1. A rectangular courtyard 3.78 m long and 5.25 m broad is to be paved exactly with square tiles all of the same size. Find the number of tiles?

A) 441 B) 450

c) 480 D None of these

**EXPLANATION****:**

3.78 meters

=378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters

=525 cm = 5 × 5 × 3 × 7

Hence Common Factors are

3 × 7 = 21cm

No of tiles = area of floor / area of one tile

= 378 X 525/ 21 X 21

= 450 tiles

2. The greatest number which on dividing 1657and 2037 leaves reminders 6and 5 respectively is:

A) 123 B) 127

C) 235 D) 305

**EXPLANATION:**

Required number = H.C.F. of (1657 – 6) and (2037 – 5)

= H.C.F. of 1651 and 2032 = 127

**Verify the answer**

13*127 = 1651+6 = 1657.

16*127 = 2032+5 = 2037.

3. The sum of two numbers is 528 and their H.C .F is 33.The number of pairs of numbers satisfying the above condition is

A) 4 B) 6

C) 8 D) 12

**EXPLANATION:**

Let the required numbers “a & b”.

Then 33a + 33b = 528

⇒ a + b = 16

Now co – primes with sum 16 are

⇒ a + b = 16

⇒ 1 + 15 = 16

⇒ 3 + 13 = 16

⇒ 5 + 11 = 16

⇒ 7 + 9 =16

The number of such pairs is 4.

4. Product of two co-prime numbers is 117.Their L.C.M should be

A) 1 B) 117 C) None of these

D) Equal to their H.C.F

**EXPLANATION:**

HCF of co-prime numbers is always 1 & Now,

HCF x LCM = Product of two numbers

⇒1 x LCM = 117

⇒ LCM = 117

5. The L. C. M of two numbers is 495 and their H.C.F is 5.If the sum of the numbers is 100, then their difference is

A) 10 B) 46

C) 70 D) 90

**EXPLANATION:**

Product of number =HCF×LCM

⇒ x (100-x) = 495 X 5

⇒ x² – 100x +495 X 5 = 0

⇒ x (x – 45) – 55( x – 45)

⇒ (x – 55)(x – 45), So 45 and 55

∴Required difference =55−45=10

6. The product of two numbers is 2028and their H.C.F is 13. The numbers of such pairs is:

A) 1 B) 2

C) 3 D) 4

**EXPLANATION:**

Let the numbers “a & b”.

HCF×LCM = Product of number

⇒13a x 13b = 2028

⇒ a x b = 12.

⇒ 1 x 12 = 12

⇒ 3 x 4 = 12

Clearly, there are 2 such pairs.

7. The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one numbers lies between 75 and 125, then that number is

A) 77 B) 88

C) 99 D) 110

**EXPLANATION:**

385÷11=35 and 35 =5 x 7,

5 and 7 are prime numbers,

So two numbers are 11×5 =55 and 11×7 = 77

With in 75 & 125 the number is 77.

8. L.C.M of two prime numbers X and Y(x > y)is 161.The value of 3y-x is:

A)-2 B)-1

C) 1 D) 2

**EXPLANATION:**

X &Y are prime no X> Y

LCM of X& Y= 161 = 7× 23

⇒X=23 Y = 7,

⇒3Y – X =

⇒3 (7) – 23= -2

9. The G.C.D of 1.08, 0.36 and 0.9 is

A) 0.03 B) 0.9

C) 0.18 D) 0.108

**EXPLANATION:**

Numbers are 1.08, 0.36 and 0.90.

