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Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions

November 9, 2023May 27, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions (Using ruler and compass only)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions (Using ruler and compass only)

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Constructions Exercise 18A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Given below are the angles x and y.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 1
Without measuring these angles, construct :
(i) ∠ABC = x + y
(ii) ∠ABC = 2x + y
(iii) ∠ABC = x + 2y
Solution:

(i) Steps of Construction :

Draw a line segment BC of any suitable length.
With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the arc x at points P and Q and arms of angle y at points R and S.
From the arc, with centre B, cut DE = PQ arc of x and EF = RS arc of y
Join BF and produce upto point A.
Thus ∠ABC = x + y

(ii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 3
Proceed in exactly the same way as in part(i)
takes DE = PQ = arc of x.
EF = PQ = arc of x and FG = RS = arc of y.
Join BG and produce it upto A.
Thus ∠ABC = x + x+ y = 2x + y
(iii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 4
Proceed in exactly the same way as in (ii)
taking DE = PQ = arc of x. and EF = RS = arc of y and FG = RS = arc of y.
4. Join BF and produce upto point A.
Thus ∠ABC = x + y + y = x + 2y

Question 2.
Given below are the angles x, y and z.
Without measuring these angles construct :
(i) ∠ABC = x + y + z
(ii) ∠ABC = 2x + y + z
(iii) ∠ABC = x + 2y + z
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 5
Solution:
(i) Steps of Construction :

Draw line segment BC of any suitable length.
With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the anlge x at the points P and Q and arms of the angle y at points R and S and arms of the angle z at the points L and M.
From the arc, with centre B, cut
DE = PQ = arc of x, EF = RS = arc of y and FG = LM = arc of z.
Join BG and produce it upto A.
Then ∠ABC = x + y + z

(ii) Proceed as in part (i) upto step 2.
3. From the arc, with centre B, cut
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 8
DE = 2PQ = 2 arc of x
EF = RS = arc of y
FG = LM = arc of z
4. Join BG and produce it upto point A
Then ∠ABC = 2x + y + z
(iii) proceed as in (i) upto step 2

3. Here cut arc DE = arc PQ = arc of x arc EF = 2 arc RS = 2 arc of y arc FG = arc LM = arc of z.
4. Join BG and produce it upto A
5. Then ∠ABC = x + 2y + z

Question 3.
Draw a line segment BC = 4 cm. Construct angle ABC = 60°.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 10

Draw a line segment BC = 4 cm
With B as centre, draw an arc of any suitable radius which cuts BC at the point D.
With D as centre, and the same radius as in step 2, draw one more arc which cuts the previous arc at the point E.
Join BE and produce it to the point A.
Thus ∠ABC = 60°

Question 4.
Construct angle ABC = 45° in which BC = 5 cm and AB = 4.6 cm.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 11

Draw a line segment BC = 5 cm
Taking B as centre, draw an arc of any suitable radius, which cuts BC at the point D.
With D as centre and the same radius, as taken in step 2, draw an arc which cuts the previous arc at point E.
With E as centre and the same radius, draw one more arc which cuts the first arc at point F.
With E and F as centres and radii equal to more than half the distance between E at F, draw arc which cut each other at point P.
Join BP to meet EF at L and produce to point O. Then ∠OBC = 90°
Draw BA, the bisector of angle OBC. [With D, L as centres and suitable radius draw two arc meeting each other at Q produced it to R]
=> ∠ABC = 45° [∴ BA is bisector of ∠OBC ∴ ∠ABC = = 45°]
From BR cut arc AB = 4.6 cm

Question 5.
Construct angle ABC = 90°. Draw BP, the bisector of angle ABC. State the measure of angle PBC.
Solution:

Draw ∠ABC = 90° (as in Ques. 4)
Draw bisector of ∠ABC
Then ∠PBC = \(\frac { 1 }{ 2 }\) (90°) = 45°

Question 6.
6. Draw angle ABC of any suitable measure.
(i) Draw BP, the bisector of angle ABC.
(ii) Draw BR, the bisector of angle PBC and draw BQ, the bisector of angle ABP.
(iii) Are the angles ABQ, QBP, PBR and RBC equal?
(iv) Are the angles ABR and QBC equal ?
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 13

Construct any angle ABC
With B as centre, draw an arc EF meeting BC at E and AB at F.
With E, F as centres draw two arc of equal radii meeting each other at the point P.
Join BP. Then BP is the bisector of ∠ABC
∠ABP = ∠PBC = \(\frac { 1 }{ 2 }\) ∠ABC
Similarly draw BR, the bisector of ∠PBC and draw BQ as the bisector of ∠ABP [With the same method as in steps 2, 3]
Then ∠ABQ = ∠QBP = ∠PBR = ∠RBC
∠ABR = \(\frac { 3 }{ 4 }\) ∠ABC and ∠QBC = \(\frac { 3 }{ 4 }\) ∠ABC
∠ABR = ∠QBC.

Constructions Exercise 18B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Draw a line segment AB of length 5.3 cm. Using two different methods bisect AB.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 14

Draw a line segment AB = 5.3 cm
With A as centre and radius equal to more than half of AB, draw arcs on both sides of AB.
With B as centre and with the same radius as taken in step 2, draw arcs on both the sides of AB.
Let the arcs intersect each other at points P and Q.
Join P and Q.
The line PQ cuts the given line segment AB at the point O.
Thus, PQ is a bisector of AB such that
OA = OB = \(\frac { 1 }{ 2 }\) AB

Second Method
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 15
Steps of Construction :

Draw the given line segment AB = 5.3 cm.
At A, construct ∠PAB of any suitable measure. Then ∠PAB = 60° construct ∠QBA = 60°
3.From AP, cut AR of any suitable length and from BQ ; cut BS = AR.
Join R and S
Let RS cut the given line segment AB at the point O.
Thus RS is a bisector of AB such that OA = OB = \(\frac { 1 }{ 2 }\) AB

Question 2.
Draw a line segment PQ = 4.8 cm. Construct the perpendicular bisector of PQ.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 16

Draw a line segment PQ = 4.8 cm.
With P as centre and radius equal than half of PQ, draw arc on both the PQ.
With Q as centre and the same radius as taken in step 2, draw arcs on both sides of PQ.
Let the arcs intersect each other at point A and B
Join A and B.
The line AB cuts the line segment PQ at the point O. Here OP = OQ and ∠AOQ = 90°. Then the line AB is perpendicular bisector of PQ.

Question 3.
In each of the following, draw perpendicular through point P to the line segment AB :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 17
Solution:
(i) Steps of Construction :

With P as centre, draw an arc of a suitable radius which cuts AB at points C and D.
With C and D as centres, draw arcs of equal radii and let these arcs intersect each other at the point Q.
[The radius of these arcs must be more than half of CD and both the arcs must be drawn on the other side]
Join P and Q
Let PQ cut AB at the point O.
Thus, OP is the required perpendicular clearly, ∠AOP = ∠BOP = 90°

(ii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 19

With P as centre, draw an arc of any suitable radius which cuts AB at points C and D.
With C and D as centres, draw arcs of equal radii. Which intersect each other at point A.
[This radius must be more than half of CD and let these arc intersect each other at the point 0]
Join P and O. Then OP is the required perpendicular.
∠OPA = ∠OPB = 90°

(iii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 20

With P as centre, draw an arc of any suitable radius which cuts AB at points C and D.
With C and D as centre, draw arcs of equal radii
[The radius of these arcs must be more than half of CD and both the arcs must be drawn on the other side.]
and let these arcs intersect each other at the point Q.
Join Q and P. Let QP cut AB at the point O. Then OP is the required perpendicular.
Clearly, ∠AOP = ∠BOP = 90°

Question 4.
Draw a line segment AB = 5.5 cm. Mark a point P, such that PA = 6 cm and PB = 4.8 cm. From the point P, draw perpendicular to AB.
Solution:
Step of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 21

Draw a line segment AB = 5.5 cm
With A as centre and radius = 6 cm, draw an arc.
With B as centre and radius = 4.8 cm draw another arc.
Let these arcs meet each other at the point P.
PA = 6 cm, PB = 4.8
With P as centre and some suitable radius draw an arc meeting AB at the points C and D.
With C as centre and radius more than half of CD, draw an arc.
With D as centre and same radius as in step 6, draw an arc.
Let these arcs meet each other at the point Q.
Join PQ.
The PQ meet AB at point O.
Then PO ⊥ AB i.e; ∠AOP = 90° = ∠POB.

Question 5.
Draw a line segment AB = 6.2 cm. Mark a point P in AB such that BP = 4 cm. Through point P draw perpendicular to AB.
Solution:
Steps of Construction :

Draw a line segment AB = 6.2 cm
Cut off BP = 4 cm
With P as centre and some radius draw arc meeting AB at the points C, D.
With C, D as centres and equal radii [each is more than half of CD] draw two arcs, meeting each other at the point O.
Join OP. Then OP is perpendicular for AB.

