Playing With Numbers Class 6 Selina Solutions

Class 6 Concise Mathematics Selina

Playing With Numbers

Complete, step-by-step solutions for Exercises 8(A) to 8(F), divisibility tests, H.C.F., L.C.M., multiple-choice questions and reasoning questions.

BODMAS Factors Prime Numbers H.C.F. L.C.M. Divisibility

Chapter Overview

This lesson organises every question into an expandable card. Open a question, study the method and compare the final answer.

FactorDivides a number exactly.
MultipleResult of multiplying a number.
H.C.F.Greatest common factor.
L.C.M.Least common multiple.

Factors Made Visual

Visual representation of factors and prime numbers from 1 to 20
Image: Number factorisation and primes visualised, HyperAnd, Wikimedia Commons, CC0.

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Exercise 8(A)

BODMAS and Simplification

Open each question to view a clear, step-by-step solution.

Question 1Simplify: 28 - 3 × 8 ÷ 6

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

28 - 3 × 8 ÷ 6
= 28 - 24 ÷ 6
= 28 - 4
= 24
Final answer: 24
Question 2Simplify: [(4 × 2) - (4 ÷ 2)] + 8

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

[(4 × 2) - (4 ÷ 2)] + 8
= [8 - 2] + 8
= 6 + 8
= 14
Final answer: 14
Question 3Simplify: 15 × 12 ÷ (5 - 2)

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

15 × 12 ÷ (5 - 2)
= 15 × 12 ÷ 3
= 15 × 4
= 60
Final answer: 60
Question 4Simplify: 32 + 48 ÷ 12 - 3 × 7

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

32 + 48 ÷ 12 - 3 × 7
= 32 + 4 - 21
= 36 - 21
= 15
Final answer: 15
Question 5Simplify: 16 - 4 of 5 ÷ 10 × 2

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

In BODMAS, of means multiplication and is evaluated before division and multiplication in this expression.

16 - 4 of 5 ÷ 10 × 2
= 16 - 20 ÷ 10 × 2
= 16 - 2 × 2
= 16 - 4
= 12
Final answer: 12
Question 6Simplify: 24 of 36 ÷ 9 - 13 × 4

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

24 of 36 ÷ 9 - 13 × 4
= 864 ÷ 9 - 52
= 96 - 52
= 44
Final answer: 44
Question 7Simplify: 19 - [18 - {12 - (7 - 5)}]

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

19 - [18 - {12 - (7 - 5)}]
= 19 - [18 - {12 - 2}]
= 19 - [18 - 10]
= 19 - 8
= 11
Final answer: 11
Question 8Simplify: 17 - [14 - {40 + 8 ÷ (7 - (6 - 3))}]

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

17 - [14 - {40 + 8 ÷ (7 - (6 - 3))}]
= 17 - [14 - {40 + 8 ÷ (7 - 3)}]
= 17 - [14 - {40 + 8 ÷ 4}]
= 17 - [14 - {40 + 2}]
= 17 - [14 - 42]
= 17 - (-28)
= 45
Final answer: 45
Question 9Simplify: 25 - [12 - {5 + 18 ÷ (4 - (5 - 3))}]

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

25 - [12 - {5 + 18 ÷ (4 - (5 - 3))}]
= 25 - [12 - {5 + 18 ÷ (4 - 2)}]
= 25 - [12 - {5 + 18 ÷ 2}]
= 25 - [12 - {5 + 9}]
= 25 - [12 - 14]
= 25 - (-2)
= 27
Final answer: 27
Question 10Simplify: 15 - [16 - {12 + 21 ÷ (9 - 2)}]

BODMAS order: Brackets, Of, Division, Multiplication, Addition and Subtraction.

15 - [16 - {12 + 21 ÷ (9 - 2)}]
= 15 - [16 - {12 + 21 ÷ 7}]
= 15 - [16 - {12 + 3}]
= 15 - [16 - 15]
= 15 - 1
= 14
Final answer: 14
Exercise 8(B)

Factors and Prime Numbers

Review factors, prime numbers and prime factorisation.

