Class 6

Students can Download Chapter 11 Algebra Ex 11.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 1.
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
Solution:
a) 17 = x + 7
An equation with variable x

b) (t – 7) > 5
An inequality

c) \(\frac{4}{2}\) = 2
No, it is a numerical equation

d) (7 × 3) – 19 = 8
No, it is a numerical equation

e) 5 × 4 – 8 = 2x
An equation with variable x

f) x – 2 = 0
An equation with variable x

g) 2m < 30
An inequality

h) 2n + 1 = 11
An equation with variable n

i) 7 = (11 × 5) – (12 × 4)
No, it is a numerical equation

j) 7 = (11 × 2 ) + p
An equation with variable p

k) 20 = 5y
An equation with variable y

l) \(\frac{3 p}{2}\) < 5 An inequality m) z + 12 > 24
An inequality

n) 20 – (10 – 5) = 3 × 5
No, it is a numerical equation

o) 7 – x = 5
An equation with variable x.

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

Solution:
a) 10y = 80
y = 10 is net a solution to the given equation because for y = 10
10y = 10 × 10 = 100, and net 80

b) 10y = 80
y = 8 is a solution to the given equation because for y = 8
10y = 10 × 8 = 80 and hence, the equation is satisfied

c) 10y = 80
y = 5 is net a solution to the given equation because for y = 5
10y = 10 × 5 = 50, and net 80.

d) 4l = 20
l = 20 is net a solution to the given equation because for l = 20
4l = 4 × 20 = 80, and net 20

e) 4l = 80
l = 80 is net a solution to the given equation because for l = 80
4l = 4 × 80 = 320, and net 20

f) 4l = 20
4l = 4 × 5 = 20 and hence, the equation because for l = 5
4l = 4 × 5 = 20 and hence, the equation is satisfied

g) b + 5 = 9
b = 5 is net a solution to the given equation because for b = 5

h) b + 5 = p
b = 9 is net a solution to the given equation because for b = 9
b + 5 = 9 + 5 = 14, and net 9

i) b + 5 = 9
b + 5 = 4 + 5 = 9 and hence, the equation is satisfied

j) h – 8 = 5
h = 13 is a solution to the given equation because for h = B
h – 8 = 13 – 8 = 5 and hence, the equation is satisfied

k) h – 8 = 5
h = 8 is net a solution to the given equation because for h = 8
h = 8 = -8 = 0, and net 5

l) h – 8 = 5
h = 0 is net a solution to the given equation because for h = 0,
h – 8 = 0 – 8 = -8, and net 5

m) p + 3 = 1
p + 3 = 3 + 3 = 6, and net 1

n) p + 3 = 1
p + 3 = 1 + 3 = 4 , and net 1

o) p + 3 = 1
p + 3 = 0 + 3 = 3, and net 1

p) p + 3 = 1
p + 3 = -1 + 3 = 2, and net 1
p + 3 = -2 + 3 = 1 and hence, the equation is satisfied.

Question 3.
Pick out the solution from the values given in the bracket next to each equation. show that the other values do not satisfy the equation.
Solution:
a) 5m = 60 (10, 5, 12, 15)
m = 12 a solution to the given equation because form = 12
5m = 5 × 12 = 60 and hence, the equation is satisfied
m = 10 is not a solution to the given equation because for m = 10.
5m = 5 × 10 = 50, and net 60
m = 5 is net a solution to the given equation because for m = 5,
5m = 5 × 5 = 25, and net 60
m = 15 is net a solution to the given equation because for m = 15
5m = 5 × 15 = 75, and net 6.

b) n + 12 = 20 (12, 8, 20, 0)
n = 8 is a solution to the given equation because for n = 8
n + 12 = 8 + 12 = 20 and hence, the equation is satisfied
n +12 = 12 + 12 = 24, and net 20
n = 20 is net a solution to the given equation because for n = 20
n + 12 = 20 + 12 = 32, and net 20

c) p – 5 = 5 (0, 10, 5 – 5)
p = 10 is a solution to the given equation because for p = 10, p – 5 = 10 – 5 = 5 and hence, the equation is satisfied.
p = 0 is net a solution to the given equation because for p = 0 .
p – 5 = 5 – 5 = 0, and net 5
p = -5 is net s solution to the given equation because for p = -5
p = 5 = -5 – 5 = -10 and net 5

