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1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
= H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
Solution:
Required number= H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
2. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
HCF is the product of all common prime factors using the least power of each common prime factor.
LCM is the product of highest powers of all prime factors.
Clearly, the numbers are (23 x 13) and (23 x 14)
∴ Larger number = (23 x 14) = 322
Solution:
The HCF of a group of numbers will be always a factor of their LCM.HCF is the product of all common prime factors using the least power of each common prime factor.
LCM is the product of highest powers of all prime factors.
Clearly, the numbers are (23 x 13) and (23 x 14)
∴ Larger number = (23 x 14) = 322
3. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together
+ 1 = 16 times
Solution:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together
+ 1 = 16 times
4. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Solution:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
5. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 - 399) = 9600.
Solution:
Greatest number of 4-digits is 9999.L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 - 399) = 9600.
6. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
Then, 37a x 37b = 4107
⇒ ab = 3
Now, co-primes with product 3 are (1, 3)
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111)
∴ Greater number = 111
Solution:
Let the numbers be 37a and 37bThen, 37a x 37b = 4107
⇒ ab = 3
Now, co-primes with product 3 are (1, 3)
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111)
∴ Greater number = 111
7. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
Then, their L.C.M. = 60x
So, 60x = 2400 or x = 40
∴ The numbers are (3 x 40), (4 x 40) and (5 x 40)
Hence, required H.C.F. = 40
Solution:
Let the numbers be 3x, 4x and 5xThen, their L.C.M. = 60x
So, 60x = 2400 or x = 40
∴ The numbers are (3 x 40), (4 x 40) and (5 x 40)
Hence, required H.C.F. = 40
8. The G.C.D. of 1.08, 0.36 and 0.9 is:
H.C.F. of 108, 36 and 90 is 18,
∴ H.C.F. of given numbers = 0.18
Solution:
Given 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
9. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Then, 13a x 13b = 2028
⇒ ab = 12
Now, the co-primes with product 12 are (1, 12) and (3, 4)
[Note: Two integers a and b are said to be co-prime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4)
Clearly, there are 2 such pairs.
Solution:
Let the numbers 13a and 13bThen, 13a x 13b = 2028
⇒ ab = 12
Now, the co-primes with product 12 are (1, 12) and (3, 4)
[Note: Two integers a and b are said to be co-prime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4)
Clearly, there are 2 such pairs.
10. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Let required number be 90k + 4, which is multiple of 7
Least value of k for which (90k + 4) is divisible by 7 is k = 4
∴ Required number = (90 x 4) + 4 = 364
Solution:
L.C.M. of 6, 9, 15 and 18 is 90Let required number be 90k + 4, which is multiple of 7
Least value of k for which (90k + 4) is divisible by 7 is k = 4
∴ Required number = (90 x 4) + 4 = 364
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