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Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles

November 9, 2023May 27, 2022 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles (Including Construction of angles)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 14 Lines and Angles (Including Construction of angles)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

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POINTS TO REMEMBER

1. POINT : A point: is a mark of position; which has no length, no breadth and no thickness. It, in general, is represented by a capital letter as shown alongside.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 1

2. LINE : A line has length, but no breadth or thickness.

“The given figure shows a line AB in which two arrow-heads in opposite directions show that can be extended infinitely in both the directions.
A line may be straight or curved but when we say a line’ it means a straight line only.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 3

3. RAY : It is a straight line which stats from a fixed point and moves in the same direction.

The given figure shows a ray\(\xrightarrow { AB }\) with fixed initial point A ‘and moving in the direction AB.

4. LINE SEGMENT : It is a straight line with its both ends fixed. The given figure shows a line segment, whose both the ends A and B are fixed.

(i) The adjoining figure shows a line AB which can be extended upto infinitey on both the sides of it.

(ii) The adjoining figure shows a ray AB with fixed end as point A and which can be extended upto infinity through point B. It is clear from the figure, that a ray is a part of a line.

(iii) The adjoining figure shows a line-segment AB with fixed ends A and B. It is clear from the figure, that a line-segment is a part of a ray as well as of a line. Also, a line segment is the shortest distance between two fixed points.

5. ANGLE : An angle is formed when two line segments or two rays have a common end-point.
The two line segments, forming an angle, are called the arms of the angle whereas their common end-point is called the vertex of the angle.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 9
The adjacent figure represents an angle ABC or ∠ABC or simply ∠B. AB and BC are the arms of the angle and their common point B is the vertex.

6. MEASUREMENT OF AN ANGLE : The unit of measuring an angle is degree. The symbol for degree is °.
Thus : 60 degree = 60°, 87 degree = 87’ and so on.
If one degree is divided into 60 equal parts, each part is called a minute ( ‘) and if one minute is further divided into 60 equal parts, each part is called a second ( ” ).
Thus, (i) r = 60′ and l’ = 60″
(ii) 9 minutes 45 seconds = 9′ 45″
(iii) 85 degrees 30 minutes 15 seconds = 85° 30′ 15″ and so on.

7. TYPES OF ANGLES :
1. Acute angle :
measures less than 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 10
2. Right angle:
measures 90°

3. Obtuse angle :
measures between 90° and 180°

4. Straight angle :
measures 180°

5. Reflex angle :
measures between 180° and 360°

8. MORE ABOUT ANGLES :
(A) Angles about a point: If a number of angles are formed about a point, their sum is always 360°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 15
In the adjoining figure :
∠AOB + ∠BOC + ∠COD +∠DOE + ∠EOA = 360°
(B) Adjacent angles : Two angles are said to be adjacent angles, if:
(i) they have a common vertex,
(ii) they have a common arm and
(iii) the other arms of the two angles lie on opposite sides of the common arm.
The adjoining figure shows a pair of adjacent angles :
(i) they have a common vertex (O),
(ii) they have a common ann (OB) and
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 16
(iii) the other arms OA and OC of the two angles are on opposite sides of the common arm OB.
(C) Vertically opposite angles : When two straight lines intersect each other four angles are formed.
The pair of angles which lie on the opposite sides of the point of intersection are called vertically opposite angles.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 17
In the adjoining figure, two straight lines AB and CD intersect each other at point 0. Angles AOD and BOC form one pair of vertically opposite angles; whereas angles AOC and BOD form another pair of vertically opposite angles.
Vertically opposite angles are always equal.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 18
i. e. ∠AOD = ∠BOC and ∠AOC = ∠BOD.
Important: In the adjoining figure, rays OX and OY meet a O to form ∠XOY (i.e. ∠a) and reflex ∠XOY (i. e. ∠b). It must be noted that ∠XOY means the smaller angle only unless it is mentioned to take otherwise.

9. COMPLEMENTARY AND SUPPLEMENTARY ANGLES
1. Two angles are called complementary angles, if their sum is one’right angle i.e. 90° Each angle is called the complement of the other.
e.g., 20″ and 70″ are complementary angles, because 20° + 70° = 90°.
Clearly, 20″ is the complement of 70° and 70° is the complement of 20°.
Thus, the complement of angle 53° = 90° – 53° = 37°.
2. Two angles are called supplementary angles, if their sum is two right angles i.e. 180″. Each angle is called the supplement of the other.
e.g., 30″ and 150° are supplementary angles because 30° + 150° = 180°.
Clearly, 30″ is the supplement of 150° and vice-versa.
Thus, the supplement of 105° = 180° – 105° = 75°.

10. Transversal : It is a straight line which cuts two or more given straight lines.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 19
In the adjoining figure, PQ cuts straight lines AB and CD, and so it is a transversal.
When a transversal cuts two given straight lines (refer the adjoining figure), the following pairs of angles are formed.
1. Two pairs of interior alternate angles : Angles marked 1 and 2 form one pair of interior alternate angles, while angles marked 3 and 4 form another pair of interior alternate angles.
2. Two pairs of exterior alternate angles : Angles marked 5 and 8 form one pair, while angles marked 6 and 7 form the other pair of exterior alternate angles.
3. Four pairs of corresponding angles : Angles marked 3 and 6; 1 and 5; 8 and 2; 7 and 4 form the four pairs of corresponding angles.
4. Two pairs of allied or co-interior or conjoined angles : Angles marked 3 and 2 form one pair and angles marked 1 and 4 form another pair of allied angles.

11. PARALLEL LINES : Two straight lines are said to be parallel, if , they do not meet anywhere; no matter how long are they produced in any direction.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 20
The adjacent figure shows two parallel straight lines AB and CD.
When two parallel lines AB and CD are cut by a transversal PQ :
(i) Interior and exterior alternate angles are equal:
i.e. ∠3 = ∠6 and ∠4 = ∠5 [Interior alternate angles]
∠1 = ∠8 and ∠2 = ∠7 [Exterior alternate angles]
(ii) Corresponding angles are equal:
i.e. ∠1 = ∠5;∠2 = ∠6;∠3 = ∠7 and ∠4 = ∠8
(iii) Co-interior or allied angles are supplementary :
i. e. ∠3 + ∠5 = 180° and ∠4 +∠6 = 180°

12. CONDITIONS OF PARALLELISM : If two straight lines are cut by a transversal such that:
(i) a pair of alternate angles are equal, or
(ii) a pair of corresponding angles are equal, or
(iii) the sum of the interior angles on the same side of the transversal is 180°, then the two straight lines are parallel to each other.
Therefore, in order to prove that the given lines are parallel, show either alternate angles are equal or, corresponding angles are equal or, the co-interior angles are supplementary.

Lines and Angles Exercise 14A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
State, true or false :
(i) A line segment 4 cm long can have only 2000 points in it.
(ii) A ray has one end point and a line segment has two end-points.
(iii) A line segment is the shortest distance between any two given points.
(iv) An infinite number of straight lines can be drawn through a given point.
(v) Write the number of end points in
(a) a line segment AB (b) arayAB
(c) alineAB
(vi) Out of \(\overleftrightarrow { AB }\) , \(\overrightarrow { AB }\) , \(\overleftarrow { AB }\) and \(\overline { AB }\) , which one has a fixed length?
(vii) How many rays can be drawn through a fixed point O?
(viii) How many lines can be drawn through three
(a) collinear points?
(b) non-collinear points?
(ix) Is 40° the complement of 60°?
(x) Is 45° the supplement of 45°?
Solution:
(i) False : It has infinite number of points.
(ii) True
(iii) True
(iv) True
(v) (a) 2 (b) 1 (c) 0
(vi) AB
(vii) Infinite
(viii) (a) 1 (b) 3
(ix) False : 40° is the complement of 50° as 40° + 50° = 90°
(x) False : 45° is the supplement of 135° not 45°.

