Class 6

Students can Download Chapter 11 Algebra Ex 11.2 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 11 Algebra Ex 11.2

Question 1.
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l
Solution:
Side of equilateral ∆ le = l
Perimeter = l + l + l + 3l

Question 2.
The side of a regular hexagon(Fig 11.10) is denoted by l. Express the perimeter of the hexagon using l.
(Hint: A regular hexagon has all its six sides equal in length)

Sides of regular hexagon = l
Perimeter = 6l

Question 3.
A cube is a three – dimensional figure as shown in Fig 11.11. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.
Solution:
length of edge = l
Number of edges = 12
Total length of the edges = Number of edges × length of one edge = 12l

Question 4.
The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure (Fig 11.12 )AB is a diameter of the circle ; C is its centre). Express the diameter of the circle (d) in terms of its radius(r).
Solution:

Diameter = AB = AC + CB = r + r = 2r
d = 2r

Question 5.
To find sum of three numbers 14, 27 and 13, we can have two ways:
a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
b) We may add 27 and 13 to get 40 and then add 14 to get 40 and then add 14 to get the sum 54. Thus, (14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers, his property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on whole numbers, in a general way, by using variables a, b and c.
Solution:
For any three whole numbers a, b, and c
(a + b) + c = a + (b + c)

Students can Download Chapter 8 Decimals Ex 8.2 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 8 Decimals Ex 8.2

Question 1.
Complete the table with the help of these boxes and use decimals to write the number.

Solution:

Question 2.
Write the numbers given in the following place value table in decimal form

Solution:

Question 3.
Write the following decimals in the place value table.
a) 0.29
b) 2.08
c) 19.60
d) 148.32
e) 200.812.
Solution:

Question 4.
Write each of the following as decimals.

Solution:

Question 5.
Write each of the following decimals in words.
a) 0.03
b) 1.20
c) 108.56
d) 10.07
e) 0.032
f) 5.008
Solution:
a) 0.03 = Zero point zero three
b) 1.20 = One point two zero
c) 108.56 = One hundred eight point five six
d) 10.07 = Ten point zero seven
e) 0.032 = Zero point zero three two
f) 5.008 = Five point zero zero eight

Question 6.
Between Which two numbers in tenths place on the number line does each of the given number lie?
a) 0.06
b) 0.45
c) 0.19
d) 0.66
e) 0.92
f) 0.57
Solution:
a) 0.06 = 0 and 0.6
b)0.45 = 0.4 and 0.5
c) 0.19 = 0.1 and 0.2
d) 0.66 = 0.6 and 0.7
e) 0.92 = 0.9 and 1.0
f) 0.57 = 0.5 and 0.6

Question 7.
Write as fractions in lowest terms.
a) 0.06
b) 0.05
c) 0.75
d) 0.18
e) 0.25
f) 0.125
g) 0.066
Solution:

Students can Download Chapter 14 Practical Geometry Ex 14.5 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 14 Practical Geometry Ex 14.5

Question 1.
Draw \(\overline{\mathbf{A B}}\) of length 7.3 cm and find its axis of symmetry.
Solution:
The below given steps will be followed to construct AB of length 7.3 cm and to find its axis of symmetry.
(1) Draw a line segment \(\overline{\mathbf{A B}}\) of 7.3 cm.
(2) Taking A as centre, draw a circle by using compasses. The radius of circle should be more than half the length of \(\overline{\mathbf{A B}}\).
(3) With the same radius as before, draw another circle using compasses while taking point B as centre. Let it cut the previous circle at C and D.
(4) Join \(\overline{\mathbf{C D}}\). \(\overline{\mathbf{C D}}\) is the axis of symmetry.

Question 2.
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Solution:

The below given steps will be followed to construct a line segment of length 9.5 cm and its perpendicular bisector.
(1) Draw a line segment \(\overline{\mathrm{PQ}}\) of 9.5 cm.
(2) Taking P as centre, draw a circle by using compasses. The radius of circle should be more than half the length of \(\overline{\mathrm{PQ}}\).
(3) With the same radius as before, draw another circle using compasses while taking point Q as centre. Let it cut the previous circle at R and S.
(4) Join RS. \(\overline{\mathrm{RS}}\) is the axis of symmetry i.e., the perpendicular bisector of line PQ.

