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11. Find the LCM of the following numbers in which one number is the factor of the other.
(a) 5, 20      (b) 6, 18      (c) 12, 48      (d) 9, 45
What do you observe in the results obtained?

(a) LCM = 2 2 5 = 20 (b) LCM = 2  3  3 = 18 (c) LCM = 2 2 2 2 3 = 48 (d) LCM = 3 3 5 = 45 Yes, it can be observed that in each case, the LCM of the given numbers is the larger number. When one number is a factor of the other number, their LCM will be the larger number.

10. Find the LCM of the following numbers :
(a) 9 and 4      (b) 12 and 5     (c) 6 and 5     (d) 15 and 4
Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

(a) LCM = 2 2 3 3 = 36 (b) LCM = 2 2 3 5 = 60 (c) LCM = 2  3 5 = 30 (d) LCM = 2 3 5 = 30 Yes, it can be observed that in each case, the LCM of the given numbers is the product of these numbers. When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3.

9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.

LCM of 18, 24 and 32 LCM = 2 2 2 2 2 3 3 We have to find the smallest 4-digit multiple of 288. It can be observed that 288 3 = 864 and 288 4 = 1152. Therefore, the smallest 4-digit number which is divisible by 18, 24, and 32 is 1152.

8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

LCM of 6, 15, 18 LCM = 2 2 3 5 = 90 Required number = 90 + 5 = 95

7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Maximum capacity of the required tanker =HCF of 403,434,465 403=13 31 434=2 7 31 465=3 5 31 HCF=31  Hence the maximum capacity of a container that can measure the diesel of the three containers the exact number of times is 31.

6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

The time period after which these lights will change = LCM of 48, 72 and 108 LCM=2 2  2 2 3 3 3 = 432 They will change together after every 432 seconds i.e., 7 min 12 seconds. Hence, they will change simultaneously at 7:07:12 am.

5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

CM of 8, 10 and 12 LCM = 2   2   2   3 5 = 120 We have to find the greatest 3-digit multiple of 120. It can be seen that 120 8 = 960 and 120 9 = 1080. Hence, the greatest 3-digit number exactly divisible by 8, 10, and 12 is 960.

4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Smallest number = LCM of 6, 8, 12 We have to find the smallest 3-digit multiple of 24. It can be seen that 24   4 = 96 and 24   5 = 120. Hence, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is 120.

3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Length =825cm=3 5 5 11  Breadth =675cm=3 3  3 5 5  Height =450cm=2 3 3 5 5  Longest tape = HCF of 825,675, and 450=  3 5 5                       =75cm  Therefore, the longest tape is 75cm.

2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Step measure of 1 st  Boy =63cm  Step measure of 2 nd  Boy =70cm  Step measure of 3 rd  Boy =77cm  LCM of 63,70,77 :

1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Weight of the two bags = 75 kg and 69 kg Maximum weight : HCF (75, 69) 75 = 3 5 5 69 = 3 23 HCF = 3 Hence maximum value of weight which can measure the weight of the fertiliser exact number of times is 3.

3. HCF of co-prime numbers 4 and 15 was found as follows by factorisation : 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

No, the answer is not correct. 1 is the correct HCF. As  4 = 2 2 1 15 = 3 5 1 1 is common in both so HCF is 1.

2. What is the HCF of two consecutive
(a) numbers?     (b) even numbers?     (c) odd numbers?

(a) 1 e.g., HCF of 2 and 3 is 1. (b) 2 e.g., HCF of 2 and 4 is 2. (c) 1 e.g., HCF of 3 and 5 is 1.

1. Find the HCF of the following numbers :
(a) 18, 48              (b) 30, 42
(c) 18, 60              (d) 27, 63
(e) 36, 84              (f) 34, 102
(g) 70, 105, 175    (h) 91, 112, 49
(i) 18, 54, 81          (j) 12, 45, 75

(a) 18, 48      18 = 2 x 3 x 3 48 = 2 x 2 x 2 x 2 x 3 HCF = 2 x 3 =6   (b) 30, 42 30 = 2 x 3 x 5 42 = 2 x 3 x 7 HCF = 2 x 3 =6   (c)  18, 60  18 = 2 x 3 x 3 60 = 2 x 2 x 3 x 5 HCF = 2 x 3 = 6   (d) 27, 63 27 = 3 x 3 x 3 63 = 3 x 3 x 7 HCF = 3 x 3 = 9   (e) 36, 84    36 = 2 x 3 x 3 x 3 84 = 2 x 2 x 3 x 7 HCF = 2 x 2 x 3 = 12   (f) 34, 102 34 = 2 x 17 102 = 2 x 3 x 17 HCF = 2 x 17 = 34   (g) 70,...

Q. Find the HCF of the following:
(i) 24 and 36         (ii) 15, 25 and 30
(iii) 8 and 12         (iv) 12, 16 and 28

(i) 24 and 36  24 = 2 2 2 3 36 = 2 2 3 3 HCF = 2 2 3 = 12 (ii) 15, 25 and 30 15 = 3 5  25 = 5 5  30 = 2 3 5 HCF = 5 (iii) 8 and 12  8 = 2 2 2 12 = 2 2 3 HCF = 2 2 = 4 (iv) 12, 16 and 28 12 = 2 2 3  16 = 2 2 2 2 28 = 2 2 7 HCF = 2 2 = 4

12. I am the smallest number, having four different prime factors. Can you find me?

Since it is the smallest number of such type, it will be the product of 4 smallest prime numbers. that is  2 3  5 7 = 210

10. Determine if 25110 is divisible by 45.
[Hint : 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9].

45 = 5  9 Factors of 5 = 1, 5 Factors of 9 = 1. 3, 9 Therefore, 5 and 9 are co-prime numbers. Since the last digit of 25110 is 0. it is divisible by 5. Sum of the digits of 25110 = 2 + 5 + 1 +1 + 0 = 9 As the sum of the digits of 25110 is divisible by 9, therefore. 25110 is divisible by 9. Since the number is divisible by 5 and 9 both, it is divisible by 45.

9. In which of the following expressions, prime factorisation has been done?
(a) 24 = 2 3 4                    (b) 56 = 7 2 2 2
(c) 70 = 2 7                    (d) 54 = 2 3 9

in factorization, we don't write composite numbers. All the factors should be prime numbers in this method.  (a) 24 = 2 3 4  As we know that 4 is a composite number.  Hence This is not a prime factorization.  b) 56 = 7 2 2 2 In this factorization, all the factors of 56 are prime numbers. There is no composite number.  Hence This is prime factorization.  c) 70 = 2 5  7  In this...

8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

3+5=8, which is divisible by 4 15+17=32, which is divisible by 4 19+21=40, which is divisible by 4

7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

2 3 4=24, which is divisible by 6 9 10 11=990, which is divisible by 6 20 21 22=9240, which is divisible by 6
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