Rational Number and Ans
REAL NUMBERS
Q.1 Determine the prime factorization of the number 556920. (1 Mark)
(Ans) 23 x 32 x 5 x 7 x 13 x 17
Explanation :
Using the Prime factorization, we have
556920 = 2 x 2 x 2 x 3 x 3 x 5 x 7 x 13 x 17 = 23 x 32 x 5 x 7 x 13 x 17
Q.2 Use Euclid’s division algorithm to find the HCF of 210 and 55. (1 Mark)
(Ans) 5
Explanation:
5 , Given integers are 210 and 55 such that 210 > 55. Applying Euclid’s division leema to 210 and 55, we get
210 = 55 x 3 + 45 ……….(1)
55 = 45 x 1 +10 ………(2)
45 = 10 x 4 + 5 ………..(3)
10 = 5 x 2 + 0 ………..(4) we consider the new divisor 10 and the new remainder 5 and apply division leema to get 10 = 5 x 2 + 0 The remainder at this stage is zero. So, the divisor at this stage or the remainder at the previous stage i.e.5 is the HCF of 210 and 55.
Q.3 The areas of three fields are 165m2 , 195m2 and 285m2respectively. From these flowers beds of equal size are to be made. If the breadth of each bed be 3 metres, what will be the maximum length of each bed? (1 Mark)
(Ans) 4m
Explanation :
The area of three fields are 165 m2, 195 m2and 285 m2. Maximum area of a flower bed = HCF of 165, 195 and 285We first find the HCF of 165 and 195. Using Euclid's algorithm, we have the following equations.
195 = 165 × 1 + 30
165 = 30 × 5+ 15
30 = 15 × 2 + 0
The remainder has now become zero, so our procedure stops.
Since the divisor at this stage is 15.
HCF (165, 195) = 15
Now we find the HCF of 15 and 285
Using Euclid's algorithm, we have the following equations :
285 = 15 × 17 + 0
The remainder has now become zero, so our procedure stops.
Since the divisor at this stage is 15.
HCF (15, 285) = 15
So HCF of 165, 195 and 285 = 15
Length of a flower bed = = 5m.
Q.4 The sum of HCF & LCM of 204 and 1190 is (1 Mark)
(Ans) 7174
Explanation :
By Euclid's algorithm, we have the following equations
1190 = 204 × 5 + 170
204 = 170 × 1 + 34
170 = 34 × 5 + 0
The remainder has now become
Q.1 Determine the prime factorization of the number 556920. (1 Mark)
(Ans) 23 x 32 x 5 x 7 x 13 x 17
Explanation :
Using the Prime factorization, we have
556920 = 2 x 2 x 2 x 3 x 3 x 5 x 7 x 13 x 17 = 23 x 32 x 5 x 7 x 13 x 17
Q.2 Use Euclid’s division algorithm to find the HCF of 210 and 55. (1 Mark)
(Ans) 5
Explanation:
5 , Given integers are 210 and 55 such that 210 > 55. Applying Euclid’s division leema to 210 and 55, we get
210 = 55 x 3 + 45 ……….(1)
55 = 45 x 1 +10 ………(2)
45 = 10 x 4 + 5 ………..(3)
10 = 5 x 2 + 0 ………..(4) we consider the new divisor 10 and the new remainder 5 and apply division leema to get 10 = 5 x 2 + 0 The remainder at this stage is zero. So, the divisor at this stage or the remainder at the previous stage i.e.5 is the HCF of 210 and 55.
Q.3 The areas of three fields are 165m2 , 195m2 and 285m2respectively. From these flowers beds of equal size are to be made. If the breadth of each bed be 3 metres, what will be the maximum length of each bed? (1 Mark)
(Ans) 4m
Explanation :
The area of three fields are 165 m2, 195 m2and 285 m2. Maximum area of a flower bed = HCF of 165, 195 and 285We first find the HCF of 165 and 195. Using Euclid's algorithm, we have the following equations.
195 = 165 × 1 + 30
165 = 30 × 5+ 15
30 = 15 × 2 + 0
The remainder has now become zero, so our procedure stops.
Since the divisor at this stage is 15.
HCF (165, 195) = 15
Now we find the HCF of 15 and 285
Using Euclid's algorithm, we have the following equations :
285 = 15 × 17 + 0
The remainder has now become zero, so our procedure stops.
Since the divisor at this stage is 15.
HCF (15, 285) = 15
So HCF of 165, 195 and 285 = 15
Length of a flower bed = = 5m.
Q.4 The sum of HCF & LCM of 204 and 1190 is (1 Mark)
(Ans) 7174
Explanation :
By Euclid's algorithm, we have the following equations
1190 = 204 × 5 + 170
204 = 170 × 1 + 34
170 = 34 × 5 + 0
The remainder has now become