Question 3.
Draw the perpendicular bisector of \(\overline{\mathrm{XY}}\) whose length is 10.3 cm.
(a) Take any point P on the bisector drawn. Examine whether PX = PY.
(b) If M is the mid point of \(\overline{\mathrm{XY}}\), what can you say about the lengths MX and XY?
Solution:

(1) Draw a line segment \(\overline{\mathrm{XY}}\) of 10.3 cm.
(2) Taking point X as centre, draw.a circle by using compasses. The radius of circle should be more than half the length of \(\overline{\mathrm{XY}}\).
(3) With the same radius as before, draw another circle using compasses while taking point Y as centre. Let it cut the previous circle at A and B.
(4) Join \(\overline{\mathrm{AB}}\) \(\overline{\mathrm{AB}}\) is the axis of symmetry.
(a) Take any point P on \(\overline{\mathrm{AB}}\). We will find that the measures of the lengths of PX and PY are same.
It is because \(\overline{\mathrm{AB}}\) the axis of symmetry. Hence, any point lying on \(\overline{\mathrm{AB}}\) will be at the same distance from both the ends of \(\overline{\mathrm{XY}}\).

(b) M is the mid-point of \(\overline{\mathrm{XY}}\). Perpendicular bisector \(\overline{\mathrm{AB}}\) will be passing through point M. Hence, length of \(\overline{\mathrm{XY}}\) is just double of \(\overline{\mathrm{MX}}\) Or, 2MX = XY

Question 4.
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Solution:

(1) Draw a line segment \(\overline{\mathrm{XY}}\) of 12.8 cm.
(2) Draw a circle, while taking point X as centre and radius more than half of XY.
(3) With same radius and taking centre as Y, again draw arcs to cut the circle at Aand B. Join AB which intersects \(\overline{\mathrm{XY}}\) at M.
(4) Taking X and Y as centres, draw two circles with radius more than half of \(\overline{\mathrm{XM}}\).
(5) With same radius and taking M as centre, draw arcs to intersect these circles at P, Q and R, S.
(6) Join PQ and RS. These are intersecting \(\overline{\mathrm{XY}}\) at T and U.
(7) Now, \(\overline{\mathrm{XT}}\) = \(\overline{\mathrm{TM}}\) = \(\overline{\mathrm{MU}}\) = \(\overline{\mathrm{UY}}\). These are 4 equal parts of \(\overline{\mathrm{XY}}\).
By measuring these line segments with the help of ruler, we will find that each is of 3.2 cm.

Question 5.
With \(\overline{\mathrm{PQ}}\) of length 6.1 cm as diameter, draw a circle.
Solution:

(1) Draw a line segment of 6.1 cm.
(2) Taking point P as centre and radius more than half of \(\overline{\mathrm{PQ}}\), draw a circle.
(3) With same radius,and taking Q as centre, draw arcs to intersect this circle at points R and S.
(4) Join RS which intersects \(\overline{\mathrm{PQ}}\) at T.
(5) Taking T as centre and with radius TP, draw a circle which will also pass through Q. It is the required circle.

Question 6.
Draw a circle with centre C and radius 3.4 cm. Draw any chord \(\overline{\mathrm{AB}}\). Construct the perpendicular bisector of \(\overline{\mathrm{AB}}\) and examine if it passes through C.
Solution:

(1) Mark any point Con the sheet.
(2) By adjusting the compasses up to 3.4 cm and by putting the pointer of the compasses at point C, turn the compasses slowly to draw the .circle. It is the required circle of 3.4 cm radius.
(3) Now, mark any chord \(\overline{\mathrm{AB}}\) in the circle.
(4) Taking A and B as centres, draw arcs on both sides of \(\overline{\mathrm{AB}}\). Let these intersect each other at D and E.
(5) Join DE, which is the perpendicular bisector of AB. When \(\overline{\mathrm{DE}}\) is extended, it will pass through point C.

Question 7.
Draw a circle with centre C and radius 3.4 cm. Draw any chord \(\overline{\mathrm{AB}}\). Construct the perpendicular bisector of \(\overline{\mathrm{AB}}\) and examine if it passes through C, if \(\overline{\mathrm{AB}}\) happens to be a diameter.
Solution:
(1) Mark any point C on the sheet.