Constructions Exercise 18C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Draw a line AB = 6 cm. Mark a point P any where outside the line AB. Through the point P, construct a line parallel to AB.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 23

Draw a line AB = 6 cm
Take any point Q on the line AB and join it with the given point P.
At point P, construct ∠CPQ = ∠PQB
Produce CP upto any point D.
Thus, CPD is the required parallel line.

Question 2.
Draw a line MN = 5.8 cm. Locate a point A which is 4.5 cm from M and 5 cm from N. Through A draw a line parallel to line MN.
Solution:
Steps of construction :

Draw a line MN = 5.8 cm
With M as centre and radius = 4.5 cm, draw an arc.
With N as centre draw another arc of radius 5 cm. These arcs intersect each other at A.
Join AM and AN.
At point A, draw ∠DAN = ∠ANM
Produce DA to any point C.
Thus CAD is the required parallel line.

Question 3.
Draw a straight line AB = 6.5 cm. Draw another line which is parallel to AB at a distance of 2.8 cm from it.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 25

Draw a straight line AB = 6.5 cm
Taking point A as centre, draw an arc of radius 2.8 cm.
Taking B as centre, drawn another arc of radius 2.8 cm.
Draw a line CD which touches the two arcs drawn.
Thus CD is the required parallel line.

Question 4.
Construct an angle PQR = 80°. Draw a line parallel to PQ at a distance of 3 cm from it and another line parallel to QR at a distance of 3.5 cm from it. Mark the point of intersection of these parallel lines as A.
Solution:
Steps of construction :

Draw ∠PQR = 80°
With P as centre draw an arc of radius 2 cm.
Again with Q as centre, draw another arc of radius 2 cm. Then BM is a line which touches the two arcs. Then BM is a line parallel to PQ.
With Q as centre, draw an arc of radius 3.5 cm. With R as centre draw another arc of radius 3.5 cm. Draw a line HC which touches these two arcs. Let these two parallel line intersect at A.

Question 5.
Draw an angle ABC = 60°. Draw the bisector of it. Also draw a line parallel to BC a distance of 2.5 cm from it.
Let this parallel line meet AB at point P and angle bisector at point Q. Measure the length of BP and PQ. Is BP = PQ ?
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 27

Draw, ∠ABC = 60°
Draw BD, the bisector of ∠ABC.
Taking B as centre, draw an arc of radius 2.5 cm.
Taking C as centre, draw another arc of radius 2.5 cm.
Draw a line MN which touches these two arcs drawn. Then MN is the required line parallel to BC.
Let this line MN meets AB at P and bisector BD at Q.
Measure BP and PQ.
By measurement we see BP = PQ.

Question 6.
Construct an angle ABC = 90°. Locate a point P which is 2.5 cm from AB and 3.2 cm from BC.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 28

Draw ∠ABC = 90°
From AB, cut BD = 3.2 cm.
Through point C, draw CH⊥BC. From CH, cut CE = 3.2. Join DE. Now DE is a line parallel to BC and at a distance of 3.2 cm from BC.
From BC cut BM = 2.5 cm.
Through point A, draw AK ⊥ AB. From AK cut AN = 2.5 cm. Join NM. Therefore NM is parallel to AB and at a distance of 2.5 cm from AB.
DE and MN intersect each other at P. Thus P is the required point which is 2.5 cm from AB and 3.2 cm from BC.

Constructions Exercise 18D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Construct a quadrilateral ABCD; if:
(i) AB = 4.3 cm, BC = 5.4, CD = 5 cm, DA = 4.8 cm and angle ABC = 75°.
(ii) AB = 6 cm, CD = 4.5 cm, BC = AD = 5 cm and ∠BCD = 60°.
(iii) AB = 8 cm, BC = 5.4 cm, AD = 6 cm, ∠A = 60° and ∠B = 75°.
(iv) AB = 5 cm, BC = 6.5 cm, CD =4.8 cm, ∠B = 75° and ∠C = 120°.
(v) AB = 6 cm = AC, BC = 4 cm, CD = 5 cm and AD = 4.5 cm.
(vi) AB = AD = 5cm, BD = 7 cm and BC = DC = 5.5 cm
Solution:
(i) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 29
Steps :

Draw AB = 4.3 cm.
At B, draw ∠PBA = 75°
Cut BC = 5.4 cm.
From C & A, draw arcs of radii 5 cm and 4.8 cm respectively which intersect at D.
Join AD and DC.
ABCD is the required quadrilateral.

(ii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 30
Steps :

Draw BC = 5 cm.
Draw ∠PCB = 60° and cut CD = 4.5 cm.
From B and D, draw arcs of radii 6 cm and 5 cm respectively which intersect at A.
Join AB and AD.
Thus ABCD is the required quadrilateral.

(iii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 31
Actual quadrilateral is constructed with the help of above rough figure.

Steps :

Draw AB = 8 cm.
At A, draw ∠PAB = 60° and cut DA = 6 cm.
At B, draw ∠QBA = 75° and cut BC = 5.4 cm.
Join DC.
Thus ABCD is the required quadrilateral.

(iv) Rough figure is as shown below.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 33
Steps :

Draw BC = 6-5 cm.
Draw ∠B = 75° and cut BA = 5 cm.
Draw ∠C = 120° and cut CD = 4.8 cm.
Join AD.
Thus ABCD is the required quadrilateral.

(v) Rough figure is as shown below.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 34
Steps :

Draw AB = 6 cm.
From A and B, draw arcs of radii 6 cm and 4 cm which cut at C.
From A and C, draw arcs of radii 4.5 cm and 5 cm respectively which intersect at D.
Join BC, CD and DA. Thus ABCD is the required quadrilateral.

(vi) Rough figure is as follow :

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 35
Actual construction is as follow (using above rough fig.)

Steps :

Draw AB = 5 cm.
From A & B draw arcs of radii 5 cm and 7cm which intersect at D.
From B & D draw arcs of radii 5.5 cm each which intersect at C.
Join AD, BD, DC and BC.
Thus ABCD is the required quadrilateral.

Question 2.
Construct a parallelogram ABCD, if :
(i) AB = 3.6 cm, BC = 4.5 cm and ∠ABC = 120°.
(ii) BC = 4.5 cm, CD = 5.2 cm and ∠ADC = 75°.
(iii) AD = 4 cm, DC = 5 cm and diagonal BD = 7 cm.
(iv) AB = 5.8 cm, AD = 4.6 cm and diagonal AC = 7.5 cm.
(v) diagonal AC = 6.4 cm, diagonal BD = 5.6 cm and angle between the diagonals is 75°.
(vi) lengths of diagonals AC and BD are 6.3 cm and 7.0 cm respectively, and the angle between them is 45°.
(vii) lengths of diagonals AC and BD are 5.4 cm and 6.7 cm respectively and the angle between them is 60°.
Solution:
(i) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 37
The above rough figure is used to construct the actual ||gm as follow :

Steps :

Draw AB = 3.6 cm.
Draw BP such that ∠B = 120°.
Cut BC = 4.5 cm.
From A, draw arc of radius 4.5 cm.
From C, draw arc of radius 3.6 cm. Which interescts first arc at D.
Join AD and CD.
Hence ABCD is the required ||gm.

(ii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 39
Steps :

Draw CD = 5.2 cm.
Draw ZCDP = 75°
Cut DA = 4.5 cm.
From A draw arc of radius 5.2 cm.
From C, draw arc of radius 4.5 cm which meets first arc at B.
Join AB and CB.
Thus ABCD is the required ||gm.

(iii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 40
Steps :

Draw AD = 4 cm.
From A, draw an arc of radius 5 cm.
From B, draw an arc of radius 4 cm.
From D, draw an arc of ardius 5 cm which intersect first arc at C.
Join AB, BD, BC and CD.
Thus ABCD is the required || gm.

(iv) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 41
opposite sides of ||gm are equal
BC = AD = 4.6 cm.
Actual figure is constructed as follow :

Steps:

Draw AB = 5.8 cm.
Draw an arc of radius 4.6 cm with centre B.
Draw an arc of radius 7.5 cm from A which intersects first arc at C.
From A, draw an arc of radius 4.6 cm.
From C, draw an arc of radius 5.8 cm which intersects first arc at D.
Join AD, CD, BC and AC.
Thus ABCD is the required //gm.

(v) Rough figure is as follow.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 43
Steps :

Draw AC = 6.4 cm.
Bisect AC at O.
Draw ∠XOC = 75° and produce XO to Y.
Cut OB = OD = 2 8 cm.
Join AB, BC, AD and CD.
Thus ABCD is the required ||gm.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 44
Steps :

Draw AC = 6.3 cm.
Bisect AC at O.
At O, draw ∠XOC = 45° and produce XO to Y.
Cut OB = OD = 3.5 cm (half the diagonal 7 cm.)
Join AB, CB, AD and CD. Thus ABCD is the required || gm.