d) \(\frac{q}{2}\) = 7 (7, 2, 10, 14)
q = 14 is a solution to the given equation because for q = 7 \(\frac{q}{2}\) = \(\frac{7}{2}\) and not 7
\(\frac{q}{2}\) = \(\frac{7}{2}\) and net 7
q = 2 is not a solution to the given equation because for q = 7
\(\frac{q}{2}\) = \(\frac{7}{2}\) and net 7
q = 2 is net a solution to the given equation because for q = 2
\(\frac{q}{2}\) = \(\frac{2}{2}\) = 1, and net 7
q = 10 is net a solution to the given equation because for qn = 10,
\(\frac{q}{2}\) = \(\frac{10}{2}\) = 5, and net 7

e) r – 4 = 0 (4, -4, 8, 0)
r = 4 is a solution to the given equation because for r = 4
r = 4 = 4 – 4 = 0 and hence, the equation is satisfied
r = -4 is net a solution to-the given equation because for r = -4
r – 4 = -4 – 4 = -8 and net 0
r = 8 is net a solution to the given equation because for r = 8
r – 4 = 8 – 4 = 4 and net 0
r = 0 is net a solution to the given equation because for r = 0
r – 4 = 0 – 4 = -4 and net 0 .

f) x + 4 = 2 (-2, 9, 2, 4)
x = -2 is a solution to the given equation because for x – 2
x + 4 = -2 + 4 = 2 and hence, the equation is satisfied
x = 0 is net a solution to the given equation because for x = 0
x + 4 = 0 + 4 = 4 and net 2
x = 2 is net a solution to the given equation because for x = 2
x + 4 = 2 + 4 = 6 and net 2
x = 4 is net a solution to the given equation because for x = 4
x + 4 = 4 + 4 = 8, and net 2.

Question 4.
a) Complete the table and by inspection of the table find the solution to the equation m +10 = 6

Solution:
By inspection, we can find that m = 6 is the solution of the above equation as for m = 6, m + 10 = 6 + 10 = 16

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

Solution:
For 5t, the table can be constructed as follows


By inspection, we can find that t = 7 is the solution of the above equation as for t = 7, 5t = 5 × 7 = 35

c) Complete the table and find the solution of the equation x/3 = 4 using the table.

Solution:
For, the table can be constructed as follows


By inscription, we can find that z = 12 is the solution of the above equation of the above equation as for z = 12, \(\frac{z}{3}\) = 4

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

Solution:

By inspection, we can find that m = 10 is the solution of the above equation as for m = 10, m – 7 = 10 – 7 = 3

Question 5.
Solve the following riddles, you may yourself construct such riddles. Who am I?

i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!
Solution:
There are 4 comers in a square
Thrice the number of comers in the square will be 3 × 4 = 12
When this result result i,e 12, is added to the number if Comes to be 34. therefore the number will be the difference of 34 and 12 i.e 34 – 12 = 22

ii) For each day of the week
Make an upcount from me
If you make mo mistake
you will get twenty three!
Solution:
23 was the result when the old number was up Counted on Saturday
22 was the result when the old number was up counted on Saturday
20 was the result ‘when the old number was up counted on Thursday
19 was the result when the old number was up counted on tuesday
18 was the result when the old number was up counted on Tuesday
17 was the result when the old number was up counted on Tuesday

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

Solution:
In a cricket taken at the 11 players Hence the number is such that when 6 is subtracted from it, the result is 11, Therefore, the number is 11 + 6 = 17

iv) Tell me who I am
I shall give a pretty clue!
you will get me back
If you take me out of twenty two!
Solution:
The number is such that when it is subtracted from 22 the result is again the number it self the number is 11, when again gives 11, when it is subtracted from 22.

Students can Download Chapter 14 Practical Geometry Ex 14.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 14 Practical Geometry Ex 14.4

Question 1.
Draw any line segment \(\overline{\mathbf{A B}}\). Mark any point M on it. Through M, draw a perpendicular to \(\overline{\mathbf{A B}}\). (use ruler and compasses)
Solution:
(1) Draw the given line segment \(\overline{\mathbf{A B}}\) and mark an point M on it.
(2) With M as centre and a convenient radius, construct an arc intersecting the line segment \(\overline{\mathbf{A B}}\) at two points C and D.

