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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 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 Mathematics class 7 ICSE Solutions – Data Handling

November 10, 2023May 23, 2022 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 21 Data Handling

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 21 Data Handling

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

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

Data Handling Exercise 21A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Consider the following numbers :
68, 76, 63, 75, 93, 83, 70, 115, 82, 105, 90, 103, 92, 52, 99, 73, 75, 63, 77 and 71.
(i) Arrange these numbers in ascending order.
(ii) What the range of these numbers?
Solution:
(i) When the above data are written in ascending order. We get,
52, 63, 63, 68, 70, 71, 73, 75, 75, 76, 77, 82, 83, 90, 92, 93, 99, 103, 105, 115
(ii) Range of given numbers = Largest number – Smallest number
= 115-52 = 48

Question 2.
Represent the following data in the form of a frequency distribution table :
16, 17, 21, 20, 16, 20, 16, 18, 17, 21, 17, 18, 19, 17, 15, 15, 19, 19, 18, 17, 17, 15, 15, 16, 17, 17, 19, 18, 17, 16, 15, 20, 16, 17, 19, 18, 19, 16, 21 and 17.
Solution:
The frequency distribution for these data will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 2

Question 3.
A die was thrown 20 times and following scores were recorded.
2, 1, 5, 2, 4, 3, 6, 1, 4, 2, 5, 1, 6, 2, 6, 3, 5, 4, 1 and 3.
Prepare a frequency table for the scores.
Solution:
The frequency table for the scores will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 3

Question 4.
Following data shows the weekly wages (in ₹) of 10 workers in a factory.
3500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 4250 and 4000
(i) Prepare a frequency distribution table.
(ii) What is the range of wages (in ₹)?
(iii) How many workers are getting the maximum wages?
Solution:
(i) The frequency table for the wages of 10 workers will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 4
(ii) Range of wages (₹) = ₹4750 – ₹3500 = ₹1250
(iii) One

Question 5.
The marks obtained by 40 students of a class are given below :
80, 10, 30, 70, 60, 50, 50, 40, 40, 20, 40, 90, 50, 30, 70, 10, 60, 50, 20, 70, 70, 30, 80, 40,20, 80, 90, 50, 80, 60, 70, 40, 50, 60, 90, 60, 40, 40, 60 and 60
(i) Construct a frequency distribution table.
(ii) Find how many students have marks equal to or more than 70?
(iii) How many students obtained marks below 40?
Solution:
(i) The frequency distribution table will be shown as below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 5
(ii)Students have marks equal to or more than 70 = 5 + 4 + 3 = 12
(iii) Students obtained marks below 40 = 2 + 3 + 3 = 8 students

Question 6.
Arrange the following data in descending order:
3.3, 3.2, 3.1, 3.7, 3.6, 4.0, 3.5, 3.9, 3.8, 4.1, 3.5, 3.8, 3.7, 3.9 and 3.4.
(i) Determine the range.
(ii) How many numbers are less than 3.5?
(iii) How many numbers are 3.8 or above?
Solution:
Descending order : 4.1, 4.0, 3.9, 3.9, 3.8, 3.8, 3.7, 3.7, 3.6, 3.5, 3.5, 3.4, 3.3, 3.2, 3.1
(i) Range = 4.1 – 3.1 = 1
(ii) Number less than 3.5 = 4
i.e., 3.4, 3.3, 3.2, 3.1
(iii) Number are 3-8 or above = 6
i.e., 3.8, 3.8, 3.9, 3.9, 4.0, 4.1

Data Handling Exercise 21B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find the mean of 53, 61, 60, 67 and 64.
Solution:
Mean of 53, 6i, 60, 67 and 64
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 6

Question 2.
Find the mean of first six natural numbers.
Solution:
First six natural numbers are : 1, 2, 3, 4, 5, 6
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 7

Question 3.
Find the mean of first ten odd natural numbers.
Solution:
First ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 8

Question 4.
Find the mean of all factors of 10.
Solution:
The factor of 10 are 2 and 5
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 9

Question 5.
Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.
Solution:
Mean of x + 3, x + 5, x + 7, x + 9 and x + 11
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 10

Question 6.
If different values of variable x are 19.8,15.4,13.7,11.71,11.8, 12.6,12.8,18.6,20.5 and 2.1, find the mean.
Solution:
19. +15.4 +13.7 +11.71 +11.8 +12.6 + 12.8 +18.6 + 20.5 +21.1
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 11