(vii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 45
Steps :

Draw BD 6.7 cm.
Bisect BD at O.
At O, draw ∠XOD = 60° and produce XO to Y.
Cut OA = OC = 2.7 cm (half the diagonals 5.4 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required ||gm.

Question 3.
Construct a rectangle ABCD ; if :
(i) AB = 4.5 cm and BC = 5.5 cm.
(ii) BC = 61 cm and CD = 6.8 cm.
(iii) AB = 5.0 cm and diagonal AC = 6.7 cm.
(iv) AD = 4.8 cm and diagonal AC = 6.4 cm.
(v) each diagonal is 6 cm and the angle between them is 45°.
(vi) each diagonal is 5.5 cm and the angle between them is 60°.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 46
Steps :

Draw BC = 5.5 cm.
At B, draw ∠XBC = 90°
Cut BA = 4.5 cm.
From A, draw an arc of radius 5.5 cm.
From C, draw an arc of radius 4 5 cm which meets first arc at D.
Join AD and CD.
Thus ABCD is the required rectangle.

(ii)
na Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 47
Steps:

Draw BC = 6.1 cm.
At C, draw ∠PCB = 90°.
Cut CD = 6.8 cm.
Draw an arc of radius 6.8 cm from B.
From D, draw an arc of radius 6.1 cm which meets the first arc at A.
Join AB and AD.
Thus ABCD is the required rectangle.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 48
Steps :

Draw AB = 5 cm.
At B, draw ∠XBA = 90°.
From A, draw an arc of radius 6.7 cm which meets XB at C.
From C, draw an arc of a radius 5 cm.
From A, draw an arc of radius equal to BC which meets first arc at D.
Join AD and CD. Thus ABCD is the required rectangle.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 49
Steps :

Draw AD = 4.8 cm.
At D, draw ∠XDA = 90°.
From A, draw an arc of radius 6-4 cm which meets DX at C.
From A, draw an arc of radius equal to DC.
From C, draw an arc of radius 4.8 cm which meets first arc at B.
Join AB and CB. Thus ABCD is the required rectangle.

(v)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 50
Steps :

Draw AC = 6 cm.
Bisect AC at O.
At O, draw ∠XOC = 45° and produce XO to Y.
Cut OB = OD = 3 cm (half the diagonal 6 cm)
Join AB, CB, AD and CD.
Thus ABCD is the required rectangle.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 51
Steps :

Draw AC = 5.5 cm.
Bisect AC at O.
At O, draw ∠XOC = 60° and produce XO to Y.
Cut OB = OA and OD = OA (half the diagonal AC).
Join AB, BC, AD and CD.
Thus ABCD is the required rectangle.

Question 4.
Construct a rhombus ABCD, if ;
(i) AB = 4 cm and ∠B = 120°.
(ii) BC = 4.7 cm and ∠B = 75°.
(iii) CD = 5 cm and diagonal BD = 8.5 cm.
(iv) BC = 4.8cm, and diagonal AC = 7cm.
(v) diagonal AC = 6 cm and diagonal BD = 5.8 cm.
(vi) diagonal AC = 4.9 cm and diagonal BD = 6 cm.
(vii) diagonal AC = 6.6 cm and diagonal BD = 5.3 cm.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 52
Steps :

Draw AB = 4 cm.
At B, draw ∠XBA = 120°
Cut BC = 4 cm.
Draw arcs of radii 4 cm each from A and C which intersect at D.
Join CD and AD.
Thus ABCD is the required rhombus.

(ii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 53
Steps :

Draw BC = 4.7 cm.
At B, draw ∠XBC = 75°
Cut BA = 4.7 cm.
From A and C, draw arcs of radii 4.7 cm each which intersect at D.
Join AD and CD.
Thus ABCD is the rhombus.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 54
Steps :

Draw CD = 5 cm.
From C & D draw arcs of radii 5 cm and 8.5 cm respectively which intersect at B.
From B and D, draw arcs of radii 5 cm each which intersect at A.
Join AB and AD.
Thus ABCD is the required rhombus.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 55
Steps :

Draw AC = 7 cm.
Draw arcs of radii 4.8 cm each from A and C which intersect at B.
From A & C again draw arcs of radii 4.8 cm each which intersect at D.
Join AB, BC, AD and CD.
Thus ABCD is the required rhombus.

(v)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 56
Steps :

Draw BD = 5.8 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3 cm (half the diagonal 6 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required rhombus.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 57
Steps :

Draw AC = 4.9 cm.
Draw perpendicular bisector XY of AC.
Cut OB = OD = 3 cm (half the diagonal 6 cm)
Join AB, BC, AD and CD.
Thus ABCD is the required rhombus.

(vii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 58
Steps :

Draw BD = 5.3 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3.3 cm (half the diagonal 6.6 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required rhombus.

Question 5.
Construct a square, if :
(i) its one side is 3.8 cm.
(ii) its each side is 4.3 cm.
(iii) one diagonal is 6.2 cm.
(iv) each diagonal is 5.7 cm.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 59
Steps :

Draw AB = 3.8 cm.
At B, draw ∠PBA = 90°.
Cut BC = 3.8 cm.
From A and C, draw arcs of radii 3.8 cm each which intersect at D.
Join AD and CD.
Thus ABCD is the required square.

(ii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 60
Steps :

Draw AB = 4.3 cm.
Draw ∠PAB = 90° at A.
Cut AD = 4.3 cm.
From B and D, draw arcs of radii 4.3 cm each which intersect at C.
Join AD, BC and CD.
Hence ABCD is the required square.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 61
Steps :

Draw BD = 6.2 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3.1 cm (half the diagonal)
Join AB, AD, BC and CD.
Thus ABCD is the required square.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 62
Steps :

Draw BD = 5.7 cm.
Draw perpendicular bisector XY of BD.
From 0, draw arcs of radii equal to OB which cuts XY at A and C.
Join AB, AD, BC and CD.
Thus ABCD is the required square.

Question 6.
Construct a quadrilateral ABCD in which ; ∠A = 120°, ∠B = 60°, AB = 4 cm, BC = 4.5 cm and CD = 5 cm.
Solution:
Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 63
Steps :

Draw AB = 4 cm.
At A, draw ∠PAB = 120°.
At B, draw ∠QBA = 60°.
From BQ, cut BC = 4.5 cm.
From C, draw an arc of radius 5 cm which meets AP at D.
Join CD.
Thus ABCD is the required quadrilateral.

Question 7.
Construct a quadrilateral ABCD, such that AB = BC = CD = 4.4 cm, ∠B = 90° and ∠C = 120°.
Solution:
Rough figure is as follow
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 64
Steps :

Draw BC = 4.4 cm.
At B, draw ∠PBC = 90°.
Cut BA = 4.4 cm.
At C, draw ∠QCB = 120°.
Cut CD = 4.4 cm.
Join AD.
Thus ABCD is the required quadrilateral.

Question 8.
Using ruler and compasses only, construct a parallelogram ABCD, in which : AB = 6 cm, AD = 3 cm and ∠DAB = 60°. In the same figure draw the bisector of angle DAB and let it meet DC at point P. Measure angle APB.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 65
Steps :

Draw AB = 6 cm.
At A draw ∠QAB = 60°.
From AQ cut AD = 3 cm.
From D, draw an arc of radius 6 cm.
From B, draw an arc of radius 3 cm which meets first arc at C.
Join CD and BC.
Thus ABCD is the required ||gm.
Bisect ∠DAB, so that bisector meets CD at P.
Join PB and measure ZAPB.
∴ ∠APB = 90°.

Question 9.
Draw a parallelogram ABCD, with AB = 6 cm, AD = 4.8 cm and ∠DAB = 45°. Draw the perpendicular bisector of side AD and let it meet AD at point P. Also draw the diagonals AC and BD ; and let they intersect at point O. Join O and P. Measure OP.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 66
Steps :

Draw AB = 6 cm.
Draw ∠PAB = 45°.
Cut AD = 4.8 cm.
From D, draw an arc of radius 6 cm.
From B, draw an arc of radius 4.8 cm which meets first arc at C.
Join BC, CD, AD.
Thus ABCD is the required ||gm.
Draw perpendicular bisector XY of AD which cuts AD at P.
Join AC and BD which intersect at O.
Join OP and measure it.
OP = 3 cm.

