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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
LCD of 12 and 18
Options- A. 36
- B. 42
- C. 12
- D. 6
- Correct Answer
- 36
- 1. H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 2 4 x 3 5 x 5 2 x 7 2 . What is the third number?
Options- A. 47628
- B. 49874
- C. 24157
- D. 42146 Discuss
- 2. 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
- 3. 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
- 4. LCM of 80, 85, 90
Options- A. 11440
- B. 11998
- C. 12240
- D. 12880 Discuss
- 5. 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. 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
- 8. 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
- 9. 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
- 10. A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 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. 35
- B. 36
- C. 37
- D. 38 Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: 47628
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: 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: 31
Explanation:
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31
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: 11/120
Explanation:
Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.
Therefore, Required sum =
Correct Answer: 37
Explanation:
If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.
So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters.
Now, number of cans of Maaza = 80/16 = 5
Number of cans of Pepsi = 144/16 = 9
Number of cans of Sprite = 368/16 = 23
Thus, the total number of cans required = 5 + 9 + 23 = 37
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