H.C.F. of 108, 36 and 90 is 18,

∴ H.C.F. of given numbers = 0.18

10. Three numbers are in the ratio of 3: 4: 5 and their L.C.M. are 2400. Their H.C.F.is:

A) 40 B) 80

C) 120 D) 200

**EXPLANATION:**

⇒ 3 x 4 x 5 x X =2400

⇒ 60x=2400

⇒ X=40

11. The smallest number which when diminished by 7, is divisible by 12,16,18,21 and 28 is

A) 1008 B) 1015

C) 1022 D) 1032

**EXPLANATION:**

L.C.M. of 12, 16, 18, 21, 28:

⇒Factors of 12 = 2 x 2 x 3

⇒Factors of 16 =2 x 2 x 2 x 2

⇒Factors of 18 = 2 x 3 x 3

⇒Factors of 21 =3 x 7

⇒Factors of 28 = 2 x 2 x 7

⇒LCM of 2 x 2 x 2 x 2 x 3 x 3 x 7 = 1008 + 7 = 1015

12. The least numbers which when divided by 5, 6, 7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A) 1677 B) 1683

C) 2523 D) 3363

**EXPLANATION:**

L.C.M of 5, 6, 7, 8 = 840

Required Number is of the form

⇒ 840k + 3

Least value of k for which

⇒ (840k+3) divisible by 9 is k = 2

Required Number

⇒ (840 x 2+3) =1683

13. The L.C.M of 22, 54, 108, 135 and 198 is:

A) 330 B) 1980

C) 5940 D) 11880

**EXPLANATION:**

⇒22 = 2 x 11

⇒54 = 2×3^{3}

⇒108 = 2^{2}×3^{3}

⇒135 = 3^{3}×5

⇒198 = 2×3^{2}×11

∴ L.C.M = 22×33×5×11=5940

14. The least multiple of 7, which leaves a remainder of 4, when divided by 6,9,15 and 18 is:

A) 74 B) 94

C) 184 D) 364

**EXPLANATION:**

L.C.M. of 6, 9, 15 and 18 is 90

Let required number be 90k + 4, which is multiple of 7

90K+4 is the required n, If K=1

90 x 1+4=94 Not multiple of 7

90 x 2+4=184 Not multiple of 7

90 x 3+4=274 Not multiple of 7

90 x 4+4=364 is multiple of 7

So least value will be 364

15. Three numbers are the ratio

1 : 2 : 3 and their H.C.F are 12. The numbers are

A) 4, 8, 12 B) 5, 10, 15

C) 10, 20, 30 D) 12, 24, 36

**EXPLANATION:**

Let the common number in x, 2x and 3x is x.

Therefore,

⇒x = 12.

⇒2x = 24

⇒3x = 36

So the numbers are 12, 24 and 36.

16. If the sum of two numbers is 55 and H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

A) 55/601 B) 601/55

C) 11/120 D) 120/11

**EXPLANATION:**

Product of numbers is

5 x 120 = 600

⇒ Factors = 5 x 5 x 3 x 2 x 2 x 2

⇒ Sum is = 55

⇒ xy = 600

⇒ x + y = 55

⇒1/x + 1/y

⇒ (x + y)/xy

⇒55/600

⇒11/120

17. The maximum numbers of students among them 1001 pens can be distributed in such way that each student gets the same numbers of pens and same numbers of pencils is:

A) 91 B) 910

C) 1001 D) 1911

**EXPLANATION:**

Required number of students

= HCF of 1001 and 910 = 91

18. Which of the following has the most numbers of divisors?

A) 99 B) 101

C) 176 D) 182

**EXPLANATION:**

99 = 1 × 3 × 3 × 11

101 = 1 × 101

176 = 1 × 2 × 2 × 2 × 2 × 11

182 = 1 × 2 × 7 × 13

Hence, 176 have the most number of divisors.

19. The H.C.F of two numbers is 11 and their L.C.M.is 7700. If one of the numbers is 275, then the other is

A) 279 B) 283

C) 308 D) 318

**EXPLANATION:**

HCF × LCM = Product of number

⇒ 11 × 7700 = 275 x K

⇒ K = 11×7700/275=308

20. The H.C.F and L.C.M. of two numbers are 84 and 21 respectively. If the ratio of the numbers is 1 : 4, then the larger of the two numbers is:

A) 12 B) 48

C) 84 D) 108

**EXPLANATION:**

⇒ HCF of numbers = 21

⇒ Numbers = 21m and 21n Where ⇒ m and n are prime to each other.

⇒ Ratio of numbers = 1 : 4

⇒ Larger number = 21 × 4 = 84

21. Six bells commence tolling together and toll at intervals of 2 4 6 8 10 and 12 seconds respectively. In 30minites, how many times do they toll together?

A) 4 B) 10

C) 15 D) 16

**EXPLANATION:**

L.C.M. of 2, 4, 6, 8, 10, and 12 is 120.

⇒ So, the bells will toll together after every 120 seconds (2 minutes). In 30 minutes, they will toll together 30/2 + 1 = 16 times

22. The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, & 12m, 95cm is

A) 15cm B) 25cm

C) 35cm D) 42cm

**EXPLANATION:**

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm

23. Find the Greatest Number that wills divide 43, 91 and 183 so as to leave the same reminder in each case

A) 4 B) 7

C) 9 D) 13

**EXPLANATION:**

Required number

= H.C.F. of (91 – 43), (183 – 91) and (183 – 43)

= H.C.F. of 48, 92 and 140 = 4**.**

24. H.C.F of 4 x 27 x 3125,

8 x 9 x 25 x 7 and

16 x 81 x 5 x 11 x 49 is

A) 101 B) 107

C) 111 D) 185

**EXPLANATION:**

⇒ 4 × 27 × 3125

=2^{2 }× 3^{3 }× 5^{5}

⇒ 8 × 9 × 25 × 7

=2^{3}×3^{2}×5^{2}×7

⇒16 × 81 × 5 × 11 × 49

=2^{4}×3^{4}×5×7^{2}×11

⇒ H.C.F = 2^{2}×3^{2}×5 = 180

26. The ratio of two numbers is 3:4 and their H.C.F is 4. Their L.C.M is

A) 12 B) 16

C) 24 D) 48

Let the numbers be 3x and 4x

⇒ Then their H.C.F = x. So, x=4

⇒The numbers are 12 and 16

⇒ L.C.M of 12 and 16 = 48

27. Two numbers are in the ratio 2: 3.If their L C M. is 48. What is sum of the numbers?

A) 28 B) 40 C) 64 D) 42

**EXPLANATION****:**

Let the numbers be 2x and 3x

LCM =48

⇒ 6x = 48 Therefore x = 8

⇒ The numbers are:

⇒ 2(8) & 3(8) = 16 and 24

⇒ Sum of two numbers

⇒ 16 + 24 = 40

28. What is the greatest number of four digits which is Divisible by 15, 25, 40 and 75?

A) 9800 B) 9600

C) 9400 D) 9200

**EXPLANATION****:**

Greatest possible number of

4-digits is 9999

⇒15 = 3 × 5

⇒25 = 5 × 5

⇒40 = 2 × 2 × 2 × 5

75⇒ = 3 × 5 × 5

⇒L C M of 15 25 40 = 600

⇒On dividing 9999 by 600

⇒ the remainder is 399.

Required number

⇒ (9999 – 399) = 9600

**SHORT CUT: **

n=16 and LCM= 600

⇒16 x 600 = 9600

29. Three numbers are in the ratio of 2: 3: 4 and their L C M is 240. Their H.C.F. is:

A.)40 B) 30 C) 20 D) 10

**EXPLANATION****:**

⇒ 2x 3x and 4x

⇒Their HCF = x and

⇒ Their LCM = 12x.

⇒ Now 12x = 240, x = 20

⇒ Hence the numbers are 40, 60 and 80, the smallest being 40.

30. What is the lowest multiple of 12, 36 and 20?

A) 160 B) 220 C) 120 D) 180

**EXPLANATION****:**

⇒12 = 2×2×3

⇒36= 2×2×3×3

⇒20=2×2×5

⇒LCM= 2×2×3×3×5= 180

31. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?

