Discussion
Home ‣ Aptitude ‣ HCF and LCM Comments
- Question
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
- Correct Answer
- 1683
ExplanationL.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
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. LCM of 80, 85, 90
Options- A. 11440
- B. 11998
- C. 12240
- D. 12880 Discuss
- 8. A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?
Options- A. 47
- B. 46
- C. 37
- D. 35 Discuss
- 9. The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:
Options- A. 91
- B. 910
- C. 1001
- D. 1911 Discuss
- 10. Find the Greatest Number that will devide 43, 91 and 183 so as to leave the same remainder in each case
Options- A. 4
- B. 7
- C. 9
- D. 13 Discuss
Search Results
Correct Answer: 60
Explanation:
60 is not a multiple of 8
Correct Answer: 308
Explanation:
Other number = = 308
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: 180
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: 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.
Correct Answer: 37
Explanation:
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.
Correct Answer: 91
Explanation:
Required number of students = H.C.F of 1001 and 910 = 91
Correct Answer: 4
Explanation:
Required Number = H.C.F of (91- 43), (183- 91) and (183-43)
= H.C.F of 48, 92, and 140 = 4
Comments
There are no comments.Programming
Copyright ©CuriousTab. All rights reserved.