LCM and HCF
LCM and HCF - Important Points
| 6. | Four different buzzers make a beep after every 30 minutes, 1 hour, 1 hour 30 minutes and 1 hour 45 minutes respectively. All buzzers beeped together at 12 noon. They will again beep together at |
|---|
A. 3:00 AM
B. 6:00 AM
C. 9:00 AM
D. 12 Midnight
View Answer Discuss Work SpaceAnswer: option c
Explanation:
Interval after buzzers will beep together = (L.C.M. of 30, 60, 90, 105) min
= 1260 min
= 1260/60 = 21 hrs
So, the buzzers will again beep together after 21 hrs i.e., at 9 AM(after 12 noon)
| 7. | HCF of 2/3, 4/5 and 6/7 is |
|---|
A. 4/105
B. 2/105
C. 16/105
D. 8/105
View Answer Discuss Work SpaceAnswer: option b
Explanation:
Given fractions = 2/3, 4/5 and 6/7
HCF of fractions = HCF of numerators / LCM of denominators
= HCF (2, 4, 6) / LCM (3, 5, 7)
= 2/105
| 8. | The sum of two number is 528 and their HCF is 33. How many pairs of such numbers are possible? |
|---|
A. 1
B. 2
C. 3
D. 4
View Answer Discuss Work SpaceAnswer: option d
Explanation:
Given, the HCF = 33
Then, let the two numbers be 33a and 33b,where a and b are coprime.
=> 33a + 33b = 528
=> a + b = 16
Hence, values of (a, b) which are coprime to each other can be (1,15), (3,13), (5,11) and (7,9).
Hence, required number of pairs is 4.
| 9. | The length, breadth and height of a room are 363m, 528m and 693m respectively. Length of the longest rod that can measure the three dimensions of the room exactly is- |
|---|
A. 33 metre
B. 36 metre
C. 29 metre
D. 30 metre
View Answer Discuss Work SpaceAnswer: option a
Explanation:
Maximum length of Rod = HCF of 363m, 528m and 693m
363 = 3 * 11 * 11
528 = 2 * 2 * 2 * 2 * 3 * 11
693 = 3 * 3 * 7 * 11
HCF of 363m, 528m and 693m = 3 * 11
= 33 metre
| 10. | What is the least number which when divided by 14, 9 and 12 leaves the same remainder 1 in each case? |
|---|
A. 253
B. 352
C. 505
D. 252
View Answer Discuss Work SpaceAnswer: option a
Explanation:
LCM and HCF Questions and Answers FAQ
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