Question 10.
Using ruler and compasses only, construct a rhombus whose diagonals are 8 cm and 6 cm. Measure the length of its one side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 67
Steps :

Draw BD = 8 cm.
Draw perpendicular bisector PQ of BD.
Cut OA = OC = 3 cm [half the diagonal 6 cm]
Join AB, AD, BC and CD.
Measure side AB which is 5 cm.
Thus ABCD is the required rhombus.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

November 10, 2023May 24, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

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Profit, Loss and Discount Exercise 8A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Megha bought 10 note-books for Rs.40 and sold them at Rs.4.75 per note-book. Find, her gain percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 1

Question 2.
A fruit-seller buys oranges at 4 for Rs.3 and sells them at 3 for Rs.4 Find his profit percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 2

Question 3.
A man buys a certain number of articles at 15 for Rs. 112.50 and sells them at 12 for Rs.108. Find ;
(i) his gain as percent;
(ii) the number of articles sold to make a profit of Rs.75.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 4

Question 4.
A boy buys an old bicycle for Rs. 162 and spends Rs. 18 on its repairs before selling the bicycles for Rs. 207. Find his gain or loss percent.
Solution:
Buying price of the old bicycle = Rs.162
Money spent on repairs = Rs. 18
Real C.P. of the bicycle= 162+18 = Rs.180
S.P. of the bicycle = Rs.207
Profit = S.P. – C.P. = 207 – 162 = Rs. 45
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 6Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 6

Question 5.
An article is bought from Jaipur for Rs. 4,800 and is sold in Delhi for Rs. 5,820. If Rs. 1,200 is spent on its transportations, etc. ; find he loss or the gain as percent.
Solution:
Cost price = Rs. 4,800
Selling Price = Rs. 5,820
Transport etc. charges = Rs. 1,200
Total cost price = Rs, 4,800 + Rs. 1,200 = Rs. 6,000
Loss = Rs. 6000 – Rs. 5820 = Rs. 180
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 7

Question 6.
Mohit sold a T.V. for Rs. 3,600 ; gaining one-sixth of its selling price. Find :
(i) the gain
(ii) the cost price of the T.V.
(iii) the gain percent.
Solution:
S.P. of T.V. = Rs. 3,600
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 8

Question 7.
By selling a certain number of goods for Rs. 5,500; a shopkeeper loses equal to one-tenth of their selling price. Find :
(i) the loss incured
(ii) the cost price of the goods
(iii) the loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 9

Question 8.
The selling price of a sofa-set is \(\frac { 4 }{ 5 }\) times of its cost price. Find the gain or the loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 10

Question 9.
The cost price of an article is \(\frac { 4 }{ 5 }\) times of its selling price. Find the loss or the gain as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 11

Question 10.
A shopkeeper sells his goods at 80% of their cost price. Find the percent gain or loses ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 12

Question 11.
The cost price of an article is 90% of its selling price. What is the profit or the loss as percent ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 13

Question 12.
The cost price of an article is 30 percent less than its selling price. Find, the profit or loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 15

Question 13.
A shop-keeper bought 300 eggs at 80 paisa each. 30 eggs were broken in transaction and then he sold the remaining eggs at one rupee each. Find, his gain or loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 16

Question 14.
A man sold his bicycle for Rs.405 losing one-tenth of its cost price, find :
(i) its cc price;
(ii) the loss percent.
Solution:
(i) Let C.P. of the bicycle = Rs. x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 17

Question 15.
A man sold a radio-set for Rs.250 and gained one-ninth of its cost price. Find ;
(i) its cost price;
(ii) the profit percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 18

Profit, Loss and Discount Exercise 8B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the selling price, if:
(i) C.P. = Rs. 950 and profit = 8%
(ii) C.P. = Rs. 1,300 and loss = 13%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 21

Question 2.
Find the cost price, if :
(i) S.P. = Rs. 1,680 and profit = 12%
(ii) S.P. = Rs. 1,128 and loss = 6%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 22

Question 3.
By selling an article for Rs.900; a man gains 20%. Find his cost price and the gain.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 24

Question 4.
By selling an article for Rs.704; a person loses 12%. Find his cost price and the loss
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 25

Question 5.
Find the selling price, if :
(i) C.P. = Rs.352; overheads = Rs.28 and profit = 20
(ii) C.P. = Rs.576; overheads = Rs.44 and loss = 16%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 26

Question 6.
If John sells his bicycle for Rs. 637, he will suffer a loss of 9%. For how much should it be sold, if he desires a profit of 5% ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 27

Question 7.
A man sells a radio-set for Rs.605 and gains 10%. At what price should he sell another radio of the same kind, in order to gain 16% ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 28

Question 8.
By selling a sofa-set for Rs.2,500; the shopkeeper loses 20%. Find his loss percent or profit percent ; if he sells the same sofa-set for Rs.3150.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 29

Question 9.
Mr. Sinha sold two tape-recorders for Rs.990 each; gaining 10% on one and losing 10% on the other. Find his total loss or gain as percent on the whole transaction.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 31

Question 10.
A tape-recorder is sold for Rs. 2,760 at a gain of 15% and a C.D. player is sold for Rs. 3,240 at a loss of 10% Find :
(i) the C.P. of the tape-recorder
(ii) the C.P. of the C.D. player.
(iii) the total C.P. of both.
(iv) the total S.P. of both
(v) the gain % or the loss % on the whole
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 33

Question 11.
Rajesh sold his scooter to Rahim at 8% loss and Rahim, in turn, sold the same scooter to Prem at 5% gain. If Prem paid Rs. 14,490 for the scooter ; find :
(i) the S.P. and the C.P. of the scooter for Rahim
(ii) the S.P. and the C.P. of the scooter for Rajesh
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 34

Question 12.
John sold an article to Peter at 20% profit and Peter sold it to Mohan at 5% loss. If Mohan paid Rs.912 for the article; find how much did John pay for it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 35

Profit, Loss and Discount Exercise 8C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A stationer buys pens at 5 for Rs.28 and sells them at a profit of 25 %. How much should a customer pay; if he buys
(i) only one pen ;
(ii) three pens ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 37

Question 2.
A fruit-seller sells 4 oranges for Rs. 3, gaining 50%. Find :
(i) C.P. of 4 oranges,
(ii) C.P. of one orange.
(iii) S.P. of one orange.
(iv) profit made by selling one orange.
(v) number of oranges brought and sold in order to gain Rs. 24.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 38

Question 3.
A man sells 12 articles for Rs. 80 gaining 33\(\frac { 1 }{ 3 }\) %. Find the number of articles bought by the man for Rs. 90.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 40

Question 4.
The cost price of 20 articles is same as the selling price of 16 articles. Find the gain percent.
Solution:
C.P. of 20 articles = S.P. of 16 articles.
Let C.P. of 1 article = Re. 1
C.P. of 20 articles = Rs.20
and C.P. of 16 articles = Rs.16
S.P. of 16 articles = Rs.20
[S.P. of 16 articles = C.P. of 20 articles]
Gain = Rs.20 – Rs.16 = Rs.4
Gain% = \(\frac { 4 }{ 16 }\) x 100
= \(\frac { 4 x 100 }{ 16 }\)
= 25%

Question 5.
The selling price of 15 articles is equal to the cost price of 12 articles. Find the gain or loss as percent.
Solution:
S.P. of 15 articles = C.P. of 12 articles
Let C.P. of 1 article = Re.1
C.P. of 12 article = Rs.12
and C.P. of 15 articles = Rs.15
S.P. of 15 articles = Rs.12
[S.P. of 15 articles = C.P. of 12 articles]
Loss = Rs.15 – Rs.12 = Rs.3
Loss% = \(\frac { 3 }{ 15 }\) x 100 = 20%

Question 6.
By selling 8 pens, Shyam loses equal to the cost price of 2 pens. Find his loss percent.
Solution:
Let C.P. of 1 pen = Re.1
C.P. of 2 pens = Rs.2
and C.P. of 8 pens = Rs.8
Loss = Rs.2 ….[Loss = C.P. of 2 Pens]
Loss% = \(\frac { 2 }{ 8 }\) x 100 = 25%

Question 7.
A shop-keeper bought rice worth Rs.4,500. He sold one-third of it at 10% profit.
If he desires a profit of 12% on the whole ; find :
(i) the selling price of the rest of the rice ;
(ii) the percentage profit on the rest of the rice.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 41

Question 8.
Mohan bought a certain number of note-books for Rs.600. He sold \(\frac { 1 }{ 4 }\) of them at 5 percent loss. At what price should he sell the remaining note-books so as to gain 10% on the whole ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 43

Question 9.
Raju sells a watch at 5% profit. Had he sold it for Rs.24 more ; he would have gained 11%. Find the cost price of the watch.
Solution:
Let C.P. of the watch = Rs.100
When profit = 5%; S.P. = Rs.(100+5) = Rs.105
When profit = 11%;
S.P. = Rs.(100 + 11) = Rs .111
Difference of two selling prices = Rs. 111 – Rs. 105 = Rs.6
When watch sold for Rs.6 more; then C.P. of the watch = Rs.100
When watch sold for Re. 1 more; then C.P. of the watch = Rs. \(\frac { 100 }{ 6 }\)
When watch sold for Rs.24 more; then C.P. of the watch = Rs. \(\frac { 100 }{ 6 }\) x 24 = Rs.400

Question 10.
A man sold a bicycle at 5% profit. If the cost had been 30% less and the selling price Rs.63 less, he would have made a profit of 30%. What is the cost price of the bicycle ?
Solution:
Let C.P. of the bicycle = Rs.100
In the I case :
When Profit = 5% ;
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 45

Question 11.
Renu sold an article at a loss of 8 percent. Had she bought it at 10% less and sold for Rs.36 more; she would have gained 20%. Find the cost price of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 46

Profit, Loss and Discount Exercise 8D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
An article is marked for Rs. 1,300 and is sold for Rs. 1,144 ; find the discount percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 48

Question 2.
The marked price of a dinning table is Rs. 23,600 and is available at a discount of 8%. Find its selling price.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 49

Question 3.
A wrist-watch is available at a discount of 9%. If the list-price of the watch is Rs. 1,400 ; find the discount given and the selling price of the watch.
Solution:
List price of the watch = Rs. 1,400
Discount = 9%
Discount = \(\frac { 1400 x 9 }{ 100 }\) = 14 x 9 = Rs. 126
S.P. = (List price – Discount) = Rs. (1400 – 126) = Rs. 1274

Question 4.
A shopkeeper sells an article for Rs. 248.50 after allowing a discount of 10%. Find the list price of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 50

Question 5.
A shop-keeper buys an article for Rs.450. He marks it at 20% above the cost price. Find :
(i) the marked price of the article.
(ii) the selling price, if he sells the articles at 10 percent discount.
(iii) the percentage discount given by him, if he sells the article for Rs.496.80
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 51

Question 6.
The list price of an article is Rs.800 and is available at a discount of 15 percent. Find :
(i) selling price of the article ;
(ii) cost price of the article, if a profit of 13\(\frac { 1 }{ 3 }\) % is made on selling it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 53

Question 7.
An article is marked at Rs. 2,250. By selling it at a discount of 12%, the dealer makes a profit of 10%. Find :
(i) the selling price of the article.
(ii) the cost price of the article for the dealer.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 55

Question 8.
By selling an article at 20% discount, a shopkeeper gains 25%. If the selling price of the article is Rs. 1,440 ; find :
(i) the marked price of the article.
(ii) the cost price of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 56

Question 9.
A shop-keeper marks his goods at 30 percent above the cost price and then gives a discount of 10 percent. Find his gain percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 57

Question 10.
A ready-made garments shop in Delhi, allows 20 percent discount on its garments and still makes a profit of 20 percent. Find the marked price of a dress which is bought by the shop-keeper for Rs.400.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 58

Question 11.
At 12% discount, the selling price of a pen is Rs. 13.20. Find its marked price. Also, find the new selling price of the pen, if it is sold at 5% discount.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 60

Question 12.
The cost price of an article is Rs. 2,400 and it is marked at 25% above the cost price. Find the profit and the profit percent, if the article is sold at 15% discount.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 61

Question 13.
Thirty articles are bought at Rs. 450 each. If one-third of these articles be sold at 6% loss; at what price must each of the remaining articles be sold in order to make a profit of 10% on the whole?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 63

Question 14.
The cost price of an article is 25% below the marked price. If the article is available at 15% discount and its cost price is Rs. 2,400; find:
(i) Its marked price
(ii) its selling price
(iii) the profit percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 64

Question 15.
Find a single discount (as percent) equivalent to following successive discounts:
(i) 20% and 12%
(ii) 10%, 20% and 20%
(iii) 20%, 10% and 5%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 65

Question 16.
Find the single discount (as percent) equivalent to successive discounts of:
(i) 80% and 80%
(ii) 60% and 60%
(iii) 60% and 80%
Solution:
(i) Successive discounts = 80% and 80%
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 68

Profit, Loss and Discount Exercise 8E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Rajat purchases a wrist-watch costing Rs. 540. The rate of Sales Tax is 8%. Find the total amount paid by Rajat for the watch.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 70
Total Amount of Watch = ₹ 540 + ₹ 43.20 = ₹ 583.20

Question 2.
Ramesh paid ₹ 345.60 as Sales Tax on a purchase of ₹ 3,840. Find the rate of Sales Tax.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 71

Question 3.
The price of a washing machine, inclusive of sales tax is ₹ 13,530/-. If the Sales Tax is 10%, find its basic (cost) price.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 72

Question 4.
Sarita purchases biscuits costing ₹ 158 on which the rate of Sales Tax is 6%. She also purchases some cosmetic goods costing ₹ 354 on which rate of Sales Tax is 9%. Find the total amount to be paid by Sarita.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 73
Total cost of cosmetic goods = ₹ 354 + ₹ 31.86 = ₹ 385.86
Total amount paid by Sarita = ₹ 167. 48 + 385.86 = ₹ 553. 34

Question 5.
The price of a T.V. set inclusive of sales tax of 9% is ₹ 13,407. Find its marked price. If Sales Tax is increased to 13%, how much more does the customer has to pay for the T.V. ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 74

Question 6.
The price of an article is ₹ 8,250 which includes Sales Tax at 10%. Find how much more or less does a customer pay for the article, if the Sales Tax on the article:
(i) increases to 15%
(ii) decreases to 6%
(iii) increases by 2%
(iv) decreases by 3%
Solution:
Price of an article = ₹ 8,250
Rate of Sales Tax = 10%
Let the list price = x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 75

Question 7.
A bicycle is available for ₹ 1,664 including Sales Tax. If the list price of the bicycle is ₹ 1,600, find :
(i) the rate of Sales Tax
(ii) the price a customer will pay for the bicycle if the Sales Tax is increased by 6%.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 78

Question 8.
When the rate of sale-tax is decreased from 9% to 6% for a coloured T.V. ; Mrs Geeta will save ₹ 780 in buying this T.V. Find the list price of the T.V.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 79

Question 9.
A shopkeeper sells an article for ₹ 21,384 including 10% sales-tax. However, the actual rate of sales-tax is 8%. Find the extra profit made by the dealer.
Solution:
Sale Price of an article including S.T. = ₹ 21384
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 80

Profit, Loss and Discount Exercise 8F – Selina Concise Mathematics Class 8 ICSE Solutions

[In this exercise, all the prices are excluding tax/VAT unless specified]
Question 1.
A shopkeeper buys an article for ₹ 8,000 and sells it for ₹ 10,000. If the rate of tax under VAT is 10%, find :
(i) tax paid by the shopkeeper
(ii) tax charged by the shopkeeper
(iii) VAT paid by the shopkeeper
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 81
(iii) VAT paid by the shopkeeper = ₹ 1000 – ₹ 800 = ₹ 200

Question 2.
A trader buys some goods for ₹ 12,000 and sells them for ₹ 15,000. If the rate of tax under VAT is 12%, find the VAT paid by the trader?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 82

Question 3.
The marked price of an article is ₹ 7,000 and is available at 20% discount. Manoj buys this article and then sold it at its marked price. If the rate of tax at each state is 10%, find the VAT paid by Manoj.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 83

Question 4.
A buys some goods for ₹ 4,000 and sold them to B for ₹ 5,000. B sold these goods to C for ₹ 6,000. If the rate of tax (under VAT) at each stage is 5%, find :
(i) VAT paid by A
(ii) VAT paid by B
Solution:
C.P. of some goods for A = ₹ 4000
C.P. of some goods for B = ₹ 5000
and C.P. for C = ₹ 6000
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 84

Question 5.
A buys an article for ₹ 8,000 and sold it to B at 20% profit. If the rate of tax under VAT is 8%, find :
(i) tax paid by A
(ii) tax charged by A
(iii) VAT paid by A
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 85

Question 6.
A shopkeeper purchases an article for ₹ 12,400 and sells it to a customer for ₹ 17,000. If the tax under VAT is 8%, find the VAT paid by the shopkeeper.
Solution:
C.P. of article = ₹ 12,400
Rate of VAT =8%
Total VAT = ₹ 12,400 x \(\frac { 8 }{ 100 }\) = ₹ 992
S.P. of the article = ₹ 17000
VAT charge 8% = ₹ 17000 x \(\frac { 8 }{ 100 }\) = ₹ 1360
Amount of VAT paid by the shopkeeper = ₹ 1360 – ₹ 992 = ₹ 368

Question 7.
A purchases an article for ₹ 7,200 and sells it to B for ₹ 9,600. B, in turn, sells the article to C for ₹ 11,000. If the tax (under VAT) is 10%, find the VAT paid by A and B.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 86

Question 8.
A manufacturer buys some goods for ₹ 60,000 and pays 5% tax. He sells these goods for ₹ 80,000 and charges tax at the rate of 12%. Find the VAT paid by the manufacturer.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Disco
VAT paid by the manufacturer = ₹ 9600 – ₹ 3000 = ₹ 6600

