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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 the sum of their cubes is 4500 .The smallest numbers is
Options- A. 4
- B. 5
- C. 6
- D. 10 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. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. LCM of 80, 85, 90
Options- A. 11440
- B. 11998
- C. 12240
- D. 12880 Discuss
- 9. 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
- 10. 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
Search Results
Correct Answer: 5
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: 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: 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.
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
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