**1. (d); HCF of two numbers is 8.**

This means 8 is a factor common to both the numbers.

âˆ´ LCM must be a multiple of 8. By going through the options, 60 cannot be the LCM since it is not a multiple of 8.

**2. (a); HCF Ã— LCM = Product of two numbers.**

4×520=52 x Second number

Second number = 4×520/52 = 40

**3. (d); HCF Ã— LCM = First number Ã— Second numberÂ **

Second number = 96×1296/864 =144

**4. (d); HCF Ã— LCM = Product of two numbers. **

Then, LCM = 216/6 = 36

**5. (a); Let the numbers be 3x and 4x. **

HCF = x = 4

LCM of 3x and 4x = 12x = 12 * 4 = 48

6. (c); Given, first number = 132.

Since, each of the two numbers is a multiple of xÂ²=12 (given), HCF = 12 and LCM = 1056 (given)

LCM Ã— HCF = First number Ã— Second number Second number = 1056Ã—12/132 = 96

**7. (d); Let the numbers be 3x and 4x respectively.**

HCF Ã— LCM = First number Ã— Second number 2028 = 3x * 4x

xÂ²= 2028/12 = 169 , x = 13

Numbers are = 39 ,52

Sum of numbers = 39 + 52 = 91

**8. (c); Let the numbers be 3x and 4x.**

LCM of 3x and 4x = 12x

But given, LCM = 240

12x = 240; x = 240/12 = 20

âˆ´Smaller number = 3x = 3 x 20 = 60Â

**9. (b); HCF Ã— LCM = Product of two number âˆ´Second number = 16×160/32 =80**

**10. (b); HCF Ã— LCM = Product of two numbers**

âˆ´LCM = 4107/37 = 111, ab * HCF=LCM

where a, b are prime factors, ab = 111/37 = 3

Prime number pairs (3, 1)

Numbers = 3Ã— HCF, 1Ã— HCF

âˆ´Numbers are 111 and 37. Hence, 111 is the greater number.

**11. (a); Capacity of three containers containing diesel is 403 l, 434 l and 465 l, respectively.**

Now, maximum capacity of the container that can measure the diesel of the three containers exactly

=HCF of quantity of three containers

= HCF (403,434,465).

So, Capacity of container = 31L

**12. (b); Time taken by A = 30/5 * h.**

Time taken by B = 30/7 * h.

Time taken by C = 30/8 * h.

Time taken by all to come together = LCM of 30/5, 30/7 & 30/8 Â

= LCM of 30,30 and 30/ HCF of 5,7 and 8 = 30/1 = 30h

**13. (b); LCM of 24, 36 and 54**

Required time = LCM of 24, 36 and 54 = 216

= 216/60 = 3 * 36/60 min = 3 min 36s

âˆ´Time when they will change simultaneously

= 10:15:00 + 3 min 36 s

= 10:18:36 am

**14. (c); Let the numbers be 48a and 48b, where a and b are coprimes,**

âˆ´48a + 48b = 384

48(a + b) = 384

a + b = 384/48 = 8 â€¦â€¦. (i)

Possible valid pairs of a and b satisfying this condition are (1, 7) and (3, 5).

âˆ´Numbers are 48 * 1 = 48 and 48 * 7 = 336 or

or, Â 48 * 3 = 144

and 48 * 5 = 240

âˆ´Required difference = 336-48 = 288

or 240-144= 96Â

**15. (c); Let the numbers be a and b. **

According to the question, a + b = 45 â€¦. (i)

Again, a – b = 1/9 * (a + b) â‡’ a-b= 1 /9 *45

= Â a – b = 5 â€¦â€¦ (ii)

Adding equ. (i) and (ii), we get 2a = 50

a = 50/2 = 25

On putting the value of a Eq. (i)

we get 25 + b = 45

b = 45 – 25 = 20

a = 25 and b = 20Â

âˆ´LCM of 25 and 20.

LCM = 5*5*4 =100Â

**16. (c); Let the numbers be 4a and 4b where a and bare coprimes.**

LCM = 4ab

(4a + 4b)/(4ab) = 7/12

1/a + 1/b = 7/12 = 1/3 + 1/4

a = 3, b = 4

âˆ´First number = (4 * 3) = 12

Second number = (4 Ã— 4) = 16

âˆ´Smaller number = 12

**17. (b); Minimum number of students = LCM of 6, 8, 10.**

LCM = 2 Ã—2Ã—2Ã—3Ã—5= 120.

