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81. The LCM of two numbers is 12 times their HCF. The sum of HCF and LCM is 403. If one number is 93, find the other = ?
Solution:
82. What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?
Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39
Solution:
Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39
83. The smallest number, which when increased by 5 is divisible by each of 24, 32, 36 and 64 is ?
⇒ 8 × 3 × 4 × 3 × 2 = 576
Required number is 576 - 5 = 571
Solution:
LCM of (24, 32, 36, 64)⇒ 8 × 3 × 4 × 3 × 2 = 576
Required number is 576 - 5 = 571
84. The LCM of two numbers is 44 times of their of HCF. The sum of the LCM and HCF is 1125. If one number is 25 then the other number is = ?
Solution:
85. The LCM of three different numbers is 120. Which of the following cannot be their HCF ?
Solution:
Since HCF is always a factor of LCM, we cannot have three numbers with HCF 35 and LCM 120.
86. The greatest number for four digits which when divided by 12, 16 and 24 leave remainder 2,6 and 14 respectively is = ?
16 - 6 = 10
24 - 14 = 10
LCM of (12, 16, 24) = 6 × 2 × 4 × 1 = 48
Greatest number of four digits = 9999
∴ When it is divided by 48 we get remainder = 15
⇒ The greatest number of 4 digits which completely divides the given number is = 9999 - 15 = 9984
∴ Number is = 9984 - 10 = 9974
Solution:
12 - 2 = 1016 - 6 = 10
24 - 14 = 10
LCM of (12, 16, 24) = 6 × 2 × 4 × 1 = 48
Greatest number of four digits = 9999
∴ When it is divided by 48 we get remainder = 15
⇒ The greatest number of 4 digits which completely divides the given number is = 9999 - 15 = 9984
∴ Number is = 9984 - 10 = 9974
87. Two numbers are in the ratio 3 : 4. The product of their HCF and LCM is 2028. The sum of the numbers is = ?
∴ HCF = x
LCM = 3 × 4 × x = 12x
given that = HCF × LCM
=> x × 12x = 2028
=> 12x2 = 2028
=> x2 = 169
=> x = 13
∴ sum of numbers
= 3x + 4x
= 7x
= 7 × 13
= 91
Solution:
Let the numbers are = 3x, 4x respectively∴ HCF = x
LCM = 3 × 4 × x = 12x
given that = HCF × LCM
=> x × 12x = 2028
=> 12x2 = 2028
=> x2 = 169
=> x = 13
∴ sum of numbers
= 3x + 4x
= 7x
= 7 × 13
= 91
88. Sum of two numbers is 384. HCF of the numbers is 48. The difference of the numbers is = ?
Let number are 48x and 48y respectively
So, possible pairs of co-prime number are (1, 7) (3, 5)
numbers are (48, 336) or (144, 240)
∴ difference between numbers is = 336 - 48
= 288 and 240 - 144 = 96
Solution:
HCF = 48Let number are 48x and 48y respectively
So, possible pairs of co-prime number are (1, 7) (3, 5)
numbers are (48, 336) or (144, 240)
∴ difference between numbers is = 336 - 48
= 288 and 240 - 144 = 96
89. The HCF and LCM of two numbers are 21 and 4641 respectively. If one of the numbers lies between 200 and 300 , the two numbers are ?
Let the numbers be 21a and 21b
Then 21a × 21b = 97461
ab = 221
Now, co-primes with product 221 are (1, 221) and (13, 17)
So, the numbers are (21 × 1, 21 × 221) and (21 × 13, 21 × 17)
Since one number lies between 200 and 300, the suitable pair is (273, 357)
Solution:
Product of number = 21 × 4641 = 97461Let the numbers be 21a and 21b
Then 21a × 21b = 97461
ab = 221
Now, co-primes with product 221 are (1, 221) and (13, 17)
So, the numbers are (21 × 1, 21 × 221) and (21 × 13, 21 × 17)
Since one number lies between 200 and 300, the suitable pair is (273, 357)
90. The LCM of two multiples of 12 is 1056. If one of the number is 132, the other number is ?
LCM = 12xy
12xy = 1056 (given)
xy = 88
∴ Possible pairs are (1, 88), (8, 11)
Possible numbers are = (12, 1056)(96, 132)
given that one number is 132
so other is 96
Solution:
Let number be 12x and 12y respectivelyLCM = 12xy
12xy = 1056 (given)
xy = 88
∴ Possible pairs are (1, 88), (8, 11)
Possible numbers are = (12, 1056)(96, 132)
given that one number is 132
so other is 96