Std 8 Mathematics · Chapter 1

1. Rational and Irrational Numbers

Solutions to Practice Sets 1.1, 1.2, 1.3 and 1.4, with number-line diagrams. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

Practice Set 1.1

1Show the following numbers on a number line. Draw a separate number line for each example.

(1) 32, 52, −32

The denominator is 2, so divide each unit on both sides of 0 into 2 equal parts.

32 = (1 + 12): count 3 parts to the right of 0.
52 = (2 + 12): count 5 parts to the right of 0.
−32 = −(1 + 12): count 3 parts to the left of 0.

Number line for (1)

(2) 75, −25, −45

The denominator is 5, so divide each unit on both sides of 0 into 5 equal parts.

75 = (1 + 25): count 7 parts to the right of 0.
−25: count 2 parts to the left of 0.
−45: count 4 parts to the left of 0.

Number line for (2)

(3) −58, 118

The denominator is 8, so divide each unit on both sides of 0 into 8 equal parts.

−58: count 5 parts to the left of 0.
118 = (1 + 38): count 11 parts to the right of 0.

Number line for (3)

(4) 1310, −1710

The denominator is 10, so divide each unit on both sides of 0 into 10 equal parts.

1310 = (1 + 310): count 13 parts to the right of 0.
−1710 = −(1 + 710): count 17 parts to the left of 0.

Number line for (4)

2Observe the number line and answer the questions.

Each unit is divided into 4 equal parts, so each small division is 14.

(1) Which number is indicated by point B?

Point B is 2 divisions to the right of −3, so B = −3 + 24 = −104 = −52.

Answer: B shows −104, that is, −52

(2) Which point indicates the number 134?

134 = 1 + 34 = 74. It is the 3rd division after 1, that is, the 7th division to the right of 0.

Answer: Point C

(3) State whether the statement, ‘the point D denotes the number 52’ is true or false.

Point D is 2 divisions after 2, so D = 2 + 24 = 104 = 52.

Answer: True

Practice Set 1.2

1Compare the following numbers. Verify the comparisons using a number line.

Rule used: if ab and cd have positive denominators, compare a × d with b × c. The same symbol (<, =, >) holds between the fractions.

(1) −7, −27 > 2, so −7 < −2.
(Of two negative numbers, the one with the bigger number part is smaller.)
−7 < −2

(2) 0, −95−95 is negative, and every negative number is less than 0.
0 > −95

(3) 87, 087 is positive, and every positive number is greater than 0.
87 > 0

(4) −54, 14A negative number is less than a positive number. (Also, the denominators are equal and −5 < 1.)
−54 < 14

(5) 4029, 14129Denominators are equal, so compare numerators: 40 < 141.
4029 < 14129

(6) −1720, −1320Denominators are equal, so compare numerators: −17 < −13.
−1720 < −1320

(7) 1512, 71615 × 16 = 240 and 12 × 7 = 84.
240 > 84
1512 > 716

(8) −258, −94−94 = −9 × 24 × 2 = −188.
−25 < −18
−258 < −94

(9) 1215, 3535 = 3 × 35 × 3 = 915.
12 > 9
1215 > 35

(10) −711, −34First compare 711 and 34: 7 × 4 = 28 and 11 × 3 = 33, so 711 < 34.
For negatives the order reverses.
−711 > −34

Verifying on a number line: mark both numbers on one number line. The number lying to the right is always the greater one. For example, in (10), −711 ≈ −0.64 lies to the right of −34 = −0.75, so −711 is greater.

Practice Set 1.3

1Write the following rational numbers in decimal form.

(1) 937

Dividing, the remainders are 9, 16, 12 and then 9 comes again, so the digits 243 repeat: 0.243243…

Answer: 937 = 0.243

(2) 1842

First simplify: 1842 = 18 ÷ 642 ÷ 6 = 37. Dividing, the remainders are 3, 2, 6, 4, 5, 1 and then 3 comes again, so the digits 428571 repeat: 0.428571428571…

Answer: 1842 = 0.428571

(3) 914

Dividing, the remainders are 9, 6, 4, 12, 8, 10, 2 and then 6 comes again, so the digits 428571 repeat: 0.6428571428571…

Answer: 914 = 0.6428571

(4) −1035

103 ÷ 5 = 20.6 and the remainder becomes 0, so the division ends. This is a terminating decimal.

Answer: −1035 = −20.6

(5) −1113

Dividing, the remainders are 11, 6, 8, 2, 7, 5 and then 11 comes again, so the digits 846153 repeat: −0.846153846153…

Answer: −1113 = −0.846153

Practice Set 1.4

1The number √2 is shown on a number line. Steps are given to show √3 on the number line using √2. Fill in the boxes properly and complete the activity.

OQ = √2, QR = 1 unit, and the arc with radius OR cuts the number line at C = √3.

The point Q on the number line shows the number √2
A line perpendicular to the number line is drawn through Q, and point R is taken on it at unit distance from Q.
Right-angled △ORQ is obtained by drawing seg OR, with l(OQ) = √2 and l(QR) = 1.

∴ by Pythagoras theorem,

[l(OR)]² = [l(OQ)]² + [l(QR)]²

= √2² + 1² = 2 + 1

= 3 ∴ l(OR) = √3

Draw an arc with centre O and radius OR. It meets the number line at C, so the point C shows the number √3.

2Show the number √5 on the number line.

Step 1: Draw a number line and mark O = 0. Mark point A showing 2, so OA = 2 units.

Step 2: Through A, draw a line perpendicular to the number line. On it, take point P such that AP = 1 unit.

Step 3: Draw seg OP. △OAP is a right-angled triangle with ∠OAP = 90°.

Step 4: By Pythagoras theorem, OP² = OA² + AP² = 2² + 1² = 4 + 1 = 5. ∴ OP = √5.

Step 5: With centre O and radius OP, draw an arc cutting the number line at C. Then OC = OP = √5.

Point C shows √5.

Answer: Point C shows √5

3★Show the number √7 on the number line.

Step 1: First show √3 on the number line exactly as in Question 1. Name that point A, so OA = √3.

Step 2: Through A, draw a line perpendicular to the number line. On it, take point P such that AP = 2 units.

Step 3: Draw seg OP. △OAP is a right-angled triangle with ∠OAP = 90°.

Step 4: By Pythagoras theorem, OP² = OA² + AP² = (√3)² + 2² = 3 + 4 = 7. ∴ OP = √7.

Step 5: With centre O and radius OP, draw an arc cutting the number line at C. Then OC = OP = √7.

Point C shows √7.

Answer: Point C shows √7

Another way: keep adding a 1-unit perpendicular each time — √2 → √3 → √4 → √5 → √6 → √7, since (√n)² + 1² = n + 1. Both methods give the same point.

2. Parallel Lines and Transversals →