Question 7.
The mean of a certain number of observations is 32. Find the resulting mean, if each observation is,
(i) increased by 3
(ii) decreased by 7
(iii) multiplied by 2
(iv) divided by 0.5
(v) increased by 60%
(vi) decreased by 20%
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 12

Question 8.
The pocket expenses (per day) of Anuj, during a certain week, from monday to Saturday were ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25. Find the mean pocket expenses per day.
Solution:
The pocket expenses (per day) during a certain week are : ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 13

∴Anuj expenses per day = ₹85.75

Question 9.
If the mean of 8, 10, 7, x + 2 and 6 is 9, find the value of x.
Solution:
The mean 8, 10, 7, x + 2 and 6 is 9
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 14

Question 10.
Find the mean of first six multiples of 3.
Solution:
The six multiples of 3 are 3, 6, 9, 12, 15, 18
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 15

Question 11.
Find the mean of first five prime numbers.
Solution:
The first five prime numbers are 2, 3, 5, 7, 11
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 16

Question 12.
The mean of six numbers :x-5,x- 1, x, x + 2, x + 4 and x + 12 is 15. Find the mean of first four numbers.
Solution:
The mean of six numbers are x – 5, x – 1,x,x + 2,x + 4 and x + 12 is 15
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 17

Question 13.
Find the mean of squares of first five whole numbers.
Solution:
First five whole numbers are 0, 1, 2, 3, 4
Then square the whole prime numbers
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 19

Question 14.
If the mean of 6, 4, 7, p and 10 is 8, find the value of p.
Solution:
The mean of 6, 4, 7, p and 10 is 8
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 20

Question 15.
Find the mean of first six multiples of 5.
Solution:
Six multiples of 5 are :
5, 10, 15, 20, 25 and 30
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 21

Question 16.
The rainfall (in mm) in a city on 7 days of a certain week is recorded as follows
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 22
Find the total and average (mean) rainfall for the week.
Solution:
The rainfall in a city on 7 days are 0.5, 2.7, 2.6, 0.5, 2, 5.8, 1.5

Question 17.
The mean of marks scored by 100 students was found to be 40, later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
Solution:
Mean of 40 observations = 100
Total sum of 40 observations = 100 × 40 = 4000
Incorrect total of 40 observation is = 4000
Correct total of 40 observations = 4000 – 83 + 53 = 3970
∴ Correct mean = \(\frac { 3970 }{ 100 }\) = 39.70

Question 18.
The mean of five numbers is 27. If one number is excluded, the mean of remaining numbers is 25. Find the excluded number.
Solution:
Mean of 5 observations = 27
Total sum of 5 observations = 27 × 5 = 135
On excluding an observation, the mean of remaining 6 observations = 25
⇒ Total of remaining 4 observations = 25 x 4 = 100
⇒ Included observation = Total mean of 5 observations – Total mean of 4 observations
= 135- 100 = 35

Question 19.
The mean of 5 numbers is 27. If one new number is included, the new mean is 25. Find the included number.
Solution:
Mean of 5 observations = 27
Total sum of 5 observations = 27 x 5 = 135
On including an observation the mean of 6 observation = 25 x 6 = 150
⇒ Included observations = Total Mean of 6 observations – Total mean of 5 observations = 150- 135 = 15

Question 20.
Mean of 5 numbers is 20 and mean of other 5 numbers is 30. Find the mean of all the 10 numbers taken together.
Solution:
The mean of 5 number = 20
Then, mean of other 5 number = 30
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 24

Question 21.
Find the median of:
(i) 5,7, 9, 11, 15, 17,2, 23 and 19
(ii) 9, 3, 20, 13, 0, 7 and 10
(iii) 18, 19, 20, 23, 22, 20, 17, 19, 25 and 21
(iv) 3.6, 9.4, 3.8, 5.6, 6.5, 8.9, 2.7, 10.8, 15.6, 1.9 and 7.6.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 25

Question 22.
Find the mean and the mode for the following data :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 29
Solution:
We prepare the table given below :

Question 23.
Find the mode of:
(i) 5, 6, 9, 13, 6, 5, 6, 7, 6, 6, 3
(ii) 7, 7, 8, 10, 10, 11, 10, 13, 14
Solution:
(i) Arranging the Numbers in ascending order : 3, 5, 5, 6, 6, 6, 6, 6, 7, 9, 13
Mostly repeated term = 6
∴ Mode = 6
(ii) Arranging the Numbers in ascending order = 7, 7, 8, 10, 10, 10, 11, 13, 14
Mostly repeated term =10
∴ Mode = 10

