2x – 2 = 0
÷ 2 ⇒ 2x² – x – 1 = 0
(x – 1) (2x + 1) = 0

x – 1 = 0 (or) 2x + 1 = 0
x = 1 (or) 2x = – 1
x = – \(\frac { 1 }{ 2 } \) ⇒ since, x ≠ – \(\frac { 1 }{ 2 } \)
∴ The value of x = 1

Question 8.
In the figure AC || BD and CE || DF. if OA = 12 cm, AB = 9cm, OC = 8 cm and EF = 4.5 cm, then find FO.
Solution:

In OBD, AC || BD
∴ \(\frac { OA }{ AB } \) = \(\frac { OC }{ CD } \) (By Thales theorem)
\(\frac { 12 }{ 9 } \) = \(\frac { 8 }{ CD } \)
∴ CD = \(\frac{9 \times 8}{12}\) = 6 CM
In ODF, CE || DF
\(\frac { OC }{ CD } \) = \(\frac { OE }{ EF } \) (By Thales theorem)
\(\frac { 8 }{ 6 } \) = \(\frac { OE }{ 4.5 } \) ⇒ OE = \(=\frac{8 \times 4.5}{6}\) = 6 cm
FO = FE + EO = 4.5 + 6 = 10.5 cm
∴ The value of FO = 10.5 cm

Question 9.
Check whether AD is the bisector of ∠A of ∆ABC in each of the following.
(i) AB = 4 cm, AC = 6 cm, BD 1.6 cm, and CD = 2.4 cm.
Solution:
\(\frac { BD }{ DC } \) = \(\frac { 1.6 }{ 2.4 } \) = \(\frac { 16 }{ 24 } \) = \(\frac { 2 }{ 3 } \) …..(1)
\(\frac { AB }{ AC } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \) ……..(2)
From (1) and (2) we get,

\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
By the converse of angle bisector theorem we have,
∴ AD is the internal bisector of ∠A

(ii) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 3 cm
Solution:
\(\frac { BD }{ DC } \) = \(\frac { 1.5 }{ 3 } \) = 0.5 …….(1)
\(\frac { AB }{ AC } \) = \(\frac { 6 }{ 8 } \) = \(\frac { 3 }{ 4 } \) …….(2)

From (1) and (2) we get,
\(\frac { BD }{ DC } \) ≠ \(\frac { AB }{ AC } \)
Hence AD is not the bisector of ∠A.

Question 10.
In a ∆ABC, AD is the internal bisector of ∠A, meeting BC at D. If AB 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.
Solution:
Given, AB = 5.6 cm, AC = 6 cm, DC = 3 cm
Let BD be x
In ∆ ABC, AD is the internal bisector of ∠A.
By Angle bisector theorem, we have,

\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \) ⇒ \(\frac { x }{ 3 } \) = \(\frac { 5.6 }{ 6 } \)
x = \(\frac{3 \times 5.6}{6}\) = 2.8 cm
∴ BC = BD + DC = 2.8 + 3 = 5.8 cm
Length of BC = 5.8 cm

Question 11.
In the figure, tangents PA and PB are drawn to a circle with centre O from an external point P. If CD is a tangent to the circle at E and AP = 15 cm, find the perimeter of ∆PCD.
Solution:
We know that the lengths of the two tangents from an exterior point to a circle are equal.
∴ CA = CE, DB = DE and PA = PB
Now, the perimeter of ∆PCD

= PC + CD + DP
= PC + CE + ED + DP
= PC + CA + DB + DP
= PA + PB = 2 PA (PB = PA)
Thus, the perimeter of APCD = 2 × 15 = 30 cm

Question 12.
ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6 cm, BC = 6.5 cm and CD = 7 cm, then find the length of AD.
Solution:
Let P, Q, R and S be the points where the circle touches the quadrilateral.
We know that the lengths of the two tangents drawn from an exterior point to a circle are equal. Thus, we have,

AP = AS, BP = BQ, CR = CQ and DR = DS
Hence, AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC
⇒ AD = AB + CD – BC
= 6 + 7 – 6.5 = 6.5
Thus, AD = 6.5 cm

Question 13.
A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.
Solution:
Let the initial position of the man be “O” and the final position be B.
In the ∆AOB,

OB² = OA² + AB²
OB² = 10² + 24²
= 100 + 576 = 676
OB = \(\sqrt { 676 }\) = 26 m
The man is at a distance of 26 m from the starting point.

Question 14.
Suppose AB, AC and BC have lengths 13, 16 and 20 respectively. If \(\frac { AF }{ FB } \) = \(\frac { 4 }{ 5 } \) and \(\frac { CE }{ EA } \) = \(\frac { 5 }{ 12 } \) Find BD and DC.
Solution:
Given that AB = 13, AC = 16 and BC = 20. Let BD = JC and DC = y.
Using Ceva’s theorem we have

\(\frac { BD }{ DC } \) × \(\frac { CE }{ EA } \) × \(\frac { AF }{ FB } \) = 1 ………(1)
Substitute the given values
\(\frac { x }{ y } \) × \(\frac { 5 }{ 12 } \) × \(\frac { 4 }{ 5 } \) = 1 ⇒ \(\frac { x }{ y } \) × \(\frac { 1 }{ 3 } \) = 1
\(\frac { x }{ y } \) = 3 ⇒ x = 3y
x – 3y = 0 ……(2)
Given BC = 20
x + y = 20 …..(3)
Subtract (2) and (3)

Question 15.
ABC is a right – angled triangle at B. Let D and E be any two points on AB and BC respectively. prove that AE² + CD² = AC² + DE².
Solution:
In the right ∆ ABE right- angled at B.
AE² = AB² + BE² …..(1)
In the right ∆ DBC,CD² = BD² + BC² ………(2)
Adding (1) and (2) we get
AE² + CD² = AB² + BE² + BD² + BC²
= (AB² + BC²) + (BC² + BD²)
AE² + CD² = AC² + DE²
Hence it is proved

[AC² = AB² + BC²]
[DE² = BE² + BD²]

III. Answer the following questions

Question 1.
A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that OA² + OC² = OB² + OD².
Solution:
Given: O is any point inside the rectangle ABCD.
To prove: OA² + OC² = OB² + OD²
Construction: Through “O” draw EF || AB.
Proof: Using Pythagoras theorem