(3) With C and D as centres and a radius greater than CM, construct two arcs. Let these be intersecting each other at E.
(4) Join EM. \(\overline{\mathrm{EM}}\) is perpendicular to AB

Question 2.
Draw a line segment \(\overline{\mathrm{AB}}\) Take any point R not on it. Through R, draw a perpendicular to \(\overline{\mathrm{AB}}\). (use ruler and set-square)
Solution:
(1) Take the given line segment \(\overline{\mathrm{PQ}}\) and mark any point R outside \(\overline{\mathrm{PQ}}\).
(2) Place a set square on \(\overline{\mathrm{PQ}}\) such that one arm of its right angle aligns along \(\overline{\mathrm{PQ}}\).

(3) Place the ruler along the edge opposite to the right P angle of the set square.
(4) Hold the ruler fixed. Slide the set square along the ruler till the point R touches the other arm of the set square.
(5) Draw a line along this edge of the set square which will be passing through R. It is the required line, which is perpendicular to \(\overline{\mathrm{PQ}}\).

Question 3.
Draw a line l and a point X on it. Through X, draw a line segment \(\overline{\mathrm{XY}}\) perpendicular to l.
Now draw a perpendicular to \(\overline{\mathrm{XY}}\) at Y. (use ruler and compasses)
Solution :
(1) Draw a line l and mark a point X on it.
(2) Taking X as centre and with a convenient radius, draw an arc intersecting line 1 at two points A and B.
(3) With A and B as centres and a radius more than AX, construct two arcs intersecting each other at Y.

(4) Join XY. \(\overline{\mathrm{XY}}\) is perpendicular to l.
Similarly, a perpendicular to \(\overline{\mathrm{XY}}\) at point Y can be drawn. The line \(\overline{\mathrm{ZY}}\) is perpendicular t0 \(\overline{\mathrm{XY}}\) at Y.

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Students can Download Chapter 14 Practical Geometry Ex 14.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 14 Practical Geometry Ex 14.3

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Question 1.
Draw any line segment \(\overline{\mathbf{P Q}}\). Without measuring \(\overline{\mathbf{P Q}}\), construct a copy of \(\overline{\mathbf{P Q}}\).
Solution:

The following steps will be followed to draw the given line segment \(\overline{\mathbf{P Q}}\) and to construct a copy of \(\overline{\mathbf{P Q}}\).
(1) Let \(\overline{\mathbf{P Q}}\) be the given line segment.
(2) Adjust the compasses up to the length of \(\overline{\mathbf{P Q}}\).
(3) Draw any line land mark a point A on it.
(4) Put the pointer on point A, and without changing the setting of compasses, draw an arc to cut the line segment at point B.
\(\overline{\mathbf{A B}}\) is the required line segment.

Question 2.
Given some line segment \(\overline{\mathbf{A B}}\), whose length you do not know, construct \(\overline{\mathbf{P Q}}\) such that the length of \(\overline{\mathbf{P Q}}\) is twice that of \(\overline{\mathbf{A B}}\).
Solution:

The following steps will be followed to construct a line segment \(\overline{\mathbf{P Q}}\) such that the length of \(\overline{\mathbf{P Q}}\) is twice that of \(\overline{\mathbf{A B}}\).
(1) Let \(\overline{\mathbf{A B}}\) be the given line segment.
(2) Adjust the compasses up to the length of \(\overline{\mathbf{A B}}\).
(3) Draw any line 1 and mark a point P on it.
(4) Put the pointer on P and without changing the setting of compasses, draw an arc to cut the line segment at point X.
(5) Now, put the pointer on point X and again draw an arc with the same radius as before, to cut the line 1 at point Q.
\(\overline{\mathbf{P Q}}\) is the required line segment.

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Students can Download Chapter 14 Practical Geometry Ex 14.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 14 Practical Geometry Ex 14.2

Question 1.
Draw a line segment of length 7.3 cm using a ruler.
Solution:
A line segment of length 7.3 cm can be drawn using a ruler as follows.
(1) Mark a point A on the sheet.
(2) Put 0 mark of ruler at point A.
(3) Mark a point B on the sheet at 7.3 cm on ruler.
(4) Join A and B.
\(\overline{\mathrm{AB}}\) is the required line segment.