A) 1108 B) 1683

C) 2007 D) 3363

**EXPLANATION****:**

L.C.M of 5, 6, 7, 8 = 840

Required Number is of the form

⇒ 840k + 3

Least value of k for which

⇒ (840k+3) divisible by 9 is k = 2

Required Number

⇒ (840 x 2+3) =1683

32. The H.C.F. of two numbers is 5 and their L.C.M. is 150. If one of the numbers is 25, then the other is:

A) 30 B) 28 C) 24 D) 20

**EXPLANATION****:**

HCF x LCM = Product of two no’s

⇒ 5 x 150 = 25 x k

⇒ k = 30

33. 504 can be expressed as a product of primes as

A) 2 × 2 × 3 × 3 × 7 × 7

B) 2 × 3 × 3 × 3 × 7 × 7

C) 2 × 3 × 3 × 3 × 3 × 7

D) 2 × 2 × 2 × 3 × 3 × 7

**EXPLANATION****:**

⇒504 = 2 × 2 × 2 × 3 × 3 × 7

34. Which of the following integers has the most number of divisors?

A.)101 B) 99 C) 182 D) 176

**EXPLANATION****:**

⇒99 = 1 × 3 × 3 × 11

⇒101 = 1 × 101

⇒176 = 1 × 2 × 2 × 2 × 2 × 11

⇒182 = 1 × 2 × 7 × 13

⇒176 have the most number of divisors.

35. The least number which should be added to 28523 that the sum is exactly divisible by 3, 5, and 7 and 8 is

A) 41 B) 42 C) 32 D) 37

**EXPLANATION****:**

LCM of 3, 5, 7 and 8 = 840

⇒28523 ÷ 840 = 33

⇒ Remainder = 803

⇒Hence the least number which should be added = 840 – 803 = 37

36. What is the least number which when doubled will be exactly divisible by 12, 18, and 21 and 30?

A) 630 B) 420

C) 720 D) 840

**EXPLANATION****:**

⇒LCM of 12 18 21 and 30

⇒LCM=2×2×3×3×5×7=1260

⇒1260 ÷ 2= 630

37. What is the greatest possible length which can be used to measure exactly the lengths 8 m, 4 m 20 cm and 12 m 20 cm?

A) 10 cm B) 30 cm

C) 25 cm D) 20 cm

**EXPLANATION****:**

⇒ 8m = 800=10 x 2 x 2 x 2 x 2 x 5

⇒ 4m+20cm=420 =10 x 2 x 3 x 7

⇒ 12m+20cm=1220=10 x 2 x 61

⇒ HCF = 10 x 2 = 20 cm

38 Which of the following fraction is the largest?

A) 7/8 B) 13/16

C) 31/40 D) 63/80

**EXPLANATION****:**

L C M of 8 16 40 and 80 = 80

⇒ 7/8 = 70/80

⇒ 13/16 = 65/80

⇒ 31/40 = 62/80

7/8 is the largest fraction.

39. The product of two 2 digit numbers is 2028 and HCF is 13. What is the LCM ?

A) 26, 78 B) 39, 52 C) 13, 156 D) 36, 68

**EXPLANATION****:**

⇒Product of two no’s=HCF × LCM

⇒2028 = 13 × L C M

⇒L C M = 2028 ÷ 13

⇒L C M = 156

40. The product of two numbers is 2028 and their HCF is 13. What is the number of such pairs?

A) 4 B) 3 C) 2 D) 1

**EXPLANATION:**

Let the numbers “a & b”.

HCF×LCM = Product of number

⇒13a x 13b = 2028

⇒ a x b = 12.

⇒ 1 x 12 = 12

⇒ 3 x 4 = 12

Clearly, there are 2 such pairs.

