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Thursday, September 2, 2021

HCF and LCM

Page No 15:

Question 1:

Which number is neither a prime number nor a composite number?

ANSWER:

1 is neither a prime number nor a composite number.

Page No 15:

Question 2:

Which of the following are pairs of co-primes?
(i) 8, 14
(ii) 4, 5
(iii) 17, 19
(iv) 27, 15

ANSWER:

Two numbers which have only 1 as a common factor are said to be co-prime or relatively prime or mutually prime numbers.
We can write 17 as 17 × 1 and 19 as 17 × 1.
Hence, 17 and 19 is a pair of co-prime numbers.

Page No 15:

Question 3:

List the prime numbers from 25 to 100 and say how many they are.

ANSWER:

There are a total of 16 prime numbers between 25 and 100 which are 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Page No 15:

Question 4:

Write all the twin prime numbers from 51 to 100.

ANSWER:

If the difference between two co-prime numbers is 2, the numbers are said to be twin prime numbers.
Hence, the twin prime numbers between 51 and 100 are 59 and 61, 71 and 73.

Page No 15:

Question 5:

Write 5 pairs of twin prime numbers from 1 to 50.

ANSWER:

If the difference between two co-prime numbers is 2 then, the numbers are said to be twin prime numbers.
Hence, the twin prime numbers from 1 to 50 are (2,3), (5,7), (11,12), (17,19) and (29,30).

Page No 15:

Question 6:

Which are the even prime numbers?

ANSWER:

There is only even prime number which is 2.

Page No 17:

Question 1:

Factorise the following numbers into primes.
(i)    32
(ii)   57
(iii) 23
(iv) 150
(v)   216
(vi) 208
(vii) 765
(viii) 342
(ix)   377
(x)    559

ANSWER:

(i)
32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
(ii)
57 = 3 × 19
(iii)
23 = 23 × 1
(iv)
150 = 2 × 75
= 2 × 3 × 25
= 2 × 3 × 5 × 5
(v)
216 =  2 × 108
=  2 × 2 × 54
=  2 × 2 × 2 × 27
=  2 × 2 × 2 × 3 × 9
=  2 × 2 × 2 × 3 × 3 × 3
(vi)
208 = 2 × 104
= 2 × 2 × 52
= 2 × 2 × 2 × 26
= 2 × 2 × 2 × 2 × 13
(vii)
765 = 3 × 255
= 3 × 3 × 85
= 3 × 3 × 5 × 17
(viii)
342 = 2 × 171
= 2 × 3 × 57
= 2 × 3 × 3 × 19
(ix)
377 = 13 × 29
(x)
559 = 13 × 43

Page No 19:

Question 1:

Find the HCF.
(i)     25, 40
(ii)    56, 32
(iii)   40, 60, 75
(iv)   16, 27
(v)    18, 32, 48
(vi)   105, 154
(vii) 42, 45, 48
(viii) 57, 75, 102
(ix)   56, 57
(x)    777, 315, 588

ANSWER:

(i)

HCF = 5

(ii)

HCF = 2 × 2 × 2= 8

(iii)

HCF = 5

(iv)

HCF = 1

(v)

HCF = 2

(vi)

HCF = 7

(vii)

HCF = 3

(viii)

HCF = 3

(ix)

HCF = 1

(x)

HCF = 3 × 7 = 21

Page No 19:

Question 2:

Find the HCF by the division method and reduce to the simplest form.