Question 2.
Construct a line segment of length 5.6 cm using ruler and compasses.
Solution:
A line segment of length 5.6 cm can be drawn using a ruler and compasses as follows.
(1) Draw a line 1 and mark a point A on this line.

(2) Place the compasses on the zero mark of the ruler. Open it to place the pencil up to the 5.6 cm mark.
(3) Place the pointer of compasses on point A and draw an arc to cut 1 at B. AB is the line segment of 5.6 cm length.

Question 3.
Construct \(\overline{\mathbf{A B}}\) of length 7.8 cm. From this, cut off \(\overline{\mathbf{A C}}\) of length 4.7 cm. Measure BC.
Solution:
(1) Draw a line land mark a point A on it.
(2) By adjusting the compasses up to 7.8 cm, draw an arc to cut 1 on B, while putting the
pointer of compasses on point A.
\(\overline{\mathbf{A B}}\) is the line segment of 7.8 cm.
(3) By adjusting the compasses up to 4.7 cm. draw an arc to cut 1 on C, while putting the pointer of compasses on point A. \(\overline{\mathbf{A C}}\) is the line segment of4.7 cm.
(4) Now, put the ruler along with this line such that 0 mark of the ruler will match with point C. On reading the position of point B, it comes to 3.1 cm, \(\overline{\mathbf{B C}}\) is 3.1 cm

Question 4.
Given \(\overline{\mathbf{A B}}\) of length 3.9 cm, construct \(\overline{\mathbf{P Q}}\) such that the length of \(\overline{\mathbf{P Q}}\) is twice that of \(\overline{\mathbf{A B}}\). Verify by measurement.

(Hint: Construct \(\overline{\mathbf{P X}}\) such that length of \(\overline{\mathbf{P X}}\) = length of \(\overline{\mathbf{A B}}\) ; then cut off \(\overline{\mathbf{X Q}}\) such that \(\overline{\mathbf{X Q}}\) also has the length of \(\overline{\mathbf{A B}}\).)
Solution:
A line segment \(\overline{\mathbf{P Q}}\) can be drawn such that the length \(\overline{\mathbf{P Q}}\) of is twice that of as follows.
(1) Draw a line 1 and mark a point P on it and let AB be the given line segment of 3.9 cm.

(2) By adjusting the compasses up to the length of AB. draw an arc to cut the line at X, while taking the pointer of compasses at point P.

(3) Again put the pointer on point X and draw an arc to cut line 1 again at Q.

\(\overline{\mathbf{P Q}}\) is the required line segment. By ruler, the length of \(\overline{\mathbf{P Q}}\) can be measured which comes to 7.8 cm.

Question 5.
Given \(\overline{\mathbf{A B}}\) of length 7.3 cm and \(\overline{\mathbf{C D}}\) of length 3.4 cm, construct a line segment \(\overline{\mathbf{X Y}}\) such that the length of \(\overline{\mathbf{X Y}}\) is equal to the difference between the lengths of
\(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{C D}}\). Verify by measurement.
Solution:
(1) Given that, \(\overline{\mathbf{A B}}\) = 7.3 cm and \(\overline{\mathbf{C D}}\) = 3.4 cm

(2) Adjust the compasses up to the length of CD and put the pointer of the compasses at A. Draw an arc to cut AB at P.

(3) Adjust the compasses up to the length of PB. Now draw a line l and mark a point X on it.

(4) Now, putting the pointer of compasses at point X, draw an arc to cut the line at Y.

\(\overline{\mathbf{X Y}}\) is the required line segment.

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Students can Download Chapter 11 Algebra Ex 11.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 11 Algebra Ex 11.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule,
Solution:
a) A pattern of letter T as
From the figure, it can be observed that it will required two matchsticks to make a T, Therefore the pattern is 2n

b) A pattern of letter Z as
From the figure, it can be observed that it will require three matchsticks two make a Z there fore, the pattern is 3n.

c) A pattern of letter U as
From the figure, it can be observed that it will require there matchsticks to make a U. Therefore the pattern 3n

d) A pattern of letter V as
From the figure it can be observed that it will required two matchsticks to make, a V, Therefore, the pattern is 2n

e) A pattern of letter E as
From the figure, it can be observed that it will required five matchsticks to make an E. Therefore, the pattern in 5n.

f) A pattern of letter S as
From the figure, it can be observed that it will require five matchsticks to make a S therefore, the pattern is 5n

g) A pattern of letter A as
From the figure, it can be observed that it will required six matchsticks to make an A, therefore, the pattern is 6n.