Question 2.
In which of the following figures, are ∠AOB and ∠AOC adjacent angles? Give, in each case, reason for your answer.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 21

Solution:
If ∠AOB and ∠AOC are adjacent angle if they have OA their common arm.
(i) In the figure, OB is their common arm
∴∠AOB and ∠AOC are not adjacent angles.
(ii) In the figure, OC is their common arm
∴∠AOB and ∠AOC also not adjacent angles.
(iii) In this figure, OA is their common arm
∴ ∠AOB and ∠AOC are adjacent angles.
(iv) In this figure, OB is their common arm
∴ ∠AOB and ∠AOC are not adjacent angles.

Question 3.
In the given figure, B AC is a straight line.
Find : (i) x (ii) ∠AOB (iii) ∠BOC
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 23
Solution:

Question 4.
Find yin the given figure.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 25
Solution:

Question 5.
In the given figure, find ∠PQR.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 27
Solution:
SQR is a straight line
∴∠SQT + ∠TQP + ∠PQR = 180°
⇒ x + 70° + 20° – x + ∠PQR = 180°
⇒ 90″ + ∠PQR = 180°
⇒ ∠PQR = 180°-90° = 90°
Hence ∠PQR = 90°

Question 6.
In the given figure. p° = q° = r°, find each.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 28
Solution:

Question 7.
In the given figure, if x = 2y, find x and y
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 30
Solution:

Question 8.
In the adjoining figure, if b° = a° + c°, find b.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 32
Solution:

Question 9.
In the given figure, AB is perpendicular to BC at B.
Find : (i) the value of x.
(ii) the complement of angle x.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 34
Solution:

Question 10.
Write the complement of:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 36

Solution:

Question 11.
Write the supplement of:
(i) 100°
(ii) 0°
(iii) x°
(iv) (x + 35)°
(v) (90 +a + b)° f
(vi) (110 – x – 2y)°
(vii) \(\frac { 1 }{ 5 }\) of a right angle
(viii) 80° 49′ 25″
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 39

Question 12.
Are the following pairs of angles complementary ?
(i) 10° and 80°
(ii) 37° 28′ and 52° 33′
(iii) (x+ 16)°and(74-x)°
(iv) 54° and \(\frac { 2 }{ 5 }\) of a right angle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 41

Question 13.
Are the following pairs of angles supplementary?
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 42
Solution:

Question 14.
If 3x + 18° and 2x + 25° are supplementary, find the value of x.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 44

Question 15.
If two complementary angles are in the ratio 1:5, find them.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 45

Question 16.
If two supplementary’ angles are in the ratio 2 : 7, find them.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 47

Question 17.
Three angles which add upto 180° are in the ratio 2:3:7. Find them.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 48

Question 18.
20% of an angle is the supplement of 60°. Find the angle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 49

Question 19.
10% of x° is the complement of 40% of 2x°. Find x
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 50

Question 20.
Use the adjacent figure, to find angle x and its supplement.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 52
Solution:

Question 21.
Find k in each of the given figures.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 54
Solution:

Question 22.
In the given figure, lines PQ, MN and RS intersect at O. If x : y = 1 : 2 and z = 90°, find ∠ROM and ∠POR.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 58
Solution:

Question 23.
In the given figure, find ∠AOB and ∠BOC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 60
Solution:

Question 24.
Find each angle shown in the diagram.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 62
Solution:

Question 25.
AB, CD and EF are three lines intersecting at the same point.
(i) Find x, if y = 45° and z = 90°.
(ii) Find a, if x = 3a, y = 5x and r = 6x.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 65
Solution:

Lines and Angles Exercise 14B – Selina Concise Mathematics Class 7 ICSE Solutions

In questions 1 and 2, given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles :
Question 1.
(i) ∠3 and ∠6
(ii) ∠2 and ∠4
(iii) ∠3 and ∠7
(iv) ∠2 and ∠7
(v) ∠4 and∠6
(vi) ∠1 and ∠8
(vii) ∠1 and ∠5
(viii) ∠1 and ∠4
(ix) ∠5 and ∠7
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 68
Solution:
(i) ∠3 and ∠6 are interior alternate angles.
(ii) ∠2 and ∠4 are adjacent angles.
(iii) ∠3 and ∠7 are corresponding angles.
(iv) ∠2 and ∠7 are exterior alternate angles,
(v) ∠4 and ∠6 are allied or co-interior angles,
(vi) ∠1 and ∠8 are exterior alternate angles.
(vii) ∠1 and ∠5 are corresponding angles.
(viii) ∠1 and ∠4 are vertically opposite angles.
(ix) ∠5 and ∠7 are adjacent angles.

Question 2.
(i) ∠1 and ∠4
(ii) ∠4 and ∠7
(iii) ∠10 and ∠12
(iv) ∠7 and ∠13
(v) ∠6 and ∠8
(vi) ∠11 and ∠8
(vii) ∠7 and ∠9
(viii) ∠4 and ∠5
(ix) ∠4 and ∠6
(x) ∠6 and ∠7
(xi) ∠2 and ∠13
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 69
Solution:
(i) ∠1 and ∠4 are vertically opposite angles.
(ii) ∠4 and ∠7 are alternate angles.
(iii) ∠10 and ∠12 are vertically opposite angles.
(iv) ∠7 and ∠13 are corresponding angles.
(v) ∠6 and ∠8 are vertically opposite angles.
(vi) ∠11 and ∠8 are allied or co-interior angles.
(vii) ∠7 and ∠9 are vertically opposite angles.
(viii) ∠4 and ∠5 are adjacent angles.
(ix) ∠4 and ∠6 are allied or co-interior angles.
(x) ∠6 and ∠7 are adjacent angles.
(xi) ∠2 and ∠13 are allied or co-interior angles.

Question 3.
In the given figures, the arrows indicate parallel lines. State which angles are equal. Give reasons.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 70
Solution:
In the figure (i),
a = b (corresponding angles)
b = c (vertically opposite angles)
a = c (alternate angles)
∴ a = b = c
(ii) In the figure (ii),
x =y (vertically opposite angles)
y=l (alternate angles)
x = I (corresponding angles)
1 = n (vertically opposite angles)
n = r (corresponding angles)
∴ x = y = l = n = r
Again m = k (vertically opposite angles)
k = q (corresponding angles)
∴ m = k = q

Question 4.
In the given figure, find the measure of the unknown angles :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 71
Solution:
a = d (vertically opposite angles)
d=f (corresponding angles)
f= 110° (vertically opposite angles)
∴ a = d = f = 110°
e + 110° = 180° (co-interior angles)
∴ e = 180°- 110° = 70°
b = c (vertically opposite angles)
b = e (corresponding angles)
e = g (vertically opposite angles)
∴ b = c = e = g = 70° ”
Hence a = 110°, b = 70°, e = 70°, d = 110°, e = 70°,f= 110° and g = 70°

Question 5.
Which pair of the dotted line, segments, in the following figures, are parallel. Give reason:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 72

Solution:
(i) In figure (i), If lines are parallel, then 120°+ 50° =180°
But there are co-interior angles
⇒170° = 180°. But it not true Hence, there are not parallel lines
(ii) In figure (ii),
∠1 = 45° (vertically opposite angles)
Lines are parallel if
∠1 + 135° = 180° (co-interior angles)
⇒45°+ 135°= 180°
⇒ 180° = 180° which is true.
Hence, the lines are parallel.
(iiii) In figure (iii),
Lines are parallel if corresponding angles are equal
If 120° =130° which is not correct
∴ Lines are not parallel.
(iv) ∠1 = 110° (vertically opposite angles)
If lines are parallel then
∠1 + 70° = 180° (co-interior angles)
⇒110° + 70°= 180°
⇒180° =180°
Which is correct.
∴ Lines are parallel.
(v) ∠1 + 100°= 180°
⇒∠1 = 180°- 100°= 80 (linear pair)
Lines l1 and l2 will be parallel If ∠1 = 70°
⇒ 80° = 70° which is not true
∴ l1 and 12 are not parallel Again, A, l3and l5 will be parallel
If 80° = 70° (corresponding angle)
Which is not true.
∴l3 and l5 are not parallel
But ∠1 = 80° (alternate angles)
⇒ 80° = 80°
Which is true
∴ l2 and l4 are parallel
(vi) Lines are parallel
If alternate angles are equal
⇒ 50° = 40°
Wliich is not ture lines are not parallel.