Question 1Write all the factors of 15, 55, 48 and 36

(i) 15: 1, 3, 5, 15

(ii) 55: 1, 5, 11, 55

(iii) 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

(iv) 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Final answer: The factor lists are shown above.
Question 2Write all the prime numbers in the given ranges

A prime number is a natural number greater than 1 that has exactly two factors: 1 and itself.

(i) Less than 25: 2, 3, 5, 7, 11, 13, 17, 19, 23

(ii) Between 15 and 35: 17, 19, 23, 29, 31

(iii) Between 8 and 76: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73

Final answer: The prime-number lists are shown above.
Question 3Write the prime numbers from 5 to 45, 2 to 32 and 8 to 48

(i) From 5 to 45: 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43

(ii) From 2 to 32: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

(iii) From 8 to 48: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Final answer: The prime-number lists are shown above.
Question 4Write the prime factors of 16, 35 and 49
(i) 16 = 2 × 2 × 2 × 2 = 24

The only prime factor of 16 is 2.

(ii) 35 = 5 × 7

The prime factors of 35 are 5 and 7.

(iii) 49 = 7 × 7 = 72

The only prime factor of 49 is 7.

Final answer: 16: 2; 35: 5 and 7; 49: 7.
Exercise 8(C)

Highest Common Factor

Compare common-factor, prime-factor and division methods.

Question 1(i)Using the common factor method, find the H.C.F. of 25 and 20

Factors of 25: 1, 5, 25

Factors of 20: 1, 2, 4, 5, 10, 20

Common factors are 1 and 5. The highest is 5.

Final answer: H.C.F. = 5
Question 1(ii)Using the common factor method, find the H.C.F. of 8, 12 and 18

Factors of 8: 1, 2, 4, 8

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 18: 1, 2, 3, 6, 9, 18

Common factors are 1 and 2. The highest is 2.

Final answer: H.C.F. = 2
Question 1(iii)Using the common factor method, find the H.C.F. of 24, 36, 45 and 60

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Factors of 45: 1, 3, 5, 9, 15, 45

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Common factors are 1 and 3. The highest is 3.

Final answer: H.C.F. = 3
Question 2(i)Using prime factorisation, find the H.C.F. of 40, 60 and 80
40 = 23 × 5
60 = 22 × 3 × 5
80 = 24 × 5

The common prime factors with the lowest powers are 22 and 5.

H.C.F. = 22 × 5 = 20
Final answer: H.C.F. = 20
Question 2(ii)Using prime factorisation, find the H.C.F. of 48, 84 and 88
48 = 24 × 3
84 = 22 × 3 × 7
88 = 23 × 11

The common prime factor with the lowest power is 22.

H.C.F. = 22 = 4
Final answer: H.C.F. = 4
Question 2(iii)Using prime factorisation, find the H.C.F. of 12, 16 and 28
12 = 22 × 3
16 = 24
28 = 22 × 7
H.C.F. = 22 = 4
Final answer: H.C.F. = 4
Question 3(i)Using the division method, find the H.C.F. of 16 and 24

Divide the larger number by the smaller number. Then divide the previous divisor by the remainder.

DividendDivisorQuotientRemainder
241618
16820

The last non-zero divisor is 8.

Final answer: H.C.F. = 8
Question 3(ii)Using the division method, find the H.C.F. of 7, 14 and 24

First find the H.C.F. of 7 and 14.

DividendDivisorQuotientRemainder
14720

Now find the H.C.F. of 7 and 24.

DividendDivisorQuotientRemainder
24733
7321
3130

The last non-zero divisor is 1.

Final answer: H.C.F. = 1
Question 3(iii)Using the division method, find the H.C.F. of 32, 56 and 46

First find the H.C.F. of 32 and 56.

DividendDivisorQuotientRemainder
5632124
322418
24830

Now find the H.C.F. of 8 and 46.

DividendDivisorQuotientRemainder
46856
8612
6230

The last non-zero divisor is 2.