Question 9.
The cost of an article is ₹ 6,000 to a distributor, he sells it to a trader for ₹ 7,500 and the trader sells it further to a customer for ₹ 8,000. If the rate of tax under VAT is 8%; find the VAT paid by the:
(i) distributor
(ii) trader
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 88

Question 10.
The marked price of an article is ₹ 10,000. A buys it at 30% discount on the marked price and sells it at 10% discount on the marked price. If the rate of tax under VAT is 5%, find the amount of VAT paid by A.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 89

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

November 10, 2023May 24, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 13 Factorisation. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Factorisation Exercise 13A – Selina Concise Mathematics Class 8 ICSE Solutions

Factorise :
Question 1.
15x + 5
Solution:
15x + 5 = 5(3x + 1)

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 16

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 18

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 20

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 22

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 25

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 27

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 29

Question 16.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 30
Solution:

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 33

Factorisation Exercise 13B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Factorise : a² + ax + ab + bx
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 34

Question 2.
Factorise : a² – ab – ca + bc
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 35

Question 3.
Factorise : ab – 2b + a² – 2a
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 36

Question 4.
Factorise : a³ – a² + a – 1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 37

Question 5.
Factorise : 2a – 4b – xa + 2bx
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 38

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 40

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 42

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 44

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 46

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 48

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 50

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 52

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 54

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 56

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 58

Question 16.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 60

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 62

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 64

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 66

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 68

Factorisation Exercise 13C – Selina Concise Mathematics Class 8 ICSE Solutions

Note : a² – b² = (a + b) (a – b)
Question 1.
Factorise : 16 – 9x²
Solution:
16 – 9x² = (4)² – (3x)² = (4 + 3x) (4 – 3x)

Question 2.
Factorise : 1 – 100a²
Solution:
1 – 100a² = (1)² – (10a)² = (1 + 10a) (1 – 10a)

Question 3.
Factorise : 4x² – 81y²
Solution:
4x² – 81y² = (2x)² – (9y)² = (2x + 9y) (2x – 9y)

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 69
Solution:

Question 5.
Factorise : (a+2b)² – a²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 71

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 73

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 75

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 77

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 79

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 81

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 83

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 85

Question 13.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 86
Solution:

Question 14.
Evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 88
Solution:

Question 15.
Evaluate :
(0.7)² – (0.3)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 90

Question 16.
Evaluate :
(4.5)² – (1.5)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 91

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 93

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 95

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 97

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 99

Factorisation Exercise 13D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 100
= (x+4) (x+2)

Question 2.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 102

Question 3.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 104

Question 4.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 106

Question 5.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 108

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 110

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 112

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 114

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 116

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 118

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 119
= (a – 1)(3a – 2)

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 121

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 123

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 125

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 127

Question 16.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 129

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 132

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 134

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 136

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 138

Question 21.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 140

Question 22.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 142

Question 23.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 144

Question 24.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 146

Question 25.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 149

Factorisation Exercise 13E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In each case find whether the trinomial is a perfect square or not:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 150
Solution:

Question 2.
Factorise completely 2 – 8x².
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 153

Question 3.
Factorise completely : 8x²y – 18y³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 154

Question 4.
Factorise completely : ax² – ay²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 155

Question 5.
Factorise completely : 25x³ – x
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 156

Question 6.
Factorise completely : a⁴ – b⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 157

Question 7.
Factorise completely : 16x⁴ – 81y⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 158

Question 8.
Factorise completely : 625 – x⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 159

Question 9.
Factorise completely : x² – y² – 3x – 3y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 160

Question 10.
Factorise completely : x² – y² – 2x + 2y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 162

Question 11.
Factorise completely : 3x² + 15x – 72
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 163

Question 12.
Factorise completely : 2a² – 8a – 64
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 164

Question 13.
Factorise completely : 5b² + 45b + 90
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 165

Question 14.
Factorise completely : 3x²y + 11xy + 6y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 166

Question 15.
Factorise completely : 5ap² + 11ap + 2a
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 167

Question 16.
Factorise completely : a² + 2ab + b² – c²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 168

Question 17.
Factorise completely : x² + 6xy + 9y² + x + 3y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 169

Question 18.
Factorise completely : 4a² – 12ab + 9b² + 4a – 6b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 170

Question 19.
Factorise completely : 2a²b² – 98b⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 171

Question 20.
Factorise completely : a² – 16b² – 2a – 8b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 172

Factorisation Exercise 13F – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 173
Solution:

Question 2.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 175
Solution:

Question 3.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 178
Solution:

Question 4.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 182
Solution:

Question 5.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 186
Solution:

Question 6.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 188

Solution:

Question 7.
Factorise xy² – xz², Hence, find the value of:

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 192

Question 8.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 193
Solution:

Question 9.
Factorise a²b – b³ Using this result, find the value of 101² x 100 – 100³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 197

Question 10.
Evaluate (using factors): 301² x 300 – 300³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 198

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable

November 10, 2023May 23, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Linear Equations in one Variable Exercise 14A – Selina Concise Mathematics Class 8 ICSE Solutions

Solve the following equations:
Question 1.
20 = 6 + 2x
Solution:
20 = 6 + 2x
20 – 6 = 2x
14 = 2x
7 = x
x = 7

Question 2.
15 + x = 5x + 3
Solution:
15 – 3 = 5x – x
12 = 4x
3 = x
x = 3

Question 3.
\(\frac { 3x+2 }{ x-6 }\) = -7
Solution:
3x + 2 = -7 (x – 6) (by cross multiplying)
3x + 2 = -7x + 42
3x + 7x = 42 – 2
10x = 40
x = 4

Question 4.
3a – 4 = 2(4 – a)
Solution:
3a – 4 = 8 – 2a
3a + 2a = 8+4
5a = 12
a = 2.4

Question 5.
3(b – 4) = 2(4 – b)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 1

Question 6.
\(\frac { x+2 }{ 9 } =\frac { x+4 }{ 11 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 2

Question 7.
\(\frac { x-8 }{ 5 } =\frac { x-12 }{ 9 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 3

Question 8.
5(8x + 3) = 9(4x + 7)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 4

Question 9.
3(x +1) = 12 + 4(x – 1)
Solution:
3(x + 1) = 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x – 4x = 12 – 4 – 3
-x = 5
x = -5

Question 10.
\(\frac { 3x }{ 4 } -\frac { 1 }{ 4 } \left( x-20 \right) =\frac { x }{ 4 } +32\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 5

Question 11.
\(3a-\frac { 1 }{ 5 } =\frac { a }{ 5 } +5\frac { 2 }{ 5 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 6

Question 12.
\(\frac { x }{ 3 } -2\frac { 1 }{ 2 } =\frac { 4x }{ 9 } -\frac { 2x }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 7

Question 13.
\(\frac { 4\left( y+2 \right) }{ 5 } =7+\frac { 5y }{ 13 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 9

Question 14.
\(\frac { a+5 }{ 6 } -\frac { a+1 }{ 9 } =\frac { a+3 }{ 4 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 8

Question 15.
\(\frac { 2x-13 }{ 5 } -\frac { x-3 }{ 11 } =\frac { x-9 }{ 5 } +1\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 13

Question 16.
6(6x – 5) – 5 (7x – 8) = 12 (4 – x) + 1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 14

Question 17.
(x – 5) (x + 3) = (x – 7) (x + 4)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 15

Question 18.
(x – 5)2 – (x + 2)2 = -2
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 16

Question 19.
(x – 1) (x + 6) – (x – 2) (x – 3) = 3
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 18

Question 20.
\(\frac { 3x }{ x+6 } -\frac { x }{ x+5 } =2\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 19

Question 21.
\(\frac { 1 }{ x-1 } +\frac { 2 }{ x-2 } =\frac { 3 }{ x-3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 20

Question 22.
\(\frac { x-1 }{ 7x-14 } =\frac { x-3 }{ 7x-26 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 22

Question 23.
\(\frac { 1 }{ x-1 } -\frac { 1 }{ x } =\frac { 1 }{ x+3 } -\frac { 1 }{ x+4 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 23

Question 24.
Solve: \(\frac { 2x }{ 3 } -\frac { x-1 }{ 6 } +\frac { 7x-1 }{ 4 } =2\frac { 1 }{ 6 }\)
Hence, find the value of ‘a’, if \(\frac { 1 }{ a }\) + 5x = 8.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 24

Question 25.
Solve: \(\frac { 4-3x }{ 5 } +\frac { 7-x }{ 3 } +4\frac { 1 }{ 3 } =0\)
Hence find the value of ‘p’ if 2p – 2x + 1 = 0
Solution:

Hence x = 8
Now, 3p – 2x + 1=0
⇒ 3p – 2 x 8 + 1 = 0
⇒ 3p – 16 + 1 =0
⇒ 3p – 15 = 0.
⇒ 3p=15
⇒ p = 5

