Discussion
Home ‣ Aptitude ‣ HCF and LCM Comments
- Question
54, 60 LCM is
Options- A. 600
- B. 540
- C. 60
- D. 54
- Correct Answer
- 540
ExplanationHere the LCM of 54, 60 is
=> 2 x 3 x 3 x 3 x 10
= 6 x 9 x 10
= 54 x 10
= 540.
- 1. Which of the following has the most number of divisors?
Options- A. 99
- B. 101
- C. 176
- D. 182 Discuss
- 2. LCD of 12 and 18
Options- A. 36
- B. 42
- C. 12
- D. 6 Discuss
- How to calculate LCD ::
- 3. 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
- 4. 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
- 5. 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
- 6. If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?
Options- A. -23
- B. 27
- C. 16
- D. -19 Discuss
- 7. The difference of two numbers is 14. Their LCM and HCF are 441 and 7. Find the two numbers ?
Options- A. 63 and 49
- B. 64 and 48
- C. 62 and 46
- D. 64 and 49 Discuss
- 8. The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, 12 m 95 cm is
Options- A. 15 cm
- B. 25 cm
- C. 35 cm
- D. 42 cm 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. 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. 13/125
- B. 14/57
- C. 11/120
- D. 16/41 Discuss
Search Results
Correct Answer: 176
Explanation:
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
Correct Answer: 36
Explanation:
LCD is nothing but Lowest or Least Common Denominator
Here LCD of 12 and 18 means LCD of two fractions with denominators 12 and 18 respectively.
Therefore, LCM of 12 & 18 = 6 x 3 x 2 = 36
The lowest common denominator or least common denominator (LCD) is the least common multiple (LCM) of the denominators of a set of fractions.
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
Correct Answer: 91
Explanation:
Required number of students = H.C.F of 1001 and 910 = 91
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: -19
Explanation:
HCF of 210 and 55 is 5
Now, 210x5 + 55P = 5
=> 1050 + 55P = 5
=> 55P = -1045
=> P = -1045/55
=> P = -19.
Correct Answer: 63 and 49
Explanation:
Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y
Difference = 14
=> 7x - 7y = 14
=> x - y = 2
product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63
Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7
The numbers are 7x and 7y
=> 63 and 49
Correct Answer: 35 cm
Explanation:
Required Length = H.C.F of 700 cm, 385 cm and 1295 c
= 35 cm.
Correct Answer: 1683
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
Correct Answer: 11/120
Explanation:
Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120
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