Question 24.
Find the mode of :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 31

Solution:
(i) Since, the frequency of number 18 is maximum
∴Mode = 18
(ii) Since, the frequency of number 41 is maximum
∴ Mode = 41

Question 25.
The heights (in cm) of 8 girls of a class are 140,142,135,133,137,150,148 and 138 respectively. Find the mean height of these girls and their median height.
Solution:
Arranging in ascending order : 133, 135, 137, 138, 140, 142, 148, 150
Here, number of girls = 8 which is even
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 32

Question 26.
Find the mean, the median and the mode of:
(i) 12, 24, 24, 12, 30 and 12
(ii) 21, 24, 21, 6, 15, 18, 21, 45, 9, 6, 27 and 15.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 33.

Question 27.
The following table shows the market positions of some brands of soap.
Draw a suitable bar graph :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 35
Solution:

Question 28.
The birth rate per thousand of different countries over a particular period of time is shown below.
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 37
Solution:

Selina Concise Mathematics class 7 ICSE Solutions – Unitary Method (Including Time and Work)

November 10, 2023May 23, 2022 by Prasanna

Selina Concise Mathematics class 7 ICSE Solutions – Unitary Method (Including Time and Work)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

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

The method in which the value of a unit (one) quantity is first calculated to get the value of any other quantity is called the unitary method.
In unitary method, we come across two types of variations :
(i) Direct-variation
(ii) Inverse-variation.

(i) Direct variation : Increase in one quantity causes increase in the other and decrease in one quantity causes decrease in the other.
(ii) Inverse variation : Increase in one quantity causes decrease in the other and decrease in one quantity causes increase in the other.
This is found in the sums of speed, work done etc.

EXERCISE 7 (A)

Question 1.
Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles ?

Answer:
Weight of 8 articles = 4.8 kg
Weight of 1 article = \(\frac { 4.8 }{ 8 }\) kg
and weight of 11 articles =\(\frac { 4.8 }{ 8 }\) x 11 kg
= 0.6 x 11 = 6.6 kg

Question 2.
6 books weigh 1 .260 kg. How many books will weigh 3.150 kg ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 1

Question 3.
8 men complete a work in 6 hours. In how many hours will 12 men complete the same work ?

Answer:
8 men can complete a work in = 6 hours
1 man can complete the work in = 6×8 hours
12 men can complete the work in = \(\frac { 6 x 8 }{ 12 }\) = 4 hours

Question 4.
If a 25 cm long candle burns for 45 minutes, how long will another candle of the same material and same thickness but 5 cm longer than the previous one, burn ?

Answer:
25 cm long candle burn in = 45 minutes
1 cm long candle will burn in = \(\frac { 45 }{ 25 }\) mintues
25 + 5 = 30 cm long candle will burn in
= \(\frac { 45 x 30 }{ 25 }\)minutes = 54 minutes

Question 5.
A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages ?

Answer:
For typing 24 pages, time is required = 80 minutes
For typing 1 page, time is required =\(\frac { 80 }{ 24 }\) minutes
and for typing 87 pages, time is required
= \(\frac { 80 x 87 }{ 24 }\) minutes = 290 minutes

Question 6.
Rs. 750 support a family for 15 days. For how many days will Rs. 2,500 support the same family ?

Answer:
Rs. 750, can support a family for = 15 days
Re. 1 will support for = \(\frac { 15 }{ 750 }\)days
and Rs. 2,500 will support for = \(\frac { 15 }{ 750 }\)x 2500 days = 50 days

Question 7.
400 men have provisions for 23 weeks. They are joined by 60 men. How long will the provisions last ?

Answer:
400 men have provisions for = 23 weeks
1 man will have provisions for = 23 x 400 weeks
and 400 + 60 = 460 men will have provisions for = \(\frac { 23 x 400 }{ 460 }\) weeks = 20weeks

Question 8.
200 men have provisions for 30 days. If 50 men left, the same provisions would last for the remaining men, in how many days?

Answer:
200 men have provisions for = 30 days
1 man will have provisions for = 30 x 200 days
200 – 50 = 150 men will have provisions
for = \(\frac { 30 x 200 }{ 150 }\) days = 40 days

Question 9.
8 men can finish a certain amount of provisions in 40 days. If 2 more men join with them, find for how many days the same amount of provisions be sufficient ?

Answer:
8 men can finish a provision in = 40 days
1 man will finish in = 40 x 8 days
8+2=10 men will finish in =\(\frac { 40 x8 }{ 10 }\)
= 32 days

Question 10.
If interest on Rs. 200 be Rs. 25 in a certain time, what will be the interest on Rs 750 for the same time ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 2

Question 11.
If 3 dozen eggs cost Rs. 90, find the cost of 3 scores of eggs. (1 score = 20)

Answer:
3 dozen = 3 x 12 = 36 eggs,
3 scores = 3 x 20 = 60
The cost of 36 eggs is = Rs. 90
The cost of 1 egg will be = Rs. \(\frac { 90 }{ 36 }\)
∴ Cost of 60 eggs will be = Rs. \(\frac { 90 x60 }{ 36 }\)
= Rs. 150

Question 12.
If the fare for 48 km is Rs. 288, what will be the fare for 36 km ?

Answer:
Fare for 48 km = Rs. 288
fare for 1 km = Rs. \(\frac { 288 x 36 }{ 48 }\) = Rs. 216

Question 13.
What will be the cost of 3.20 kg of an item, if 3 kg of it costs Rs. 360 ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 4

Question 14.
If 9 lines of a print, in a column of a book contains 36 words. How many words will a column of 51 lines cqntain ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 5

Question 15.
125 pupil have food sufficient for 18 days. If 25 more pupil join them, how long will the food last now ? What assumption have you made to come to your answer ?

Answer:
Pupils in the beginning = 125
More pupils joined = 25
Total pupils = 125 + 25 = 150
Food is sufficient for 125 pupils for = 18 days
Food will be sufficient for 1 pupil for = 18 x 125 days (less pupil more days)
and food will be sufficient for 150 pupils = \(\frac { 18 x 125 }{ 150 }\) days (more pupil more days)
= \(\frac { 18 x 5 }{ 6 }\) 15 days

Question 16.
A carpenter prepares a new chair in 3 days, working 8 hours a day. Atleast how many hours per day must he work in order to make the same chair in 4 days ?

Answer:
A chair is completed in 3 days working per day = 8 hours
Then their will be completed in 1 day working for = 8 x 3 hours per day (less days more hours)
and it will be completed in 4 days working for = \(\frac { 8 x 3 }{ 4 }\)= 6 hours per day.

Question 17.
A man earns ₹5,800 in 10 days. How much will he earn in the month of February of a leap year?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 6

Question 18.
A machine is used for making rubber balls and makes 500 balls in 30 minutes. How many balls will it make in 3\(\frac { 1 }{ 2 }\) hours?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 7

Question 19.
In a school’s hostel mess, 20 children consume a certain quantity of ration in 6 days. However, 5 children did not return to the hostel after holidays. How long will the same amount of ration last now?

Answer:
Total number of children = 20
20 children consume a certain quantity of ration in = 6 days
1 children consume a certain quantity of ration in = 6 x 20 days
As 5 children did not return to the hostel after holidays.
Then number of children in hostel = 20-5 = 15
Hence, 15 children consume certain quantity 6×20
of ration in = \(\frac { 6 x 20 }{ 15 }\) days = 8 days

EXERCISE 7 (B)

Question 1.
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 8

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 9

Question 2.
3\(\frac { 1 }{ 2 }\)m of cloth costs Rs. 168 ; find the cost of 4\(\frac { 1 }{ 3 }\)m of the same cloth.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 10

Question 3.
A wrist watch loses 10 sec in every 8 hours; in how much time will it lose 15 sec. ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 12

Question 4.
In 2 days and 20 hours, a watch gains 20 sec ; find how much time will the watch take to gain 35 sec. ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 13

Question 5.
50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?

Answer:
Land is same in both the cases.
Now 50 men can mow land in = 3 days
∴ 1 man will mow it in = 3 x 50 days
and 15 men will mow it in = \(\frac { 3 x 50 }{ 15 }\) = 10 days

Question 6.
The wages of 10 workers for a six days week are Rs, 1,200. What are the one day wages: (i) of one worker ? (ii) of 4 workers?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 14

Question 7.
If 32 apples weigh 2 kg 800 g. How many apples will there be in a box, containing 35 kg of apples ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 15

Question 8.
A truck uses 20 litres of diesel for 240 km. How many litres will be needed for 1200 km?