Final answer: H.C.F. = 2
Question 4(i)Using a suitable method, find the H.C.F. of 45, 75 and 135
45 = 32 × 5
75 = 3 × 52
135 = 33 × 5
H.C.F. = 3 × 5 = 15
Final answer: H.C.F. = 15
Question 4(ii)Using a suitable method, find the H.C.F. of 66, 33 and 132
66 = 2 × 3 × 11
33 = 3 × 11
132 = 22 × 3 × 11
H.C.F. = 3 × 11 = 33
Final answer: H.C.F. = 33
Question 4(iii)Using a suitable method, find the H.C.F. of 24, 36, 60 and 132
24 = 23 × 3
36 = 22 × 32
60 = 22 × 3 × 5
132 = 22 × 3 × 11
H.C.F. = 22 × 3 = 12
Final answer: H.C.F. = 12
Question 5Find the greatest number that divides 180, 225 and 315 completely

The required greatest number is the H.C.F.

180 = 22 × 32 × 5
225 = 32 × 52
315 = 32 × 5 × 7
H.C.F. = 32 × 5 = 45
Final answer: 45
Question 6Show that 45 and 56 are co-prime numbers

Two numbers are co-prime when their H.C.F. is 1.

45 = 32 × 5
56 = 23 × 7

There is no common prime factor, so their H.C.F. is 1.

Final answer: 45 and 56 are co-prime.
Exercise 8(D)

Lowest Common Multiple

Learn listing, prime-factorisation and common-division methods.

Question 1(i)Using the common multiple method, find the L.C.M. of 8, 12 and 24

Multiples of 8: 8, 16, 24, 32, 40, 48, ...

Multiples of 12: 12, 24, 36, 48, ...

Multiples of 24: 24, 48, 72, ...

The smallest common multiple is 24.

Final answer: L.C.M. = 24
Question 1(ii)Using the common multiple method, find the L.C.M. of 10, 15 and 20

Multiples of 10: 10, 20, 30, 40, 50, 60, ...

Multiples of 15: 15, 30, 45, 60, ...

Multiples of 20: 20, 40, 60, 80, ...

The smallest common multiple is 60.

Final answer: L.C.M. = 60
Question 1(iii)Using the common multiple method, find the L.C.M. of 3, 6, 9 and 12

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, ...

Multiples of 6: 6, 12, 18, 24, 30, 36, ...

Multiples of 9: 9, 18, 27, 36, ...

Multiples of 12: 12, 24, 36, ...

The smallest common multiple is 36.

Final answer: L.C.M. = 36
Question 2(i)Find the L.C.M. of 18, 24 and 96 by prime factorisation and common division

Prime factor method

18 = 2 × 32
24 = 23 × 3
96 = 25 × 3
L.C.M. = 25 × 32 = 288

Common division method

Prime divisorNumbers after division
29, 12, 48
29, 6, 24
29, 3, 12
29, 3, 6
29, 3, 3
33, 1, 1
31, 1, 1
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
Final answer: L.C.M. = 288
Question 2(ii)Find the L.C.M. of 14, 21 and 98 by prime factorisation and common division

Prime factor method

14 = 2 × 7
21 = 3 × 7
98 = 2 × 72
L.C.M. = 2 × 3 × 72 = 294

Common division method

Prime divisorNumbers after division
27, 21, 49
37, 7, 49
71, 1, 7
71, 1, 1
L.C.M. = 2 × 3 × 7 × 7 = 294
Final answer: L.C.M. = 294
Question 2(iii)Find the L.C.M. of 34, 85 and 51 by prime factorisation and common division

Prime factor method

34 = 2 × 17
85 = 5 × 17
51 = 3 × 17
L.C.M. = 2 × 3 × 5 × 17 = 510

Common division method

Prime divisorNumbers after division
217, 85, 51
317, 85, 17
517, 17, 17
171, 1, 1
L.C.M. = 2 × 3 × 5 × 17 = 510
Final answer: L.C.M. = 510
Question 3Find the least number divisible by 15, 25, 40 and 50

The required number is the L.C.M.