41. N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N?

A) 4 B) 3 C) 6 D) 5

**EXPLANATION****:**

⇒ 4665 – 1305= 3360

⇒ 6905 – 4665 = 2240

⇒ 6905 – 1305 = 5600

⇒ H C F of 3360 2240 5600 is1120

⇒Sum of digits N=(1+1+2+0)=4

42. A boy divided the numbers 7654, 8506 and 9997 by a certain largest number and he gets same remainder in each case. What is the common remainder?

A) 156 B) 199 C) 211 D) 231

**EXPLANATION****:**

⇒ 9997 – 7654 = 2343

⇒ 9997 – 8506 = 1491

⇒ 8506 – 7654 = 852

⇒ HCF of 2343 1491 852 = 213

⇒ 7654 ÷ 213, Reminder = 199

⇒ 8506 ÷ 213, Reminder = 199

⇒ 9997 ÷ 213, Reminder = 199

43. Find the greatest common divisor of 24 and 16

A) 6 B) 2 C) 4 D) 8

**EXPLANATION****:**

⇒ 16 = 2 x 2 x 2 x 2

⇒ 24 = 2 x 2 x 2 x 3

So HCF of 24 & 16 is 8

44. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all Starting at the same point. After what time will them in at the starting point?

A) 36 minutes 22 seconds

B) 46 minutes 22 seconds

C) 36 minutes 12 seconds

D) 46 minutes 12 seconds

**EXPLANATION****:**

⇒LCM of 252 308 & 198

⇒ 252 = 2 x 2 x 3 x 3 x 7

⇒ 308 = 2 x 2 x 7 x 11

⇒ 198 = 2 x 3 x 3 x 11

⇒ 2 x 2 x 3 x 3 x 7 x 11 = 2772 sec ⇒ 46 minutes 12 seconds

45. The ratio of two numbers is 4 : 5. If the HCF of these numbers is 6, what is their LCM?

A) 30 B) 60 C) 90 D) 120

**EXPLANATION****:**

⇒ LCM x HCF = Product of two nos

⇒ HCF of 4x and 5x is x

⇒ LCM x 6 = 4(6) x 5(6)

⇒ LCM x 6 = 24 x 30

⇒ LCM = 24 x 5

⇒ LCM = 120

46. What is the HCF of 2.04, 0.24 and 0.8?

A) 1 B) 2 C) 0.02 D) 0.04

**EXPLANATION****:**

⇒ HCF of 204, 24 and 80 = 4

⇒ HCF of 2.04, 0.24 and 0.8 = 0.04

47. If HCF of two numbers is 11 and the product of these numbers is 363, what is the greater number?

A) 9 B) 22 C) 33 D) 11

**EXPLANATION****:**

Let the numbers be 11a and 11b

HCF x LCM = Product of two no’s

⇒ 11a × 11b = 363

⇒ a x b = 3

⇒11a × 11b = 363

⇒ (11 × 1) X (11 × 3)

⇒ (11 X 33)

The greater number = 33

48. What is the greatest number which on dividing 1223 and 2351 leaves remainders 90 and 85 respectively?

A) 1133 B) 127 C) 42 D) 1100

**EXPLANATION****:**

⇒ 1223 – 90 = 1133

⇒ 2351 – 85 = 2266

⇒ HCF of 1133 2266 is 1133

49. What is the least multiple of 7 which leaves a Remainder of 4 when divided by 6, 9, 15 and 18?

A) 364 B) 350 C) 343 D) 371

**EXPLANATION:**

L.C.M. of 6, 9, 15 and 18 is 90

Let required number be 90k + 4, which is multiple of 7

90K+4 is the required n, If K=1

90 x 1+4=94 Not multiple of 7

90 x 2+4=184 Not multiple of 7

90 x 3+4=274 Not multiple of 7

90 x 4+4=364 is multiple of 7

So least value will be 364

50. Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers?