(2) By adjusting the compasses up to 3.4 cm and by putting the pointer of the compasses at point C, turn the compasses slowly to draw the circle. It is the required circle of 3 .4 cm radius.
(3) Mark any diameter \(\overline{\mathrm{AB}}\) in the circle.
(4) Now, taking Aand B as centres, draw arcs on both sides of \(\overline{\mathrm{AB}}\) taking radius more than \(\overline{\mathrm{AB}}\). Let these intersect each other at D and E.
(5) Join DE, which is the perpendicular bisector of AB. It can be observed that \(\overline{\mathrm{DE}}\) is passing through the centre C of the circle.

Question 8.
Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Solution:
1) Mark any point C on the sheet. Now, by adjusting the compasses up to 4 cm and by putting the pointer of compasses at point C, turn the compasses slowly to draw the circle. It is the required circle of 4 cm radius.
(2) Take any two chords \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) in the circle.

(3) Taking A and B as centres and with radius more than half of \(\overline{\mathrm{AB}}\), draw arcs on both sides of AB, intersecting each other at E, F. Join EF which is the perpendicular bisector of AB.
(4) Taking C and D as centres and with radius more than half of \(\overline{\mathrm{CD}}\), draw arcs o n both sides of CD, intersecting each other at G, H. Join GH which is the perpendicular bisector of CD. Now, we will find that when EF and GH are extended, they meet at the centre of the circle i.e., point O.

Question 9.
Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of \(\overline{\mathbf{O A}}\) and \(\overline{\mathbf{O B}}\). Let them meet at P. Is PA = PB ?
Solution:
(1) Draw any angle whose vertex is O.
(2) With a convenient radius, draw arcs on both rays of this angle while taking O as centre. Let these points be A and B.

(3) Taking O and A as centres and with radius more than half of OA, draw arcs on both sides of OA. Let these be intersecting at C and D. Join CD.
(4) Similarly, we can find the perpendicular bisector \(\overline{\mathbf{E F}}\) and \(\overline{\mathbf{O B}}\). These perpendicular bisectors \(\overline{\mathbf{C D}}\) and \(\overline{\mathbf{E F}}\) will intersect each other at P.
Now, PA and PB can be measured. These are equal in length.

Students can Download Chapter 9 Data Handling Ex 9.3 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 9 Data Handling Ex 9.3

Question 1.
The bar graph given along side shows the amount of wheat purchased by government during the year 1998 – 2002.

Read the bar graph and write down your observations. In which year was.
Solution:
a) the wheat production maximum?
The wheat production was maximum in year 2002

b) the wheat production minimum?
The wheat production was minimum in year 1998.

Question 2.
Observe this bar graph which is showing the sale of shirts in ready made shop from Monday to Saturday.

Now answer the following questions:
Solution:
(a) What information does the above bar graph give?
This bar graph represents the number of shirts sold from Monday to Saturday,

(b) What is the scale chosen on the horizontal line representing number of shirts?
1 unit = 5 shirts

(c) On which day was the maximum number of shirts sold? How many shirts were sold on that day?
The number of shirts sold on Saturday was the maximum, i.e., 60.

(d) On which day was the minimum number shirts sold?
The number of shirts sold on Tuesday was the minimum, i.e., 10.

(e) How many shirts were sold on Thursday?
35 shirts were so Id on Tuesday

Question 3.
Observe this bar graph which show s the marks obtained by Aziz in half – yearly examination in different subject.

Answer the given question.
Solution:
a) What information does the bar graph give?
The graph shows the marks obtained by Aziz different subjects

b) Name the subject in which Aziz scored maximum marks.
in Hindi, Aziz scored maximum marks i.e. 80.

c) Name the subject in which he has scored minimum marks.
in social studies, Aziz scored minimum marks i.e. 40.

d) State the name of the subjects and marks obtained in each of them.
Hindi – 80,
english – 70,
Science – 50,
Social – 40.