15 = 3 × 5
25 = 52
40 = 23 × 5
50 = 2 × 52
L.C.M. = 23 × 3 × 52 = 600
Final answer: 600
Question 4Find the smallest number which leaves remainder 7 when divided by 18, 36 and 48

First find the L.C.M. of 18, 36 and 48.

18 = 2 × 32
36 = 22 × 32
48 = 24 × 3
L.C.M. = 24 × 32 = 144
Required number = 144 + 7 = 151
Final answer: 151
Question 5Find the smallest number which, when increased by 11, is divisible by 16, 24, 40 and 45

First find the L.C.M. of 16, 24, 40 and 45.

16 = 24
24 = 23 × 3
40 = 23 × 5
45 = 32 × 5
L.C.M. = 24 × 32 × 5 = 720
Required number = 720 - 11 = 709
Final answer: 709
Question 6Find the L.C.M. and H.C.F. of 24 and 30. Is the L.C.M. divisible by the H.C.F.?
24 = 23 × 3
30 = 2 × 3 × 5
H.C.F. = 2 × 3 = 6
L.C.M. = 23 × 3 × 5 = 120
120 ÷ 6 = 20, with remainder 0
Final answer: L.C.M. = 120, H.C.F. = 6; yes, the L.C.M. is completely divisible by the H.C.F.
Question 7The H.C.F. and L.C.M. are 50 and 300. One number is 150. Find the other number

Use the relation: Product of two numbers = H.C.F. × L.C.M.

150 × second number = 50 × 300
Second number = (50 × 300) ÷ 150
Second number = 100
Final answer: 100
Question 8The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.

Product of two numbers = H.C.F. × L.C.M.

432 = H.C.F. × 72
H.C.F. = 432 ÷ 72 = 6
Final answer: H.C.F. = 6
Question 9The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.

Product of two numbers = H.C.F. × L.C.M.

19,200 = 40 × L.C.M.
L.C.M. = 19,200 ÷ 40 = 480
Final answer: L.C.M. = 480
Question 10Find the smallest number divisible by 12, 15, 18, 24 and 36

The required number is their L.C.M.

12 = 22 × 3
15 = 3 × 5
18 = 2 × 32
24 = 23 × 3
36 = 22 × 32
L.C.M. = 23 × 32 × 5 = 360
Final answer: 360
Exercise 8(E)

Divisibility Tests

Use the final digit, final two or three digits, digit sums and alternating sums.

Question 1Which of 352, 523 and 496 are divisible by 2?

A number is divisible by 2 when its last digit is even.

352 ends in 2, so it is divisible by 2.

523 ends in 3, so it is not divisible by 2.

496 ends in 6, so it is divisible by 2.

Final answer: 352 and 496
Question 2Which of 222, 532 and 678 are divisible by 4?

A number is divisible by 4 when the number formed by its last two digits is divisible by 4.

22 is not divisible by 4.

32 is divisible by 4.

78 is not divisible by 4.

Final answer: 532
Question 3Which of 324, 2536 and 92760 are divisible by 8?

A number is divisible by 8 when the number formed by its last three digits is divisible by 8.

324 is not divisible by 8.

536 ÷ 8 = 67, so 2536 is divisible by 8.

760 ÷ 8 = 95, so 92760 is divisible by 8.

Final answer: 2536 and 92760
Question 4Which of 221, 543 and 28492 are divisible by 3?

A number is divisible by 3 when the sum of its digits is divisible by 3.

221: 2 + 2 + 1 = 5
543: 5 + 4 + 3 = 12
28492: 2 + 8 + 4 + 9 + 2 = 25
Final answer: 543
Question 5Which of 1332, 53247 and 4968 are divisible by 9?

A number is divisible by 9 when the sum of its digits is divisible by 9.

1332: 1 + 3 + 3 + 2 = 9
53247: 5 + 3 + 2 + 4 + 7 = 21
4968: 4 + 9 + 6 + 8 = 27
Final answer: 1332 and 4968
Question 6Which of 324, 2010 and 33278 are divisible by 6?