Hence, the minimum number of students = 120

**18. (a); Let the number be 2x, 3x and 4x respectively. **

âˆ´HCF = x = 12

âˆ´Numbers 2 * 12 = 24, 3 * 12 = 36, 4 * 12 = 48

LCM of 24, 36, 48

âˆ´LCM = 2 Ã— 2 Ã—2Ã—2Ã—3Ã—3=144

**19. (d); Given,**

or 27^{333} + 1^{333} and 27^{334} + 1^{334}

Now, x^{m} + a^{m} is divisible by (x + a) when m is odd.

27^{ 333} + 1^{333} is divisible by (27 + 1) =28.

Similarly, 27^{334} + 1^{334} is never divisible by (x + a).

So, the greatest common divisor betweenÂ Â is 1.

**20. (b); Maximum quantity in each can**

=HCF of (21, 42 and 63) L = 21L

By Division Method

HCF = 21 L

âˆ´Least number of cans 21/21 + 42

/21 + 63/21 =1+2+3=6cans.

**21. (c); Here, first number = 2 Ã—44 = 88.**

HCF = 44 and LCM = 264.

**By the formula **

1st number Ã— 2nd number = HCF Ã— LCM â‡’ 88 Ã— 2nd number = 44 Ã— 264 â‡’ 2nd number 264 x 44/ 88

â‡’ 2nd number = 132

**22. (b); Let the common ratio = x **

Then, numbers are 5x and 6x.

Now, HCF of these two numbers is x.

By the technique

LCM Ã— HCF = Product of two numbers.

â‡’ 480* x = 5x*6x

â‡’ 480x = 30x^{2}

âˆ´ x = 16

âˆ´ HCF is 16

**23. (d);** We know that, LCM of two coprimes is equal to their product. Hence, LCM = 117

**24. (c); If the numbers be 2x and 3x, then LCM of 2x and 2x = 6x **

LCM = 48

âˆ´ 6x=48â‡’ x = 48/6 = 8

âˆ´ The numbers are (8 * 2 = 16) and (8 * 3 = 24) respectively.

âˆ´Sum = 16 + 24 = 40

**25. (d); Let the numbers be 4x and 5x. **

âˆ´HCF=8=x

First number = 8 Ã— 4 = 32

Second number = 8Ã—5=40

âˆ´LCM of 32, 40=160

**26. (b); Let the number be 2a and 2b, where a and b are coprime.**

âˆ´Â LCM = 2ab

2ab = 84

ab = 42 = 6 * 7

âˆ´Numbers are 12 and 14.

âˆ´Sum 12 + 14 = 26

**27. (d); By the technique **

HCF Ã— LCM = First number Ã— Second number

âˆ´ Second number = 8×48/24= 16

**28. (c); Let the larger number be a.**

Smaller number = a – 2

HCF Ã— LCM = Product of two numbers.

24 = a(a – 2) â‡’ a^{2} – 2a – 24 = 0

â‡’ a^{2} – 6a + 4a – 24 = 0

â‡’ a(a – 6) + 4(a – 6) = 0

â‡’ a = 6,-4

But aâ‰ -4 Â âˆ´a = 6

**29. (c); Let the numbers are ax and bx **

where x =HCF=15

ATQ,

ax Ã— bx = 6300

ab = 6300/225 = 28

âˆ´possible pairs â‡’ 28 Ã—1, Â 7Ã— 4

Hence there are two pairs.

**30. (d); Let LCM be x and HCF be y.**

According to the question.

LCM = 20 Ã— HCF

i.e., x = 20y and x + y = 2520

Putting the value of x, we get

20y + y = 2520â‡’ 21y = 2520â‡’ y = 2520/21 = 120

âˆ´LCM =x = 120 x 20 = 2400

LCM Ã— HCF = Product of two numbers

2400 * 120 =48* x, x= 600

**31. (b); Given fractions = 2/4, 4/5 and 6/7Â **

HCF of fractions = HCF of numerators/LCM of denominators

= (HCF(2,4,6))/(LCM(3,5,7)) = 2/105

**32. (a); LCM of 28 and 42**

âˆ´LCM = 2 Ã—2Ã—7Ã—3=84

HCF of 28 and 42

By Division method

âˆ´ HCF = 14

âˆ´ Ratio = LCM/HCF = 84/6 =6/1=6:1

**33. (a); LCM = (xÂ²+6x+8)(x+1) or (x + 4)(x + 2)(x + 1) **

HCF = (x+1)

1st expression = x^{2} + 3x + 2 or (x + 1)(x + 2)

As we know that, product of two expressions = LCM Ã— HCF

(x + 1)(x + 2) * 2nd expression = (x + 4)(x + 2)(x + 1)(x + 1)

2nd expressions = (x + 4)(x + 2)(x + 1)(x + 1)/(x + 1)(x + 2)) = (x + 4)(x + 1)

= x^{2} + 4x + x + 4 = x^{2} + 5x + 4

**34. (c); Let the numbers be 3x and 4x.**

LCM of 3x and 4x = 12x.