Question 6.
In the given figures, the directed lines are parallel to each other. Find the unknown angles.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 74

Solution:

Question 7.
Find x. y and p is the given figures
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 82
Solution:

Question 8.
Find x in the following cases :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 85

Lines and Angles Exercise 14C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Using ruler and compasses, construct the following angles :
(i)30°
(ii)15°
(iii) 75°
(iv) 180°
(v) 165°
(vi) 22.5°
(vii) 37.5°
(viii) 67.5°
Solution:
(i) 30°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P and with same radius cut off the arc at Q.
(iv) Now with centre P and Q draw two arcs intersecting each other at R.
(v) Join BR and produce it to A, forming ZABC
= 30°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 90

(ii) (15°)
Steps of Construction:
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P and with same radius cut off the arc at Q.
(iv) Taking P and Q as curves, draw two arcs intersecting each other at D nnd join BD.
(v) With centre P and R, draw two more arcs intersecting each other at S.
(vi) Join BS and produce it to A.
Then ∠ABC = 15°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 91

(iii) 75°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius draw an arc and cut off PQ, then QR of the same radius.
(iii) With centre Q and R, draw two arcs intersecting each other at S.
(iv) Join SB.
(v) With centre Q and D draw two arcs intersecting each other at T.
(vi) Join BT and produce it to A.
Then ∠ABC = 75°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 92

(iv) 180°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius draw arc meeting BC at P.
(iii) With centre P and with same radius cut of arcs PQ, QR and then RS.
(iv) Join BS and produce it to A.
Then ∠ABC = 180°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 93

(v) 165°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius draw an arc meeting BC at P.
(iii) With centre P and same radius cut off arcs PQ, QR and then RS.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 94
(iv) Join SB.
(v) With centres R and S, draw two arcs intersecting each other at M.
(vi) With centre T and S draw two arcs intersecting each other at L.
(vi) Join BL and produce it to A. Then ∠ABC = 165°

(vi) 22.5°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 95
(iii) With centre P and some radius, cut off arcs PQ.
(iv) Bisect arc PQ at R and join BR.
(v) Bisect arc QR at S and join BS.
(vii) Now bisect arc PR at T.
(viii) Join BT and produce it to A.
Then ∠ABC = 22 \(\frac { 1 }{ 2 }\) ° or 22.5°.

(vii) 37.5°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
(iii) With centre P and same radius cut off arcs PQ and QR.
(iv) Now bisect arc QR at S and again bisect arc QS at T.
(v) Bisect arc PT at K.
(vi) Join BK and produce it to A.
Then, ∠ABC – 37 \(\frac { 1 }{ 2 }\) °or 37-5.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 96

(viii) 67.5°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
(iii) With centre P and with same radius, cut arcs PQ and then QR.
(iv) Bisect arc QR at K and again bisect arc QK at S.
(v) Bisect again arc SQ at T.
(vi) Join BT and produce it to A.
Then ∠ABC = 67\(\frac { 1 }{ 2 }\) ° or 67.5°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 97

Question 2.
Draw ∠ABC = 120°. Bisect the angle using ruler and compasses only. Measure each 1 angle so obtained and check whether the angles obtained on bisecting ∠ABC are equal or not.
Solution:
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 98
(iii) With centre P and with same radius, cut arcs PQ and QR.
(iv) Join BR and produce it to A.
Then ∠ABC = 120°
(v) With centres P and R, draw two arcs intersecting each other at S.
(vi) Join BS and produce it to D. BD is the bisector of ∠ABC.
On measuring each angle, it is of 60° each. Yes, both angles are equal in measure.

Question 3.
Draw a line segment PQ = 6 cm. Mark a point A in PQ so that AP = 2 cm. At point A, construct angle QAR = 60°.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 6 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 99
(ii) Mark a point A on PQ so that AP = 2 cm.
(iii) With centre A and some suitable radius draw an arc meeting AQ at C.
(iv) With centre C and with same radius, cut arc CB.
(v) Join AB and produce it to R.
Then ∠QAR = 60°

Question 4.
Draw a line segment AB = 8 cm. Mark a point P in AB so that AP = 5 cm. At P, construct angle APQ = 30°.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 8 cm.
(ii) Mark a point P in AB such that AP = 5 cm.
(iii) With centre P and some suitable radius, draw an arc meeting AB in L.
(iv) With centre L and same radius cut arc LM.
(v) Bisect arc LM at N.
(vi) Join PN and produce it to Q.
Then ∠APQ = 30°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 100

Question 5.
Construct an angle of 75° and then bisect it.
Solution:
Steps of Construction :
(i) Draw a line segment BC.
(ii) At B, draw an angle ABC equal to 75°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 101
(iii) With centres P and T, draw arcs intersecting each other at L.
(iv) Join BL and produce it to D. Then BD bisects ∠ABC.

Question 6.
Draw a line segment of length 6 .4 cm. Draw its perpendicular bisector.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6.4 cm.
(ii) With centres A and B and with some suitable radius, draw arcs intersecting each other at S and R.
(iii) Join SR intersecting AB at Q. Then PQR is the perpendicular bisector of line segment AB
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 102

Question 7.
Draw a line segment AB = 5.8 cm. Mark a point P in AB such that PB = 3.6 cm. At P, draw perpendicular to AB.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) Mark a point P in AB such that PB = 3.6 cm.
(iii) With centre P and some suitable radius draw an arc meeting AB in L.
(iv) With centre L and same radius cut arcs LM and then as N.
(v) Bisect arc MN at S.
(vi) Join PS and produce it to Q. Then PQ is perpendicular to AB at P.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 103

Question 8.
In each case, given below, draw a line through point P and parallel to AB :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 104
Solution:
Steps of construction :
(i) From P. draw a line segment meeting AB at
(ii) With centre Q and some suitable radius draw an arc CD.
(iii) With centre P and same radius draw another arc meeting PQ at E.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 105
(iv) With centre E and radius equal to CD, cut this arc at F

(v) Join PF and produce it to both sides to L and M. Then line LM is parallel to given line AB.

Selina Concise Physics Class 8 ICSE Solutions – Sound

November 9, 2023May 27, 2022 by Raju

Selina Concise Physics Class 8 ICSE Solutions – Sound

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 7 Sound. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Physics Chapter 7 Sound

SOUND: “Is energy which produces in us the sensation of hearing.” It is produced by vibration of body.
Sound needs a medium for its propagation. Sound cannot travel in vacuum.
Speed of sound is maximum in solids. 5000 ms^-1 in steel, in water 1500 ms^-1 and in air it is least 330 ms^-1 nearly.
When a body vibrates, the particles of medium also start vibrating and K.E. of particles changes into potential energy and P.E. into
K.E. This is why sound in energy.
Sound travels in a medium in the form of wave.
Longitudinal wave : When the particles of medium move in the direction of motion of wave by forming compression and rarefaction.
AMPLITUDE : “The maximum displacement of the particle of medium on either side of mean position.”
TIME PERIOD: “The time taken by a particle of medium to complete its one vibration” “t”
FREQUENCY: “The number of vibrations made by a particle of
the medium in one second. ƒ measured in Hertz (Hz)
FREQUENCY ƒ = 1/ t or t = 1 / ƒ
WAVE LENGTH: “The distance travelled by the wave in one one time period of vibration of particle of medium.”
Or
“The distance between two consecutive compressions or between two consecutive rarefactions.” It is denoted by ‘ λ ’ and S.I. unit of wave length is metre (m).
CHARACTERISTICS OF SOUND :
(i) Loudness.
(ii) Pitch (or shrillness).
(iii) Quality (or timbre or wave form).
LOUDNESS : is the characteristic of sound by virtue of which a loud sound can be distinguished from a faint sound, both having same frequency and same wave form.
It depend on: (i) Amplitude of wave (ii) Surface area of vibrating body (ii) Distance from the source of sound (iv) Sensitivity of listener: Unit of loudness is (dB) decibel.
PITCH: It depends on number of vibrations per second or frequency : more frequency is high pitch shrilled sound and low frequency is flat sound.
QUALITY: is the characteristic which distinguishes two sounds’of the same pitch and same loudness. It depends on wave form.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) When sound propagates in air, it does not carry energy with it.
Answer. False.

(b) In a longitudinal wave, compression and rarefaction are formed.
Answer. True.

(c) The distance from one compression to nearest rarefaction is called wavelength.
Answer. False.

(d) The frequency is measured in second.
Answer. False.

(e) The quality of a sound depends on the amplitude of wave.
Answer. False.

(f) The pitch of sound depends on frequency.
Answer. True.

(g) Decibel is the unit of pitch of a sound.
Answer. False.

2. Fill in the blanks

(a) The time period of a wave is 2 s. Its frequency is 0.5 S^-1.
(b) The pitch of a stringed instrument is increased by increasing tension in string.
(c) The pitch of a flute is decreased by increasing length of air column.
(d) Smaller the membrane, higher is the pitch.
(e) If a drum is beaten hard, its loudness increases.
(f) A tuning fork produces sound of single frequency.

3. Match the following

4. Select the correct alternative

(a) Sound can not travel in

solid
liquid
gas
vacuum

(b) When sound travels in form of a wave

the particles of medium move from the source to the listener
the particles of medium remains stationary
the particles of medium start vibrating up and down
the particles of medium transfer energy without leaving their mean positions.

(c) The safe limit of loudness of audible sound is

0 to 80 dB
above 80 dB
120 dB
above 120 dB

(d) The unit of loudness is

cm
second
hertz
decibel

(e) In a piano, pitch is decreased by

using thicker string
increasing tension
reducing length of string
striking it hard Ans.

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

November 9, 2023May 27, 2022 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 7 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Integers Exercise 1A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Evaluate:

427 x 8 + 2 x 427
394 x 12 + 394 x (-2)
558 x 27 + 3 x 558

Solution:

427 x 8 + 2 x 427 = 427 x (8 + 2) (Distributive property)
= 427 x 10
= 4270
394 x 12 + 394 x (-2) = 394 x (12-2) (Distributive property)
= 394 x 10
= 3940
558 x 27 + 3 x 558 = 558 x (27 + 3) (Distributive property)
= 558 x 30
= 16740

Question 2.
Evaluate:

673 x 9 + 673
1925 x 101 – 1925

Solution:

673 x 9 + 673 = 673 x (9 + 1) (Distributive property) = 673 x 10 = 6730
1925 x 101 – 1925 = 1925 x (101 – 1) (Distributive property) = 1925 x 100 = 192500

Question 3.
Verify:

37 x {8 +(-3)} = 37 x 8 + 37 x – (3)
(-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19
{7 – (-7)} x 7 = 7 x 7 – (-7) x 7
{(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)

Solution:

37 x {8 + (-3)} = 37 x 8 + 37 x – (3)
L.H.S. = 37 x {8 + (-3)}
= 37 x {8-3}
= 37 x {5}
= 37 x 5
= 185
R.H.S. = 37 x 8 + 37 – 3
= 37 x (8 – 3)
= 37 x 5
= 185
Hence, L.H.S. = R.H.S.
(-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19
L.H.S. = (-82) x {(_4) + 19}
= (-82) x {-4 + 19}
= (-82)x {15}
= -82 x 15
=-1230
R.H.S. = (-82) x (-4) + (-82) x 19
= -82 x (-4 + 19)
= -82 x 15
=-1230
Hence, L.H.S. = R.H.S.
{7 – (-7)}. x 7 = 7 x 7 – (-1) x 7
L.H.S. = {7 – (-7)} x 7
= {7 + 7} x 7
= {14} x 7
= 14 x 7
= 98
R.H.S. = 7 x 7 – (-7) x 7
=7 x 7+7 x 7 =
7 x (7 + 7)
= 7 x (14)
= 98
Hence, L.H.S. = R.H.S.
{(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)
L.H.S. = {(-15)-8} x-6
= {-15-8} x-6
= {-23} x-6
= -23 x- 6
= 138
R.H.S. = (-15) x (-6) – 8 x (-6)
= -6 x (-15-8)
= -6 x -23
= 138
Hence, L.H.S. = R.H.S.

Question 4.
Evaluate:

15 x 8
15 x (-8)
(-15) x 8
(-15) x -8

Solution:

15 x 8= 120
15 x (-8) = -120
(-15) x 8 = -120
(-15) x -8 = 120
(Since the number of negative integers in the product is even)

Question 5.
Evaluate:

4 x 6 x 8
4 x 6 x (-8)
4 x (-6) x 8
(-4) x 6 x 8
4 x (-6) x (-8)
(-4) x (-6) x 8
(-4) x 6 x (- 8)
(-4) x (-6) x (-8)

Solution:

4 x 6 x 8 = 192
4 x 6 x (-8) = -192
(It have one negative factor)
4 x (-6) x 8 = -192
(It have one negative factor)
(-4 )x 6 x 8 = -192
(It have one negative factor)
4 x (-6) x (-8) = 192
(It have two negative factors)
(-4) x (-6) x 8 = 192
(It have two negative factors)
(-4) x 6 x (-8) = 192
(It have two negative factors)
(-4) x (-6) x (-8) = -192
(It have three negative factors)

Question 6.
Evaluate:

2 x 4 x 6 x 8
2 x (-4) x 6 x 8
(-2) x 4 x (-6) x 8
(-2) x (-4) X 6 x (-8)
(-2) x (-4) x (-6) x (-8)

Solution:

2 x 4 x 6 x 8 = 384
2 x (-4) x 6 x 8 = -384
(Number of negative integer in the product is odd)
(-2) x 4 x (-6) x 8 = 384
(Number of negative integer in the product is even)
(-2) x (-4) x 6 x (-8) = -384
(Number of negative integer in the product is odd)
(-2) x (-4) x (-6) x (-8) = 384
(Number of negative integer in the product is even)

Question 7.
Determine the integer whose product with ‘-1’ is:

-47
63
-1
0

Solution:

-1 x 47 = -47
Hence, integer is 47
-1 x -63 = 63
Hence, integer is -63
-1 x 1 = -1
Hence, integer is 1
-1 x 0 = 0
Hence, integer is 0

Question 8.
Eighteen integers are multiplied together. What will be the sign of their product, if:

15 of them are negative and 3 are positive?
12 of them are negative and 6 are positive?
9 of them are positive and the remaining are negative?
all are negative?

Solution:

Since out of eighteen integers, 15 of them are negative, which is odd number. Hence, sign of product will be negative (-).
Since out of eighteen integers 12 of them are negative, which is even number. Hence sign of product will be positive (+).
Since out of eighteen integers 9 of them are negative, which is odd number. Hence, sign of product will be negative (-).
Since all are negative, which is even number. Hence sign of product will be positive (+).

Question 9.
Find which is greater?