Students can Download Chapter 10 Mensuration Ex 10.2 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 10 Mensuration Ex 10.2

Question 1.
Find the areas of the following figures by counting square

Solution:
(a) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.
(b) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be 5 square units.
(c) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this
figure will be 4 square units.
(d) The figure contains 8 fully filled squares only. Therefore, the area of this figure will be 8 square units.
(e) The figure contains 10 fully filled squares only. Therefore, the area of this figure will be 10 square units.
(f) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.
(g) The figure contains 4 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 6 square units.
(h) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be 5 square units.
(i) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.
(j) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.
(k) The figure contains 4 fully filled squares and 2 half-filled squares. Therefore, the area of this figure will be 5 square units.
(l) From the given figure, it can be observed that,

Total area = 2 + 6 = 8 square units

(m) From the given figure, it can be observed that,

Total area = 5 + 9 = 14 square units
(n) From the given figure, it can be observed that,

Total area = 8 + 10 = 18 square units

Students can Download Chapter 10 Mensuration Ex 10.1 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 10 Mensuration Ex 10.1

Question 1.
Find the perimeter of each of the following figures:

Solution :
Perimeter of a polygon is equal to the sum of the lengths of all sides of that polygon.
(a) Perimeter = (4 + 2 +1 + 5) cm = 12 cm
(b) Perimeter = (23 + 35 + 40 + 35) cm = 133 cm
(c) Perimeter = (15 + 15 + 15 + 15) cm = 60 cm
(d) Perimeter = (4 + 4 + 4 + 4 +4) cm = 20 cm
(e) Perimeter = (1 + 4 + 0.5 + 2.5 + 2.5 + 0.5+ 4) cm = 15 cm
(f) Perimeter = (1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4) = 52 cm

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution:
Length (l) of rectangular box = 40 cm
Breadth (b) of rectangular box = 10 cm
Length o f tape required = perimeter of rectangular box
= 2 ( l + b)
= 2(40 + 10) = 100 cm

Question 3.
A table – top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table – top?
Solution:
Length (l) of the top of the table = 2 m 25 cm = 2 + 0.25
= 2.25 m
Breadth (b) of the top of the table = 1 m 50 cm = 1 + 0.50 = 1.50 m
Perimeter of table – top = 2 (l + b)
= 2 × (2.25 + 1.50)
= 2 × 3.75 = 7.5 m
The perimeter of the table-top is 7.5 m

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution:
Length (l) of the photograph = 32cm
Breadth (b) of the photograph = 21 cm
Perimeter of photograph = 2 × (l + b)
= 2 × (32 + 21)
= 2 × (53)
= 106 cm
Length of wooden strip required is 106 cm.

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution:
Given,
Length of the Land (l) = 0.7 km
Breadth of the land (b) = 0.5 km.
perimeter of the land = 2 × (l + b)
= 2 × (0.7 + 0.5)
= 2 × (1.2)
= 2.4 km.
The length of the wire required to fenced the land is given by 4 × 2.4 = 9.6 km.

Question 6.
Find the perimeter of each of the following Shapes:
a) A triangle of sides 3 cm, 4 cm and 5 cm.
b) An equilateral triangle of side 9 cm.
c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Solution:
a) perimeter of a ∆le = (AB + BC + CA) cm
= (3 + 4 + 5) cm.
= 12 cm

b) perimeter of an equilateral triangle = 3 × side of ∆le
= (3 × 9 ) cm
= 27 cm.

c) perimeter of an isosceles triangle = 8 + 8 + 6 = 22 cm.

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm, and 15 cm.
Solution:
Perimeter of triangle = sum of lengths of all sides of the triangle
∴ Perimeter of ∆le = 10 + 14 + 15 = 39 cm.

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8m.
Solution:
Perimeter of regular hexagon = 6 × side of regular hexagon
= 6 × 8 m
∴ Perimeter of regular hexagon = 48 m

Question 9.
Find the side of the square whose perimeter is 20m.
Solution:
Perimeter of square = 4 × side 20
= 4 × side

∴ Side of the square = 5 m

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter of regular pentagon = 5 × side of regular pentagon.
100 = 5 × side

∴ Side of the regular pentagon = 20 cm.

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
a) a square?
b) an equilateral triangle?
c) a regular hexagon?
Solution:
a) perimeter a square = 4 × side of the square
= 30 = 4 × side of the square

Side of the square = 7.5 cm.

b) perimeter an equilateral triangle = 3 × side of the ∆le
30 = 3 × side of the ∆le

Side of the equilateral ∆le = 10 cm.

c) perimeter of a regular hexagon = 6 × side of the regular hexagon
30 = 6 × side

Side of the regular hexagon = 5 cm.