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Wgich are these? Why does this happen?
Solution:
It is known that L required only two matchsticks Therefore the pattern for L is 2n Among all the letters given above in question 1, only T and V are the two letters Which require two matchsticks Hence (a) and (b).

Question 3.
Cadets are marching in a parade. Therefore are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution:
Let number of rows be n
Number of cadets in one row = 5
Total number of cadets = number of cadets in a row x numbers of rows= 5

Question 4.
If there are 50 managoes in a box, how will you write he total number of managoes in terms of the number of-boxes?(Use b for the number of boxes).
Solution:
Let the number of boxes b b
Number of manager in a box = 50
Total number of managoes = Numbers of mangoes in a box × numbers of boxes = 50b

Question 5.
The teacher distributes 5 pencils per students can you tell hoe many pencils are needed given the number of students? (use S for the number of students)
Solution:
Let the number of students be S pencils given to each student = 5
Total number of pencils = Number of pencils given to each students × number of students,
= 5s

Question 6.
A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes?(Use t for flying time in minutes).
Solution:
Let the flying time be t minutes
Distance covered in one minute = 1 km
Distance covered n t minutes = Distance covered in one minute × Flying time
= 1 × t = t km

Question 7.
Radha is drawing a dot Rangoli ( a beautiful pattern of lines joining dots with chalk powder, she has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? if there are 10 rows?

Solution:
Number of dots in 1 row = 9
Number of rows = r
Total number of dots in r rows = Number of rows × number of dots in a row
= 9r
Number of dots in 8 rows = 8 × 9 = 72
Number of dots in 10 rows = 10 × 9 = 90

Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. can you write Leele’s age in terms of Radha’s age? Take Radha’s age to be x years.
Solution:
Let Radha’s age be x years
Leele’s age = Radha’s age – 4
= (x – 4) years.

Question 9.
Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain, if the number of laddus mother gave away is l, how many laddus did she make?
Solution:
Number of laddus given away = 1
Number of laddus remaining = 5
Total number of laddus = Number of laddus given away + numbers of laddus remaining
1 + 5

Question 10.
Orange are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller oxes and still 10 oranges remain outside, if the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Solution:
Number of oranges in one small box = x
Number of orange in two small boxes = 2x
Number of orange left = 10.
Number of orange in the large box =
Number of orange in two small boxes + Number of orange left = 2x + 10

Question 11.
a) Look at the following matchstick pattern of squares (Fig 11.6). The squares are
not separate. Two neighbouring squares have a common matchstick. observe the patterns and find the rule that gives the number of matchsticks

in terms of the number of squares( Hint: if you remove the vertical stick at the end. you will get a pattern of Cs.)
b) Fig 11.7 gives a matchstick pattern of triangles. As in Exercise 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.

Solution:
a) If can be observed that in the given matchsticks pattern, the number of matchsticks are 4, 7, 10 and 13. which is 1 more than thrice of the numbers of square in the pattern. Hence, the pattern is 3n +1 Where n is the number of squares.
b) It can be observed that in the given matchsticks pattern, the number of matchstick pattern the number are 3, 5, 7 and 9. which is 1 more than twice of the number of ∆ les in the pattern
Hence, the pattern is 2n + 1 Where n is the number of ∆les.

Students can Download Chapter 12 Ratio and Proportion Ex 12.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.3

Question 1.
If the cost of 7 m of cloth is Rs 294, Find the cost of 5 m of cloth.
Solution:
Cost of 7m cloth = Rs 294
Cost of 1 m cloth = \(\frac{294}{7}\) = Rs 42
Therefore, Cost of 5m cloth = 42 × 5 = Rs 210

Question 2.
EKta earns Rs 1500 in 10 days. How much will she earn in 30 days?
Solution:
EKta earns money in 10 days = Rs 1500

Therefore, Money earned in 30 days = 150 × 30 = Rs 4500.