(8 + 10) x 15 or 8 + 10 x 15
12 x (6 – 8) or 12 x 6 – 8
{(-3) – 4} x (-5) or (-3) – 4 x (-5)

Solution:

(8 + 10) x 15 or 8 + 10 x 15
(8 + 10) x 15 = 18 x 15 = 270
8 + 10 x 15 = 8 + 150 = 158
∴(8 + 10) x 15 > 8 + 10 x 15
12 x (6 – 8) or 12 x 6 – 8
12 x (6 – 8) = 12 (-2) = -24
12 x 6 – 8 = 72 – 8 = 64
∴12 x 6 – 8 > 12 x (6-8)
{(-3) – 4} x (-5) or (-3) – 4 x (-5)
{(-3) – 4} x (-5) = {-3 – 4} x (-5) = -7 x -5 = 35
(-3) – 4 x (-5) = -7 x (-5) = 35
∴{(-3) – 4} x (-5) = (-3) – 4 x (-5)

Question 10.
State, true or false :

product of two integers can be zero.
product of 120 negative integers and 121 positive integers is negative.
a x (b + c) = a x b + c
(b – c) x a=b – c x a

Solution:

False.
False.
Correct : Since 120 integers are even numbers, hence product will be positive and for 121 integers are positive in numbers, hence product will be positive.
False.
Correct :a x (b + c) ≠ a x b + c
ab + ac ≠ ab + c
False.
Correct: (b – c) x a ≠ b – c x a
ab – ac ≠ b – ca

Integers Exercise 1B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Divide:
(i) 117 by 9
(ii) (-117) by 9
(iii) 117 by (-9)
(iv) (-117) by (-9)
(v) 225 by (-15)
(vi) (-552) ÷ 24
(vii) (-798) by (-21)
(viii) (-910) ÷ – 26

Solution :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 3

Question 2.
Evaluate:
(i) (-234) ÷ 13
(ii) 234 ÷ (-13)
(iii) (-234) ÷ (-13)
(iv) 374 ÷ (-17)
(v) (-374) ÷ 17
(vi) (-374) ÷ (-17)
(vii) (-728) ÷ 14
(viii) 272 ÷ (-17)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 4
Question 3.
Find the quotient in each of the following divisions:
(i) 299 ÷ 23
(ii) 299 ÷ (-23)
(iii) (-384) ÷ 16
(iv) (-572) ÷ (-22)
(v) 408 ÷ (-17)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 5

Question 4.
Divide:
(i) 204 by 17
(ii) 152 by-19
(iii) 0 by 35
(iv) 0 by (-82)
(v) 5490 by 10
(vi) 762800 by 100

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 7

Question 5.
State, true or false :

0 ÷ 32 = 0
0 ÷ (-9) = 0
(-37) ÷ 0 = 0
0 ÷ 0 = 0

Solution:

True.
True.
False.
Correct: It is not meaningful (defined)
False.
Correct: It is not defined.

Question 6.
Evaluate:
(i) 42 ÷ 7 + 4
(ii) 12+18 ÷ 3
(iii) 19 – 20 ÷ 4
(iv) 16 – 5 x 3+4
(v) 6 – 8 – (-6) ÷ 2
(vi) 13 -12 ÷ 4 x 2
(vii) 16 + 8 ÷ 4- 2 x 3
(viii) 16 ÷ 8 + 4 – 2 x 3
(ix) 16 – 8 + 4 ÷ 2 x 3
(x) (-4) + (-12) ÷ (-6)
(xi) (-18) + 6 ÷ 3 + 5
(xii) (-20) x (-1) + 14 – 7

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers image - 1

Integers Exercise 1C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Evaluate:
18-(20- 15 ÷ 3)
Solution:
18-(20- 15 ÷ 3)
= 18 – \(\left( 20\quad -\frac { 15 }{ 5 } \right)\)
= 18 – (20 – 5)
= 18 – 20 + 5
= 18 + 5 – 20
= 23 – 20
= 3

Question 2.
-15+ 24÷ (15-13)
Solution:
-15+ 24÷ (15- 13)
= -15 + 24 ÷ 2
= -15 + 12
= -3

Question 3.
35 – [15 + {14-(13 + \(\overline { 2-1+3 }\))}]
Solution:
35- [15 + {14-(13 + \(\overline { 2-1+3 }\))}]
= 35-[15+ 14-(13+4)]
= 35 — [15 + 14 – (13 + 4}]
= 35-{15 + 14-17]
= 35-15-14+ 17
= 35 + 17-15-14
= 52 – 29
= 23

Question 4.
27- [13 + {4-(8 + 4 – \(\overline { 1+3 }\))}]
Solution:
27- [13 + {4-(8 + 4 – \(\overline { 1+3 }\))}]
= 27-[13 +{4-(8+ 4-4)}]
= 27-[13 + {4-8}]
= 27 – [13 + (-4)]
= 21 – [9]
= 27-9
= 18

Question 5.
32 – [43-{51 -(20 – \(\overline { 18 -7 }\))}]
Solution:
32 – [43 – {51 – (20 – \(\overline { 18 -7 }\))}]
= 32-[43 – {51 -(20- 11)}]
= 32-[43-{51 -9}]
= 32-[43 -42]
= 32-1
=31

Question 6.
46-[26-{14-(15-4÷ 2 x 2)}]
Solution:
46 – [26 – {14 – (15 – 4 ÷ 2 x 2)}]
= 46-[26- {14-(15-2 x 2)}]
= 46-[26- {14-(15 -4)}]
= 46-[26- {14- 11}]
= 46 – [26 – 3]
= 46 – 23
= 23

Question 7.
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
Solution:
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
= 45-[38- {60 ÷ 3-(6-3)÷ 3}]
= 45-[38 -{20-3 ÷ 3}]
= 45-[38- {20-1}]
= 45-[38- 19]
= 45-19
= 26

Question 8.
17- [17 — {17 — (17 – \(\overline { 17 -17 }\))}]
Solution:
17- [17-{17-(17 –\(\overline { 17 -17 }\))}]
= 17-[17-{17-(17-0)}]
= 17 – [17 – {17 — 17}]
= 17 — [17 — 0]
= 17-17
= 0

Question 9.
2550 – [510 – {270 – (90 – \(\overline { 80 + 7 }\))}]
Solution:
2550- [510-{270-(90-\(\overline { 80 + 7 }\))}]
= 2550 – [510 – {270 – (90 – 87)}]
= 2550 -[510- {270 -3}]
= 2550-[510-267]
= 2550 – 243
= 2307

Question 10.
30+ [{-2 x (25-\(\overline { 13 -3 }\))}]
Solution:
30+ [{-2 x (25-\(\overline { 13 -3 }\))}]
= 30 + [{-2 x (25 – 10)}]
= 30 + [{-2 x 15}]
= 30 + [-30]
= 30-30
= 0

Question 11.
88-{5-(-48)+ (-16)}
Solution:
88- {5-(-48)+ (-16)}
=88 – \(\left\{ 5-\frac { (-48) }{ -16 } \right\}\)
= 88 – {5-3}
= 88 – 2
= 86

Question 12.
9 x (8-\(\overline { 3 +2 }\)) – 2 (2 + \(\overline { 3 +3 }\))
Solution:
9 x (8-\(\overline { 3 +2 }\)) -2(2 + \(\overline { 3 +3 }\))
= 9 x (8 – 5) – 2(2 + 6)
= 9 x 3 – 2 x 8
= 27- 16
= 11

Question 13.
2 – [3 – {6 – (5 – \(\overline { 4 -3 }\))}]
Solution:
2 – [3 – {6 – (5 – \(\overline { 4 -3 }\))}]
⇒ 2 – [3 – {6 – (5 – 1)}]
⇒ 2 – [3 – {6 – 4}]
⇒2 – (3 – 2)
⇒2-1 = 1