(i) $\frac{275}{525}$

(ii) $\frac{76}{133}$

(iii) $\frac{161}{69}$

ANSWER:

(i)

HCF = 25
$∴ \frac{275}{525} = \frac{275 \div 25}{525 \div 25} = \frac{11}{21}$
(ii)

HCF = 19
$∴ \frac{76}{133} = \frac{76 \div 19}{133 \div 19} = \frac{4}{7}$
(iii)

HCF = 23
$∴ \frac{161}{69} = \frac{161 \div 23}{69 \div 23} = \frac{7}{3}$

Page No 21:

Question 1:

Find the LCF.
(i)     12, 15
(ii)    6, 8, 10
(iii)   18, 32
(iv)   10, 15, 20
(v)    45, 86
(vi)   15, 30, 90
(vii) 105, 195
(viii) 12, 15, 45
(ix)   63, 81
(x)    18, 36, 27

ANSWER:

(i)

LCM = 3 × 5 × 4 = 60

(ii)

LCM = 2 × 3 × 4 × 5 = 120

(iii)

LCM = 2 × 9 × 16 = 288

(iv)

LCM = 2 × 2 × 3 × 5 = 60

(v)

LCM = 45 × 86 = 3870

(vi)

LCM = 2 × 3 × 3 × 5 = 90

(vii)

LCM = 3 × 5 × 7 × 13 = 1365

(viii)

LCM = 2 × 2 × 3 × 3 × 5 = 180

(ix)

LCM = 3 × 3 × 3 × 3 × 5 = 567

(x)

LCM = 2 × 2 × 3 × 3 × 3 = 108

Page No 21:

Question 2:

Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers.
(i) 32, 37
(ii) 46, 51
(iii) 15, 60
(iv) 18, 63
(v) 78, 104

ANSWER:

(i)

HCF = 1
LCM = 32 × 37 = 1184
Product of two numbers = 32 × 37 = 1184
Product of HCF and LCM = 1 × 1184 = 1184

(ii)

HCF = 1
LCM = 46 × 51 = 2346
Product of two numbers = 46 × 51= 2346
Product of HCF and LCM = 1 × 2346 = 2346

(iii)

HCF = 3 × 5 = 15
LCM = 3 × 5 × 4 = 60
Product of two numbers = 15 × 60 = 900
Product of HCF and LCM = 15 × 60 = 900

(iv)

HCF = 3 × 3 = 9
LCM = 3 × 3 × 2 × 7 = 126
Product of two numbers = 18 × 63 = 1134
Product of HCF and LCM = 9 × 126 = 1134
(v)

HCF = 2 × 13 = 26
LCM = 2 × 13 × 3 × 4 = 312
Product of two numbers = 78 × 104 = 8112
Product of HCF and LCM = 26 × 312 = 8112

Page No 23:

Question 1:

Choose the right option.
(i) The HCF of 120 and 150 is ................... .
(1) 30
(2) 45
(3) 20
(4) 120

(ii) The HCF of this pair of numbers is not 1.
(1) 13, 17
(2) 29, 20
(3) 40, 20
(4) 14, 15

ANSWER:

(i)

HCF = 2 × 3 × 5 = 30
Hence, the correct answer is option (1).
(ii)
40 = 2 × 2 × 2 × 5
20 = 2 × 2 × 5
The HCF of 20 and 40 is 2 × 2 × 5 or 20.
Hence, the correct answer is option (3).

Page No 23:

Question 2:

Find the HCF and LCM.
(i) 14, 28
(ii) 32, 16
(iii) 17, 102, 170
(iv) 23, 69
(v) 21, 49, 84

ANSWER:

(i)

HCF = 2 × 7 = 14
LCM = 2 × 7 × 2 = 28

(ii)

HCF = 2 × 2 × 2 × 2 = 16
LCM = 2 × 2 × 2 × 2 × 2 = 32

(iii)

HCF = 17
LCM = 17 × 2 × 3 × 5 = 510

(iv)

HCF = 23
LCM = 23 × 3 = 69

(v)

HCF = 7
LCM = 3 × 4 × 7 × 7 = 588

Page No 23:

Question 3:

Find the LCM.
(i) 36, 42
(ii) 15, 25, 30
(iii) 18, 42, 48
(iv) 4, 12, 20
(v) 24, 40, 80, 120

ANSWER:

(i)

LCM = 2 × 2 × 3 × 3 × 7 = 252

(ii)

LCM = 2 × 3 × 5 × 5 = 150

(iii)

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 1008

(iv)

LCM = 2 × 2 × 3 × 5 = 60

(v)

LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240

Page No 23:

Question 4:

Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

ANSWER:

LCM of 8, 9, 10, 15, 20 is given by

LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360
Hence, 365 is the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

Page No 23:

Question 5:

Reduce the fractions $\frac{348}{319}$ , $\frac{221}{247}$ , $\frac{437}{551}$ to the lowest terms.