Question 3.
If it has rained 276 mm in the last 3 days. How many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Solution:
Measure of rain in 3 days = 276 mm
Measure of rain in 1 day = \(\frac{276}{3}\) = 92 mm
Therefore, measure of rain in 7 days = 92 × 7 = 644 mm

Question 4.
cost of 5 kg of Wheat is Rs 30.50,
a) What will be the cost of 8 kg of wheat?
b) What quantity of wheat can be purchased in Rs 61?
Solution:
a) Cost of 5 kg Wheat = Rs 30.50
Cost of 1kg Wheat = \(\frac{30.50}{5}\) = Rs 6.10
Therefore, Cost of 8 kg Wheat = 6.10 × 8 = Rs. 48.80

b) Wheat purchase in Rs 30.50 = 5 kg
Wheat purchase in Rs 1 = \(\frac{5}{30.50}\) kg

Question 5.
The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how degree will the temperature drop in the next ten days?
Solution:
Temperature drop in 30 days = 15°c
Temperature drop in 1 days = \(\frac{15}{30}=\left(\frac{1}{2}\right)^{\circ} \mathrm{C}\)

Thus, these will be a temperature drop of 5°c on the next ten days.

Question 6.
Shaina pays Rs 7500 as rent for 3 months. How much does she has to pay for a Whole year, if the rent per month remains same?
Solution:
Rent for 3 months = Rs 7500
Rent for 1 months = \(\frac{7500}{3}\) = Rs 2500
Therefore, rent for 12 months = 2500 × 12 = 30.000
Thus, she has to pay Rs 30.000 for a whole year.

Question 7.
Cost of 4 dozens bananas is Rs 60. How many bananas can be purchase for Rs 12.50?
Solution:
Numbers of bananas bought in Rs 60 = 4 × 12 = 48
Numbers of bananas bought in Rs 1 = \(\frac{48}{60}\)

Thus, 10 bananas can be purchase for Rs. 12.50

Question 8.
The weight of 72 books is 9 kg. What is the weight of 40 such books?
Solution:
Weight of 72 books = 9 kg


Thus, the weight of 40, Such books is 5 kg

Question 9.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Solution:
Diesel required for 594 km = 108 litres

Thus, 300 litres diesel will be required by the truck to cover a distance of 1650 km

Question 10.
Raju purchases 10 pens for Rs 150 and manish buys 7 pens for Rs 84. Can you say Who got the pens cheaper?
Solution:
Raju purchase 10 pens for Rs 150

Manish purchase 7 pens for Rs 84

Therefore, manish got the pens cheaper

Question 11.
Anish made 42 runs in 6 overs and an up made 63 runs in 7 overs. Who made more runs per over?
Solution:
Runs made by Anish in 6 overs = 42

Runs made by Anup in 7 overs = 63

Clearly Anup made more runs per over.

Students can Download Chapter 14 Practical Geometry Ex 14.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

Question 1.
Draw ∠POQ of measure 75° and find its line of symmetry.
Solution:

The below given steps will be followed to construct an angle of 75° and its line of symmetry
(1) Draw a line l and mark two points O and Q on it, as shown in the figure. Draw an arc of convenient radius, while taking point O as centre. Let it intersect line l at R
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join OU. Let it intersect the arc at V. Now, taking S and V as centres, draw arcs with radius more than \(\frac{1}{2}\)SV. Let those intersect each other at P. Join OP, which is the ray making 75° with the line l.
(6) Let this ray be intersecting our major arc at point W. Now, taking R and W as centres draw arcs with radius more than \(\frac{1}{2}\)RW in the interior of angle of 75°. Let these be intersecting each other at X. Join OX.
OX is the line of symmetry for ∠POQ = 75°.

Question 2.
Draw an angle of measure 147° and construct its bisector.
Solution:
The below given steps will be followed to construct an angle of 147° measure and its bisector.
(1) Draw a line l and mark a point O on it. Place the centre of the protractor at point O and the zero edge along line l.
(2) Mark a point A at 147°. Join O A. OA is the required ray making 147° with line l.
(3) Draw an arc of convenient radius, while taking point O as centre. Let it intersect both rays of angle 147° at point A and B.
(4) Taking A and B as centres, draw arcs of radius more than \(\frac{1}{2}\)AB in the interior of angle of 147°. Let those intersect each other at C. Join OC.
OC is the required bisector of 147° angle.

