Class 7

Students can Download Chapter 12 Algebraic Expressions Ex 12.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
i) Subtraction of z from y.
Solution:
y – z

ii) One-half of the sum of numbers x and
Solution:

iii) The number z multiplied by itself.
Solution:
Z × Z = Z²

iv) One- fourth of the product of numbers p and q.
Solution:

v) Numbers x and y both squared and added.
Solution:
x² + y²

vi) Number 5 added to three times the product of numbers m and n.
Solution:
3mn + 5

vii) Product of numbers y and z subtracted from 10.
Solution:
10 – yz

viii) Sum of numbers a and b subtracted from their product.
Solution:
ab, – {a + b)

Question 2.
i) Identify the terms and their factors in the following expression show the terms and factors by tree diagrams.
Solution:

c) y – y³
Terms: y; -y³

Factors : y ; -1, y,

d) 5xy² + 7x²y
Terms : 5xy² ; 7x²y
Factors : 5, x, y, y; 7, x, x, y

e) -ab + 2b² – 3a²
Terms : -ab ; 2b²; 3a²
Factors : 1, a, b ; 2, b, b ; -3, a, a

ii) Identify terms and factors in the expressions given below :

a) -4x + 5
Terms : -4x ; 5
Factors : -4, x; 5

b) -4x + 5y
Terms : -4x ; 5y
Factors :-4, x; 5, y

c) 5y + 3y²
Terms : 5y ; 3y²
Factors : 5, y ; 3, y, y

d) xy + 2x²y²
Terms : xy ; 2x²y²
Factors : x, y ; 2, x, x, y, y

e) pq + q
Terms : pq ; q
Factors : p, q ; q

f) 1.2 ab – 2.4b + 3.6a
Terms : 1.2 ab ; 2.4b ; 3.6a
Factors : 1.2, a, b ; -2.4, b ; 3.6, a

g)

Solution:

h) 0.1 p² + 0.2 q²
Terms : 0.1p²; 0.2q²
Factors : 0.1, p, p ; 0.2, q, q

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions :
i) 5 – 3t²
Term other than constant = -3/2
numerical co-efficient = -3

ii) 1 + t + t² + t³
Term other than constant = t, t², t³
Numerical co-efficients = 1, 1, 1.

iii) x + 2xy + 3y
Terms other than constants = x, 2xy + 3y
Numerical co-efficients = 1, 2, 3

iv) 100m + 1000n
Terms other than constants = 100m, 1000n
Numerical co-efficients = 100, 1000.

v) -p²q² + 7pq
Terms other than constants = -p² q², 7 pq
Numerical co-efficients = -1, 7

vi) 1.2a + 0.8b
Terms other than constants = 1.2a, 0.8b
Numerical co-efficients = 1.2, 0.8

vii) 3.14 r²
Terms other than constants = 3.14r²
Numerical co-efficients = 3.14

viii) 2 (l + b)
Terms other than constants = 2l, 2b
(∵ 2(l + b) = 2l + 2b)
Numerical co-efficients = 2, 2

ix) 0.1 y + 0.01 y²
Terms other than constants = 0.1 y, 0.01 y²
Numerical co-efficients = 0.1, 0.01

Question 4.
a) Identify terms which contain x and give the coefficient of x.
i) y²x + y
Terms which contains x — y²
co-efficient of x — y²

ii) 13y² – 8yx
Terms which contains x — (x – 8yx)
co-efficient of x — (-8y)

iii) x + y + 2
Terms which contains x —- x
co-efficient of x —- 1

iv) 5 + z + zx
Terms which contains x — x
co-efficient of x — z

v) 1 + x + xy
Terms which contains x — x and xy
co-efficient of x — 1 y

vi) 12xy² + 25 — 12xy²
Terms which contains x
co-efficient of x — 12y²

vii) 7x + xy²
7x Terms which contains x
co-efficient of x —- 7
Terms which contains x — xy²
y² co-efficient of

b) Identify terms which contain y² and give the coefficient of y’.
i) 8 – xy²
Terms which contains y²
xy² co-efficient of y² — x

ii) 5y² + 7x
Terms which contains y²
co-efficient of y² — 5

iii) 2x² y – 15xy² + 7y²
Terms which contains y² — (-15xy²)
co-efficient of y² — (-15x)
Terms which contain y² — 7y²
co-efficient of y² —- 7

Question 5.
Classify into monomials, binomials and trinomials.
i) 4y – 7z
This expression is binomial because it contains two terms. That is 4 y and -7z.
ii) y²
This is monomial expression because it contains only one term y².
iii) x + y – xy
This is trinomial expression because it contains three terms x, y and -xy.
iv) 100
This is a monomial expression because it contains only one term 100.
v) ab – a – b
This is trinomial expression because it contains 3 terms ab, -a and -b.
vi) 5 – 3t
This is binomial expression because it contains 2 terms 5 and -3t.
vii) 4p²q – 4pq²
This is binomial expression because it contains two terms 4p²q and – 4 pq²
viii) 7mn
This is monomial expression because it contains only one term 7mn.
ix) z² – 3z + 8
This is trinomial expression because it contains three terms z², -3z and 8.
x) a² + b²
This is binomial expression because it contains two terms a² and b².
xi) z² + z
This is binomial expression because it contains 2 terms z² and z.
xii) 1 + x + x²
This is trinomial expression because it contains 3 terms 1, x and x².