Now, LCM = 84

Then, 12x = 84â‡’ x = 84/12 = 7

âˆ´Greatest number = 4x = 4 * 7 = 28

**35. (b); Let the number be 29a and 29b, respectively where a and b are coprimes **

LCM of 29a and 29b = 29ab

Now, 29ab = 4147

âˆ´ab = 4147/29 = 143

Thus, ab = 11 * 13

First number = (29 * 11) = 319

Second number = (29Ã—13) = 377

âˆ´Sum of numbers = 319 +377 = 696

**36. (b); Factors of 11 and 385 are 11 = 11 * 1; 385 = 11 * 5 * 7 **

âˆ´ LCM = 11Ã—5Ã—7=385

HCF=11

First number = 11 Ã— 5 = 55.

Second number = 11 Ã— 7 = 77 â‡’ (11,385) or (55, 77)

**37. (d); HCF of the two digit numbers = 16 **

Hence, let the numbers be 16a and 16b.

where, a and b are coprimes.

Now, HCF Ã— LCM = Product of two numbers.

â‡’Â 16a * 16b = 16 * 480 â‡’ ab = (16 * 480)/(16 * 16) = 30

Possible pairs of a and b satisfying the condition ab = 30 are (3, 10), (5, 6), (1, 30), (2, 15). Since the numbers are of 2 digit each.

Hence, required pair is (5, 6).

First number = 16 * 5 = 80

Second number = 16 * 6 = 96

**38 . (d); Greatest number that can exactly divide 200 and 320 HCF of 200 and 320 = 40**

Hence the greatest number is 40.

**39. (d);** Required number = HCF of {(1868-1356), (2764-1868), (2764-1356)} = HCF of (512, 896, 1408)

Hence, required number is 128.

**40. (c); LCM of 5, 6, 7, 8 = 35 * 24 = 840 **

Required number = 840x + 3 , such that it is exactly divisible by 9.

By hit and Trial for x = 2, it is divisible by 9.

Required number = 840x + 3 = 840 * 2 + 3 = 1683 number

(these type of questions can be solved with the help of given options)

(Out of all the given options, only 1683 is divisible by 9.)

**41. (a); LCM of 12, 16 and 24**

=2Ã—2Ã—2Ã—2Ã—3=48

âˆ´ Greatest four digit numbers divisible by 48 9999-15-9984

âˆ´ Required number = 9984-10-9974

(10 is the difference of each remainder)

**42. (c); LCM of 9, 10 and 15**

âˆ´Â LCM=2Ã—3Ã—3Ã—5=90

âˆ´ Required number = 46-7 = 39

**43. (a); Required number **

=HCF of [| 25 – 73|,|73-97|,|97-25|]

=HCF of (48, 24, 72}

HCF=2Ã—2Ã—2Ã—3=24

âˆ´HCF = 24

âˆ´Largest number = 24

**44. (c); LCM of 4, 5, 6, 7 and 8 = 840 **

Required number = 840x + 2

By Hit and Trial

Putting x = 3. we get = 840x + 2 = 840 * 3 + 2 = 2522

2522 is a multiple of 13.

**45. (a); Greatest number **

= HCF of [(307-3), (330 â€“ 7)]

= HCF of (304, 323)

âˆ´ Required number = 19

**46. (b); Least six digit number is 100000**

LCM of 4, 6, 10, 15

âˆ´ LCM=2Ã—2Ã—3Ã—5=60

âˆ´ Required number

= 100000 + (60-40) +2 = 100022

âˆ´Â Sum of the digits of N=1+0+0+0+2+2=5

**47. (d); LCM of 10, 16, 24**

âˆ´ LCM of 22 x 22 x 5 x 3

[powers must be equal for number to be perfect square]

âˆ´Â Required number = 2^{2} x 2^{2} x 5^{2} x 3^{2} = 4Ã—4Ã—25Ã—9=3600

**48. (c); By the technique **

Required number = HCF of [(989-5), (1327-7)] = HCF of (984, 1320)

âˆ´ HCF = 24

âˆ´Â Required number = 24

**49. (b); LCM of 15, 18, 21, 24**

LCM=2Ã—2Ã—2Ã—3Ã—3Ã—5Ã— 7 = 2520

Largest number of four digit = 9999

Required number = 9999-2439 – 4 = 7556

**50. (d); Required number = HCF of [(122-2), (243-3)] i.e., HCF of (120, 240) **

By Division Method

âˆ´Â Required number = 120