Question 3.
Draw a right angle and construct its bisector.
Solution:

The below given steps will be followed to construct a right angle and its bisector.
(1) Draw a line l and mark a point P on it. Draw an arc of convenient radius, while taking point P as centre. Let it intersect line l at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centres, draw arcs of same radius to intersect each other at U.
(5) Join PU. PU is the required ray making 90° with line l. Let it intersect the major arc at point V.
(6) Now, taking R and V as centres, draw arcs with radius more than \(\frac{1}{2}\)RV to intersect each other at W. Join PW.
PW is the required bisector of this right angle.

Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Solution:

The below given steps will be followed to construct an angle of 153° measure and its bisector.
(1) Draw a line l and mark a point O on it. Place the centre of the protractor at point O and the zero edge along line 1.
(2) Mark a point A at 153°. Join OA. O A is the required ray making 153° with line l.
(3) Draw an arc of convenient radius, while taking point O as centre. Let it intersect both rays of angle 153° at point A and B.
(4) Taking A and B as centres, draw arcs of radius more than \(\frac{1}{2}\)AB in the interior of angle of 153°. Let those intersect each other at G. Join OC.
(5) Let OC intersect the major arc at point D. Now, with radius more than \(\frac{1}{2}\)AD, draw arcs while taking A and D as centres, and D and B as centres. Let these be intersecting each other at point E and F respectively. Join OE, OF.
OF, OC, OE are the rays dividing 153° angle in 4 equal parts.

Question 5.
Construct with ruler and compasses, angles of following measures :
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°
Solution:
(a) 60°

The below given steps will be followed to construct an angle of 60°.
(1) Draw a line l and mark a point P on it. Now, taking P as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
(3) Join PR Which is the required ray making 60° with line l.

(b) 30°

The below given steps will be followed to construct an angle of 30°.
(1) Draw a line l and mark a point P on it. Now taking P as centre and with convenient radius, draw an arc of a circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
(3) Now, taking Q and R as centre and with radius more than RQ, draw arcs to intersect each other at S. Join PS which is the required ray making 30° with line l.

(c) 90°

The below given steps will be followed to construct an angle of 90°.
(1) Draw a line l and mark a point P on it. Now taking P as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
(3) Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S (see figure).
(4) Taking R and S as centre, draw an arc of same radius to intersect each other at T.
(5) Join PT, which is the required ray making 900 with line l.

(d) 120°

The below given steps will be followed to construct an angle of 120°.
(1) Draw a line 1 and mark a point P on it. Now taking P as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
(3) Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S (see figure).
(4) Join PS, which is the required ray making 120° with line l.

(e) 45°

The below given steps will be followed to construct an angle of 45°.
(1) Draw a line l and mark a point P on it. Now taking P as centre and with a convenient draw an arc of a circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
(3) Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S (see figure).
(4) Taking R and S as centres, draw arcs of same radius to intersect each other at T.
(5) Join PT. Let it intersect the major arc at point U.
(6) Taking Q andU as centres, draw arcs with radius more than QU to intersect each other at V. Join PV.
PV is the required ray making 45° with the given line l.

(f) 135°

The below given steps will be followed to construct an angle of 135°.
(1) Draw a line l and mark a point P on it. Now taking P as centre and with a convenient radius, draw a semi circle which intersects line l at Q and R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously .drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw arcs of same radius to intersect each other at U.
(5) Join PU. Let it intersect the arc at V. Now taking Q and V as centres and with radius more than QV, draw arcs to intersect each other at W.
(6) Join PW which is the required ray making 135° with line l.

Question 6.
Draw an angle of measure 45° and bisect it.
Solution:

The below given steps will be followed to construct.an angle of 45° and its bisector.
(1) ∠POQ of 45° measure can be formed on a line l by using the protractor.
(2) Draw an arc of a convenient radius, while taking point O as centre. Let it intersect both rays of angle 45° at point A and B.
(3) Taking A and B as centres, draw arcs of radius more than \(\frac{1}{2}\)AB in the interior of angle of 45°. Let those intersect each other at C. Join OC.
OC is the required bisector of 45° angle.