Integers Exercise 1D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
The sum of two integers is -15. If one of them is 9, find the other.
Solution:
Sum of two integers = -15
One integer = 9
∴ Second integer = -15 – 9
= -(15 + 9)
= -24

Question 2.
The difference between an integer and -6 is -5. Find the values of x.
Solution:
The difference between an integer
= x-(-6) = -5
∴ Value of
⇒ x – (-6) = -5
⇒ x + 6 = -5
x = -5 – 6
x = -11

Question 3.
The sum of two integers is 28. If one integer is -45, find the other.
Solution:
Sum of two integers = 28
One integer = -45
∴ Second integer = 28 – (-45)
= 28 + 45
= 73

Question 4.
The sum of two integers is -56. If one integer is -42, find the other.
Solution:
Sum of two integers = -56
One integer = -42
∴Second integer = -56 – (-42)
= -56+ 42
=-14

Question 5.
The difference between an integer x and (-9) is 6. Find all possible values ofx.
Solution:
The difference between an integer x – (-9) = 6 or -9 – x = 6
∴ Value of x
⇒ x – (-9) = 6 or ⇒ -9 – x = 6
⇒ x + 9 = 6 or Answer-x = 6 + 9
⇒ x = 6 – 9 or ⇒ -x = 15
⇒x = -3 or ⇒ x = -15
Hence, possible values ofx are -3 and -15.

Question 6.
Evaluate:

(-1) x (-1) x (-1) x ….60 times.
(-1) x (-1) x (-1) x (-1) x …. 75 times.

Solution:

1 (because (-1) is multiplied even times.)
-1 (because (-1) is multiplied odd times.)

Question 7.
Evaluate:

(-2) x (-3) x (-4) x (-5) X (-6)
(-3) x (-6) x (-9) x (-12)
(-11) x (-15) + (-11) x (-25)
10 x (-12) + 5 x (-12)

Solution:

(-2) x (-3) x (-4) x (-5) x (-6)
⇒ 6 x 20 x (-6) = 120 x (-6)
= -720
(-3) x (-6) x (-9) x (-12)
⇒ 18 x 108
= 1944
(-11) x (-15) + (-11) x (-25)
⇒ 165 + 275
= 440
10 x (-12) + 5 x (-12)
⇒ -120-60
= -180

Question 8.

If x x (-1) = -36, is x positive or negative?
If x x (-1) = 36, is x positive or negative?

Solution:

x x (-1) = -36
-lx = -36
x = \(\frac { -36 }{ -1 }\)
x = 36
∵ x = 36
∴ It is a positive integer.
x x (-1) = 36
-1x = 36
x = \(\frac { 36 }{ -1 }\)
x = -36
∵x = -36
∴It is a negative integer.

Question 9.
Write all the integers between -15 and 15, which are divisible by 2 and 3.
Solution:
The integers between -15 and 15 are :
-12, -6, 0, 6 and 12
That are divisible by 2 and 3.

Question 10.
Write all the integers between -5 and 5, which are divisible by 2 or 3.
Solution:
The integers between -5 and 5 are :
-4, -3, -2, 0, 0, 2, 3 and 4
That are divisible by 2 or 3.

Question 11.
Evaluate:

(-20) + (-8) ÷ (-2) x 3
(-5) – (-48) ÷ (-16) + (-2) x 6
16 + 8 ÷ 4- 2 x 3
16 ÷ 8 x 4 – 2 x 3
27 – [5 + {28 – (29 – 7)}]
48 – [18 – {16 – (5 – \(\overline { 4 +1 }\))}]
-8 – {-6 (9 – 11) + 18 = -3}
(24 ÷ \(\overline { 12 -9 }\) – 12) – (3 x 8 ÷ 4 + 1)

Solution:
We know that, if these type of expressions that has more than one fundamental operations, we use the rule of DMAS i.e., First of all we perform D (division), then M (multiplication), then A (addition) and in the last S (subtraction).

(-20) + (-8) ÷ (-2) x 3
⇒ -20 + 4 x 3
⇒ -20+ 12
=-8
(-5) – (-48) ÷ (-16) + (-2) x 6
⇒ (-5) – 3 + (-2) x 6
⇒ -5 – 3 – 12
⇒ -8- 12
= -20
16 + 8 ÷ 4 – 2 x 3
⇒ 16 + 2 – 2 x 3
⇒16 + 2 – 6
⇒ 18-6
= 12
16 ÷ 8 x 4 – 2 x 3
⇒ 2 x 4 – 2 x 3
⇒ 8 – 6
= 2
27 – [5 + {28 – (29 – 7)}]
⇒ 27 – [5 + {28 – 22}]
⇒ 27 – [5 + 6]
⇒ 27 — 11
= 16
48-[18-{16-(5 – \(\overline { 4 +1 }\))}]
⇒ 48-[18-{16-(5-5)}]
⇒ 48-[18- {16-0)}]
⇒ 48-[18- 16]
⇒ 48 – 2
= 46
-8 – {-6 (9 – 11) + 18 ÷ -3}
⇒ -8 – {-6 (-2) – 6}
⇒ -8- {12-6}
⇒ -8 – {6}
⇒ -8-6
= -14
(24 ÷ \(\overline { 12 -9 }\) – 12) – (3 x 8 = 4 + 1)
⇒ (24 ÷ 3-12)-(3 x 2 + 1)
⇒ (8- 12)-(6+ 1)
⇒ —4 — 7
= —11

Question 12.
Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.
Solution:
Required number = (Sum of all integers between 20 and 30 – Integers between 20 and 30)
(20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) – (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 )
⇒ 20 + 30 = 50
∴ Required number = 50

Question 13.
Add the product of (-13) and (-17) to the quotient of (-187) and 11.
Solution:
(-13) x (-17)+ (-187- 11)
⇒ (-13) x (-17) + (-17)
⇒ 221 – 17 = 204

Question 14.
The product of two integers is-180. If one of them is 12, find the other.
Solution:
The product of two integers = -180 One integer = 12
∴ Second integer = -180 – 12 = -15

Question 15.

A number changes from -20 to 30. What is the increase or decrease in the number?
A number changes from 40 to -30. What is the increase or decrease in the number?

Solution:

∵A number changes from = -20 to 30
⇒ -20 – 30 = -50
∴-50, it will be increases.
∵A number changes from = 40 to -30
⇒ 40 – (-30)
40 + 30 = 70
∴70, it will be decreases

Selina Concise Biology Class 7 ICSE Solutions – Plant And Animal Tissues

November 9, 2023May 27, 2022 by Raju

Selina Concise Biology Class 7 ICSE Solutions – Plant And Animal Tissues

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

A PlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Biology. You can download the Selina Concise Biology ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Biology for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Biology Chapter 1 Plant And Animal Tissues