Answer:
For 240 km, diesel is needed = 20 litres
∴ for 1 km, diesel will be needed 20

Question 9.
A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men ?

Answer:
1200 men has provision for = 15 days
1 man will have that provision for = 15 x 1200 days
∴1200 + 600 = 1800 men will has that provisions for =\(\frac { 15 x 1200 }{ 1800 }\)days
= 10 days

Question 10.
A camp has provisions for 60 pupil for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupil ?

Answer:
60 pupil have provision for = 18 days 1 pupil will have provision for = 18 x 60 days (less pupils more days)
and 72 pupils will have provision for = \(\frac { 18 x 60 }{ 72 }\) days
= 15 days.

EXERCISE 7 (C)

Question 1.
A can do a piece of work in 6 days and B can do it in 8 days. How long will they take to complete it together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 16

Question 2.
A and B working together can do a piece of work in 10 days B alone can do the same work in 15 days. How long will A alone take to do the same work ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 17

Question 3.
A can do a piece of work in 4 days and B can do the same work in 5 days. Find, how much work can be done by them working together in : (i) one day (ii) 2 days.
What part of work will be left, after they have worked together for 2 days ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 18

Question 4.
A and B take 6 hours and 9 hours respectively to complete a work. A works for 1 hour and then B works for two hours.
(i) How much work is done in these 3 hours ?
(ii) How much work is still left ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 19

Question 5.
A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long will they take to do it working together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 20

Question 6.
Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. How long will it take to fill the empty cistern, if all of them are opened together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 21

Question 7.
Mohit can complete a work in 50 days, whereas Anuj can complete the same work in 40 days.
Find:
(i) work done by Mohit in 20 days.
(ii) work left after Mohit has worked on it for 20 days.
(iii) time taken by Anuj to complete the remaining work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 22

Question 8.
Joseph and Peter can complete a work in 20 hours and 25 hours respectively.
Find :
(i) work done by both together in 4 hrs.
(ii) work left after both worked together for 4 hrs.
(iii) time taken by Peter to complete the remaining work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 24

Question 9.
A is able to complete \(\frac { 1 }{ 3 }\) of a certain work in 10 hrs and B is able to complete\(\frac { 2 }{ 5 }\) of the same work in 12 hrs.
Find:
(i) how much work can A do in 1 hour ?
(ii) how much work can B do in 1 hour ?
(iii) in how much time will the work be completed, if both work together.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 25

Question 10.
Shaheed can prepare one wooden chair in 3 days and Shaif can prepare the same chair in 4 days. If they work together, in how many days will they prepare :
(i) one chair ?
(ii)14 chairs of the same kind?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 26

Question 11.
A, B and C together finish a work in 4 days. If A alone can finish the same work in 8 days and B in 12 days, find how long will C take to finish the work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 27

Selina Concise Mathematics class 7 ICSE Solutions – Exponents (Including Laws of Exponents)

November 10, 2023May 23, 2022 by Prasanna

Selina Concise Mathematics class 7 ICSE Solutions – Exponents (Including Laws of Exponents)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

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

EXERCISE 5 (A)

Question 1.
Find the value of:
(i) 6²
(ii) 7³
(iii) 4⁴
(iv) 5⁵
(v) 8³
(vi) 7⁵

Solution:
(i) 6² = 6 x 6 = 36
(ii) 7³ = 7 x 7 x 7 = 343
(iii) 4⁴ = 4 x 4 x 4 x 4 = 256
(iv) 5⁵= 5 x 5 x 5 x 5 x 5 = 3125
(v) 8³ = 8 x 8 x 8 = 512
(vi) 7⁵= 7 x 7 x 7 x 7 x 7 =16807

Question 2.
Evaluate:
(i) 2³ x 4²
(ii) 2³ x 5²
(iii) 3³ x 5²
(iv) 2² x 3³
(v) 3² x 5²
(vi) 5³ x 2⁴
(vii) 3²x 4²
(ix) (5 x 4)²