A number is divisible by 6 when it is divisible by both 2 and 3.

324: It is even, and 3 + 2 + 4 = 9. Therefore, it is divisible by 6.

2010: It is even, and 2 + 0 + 1 + 0 = 3. Therefore, it is divisible by 6.

33278: It is even, but 3 + 3 + 2 + 7 + 8 = 23, which is not divisible by 3.

Final answer: 324 and 2010
Question 7Which of 5080, 66666 and 755 are divisible by 5?

A number is divisible by 5 when its last digit is 0 or 5.

5080 ends in 0.

66666 ends in 6.

755 ends in 5.

Final answer: 5080 and 755
Question 8Which of 9990, 0 and 847 are divisible by 10?

A number is divisible by 10 when its last digit is 0.

9990 ends in 0.

0 is divisible by every non-zero integer, including 10.

847 ends in 7.

Final answer: 9990 and 0
Question 9Which of 5918, 68717 and 3882 are divisible by 11?

For divisibility by 11, find the difference between the sums of digits in alternating places. The difference must be 0 or a multiple of 11.

5918: (8 + 9) - (1 + 5) = 17 - 6 = 11
68717: (7 + 7 + 6) - (1 + 8) = 20 - 9 = 11
3882: (2 + 8) - (8 + 3) = 10 - 11 = -1
Final answer: 5918 and 68717
Question 10Which of 960, 8295 and 10243 are divisible by 15?

A number is divisible by 15 when it is divisible by both 3 and 5.

960: Ends in 0 and its digit sum is 15. It is divisible by 15.

8295: Ends in 5 and its digit sum is 24. It is divisible by 15.

10243: Does not end in 0 or 5. It is not divisible by 15.

Final answer: 960 and 8295
Exercise 8(F)

Application Questions

Apply H.C.F., L.C.M. and divisibility ideas to word problems.

Question 1(i)Find the smallest number completely divisible by 28 and 42

The required number is the L.C.M.

28 = 22 × 7
42 = 2 × 3 × 7
L.C.M. = 22 × 3 × 7 = 84
Final answer: 84
Question 1(ii)Find the largest number that divides 28 and 42 completely

The required number is the H.C.F.

28 = 22 × 7
42 = 2 × 3 × 7
H.C.F. = 2 × 7 = 14
Final answer: 14
Question 2Take two numbers divisible by 8. Test whether their sum and difference are also divisible by 8

Take 16 and 24.

Sum = 16 + 24 = 40; 40 ÷ 8 = 5
Difference = 24 - 16 = 8; 8 ÷ 8 = 1

Both the sum and the difference are divisible by 8.

Final answer: Yes, both are divisible by 8.
Question 3What is the H.C.F. of two consecutive numbers, consecutive even numbers and consecutive odd numbers?

(i) Consecutive numbers: H.C.F. = 1. Example: H.C.F. of 7 and 8 is 1.

(ii) Consecutive even numbers: H.C.F. = 2. Example: H.C.F. of 8 and 10 is 2.

(iii) Consecutive odd numbers: H.C.F. = 1. Example: H.C.F. of 7 and 9 is 1.

Final answer: 1, 2 and 1, respectively.
Question 4Find the smallest 3-digit number exactly divisible by 6, 8 and 12
6 = 2 × 3
8 = 23
12 = 22 × 3
L.C.M. = 23 × 3 = 24

The multiples near 100 are 96 and 120. The first 3-digit multiple is 120.

Final answer: 120
Question 5Find the greatest 3-digit number exactly divisible by 6, 8 and 12

The L.C.M. of 6, 8 and 12 is 24.

999 ÷ 24 = 41 remainder 15
Greatest required number = 999 - 15 = 984
Final answer: 984
Question 6Find the L.C.M. of 140 and 168, then use it to find their H.C.F.
140 = 22 × 5 × 7
168 = 23 × 3 × 7
L.C.M. = 23 × 3 × 5 × 7 = 840

Use: Product of two numbers = H.C.F. × L.C.M.

