Chapter 9 – Rational Numbers
Page No 182:
Question 1:
List five rational numbers between:
(i) − 1 and 0 (ii) − 2 and − 1
(iii) 
(iv) 
(iv) Answer:
(i) −1 and 0
(ii) −2 and −1
Five rational numbers are
(iii) 
Five rational numbers are
(iv) 
Five rational numbers are
Question 2:
Write four more rational numbers in each of the following patterns:
(i) 
(ii) 
(ii)
(iii) 
(iv) 
(iv) Answer:
(i) 
It can be observed that the numerator is a multiple of 3 while the denominator is a multiple of 5 and as we increase them further, these multiples are increasing. Therefore, the next four rational numbers in this pattern are
(ii)
The next four rational numbers in this pattern are
(iii)
The next four rational numbers in this pattern are
(iv) 
The next four rational numbers in this pattern are
Page No 183:
Question 3:
Give four rational numbers equivalent to:
(i) 
(ii) 
(iii) 
(ii) (iii) Answer:
(i) 
Four rational numbers are
(ii) 
Four rational numbers are
(iii) 
Four rational numbers are
Question 4:
Draw the number line and represent the following rational numbers on it:
(i) 
(ii) 
(ii)
(iii) 
(iv) 
(iv) Answer:
(i) 
This fraction represents 3 parts out of 4 equal parts. Therefore, each space between two integers on number line must be divided into 4 equal parts.
can be represented as
(ii) 
This fraction represents 5 parts out of 8 equal parts. Negative sign represents that it is on the negative side of number line. Therefore, each space between two integers on number line must be divided into 8 equal parts.
can be represented as
(iii) 
This fraction represents 1 full part and 3 parts out of 4 equal parts. Negative sign represents that it is on the negative side of number line. Therefore, each space between two integers on number line must be divided into 4 equal parts.
can be represented as
(iv) 
This fraction represents 7 parts out of 8 equal parts. Therefore, each space between two integers on number line must be divided into 8 equal parts.
can be represented as
Question 5:
The points P, Q, R, S, T, U, A and B on the number line are such that,
TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.
Answer:
Distance between U and T = 1 unit
It is divided into 3 equal parts.
TR = RS = SU =
R = 
S = 
Similarly,
AB = 1 unit
It is divided into 3 equal parts.
P = 
Q = 
Question 6:
Which of the following pairs represent the same rational number?
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(vi) 
(v) (vi)
(vii) 
Answer:
(i) 
As
, therefore, it does not represent same rational numbers.
, therefore, it does not represent same rational numbers.
(ii) 
Therefore, it represents same rational numbers.
(iii) 
Therefore, it represents same rational numbers.
(iv) 
Therefore, it represents same rational numbers.
(v) 
Therefore, it represents same rational numbers.
(vi) 
As
, therefore, it does not represent same rational numbers.
, therefore, it does not represent same rational numbers.
(vii) 
Question 7:
Rewrite the following rational numbers in the simplest form:
(i) 
(ii) 
(ii)
(iii) 
(iv) 
(iv) Answer:
(i) 
(ii) 
(iii) 
(iv) 
Question 8:
Fill in the boxes with the correct symbol out of >, <, and =
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(vi) 
(v) (vi)
(vii) 
Answer:
(i)
As −15 < 14,
Therefore,
(ii)
As −28 < −25
Therefore, 
(iii) Here, 
Therefore, 
(iv)
As −32 > −35,
Therefore, 
(v)
As −4 < −3,
Therefore, 
(vi) 
(vii) 
Page No 184:
Question 9:
Which is greater in each of the following?
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(v) Answer:
(i) 
By converting these into like fractions,
As 15 > 4, therefore, 
is greater.
is greater.
(ii) 
(iii)
By converting these into like fractions,
(iv) 
(v) 
By converting these into like fractions,
Question 10:
Write the following rational numbers in ascending order:
(i) 
(ii) 
(iii) 
(ii) (iii) Answer:
(i) 
As −3 < −2 < −1,
(ii) 
By converting these into like fractions,
As −12 < −3 < −2,
(iii) 
By converting these into like fractions,
As −42 < −21 < −12,
Page No 190:
Question 1:
Find the sum:
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(vi) 
(v) (vi)
(vii) 
Answer:
(i)
45+-11 4 = 45-11 4 = 16-5520 = -39 20(ii) 
L.C.M of 3 and 5 is 15.
(iii) 
L.C.M of 10 and 15 is 30.
(iv) 
L.C.M of 11 and 9 is 99.
(v) 
L.C.M of 19 and 57 is 57.
(vi) 
(vii) 
= 
=
L.C.M of 3 and 5 is 15.
Question 2:
Find
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(v) Answer:
(i) 
L.C.M of 24 and 36 is 72.
(ii) 
L.C.M of 63 and 7 is 63.
(iii) 
L.C.M of 13 and 15 is 195.
(iv) 
L.C.M of 8 and 11 is 88.
(v) 
L.C.M of 9 and 1 is 9.
Question 3:
Find the product:
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(vi) 
(v) (vi) Answer:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Question 4:
Find the value of:
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(vi) 
(v) (vi)
(vii) 
Answer:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
Question 3:
Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of
triangle is this?