ANSWER:

$\frac{348}{319} = \frac{348 \div 29}{319 \div 29} = \frac{12}{11}$
$\frac{221}{247} = \frac{221 \div 13}{247 \div 13} = \frac{17}{19}$
$\frac{437}{551} = \frac{437 \div 19}{551 \div 19} = \frac{23}{29}$

Page No 23:

Question 6:

The LCM and HCF of two numbers are 432 and 72 respectively. If one of the numbers is 216, what is the other ?

ANSWER:

Let the other number be x.
Now, HCF × LCM = Product of two numbers
⇒ 72 × 432 = x × 216
$\Rightarrow x = \frac{72 \times 432}{216}$
= 144
Hence, the other number is 144.

Page No 23:

Question 7:

The product of two two-digit numbers is 765 and their HCF is 3. What is their LCM?

ANSWER:

HCF × LCM = Product of two numbers
⇒ 3 × LCM = 765
$\Rightarrow LCM = \frac{765}{3}$
= 255
Hence, the LCM of the two numbers is 255.

Page No 23:

Question 8:

A trader has three bundles of string 392 m, 308 m and 490 m long. What is the great est length of string that the bundles can be cut up into without any left over string?

ANSWER:

The greatest length of string that the bundles can be cut up into without any left over string is given by the HCF of 392, 308 and 490.

HCF = 2 × 7 = 14
Hence, the greatest length of string that the bundles can be cut up into without any left over string is 14 m.

Page No 23:

Question 9:

Which two consecutive even numbers have an LCM of 180?

ANSWER:

Let us suppose the two consecutive even numbers be 2x and 2x + 2.
Now, product of two numbers = HCF × LCM
⇒ (2x)(2x + 2) = 2 × 180 (HCM of two even number is 2)
⇒ (x)(2x + 2) = 180
⇒ 2x² + 2x = 180
⇒ 2x² + 2x − 180 = 0
⇒ x² + x − 90 = 0
⇒ (x − 9)(x + 10) = 0
⇒ x − 9 = 0 or x + 10 = 0
⇒ x = 9 or x = −10 (Neglecting)
Hence, the two consecutive even numbers are18 and 20.

Saturday, August 28, 2021

Ch.4 Angles and Pairs of angles

04.Angles and Pairs of Angles

TextBook Page No 25:

PRACTICE SET 15

Question 1:

Observe the figure and complete the table for ∠AWB.

Points in the interior
Points in the exterior
Points on the arms of the angles

ANSWER:

Points in the interior	R, C, N, X
Points in the exterior	T, U, Q, V, Y
Points on the arms of the angles	A, W, G, B

Page No 25:

Question 2:

Name the pairs of adjacent angles in the figures below.

ANSWER:

Two angles which have a common vertex, a common arm and separate interiors are said to be adjacent angles.
The pairs of adjacent angles are given below:
∠ANB and ∠BNC,
∠BNC and ∠ANC,
∠ANC and ∠ANB,
∠PQR and ∠PQT

Page No 25:

Question 3:

Are the following pairs adjacent angles? If not, state the reason.