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solution:
Given AB = 12 cm, and BC = 14 cm, CA = ?
Perimeter of triangle = sum of all sides of the triangles
36 = AB + BC + CA
36 = 26 + CA

CA = 36 – 26 = 10 cm.
CA = 10 cm.
Hence, the third side of the triangle is 10 cm.

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.
Solution:
Length of fence required is equal to the perimeter of the square park
= 4 × side
= 4 × 250
= 1000 m
Cost for fencing 1 m of square park = Rs. 20
Cost for fencing 1000m of square park = 1000 × 20
= Rs. 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per metre.
Solution:
Given:-
Length of rectangular park (l) = 175 cm
Breadth of rectangular park (b) = 125 cm
Length of wire required for fencing the park = Perimeter of the park
= 2 × (l + b)
= 2 × (175 + 125)
= 2 × 300 = 600 m
Cost for fencing 1 m of the park = Rs 12.
Cost for fencing 600m of the square park = 600 × 12
= Rs. 7200

Question 15.
Sweety runs around a square park of side 75m. Bulbul runs around a rectangular park with length 60m and breadth 45m. Who covers less distance?
Solution:
The distance covered by sweety = 4 × rectangular square park
= 4 × 75 m = 300 m
The distance covered by bulbul = perimeter of rectangular park
= 2 × (60 + 45)
= 2 × 105
= 210 m
Therefore, bulbul covers the less distance than sweety.

Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?

Solution:
a) perimeter of square = 4 × 25 = 100 cm.
b) perimeter of rectangle = 2 × (10 + 40) = 2 × 50 = 100 cm
c) perimeter of rectangle = 2 × (20 + 30) = 100 cm.
d) perimeter of ∆le = 30 + 30 + 40 = 100 cm
Imfered:- The above figures have the same perimetre .

Question 17.
Avneet buys 9 square paving slabs, each with a side of \(\frac{1}{2}\) m. He lays them in the form of a square.

Solution:
a) What iis the perimeter of his arrangement [ Fig 10.7 (i)] ?

b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig 10.7(ii)]?
perimeter of cross = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 m

c) Which has greater perimeter?
The arrangement in the form of a cross has a greater perimeter.

d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken?
Arrangements with perimeter greater than 10m can not be determined.

Students can Download Chapter 8 Decimals Ex 8.6 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 8 Decimals Ex 8.6

Question 1.
Subtract:-
Solution:
a) Rs 18.25 from Rs 20.75

b) Rs 202.54 m from Rs 250 m

 

c) Rs 5.36 from Rs 8.40

d) 2.051 km from 5.20 km

 

e) 0.314 kg from 2.107 kg

Question 2.
Find the value of:-
Solution:
a) 9.756 – 6.28

b) 21.05 – 15.27

c) 18.5 – 6.79

d) 11.6 – 9.847

 

Question 3.
Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solution:

Therefore he will get back Rs. 14.35

Question 4.
Rani had Rs 18.50. She bought one ice cream for Rs 11.75. How much money does she have now?
Solution:

Question 5.
Tina had 20m 5cm long cloth, she cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Solution:
Length of cloth = 20 m 5 cm = 20.05 m
Length of cloth cut so as to make a curtain = 4 m 50 cm = 4.50 m
The length of the left with her will be the difference of these two
Hence, the length of the cloth left with her is

15.55 m cloth will be remaining

Question 6.
Namitha travels 20 km 50 m every day. Out of this she travels 10 km 200m by bus
and the rest by auto. How much distance does she travel by auto?
Solution:
Total distance traveled by namita = 20 km 50m = 20.050 km
Distance traveled by bus = 10 km 200 m = 10.200 km
Distance traveled by auto = Total distance traveled – Distance traveled

Question 7.
Aaksha bought vegetables weighing 10 kg. out of this, 3 kg 500 g is onions, 2 kg 75g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Solution:
Total Weight of the vegetable = 10.000 kg
Weight of the onions = 3 kg 500 g = 3.500 kg
Weight of the tomatoes = 2 kg 75 g = 2.075 kg
Weight of potatoes = Total weight of vegetables bought
(Weight of onions + weight of tomatoes )
= 10000 – ( 3500 + 2075)

Hence, the weight of the potatoes was 4425 kg.