In the right ∆OEA,
∴ OA²= OE² + AE² …(1)
(By Pythagoras theorem)
In the right ∆OFC,
OC² = OF² + FC² …(2)
(By Pythagoras theorem)
OB² = OF² + FB² … (3)
(By Pythagoras theorem)
In the right ∆OED,
OD² = OE² + ED² … (4)
(By Pythagoras theorem)
By adding (3) and (4) we get
OB² + OD² = OF² + FB² + OE² + ED²
= (OE² + FB²) + (OF² + ED²)
= (OE² + EA²) + (OF² + FC²)
[FB = EA and ED = FC]
OB² + OD² = OA² + OC² using (1) and (2)

Question 2.
A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water 40 cm away from the original position of the stem. How much of the stem was below the water surface originally?
Solution:
Let O be the bottom of the stem immersed in water.
Let B be the lotus, AB be the length of the stem above the water surface
AB = 20 cm
Let OA be the length of the stem below the water surface
Let OA = x cm
Let C be the point where the lotus touches the water surface when the wind blow.
OC = OA + AB
OC = x + 20 cm
In ∆ AOC, OC² = OA² + AC²
(x + 20)² = x² + 40²
x² + 400 + 40x = x² + 1600

40x = 1600 – 400
40x = 1200
x = \(\frac { 1200 }{ 40 } \) = 30 cm
The stem is 30 cm below the water surface.

Question 3.
In the figure, DE || BC and \(\frac { AD }{ BD } \) = \(\frac { 3 }{ 5 } \), calculate the value of

Solution:
(i) Given, DE || BC and \(\frac { AD }{ BD } \) = \(\frac { 3 }{ 5 } \)
Let AD = 3k and BD = 5k; AB = 3k + 5k = 8k
In ∆^le ABC and ∆ADE
∠A = ∠A (common angle)
∠ABC = ∠ADE (corresponding angle)
Since DE || BC
∴ ∆ ABC ~ ∆ ADE

(ii) Let Area of ∆ ADE be 9k and Area of ∆ ABC be 64 k
Area of ∆ BCED = Area of ∆ ABC – Area of ∆ ADE
= 64 k – 9 k
= 55 k

Question 4.
A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?
Solution:
Let in AB be “x”, BE be “z” and BC be “y”
In the right ∆ AEB,
AB² = AE² + BE²
x² = 16² + Z² …….(1)
In the right ∆ BEC,
BC² = EC² + BE²
y² = 81² + z² ……(2)

In the right ∆ ACD,
AC² = AD² + DC²
97² = x² + y² ….(3)
Add (1) and (2) ⇒ x² + y² = 162 + z² + 81² + z²
x² + y² = 2z² + 16² + 81²
97² = 2z² + 16² + 81² (from 3)
9409 = 2z² + 256 + 6561
= 2z² + 6817
2z² = 9409 – 6817 = 2592
z² = \(\frac { 2592 }{ 2 } \) = 1296
z = \(\sqrt { 1296 }\) = 36
∴ Length of cross bar BD = 2 × BE = 2 × 36 = 72 cm

Question 5.
Find the unknown values in each of the following figures. All lengths are given in centimetres (All measures are not in scale)

Solution:
(i) In ∆ ABC and ∆ ADE,
∠A = ∠A (common angle)
∠ABC = ∠ADE (corresponding angle) [BC || DE]
∆ ABC ~ ∆ ADE (By AAA similarity)

In ∆ EAG and ∆ ECF,
∠E || ∠E (common angle)
∠ECF = ∠EAG (corresponding angle)
Given CF || AG
∆ EAG ~ ∆ ECF
\(\frac { EC }{ EA } \) = \(\frac { CF }{ AG } \) ⇒ \(\frac { 8 }{ x+8 } \) = \(\frac { 6 }{ y } \) ⇒ \(\frac { 8 }{ 12 } \) = \(\frac { 6 }{ y } \) (x = 4)
8y = 6 × 12
y = \(\frac{12 \times 6}{8}\) = 9 cm
∴ The value of x = 4 cm and y = 9 cm.

(ii) In ∆ HBC and ∆ HFG,
∠H = ∠H (common angle)
∠HFG = ∠HBC (corresponding angle)
Given FG || BC
\(\frac { HF }{ HB } \) = \(\frac { FG }{ BC } \) ⇒ \(\frac { 4 }{ 10 } \) = \(\frac { x }{ 9 } \) ⇒ 10x = 36
x = 3.6 cm
In ∆ FBD and ∆ FHD,
∠BFD = ∠HFG (vertically opposite angle)
∠FBD = ∠FHG (Alternate angles)
By AA similarity
∆ FBD ~ ∆ FHG
\(\frac { FG }{ FD } \) = \(\frac { FH }{ FB } \) ⇒ \(\frac { x }{ 3+y } \) = \(\frac { 4 }{ 6 } \)
4 (3 + y) = 3.6 × 6
3 + y = \(\frac{3.6 \times 6}{4}\) = 5.4 ⇒ y = 5.4 – 3 = 2.4 cm
In ∆ AEG and ∆ ABC,
∠A = ∠A (common angle)
∠AEG = ∠ABC (corresponding angles)
Given EG || BC
\(\frac { AE }{ AB } \) = \(\frac { EG }{ BC } \)
\(\frac { z }{ z+5 } \) = \(\frac { x+y }{ 9 } \) ⇒ \(\frac { z }{ z+5 } \) = \(\frac { 3.6+2.4 }{ 9 } \)
\(\frac { z }{ z+5 } \) = \(\frac { 6 }{ 9 } \) ⇒ \(\frac { z }{ z+5 } \) = \(\frac { 2 }{ 3 } \)
3z = 2z + 10 ⇒ 3z – 2z = 10 ⇒ z = 10
∴ The values of x = 3.6 cm, y = 2.4 cm and z = 10 cm.

Question 6.
The internal bisector of ∠A of AABC meets BC at D and the external bisector of ∠A meets BC produced at E. Prove that \(\frac { BD }{ BE } \) = \(\frac { CD}{ CE } \)
Solution:
Given: In ∆ ABC, AD is the internal bisector of ∠A meets BC at D. AE is the external bisector of ∠A meets BC produced to E.
To Proof: \(\frac { BD }{ BE } \) = \(\frac { CD }{ CE } \)

Proof: In ∆ ABC, AD is the internal bisector of ∠A.
By ABT theorem we have,
∴ \(\frac { BE }{ CD } \) = \(\frac { AB }{ AC } \) ……….(1)
In ∆ABC, AE is the external bisector of ∠A.
By ABT theorem we have,
∴ \(\frac { BE }{ CE } \) = \(\frac { AB }{ AC } \) ……….(2)
From (1) and (2) we get
\(\frac { BD }{ CD } \) = \(\frac { BE }{ CE } \) ⇒ ∴ \(\frac { BD }{ BE } \) = \(\frac { CD }{ CE } \)

Question 7.
In a quadrilateral ABCD, the bisectors of ∠B and ∠D intersect on AC at E. Prove that \(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
Solution:
Given: ABCD is a quadrilateral. BE is the bisector of ∠B intersecting AC at E, DE is the bisector of ∠D intersecting AC at E.
To proove: \(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
Proof: In ∆ABC, BE is the internal bisector of ∠D.
By Angle bisector theorem we have,
\(\frac { AE }{ EC } \) = \(\frac { AB }{ BC } \) ……..(1)

In ∆ ACD, DE is the internal bisector of ∠C.
By Angle bisector theorem we have,
∴ \(\frac { AE }{ EC } \) = \(\frac { AD }{ DC } \) ……….(2)
From (1) and (2) we get,
\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)

Question 8.
ABCD is a quadrilateral with AB parallel to DC. A line drawn parallel to AB meets AD at P and BC at Q. Prove that \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)
Solution:
Given: ABCD is a quadrilateral. AB || DC.
The line PQ intersect AD at P and BC at Q

To prove: \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)
Proof: In the ∆ABC, OQ || AB
By BPT, theorem we have, \(\frac { AO }{ OC } \) = \(\frac { BQ}{ QC } \) ……………(1)
In the ∆ACD, PO || DC
By BPT, theorem we have, \(\frac { AO }{ OC } \) = \(\frac { AP }{ PD } \) ……..(2)
From (1) and (2) we get, \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)

Question 9.
D is the midpoint of the side BC of AABC. If P and Q are points on AB and on AC such that DP bisects ∠BDA and DQ bisects ∠ADC, then prove that PQ || BC.
Solution:
In ∆ABD,DP is the angle bisector of ∠BDA.
∴ \(\frac { AP }{ PB } \) = \(\frac { AD }{ BD } \) (angle bisector theorem) ……..(1)
In ∆ADC, DQ is the bisector of ∠ADC
∴ \(\frac { AQ }{ QC } \) = \(\frac { AD }{ DC } \) (angle bisector theorem) ……….(2)

But, BD = DC (D is the midpoint of BC)
Now (2) ⇒ ∴ \(\frac { AQ }{ QC } \) = \(\frac { AD }{ BD } \)
From (1) and (3) We get,
∴ \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)
Thus PQ || BC (converse of thales theorem)

Question 10.
ABCD is a trapezium with AB || DC. The diagonal AC and BD intersect at E. If ∆AED ~ ∆BEC. Prove that AD = BC.
Solution:
By given data ABCD is a trapezium with AB || DC.
In ∆ ECD and ∆ ABE

∠EDC = ∠EBA
∠ECD = ∠EAB
∴ ∆ DEC ~ ∆ BEA by AA – similarity
\(\frac { DE }{ BE } \) = \(\frac { EC }{ EA } \) = \(\frac { DC }{ BA } \)
\(\frac { DE }{ BE } \) = \(\frac { EC }{ EA } \)
\(\frac { DE }{ EC } \) = \(\frac { BE }{ EA } \) ……….(1)

Also given ∆ DEA ~ ∆ CEB
\(\frac { DE }{ CE } \) = \(\frac { EA }{ EB } \) = \(\frac { DA }{ CB } \)
\(\frac { DE }{ CE } \) = \(\frac { EA }{ EB } \) …………(2)
From (1) and (2) we get
\(\frac { BE }{ EA } \) = \(\frac { EA }{ EB } \) ⇒ EB² = EA²
∴ EB = EA
Substitute in (2) we get
\(\frac { EA }{ EA } \) = \(\frac { DA }{ CB } \)
1 = \(\frac { DA }{ CB } \) ⇒ ∴ AD = DC Hence it is proved

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.5 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.5

Miscellaneous Practice Problems

Question 1.
The maximum speed of some of the animals are given below:
the Elephant = 20 km/h; the
Lion = 80 km/h;
the Cheetah =100 km/h
Find the following ratios of their speeds in simplified form and find which ratio is the least?
(i) the Elephant and the Lion
(ii) the Lion and the Cheetah
(iii) the Elephant and the Cheetah
Solution:
(i) The Elephant: the Lion
= 20 : 80 = \(\frac{20}{80}\) = \(\frac{1}{4}\) = 1 : 4

(ii) the Lion : the Cheetah
= 80 : 100 = \(\frac{80}{100}\) = \(\frac{4}{5}\) = 4 : 5

(iii) the Elephant: the Cheetah
= 20 : 100 = \(\frac{20}{100}\) = \(\frac{1}{5}\) = 1 : 5
The ratio of Elephant to Cheetah is the least.

Question 2.
A particular high school has 1500 students 50 teachers and 5 administrators. If the school grows to 1800 students and the ratios are maintained, then find the number of teachers and administrators.
Solution:
Administrators : teachers : students = 5 : 50 : 1500 = 1 : 10 : 300
If the school grows to 1800 students then 10 parts = teachers
1 part = administrators
300 parts = 1800
1 part = \(\frac{1800}{300}\) = 6
10 parts = 6 × 10 = 60
So, if the school grows to 1800 students the new ratio is administrators : teachers: students
6 : 60 : 1800

Question 3.
I have a box which has 3 green, 9 blue, 4 yellow, 8 orange coloured cubes in it.
(a) What is the ratio of orange to yellow cubes?
(b) What is the ratio of green to blue cubes?
(c) How many different ratios can be formed, when you compare each colour to any one of the other colours?
Solution:
Number of green cubes = 3
Number of blue cubes = 9
Number of yellow cubes = 4
Number of orange cubes = 8
(a) Ratio of orange to yellow cubes \(\frac{\text { Number of orange cubes }}{\text { Number of yellow cubes }}=\frac{8}{4}=\frac{2}{1}=2: 1\)
Ratio of orange to yellow cubes = 2 : 1
(b) \(\frac{\text { Number of green cubes }}{\text { Number of blue cubes }}=\frac{3}{9}=\frac{1}{3}\)
Ratio of green to blue cubes = 1 : 3
(c) The ratios can be Orange : Yellow, Orange: blue, Orange : green, Yellow : Orange, yellow : blue, yellow : green, blue : green, blue : orange, blue : yellow, green : orange, green : yellow, green : blue. Thus 12 ratios can be formed.