H.C.F. = (140 × 168) ÷ 840 = 28
Final answer: L.C.M. = 840 and H.C.F. = 28
Question 7Find the H.C.F. of 108 and 450, then use it to find their L.C.M.
108 = 22 × 33
450 = 2 × 32 × 52
H.C.F. = 2 × 32 = 18
L.C.M. = (108 × 450) ÷ 18 = 2700
Final answer: H.C.F. = 18 and L.C.M. = 2700
Question 8Take any two numbers and show that the L.C.M. is divisible by the H.C.F.

Take 12 and 18.

12 = 22 × 3
18 = 2 × 32
H.C.F. = 2 × 3 = 6
L.C.M. = 22 × 32 = 36
36 ÷ 6 = 6, with remainder 0
Final answer: The L.C.M. 36 is completely divisible by the H.C.F. 6.
Question 9Find the L.C.M. and H.C.F. of 16, 32 and 96. Show that the L.C.M. is divisible by the H.C.F.
16 = 24
32 = 25
96 = 25 × 3
H.C.F. = 24 = 16
L.C.M. = 25 × 3 = 96
96 ÷ 16 = 6, with remainder 0
Final answer: L.C.M. = 96 and H.C.F. = 16; the L.C.M. is divisible by the H.C.F.
Multiple Choice Questions

Objective Practice

Check each option, then open the answer to understand the reasoning.

Question 1A number is divisible by both 5 and 8. It must be divisible by:
A20
B15
C40
D80

Because 5 and 8 are co-prime, their L.C.M. is 5 × 8 = 40.

Final answer: Option C: 40
Question 2Every number other than 0 has an infinite number of:
APrime factors
BFactors
CMultiples
DNone of these

A non-zero number has finitely many factors but infinitely many multiples.

Final answer: Option C: Multiples
Question 3Find the value of 18 - (15 ÷ (12 - 9))
A13
B23
C1
DNone of these
18 - (15 ÷ (12 - 9)) = 18 - (15 ÷ 3) = 18 - 5 = 13
Final answer: Option A: 13
Question 4How many common factors do 60 and 75 have?
A2
B3
C4
D1

Common factors of 60 and 75 are 1, 3, 5 and 15. There are 4 common factors.

Final answer: Option C: 4
Question 5How many natural numbers have exactly one factor?
A0
B1
C2
DNone of these

Only the number 1 has exactly one factor, which is 1 itself.

Final answer: Option B: 1
Question 6Which operation should be performed first in 25 × 3 + (16 ÷ 4) - 10 ÷ 2 + 3?
A25 × 3
B16 ÷ 4
C10 ÷ 2
D2 + 3

BODMAS requires the operation inside brackets to be completed first.

Final answer: Option B: 16 ÷ 4
Question 7The H.C.F. and L.C.M. of 12 and 17 are respectively:
A1 and 17
B12 × 17 and 12 + 17
C12 and 1
D1 and 12 × 17

12 and 17 are co-prime. Their H.C.F. is 1 and their L.C.M. is 12 × 17.

Final answer: Option D: 1 and 12 × 17
Question 8The L.C.M. of two numbers is 56 and their H.C.F. is 4. Their product is:
A56
B56 ÷ 4
C56 × 4
DNone of these

Product of the two numbers = H.C.F. × L.C.M. = 4 × 56.

Final answer: Option C: 56 × 4
Question 9The H.C.F. and L.C.M. of 2 and 6 are:
A1 and 12
B1 and 6
C2 and 6
D2 and 12
2 = 2 and 6 = 2 × 3. Therefore, H.C.F. = 2 and L.C.M. = 6.
Final answer: Option C: 2 and 6
Question 10The H.C.F. and L.C.M. of 7 and 15 are:
A1 and 105
B7 and 75
C1 and 15
D1 and 7

7 and 15 are co-prime. H.C.F. = 1 and L.C.M. = 7 × 15 = 105.

Final answer: Option A: 1 and 105
Statement I-II Questions

Statement Analysis

Evaluate each statement independently before choosing the option.

