Real Numbers – Class 10 Mathematics (Lecture 7)
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If the HCF of 408 and 1032 is expressible in the form \(1032m – 408 \times 5\), find \(m\).
Solution:
First, find HCF(408, 1032) using Euclidean algorithm:
1. \(1032 = 408 \times 2 + 216\)
2. \(408 = 216 \times 1 + 192\)
3. \(216 = 192 \times 1 + 24\)
4. \(192 = 24 \times 8 + 0\)
HCF = 24
Given: \(1032m – 408 \times 5 = 24\)
\(1032m – 2040 = 24\)
\(1032m = 24 + 2040 = 2064\)
\(m = \frac{2064}{1032} = 2\)
Answer: \(m = 2\)
Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3 respectively.
Solution:
Let the number be \(d\).
When dividing 280, remainder is 4 ⇒ \(280 – 4 = 276\) is divisible by \(d\)
When dividing 1245, remainder is 3 ⇒ \(1245 – 3 = 1242\) is divisible by \(d\)
So \(d\) must divide both 276 and 1242.
We need the largest such \(d\) = HCF(276, 1242)
Find HCF:
\(276 = 2^2 \times 3 \times 23\)
\(1242 = 2 \times 3^3 \times 23\)
HCF = \(2 \times 3 \times 23 = 138\)
Answer: 138
Find the greatest number that divides 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Solution:
Let the number be \(d\).
445 leaves remainder 4 ⇒ \(445 – 4 = 441\) is divisible by \(d\)
572 leaves remainder 5 ⇒ \(572 – 5 = 567\) is divisible by \(d\)
699 leaves remainder 6 ⇒ \(699 – 6 = 693\) is divisible by \(d\)
So \(d\) must divide 441, 567, and 693.
We need the largest such \(d\) = HCF(441, 567, 693)
Find HCF:
\(441 = 3^2 \times 7^2\)
\(567 = 3^4 \times 7\)
\(693 = 3^2 \times 7 \times 11\)
HCF = \(3^2 \times 7 = 9 \times 7 = 63\)
Answer: 63
Find the HCF of 65 and 117 and express it in the form \(65m + 117n\).
Solution:
First, find HCF(65, 117):
\(65 = 5 \times 13\)
\(117 = 3^2 \times 13\)
HCF = 13
Now express 13 as \(65m + 117n\):
Using Euclidean algorithm:
1. \(117 = 65 \times 1 + 52\)
2. \(65 = 52 \times 1 + 13\)
3. \(52 = 13 \times 4 + 0\)
From step 2: \(13 = 65 – 52 \times 1\)
From step 1: \(52 = 117 – 65 \times 1\)
Substitute: \(13 = 65 – (117 – 65 \times 1) \times 1\)
\(13 = 65 – 117 + 65\)
\(13 = 65 \times 2 – 117 \times 1\)
So, \(13 = 65 \times 2 + 117 \times (-1)\)
Thus, \(m = 2\), \(n = -1\)
Answer: HCF = 13, expressed as \(65 \times 2 + 117 \times (-1)\)
If \(A = 2n + 13\), \(B = n + 7\) where \(n\) is a natural number, then HCF of \(A\) and \(B\) is:A) 2 B) 1 C) 3 D) 4
Solution:
Let’s find HCF using Euclidean algorithm concept:
Consider \(A – 2B = (2n + 13) – 2(n + 7) = 2n + 13 – 2n – 14 = -1\)
So, \(A – 2B = -1\) or \(2B – A = 1\)
This means any common divisor of \(A\) and \(B\) must divide 1.
Therefore, HCF(\(A, B\)) = 1
Answer: B) 1
Check with example: If \(n = 1\), \(A = 15\), \(B = 8\), HCF = 1
If \(n = 2\), \(A = 17\), \(B = 9\), HCF = 1
There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:A) 22 B) 16 C) 36 D) 21
Solution:
We need to find the maximum number of students per section (HCF) so that sections are equal.
HCF(576, 448):
\(576 = 2^6 \times 3^2\)
\(448 = 2^6 \times 7\)
HCF = \(2^6 = 64\)
Number of boys sections = \(\frac{576}{64} = 9\)
Number of girls sections = \(\frac{448}{64} = 7\)
Total sections = \(9 + 7 = 16\)
Answer: B) 16
Three farmers have 490 kg, 588 kg and 882 kg of wheat respectively. Find the maximum capacity of a bag so that the wheat can be packed in exact number of bags:A) 98 kg B) 290 kg C) 200 kg D) 350 kg
Solution:
We need HCF(490, 588, 882) to find maximum bag capacity.
Prime factorisation:
\(490 = 2 \times 5 \times 7^2\)
\(588 = 2^2 \times 3 \times 7^2\)
\(882 = 2 \times 3^2 \times 7^2\)
HCF = \(2 \times 7^2 = 2 \times 49 = 98\)
Answer: A) 98 kg
Priya takes 18 minutes and Harish takes 12 minutes to drive one round of a circular field. If they start together in the same direction, after how many minutes will they meet?A) 36 minutes B) 18 minutes C) 6 minutes D) They will not meet
Solution:
They will meet when both have completed integer number of rounds.
Find LCM(18, 12) to get the time when they coincide:
\(18 = 2 \times 3^2\)
\(12 = 2^2 \times 3\)
LCM = \(2^2 \times 3^2 = 4 \times 9 = 36\) minutes
In 36 minutes:
Priya completes \(36 \div 18 = 2\) rounds
Harish completes \(36 \div 12 = 3\) rounds
Answer: A) 36 minutes
There are 312, 260 and 156 students in classes X, XI and XII respectively. Buses are to be hired with equal number of students in each bus. Find the maximum number of students per bus:A) 52 B) 56 C) 48 D) 63
Solution:
We need HCF(312, 260, 156) to find maximum students per bus.
Prime factorisation:
\(312 = 2^3 \times 3 \times 13\)
\(260 = 2^2 \times 5 \times 13\)
\(156 = 2^2 \times 3 \times 13\)
HCF = \(2^2 \times 13 = 4 \times 13 = 52\)
Answer: A) 52