Question 4.
A gets double of what B gets and B gets double of what C gets. Find A : B and B : C and verify whether the result is in proportion or not.
Solution:
Let x be the part owned by C then A : B : C = 2(2x) : 2x : x = 4x : 2x : x
A : B = 4x : 2x = 2 : 1
B : C = 2x : x = 2 : 1
A : B : : B : C. i.e, They are in proportion.

Question 5.
The ingredients required for the preparation of Ragi Kali, a healthy dish of Tamilnadu is given below.

(a) If one cup of ragi flour is used then, what would be the amount of raw rice required?
(b) If 16 cups of water are used, then how much ragi flour should be used?
(c) Which of these ingredients cannot be expressed as a ratio? Why?
Solution:
(i) \(\frac{1}{4}\) cup
(ii) 8 cups
(iii) Ragi flour, Raw rice, and water are in one unit. Sesame oil and salt are in different units. These different units cannot be compared and cannot be expressed as a ratio because the two quantities of a ratio should be in the same unit.

Question 6.
Antony brushes his teeth in the morning and night on all days of the week. Shabeen brushes her teeth only in the morning. What is the ratio of the number of times they brush their teeth in a week?
Solution:
Number of times Antony brushes a day = 2
Number of times Antony brushes a week = 2 × 7 = 14
Number of times Shabeen brushes a day = 1
Number of times Shabeen brushes a week = 1 × 7 = 7
Number of times Antony brushes : Number of times Antony brushes = 14 : 7 = 2 : 1
The required ratio = 2 : 1

Question 7.
Thirumagal’s mother wears a bracelet made of 35 red beads and 30 blue beads. Thirumagal wants to make smaller bracelets using the same two coloured beads in the same ratio. In how many different ways can she make the bracelets?

Solution:
Red : blue = 35 : 30 = 7 : 6
Different ways (i) 7 : 6
(ii) 14 : 12;
(iii) 21 : 18;
(iv) 28 : 24

Question 8.
Team A wins 26 matches out of 52 matches. Team B wins three fourth of 52 matches played. Which team has a better winning record?
Solution:
Team A = \(\frac{26}{52}\) = \(\frac{1}{2}\)
Team B = \(\frac{3}{4}\) × 52 = 39
Team B has a better winning record.

Question 9.
In a school excursion, 6 teachers and 12 students from 6th standard and 9 teachers and 27 students from 7th standard, 4 teachers and 16 students from 8th standard took part. Which class has the least teacher to student ratio?
Solution:
Std VI – teachers: students = 6 : 12 = 1 : 2
Std VII – teachers : students = 9 : 27 = 1 : 3
Std VIII – teachers : students = 4 : 16 = 1 : 4
Std VIII has the least ratio.

Question 10.
Fill the boxes using any set of suitable numbers 6 : ___ : : ___ : 15
Solution:
6 : ……. = …….. : 15
Product of the extremes = 6 × 15 = 90
Set of suitable numbers
1 and 90, 2 and 45, 3 and 30, 5 and 18, 6 and 15

Question 11.
From your school diary, write the ratio of the number of holidays to the number of working days in the current academic year.
Solution:
Number of holidays = 145
Number of working days = 220
Holidays : working days = 145 : 220
= \(\frac{145}{220}\)
= \(\frac{29}{44}\)
= 29 : 44

Question 12.
If the ratio of Green, Yellow and Black balls in a bag is 4 : 3 : 5, then
(a) Which is the most likely ball that you can choose from the bag?
(b) How many balls in total are there in the bag if you have 40 black balls in it?
(c) Find the number of green and yellow balls in the bag.
Solution:
Green : Yellow : Black = 4 : 3 : 5
(i) Blackballs;
(ii) 96 balls (32 + 24 + 40);
(iii) green balls = 32
yellow balls = 24

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.4 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.4

Question 1.
Fill in the blanks.
(i) If the cost of 3 pens is Rs 18, then the cost of 5 pens is ……..
(ii) If Karkuzhali earns Rs 1800 in 15 days, then she earns Rs 3000 in …….. days
Solution:
(i) ₹ 30
Hint: \(5 \times \frac{18}{3}\) = 5 × 6 = ₹ 30
(ii) 25 Days
Hint:
\(\frac{1800}{3000}=\frac{15}{x}\)
⇒ x = \(\frac{15 \times 3000}{1800}\) = 25 days

Question 2.
Say True or False.
(i) If the weight of 40 books is 8 kg, then the weight of 15 books is 3 kg.
(ii) A car travels 90 km in 3 hours with constant speed. It will travel 140 km in 5 hours at the same speed.
Solution:
(i) True
Hint: Weight of 1 book = \(\frac{8}{40}=\frac{1}{5} \mathrm{kg}\)
Flence Weight of 15 books = \(\frac{1}{5} \times 15=3 \mathrm{kg}\)
(ii) False
1 hour the car travels = \(\frac{90}{3}\) = 30 km
In 5 hours the car travels = 30 × 5 = 150 km

Question 3.
If a person reads 20 pages of a book in 2 hours, how many pages will he read in 8 hours at the same speed?
Solution:
In 2 hours, pages read = 20
In 1 hour, pages read = \(\frac{20}{2}\) = 10
In 8 hours, pages read = 10 × 8
= 80 pages

Question 4.
The cost of 15 chairs is ₹ 7500. Find the number of such chairs that can be purchased for ₹ 12,000?
Solution:
Cost of 15 chairs = Rs 7500
Cost of 1 chair = Rs \(\frac{7500}{15}\) = Rs 500
Number of chairs that can be purchased for Rs 12000 = \(\frac{12000}{500}\) = 24 chairs