A) 47 B) 43 C) 53 D) 51

**EXPLANATION****:**

Since the numbers are co-prime, their HCF = 1

⇒ HCF of 119 and 391 = 17

⇒ First Number = 119 ÷ 17 = 7

⇒ Last Number = 391 ÷ 17 = 23

⇒ Sum of the three numbers

⇒ 7 + 17 + 23 = 47

51. Reduce 4329 ÷ 4662 to its lowest terms

A) 7 ÷ 13 B) 13 ÷ 17

C) 13 ÷ 14 D) 7 ÷ 12

**EXPLANATION****:**

⇒ HCF of 4329 and 4662 = 333

⇒ 4329 ÷ 333 = 13

⇒ 4662 ÷ 333 = 14

⇒ Hence, 4329 ÷ 4662 =13 ÷ 14

52. What is the greatest number which divides 24, 28 and 34 and leaves the same remainder in each case?

A) 1 B) 2 C) 3 D) 4

**EXPLANATION****:**

Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder

⇒ 34 – 24 = 10

⇒ 34 – 28 = 6

⇒ 28 – 24 = 4

⇒ HCF of 10, 6, 4 = 2

53. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. How many times will they ring together in 60 minutes?

A) 31 B) 15 C) 16 D) 30

**EXPLANATION****:**

⇒Hence, all the six bells will ring together in every 2 minutes

Hence, number of times they will ring together in 60 minutes

⇒LCM of 4, 8, 10, 12, 15 & 20 = ⇒120

⇒120 seconds = 2 minutes

⇒ 1+60 ÷ 2 = 31

54. What is the least number which when divided by 8, 12, 15 and 20 leaves in each case a remainder of 5?

A) 125 B) 117 C) 132 D) 112

**EXPLANATION****:**

⇒LCM of 8 12 15and 20 is 120

⇒120 + 5 = 125

55. The HCF of two numbers is 23 and the other two Factors of their LCM are 13 and 14. What is the largest number?

A) 312 B) 282 C) 299 D) 322

**EXPLANATION****:**

The numbers are

⇒ (23 x 13) and (23 x 14)

⇒Larger number (23 x 14) = 322.

56. What is the smallest number which when diminished by 12, is divisible 8, 12, 22 and 24?

A) 276 B) 264 C) 272 D) 268

**EXPLANATION****:**

⇒ (LCM of 8, 12, 22 and 24) + 12 = 264 + 12 = 276

57. What is the HCF of 1/3, 2/3, and 1/4?

A) 2/3 B) 1/3 C) 1/ 4 D) 1/12

**EXPLANATION****:**

H C F of fractions = H C F of Numerator / L C M of Denominator

⇒ LCM of denominators is 12

⇒ HCF of numerators is 1.

⇒ HCF of the 3 fractions is 1/12

58. What is the LCM of 2/3, 5/6, and 4/9?

A) 3/10 B) 3/20

C) 10/ 3 D) 20/3

**EXPLANATION****:**

LCM of fractions = L C M of Numerator / H C F of Denominator

⇒ LCM of 2 5 4 = 20

⇒ HCF of 3 6 9 = 3

⇒ LCM of given fractions= 20/3

59. A rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square tiles, all of the same size. What is the largest size of the tile which could be used for the purpose?

A) 14 cms B) 21 cms

C) 42 cms D) None of these

**EXPLANATION****:**

⇒ 3.78 m =378 cm = 2×3×3×3×7

⇒ 5.25 m=525 cm = 5×5×3×7

⇒ Common factors are3 & 7

⇒ Hence LCM = 3 × 7 = 21

⇒ Largest square tiles = 21 cm.

60. A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn’t want to mix any two drinks in a can. What is the least number of cans required?

A) 35 B) 36 C) 37 D) 38

**EXPLANATION****:**

⇒HCF of 80 144 & 368 = 16 liters

⇒ Maaza = 80 ÷ 16 = 5

⇒Pepsi = 144 ÷ 16 = 9

⇒Sprite = 368 ÷ 16 = 23

⇒ Total number of cans required = 5 + 9 + 23 = 37

61. A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square miles of equal size required to cover the entire floor of the room.

A) 107 B) 117 C) 127 D) 137

**EXPLANATION****:**

⇒ 6 m + 24 cms = 624 cms

⇒ 4 m + 32 cms = 432 cm

⇒HCF of 624 and 432 = 48

⇒ Tiles required

= (624 x 432) ÷ (48 x 48)

= 13 x 9 = 117

62. The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:

A) 11115 B) 15110

C) 15130 D) 15310

**EXPLANATION****:**

LCM of 48 60 72 108 140 =15120

Similarity between difference of numbers and their respective Remainder 10

Common difference from LCM: 15120-10=15110

63. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

A) 534 B) 486 C) 544 D) 548

**EXPLANATION****:**

⇒L.C.M. of (12, 15, 20, 54) + 8

⇒ 540 + 8

⇒ 548

64. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A) 124 B) 100 C) 111 D) 175

**EXPLANATION****:**

⇒Let numbers 37a and 37b.

⇒HCF x LCM = Product of no’s

⇒37a x 37b = 4107

⇒ a x b = 3.

⇒ Required numbers are

⇒ (37 x 1) & (37 x 3)

⇒ Greater number = 111.

65. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A) 10 B) 14 C) 23 D) 30

**EXPLANATION****:**

⇒ L.C.M. of 5, 6, 4 and 3 = 60

⇒2497 ÷ 60, remainder is 37

⇒ (60 – 37) = 23

66. If the product of two numbers is 2496 and HCF is 8, then the ratio of HCF and LCM is

A) 1:32 B) 39:1

C) 1:39 D) 4:63

**EXPLANATION****:**

HCF x LCM = Product of no’s

⇒ 8 x LCM = 2496

⇒ LCM = 2496 ÷ 8 = 312

⇒ HCF : LCM = 8 : 312 = 1: 39

67. The greatest possible length which can be used to measure exactly the lengths 1m 92cm,3m 84cm ,23m 4cm

A) 23 B) 32 C) 36 D) 34

**EXPLANATION****:**

⇒ 1m + 92 cms = 192 cms

⇒ 3m + 84 cms = 384 cms

⇒ 23m + 4 cms = 2304 cms

⇒ HCF of 192 384 2304

⇒ 192 = 4^{2}×2^{2}×3

⇒ 384 = 4^{2}×2^{2}×6

⇒ 2304 = 4^{2}×2×6^{2}

⇒ HCF = 4^{2}×2 =
16×2 = 32

68. HCF of 4/3, 8/6, 36/63 and 20/42

A) 4/126 B) 4/8

C) 4/36 D) 4/42

**EXPLANATION****:**

HCF of fractions = HCF of Numerator / LCM of Denominator

⇒ HCF of 4 8 36 20 = 4

⇒ LCM of 3 6 63 42 = 126

69. Find the LCM of 3/8, 9/32, 33/48, 18/72

A) 3/8 B) 8/33

C) 198/8 D) 8/3

**EXPLANATION****:**

LCM of fractions =

⇒ LCM of 3 9 33 18 = 198

⇒ HCF of 8 32 48 72 = 8

70. The HCF of 2511 and 3402 is

A) 31 B) 42 C) 76 D) 81

**EXPLANATION****:**

⇒ 2511 = 81×31

⇒ 3402 = 81×42

⇒ Hence HCF is 81

71 .A gardeners had a number of shrubs to plant in rows. At first he tried to plant 8, then 12 and then 16 in a row but he had always 3 shrubs left with him. On trying 7 he had none left. Find the total number of shrubs.

A) 154 B) 147

C) 137 D) 150

**EXPLANATION****:**

⇒ L.C.M of 8 12 16 = 48

⇒ 48 x 1 + 3 = 5 not divisible by 7

⇒48 x 2 + 3 = 99not divisible by 7

⇒48 x 3 + 3 = 147 Divisible by 7 Required number = 147

72. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?

