Question 1: Use prime factorization method to find the H.C.F. of the following:
i) ii)
iii)
iv) v)
vi)
Note: Prime Factorization Method:
Step 1: Express each one of the given numbers as a product of prime factors
Step 2: The product of terms containing least power of common prime factors gives the HCF of the given numbers.
Answer:
i)
Therefore HCF
ii)
Therefore HCF
iii)
Therefore HCF
iv) iv)
Therefore HCF
v)
Therefore HCF
vi)
Therefore HCF
Question 2: Find the HCF of the following numbers by using the long division method:
i) ii)
ii)
iv) v)
vi)
Answer:
i)
Therefore HCF
ii)
Therefore HCF
iii)
Therefore HCF
iv)
Therefore HCF of 1445 and 1190 = 85
Therefore HCF
v)
Therefore HCF of and
Therefore HCF
vi)
HCF of and
Therefore HCF
Question 3: Reduce each of the following fractions to its lowest terms:
i) ii)
iii)
Answer:
i)
Therefore
ii)
Therefore
iii)
Therefore
Question 4. Which of the following numbers are co-primes?
i) ii)
iii)
Answer:
Note: The natural numbers are said to be co-primes if the HCF of the numbers is .
You can calculate the HCF by using any of the two methods…prime factorization or long division method.
i)
HCF of
Hence, and
are co-primes.
ii)
HCF of
Hence, and
are co-primes.
iii)
HCF of
Hence, and
are not co-primes.
Question 5: Find the greatest number that exactly divides and
.
Answer:
We need to find the HCF of and
.
Hence the HCF of the three numbers is . That means,
is the largest number that would divide all the three numbers perfectly.
Question 6: Find the largest number which can divide and
leaving the remainders
and
respectively.
Answer:
The largest number should divide and
or
and
perfectly. Now find the HCF of
and
.
Hence is the number that can divide
and
leaving the remainders
and
respectively.
Question 7: Find the largest number which can divide and
leaving the same remainder
in each case.
Answer:
The largest number should divide and
or
and
perfectly. Now find the HCF of
and
.
HCF is . Hence
is the largest number which can divide
and
leaving the same remainder
in each case.
Question 8: Find the greatest number that will divide and
so as to leave the same remainder in each case.
Answer:
Required number
HCF of
and
HCF of
.
HCF is .
The required number is that will divide
and
so as to leave the same remainder in each case.
Question 9: Three pieces of timber m
cm,
m
cm and
m
cm have to be divided into planks of same lengths. What is the greatest possible length of each plank?
Answer:
The lengths are cm,
cm and
cm. Find HCF.
HCF is . Hence greatest length is
cm.
Question 10: An NGO wishes to distribute pencils and
erasers among poor children in such a way that each child gets the same number of pencils and the same number of erasers. Find the maximum number of children among whom these pencils and erasers are distributed.
Answer:
Take HCF of and
.
HCF .
The maximum number of children among whom these pencils and erasers are distributed
Question 11: Find the greatest possible length of a rope which can be used to measure exactly the lengths m
cm,
m
cm and
m
cm.
Answer:
Lengths are cm,
cm and
cm.
HCF of and
Hence HCF . Therefore the length of the rope is
cm.
Question 12: Three different containers contain different quantity of mixture of milk and water whose measurements are kg,
kg, and
kg. What biggest measure must be there to measure all different quantities an exact number of times.
Answer:
Required Number is the HCF of and
.
HCF . Hence
liter measure would be required to measure the three given quantities.
Question 13: Find the least number of square tiles required to pave the ceiling of a room m
cm long and
m
cm broad.
Answer:
The dimensions of the ceiling are cm and
cm.
The required number of times HCF of
and
.
HCF
Dimensions of tiles cm.
Therefore the number of tiles
Question 14: A school admitted boys and
girls in a year. The authorities decided to divide these students into classes in such a way that each class gets the same number of boys and the same number of girls. Find the maximum number of classes that can be formed.
Answer:
Required number is the HCF of and
.
HCF .
Number of classes
Size of the class students
Question 15: In a training camp, there were boys and
girls. The coordinator instructed the pupils to form separate teams for boys and girls such that each team includes equal number of pupils. Find the maximum number of pupils that each team can have and also the number of teams so formed.
Answer:
Required number is HCF of and
.
HCF .
Number of teams .
Number of pupil in each team
.