Question 5.
A car covers a distance of 125 km in 5 kg of LP Gas. How much distance will it cover in 3 kg of LP Gas?
Solution:
In 5 kg of LPG gas, distance covered = 125 km
In 1 kg of LPG gas, distance covered = \(\frac{125}{5}\) = 25 km
In 3 kg of LPG gas, distance covered = 3 × 25 km = 75 km

Question 6.
Cholan walks 6 km in 1 hour at a constant speed. Find the distance covered by him in 20 minutes at the same speed.
Solution:
In 1 hour (60 minutes), distance covered = 6 km
In 1 minute, distance covered = \(\frac{6 km}{60 min}\) = \(\frac{6000 m}{60}\) = 100 m
In 20 minutes, distance covered = 20 × 100 m = 2000 m = 2 km

Question 7.
The number of correct answers given by Kaarmugilan and Kavitha in a quiz competition are in the ratio 10 : 11. If they had scored a total of 84 points in the competition, then how many points did Kavitha get?
Solution:
Total points scored = 84 Ratio = 10 : 11
Sum of the ratio = 10 + 11 = 21
21 parts = 84 points
1 part = \(\frac{84}{21}\) = 4 points
Kavitha = 11 parts
Kaarmugilan = 10 parts
Foints scored by Kavitha = 11 parts = 11 × 4 points = 44 points

Question 8.
Karmegam made 54 runs in 9 overs and Asif made 77 runs in 11 overs. Whose run rate is better? (run rate = ratio of runs to overs)
Solution:
Karmegam Runs made in 9 overs = 54
Runs made in 1 over = \(\frac{54}{9}\) = 6 runs
Asif Runs made in 11 overs = 77
Runs made in 1 over = \(\frac{77}{11}\) = 7 runs
∴ Asif’s run rate is better than Karmegam.

Question 9.
You purchase 6 apples for Rs 90 and your friend purchases 5 apples for Rs 70. Whose purchase is better?
Solution:
Myself
Cost of 6 apples = Rs 90
Cost of 1 apple = \(\frac{Rs 90}{6}\) = Rs 15
Friend’s purchase
Cost of 5 apples = Rs 70
Cost of 1 apple = \(\frac{70}{5}\) = Rs 14
∴ Friend’s purchase is better than mine.

Objective Type Questions

Question 10.
If a Barbie doll costs ₹ 90, then the cost of 3 such dolls is ₹ _____
(a) 260
(b) 270
(c) 30
(d) 93
Solution:
(b) 270
Hint:
Cost of 3 dolls = 90 × 3 = ₹ 270

Question 11.
If 8 oranges cost Rs 56, then the cost of 5 oranges is Rs …….
(a) 42
(b) 48
(c) 35
(d) 24
Solution:
(c) 35

Question 12.
If a man walks 2 km in 15 minutes, then he will walk _____ km in 45 minutes.
(a) 10
(b) 8
(c) 6
(d) 12
Solution:
(c) 6
Hint:
1 min he walks = \(\frac{2}{15}\) km
45 min he walks = \(\frac{2}{15}\) × 45 = 6 km.

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – \(\frac { 3 }{ 17 } \) = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = \(\frac { 3 }{ 5 } \)
Slope = 0

(ii) 7x – \(\frac { 3 }{ 17 } \) = 0 (Comparing with y = mx + c)
7x = \(\frac { 3 }{ 17 } \)
Slope is undefined

Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)

Question 3.
Check whether the given lines are parallel or perpendicular
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 and \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 ; \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
Slope of the line (m₁) = \(\frac { -a }{ b } \)
= – \(\frac { 1 }{ 3 } \) ÷ \(\frac { 1 }{ 4 } \) = –\(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
Slope of the line (m₂) = – \(\frac { 2 }{ 3 } \) ÷ \(\frac { 1 }{ 2 } \) = –\(\frac { 2 }{ 3 } \) × \(\frac { 2 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
m₁ = m₂ = – \(\frac { 4 }{ 3 } \)
∴ The two lines are parallel.

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m₁) = \(\frac { -5 }{ 23 } \)
Slope of the line (m₂) = \(\frac { -23 }{ -5 } \) = \(\frac { 23 }{ 5 } \)
m₁ × m₂ = \(\frac { -5 }{ 23 } \) × \(\frac { 23 }{ 5 } \) = -1
∴ The two lines are perpendicular

Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = \(-\frac{(p+3) x}{12}+1\) (Comparing with y = mx + c)
Slope of the second line (m₁) = \(\frac { -(p+3) }{ 12 } \)
Slope of the second line 12x – 7y = 16
(m₂) = \(\frac { -a }{ b } \) = \(\frac { -12 }{ -7 } \) = \(\frac { 12 }{ 7 } \)
Since the two lines are perpendicular
m₁ × m₂ = -1
\(\frac { -(p+3) }{ 12 } \) × \(\frac { 12 }{ 7 } \) = -1 ⇒ \(\frac { -(p+3) }{ 7 } \) = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4

Question 5.
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Solution:
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line QR = \(\frac { 4+2 }{ -5-3 } \) = \(\frac { 6 }{ -8 } \) = \(\frac { 3 }{ -4 } \) ⇒ – \(\frac { 3 }{ 4 } \)
Slope of its parallel = – \(\frac { 3 }{ 4 } \)
The given point is p(-5, 2)
Equation of the line is y – y₁ = m(x – x₁)
y – 2 = – \(\frac { 3 }{ 4 } \) (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0

Question 6.
Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Let the vertices A (6, 7), B (2, -3), D (6, -2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-7 }{ 2-6 } \) = \(\frac { -10 }{ -4 } \) = \(\frac { 5 }{ 2 } \)
Slope of its perpendicular (CD) = – \(\frac { 2 }{ 5 } \)
Equation of the line CD is y – y₁ = m(x – x₁)
y + 2 = –\(\frac { 2 }{ 5 } \) (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0

Question 7.
A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 3+2 }{ 12-10 } \) = \(\frac { 5 }{ 2 } \)
Slope of the altitude AD is – \(\frac { 2 }{ 5 } \)
Equation of the altitude AD is
y – y₁ = m (x – x₁)
y – 0 = – \(\frac { 2 }{ 5 } \) (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B

Slope of AC = \(\frac { 3-0 }{ 12+3 } \) = \(\frac { 3 }{ 15 } \) = \(\frac { 1 }{ 5 } \)
Slope of the altitude AD is -5
Equation of the altitude BD is y – y₁= m (x – x₁)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0

Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.

Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { -4-2 }{ 6+4 } \) = \(\frac { -6 }{ 10 } \) = – \(\frac { 3 }{ 5 } \)
Slope of the ⊥^r AB is \(\frac { 5 }{ 3 } \)
Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
= (\(\frac { -4+6 }{ 2 } \),\(\frac { 2-4 }{ 2 } \)) = (\(\frac { 2 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y₁ = m(x – x₁)
y + 1 = \(\frac { 5 }{ 3 } \) (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0

Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Solution:
Given lines are.

x = \(\frac { 43 }{ 43 } \) = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = \(\frac { 3 }{ 3 } \) = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0

Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Solution:
Given lines are.

Substitute the value of x = \(\frac { 16 }{ 7 } \) in (2)
3 × \(\frac { 16 }{ 7 } \) + 2y = 10 ⇒ 2y = 10 – \(\frac { 48 }{ 7 } \)
2y = \(\frac { 70-48 }{ 7 } \) ⇒ 2y = \(\frac { 22 }{ 7 } \)
y = \(\frac{22}{2 \times 7}\) = \(\frac { 11 }{ 7 } \)
The point of intersect is (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
7 (\(\frac { 16 }{ 7 } \)) + 4 (\(\frac { 11 }{ 7 } \)) + k = 0 ⇒ 16 + \(\frac { 44 }{ 7 } \) + k = 0
\(\frac { 112+44 }{ 7 } \) + k = 0 ⇒ \(\frac { 156 }{ 7 } \) + k = 0
k = – \(\frac { 156 }{ 7 } \)
Equation of the line is 7x + 4y – \(\frac { 156 }{ 7 } \) = 0
49x + 28y – 156 = 0

Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Solution:
The given lines are.


Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is

Substitute the value of y = \(\frac { 9 }{ 11 } \) in (6)
– x + 2 (\(\frac { 9 }{ 11 } \)) = 3 ⇒ -x + \(\frac { 18 }{ 11 } \) = 3
-x = 3 – \(\frac { 18 }{ 11 } \) = \(\frac { 33-18 }{ 11 } \) = \(\frac { 15 }{ 11 } \)
x = – \(\frac { 15 }{ 11 } \)
The point of intersection is (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \))
Equation of the line joining the points (0, -2) and (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \)) is


31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0

Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)

x = \(\frac { 63 }{ 28 } \) = \(\frac { 9 }{ 4 } \)
Substitute the value of x = \(\frac { 9 }{ 4 } \) in (2)
4 (\(\frac { 9 }{ 4 } \)) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (\(\frac { 9 }{ 4 } \),0)
Mid point of the points (5, -4) and (-7, 6)

Equation of the line joining the points (\(\frac { 9 }{ 4 } \),0) and (-1,1)

-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0

Students can download Maths Chapter 2 Introduction to Algebra Ex 2.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Ex 2.3

Miscellaneous Practice Problems

Question 1.
Complete the following pattern.
9 – 1 =
98 – 21 =
987 – 321 =
9876 – 4321 =
98765 – 54321 =
What comes next?
Solution:
9 – 1 = 8
98 – 21 = 77
987 – 321 = 666
9876 – 4321 = 5555
98765 – 54321 = 44444
Next will be 987654 – 654321 = 333333

Question 2.
A piece of wire is ‘12s’ cm long. What will be the length of the side if it is formed as
(i) an equilateral triangle.
(ii) a square?
Solution:
(i) An equilateral triangle has 3 equal sides.
Length of the wire = ‘12s’ cm
Length of each side = \(\frac{12 s}{3}\) cm.
Length of each side of the triangle = 4s cm
(ii) A square has four equal sides.
Length of each side \(\frac{12 s}{4}\) cm
Length of each side of the square 3s cm

Question 3.
Identify the value of the shapes and figures in the table given below and verify their addition horizontally and vertically.

Solution:

Question 4.
The table given below shows the results of the matches played by 8 teams in a kabaddi championship tournament.

Find the value of all the variables in the table given above.
Solution:
k = 3, m = 1, n = 10, a = 9, b = 6, c = 4, x = 4, y = 9

Challenging Problems

Question 5.
Gopal is 8 years younger than Karnan. If the sum of their ages is 30, how old is Karnan?
Solution:
Let the age of Kaman be x years
Gopal’s age = x – 8
Ace to the problem, x + x – 8 = 30
2x – 8 = 30
2x = 30 + 8
2x = 38
x = \(\frac{38}{2}\)
x = 19
Age of Kaman = 19 years

Question 6.
The rectangles made of identical square blocks with varying lengths but having only two square blocks as width are give below

(i) How many small size squares are there in each of the rectangles P,Q, R and S?
(ii) Fill in the boxes.

Solution:
(i) P = 2; Q = 8; R = 6; S = 10

Question 7.
Find the variables from the clues given below and solve the crossword puzzle.


Solution:

Students can download Maths Chapter 2 Introduction to Algebra Ex 2.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Ex 2.2

Question 1.
Find in the blanks.
(i) The algebraic statement of ‘f’ decreased by 5 is _______
(ii) The algebraic statement of ‘s’ divided by 5 is _______
(iii) The verbal statement of ‘2m – 10’ is _______
(iv) If A’s age is ‘n’ years now, 7 years ago A’s age was ______
(v) If ‘p – 5’ gives 12 then ‘p’ is ______
Solution:
(i) f – 5
(ii) \(\frac { s }{ 5 }\)
(iii) 10 less than 2 times m (or) Take away 10 from the product of 2 and m
(iv) n – 7
(v) 17
Hint: p – 5 = 12 ⇒ n = 12 + 5 = 17

Question 2.
Say True or False.
(i) 10 more to three times ‘c’ is ‘3c + 13’.
(ii) If the cost of 10 rice bags is ‘t’, then the cost of 1 rice bag is \(\frac { t }{ 10 }\)
(iii) The statements ‘x’ divided by 3 and 3 divided by ‘x’ are the same.
(iv) The product of ‘q’ and 20 is ’20q’
(v) 7 less to 7 times ‘y’ is ‘7 – 7y’
Solution:
(i) False
Hint: 3c + 10
(ii) True
(iii) False
(iv) True
(v) False
Hint: 7y – 7

Question 3.
Express the following verbal statement to an algebraic statement.
(i) ‘t’ is added to 100
(ii) 4 times ‘q’
(iii) 8 is reduced by ‘y’
(iv) 56 added to 2 times ‘x’
(v) 4 less to 9 times of ‘y’
Solution:
(i) t + 100
(ii) 4q
(iii) 8 – y
(iv) 2x + 56
(v) 9y – 4

Question 4.
Express the following algebraic statement to a verbal statement.
(i) x ÷ 3
(ii) 5n – 12
(iii) 11 + 10x
(iv) 70s
Solution:
(i) x divided by 3.
(ii) 12 less to 5 times n.
(iii) 11 added to 10 times x
(iv) 7 times s.