(i) ∠PMQ and ∠RMQ (ii) ∠RMQ and ∠SMR

(iii) ∠RMS and ∠RMT (iv) ∠SMT and ∠RMS

ANSWER:

Two angles which have a common vertex, a common arm and separate interiors are said to be adjacent angles
(i)
In ∠PMQ and ∠RMQ, M is the common vertex and MQ is the common arm.
Therefore, ∠PMQ and ∠RMQ are adjacent angles.
(ii)
The angles ∠RMQ and ∠SMR have a common vertex M, but don't have common arm.
Therefore, ∠RMQ and ∠SMR are not adjacent angles.
(iii)
The angles ∠RMS and ∠RMT have a common vertex M, but don't have common arm.
Therefore, ∠RMS and ∠RMT are not adjacent angles.
(iv)
In ∠SMT and ∠RMS, M is the common vertex and SM is the common arm.
Therefore, ∠SMT and ∠RMS are adjacent angles.

Page No 26:

PRACTICE SET 16

Question 1:

The measures of some angles are given below. Write the measures of their complementaryangles.

(i) 40^° (ii) 63^° (iii) 45^°(iv) 55^° (v) 20^° (vi) 90^° (vii) x^°

ANSWER:

(i)
Let the measure of the complementary angle be a.
40 + a = 90
∴ a = 50°
Hence, the measure of the complement of an angle of measure 40° is 50°
(ii)
Let the measure of the complementary angle be a.
63 + a = 90
∴ a = 27°
Hence, the measure of the complement of an angle of measure 63° is 27°
(iii)
Let the measure of the complementary angle be a.
45 + a = 90
∴ a = 45°
Hence, the measure of the complement of an angle of measure 45° is 45°
(iv)
Let the measure of the complementary angle be a.
55 + a = 90
∴ a = 35°
Hence, the measure of the complement of an angle of measure 55° is 35°
(v)
Let the measure of the complementary angle be a.
20 + a = 90
∴ a = 70°
Hence, the measure of the complement of an angle of measure 20° is 70°
(vi)
Let the measure of the complementary angle be a.
90 + a = 90
∴ a = 0°
Hence, the measure of the complement of an angle of measure 00° is 0°
(vii)
Let the measure of the complementary angle be a.
x + a = 90
∴ a = (90 − x)°
Hence, the measure of the complement of an angle of measure x° is (90 − x)°

Page No 26:

Question 2:

(y $-$ 20)^°and (y + 30)^° are the measures of complementary angles. Find the measure of each angle.

ANSWER:

Sum of two complementary angles is 90^°
∴ (y $-$ 20)^° + (y + 30)^° = 90^°
⇒ y $-$ 20 + y + 30 = 90
⇒ 2y + 10 = 90
⇒ 2y = 80
⇒ y = 40

substitute the value of Y

y - 20

= 40 - 20

= 20

------------

y + 30

40 + 30

= 70

Hence, the measure of the two angles are 20^° and 70^°.

Page No 27:

PRACTICE SET NO.17

Question 1:

Write the measures of the supplements of the angles given below.

(i) 15^° (ii) 85^°(iii) 120^° (iv) 37^°(v) 108^° (vi) 0^°(vii) a^°

ANSWER:

(i)
Let the measure of the supplementary angle be a.
15 + a = 180
∴ a = 165°
Hence, the measure of the supplement of an angle of measure 15° is 165°.
(ii)
Let the measure of the supplementary angle be a.
85 + a = 180
∴ a = 95°
Hence, the measure of the supplement of an angle of measure 85° is 95°.
(iii)
Let the measure of the supplementary angle be a.
120 + a = 180
∴ a = 60°
Hence, the measure of the supplement of an angle of measure 120° is 60°.
(iv)
Let the measure of the supplementary angle be a.
37 + a = 180
∴ a = 143°
Hence, the measure of the supplement of an angle of measure 37° is 143°.
(v)
Let the measure of the supplementary angle be a.
108 + a = 180
∴ a = 72°
Hence, the measure of the supplement of an angle of measure 108° is 72°.
(vi)
Let the measure of the supplementary angle be a.
0 + a = 180
∴ a = 180°
Hence, the measure of the supplement of an angle of measure 0° is 180°.
(vii)
Let the measure of the supplementary angle be x.
a + x = 180
∴ x = (180 − a)°
Hence, the measure of the supplement of an angle of measure a° is (180 − a)°.