Synopsis

The unit of level of organisation is independent in its mode of existence and activity.
All multi cellular organisms start their life as a single cell.
Plant tissues are basically of two type
1. meristematic
2. permanent or non-dividing
The permanent – plant tissues are further of three types
1. protective
2. supportive: parenchyma, collenchyma, sclerenchyma
3. conductive: xylem, phloem
Parenchymal cells have thin-walled cells and usually with a vacuole.
Potatoes mainly are composed of parenchymal cells.
Collenchyma are parenchymatous cells which are elongated and are thick at the comers. This helps to support the parts of the plant.
Sclerenchyma tissue is formed of long, narrow and thick cells. This provides strength to the plant parts.
Xylem is formed of thick-walled, tubular and often dead cells. They transport water and minerals absorbed by roots.
Old xylem forms the wood.
Phloem is formed of living tubular cells which provide a passage for the downward transport of food.
The four major groups of animal tissues
1. epithelial tissue
2. connective tissue
3. muscular tissue
4. nervous tissue
The epithelial tissue is further of four types:
1. squamous epithelium (protective)
2. cuboidal epithelium (absorption)
3. columnar epithelium (secretory)
4. ciliated epithelium (movement of substances)
Supportive connective tissue consists of
1. Cartilage
2. Bone
Fibrous connective tissue:
It packs and binds most of the organs. It is of the following types.
1. areolar tissue: binds skin to underlying tissue.
2. adipose tissue: filled with fat.
3. tendon: connect muscles to bones.
4. ligaments: connect bone to another bone.
Fluid connective tissue consists of
1. Blood
2. Lymph
The liquid part of the blood is called plasma and the cellular part includes:
1. red blood cells
2. white blood cells
3. platelets.
Three distinct kinds of muscles are
1. striated or skeletal
2. unstriated or smooth
3. cardiac or heart.
A nerve cell is formed of a cell body called cyton and one or more elongated hair-like extensions called dendrites. The longest dendrite is called axon.
Systems of the body with their primary vital function.Skeletal system: support and protection
1. Muscular system: movement
2. Digestive system: nutrition
3. Respiratory system: exchange of gases
4. Circulatory system: transport of materials
5. Excretory system: waste removal
6. Nervous system: sensation and co-ordination
7. Reproductive system: continuation of race.

Review Questions

MULTIPLE CHOICE QUESTIONS

1. Put a tick (✓) against the most appropriate alternative in the following statements.

(i) A group of similar cells to perform a specific function forms a
(a) organ
(b) species
(c) organ system
(d) tissue

(ii) The fine branches given out from the cell body of a nerve cell are
(a) dendrites
(b) cyton
(c) axon
(d) neurons

(iii) Fluid connective tissue of humans is
(a) blood and cartilage
(b) lymph and plasma
(c) blood and lymph
(d) stroma and matrix

Short Answer Questions

Question 1.
1. Define the following terms:

Tissue
Organ

Answer:

Tissue: A group of similar cells which perform a specific function.
example: Muscular tissue in animals.
Organ: The different type of tissues which group together to function in a co-ordinated manner.
example: liver

2. Answer the following:

Question 2(i).
What is a meristematic tissue ? How is it different from permanent tissues ?
Answer:
Plant tissues are classified into two types:

Meristematic tissue
Permanent or non-diving tissue

Meristematic tissues are the plant tissues which are made up of actively dividing cells. These tissues actively divide and lead to the growth of the plant body. They are found at the growth points of the plant like tips of root, stem and branches etc.

Cells are small with thin cell walls.
Cells have large and conspicuous nuclei.
Cells have no vacuoles.
Cells are actively dividing type cells.

Difference between Meristematic and permanent
Meristematic tissue :

Meristematic tissue is present at the tip of the root and stem and in between the xylem and phloem. Form apical meristematic tissue when present at the tips. It is in the form of cambium in between the xylem and phloem.
Meristematic cells divide and form other types of tissues. The cells are thin walled.
Meristematic cells may be intercalary as in case of monocots.
The cells are small and isodiametric, vacuoles are small or absent.
Respiratory and biosynthetic activities maximum.
The cells are immature and mitochondria simple.
Proplastids act as plastids.

Permanent tissue

Permanent tissue may be simple as parenchyma, collenchyma or sclerenchyma and it may be complex as xylem and phloem.
These are made up of more than one kind of cells. These perform a common function Xylem and phloem form vascular system of the plant. These cells do not have the power to divide.
These cells may act as epidermis cortex or grit cells. Sclerenchyma gives strength.
Living cells of permanent tissue have vacuoles. The cells are large and of different shapes.
Both these activities are low.
The cells fully mature, mitochondria fully developed.
Living cells have plastids.

Question 2(ii).
Which living material would you take to demonstrate meristematic tissue ?
Answer:
Green gram seeds can be used to demonstrate meristematic tissue which when soaked in a petridish stuffed with wet cotton and left for 3-4 days would sprout out. These sprouted seeds have roots developing whose root tips have meristematic tissue.

Question 2(iii).
What is the function of meristematic tissue ?
Answer:
The meristematic tissue have the primary role in the growth of the plant tissue as it consists of active dividing cells

Question 3.
State whether the following statements are True or False.

(i) A tissue is formed of only one type of cells.
Ans. True

(ii) Only one type of tissue forms an organ.
Ans. False.
Correct: Two or more types of tissue form an organ.

(iii) Permanent tissue is made up of undifferentiated and dividing Cells.
Ans. False.
Correct: Meristematic tissue is made up of undifferentiated and dividing cells.

(iv) Meristematic tissue is found at growing tips of a plant.
Ans. True

(v) Phloem is formed of dead tubular cells.
Ans. False.
Correct: Phloem is formed of living tubular cells.

Question 4.
Fill in the blanks by selecting suitable words from the list given below:
“Thin – walled, collenchyma, vascular, tissues, conducting”

A group of different tissues working together to perform a function is called an organ.
Xylem and phloem form the conducting tissue.
Conducting tissue is also called vascular tissue.
Cells are elongated and thick at the comers in collenchyma tissue.
Parenchyma is composed of large thin-walled cell

Question 5.
Match the items given is column A with those given in column B:

Column A
(i) Fibrous connective tissue
(ii) Fluid connective tissue
(iii) Supportive connective tissue
(iv) Ligament
(v) Tendon
Column B
(a) blood
(b) cartilage
(iii) Supportive connective tissue
another bone.
(d) areolar tissue
(e) connects a muscle
with a bone.
Selina Concise Biology Class 7 ICSE Solutions - Plant And Animal Tissues 1

Question 6.
How do you rank the following among cells, tissues, organs, or organism ?

Amoeba : organism
Euglena: organism
Skin : organ
Lungs : organ
Neuron : tissue
Cardiac muscles: Ti1ue

Question 7.
Each of the tissues listed in Column A is related to one of the functions given in Column B. Match the lines correct pairs by drawing

Question 8.
Name the kind of tissue that

Carries oxygen around your body — Blood tissue.
Brings about movements in animals — muscular tissue.
Transports food to different parts of plant— phloem.
Transports water in plants — xylem.
Supports an animal’s body — connective tissue (supportive)
Binds different tissues together — Fibrous connective tissue.
Conducts messages from one part of the body to another — nervous tissue.

Question 9.
Based on the following information, identify the three types of epithelial tissue in the figures given below :

(i) Cuboidal epithelium : It consists of a single layer of cuboidal cells.

(ii) Columnar epithelium: It is composed of tall, cylindrical cells with oval nuclei usually placed at the base of the cells.

(iii) Ciliated epithelium : It consists of cells being hair-like cilia on their free surface.
Answer:
(i) fig. b (ii) fig. a (iii) fig. c

Question 10.
Write three differences between the two principal vascular tissues found in plants.
Answer:
Xylem

Transports water and minerals absorbed by the roots to other plant parts.
Consists mainly of dead cells.
Conduction is unidirectional i.e. only upwards from the roots.

Phloem

Conducts food manufactured in the leaves to other plant parts.
Consists mainly of living cells.
Bidirectional conduction i.e. both upwards and downwards from the leaves.