Question 11Statement 1: 4048 is divisible by 11. Statement 2: (8 + 0) - (4 + 4) = 0.
ABoth statements are true.
BBoth statements are false.
CStatement 1 is true and Statement 2 is false.
DStatement 1 is false and Statement 2 is true.
(8 + 0) - (4 + 4) = 8 - 8 = 0

Because the alternating-sum difference is 0, 4048 is divisible by 11. Both statements are true.

Final answer: Option A: Both statements are true.
Question 12Statement 1: Among 6, 12 and 216, only 216 is a multiple of 8. Statement 2: Every non-zero number has finitely many multiples.
ABoth statements are true.
BBoth statements are false.
CStatement 1 is true and Statement 2 is false.
DStatement 1 is false and Statement 2 is true.
216 ÷ 8 = 27

6 and 12 are not multiples of 8, but 216 is. A non-zero number has infinitely many multiples, so Statement 2 is false.

Final answer: Option C: Statement 1 is true and Statement 2 is false.
Assertion-Reason Questions

Reasoning Practice

Use definitions and divisibility rules to test both parts.

Question 13Assertion: 1848 is divisible by 22. Reason: (8 + 8) - (4 + 1) is divisible by 11.
AAssertion is true; Reason is false.
BAssertion is false; Reason is true.
CBoth Assertion and Reason are true.
DBoth Assertion and Reason are false.

A number divisible by 22 must be divisible by both 2 and 11.

1848 is even, so it is divisible by 2.

(8 + 8) - (4 + 1) = 16 - 5 = 11

The alternating-sum difference is divisible by 11, so 1848 is divisible by 11 and therefore by 22.

Final answer: Option C: Both Assertion and Reason are true.
Question 14Assertion: 10 and 11 are co-prime. Reason: Two numbers are co-prime when their greatest common divisor is 1.
AAssertion is true; Reason is false.
BAssertion is false; Reason is true.
CBoth Assertion and Reason are true.
DBoth Assertion and Reason are false.
10 = 2 × 5 and 11 is prime

They have no common factor other than 1, so their H.C.F. is 1. The assertion and reason are both true.

Final answer: Option C: Both Assertion and Reason are true.

Rapid Revision

Divisible by 2Last digit is even.
Divisible by 3Digit sum is divisible by 3.
Divisible by 4Last two digits form a multiple of 4.
Divisible by 5Last digit is 0 or 5.
Divisible by 6Divisible by both 2 and 3.
Divisible by 8Last three digits form a multiple of 8.
Divisible by 9Digit sum is divisible by 9.
Divisible by 11Alternating-sum difference is 0 or a multiple of 11.

Frequently Asked Questions

What is the BODMAS rule?

BODMAS gives the order for simplifying an expression: Brackets, Of, Division, Multiplication, Addition and Subtraction. Operations of equal priority are completed from left to right.

What is a factor?

A factor divides a number exactly without leaving a remainder. For example, 1, 2, 3 and 6 are factors of 6.

What is a multiple?

A multiple is obtained by multiplying a number by a whole number. For example, 6, 12, 18 and 24 are multiples of 6.

What is a prime number?

A prime number is a natural number greater than 1 with exactly two factors: 1 and the number itself.

What is H.C.F.?

H.C.F. is the greatest number that divides all the given numbers exactly. It is also called the greatest common divisor.

What is L.C.M.?

L.C.M. is the smallest positive number that is exactly divisible by all the given numbers.

What is the relation between H.C.F., L.C.M. and two numbers?

For two positive integers, the product of the numbers equals their H.C.F. multiplied by their L.C.M.

What are co-prime numbers?

Two numbers are co-prime when their H.C.F. is 1. The numbers themselves do not both need to be prime.

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Shaleen Shekhar

I'm curious about how things work and obsessed with making complex ideas simple. Whether it's science, AI, technology, or digital marketing, I enjoy exploring, creating, and sharing knowledge that actually helps people. Always learning, always building, and always looking for the next big idea.

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