Students can Download Chapter 11 Algebra Ex 11.4 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 11 Algebra Ex 11.4

Question 1.
Answer the following:
a) Take Sarita’s present age to be y years
Solution:
i) What will be her age 5 years from now?
Saritha’s present age + 5
= y + 5

ii) What was her age 3 years back
3 years ago, Saritha’s age = Saritha’s present age – 3
y – 3

iii) Sarita’s grandfather is 6 times her age. what is the age of her grandfather?
Grand father’s age = 6 × Sarita’s present age = 6y

iv) Grandmother is 2 years younger than grandfather. What is grandfather’s age?
Solution:
Grand father’s age = Grand father’s present age – 2 = 6y – 2

v) Saritha’s father’s age is 5 years more than 3 times Saritha’s age. What is her
father’s age?
Father ’s age = 5 + 3 x saritha’s persent age = 5 + 3y

b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?
Solution:
Length = 3 × Breadth – 4
l = (3b – 4) metres

c) A rectangular box has height h cm. its length is 5 times the height and breadth is 10 cm less than the length. Express he length and the breadth of the box in terms of the height.
Solution:
Length = 5 × Height
l = 5h cm
Breadth = 5 × Height – 10
b = (5h – 10)

d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind, of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps usings.

Solution:
She at which Beena is = (step at which Meen is) + 8
= S + 8
Step at which Leena is = (steps at which Meena is ) – 7
= S = 7
Total steps = 4 × (step at which meena is) – 7
= S – 7
Total steps = 4 × (steps at which Meena is ) – 10
= 4s – 10

e) A bus travels at v km per hour. It is going from Daspur to Beespur, After the bus travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Solution:
Speed = vkm/hr
Distance travelled in 5hrs = 5 × v = 5v km
Total distance between Daspur and Beespur = (5v + 20) km

Question 2.
Change the following statements using expressions into statements in ordinary language.
(For example, Give salim scores r runs in a cricket match, Nalin scores (r +15) runs. In ordinary language – Nalin scores 15 runs more than salim)
a) A note book costs Rs P.A book costs Rs 3 p.
Solution:
A book costs three times the cost of a note book

b) Tony puts q marbles on the table. He has 8q marbles in his box.
Solution:
Tony’s box contains 8 times the numbers of marbles on the table

c) Our class has n students. The school has 20n students.
Solution:
Total number of students in the school is 20 times that of our class.

d) Jaggu is Z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.
Solution:
Jaggu’s uncle is 4 times older than jaggu and jaggu’s aunt is 3 years younger than his uncle.

e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Solution:
The total number of close is 5 times the number of rows.

Question 3.
a) Given Munnurs age to be x years, can you guess what (9x – 2) may show?
(Hint: Think of Munnu’s younger brother.)
Can you guess what (x + 4) may shows? What (3x + 7) may show?
Solution:
(x – 2) represents that the person whose age is (x – 2) years, is 2 years younger to munnu
(X + 4) represents that the person, Whose age is (x + 4) Years, is 4 years elder to munnu
(3x+7) represents that the person whose age is (3x + 7) years, is elder to munnu and his age is 7 years more than three times of the age of munnu

b) Given sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate? y + 7, y – 3, y + 4\(\frac{1}{2}\), y – 2\(\frac{1}{2}\)
Solution:
in future
After m years from now, sara’s age will be (y + h) years in past n years ago , sara’s age was (y – n) years
(y + 7) represents that the person, whose age is (y + 7) years, is 7 years elder to sara (y – 3) represents that the person, whose age is (y – 3) years, is 3 years younger to sara
(y + 41/2) represents that the person

Whose age is (y + 4\(\frac{1}{2}\) ) years, is 4 1/2 years elder to sara (y – 2 1/2) represents that the person
Whose age is (y – 2 1/2) years, is 2 1 /2 years younger to sara

c) Given n students in the class like football, What may 2n show? What may \(\frac{n}{2}\) show?
(Hint: .Think of games other than football)
Solution:
2n may represent the number of students who like either football or some other game such as cricket Where as 11/2 represents the number of students who like Cricket, out of the total number of students who like football.

