Discussion
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- Question
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
- Correct Answer
- 37
ExplanationIf 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
- 1. 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
- 2. 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
- 3. 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
- 5. LCD of 12 and 18
Options- A. 36
- B. 42
- C. 12
- D. 6 Discuss
- 6. The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is
Options- A. 12
- B. 48
- C. 84
- D. 108 Discuss
- 7. 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
- 8. A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.
Options- A. 107
- B. 117
- C. 127
- D. 137 Discuss
- 9. 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
- 10. Find the lowest common multiple of 24, 36 and 40.
Options- A. 120
- B. 240
- C. 360
- D. 480 Discuss
Search Results
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: 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: -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: 36
Correct Answer: 84
Explanation:
Let the numbers be x and 4x. Then,
Hence Larger Number = 4x = 84
Correct Answer: 180
Correct Answer: 117
Explanation:
Let us calculate both the length and width of the room in centimeters.
Length = 6 meters and 24 centimeters = 624 cm
width = 4 meters and 32 centimeters = 432 cm
As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room.
Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48
Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117
Correct Answer: 47628
Correct Answer: 360
Explanation:
2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
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