Question 26.
Solve: \(0.25+\frac { 1.95 }{ x } =0.9\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 28

Question 27.
Solve: \(5x-\left( 4x+\frac { 5x-4 }{ 7 } \right) =\frac { 4x-14 }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 29

Linear Equations in one Variable Exercise 14B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fifteen less than 4 times a number is 9. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
15 less than 4 times the number = 4x-15
According to the statement :
4x – 15 = 9
⇒ 4x = 9 + 15
⇒ 4x = 24
⇒ x = 6

Question 2.
If Megha’s age is increased by three times her age, the result is 60 years. Find her age
Solution:
Let Megha’s age = x years
Three times Megha’s age = 3x years
According to the statement :
x + 3x = 60
=> 4x = 60
=> x = 15
Megha’s age = 15 years

Question 3.
28 is 12 less than 4 times a number. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
12 less than 4 times the number = 4x – 12
According to the statement
4x – 12 = 28
=> 4x = 28 + 12
=> 4x = 40
x = 10
Required number = 10

Question 4.
Five less than 3 times a number is -20. Find the number.
Solution:
Let the required number = x
3 times the number = 3x
5 less than 3 times the number = 3x – 5
According to statement :
3x – 5 = -20
=> 3x = -20 + 5
=> 3x = -15
=> x = -5
Required number = -5

Question 5.
Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she ?
Solution:
Let Neetu’s age = x years
3 times Neetu’s age = 3x years
Fifteen more than 3 times Neetu’s age = (3x + 15) years
4 times Neetu’s age = 4x
According to the statement :
4x = 3x + 15
=> 4x – 3x = 15
=> x = 15
Neetu’s age = 15 years

Question 6.
A number decreased by 30 is the same as 14 decreased by 3 times the number; Find the number.
Solution:
Let the required number = x
The number decreased by 30 = x – 30
14 decreased by 3 times the number = 14 – 3x
According to the statement :
x – 30 = 14 – 3x
=> x + 3x = 14 + 30
=> 4x = 44
x = 11
Required number =11

Question 7.
A’s salary is same as 4 times B’s salary. If together they earn Rs.3,750 a month, find the salary of each.
Solution:
Let B’s salary = Rs. x
A’s salary = Rs. 4x
According to the statement :
x + 4x = 3750
=> 5x = 3750
=> x = 750
4x = 750 x 4 = 3000
A’s salary = Rs. 3000
B’s salary = Rs. 750

Question 8.
Separate 178 into two parts so that the first part is 8 less than twice the second part.
Solution:
Let first part = x
Second part = 178 – x
According to the problem :
First Part = 8 less than twice the second part
x = 2(178 – x) – 8
=> x = 356 – 2x – 8
=> x+2x = 356 – 8
=> 3x = 348
=> x = 116
First Part = 116
=> Second Part = 178 – x = 178 – 116 = 62
First Part = 116
=> Second Part = 62

Alternative Method :
Let Second part = x
First part = 2x – 8
According to the problem :
x + 2x – 8 = 178
=> x + 2x = 178 + 8
=> 3x = 186
=> x = 62
First part = 2x – 8 = 2 x 62 – 8 = 124 – 8 = 116
First part = 116
Second part = 62

Question 9.
Six more than one-fourth of a number is two-fifth of the number. Find the number.
Solution:
Let the required number = x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 30
x = 40
Required number = 40

Question 10.
The length of a rectangle is twice its width. If its perimeter is 54 cm; find its length.
Solution:
Let width of the rectangle = x cm
Length of the rectangle = 2x cm
Perimeter of the rectangle = 2 [Length + Width] = 2 [2x + x] = 2 x 3x = 6x cm
Given perimeter = 54 cm
6x = 54
=> x = 9
Length = 2x = 2 x 9 = 18 cm

Question 11.
A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the origi¬nal rectangle.
Solution:
Let width of the original rectangle = x cm
Length of the original rectangle = (2x – 5)cm
Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm
New width of the rectangle = (x + 2) cm
New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm
Given; new perimeter = 74 cm
6x – 16 = 74
=> 6x = 74 + 16
=> 6x = 90
=>x = 15
Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm
Width of the original rectangle = x = 15 cm

Question 12.
The sum of three consecutive odd numbers is 57. Find the numbers.
Solution:
Let the three consecutive odd numbers be x, x+2, x+4.
According to the statement :
x + x + 2 + x + 4 = 57
=> x + x + x = 57 – 2 – 4
=> 3x = 51
=> x = 17
Three consecutive odd numbers are 17, 19, 21

Question 13.
A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Solution:
Let present age of the son = x years
present age of the man = 3x years
In 12 years :
Son’s age will be = (x + 12) years
The man’s age will be = (3x + 12) years
According to the statement :
3x + 12 = 2(x + 12)
=> 3x + 12 = 2x + 24
=> 3x – 2x = 24 – 12
=> x = 12
3x = 3 x 12 = 36
Hence, present age of the man = 36 years
Present age of the son = 12 years.

Question 14.
A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?
Solution:
Man’s age = 42 years
Son’s age = 12 years
Let after x years the age of the son will be half the age of the man.
Man’s age after x years = (42 + x) years
Son’s age after x years = (12 + x) years
According to the statement :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 31
Hence after 18 years, the age of the son will be half the age of the man

Question 15.
A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 32

Question 16.
The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.
Solution:
Let one number = x
Second number = x + 3 [Difference of two numbers is 3]
According to the statement :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 34
One number = 10
Second number = x + 3 = 10 + 3 = 13

Question 17.
Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.
Solution:
Let two consecutive natural numbers = x, x+1
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 35
Two consecutive numbers are 24 and 25

Question 18.
Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers.
Solution:
Let the three consecutive whole numbers be x, x + 1 and x + 2
According to the statement:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 36
x = 50
x + 1 = 50+1 = 51
x + 2 = 50 + 2 = 52
Three consecutive whole numbers are 50, 51 and 52

Question 19.
If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.
Solution:
Let x be added to each number, then the numbers will be 5 + x, 11 + x, 15 + x and 31 + x
According to the condition
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 37
1 should be added

Question 20.
The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.
Solution:
Let present age of son = x year
Then age of his father = 2x
8 years hence,
Age of son = (x + 8) years and age of father = (2x + 8) years
According to the condition,
\(\frac { 2x+8 }{ x+8 } =\frac { 7 }{ 4 }\)
=> 8x + 32 = 7x + 56
=> 8x – 7x = 56 – 32
=> x = 24
Present age of son = 24 years
and age of father = 2x = 2 x 24 = 48 years
Hence age of man = 48 years and age of his son = 24 years

Linear Equations in one Variable Exercise 14C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Solve:
(i) \(\frac { 1 }{ 3 } x-6=\frac { 5 }{ 2 }\)
(ii) \(\frac { 2x }{ 3 } -\frac { 3x }{ 8 } =\frac { 7 }{ 12 }\)
(iii) (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1) = 0
(iv) \(\frac { 1 }{ 10 } -\frac { 7 }{ x } =35\)
(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
(vi) x + 7 – \(\frac { 8x }{ 3 } =\frac { 17x }{ 6 } -\frac { 5x }{ 8 }\)
(vii) \(\frac { 3x-2 }{ 4 } -\frac { 2x+3 }{ 3 } =\frac { 2 }{ 3 } -x\)
(viii) \(\frac { x+2 }{ 6 } -\left( \frac { 11-x }{ 3 } -\frac { 1 }{ 4 } \right) =\frac { 3x-4 }{ 12 }\)
(ix) \(\frac { 2 }{ 5x } -\frac { 5 }{ 3x } =\frac { 1 }{ 15 }\)
(x) \(\frac { x+2 }{ 3 } -\frac { x+1 }{ 5 } =\frac { x-3 }{ 4 } -1\)
(xi) \(\frac { 3x-2 }{ 3 } +\frac { 2x+3 }{ 2 } =x+\frac { 7 }{ 6 }\)
(xii) \(x-\frac { x-1 }{ 2 } =1-\frac { x-2 }{ 3 }\)
(xiii) \(\frac { 9x+7 }{ 2 } -\left( x-\frac { x-2 }{ 7 } \right) =36\)
(xiv) \(\frac { 6x+1 }{ 2 } +1=\frac { 7x-3 }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 38

Question 2.
After 12 years, I shall be 3 times as old as 1 was 4 years ago. Find my present age.
Solution:
Let present age = x years
According to question,
(x + 12) = 3(x – 4)
x + 12 = 3x – 12
2x = 24
=> x = 12 years
Present age = 12 years

Question 3.
A man sold an article for 7396 and gained 10% on it. Find the cost price of the article
Solution:
S.P. of article = ₹ 396
Gain = 10%
Let cost price = ₹ x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 52
Cost price of an article = ₹ 360