Solution:
(i) 2³ x 4²
= 2 x 2 x 2 x 4 x 4
= 8 x 16
= 128
(ii) 2³ x 5²
= 2 x 2 x 2 x 5 x 5
= 8 x 25
= 200
(iii) 3³ x 5²
=3 x 3 x 3 x 5 x 5
= 27 x 25
= 675
(iv) 2² x 3³
= 2 x 2 x 3 x 3 x 3
= 4 x 27
= 108
(v) 3² x 5³
=3 x3 x 5 x 5 x 5
= 9 x 125
= 1125
(vi) 5³ x 2⁴
= 5 x 5 x 5 x 2 x 2 x 2 x 2
= 125 x 16
= 2000
(vii) 3² x 4²
=3 x 3 x 4 x 4
= 9 x 16
=144
(viii) (4 x 3)³
=4 x 4 x 4 x 3 x 3 x 3
= 64 x 27
= 1728
(ix) (5 x 4)²
=5 x 5 x 4 x 4
= 25 x 16
= 400

Question 3.
Evaluate:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 39

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 40

Question 4.
Evaluate :
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 1

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 2

Question 5.
Which is greater :
(i) 2³ or 3²
(ii) 2⁵ or 5²
(iii) 4³ or 3⁴
(iv) 5⁴ or 4⁵

Solution:
(i) 2³ or 3³
Since, 2³ = 2 x 2 x 2 = 8
and, 3² = 3 x 3 = 9
∵9 is greater than 8 ⇒ 3² > 2³
(ii) 2⁵ or 5²
Since, 2⁵ = 2 x 2 x 2 x 2 x 2 = 32
and, 5² = 5 x 5 = 25
∵32 is greater than 25 ⇒ 2³⁵ > 5³²
(iii) 4³ or 3⁴
Since, 4³ = 4 x 4 x 4 = 64
and, 3⁴ = 3 x 3 x 3 x 3 = 81
∵ 81 is greater than 64 ⇒ 3⁴ > 4³
(iv) 5⁴ or 4⁵
Since, 5⁴ = 5 x 5 x 5 x 5 = 625
and, 4⁵= 4 x 4 x 4 x 4 x 4= 1024
∵ 1024 is greater than 625 ⇒ 4⁵ > 5⁴

Question 6.
Express each of the following in exponential form :
(i) 512
(ii) 1250
(iii) 1458
(iv) 3600
(v) 1350
(vi) 1176

Solution:
(i) 512
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 4

(ii) 1250
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 6

(iii) 1458
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 7

(iv) 3600
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 8

(v) 1350
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 9

(vi) 1176
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 10

Question 7.
If a = 2 and b = 3, find the value of:
(i) (a + b)²
(ii) (b – a)³
(iii) (a x b)a (iv) (a x b)b

Solution:
(i) (a + b)²
= (2 + 3)² = (5)² = 5 x 5 = 25

(ii) (b – a)²
= (3 – 2)²= (1)³
= 1 x 1 x 1 = 1

(iii) (a x b)^a
= (2 x 3)² – (6)²
= 6 x 6 = 36

(iv) (a x b)^b
= (2 x 3)³ = (6)³ = 6 x 6 x 6 = 216

Question 8.
Express:
(i) 1024 as a power of 2.
(ii) 343 as a power of 7.
(iii) 729 as a power of 3.
Solution:
(i) 1024 as a power of 2.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 11

(ii) 343 as a power of 7.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 12

(iii) 729 as a power of 3.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 13

Question 9.
If 27 x 32 = 3^x x 2^y; find the values of x and y.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 14

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 15

Question 10.
If 64 x 625 = 2^a x 5^b; find :
(i) the values of a and b.
(ii) 2^b x 5^a

Solution:
(i) the values of a and b.
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 16

(ii) 2^b x 5^a
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 17

EXERCISE 5 (B)

Question 1.
Fill in the blanks:
In 5² = 25, base = ……… and index = ……….
If index = 3x and base = 2y, the number = ………

Solution:
(i) In 5² = 25, base = 5 and index = 2
(ii) If index = 3x and base = 2y, the number = 2y^3x

Question 2.
Evaluate:
(i) 2⁸ ÷ 2³
(ii) 2^3÷ 2⁸
(iii) (2⁶)⁰
(iv) (3^o)⁶
(v) 8³ x 8^-5 x 8⁴
(vi) 5⁴ x 5³ + 5⁵
(vii) 5⁴ ÷ 5³ x 5⁵
(viii) 4⁴ ÷ 4³ x 4⁰
(ix) (3⁵ x 4⁷ x 5⁸)⁰

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 18

Question 3.
Simplify, giving Solutions with positive index:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 20

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 21

Question 4.
Simplify and express the Solution in the positive exponent form :
Selina Concise Mathematics class 7 ICSE Solutions - Exponents (Including Laws of Exponents) image - 27