Question 7.
Draw an angle of measure 135° and bisect it.
Solution:

The below given steps will be followed to construct an angle of 135° and its bisector.
(1) ∠POQ of 135° measure can be formed on a line l by using the protractor.
(2) Draw an arc of a convenient radius, while taking point O as centre. Let it intersect both rays of angle 135° at point A and B.
(3) Taking A and B as centres, draw arcs of radius more than \(\frac{1}{2}\) AB in the interior of angle of 135°. Let those intersect each other at C. Join OC.
OC is the required bisector of 135° angle.

Question 8.
Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Solution:
The below given steps will be followed to construct an angle of 70° measure and its copy.
(1) Draw a line l and mark a point O on it. Place the centre of the protractor at point O and the zero edge along line l.
(2) Mark a point A at 70°. Join OA. OA is the ray making 70° with line l. Draw an arc of convenient radius in the interior of 70° angle, while taking point O as centre. Let it intersect both rays of angle 70° at point B and C.
(3) Draw the line m and mark a point P on it. With the same radius as used before, again draw an arc while taking point P as centre. Let it cut the line m at point D.
(4) Now, adjust the compasses up to the length of BC. With this radius, draw an arc when taking D as centre, which will intersect the previously drawn arc at point E.
(5) Join PE. PE is the required ray which makes the same angle (i.e. 70°) with line m

Question 9.
Draw an angle of 40°. Copy its supplementary angle.
Solution:
The below given steps will be followed to construct an angle of 40° measure and the copy of its supplementary angle.
(1) Draw a line segment \(\overline{\mathrm{PQ}}\) and mark a point O on it. Place the centre of the protractor at point O and the zero edge along line segment \(\overline{\mathrm{PQ}}\).
(2) Mark a point A at 40°. Join OA. OA is the required ray making 40° with \(\overline{\mathrm{PQ}}\) . ∠POA is the supplementary angle of 40°.
(3) Draw an arc of convenient radius in the interior of ∠POA, while taking point O as centre. Let it intersect both rays of ∠POA at point B and C.

(4) Draw a line m and mark a point S on it. With the same radius as used before, again draw an arc while taking point S as centre. Let it cut the line m at point T.
(5) Now, adjust the compasses up to the length of BC. With this radius, draw an arc while taking T as centre, which will intersect the previously drawn arc at point R.
(6) Join RS. RS is the required ray which makes the same angle with line m, as the supplementary of 40° is 140°.

Students can Download Chapter 9 Data Handling Ex 9.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 9 Data Handling Ex 9.4

Question 1.
A survey of 120 school students was done to find which activity they prefer to do in their free time.

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students.
Which activity is preferred by most of the students other than playing?
Solution:
By taking a scale of 1 unit length = 5 students a bar graph of the above given data can be drawn as follows.

The activity that is preferred by most of the students other than playing is reading story books.

Question 2.
The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

Draw a bar graph to represent the above information choosing the scale of your choice.
Solution:
By taking a scale of 1 unit length = 10 books, a bar graph of the above given data can be drawn as follows

Question 3.
Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.

a) In which year was the maximum number of by cycles manufactured?
b) In which year was the minimum number of bicycles manufactured? Solution :
By taking a scale of 1 unit length = 100
Bicycles, a bar group of the above given data can be drawn as follows.

a) The number of bicycles manufactured in 2002 was the maximum i.e. 1200
b) The number of bicycles manufactured in 1999 was the minimum .i.e. 600.

Question 4.
Number of persons in various age groups in a town is given in the following table.

Draw a bar graph to represent the above information and answer the following questions. (take 1 unit length = 20 thousands)
a) Which two age groups have same population?
b) All persons n the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Solution :
By taking a scale of 1 unit length = 20 thousands, a bar group of the above given data can be drawn as follows.

a) 30 – 44 and 45 are the two age groups which have the same population
b) It can be sinerred that senior citizens are the people who are either from age group 60 – 74 or from age group 45 and above.
Hence number of senior citizens = 80,000 + 40,000
= 1 lakhs 20 Thousand.