The L.C.M. of two numbers is 14560 and their H.C.F. is 13. If one of them is 416, the other is
Answer: B
\[ \text{H.C.F.} \times \text{L.C.M.} = \text{product} \]
\[ 13 \times 14560 = 416 \times n \]
\[ n = \frac{189{,}280}{416} = 455 \]
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3 birds fly along the circumference of a jungle. They complete one round in 27 minutes, 45 minutes and 63 minutes respectively. Since they start together, when will they meet again at the starting position?
Answer: A
They meet at the start again after a whole number of laps each, so take the L.C.M. of the lap times.
\[ 27 = 3^{3}, \quad 45 = 3^{2}\cdot 5, \quad 63 = 3^{2}\cdot 7 \]
\[ \text{L.C.M.} = 3^{3} \cdot 5 \cdot 7 = 945 \text{ minutes} \]
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What is HCF of 36/75, 48/150 , 72/135?
Answer: A
For fractions the rule inverts between the two halves:
\[ \text{HCF of fractions} = \frac{\operatorname{HCF}(\text{numerators})}{\operatorname{LCM}(\text{denominators})} \]
Numerators first:
\[ \operatorname{HCF}(36,48,72) = 12 \]
Then the denominators, through their prime factors:
\[ 75 = 3 \cdot 5^{2}, \quad 150 = 2 \cdot 3 \cdot 5^{2}, \quad 135 = 3^{3} \cdot 5 \]
\[ \operatorname{LCM} = 2 \cdot 3^{3} \cdot 5^{2} = 1350 \]
\[ \therefore\; \text{HCF} = \frac{12}{1350} \]
Taking the LCM of the numerators and the HCF of the denominators instead gives the LCM of the fractions, which is the usual confusion.
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Rajesh had to arrange his books in uniform groups. He makes groups of 4 books each. But 3 books are left. He tries it with groups of 5 books each. But still 3 books are left. 3 books are still left when he tried with groups of 9 or 10 books each. How many books does he have?
Answer: D
Three books are left over in every arrangement, so the total is 3 more than a common multiple of all four group sizes:
\[ N = \operatorname{LCM}(4,5,9,10)\,k + 3 \]
Taking prime factors, \( 4 = 2^{2} \), \( 5 = 5 \), \( 9 = 3^{2} \), \( 10 = 2 \cdot 5 \):
\[ \operatorname{LCM} = 2^{2} \cdot 3^{2} \cdot 5 = 180 \]
\[ \therefore\; N = 180k + 3, \quad\text{smallest such } N = 183 \]
The other options are all multiples of the LCM with nothing left over, so each would divide exactly and leave no remainder at all.
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