Saved Bookmarks
| 1. |
HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\) is:1. 502. \(1 \over 150\)3. \(3 \over 50\)4. \(3 \over 150\) |
|
Answer» Correct Answer - Option 4 : \(3 \over 150\) Given: We have to find the HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\) Concept Used: HCF of fraction = (HCF of numerator)/(LCM of denominator) Calculation: HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\) ⇒ \({HCF\ (9,\ 3,\ 6,\ 12) \over LCM\ (10,\ 25,\ 15,\ 5)}\) HCF of numerators is 3 LCM of denominators is 150 Required HCF is 3/150 ∴ HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\) is (3/150). We can make (3/150) fraction into a small form i.e (1/50), But we have to see options also. In the given option (1/50) is not present. If (1/50) will be present in place of (3/150), then we can mark (1/50) as the answer. Which is totally correct. Sometimes we have to check options also to mark the correct answers. In this case, we are marking (3/150) according to the options provided. |
|