Page No 27:

Question 2:

The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles.

m∠B = 60^° m∠N = 30^°m∠Y = 90^° m∠J = 150^°
m∠D = 75^°m∠E = 0^° m∠F = 15^° m∠G = 120^°

ANSWER:

If the sum of the measures of two angles is 90° they are known as complementary angles.
Hence,the pairs of complementary angles are ∠B and ∠N, ∠D and ∠F, ∠Y and ∠E.
If the sum of the measures of two angles is 180° they are known as supplementary angles.
Hence, the pairs of supplementary angles are ∠B and ∠G, ∠N and ∠J.

Page No 27:

Question 3:

In ∆XYZ, m∠Y = 90^°. What kind of a pair do ∠X and ∠Z make?

ANSWER:

In ∆XYZ,
∠X + ∠Y + ∠Z = 180^° (Angle Sum property of triangle
⇒ ∠X + 90^° + ∠Z = 180^°
⇒ ∠X + ∠Z = 90^°
Since, the sum of the measure of the two angles is 90^°.
Hence, ∠X and ∠Z are complementary angles.

Page No 27:

Question 4:

The difference between the measures of the two angles of a complementary pair is 40^°. Find the measures of the two angles.

ANSWER:

Let the measure of the first angle a.
Then, the measure of the other angle a + 40°
Now, a + a + 40 = 90
⇒ 2a = 50
⇒ a = 25°
Hence, the measure of the two angles are 25° and 65°.

Page No 27:

Question 5:

$□$ PTNM is a rectangle. Write the names of the pairs of supplementary angles.

ANSWER:

If the sum of the measures of two angles is 180° they are known as supplementary angles.
The measure of all the angles of a rectangle is 90°.
Hence, the pairs of supplementary angles are ∠P and ∠M, ∠T and ∠N, ∠P and ∠T, ∠M and ∠N, ∠P and ∠N, ∠M and ∠T.

Page No 27:

Question 6:

If m∠A = 70^°, what is the measure of the supplement of the complement of ∠A?

ANSWER:

Let the measure of the complementary angle be a.
70 + a = 90
∴ a = 20°
Let the measure of the supplementary angle of 20° be x.
20 + x = 180
∴ x = 160°
Hence, the measure of the supplement of the complement of ∠A is 160°.

Page No 27:

Question 7:

If ∠A and ∠B are supplementary angles and m∠B = (x + 20)^°, then what would be m∠A?

ANSWER:

Let the measure of the supplementary angle of ∠B be a.
(x + 20)^° + a = 180
∴ a = (160 − x)°
Hence, the measure of ∠A is (160 − x)°.

Page No 28:

Practice SetNo.18

Question 1:

Name the pairs of opposite rays in the figure alongside.

ANSWER:

Two rays which have a common origin and form a straight line are said to be opposite rays.
Hence, the pairs of opposite rays are ray PL & ray PM and ray PN & ray PT.

Page No 28:

Question 2:

Are the ray PM and PT opposite rays? Give reasons for your answer.

ANSWER:

Ray PM and PT are not opposite rays because they do not form a straight line.

Page No 29:

PRACTICE SET NO.19

Question 1:

Draw the pairs of angles as described below.If that is not possible, say why.

(i) Complementary angles that are not adjacent.

(ii) Angles in a linear pair which are not supplementary.

(iii) Complementary angles that do not form a linear pair.

(iv) Adjacent angles which are not in a linear pair.

(v) Angles which are neither complementary nor adjacent.

(vi) Angles in a linear pair which are complementary.

ANSWER:

(i)

(ii)

If the sum of the measures of two angles is 180° they are known as supplementary angles.
The sum of the measures of the angles in a linear pair is 180°.
Therefore, angles in a linear pair are always supplementary.

