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CTET Paper II (Mathematics and Science) 27 Dec 2021 Solved Paper

Question : 33 of 150

Marks: +1, -0

b = 2

3^{2} × 5 \text{ and}

c = 2^{5} × 3^{2} × 5^{2},

\frac{LCM of a, b and c}{HCF of a, b and c}

is equal to

Solution:

Concept:

LCM uses the highest power of each prime factor, while HCF uses the lowest power of common prime factors.

Explanation:

Given:

a = 2^4 \times 3 \times 5^3

b = 2^3 \times 3^2 \times 5

c = 2^5 \times 3^2 \times 5^2

Find LCM: take maximum exponent for each prime:

2^5

3^2

5^3

. So LCM =

2^5 \times 3^2 \times 5^3

Find HCF: take minimum exponent common to all:

2^3

(since all have at least

2^3

3^1

5^1

. So HCF =

2^3 \times 3 \times 5

Divide LCM by HCF:

\frac{2^5 \times 3^2 \times 5^3}{2^3 \times 3 \times 5} = 2^{2} \times 3 \times 5^{2} = 4 \times 3 \times 25 = 300

Answer:

300 (Option C)

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