4. LCM & HCF

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

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

∴ 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 × LCM = Product of two numbers.

4×520=52 x Second number
Second number = 4×520/52 = 40

3. (d); HCF × LCM = First number × Second number 

Second number = 96×1296/864 =144

4. (d); HCF × 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 x²=12 (given), HCF = 12 and LCM = 1056 (given)

LCM × HCF = First number × Second number Second number = 1056×12/132 = 96

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

HCF × LCM = First number × Second number 2028 = 3x * 4x

x²= 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

∴Smaller number = 3x = 3 x 20 = 60 

9. (b); HCF × LCM = Product of two number ∴Second number = 16×160/32 =80

10. (b); HCF × LCM = Product of two numbers

∴LCM = 4107/37 = 111, ab * HCF=LCM

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

Prime number pairs (3, 1)

Numbers = 3× HCF, 1× HCF

∴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  

= 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

∴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,

∴48a + 48b = 384

48(a + b) = 384

a + b = 384/48 = 8 ……. (i)

Possible valid pairs of a and b satisfying this condition are (1, 7) and (3, 5).
∴Numbers are 48 * 1 = 48 and 48 * 7 = 336 or

or,  48 * 3 = 144

and 48 * 5 = 240

∴Required difference = 336-48 = 288

or 240-144= 96 

15. (c); Let the numbers be a and b.
According to the question, a + b = 45 …. (i)

Again, a – b = 1/9 * (a + b) ⇒ a-b= 1 /9 *45

=  a – b = 5 …… (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 = 20 

∴LCM of 25 and 20.

LCM = 5*5*4 =100 

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

∴First number = (4 * 3) = 12

Second number = (4 × 4) = 16

∴Smaller number = 12

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

LCM = 2 ×2×2×3×5= 120.

Hence, the minimum number of students = 120

18. (a); Let the number be 2x, 3x and 4x respectively.
∴HCF = x = 12

∴Numbers 2 * 12 = 24, 3 * 12 = 36, 4 * 12 = 48

LCM of 24, 36, 48

∴LCM = 2 × 2 ×2×2×3×3=144

19. (d); Given,

or 27³³³ + 1³³³ and 27³³⁴ + 1³³⁴

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

27 ³³³ + 1³³³ is divisible by (27 + 1) =28.

Similarly, 27³³⁴ + 1³³⁴ is never divisible by (x + a).

So, the greatest common divisor between   is 1.

20. (b); Maximum quantity in each can

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

By Division Method

HCF = 21 L

∴Least number of cans 21/21 + 42
/21 + 63/21 =1+2+3=6cans.

21. (c); Here, first number = 2 ×44 = 88.

HCF = 44 and LCM = 264.

By the formula

1st number × 2nd number = HCF × LCM ⇒ 88 × 2nd number = 44 × 264 ⇒ 2nd number 264 x 44/ 88

⇒ 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 × HCF = Product of two numbers.

⇒ 480* x = 5x*6x

⇒ 480x = 30x²

∴ x = 16

∴ 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

∴ 6x=48⇒ x = 48/6 = 8

∴ The numbers are (8 * 2 = 16) and (8 * 3 = 24) respectively.
∴Sum = 16 + 24 = 40

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

∴HCF=8=x

First number = 8 × 4 = 32

Second number = 8×5=40

∴LCM of 32, 40=160

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

∴ LCM = 2ab

2ab = 84

ab = 42 = 6 * 7

∴Numbers are 12 and 14.

∴Sum 12 + 14 = 26

27. (d); By the technique

HCF × LCM = First number × Second number

∴ Second number = 8×48/24= 16

28. (c); Let the larger number be a.
Smaller number = a – 2

HCF × LCM = Product of two numbers.
24 = a(a – 2) ⇒ a² – 2a – 24 = 0

⇒ a² – 6a + 4a – 24 = 0

⇒ a(a – 6) + 4(a – 6) = 0

⇒ a = 6,-4

But a≠-4   ∴a = 6

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

where x =HCF=15

ATQ,

ax × bx = 6300

ab = 6300/225 = 28

∴possible pairs ⇒ 28 ×1,  7× 4

Hence there are two pairs.

30. (d); Let LCM be x and HCF be y.
According to the question.

LCM = 20 × HCF

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

Putting the value of x, we get

20y + y = 2520⇒ 21y = 2520⇒ y = 2520/21 = 120

∴LCM =x = 120 x 20 = 2400

LCM × HCF = Product of two numbers

2400 * 120 =48* x, x= 600

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

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

∴LCM = 2 ×2×7×3=84

HCF of 28 and 42

By Division method

∴ HCF = 14

∴ Ratio = LCM/HCF = 84/6 =6/1=6:1

33. (a); LCM = (x²+6x+8)(x+1) or (x + 4)(x + 2)(x + 1)

HCF = (x+1)

1st expression = x² + 3x + 2 or (x + 1)(x + 2)

As we know that, product of two expressions = LCM × 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)

= x² + 4x + x + 4 = x² + 5x + 4

34. (c); Let the numbers be 3x and 4x.
LCM of 3x and 4x = 12x.

Now, LCM = 84

Then, 12x = 84⇒ x = 84/12 = 7

∴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

∴ab = 4147/29 = 143

Thus, ab = 11 * 13

First number = (29 * 11) = 319

Second number = (29×13) = 377

∴Sum of numbers = 319 +377 = 696

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

∴ LCM = 11×5×7=385

HCF=11

First number = 11 × 5 = 55.

Second number = 11 × 7 = 77 ⇒ (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 × LCM = Product of two numbers.

⇒ 16a * 16b = 16 * 480 ⇒ 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

=2×2×2×2×3=48

∴ Greatest four digit numbers divisible by 48 9999-15-9984

∴ Required number = 9984-10-9974

(10 is the difference of each remainder)

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

∴ LCM=2×3×3×5=90

∴ Required number = 46-7 = 39

43. (a); Required number

=HCF of [| 25 – 73|,|73-97|,|97-25|]

=HCF of (48, 24, 72}

HCF=2×2×2×3=24

∴HCF = 24

∴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 – 7)]

= HCF of (304, 323)

∴ Required number = 19

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

LCM of 4, 6, 10, 15

∴ LCM=2×2×3×5=60

∴ Required number

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

∴ Sum of the digits of N=1+0+0+0+2+2=5

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

∴ LCM of 22 x 22 x 5 x 3

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

∴ Required number = 2² x 2² x 5² x 3² = 4×4×25×9=3600

48. (c); By the technique

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

∴ HCF = 24

∴ Required number = 24

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

LCM=2×2×2×3×3×5× 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

∴ Required number = 120