(iii)

(iv)

(v)

(vi)
If the sum of the measures of two angles is 180° they are known as supplementary angles.
The sum of the measures of the angles in a linear pair is 180°.
Therefore, angles in a linear pair are always supplementary.

Page No 30:

PRACTICE SET NO.20

Question 1:

Lines AC and BD intersect at point P. m∠APD = 47^° .Find the measures of ∠APB, ∠BPC, ∠CPD.

ANSWER:

In the given figure,
∠DPA + ∠APB = 180^∘ (Linear Pair angles)
⇒ 47^∘ + ∠APB = 180^∘
⇒ ∠APB = 133^∘
Now,
∠APD = ∠BPC = 47^∘ (Vertically opposite angles)
∠APB = ∠DPC = 133^∘ (Vertically opposite angles)
Hence, the measures of ∠APB, ∠BPC, ∠CPD are 133^∘, 47^∘ and 133^∘ respectively.

Page No 30:

Question 2:

Lines PQ and RS intersect at point M. m∠PMR = x^° What are the measures of ∠PMS, ∠SMQ and ∠QMR?

ANSWER:

In the given figure,
∠RMP + ∠PMS = 180^∘ (Linear Pair angles)
⇒ x^∘ + ∠PMS = 180^∘
⇒ ∠PMS = (180 − x)^∘
Now,
∠PMR = ∠SMQ = x^∘ (Vertically opposite angles)
∠PMS = ∠RMQ = (180 − x)^∘ (Vertically opposite angles)
Hence, the measures of ∠PMS, ∠SMQ and ∠QMR are (180 − x)^∘, x^∘ and (180 − x)^∘ respectively.

Page No 33:

PRACTICE SET NO.21

Question 1:

∠ACD is an exterior angle of $∆$ ABC.The measures of ∠A and ∠B are equal. If m∠ACD = 140^°, find the measures of the angles ∠A and ∠B.

ANSWER:

∠A + ∠B = ∠ACD (Exterior angle property)
⇒ 2∠A = 140^∘ (∵∠A = ∠B)
⇒ ∠A = 70^∘
Hence, the measures of ∠A and ∠B are 70^∘ and 70^∘ respectively.

Page No 33:

Question 2:

Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

ANSWER:

In the given figure,
∠BOC = ∠FOE = 4y (Vertically opposite angles)
∠EOD = ∠AOB = 8y (Vertically opposite angles)
∠AOF = ∠COD = 6y (Vertically opposite angles)
Now, ∠AOB + ∠BOC + ∠COD = 180^∘ (Linear Pair angles)
⇒ 8y + 4y + 6y = 180^∘
⇒ 18y = 180^∘
⇒ y = 10^∘
Therefore,
∠BOC = 4y
= 40^∘
∠EOD = 8y
= 80^∘
∠AOF = 6y
= 60^∘
Hence, the measures of ∠BOC, ∠EOD, ∠AOF are 40^∘, 80^∘ and 80^∘ respectively.

Page No 33:

Question 3:

In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are
(3x

-

17)^° and (8x + 10)^°respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

ANSWER:

Given:
∠ACB = (3x $-$ 17)^∘
∠ACD = (8x + 10)^∘
Now, ∠ACB + ∠ACD = 180^∘ (Linear Pair angles)
⇒ 3x − 17+ 8x + 10 = 180
⇒ 11x = 187
⇒ x = 17
Therefore,
∠ACB = (3x $-$ = (51 $-$ = 34^∘
∠ACD = (8x + 10)^∘
= (136 + 10)^∘
= 146^∘
Now, ∠A + ∠B = ∠ACD (Exterior angle property)
⇒ 2∠A = 146^∘ (∵∠A = ∠B)
⇒ ∠A = 73^∘
Hence, the measures of ∠ACB, ∠ACD, ∠A and ∠B are 146^∘, 34^∘, 73^∘ and 73^∘ respectively.