A) 630 B) 360 C) 603 D) 306

**EXPLANATION****:**

⇒ 12 = 2 × 2 × 3

⇒ 18 = 2 × 3 × 3

⇒ 21 = 3 × 7

⇒ 30 = 2 × 3 × 5

⇒ L C M of 12 18 21 30

⇒ 2 × 3 × 2 × 3 × 7 × 5 = 1260

⇒ required number

⇒ 1260 ÷ 2 = 630

73. If the L.C.M of x and y is z, their H.C.F is

A) xy/z B) xyz

C) (x + y)/z D)mz/xy

**EXPLANATION****:**

⇒Their HCF =xy/z

74. Two containers contain 50 and 125 liters of water respectively. Find the maximum capacity of a container which can measure the water in each container an exact number of times (in liters)

A) 25 B) 11 C) 12 D) 15

**EXPLANATION****:**

⇒ HCF of 50 and 125 = 25

75. Two baskets contain 183 and 242 Apples respectively, which are distributed in equal number among children. Find the largest number of apples that can be given, so that 3 apples are left over from the first basket and 2 from the second.

A) 45 B) 40 C) 60 D) 56

**EXPLANATION****:**

⇒ 183-3 = 180

⇒ 142-2 = 240

⇒ HCF of 240, 180 = 60

76. A person has 3 bars whose length are 12 16 24m respectively. He want to cuts the longest possible pieces, all of the same length from each of the 3 bars, what is the length of the each piece, if he is cut without any wastage

A) 12m B) 20m C) 6m D) 4m

**EXPLANATION****:**

⇒ HCF of 12, 16, 24 = 4m

77. Four bells commence tolling together and toll at the intervals of 3,9,12, 15 seconds resp. In 60 minutes how many times they will toll together.

A) 20 B) 21 C) 24 D) 30

**EXPLANATION****:**

⇒LCM of 3, 9, 12, 15

⇒180 Seconds or 3 minutes

⇒ 60m = 60 ÷ 3 = 20

⇒ 20+1 = 21

78. Five bells commence tolling together and tolls at intervals 3 6 9 12 & 15 seconds respectively. In 45 mints how many times do they toll together?

A) 20 B) 15 C) 10 D) 25

**EXPLANATION****:**

⇒ LCM of 3 6 9 12 & 15 is

⇒ 180 seconds or 3 Minutes

⇒ 45 ÷ 3 = 15 minutes

79. In a seminar the number of participants in Technology, Economics and Science are 150, 90 and 180 respectively. Find the minimum number of rooms required, where in each room the same number of participants are to be seated and all of them being in the same subject.

A) 27 B) 32 C) 30 D) 25

**EXPLANATION****:**

⇒ HCF of 150, 90 and 180 = 30

⇒ No of participants can be seated in each room = 30

80. Naren, Suraj and Praveen start running around a circular stadium and complete one round in 12s, 9s and 15s respectively. In how much time will they meet again at the starting point?

A) 2m B) 2.30m C) 3m D) 3.30m

**EXPLANATION****:**

⇒ LCM of 12, 9, 15 =180s = 3m

81. Three boxes of lengths 60m, 30m and 45m are to be cut into pieces of equal length. What is the greatest possible length of each piece?

A) 15m B) 18m C) 5m D) 3m

**EXPLANATION****:**

⇒HCF of 60 30 & 45m = 15m

82. In a college, all the students can stand in a row, so that each row has 9, 7 and 12 students. Find the least no of students in the school?

A) 145 B) 265 C) 186 D) 252

**EXPLANATION****:**

⇒LCM of 9 7 & 12 = 252

32. The HCF of 3 different no is 17, which of the following cannot be their LCM?

A) 540 B) 289 C) 340 D) 425

**EXPLANATION****:**

⇒ 540 ÷ 17 = 31.76

⇒ 289 ÷ 17 = 17

⇒ 340 ÷ 17 = 20

⇒ 425 ÷ 17 = 25

⇒ 540 is not divisible by 17