Discussion
Home ‣ Aptitude ‣ HCF and LCM Comments
- Question
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
- Correct Answer
- 180
- 1. Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 12. The numbers are
Options- A. 4, 8, 12
- B. 5, 10, 15
- C. 10, 20, 30
- D. 12, 24, 36 Discuss
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. The product of two numbers is 4107. If the H.C.F of these numbers is 37, then the greater number is
Options- A. 101
- B. 107
- C. 111
- D. 185 Discuss
- 7. The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275 , then the other is
Options- A. 279
- B. 283
- C. 308
- D. 318 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. The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is
Options- A. 1677
- B. 1683
- C. 2523
- D. 3363 Discuss
- 10. 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
Search Results
Correct Answer: 12, 24, 36
Explanation:
Let the required numbers be x, 2x, 3x. Then, their H.C.F =x. so, x= 12
Therefore, The numbers are 12, 24, 36
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
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: 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: 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: 111
Explanation:
Let the numbers be 37a and 37b. Then , 37a x 37b =4107 => ab = 3.
Now co-primes with product 3 are (1,3)
So, the required numbers are (37 x 1, 37 x 3) i.e, (37,111).
Therefore, Greater number = 111.
Correct Answer: 308
Explanation:
Other number = = 308
Correct Answer: 60
Explanation:
60 is not a multiple of 8
Correct Answer: 1683
Explanation:
L.C.M of 5, 6, 7, 8 = 840
Therefore, Required Number is of the form 840k+3.
Least value of k for which (840k+3) is divisible by 9 is k = 2
Therefore, Required Number = (840 x 2+3)=1683
Correct Answer: 31
Explanation:
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31
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