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Maths 1 Marks Question

Real Numbers · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 17 questions.

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17 questions · timed · auto-graded

Question 11 Mark
Given that $HCF (306, 657) = 9,$ find $LCM (306, 657).$
Answer
As we know that, $HCF$ $\times$ $LCM =$ Product of two numbers
$LCM (306, 657) =$ $\frac{306 \times 657}{HCF(306,657)}$ = $\frac{306 \times 657}{9}$ = 22338.
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Question 21 Mark
Find the $LCM$ and $HCF$ of $12, 15$ and $21$ integers by applying the prime factorisation method.
Answer
$12, 15$ and $21$
$12 = 2^2\times 3$
$15 = 3 \times 5$
$21 = 3 \times 7$
$HCF$ $= 3$
$LCM$ $= 2^2\times 3 \times 5 \times 7 = 420$
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Question 31 Mark
Find the $LCM$ and $HCF$ of $8, 9$ and $25$ integers by applying the prime factorisation method.
Answer
$8,9 \text { and } 25$
$ 8=2 \times 2 \times 2=2^3 $
$ 9=3 \times 3=3^2 $
$ 25=5 \times 5=5^2 $
$ \text { HCF }=1 $
$ \text { LCM }=2^3 \times 3^2 \times 5^2=1800$
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Question 41 Mark
Find the $LCM$ and $HCF$ of $17, 23$ and $29$ integers by applying the prime factorisation method.
Answer
Given numbers are: $17, 23$ and $29$
Since the three numbers are prime, we have
$17 = 1 \times 17$
$23 = 1 \times 23$
$29 = 1 \times 29$
$\Rightarrow$ $HCF = 1$
and $LCM = 17 \times 23 \times 29 = 11339$
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Question 51 Mark
Find the $LCM$ and $HCF$ of $510$ and $92$ pairs of integers and verify that $LCM$ $\times$ $HCF =$ product of the two numbers.
Answer
$510$ and $92$
$510= 2 \times 3 \times 5\times 17$
$92 = 2 \times 2 \times 23$
$HCF = 2$
$LCM = 2 \times 2 \times 3 \times 5 \times 17 \times 23 = 23460$
Product of two numbers $510$ and $92 = 510 \times 92 = 46920$
$HCF \times LCM = 2 \times 23460 = 46920$
Hence, product of two numbers $= HCF \times LCM$
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Question 61 Mark
Find the $LCM$ and $HCF$ of $26$ and $91$ pairs of integers and verify that $LCM$ $\times$ $HCF =$ product of the two numbers.
Answer
$26$ and $91$
$26= 2 \times 13$
$91 = 7 \times 13$
$HCF = 13$
$LCM = 2 \times 7 \times 13 = 182$
Product of two numbers $26$ and $91 = 26 \times 91 = 2366$
$HCF \times LCM = 13 \times 182 = 2366$
Hence, product of two numbers $= HCF \times LCM$
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Question 81 Mark
Express $5005$ as a product of its prime factors.
Answer
$5005 = 5$ $\times$ $7$ $\times$ $11$ $\times$ $13$
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Question 91 Mark
Express $3825$ as a product of its prime factors.
Answer
$3825 = 3$ $\times$ $3$ $\times$ $5$ $\times$ $5$ $\times$ $17 = 3^2\times5^2$ $\times$ $17$
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Question 101 Mark
Express $156$ as the product of its prime factors.
Answer

So, the factors of $156$ are $2$ $\times$ $2$ $\times$ $3$ $\times$ $13$
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Question 111 Mark
Express $140$ as a product of its prime factors.
Answer
$140 = 2$ $\times$ $2$ $\times$ $5$ $\times$ $7 = 22$ $\times$ $5$ $\times$ $7$
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Question 131 Mark
In $H.C.F.$ $(a, b) \times$ $L.C.M$. $(a, b)=a \times b \quad a=420$; and $b=130$ then what is $L.C.M. ?$
Answer
$5460$
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Question 171 Mark
If p, q, rare prime numbers then what will be their LCM?
Answer
The LCM of three distinct prime numbers p,q, and r is their product, p⋅q⋅r. This is because prime numbers have only two factors: 1 and themselves. Therefore, they share no common factors other than 1.
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