Students can Download Chapter 11 Algebra Ex 11.3 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 11 Algebra Ex 11.3

Question 1.
Make up as many expressions with numbers ( no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
(Hint: Three possible expressions are 5 + (8 – 7), 5 – (8 – 7),(5 × 8) + 7; make the other expressions.)
Solution:
Many expressions can be formed by using the 3 numbers 5, 7 and 8 Some of these are as follows 5 × (8 – 7)
5 × (8 + 7)
(8 + 5) × 7
(8 – 5) × 7
(7 + 5) × 8
(7 – 5) × 8

Question 2.
Which out of the following are expressions with numbers only?
a) y + 3
b) (7 × 20) – 8z
c) 5(21 – 7) + 7 × 2
d) 5
e) 3x
f) 5 – 5n
g) (7 × 20) – (5 × 10) – 45 + p
Solution:
It can be observed that the expressions in alternatives (c) and (d) are formed by using numbers only.

Question 3.
Identify the operations(addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
a) z + 1, z – 1, y + 17, y – 17
Solution:
Addition as 1 is added to z
Subtraction as 1 is subtracted from z
Addition as 17 is added to y
Subtraction as 17 is subtracted from y

b) 17y, \(\frac{y}{17}\), 5z
Multiplication as y is multiplied as y is multiplied with 17
Division as y is divided by 17
Multiplication as z is multiplied with 5
Multiplication as z is multiplied with 5

c) 2y + 17, 2y – 17
Multiplication and addition
y is multiplied with 2, and 17 is added to the result

d) 7m, -7m + 3, -7m – 3
Multiplication with 2 and 17 is subtracted from the result
Multiplication and subtractions m is multiplied by 7 and 3 is subtracted from the result

Question 4.
Give expressions for the following cases.
Solution:
a) 7 added to p
P + 7

b) 7 subtracted from p
P – 7

c) p multiplied by 7
7p

d) p divided by 7
P

e) 7 subtracted from -m
-m – 7

f) -p multiplied by 5
– 5p

g) -p divided by 5
\(\frac{-p}{5}\)

h) p multiplied by -5.
-5p

Question 5.
Give expressions in the following cases.
Solution:
a) 11 added to 2m
2m + 11

b) 11 subtracted from 2m
2m – 11

c) 5 times y to which 3 is added
5y + 3

d) 5 times y from which 3 is subtracted
5y – 3

e) y is multiplied by -8
-8y

f) y is multiplied by -8 and then 5 is added to the result
-8y + 5

g) y is multiplied by 5 and the result is subtracted from 16
16 – 5y

h) y is multiplied by -5 and the result is added to 16
-5y + 16

Question 6.
a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
Solution:
t + 4, +t – 4, 4 +, \(\frac{t}{4}\), \(\frac{4}{t}\), -4 – t, 4 + t

b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution:
2y + 7, 2y – 7, 7y + 2….

Students can Download Chapter 5 Understanding Elementary Shapes Ex 5.3 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 5 Understanding Elementary Shapes Ex 5.3

Question 1.
Match the following :

Solution:
i) Straight angle is of 180° and half of a revolution is 180°. Hence. (i) – (c)
ii) Right angle is of 90° and one – fourth of a revolution is 90°. Hence, (ii) – (d)
iii) Acute angles are the angles less than 90°. Also, less than one fourth of a revolution is the angle less than 90°. Hence, (iii) – (a)
iv) Obtuse angles are the angles greater than 90° but less than 180°. Also between 1/4 and 1/2 of a revolution is the angle whose measure lies between 90° and 180°. Hence, (iv) – (e)
v) Reflex angles are the angles greater than 180° but less than 180° but less than 360°, Also more than half a revolution is the angle whose measure is greater than 180°.
Hence (V) ⟷ (b)

Question 2.
Classify each one of the following angles as right, straight, acute, obtuse or reflex:

Solution:
a) A cute angle as its measure is less than 90°
b) Obtuse angle as its measure is more than 90° but less than 180°
c) Right angle as its measure is 90°
d) Reflex angle as its measure is more than 180° but lass than 360°
e) Straight angle as its measure is 180°
f) Acute angle as its measure is less than 90°