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- Question
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
- Correct Answer
- 77
ExplanationProduct 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
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. The L.C.M of 22, 54, 108, 135 and 198 is
Options- A. 330
- B. 1980
- C. 5940
- D. 11880 Discuss
- 6. Three numbers are in the ratio 1 : 2 : 3 and the sum of their cubes is 4500 .The smallest numbers is
Options- A. 4
- B. 5
- C. 6
- D. 10 Discuss
- 7. H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is
Options- A. 180
- B. 360
- C. 540
- D. 1260 Discuss
- 8. If HCF of two numbers is 8,which of the following can never be their LCM?
Options- A. 32
- B. 48
- C. 60
- D. 152 Discuss
- 9. A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is ?
Options- A. 63
- B. 31
- C. 16
- D. 27 Discuss
- 10. LCM of 80, 85, 90
Options- A. 11440
- B. 11998
- C. 12240
- D. 12880 Discuss
Search Results
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: -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: 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: 117
Explanation:
H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117
Correct Answer: 5940
Explanation:
22 = 2 x 11
54 =
108 =
135 =
198 =
Correct Answer: 5
Correct Answer: 180
Correct Answer: 60
Explanation:
60 is not a multiple of 8
Correct Answer: 31
Explanation:
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31
Correct Answer: 12240
Explanation:
LCM of (80, 85, 90) can be found by prime factorizing them.
80 ? 2 × 2 × 2 × 2 × 5
85 ? 17 × 5
90 ? 2 × 3 × 3 × 5
L.C.M of (80,85,90) = 2 × 2 x 2 × 2 × 3 × 3 × 5 × 17
= 16 x 9 x 85
= 144 x 85
= 12240
L.C.M of (80,85,90) = 12240.
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