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61. The sum of two numbers is 528 and their HCF is 33. The number of pairs of numbers satisfying the above conditions is = ?
Then 33a + 33b = 528
⇒ a + b = 16
Now co - primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9)
∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9)
The number of such pairs is 4.
Solution:
Let the required numbers be 33a and 33b.Then 33a + 33b = 528
⇒ a + b = 16
Now co - primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9)
∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9)
The number of such pairs is 4.
62. The product of two numbers is 2008 and their HCF is 13. The number of such pairs is = ?
Then, 13a × 13b = 2028
ab = 12
Now, co - primes with product 12 are (1, 12) and (3, 4)
So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
Clearly, there are 2 such pairs
Solution:
Let the numbers be 13a and 13bThen, 13a × 13b = 2028
ab = 12
Now, co - primes with product 12 are (1, 12) and (3, 4)
So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
Clearly, there are 2 such pairs
63. HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is = ?
Let the numbers are 7x and 7y
LCM = 7xy
7xy = 140 (given)
xy = 20
⇒ Possible co-prime factors of xy
= (1, 20), (4, 5)
⇒ Numbers are between 20 and 45
∴ Required number are = 4 × 7 = 28 and 5 × 7 = 35
⇒ Sum of numbers are = 28 + 35 = 63
Solution:
HCF of numbers = 7Let the numbers are 7x and 7y
LCM = 7xy
7xy = 140 (given)
xy = 20
⇒ Possible co-prime factors of xy
= (1, 20), (4, 5)
⇒ Numbers are between 20 and 45
∴ Required number are = 4 × 7 = 28 and 5 × 7 = 35
⇒ Sum of numbers are = 28 + 35 = 63
64. The least number which when divided by 4, 6, 8 and 9 leave zero remainder in each case and when divided by 13 leaves a remainder of 7 is = ?
= 2 × 2 × 3 × 2 × 3
= 72
∴ Required result should be = 72
Solution:
LCM of (4, 6, 8, 9)= 2 × 2 × 3 × 2 × 3
= 72
∴ Required result should be = 72
65. The ratio of two numbers is 13 : 15 and their LCM is 39780 . The numbers are = ?
Then, their LCM = 195x
So, 195x = 39780
x = 204
∴ The numbers are 2652 and 3060
Solution:
Let the numbers be 13x and 15xThen, their LCM = 195x
So, 195x = 39780
x = 204
∴ The numbers are 2652 and 3060
66. The LCM and ratio of four numbers are 630 and 2 : 3 : 5 : 7 respectively. The difference between the greatest and least numbers is = ?
Then their LCM = (2 × 3 × 5 × 7)x = 210x.[∴ 2, 3, 5, 7 are prime numbers]
So, 210x = 630 or x = 3
∴ The numbers are 6, 9, 15 and 21
Required difference = 21 - 6 = 15
Solution:
Let the numbers be 2x, 3x, 5x and 7x respectivelyThen their LCM = (2 × 3 × 5 × 7)x = 210x.[∴ 2, 3, 5, 7 are prime numbers]
So, 210x = 630 or x = 3
∴ The numbers are 6, 9, 15 and 21
Required difference = 21 - 6 = 15
67. The number nearest to 10000, which is exactly divisible by each of 3, 4, 5, 6, 7 and 8 is = ?
⇒ divided 10000 by LCM
⇒ we get remainder = 760
Now two possibilities are
= 10000 - 760 = 9240
or 10000 + (840 - 760) = 10080
So, nearest number is = 10080
Solution:
LCM of (3, 4, 5, 6, 7, 8) = 3 × 4 × 5 × 7 × 2 = 840⇒ divided 10000 by LCM
⇒ we get remainder = 760
Now two possibilities are
= 10000 - 760 = 9240
or 10000 + (840 - 760) = 10080
So, nearest number is = 10080
68. The HCF and LCM of two 2-digit number are 16 and 480 respectively. The numbers are ?
∴ Let number are 16x and 16y
⇒ 16xy = 480
⇒ xy = 30
∴ Possible pairs = (1, 30), (2, 15), (6, 5)
Possible numbers are = (16, 480), (32, 240), (80, 96)
∴ (80, 96) is the answer in the given options of 2 digits.
Solution:
HCF = 16∴ Let number are 16x and 16y
⇒ 16xy = 480
⇒ xy = 30
∴ Possible pairs = (1, 30), (2, 15), (6, 5)
Possible numbers are = (16, 480), (32, 240), (80, 96)
∴ (80, 96) is the answer in the given options of 2 digits.
69. A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is = ?
9 - 8 = 1
8 - 7 = 1
∴ LCM of (10, 9, 8) = 5 × 2 × 9 × 4 = 360
∴ Required result = 360 - 1 = 359
Solution:
10 - 9 = 19 - 8 = 1
8 - 7 = 1
∴ LCM of (10, 9, 8) = 5 × 2 × 9 × 4 = 360
∴ Required result = 360 - 1 = 359
70. The HCF and LCM of two numbers are 12 and 336 respectively. If one of the numbers is 84, the other is = ?
Solution: