Discussion
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- Question
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
- Correct Answer
- 1015
ExplanationRequired Number = (L.C.M of 12, 16, 18,21,28)+7
= 1008 + 7
= 1015
- 1. 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
- 2. 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 Discuss
- 3. 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
- 4. 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
- 5. The L.C.M of 22, 54, 108, 135 and 198 is
Options- A. 330
- B. 1980
- C. 5940
- D. 11880 Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. The ratio of two numbers is 3 : 4 and their H.C.F is 4. Their L.C.M is
Options- A. 12
- B. 16
- C. 24
- D. 48 Discuss
- 10. The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is
Options- A. 77
- B. 88
- C. 99
- D. 110 Discuss
Search Results
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: 117
Explanation:
H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117
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: -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 = = = =
Correct Answer: 48
Explanation:
Let the numbers be 3x and 4x . Then their H.C.F = x. So, x=4
Therefore, The numbers are 12 and 16
L.C.M of 12 and 16 = 48
Correct Answer: 77
Explanation:
Product of numbers = 11 x 385 = 4235
Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35
Now, co-primes with product 35 are (1,35) and (5,7)
So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)
Since one number lies 75 and 125, the suitable pair is (55,77)
Hence , required number = 77
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