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 1 Knowing Our Numbers Objective Type Questions

November 9, 2023May 27, 2022 by Prasanna

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 1 Knowing Our Numbers Objective Type Questions

Mental Maths
Question 1.
Fill in the blanks:
(i) The digit …………… has the highest place value in the number 2309.
(ii) The digit …………… has the highest face value in the number 2039.
(iii) The digit …………… has the lowest place value in the number 2039.
(iv) Both Indian and International systems of numeration have …………… period in common.
(v) In the International system of numeration, commas are placed from …………… after every …………… digits.
(vi) The bigger number from the numbers 57,631 and 57,361 is ……………
(vii) 1 crore = …………… million
(viii)The smallest 4-digit number with 3 different digits is ……………
(ix) The greatest 4-digit number with 3 different digits is ……………
(x) 15 km 300 m = …………… m
(xi) 7850 cm = …………… m …………… cm
(xii) The number 5079 when estimated to the nearest hundreds is ……………
Solution:
(i) The digit 2 has the highest place value in the number 2309.
(ii) The digit 9 has the highest face value in the number 2039.
(iii) The digit 0 has the lowest place value in the number 2039.
(iv) Both Indian and International systems of numeration have ones period in common.
(v) In the International system of numeration, commas are placed from right after every 3 digits.
(vi) The bigger number from the numbers 57,631 and 57,361 is 57,631.
(vii) 1 crore = 10 million (viii)The smallest 4-digit number with 3 different digits is 1002.
(ix) The greatest 4-digit number with 3 different digits is 9987.
(x) 15 km 300 m = 15300 m
(xi) 7850 cm = 78 m 50 cm
(xii) The number 5079 when estimated to the nearest hundreds is 5100.

Question 2.
State whether the following statements are true (T) or false (F):
(i) The difference between the place value and the face of the digit 7 in the number 2701 is 693.
(ii) The smallest 4-digit number -1 = the greatest 3-digit number.
(iii) The place of a digit is independent of whether the number is written in the Indian system or International system of numeration.
(iv) In the International system, a number having less number of digits is always smaller than the number having more number of digits.
(v) The estimated value of 9999 to the nearest tens is 10000.
Solution:
(i) The difference between the place value and the face of the digit 7 in the number 2701 is 693. True
(ii) The smallest 4-digit number-1 = the greatest 3-digit number. True
(iii) The place of a digit is independent of whether the number is written in the Indian system or International system of numeration.
True
(iv) In the International system, a number having less number of digits is always smaller than the number having more number of digits.
True
(v) The estimated value of 9999 to the nearest

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 17):
Question 3.
The face value of the digit 5 in the number 36503 is
(a) 5
(b) 503
(c) 500
(d) none of these
Solution:
The place value of 5 at hundred’s place
= 5 × 100 = 500 (c)

Question 4.
The difference between the place values of 6 and 3 in 76834 is
(a) 3
(b) 5700
(c) 5930
(d) 5970
Solution:
The place value of 6 at thousand’s place
= 6 × 1000 = 6,000
The place value of 3 at ten’s place
= 3 × 10 = 30
The difference between the place value of 6 and 3 = 6000 – 30 = 5970 (d)

Question 5.
The sum of the place values of all the digits in 5003 is
(a) 8
(b) 53
(c) 5003
(d) 8000
Solution:
The place value of 3 at one’s place
=3 × 1 = 3
The place value of 0 at ten’s place = 0 × 10 = 0
The place value of 0 at hundred’s place = 0 × 100 = 0
The place value of 5 at thousand’s place = 5 × 1000 = 5000
The sum of the place value of all the digits
in 5003 = 3 + 0 + 0 + 5000 = 5003 (c)

Question 6.
The total number of 4-digit numbers is
(a) 9000
(b) 9999
(c) 10000
(d) none of these
Solution:
The greatest 3-digit number = 999 The greatest 4-digit number = 9999 .’. The total number of 4-digit numbers
= 9999 – 999 = 9000 (a)

Question 7.
The product of the place values of two-threes in 73532 is
(a) 9000
(b) 90000
(c) 99000
(d) 1000
Solution:
The place value of 3 at ten’s place = 3 × 10 = 30
The place value of 3 at thousand’s place = 3 × 1000 = 3000
The product of place value of two threes = 30 × 3000 = 90000 (b)

Question 8.
The smallest 4-digit number having distinct digits is
(a) 1234
(b) 1023
(c) 1002
(d) 3210
Solution:
The smallest 4-digit number having distinct digits is 1002. (c)

Question 9.
The largest 4-digit number having distinct digits is
(a) 9999
(b) 9867
(c) 9786
(d) 9876
Solution:
The largest 4-digit number having distinct digits is 9867. (b)

Question 10.
The largest 4-digit number is
(a) 9999
(b) 9876
(c) 9990
(d) none of these
Solution:
The largest 4-digit number is 9999. (a)

Question 11.
The difference between the largest number of 3-digit and the largest number of 3-digit with distinct digits is
(a) 0
(b) 10
(c) 12
(d) 14
Solution:
The largest number of 3-digit = 999
The largest number of 3-digit with distinct digits = 987
∴ Their difference = 999 – 987 = 12 (c)

Question 12.
If we write natural numbers from 1 to 100, the number of times the digit 5 has been written is
(a) 11
(b) 15
(c) 19
(d) 20
Solution:
If we write natural numbers from 1 to 100, the number of times the digit 5 has been written is 20. (d)

Question 13.
The number 28,549 when rounded off to the nearest hundreds is
(a) 28,000
(b) 28,500
(c) 28,600
(d) 29,000
Solution:
28,549
The digit at tens place is 4, which is less than 5.
Hence, the rounded off number to nearest hundreds = 28,500. (b)

Question 14.
The smallest natural number which when rounded off to the nearest hundreds as 500 is
(a) 499
(b) 501
(c) 450
(d) 549
Solution:
The smallest natural number which when rounded off to the nearest hundreds as 500 is 450. (c)
This is so because the digit at tens place is 5, which is equal to 5.

Question 15.
The greatest natural number which when rounded off to the nearest hundreds as 500 is
(a) 549
(b) 599
(c) 450
(d) none of these
Solution:
The greatest natural number which when rounded off to the nearest hundreds as 500 is 549. (a)
This is so because the digit at tens place is 4, which is less than 5.

Question 16.
The greatest 5-digit number formed by the digits 3, 0, 7 is
(a) 33077
(b) 77730
(c) 77330
(d) none of these
Solution:
The greatest 5-digit number formed by the digits 3, 0, 7 is 77730. (b)

Question 17.
In the International place value system, we write 1 billion for
(a) 10 lakh
(b) 1 crore
(c) 10 crore
(d) 100 crore
Solution:
In the International place value system, we write 1 billion for 100 crore. (d)

Value Based Questions

Question 1.
The distance between Anu’s home and her school is 4 km 850 m. Everyday she cycles both ways. Find the distance covered by her in a week. (Sunday being a holiday).
What are the advantages of cycling?
Solution:
Distance between Anu’s home and her school = 4 km 850 m = 4 x 1000+ 850 = 4850 m Distance travelled by Anu per day = 4850 m x 2 = 9700 m Since, in a week there are 7 days but Sunday is off.
Hence, distance travelled by Anu for 6 days (a week) = 9700 × 6 = 58200 m
= 58 km 200 m
Cycling is good for health and it saves fuel and helps in reducing pollution.

Higher Order Thinking Skills (HOTS)

Question 1.
Is there any digit whose place value is always equal to its face value irrespective of its position in any number?
Solution:
Yes, the digit is 0.

Question 2.
Write all 4-digit numbers that can be formed with the digits2 and 5, using both digits equal number of time. Also find their sum.
Solution:
Possible numbers are : 2255, 2552, 2525, 5225, 5252, 5522
and their sum = 2255 + 2552 + 2525 + 5225 + 5252 + 5522 = 23331

Thousand	Hundred	Tens	Ones
2	2	5	5
2	5	5	2
2	5	2	5
5	2	2	5
5	2	5	2
5	5	2	2

Question 3.
What is the difference between the smallest 6-digit number with five different digits and the greatest 5-digit number with four different digits?
Solution:
The smallest 6-digit number with five different digits = 100234.
The greatest 5-digit number with four different digits = 99876.
Their difference = 100234 – 99876 = 358

Question 4.
How many times does the digit 3 occur at tert’s place in natural numbers from 100 to 1000?
Solution:
90 times i.e. 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and upto 983, 993