Answer:
The steps of construction are as follows.
(i) Draw a line segment QR of length 3.5 cm.
(ii) Taking point Q as centre, draw an arc of 4 cm radius.
(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point P.
(iv) Join P to Q and R.
ΔPQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR), therefore, ΔPQR is an isosceles triangle.
Question 4:
Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Answer:
The steps of construction are as follows.
(i) Draw a line segment BC of length 6 cm.
(ii) Taking point C as centre, draw an arc of 6.5 cm radius.
(iii) Taking point B as centre, draw an arc of radius 2.5 cm to meet the previous arc at point A.
(iv) Join A to B and C.
ΔABC is the required triangle. ∠B can be measured with the help of protractor. It comes to 90º.
Page No 200:
Question 1:
Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Answer:
The rough sketch of the required ΔDEF is as follows.
The steps of construction are as follows.
(i)Draw a line segment DE of length 5 cm.
(ii) At point D, draw a ray DX making an angle of 90° with DE.
(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F.
(iv) Join F to E. ΔDEF is the required triangle.
Question 2:
Construct an isosceles triangle in which the lengths of each of its equal sides
is 6.5 cm and the angle between them is 110°.
Answer:
An isosceles triangle PQR has to be constructed with PQ = QR = 6.5 cm. A rough sketch of the required triangle can be drawn as follows.
The steps of construction are as follows.
(i) Draw the line segment QR of length 6.5 cm.
(ii) At point Q, draw a ray QX making an angle 110° with QR.
(iii) Taking Q as centre, draw an arc of 6.5 cm radius. It intersects QX at point P.
(iv) Join P to R to obtain the required triangle PQR.
Question 3:
Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Answer:
A rough sketch of the required triangle is as follows.
The steps of construction are as follows.
(i) Draw a line segment BC of length 7.5 cm.
(ii) At point C, draw a ray CX making 60º with BC.
(iii) Taking C as centre, draw an arc of 5 cm radius. It intersects CX at point A.
(iv) Join A to B to obtain triangle ABC.
Page No 202:
Question 1:
Construct ΔABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
Answer:
A rough sketch of the required ΔABC is as follows.
The steps of construction are as follows.
(i)Draw a line segment AB of length 5.8 cm.
(ii)At point A, draw a ray AX making 60º angle with AB.
(iii) At point B, draw a ray BY, making 30º angle with AB.
(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of these two rays.
This is the required triangle ABC.
Question 2:
Construct ΔPQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.
(Hint: Recall angle sum property of a triangle).
Answer:
A rough sketch of the required ΔPQR is as follows.
In order to construct ΔPQR, the measure of ∠RPQ has to be calculated.
According to the angle sum property of triangles,
∠PQR + ∠PRQ + ∠RPQ = 180º
105º + 40º + ∠RPQ = 180º
145º + ∠RPQ = 180º
∠RPQ = 180° − 145° = 35°
The steps of construction are as follows.
(i) Draw a line segment PQ of length 5 cm.
(ii) At P, draw a ray PX making an angle of 35º with PQ.
(iii) At point Q, draw a ray QY making an angle of 105º with PQ.
(iv)Point R has to lie on both the rays, PX and QY. Therefore, R is the point of intersection of these two rays.
This is the required triangle PQR.
Question 3:
Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E =
110° and m∠F = 80°. Justify your answer.
Answer:
Given that,
m∠E = 110° and m∠F = 80°
Therefore,
m∠E + m∠F = 110° + 80° = 190°
However, according to the angle sum property of triangles, we should obtain
m∠E + m∠F + m∠D = 180°
Therefore, the angle sum property is not followed by the given triangle. And thus, we cannot construct ΔDEF with the given measurements.
Also, it can be observed that point D should lie on both rays, EX and FY, for constructing the required triangle. However, both rays are not intersecting each other. Therefore, the required triangle cannot be formed.
Page No 203:
Question 1:
Construct the right angled ΔPQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
Answer:
A rough sketch of ΔPQR is as follows.
The steps of construction are as follows.
(i) Draw a line segment QR of length 8 cm.
(ii) At point Q, draw a ray QX making 90º with QR.
(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P.
(iv) Join P to R. ΔPQR is the required right-angled triangle.
Question 2:
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Answer:
A right-angled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to be constructed. A rough sketch of ΔABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment BC of length 4 cm.
(ii) At point B, draw a ray BX making an angle of 90º with BC.
(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A.
(iv) Join A to C to obtain the required ΔABC.
Question 3:
Construct an isosceles right-angled triangle ABC, where, m∠ACB = 90° and AC = 6 cm.
Answer:
In an isosceles triangle, the lengths of any two sides are equal.
Let in ΔABC, AC = BC = 6 cm. A rough sketch of this ΔABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment AC of length 6 cm.
(ii) At point C, draw a ray CX making an angle of 90º with AC.
(iii) Taking point C as centre, draw an arc of 6 cm radius to intersect CX at point B.
(iv) Join A to B to obtain the required ΔABC.
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