Discussion
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- Question
Product of two co-prime numbers is 117. Their L.C.M should be
Options- A. 1
- B. 117
- C. Equal to their H.C.F
- D. cannot be calculated
- Correct Answer
- 117
ExplanationH.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117
- 1. The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above condition is
Options- A. 4
- B. 6
- C. 8
- D. 12 Discuss
- 2. The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is
Options- A. 10
- B. 46
- C. 70
- D. 90 Discuss
- 3. The L.C.M of 22, 54, 108, 135 and 198 is
Options- A. 330
- B. 1980
- C. 5940
- D. 11880 Discuss
- 4. A rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?
Options- A. 14 cms
- B. 21 cms
- C. 42 cms
- D. None of these Discuss
- 5. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 6. The G.C.D of 1.08, 0.36 and 0.9 is
Options- A. 0.03
- B. 0.9
- C. 0.18
- D. 0.108 Discuss
- 7. The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is
Options- A. 1008
- B. 1015
- C. 1022
- D. 1032 Discuss
- 8. L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :
Options- A. -2
- B. -1
- C. 1
- D. 2 Discuss
- 9. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Options- A. 74
- B. 94
- C. 184
- D. 364 Discuss
- 10. 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:
Options- A. 55/601
- B. 601/55
- C. 11/120
- D. 120/11 Discuss
Search Results
Correct Answer: 4
Explanation:
Let the required numbers be 33a and 33b.
Then 33a +33b= 528 => a+b = 16.
Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).
Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)
The number of such pairs is 4
Correct Answer: 10
Explanation:
Let the numbers be x and (100-x).
Then,
=>
=> (x-55) (x-45) = 0
=> x = 55 or x = 45
The numbers are 45 and 55
Required difference = (55-45) = 10
Correct Answer: 5940
Explanation:
22 = 2 x 11
54 =
108 =
135 =
198 =
Correct Answer: 21 cms
Explanation:
3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7
5.25 meters=525 cm = 5 × 5 × 3 × 7
Hence common factors are 3 and 7
Hence LCM = 3 × 7 = 21
Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.
Correct Answer: 2
Explanation:
Let the numbers 13a and 13b.
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 coprime 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.
Correct Answer: 0.18
Explanation:
Given numbers are 1.08 , 0.36 and 0.90
H.C.F of 108, 36 and 90 is 18 [ G.C.D is nothing but H.C.F]
Therefore, H.C.F of given numbers = 0.18
Correct Answer: 1015
Explanation:
Required Number = (L.C.M of 12, 16, 18,21,28)+7
= 1008 + 7
= 1015
Correct Answer: -2
Explanation:
H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161.
Let the numbers be a and b . Then , ab= 161.
Now, co-primes with product 161 are (1, 161) and (7, 23).
Since x and y are prime numbers and x >y , we have x=23 and y=7.
Therefore, 3y-x = (3 x 7)-23 = -2
Correct Answer: 364
Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
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.
Correct Answer: 11/120
Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = = = =
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