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 zero, so our procedure stops. Since the divisor at this stage is 34. HCF (1190, 204) = 34

L.C.M (204, 1190)

HCF + LCM = 34 + 7140 = 7174

Q.5 The L.C.M of 2261 and 2527 is (1 Mark)

(Ans)

Explanation :

We know that a common factor of given number divides each one of the given numbers exactly.The greatest numbers which divides each of the numbers 2261 and 2527 exactly is the HCF of 2261 and 2527. By Euclid's algorithm, we have the following equations

2527 = 2261 × 1 + 266

2261 = 266 × 8 + 133

266 = 133 × 2 + 0

The remainder has how become zero, so our procedure stops. Since thedivisor at this stage is 133. 133 is the HCF of 2527 and 2261

HCF (2527, 2261) = 133

LCM (2527, 2261)

Q.6 The LCM of two numbers is 14 times their HCF. The Sum of LCM and HCF is 600. If one number is 280 then the other number. (1 Mark) (Ans) 80

Explanation :

According to the question LCM + H.C.F = 600

Since L.C.M = 14 × H.C.F

14 × H.C.F + H.C.F = 600

15 H.C.F = 600

H.C.F = 40

From first L.C.M = 600 – H.C.F = 600 – 40 = 560

We know that H.C.F (a, b) × L.C. M (a, b) = a× b.

Other number [pic]

Q.7 If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to: (1 Mark) (Ans)

Explanation :

Let the numbers be a and b. Then, a + b = 55 and ab = 5 × 120 = 600. Required sum [pic]

Q.8 Which of the following is a non terminating repeating decimal? (1 Mark) (Ans)

Explanation :

343 = 7 × 7× 7 which cannot be written in the form 2n× 5mwhere n, m are non- negative integers. Will have a non-terminating repeating decimal expansion.

Q.9 Which of the following is a rational number? (1 Mark) Ans)

Explanation :

[pic][pic]

= [pic]is a rational number.

The number given is option (a), (c) and (d) are non-terminating, non-recurring decimals. So therefore irrational numbers.

Q.10 If P is a prime number, then...