Question 5.
The teacher asked two students to write the algebraic statement for the verbal statement “8 more than a number”. Vetri wrote ‘8 + x’ but Maran wrote ‘8x’. Who gave the correct answer?
Solution:
Let the number be x; 8 more than the number = 8 + x.
Vetri gave the correct answer as 8 + x.

Question 6.
Answer the following questions.
(i) If ‘n’ takes the value 3 then find the value of ‘n + 10’?
(ii) If ‘g’ is equal to 300 what is the value of ‘g – 1’ and ‘g + 1’?
(iii) What is the value of ‘s’, if ‘2s – 6’ gives 30?
Solution:
(i) n = 3
n + 10 = 3 + 10 = 13

(ii) g = 300
g – 1 = 300 – 1
= 299
g + 1 = 300 + 1
=301

(iii) 2s – 6 = 30
2s = 30 + 6
2s = 36
s = 36/2
s = 18

Question 7.
Complete the table and find the value of ‘k’ for which ‘k/3’ gives 5.

Solution:

\(\frac{k}{3}\) = 5
k = 15

Question 8.
The value of ‘y’ in y + 7 = 13 is
(a) y = 5
(b) y = 6
(c) y = 7
(d) y = 8
Solution:
(b) y = 6
Hint: y = 13 – 7 = 6

Question 9.
6 less to ‘n’ gives 8 is represented as
(a) n – 6 = 8
(b) 6 – n = 8
(c) 8 – n = 6
(d) n – 8 = 6
Solution:
(a) n – 6 = 8

Question 10.
The value of ‘c’ for which \(\frac{3c}{4}\) gives 18 is
(a) c = 15
(b) c = 21
(c) c = 24
(d) c = 27
Solution:
(c) c = 24

Students can download Maths Chapter 2 Introduction to Algebra Ex 2.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Ex 2.1

Question 1.
Fill in the blanks:
(i) The letters a, b, c, .., x, y, z are used to represent _____
(ii) A quantity that takes _____ values is called a variable.
(iii) If there are 5 students on a bench, then the number of students in ‘n’ benches is ‘5 × n’. Here _____ is a variable.
Solution:
(i) Variables
(ii) Different
(iii) n

Question 2.
Say True or False.

(iii) If there are 11 players in a team, then there will be ’11 + q’ players in ‘q’ teams.
Solution:
(i) False
(ii) True
(iii) False

Question 3.
Draw the next two patterns and complete the table

Solution:

Question 4.
Use a variable to write the rule, which gives the number of ice candy sticks required to make the following patterns.

Solution:
(a) No of ice candy sticks used = 3 × n = 3n
(b) No of ice candy sticks used = 4 × n = 4n

Question 5.
The teacher forms groups of five students in a class. How many students will be there in ‘p’ groups?
Solution:
No of students in a group = 5
No of students in ‘p’ groups = 5 × p = 5p

Question 6.
Arivazhagan is 30 years younger than his father. Write Arivazhagan’s age in terms of his father’s age.
Solution:
Let Arivazhagan’s father’s age be x years
According to the problem,
Arivazhagan’s age = (x – 30) years

Question 7.
If ‘u’ is an even number, how would you represent
(i) the next even number?
(ii) the previous even number?
Solution:
(i) Difference between two even numbers = 2
Given that ‘u’ is an even number.
Next, the even number is u + 2.
(ii) Previous even number is u – 2.

Objective Type Questions

Question 8.
Variable means that it
(a) can take only a few values
(b) has a fixed value
(c) can take different values
(d) can take only 8 values
Solution:
(c) can take different values

Question 9.
‘6y’ means
(a) 6 + y
(b) 6 – y
(c) 6 × y
(d) \(\frac{6}{7}\)
Solution:
(c) 6 × y

Question 10.
Radha is ‘x’ years of age now. 4 years ago, her age was
(a) x – 4
(b) 4 – x
(c) 4 + x
(d) 4x
Solution:
(a) x – 4

Question 11.
The number of days in ‘w’ weeks is
(a) 30 + w
(b) 30w
(c) 7 + w
(d) 7w
Solution:
(d) 7w

Question 12.
The value of ‘x’ in the circle is

(a) 6
(b) 8
(c) 21
(d) 22
Solution:
(d) 22

Students can download Maths Chapter 1 Numbers Ex 1.6 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.6

Miscellaneous Practice Problems

Question 1.
Try to open my locked suitcase which has the biggest 5 digit odd number as the password comprising the digits 7, 5, 4, 3 and 8. Find the password.
Solution:
87543

Question 2.
As per the census of 2001, the population of four states are given below. Arrange the states in ascending and descending order of their population.

Solution:
Ascending Order: 6,85,48,437 < 7,21,47,030 < 7,26,26,809 < 9,12,76,115
Descending Order 9,12,76,115 > 7,26,26,809 > 7,21,47,030 > 6,85,48,437

Question 3.
Study the following table and answer the questions.

(i) How many tigers were there in 2011?
(ii) How many tigers were less in 2008 than in 1990?
(iii) Did the number of tigers increase or decrease between 2011 and 2014? If yes, by how much?
Solution:
(i) 1706
(ii) 2100
(iii) Yes, 2226 – 1706 = 520 tigers increased from 2011 to 2014

Question 4.
Mullaikodi has 25 bags of apples. In each bag, there are 9 apples. She shares them equally amongst her 6 friends. How many apples does each get? Are there any apples left over?
Solution:
No of apple bags = 25
Apples in each bag = 9
Total no of apples = 25 × 9 = 225
Apples shared among her 6 friends = 225 ÷ 6
So, among her 6 friends, each of them gets 37 apples and 3 apples are leftover.

Question 5.
Poultry has produced 15472 eggs and fits 30 eggs in a tray. How many trays do they need?
Solution:
No of eggs produced = 15472
No of eggs fits in a tray = 30
No of trays required = 15472 ÷ 30
Trays required = 515 + 1 (to fit the remaining 22 eggs) = 516
Quotient = 515
Remainder = 22

Challenging Problems

Question 6.
Read the table and answer the following questions.

(i) Write the Canopus star’s diameter in words, in the Indian and the International System.
(ii) Write the sum of the place values of 5 in Sirius star’s diameter in the Indian System.
(iii) Eight hundred sixty four million seven hundred thirty. Write in Indian System.
(iv) Write the diameter in words of Arcturus star in the International System
(v) Write the difference of the diameters of Canopus and Arcturus stars in the Indian and the International Systems.
Solution:
(i) 2,59,41,900
25,941,900
Indian System: Two crores fifty-nine lakh forty-one thousand nine hundred.
International System: Twenty-five million nine hundred forty-one thousand nine hundred.
(ii) 5,50,500
(iii) 864,000,730 (86,40,00,730)
(iii) Eighty-six crore forty lakh seven hundred thirty.
(iv) Nineteen million eight hundred eighty-eight thousand eight hundred. (19,888,800)
(v) Indian System: 60,53,100 – Sixty lakh fifty-three thousand one hundred International System: 6,053,100 – Six million fifty-three thousand one hundred.

Question 7.
Anbu asks Arivu Selvi to guess a five-digit odd number. He gives the following hints.
The digit in the 1000s place is less than 5
The digit in the 100s place is greater than 6
The digit in the 10s place is 8.
What is Arivu Selvi’s answer? Does she give more than one answer?
Solution:
There are more than one answers.
One of them is 54781
Some of the other numbers maybe 64783, 74785, 84787 and so on.

Question 8.
A Music concert is taking place in a stadium. A total of 7,689 chairs are to be put in rows of 90.
(i) How many rows will there be?
(ii) Will there be any chairs left over?
Solution:
Total no of chairs to be put = 7,689
Chairs in each row = 90
No of rows = 7689 ÷ 90
Hence, 84 + 1 = 85 rows are required to fill 7650 chairs
Chairs left over = 79 (If the no of rows = 84)

Question 9.
Round off the seven-digit number 29,75,842 to the nearest lakhs and ten lakhs. Are they the same?
Solution:
Yes. Both are same (30,00,000)

Question 10.
Find the 5 or 6 or 7 digit numbers from a newspaper or a magazine to get a rounded number to the nearest ten thousand.
Solution:
(i) 14276 \(\simeq\) 10000
(ii) 1,86945 \(\simeq\) 1,90000

Students can download Maths Chapter 1 Numbers Ex 1.5 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.5

Question 1.
Fill in the blanks.
(i) The difference between the smallest natural number and the smallest whole number is ……….
(ii) 17 × …….. = 34 × 17
(iii) When ……… is added to a number, it remains the same.
(iv) Division by ………. is not defined.
(v) Multiplication by ……… leaves a number unchanged.
Solution:
(i) 1
(ii) 34
(iii) Zero
(iv) 0
(v) 1

Question 2.
Say True or False.

1. 0 is the identity for multiplication of whole numbers.
2. The Sum of two whole numbers is always less than their product.
3. Both addition and multiplication are associative for whole numbers.
4. Both addition and multiplication are commutative for whole numbers.
5. Multiplication is distributive over addition for whole numbers.

Solution:

1. False
2. False
3. True
4. True
5. True

Question 3.
Name the property being illustrated in each of the cases given below.

1. 75 + 34 = 34 + 75
2. (12 × 4) × 8 = 12 × (4 × 8)
3. 50 + 0 = 50
4. 50 × 1 = 50
5. 50 × 42 = 50 × 40 + 50 × 2

Solution:

1. Commutativity for addition
2. Associativity for multiplication
3. Zero is the additive identity
4. One is the multiplicative identity
5. Distributivity of multiplication over addition

Question 4.
Use the properties of whole numbers and simplify.
(i) 50 × 102
(ii) 500 × 689 – 500 × 89
(iii) 4 × 132 × 25
(iv) 196 + 34 + 104
Solution:
(i) 50 × 102
= 50 × (100 + 2)
= (50 × 100) + (50 × 2)
= 5000 + 100 = 5100

(ii) 500 × 689 – 500 × 89
= 500 × (689 – 89)
= 344500 – 44500
= 300000
= 500 × (689 – 89)
= 500 × 600
= 3,00000

(iii) (4 × 132) × 25
= 4 × (132 × 25)
= (4 × 132) × 25
= 528 × 25
= 13200
= 4 × (132 × 25)
= 4 × 3300
= 13200

(iv) (196 + 34) + 104 = 196 + (34 + 104)
(196 + 34) + 104 = 230 + 104 = 334
196 + (34 + 104) = 196 + 138 = 334

Objective Type Questions

Question 5.
(53 + 49) × 0 is
(a) 102
(b) 0
(c) 1
(d) 53 + 49 × 0
Solution:
(b) 0

Question 6.
\(\frac{59}{1}\) is
(a) 1
(b) 0
(c) \(\frac{1}{59}\)
(d) 59
Solution:
(d) 59

Question 7.
The product of a non-zero whole number and its successor is always
(a) an even number
(b) an odd number
(c) zero
(d) none of these
Solution:
(a) an even number

Question 8.
The whole number that does not have a predecessor is
(a) 10
(b) 0
(c) 1
(d) none of these
Solution:
(b) 0

Question 9.
Which of the following expressions is not zero?
(a) 0 × 0
(b) 0 + 0
(c) \(\frac{2}{0}\)
(d) \(\frac{0}{2}\)
Solution:
(c) \(\frac{2}{0}\)
Dividing by 0 is not defined.

Question 10.
Which of the following is not true?
(a) (4237 + 5498) + 3439 = 4237 + (5498 + 3439)
(b) (4237 × 5498) × 3439 = 4237 × (5498 × 3439)
(c) 4237 + 5498 × 3439 = (4237 + 5498) × 3439
(d) 4237 × (5498 + 3439) = (4237 × 5498) + (4237 × 3439)
Solution:
(c) 4237 + 5498 × 3439 = (4237 + 5498) × 3439