Question 4.
The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Solution:
Let the first number = x
and the second number = y
According to question,
x + y = 4500 ……(i)
and 10% x = 12.5% y
i.e. 10x = 12.5y
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 53
x = 2500
Hence, the numbers are 2500 and 2000

Question 5.
The sum of two numbers is 405 and their ratio is 8 : 7. Find the numbers.
Solution:
Let the first number = x
and the second number = 7
According to the question, x + y = 405 ……..(i)
and the numbers are in the ratio 8 : 7
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 54

x = 189
Hence, the numbers are 189 and 216

Question 6.
The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
Solution:
Ratio in the present ages of A and B = 7 : 5
Let age of A = 7x years
Let age of B = 5x years
10 years hence,
Then age of A = 7x + 10 years
and age of B = 5x + 10 years
According to the condition,
\(\frac { 7x+10 }{ 5x+10 } =\frac { 9 }{ 7 }\)
By crossing multiplication
7(7x + 10) = 9(5x + 10)
=> 49x + 70 = 45x + 90
=> 49x – 45x = 90 – 70
=> 4x = 20
=> x = 5
Present age of A = 7x = 7 x 5 = 35 years
and present age of B = 5x = 5 x 5 = 25 years

Question 7.
Find the number whose double is 45 greater than its half.
Solution:
Let the required number = x
Double of it = 2x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 56
Required number = 30

Question 8.
The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution:
Let first number = x
and The second number = x + 1
According to the condition,
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 57
First number = 15
and second number = 15 + 1 = 16
Hence, the numbers are 15, 16

Question 9.
Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Solution:
Let the required number = x
5 times of it = 5x
Twice of it = 2x
According to the condition,
5x – 5 = 2x + 4
=> 5x – 2x = 4 + 5
=> 3x = 9
=> x = 3
Required number = 3

Question 10.
The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes \(\frac { 2 }{ 3 }\). Find the original fraction.
Solution:
Let denominator of the original fraction = x
Then numerator = x – 5
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 58 .

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

November 10, 2023May 23, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

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Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square of :
(i) 59
(ii) 63
(iii) 15
Solution:
(i) Square of 59= 59 x 59 = 3481
(ii) Square of 6.3 = 6.3 x 6.3 = 39.69
(iii) Square of 15 = 15 x 15 = 225

Question 2.
By splitting into prime factors, find the square root of :
(i) 11025
(if) 396900
(iii) 194481
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -1

Question 3.
(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be mutliplied so that the product is a perfect square?
Solution:

On grouping the prime factors of 2592 as shown; on factor i.e. 2 is left which cannot be paired with equal factor.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -4
The given number should be multiplied by 2 to make the given number a perfect square.

On grouping the prime factors of 12748 as shown; one factor i.e. 3187 is left which cannot be paired with equal factor.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -6
The given number should be multiplied by 3187.

Question 4.
Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -7

Question 5.
Find the square root of :
(i) 0.1764
(ii) \(96\frac { 1 }{ 25 }\)
(iii) 0.0169
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -8

Question 6.
Evaluate
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -9

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -11

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -14
Solution:

Question 8.
A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all ₹ 1,296
Solution:
Let the number of days he had spent = x
Number of rupees spent in each day = x
Total money spent = x x x = x² = 1,296 (given)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -18

Question 9.
Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.
Solution:
Total number of students = 745
Students left after standing in arrangement = 16
No. of students who were to be arranged = 745 – 16 = 729
The number of rows = no. of students in each row
No. of rows = √729
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -19

Question 10.
13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -20

Question 11.
Find the smallest perfect square divisible by 3, 4, 5 and 6.
Solution:
L.C.M. of 3, 4, 5, 6 = 2 x 2 x 3 x 5 = 60
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -21
in which 3 and 5 are not in pairs L.C.M. = 2 x 3 x 2 x 5 = 60
We should multiple it by 3 x 5 i.e. by 15
Required perfect square = 60 x 15 = 900

Question 12.
If √784 = 28, find the value of:
(i) √7.84 + √78400
(ii) √0.0784 + √0.000784
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -22

Squares and Square Roots Exercise 3B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -23

Question 2.
Find the square root of:
(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -26

Question 3.
Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -28

Required square root = 0.78 upto two places of decimals.

Question 4.
Find the square root of each of the following correct to two decimal places:
(i) \(3\frac { 4 }{ 5 }\)
(ii) \(6\frac { 7 }{ 8 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -31

Question 5.
For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -33

Question 6.
For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Solution:
(i) 511
Taking square root of 511, we find that 27 has been left We see that 511 is greater than (22)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -35
On adding the required number to 511, we get (23)² i.e., 529
So, the required number = 529 – 511 = 18
(ii) 7172
Taking square root of 7172, we find that 116 has been left
We see that 7172 is greater than (84)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -36
Taking square root of 55078, we find that 322 has been left
We see that 55078 is greater than (234)²
On adding the required number to 55078, we get (235)² i.e., 55225
Required number = 55225 – 55078 = 147

Question 7.
Find the square root of 7 correct to two decimal places; then use it to find the value of \(\sqrt { \frac { 4+\sqrt { 7 } }{ 4-\sqrt { 7 } } }\) correct to three significant digits.
Solution:
√7 = 2.645 = 2.65
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -37

Question 8.
Find the value of √5 correct to 2 decimal places; then use it to find the square root of \(\sqrt { \frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } }\) correct to 2 significant digits.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -38

Question 9.
Find the square root of:
(i) \(\frac { 1764 }{ 2809 }\)
(ii) \(\frac { 507 }{ 4107 }\)
(iii) \(\sqrt { 108\times 2028 }\)
(iv) 0.01 + √0.0064
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -40

Question 10.
Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.
Solution:
Square root of 7.832
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -42
√7.832 = 2.80 upto two decimal places
= 2.8 upto two significant places

Question 11.
Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.
Solution:
Clearly, if 49 is subtracted from 1205, the number will be a perfect square.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -43

Question 12.
Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -44

Question 13.
Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Solution:
Clearly; if 12 is subtracted from 2037, the remainder will be a perfect square.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -45

Question 14.
Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -46

Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699
Solution:
We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9
So, the following numbers can be squares : 3051, 5684, and 50699 i.e., (i), (iii), and (v)

Question 2.
Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64
Solution:
The square of the following numbers will have 1 at their units place as (1)² = 1, (9)² = 81
81 and 139 i.e., (i) and (iii)

Question 3.
Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Solution:
The square of the following numbers will not have 1 at their units place : as only (1)² = 1, (9)² = 81 have 1 at then units place
322, 572, 2652 i.e., (i), (ii) and (v)

Question 4.
Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98
Solution:
The squares of the following numbers, Will not have 6 at their units place as only (4)² = 16, (6)² = 36 has but its units place 35, 23 and 98 i.e., (i), (ii), and (v)

Question 5.
Which of the following numbers will have 6 at their unit’s place :
(i) 262
(ii) 492
(iii) 342
(iv) 432
(v) 2442
Solution:
The following numbers have 6 at their units place as (4)² = 16, (6)² = 36 has 6 at their units place 262, 342, 2442 i.e., (i), (iii) and (v)

Question 6.
If a number ends with 3 zeroes, how many zeroes will its square have ?
Solution:
We know that if a number ends with n zeros, then its square will have 2n zeroes at their ends
A number ends with 3 zeroes, then its square will have 3 x 2 = 6 zeroes

Question 7.
If the square of a number ends with 10 zeroes, how many zeroes will the number have ?
Solution:
We know that if a number ends with n zeros Then its square will have 2n zeroes Conversely, if square of a number have 2n zeros at their ends then the number will have n zeroes
The square of a number ends 10 zeroes, then the number will have \(\frac { 10 }{ 2 }\) = 5 zeroes

Question 8.
Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Solution:
No, it is not possible for the square of a number, to have 5 zeroes which is odd because the number of zeros of the square must be 2n zeroes i.e., even number of zeroes.

Question 9.
Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Solution:
A number having 2,3,7 or 8 at the unit place is never a perfect square.

Question 10.
State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(iii) 76
(iv) 75
Solution:
(i) 23 – odd
(ii) 54 – even
(iii) 76 – odd
(iv) 75 – even

Question 11.
Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Solution:
No, number has an even number of zeroes.

Question 12.
Evaluate:
(i) 37² – 36²
(ii) 85² – 84²
(iii) 101² – 100²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -47

Question 13.
Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53
Solution:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23
= Sum of first 12 odd natural numbers = 122 = 144
(ii) 1+3 + 5 + 7 + 9 + ……….. + 39 + 41
= Sum of first 21 odd natural numbers = 212 = 441
(iii) 1 + 3 + 5 + 7 + 9 + ……………. + 51 + 53
= Sum of first 27 odd natural number = 272 = 729

Question 14.
Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Solution:
The three sets of Pythagorean triplets such that each set has numbers less than 30 are 3, 4 and 5; 6, 8 and 10; 5, 12 and 13
Proof:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -49