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Question 1 of 5
1. Question
The HCF of two natural numbers m and n is 24 and their product is 552. How many sets of values of m and n are possible?
Correct
Sol.(d)
Given HCF of two natural numbers m and n = 24
and their product, m × n = 552
LCM of two natural numbers = Product of m and n / HCF of m and n = 552/24 = 23
Here, no set of m and n is possible satisfying the given conditions.
Incorrect
Sol.(d)
Given HCF of two natural numbers m and n = 24
and their product, m × n = 552
LCM of two natural numbers = Product of m and n / HCF of m and n = 552/24 = 23
Here, no set of m and n is possible satisfying the given conditions.
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Question 2 of 5
2. Question
What is the highest common factor of 2x3 + x2 – x – 2 and 3x3 – 2x2 + x – 2 ?
Correct
Sol.(a)
Let f(x) = 2x3 + x2 – x – 2 = (x – 1)(2x2 + 3x + 2)
and g(x)= 3x3 – 2x2 + x – 2 = (x -1)(3x2 + x + 2)
Hence, the highest common factor of f(x) and g(x) is (x – 1).
Incorrect
Sol.(a)
Let f(x) = 2x3 + x2 – x – 2 = (x – 1)(2x2 + 3x + 2)
and g(x)= 3x3 – 2x2 + x – 2 = (x -1)(3x2 + x + 2)
Hence, the highest common factor of f(x) and g(x) is (x – 1).
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Question 3 of 5
3. Question
For any integer n, what is HCF (22n + 7,33n + 10) equal to?
Correct
Sol.(b)
HCF of (22n + 7, 33n + 10) is always 1.
Illustration: For n = 1, HCF (29, 43) = 1
For n = 2, HCF (51, 76) = 1
For n= 3, HCF (73, 109) = 1
Incorrect
Sol.(b)
HCF of (22n + 7, 33n + 10) is always 1.
Illustration: For n = 1, HCF (29, 43) = 1
For n = 2, HCF (51, 76) = 1
For n= 3, HCF (73, 109) = 1
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Question 4 of 5
4. Question
In a fire range, 4 shooters are firing at the respective targets. The first, the second, the third and fourth shooters hit the target once in every 5 s, 6 s, 7s and 8 s, respectively. If all of them hit their target at 9 : 00 am, when will they hit their target together again?
Correct
Sol.(c)
Time after which they will hit the target again together
= LCM (5, 6, 7 and 8)
= 5 × 3 × 7 × 2 × 4
= 840 s
They will hit target together = 840/60 = 14 min.
So, they will hit together again at 9:14 am.
Incorrect
Sol.(c)
Time after which they will hit the target again together
= LCM (5, 6, 7 and 8)
= 5 × 3 × 7 × 2 × 4
= 840 s
They will hit target together = 840/60 = 14 min.
So, they will hit together again at 9:14 am.
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Question 5 of 5
5. Question
What is the HCF of 8(x5 – x3 + x) and 28 (x6 + 1) ?
Correct
Sol. (a)
Let p(x) = 8(x5 – x3 + x) = 4 × 2 × x(x4 – x2 + 1)
and q(x) = 28(x6 + 1) = 7 × 4[(x2)3 + (1)3]
= 4 × 7 × (x2 + 1)(x4 – x2 + 1)
Hence, HCF of p(x) and q(x) = 4( x4 – x2 + 1)
Incorrect
Sol. (a)
Let p(x) = 8(x5 – x3 + x) = 4 × 2 × x(x4 – x2 + 1)
and q(x) = 28(x6 + 1) = 7 × 4[(x2)3 + (1)3]
= 4 × 7 × (x2 + 1)(x4 – x2 + 1)
Hence, HCF of p(x) and q(x) = 4( x4 – x2 + 1)
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