Question 6.
State whether a given pair of terms is of like or unlike terms.
i) 1, 100
like terms
ii)
like terms
iii) -29x, -29y
unlike terms
iv) 14xy, 42yx
like terms
v) 4m²p, 4mp²
unlike terms
vi) 12xz, 12x²z² unlike terms

Question 7.
Identify like terms in the following :
a) -xy², -4yx², 8x², 2xy², 7y, – 11x², 100x, – 11yx, 20x²y, -6x², y, 2xy, 3x
Solutions:
Like terms are
i) -xy², 2xy²
ii) -4yx², 20x²y
iii) 8x², -11x², -6x²
iv) 7y, y
v) -100x, 3x
vi) -11yx, 2xy

b) 10pq, 7p, 8q, -p²q², -7qp, – 100q, -23, 12q²p², -5p², 41, 2405p, 75qp, 13p²q, qp², 701p²
Solution:
like terms are
i) 10pq, – 7qp, 78qp,
ii) 7p, 2405p,
iii) 8q, -100q
iv) -23, 41,
v) 12q²p², -p²q²,
vi) -5p², 701p²,
vii) 13p²q, qp²

Students can Download Chapter 10 Practical Geometry Ex 10.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 1.
Construct ∆ DEF such that DE = 5 cm, DF = 3 cm and ∠EDF = 90°.
Solution:

Steps of Construction

1. Draw a line segment DE of length 5 cm. At D draw DM making 90° with DE.
2. With ‘D’ as centre, draw an arc of radius 3 cm, It cuts DM at the point F.
3. Join FE. The required triangle is obtained. That is ∆ DEF.

Question 2.
Construct an Isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110°.

Solution:
Steps of Construction

1. Draw a line segment AB of length 6.5 cm. At A draw AM making 110° with AB using a protractor.
2. With ‘A’ as centre draw an arc of radius 6.5 cm and it cuts AM at C.
3. Join BC. The required triangle is ∆ ABC.

Question 3.
Construct ∆ ABC with BC = 7.5 cm. AC = 5 cm and ∠C = 60°.
Solution:

Steps of Construction

1. Draw a line segment BC of length 7.5 cm. At C draw CM making 60° with BC.
2. With ‘C’ as centre, draw an arc of radius 5cm, it cuts at CM at A.
3. Join AB. ∆ ABC is the required triangle.

Students can Download Chapter 7 Congruence of Triangles Ex 7.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1

Question 1.
Complete the following statements :
a) Two line segments – are congruent if ___.
Solution:
They have the same measure, (length)

b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is ___
Solution:
70°

c) When we write ∠A = ∠B, we actually mean ___
m < A ≅ m

Question 2.
Give any two real-life examples for congruent shapes.
Solution:
Two Rs 10/- notes and two ten rupees coins.

Question 3.
If ∆ ABC ≅ ∆ FED under the correspondence ABC ⟷ FED, write all the corresponding congruent parts of the triangles.

Solution:
The ∆ ABC ≅ ∆ FED then the corresponding vertices A and F, B and E, C and D. The corresponding sides are and FE, BC and ED, CA, and DF. The corresponding angles are ∠A and ∠F, ∠B and ∠E, and ∠C and ∠D.

Question 4.
If ∆ DEF ≅ ∆ BAC, write the part(s) of ∆ BCA that correspond to
Solution:

i) ∠E
∠E ≅ ∠C,

ii) \(\overline{\mathbf{E} \mathbf{F}}\)
\(\overline{\mathbf{E} \mathbf{F}}\) ≅ \(\overline{\mathbf{C} \mathbf{A}}\)

iii) ∠F
∠F ≅ ∠A

iv) \(\overline{\mathbf{D F}}\)
\(\overline{\mathbf{D F}}\) ≅ \(\overline{\mathbf{B A}}\)

Students can Download Chapter 15 Visualising Solid Shapes Ex 15.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. At tempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions).

Question 2.
Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these!)

Question 3.
Examine if the following are true statements :
i) The cube can cast a shadow in the shape of a rectangle.
Solution:
True

ii) The cube can cast a shadow in the shape of a hexagon.
Solution:
False.

Students can Download Chapter 10 Practical Geometry Ex 10.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

Question 1.
Construct ∆ XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Solution:

Steps of Construction

1. Draw a line segment YZ of length 5 cm with Y as centre, draw an arc of radius 4.5 cm.
2. With Z as centre, draw an arc of radius 6cms which cuts the first arc. Mark the point of intersection of arcs as X.
3. Join XY and XZ. ∆ XYZ is the required triangle.

Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Constructions

1. Draw a line segment BC of length 5.5 cms. with ‘B’ as centre draw an arc of radius 5.5 cm.
2. With ‘C’ as centre, Cut the first arc with radius 5.5 cms. Mark the point of intersection of arcs as ‘A’
3. Join AB and AC. ∆ ABC is the required equilateral triangle.

Question 3.
Draw ∆ PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this ?
Solution:

Steps of Construction

1. Draw a line segment QR of length 3.5 cm with Q as centre, draw an arc of radius 4 cm.
2. With R as centre, cut the first arc with a radius 4 cm. Mark the point of intersection of arcs as P.
3. Join PQ & PR A PQR is an isosceles triangle PQ = PR.

Question 4.
Construct ∆ ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure <B.
Solution:
Steps of Construction

1. Draw a line segment BC of length 6 cm with ‘B’ as centre, draw an arc of radius 2.5 cm.
2. With ‘C’ as centre cut the first arc with a radius 6.5 cm.
3. Mark the point of intersection of arcs as A. Join AB & AC. ∆ ABC is the required triangle.

Students can Download Chapter 15 Visualising Solid Shapes Ex 15.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3

Question 1.
What cross-sections do you get when you give a
i) vertical cut
ii) horizontal cut to the following solids?
a) A brick
b) A round apple
c) A die
d) A circular pipe
e) An ice cream cone
Solution:

Students can Download Chapter 10 Practical Geometry Ex 10.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.1

Question 1.
Draw a line, say AB, take a point C out-side it. Through C, draw a line parallel to AB using ruler and compasses only.
Solution:

Steps of Construction

1. Draw a line AB. Take a point C outside it. Take any point D on AB and join C to D.
2. With D as centre and a convenient radius, draw an arc cutting AB at F and CD at E.
3. Now with C as centre and the same radius as in step 2, draw an arc GH cutting CD at I.
4. Place the pointed tip of the compasses at F and adjust the opening so that the pencil tip is at E.
5. With the same opening as in step 4 and with I as centre, draw an arc cutting the are GH at J.
6. Now Join CJ to draw a line ‘KL’. KL is the required parallel line.

Question 2.
Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
Solution:

Steps of Construction

1. Draw a line l, take any point A on line l.
2. Construct an angle of 90° at point A of line l and draw a line AL perpendicular to line l.
3. Mark a point X on AL such that AX = 4 cm
4. At X construct an angle of 90° and draw a line XC perpendicular to line AL.
5. The line XC (line m) is the required line through X such that m || l.

Question 3.
Let l be a line and P be a point on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
Solution:

Steps of constructions

1. Draw a line l and take a point P, not on it. Take any point Q on l. Join Q to P
2. Draw a line m parallel to the line l, as shown in the figure. line m || line l.
3. Join P to any point Q on l.
4. Choose any point R on m & Join R to Q.
5. Through R, draw a line n parallel to the line PQ. Let the line n meet the line l at S.
6. The shape enclosed by the two sets of parallel lines is a parallelogram.

Students can Download Chapter 15 Visualising Solid Shapes Ex 15.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2

Question 1.
Use isometric dot paper and make an isometric sketch for each one of the given shapes:
Solution:

Question 2.
The dimensions of a cuboid are 5 cm, 3cm and 2 cm. Draw three different isometric sketches of this cuboid.
Solution:

Question 3.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid.
Solution:

Question 4.
Make an oblique sketch for each one of the given isometric shapes :
Solution:

Question 5.
Give
(i) an oblique sketch and
(ii) an isometric sketch for each of the following :
a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique ?)
Solution:

b) A cube with an edge 4 cm long.
Solution:

Students can Download Chapter 15 Visualising Solid Shapes Ex 15.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.1

Question 1.
Identify the nets which can be used to make cubes (cut out copies of the nets and try it):
Solution:

In the given figures ii), iii), iv) and vi) which can be used to make cubes.

Question 2.
Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them.

Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.
Solution:

Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.

Question 3.
Can this be a net for a die? Explain your answer.
Solution:
No, It is not a net for a die. Because one pair of opposite faces will have 1 and 4 on them whose total is not equal to 7 and another pair of opposite faces will have 3 and 6 on them, its total is also not equal to 7.

Question 4.
Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there on the net here? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.)
Solution:

In the net given here three faces are shown.

Question 5.
Match the nets with appropriate solids:

Solution:
a – (ii)
b – (iii)
c – (iv)
d – (i)

Students can Download Chapter 9 Rational Numbers Ex 9.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

Question 1.
Find the sum :

i)

Solution:

ii)

Solution:
L.C.M of 3 and 5 is 15

iii)

Solution:
LCM of 10 and 15 is 30

iv)

Solution:
LCM of 11 and 9 is 99

v)

Solution:
L.C.M of 19 and 57 is 57

vi)

Solution:

vii)

Solution:

L.C.M of 3 and 5 is 15

Question 2.
Find:
i)

Solution:

ii)

Solution:

iii)

Solution:

L.C.M of 13 and 15 is 195

iv)

Solution:

v)

Solution:

L.C.M. of 9 and 1 is 9
Thus,

Question 3.
Find the product:
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

Question 4.
Find the value of:
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

vii)

Solution: