CBSE Class 10 Mathematics Real Numbers Assignment Set 10

Very Short Answer Type Questions

Question 1. Find the prime factors of 1080.
Answer:
Divide repeatedly by primes: 1080 ÷ 2 = 540, 540 ÷ 2 = 270, 270 ÷ 2 = 135, 135 ÷ 3 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime.
\( 1080 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^3 \times 3^3 \times 5 \)
The prime factors of 1080 are 2, 3 and 5.
In simple words: 1080 is built from the primes 2, 3 and 5.

Teacher's Note:
1. Divide by the smallest prime each time.
2. List each prime once when asked for prime factors.

Question 2. Numbers p, q and r are missing in the following factorisation. Find them.

Answer:
Going upward from 17, each number is the one below it multiplied by the prime on its left (2).
r = 17 × 2 = 34, q = r × 2 = 68, p = q × 2 = 136.
p = 136, q = 68, r = 34
In simple words: Multiply by 2 at every step going up.

Teacher's Note:
1. Each box is the product of the box below it and the prime shown beside it.
2. Check: 136 ÷ 2 ÷ 2 ÷ 2 = 17.

Question 3. Find two irrational numbers lying between \( \sqrt{2} \) and \( \sqrt{4} \).
Answer:
Since \( \sqrt{4} = 2 \) and \( \sqrt{2} \approx 1.414 \), we need irrational numbers between 1.414 and 2.
\( \sqrt{3} \approx 1.732 \) and \( \sqrt{3.5} \approx 1.871 \) both lie in this range, and both are irrational because 3 and 3.5 are not squares of rational numbers.
Two such numbers are \( \sqrt{3} \) and \( \sqrt{3.5} \).
In simple words: Take the square roots of numbers between 2 and 4 that are not perfect squares.

Teacher's Note:
1. First write \( \sqrt{4} = 2 \).
2. Check each answer by its decimal value.

Question 4. Find two irrational numbers in the range of 2 to 3.5.
Answer:
Take square roots of non-square numbers between 4 and 12.25: \( \sqrt{7} \approx 2.646 \) and \( \sqrt{10} \approx 3.162 \).
Both lie between 2 and 3.5 and are irrational.
Two such numbers are \( \sqrt{7} \) and \( \sqrt{10} \).
In simple words: Pick non-square numbers whose roots fall in the range.

Teacher's Note:
1. 2² = 4 and 3.5² = 12.25.
2. Any non-square integer between 4 and 12 will do.

Question 5. There is a number 6ⁿ, where n is a natural number. Check if there is any value of n ∈ N for which 6ⁿ is divisible by 7.
Answer:
\( 6^n = (2 \times 3)^n = 2^n \times 3^n \).
The prime factorisation of 6ⁿ contains only the primes 2 and 3. By the uniqueness in the Fundamental Theorem of Arithmetic, 7 is not a factor.
No value of n makes 6ⁿ divisible by 7.
In simple words: Powers of 6 never contain the prime 7.

Teacher's Note:
1. Use the uniqueness of prime factorisation.
2. Divisibility by 7 needs 7 in the factorisation.

Question 6. Without dividing the number \( \frac{7}{80} \), express it in decimal form.
Answer:
\( 80 = 2^4 \times 5 \). Multiply the numerator and denominator by \( 5^3 \) to make the denominator \( 2^4 \times 5^4 = 10^4 \).
\( \frac{7}{80} = \frac{7 \times 5^3}{2^4 \times 5^4} = \frac{875}{10^4} = 0.0875 \)
0.0875
In simple words: Make the denominator a power of 10 and read off the decimal.

Teacher's Note:
1. Make the powers of 2 and 5 equal.
2. Count the zeros in the power of 10 for the decimal places.

Question 7. Find the least possible positive value of P, if the square of a number P is the sum of the squares of the other two numbers Q and R, where 5Q = 12R and Q, R are positive integers.
Answer:
We are given \( P^2 = Q^2 + R^2 \) and \( \frac{Q}{R} = \frac{12}{5} \).
Since Q and R are positive integers in the ratio 12 : 5, the least values are Q = 12 and R = 5.
\( P^2 = 12^2 + 5^2 = 144 + 25 = 169 \), so P = 13.
In simple words: 12, 5, 13 is a Pythagorean triple.

Teacher's Note:
1. The least integers in a ratio come from the lowest form of the ratio.
2. Check: 5Q = 60 = 12R.

Question 8. Find a number which is greater than 3 and when it gets divided by 4, 5 and 6 always leaves 3 as remainder.
Answer:
The number is (LCM of 4, 5 and 6) + 3 = 60 + 3 = 63.
Check: 63 = 4 × 15 + 3 = 5 × 12 + 3 = 6 × 10 + 3.
In simple words: A common remainder r means the number is LCM + r.

Teacher's Note:
1. The remainder 3 is less than each divisor 4, 5, 6.
2. Add the common remainder to the LCM.

Question 9. For two numbers 150 and 100, it is given that L.C.M. (150, 100) = 300. Find H.C.F. (150, 100).
Answer:
HCF × LCM = product of the two numbers, so \( \text{HCF} = \frac{150 \times 100}{300} = 50 \).
HCF (150, 100) = 50
In simple words: Divide the product by the LCM.

Teacher's Note:
1. The relation HCF × LCM = a × b works for two numbers.
2. Check with prime factors: 150 = 2 × 3 × 5², 100 = 2² × 5²; HCF = 2 × 5² = 50.

Question 10. Find the HCF and LCM of the given fractions \( \frac{8}{9}, \frac{16}{81}, \frac{2}{3} \) and \( \frac{10}{27} \).
Answer:
HCF of fractions = \( \dfrac{\text{HCF of numerators}}{\text{LCM of denominators}} = \dfrac{\text{HCF}(8, 16, 2, 10)}{\text{LCM}(9, 81, 3, 27)} = \dfrac{2}{81} \).
LCM of fractions = \( \dfrac{\text{LCM of numerators}}{\text{HCF of denominators}} = \dfrac{\text{LCM}(8, 16, 2, 10)}{\text{HCF}(9, 81, 3, 27)} = \dfrac{80}{3} \).
HCF = \( \frac{2}{81} \), LCM = \( \frac{80}{3} \)
In simple words: Use HCF of tops over LCM of bottoms, and LCM of tops over HCF of bottoms.

Teacher's Note:
1. Write each fraction in lowest terms first (all are already).
2. Remember the two formulas for fractions.

Short Answer Type Questions

Question 11. Find the number which is nearest to 100000 and greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Answer:
\( 8 = 2^3,\ 15 = 3 \times 5,\ 21 = 3 \times 7 \), so LCM = \( 2^3 \times 3 \times 5 \times 7 = 840 \).
The required number is a multiple of 840 just above 100000. 100000 ÷ 840 = 119.05…, so the next multiple is 840 × 120 = 100800.
100800
In simple words: Find the LCM, then the first multiple of it above 100000.

Teacher's Note:
1. Divide 100000 by 840 and round the quotient up.
2. Check: 100800 ÷ 8 = 12600, ÷ 15 = 6720, ÷ 21 = 4800.

Question 12. Prove that \( 3 - \sqrt{5} \) is an irrational number.
Answer:
Assume, to the contrary, that \( 3 - \sqrt{5} = \frac{a}{b} \), where a and b are coprime integers and b ≠ 0.
Then \( \sqrt{5} = 3 - \frac{a}{b} = \frac{3b - a}{b} \). Since a and b are integers, \( \frac{3b - a}{b} \) is rational, so \( \sqrt{5} \) would be rational.
But \( \sqrt{5} \) is irrational, a contradiction. Hence \( 3 - \sqrt{5} \) is irrational.
In simple words: If 3 − √5 were rational, √5 would be rational too.

Teacher's Note:
1. Use \( \sqrt{5} \) is irrational as a known fact.
2. Write the assumption, the contradiction and the conclusion.

Question 13. In a school 437 girls and 342 boys have been divided into groups, so that each group has the same number of students and no group has boys and girls mixed. What is the least number of groups needed?
Answer:
To keep groups as large as possible (fewest groups), the group size must be the HCF of 437 and 342.
437 = 342 × 1 + 95, 342 = 95 × 3 + 57, 95 = 57 × 1 + 38, 57 = 38 × 1 + 19, 38 = 19 × 2 + 0, so HCF = 19.
Number of groups = \( \frac{437}{19} + \frac{342}{19} = 23 + 18 = 41 \)
41 groups
In simple words: Largest equal group size is the HCF; then count the groups.

Teacher's Note:
1. Girls and boys are grouped separately, so add the groups for each.
2. Fewest groups means largest possible group.

Question 14. Pooja multiplied a number 484 with a certain number to obtain the result 3823a. Find the value of a.
Answer:
\( 484 = 2^2 \times 11^2 \), so the result 3823a must be divisible by 2 and 11. Hence a is even: 0, 2, 4, 6 or 8.
The candidates are 38230, 38232, 38234, 38236, 38238. Test divisibility by 11: for 38236, (3 + 2 + 6) − (8 + 3) = 0, so it is divisible by 11. For the other candidates the difference is not a multiple of 11.
Check: 38236 ÷ 484 = 79, so 484 × 79 = 38236.
a = 6
In simple words: The product must be divisible by both 2 and 11.

Teacher's Note:
1. Use the divisibility rule of 11: difference of alternate digit sums.
2. Confirm by dividing 38236 by 484.

Question 15. If the first 100 multiples of 10 are multiplied with each other, find the number of zeroes at the end of the product.
Answer:
The product is \( 10 \times 20 \times 30 \times \cdots \times 1000 = 10^{100} \times (1 \times 2 \times 3 \times \cdots \times 100) = 10^{100} \times 100! \).
Trailing zeros of 100! come from pairs of 2 and 5. The number of 5s is 20 + 4 = 24 (multiples of 5 and of 25), and there are many more 2s. So 100! ends in 24 zeros.
Total zeros = 100 + 24 = 124.
In simple words: Each 10 gives one zero, and 100! adds 24 more.

Teacher's Note:
1. Count the factors of 5 in 100!.
2. Multiples of 25 give an extra factor of 5.

Question 16. Three people running around a rectangular track can complete one turn in 2, 4 and 5.5 hours respectively. When will they meet at the starting point?
Answer:
Convert to minutes: 120, 240 and 330 minutes.
\( 120 = 2^3 \times 3 \times 5,\ 240 = 2^4 \times 3 \times 5,\ 330 = 2 \times 3 \times 5 \times 11 \), so LCM = \( 2^4 \times 3 \times 5 \times 11 = 2640 \) minutes.
2640 ÷ 60 = 44 hours after the start.
In simple words: They meet again after the LCM of the three times.

Teacher's Note:
1. Convert 5.5 hours to minutes to work with whole numbers.
2. Change the final answer back to hours.

Question 17. Find the value of \( \left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2008}\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2007}\right) - \left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2008}\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2007}\right) \).
Answer:
Let \( p = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2007} \). Then the first bracket is \( p + \frac{1}{2008} \), the second is \( 1 + p \), the third is \( 1 + p + \frac{1}{2008} \) and the fourth is p.
Expression = \( \left(p + \frac{1}{2008}\right)(1 + p) - \left(1 + p + \frac{1}{2008}\right)p \)
\( = p + p^2 + \frac{1 + p}{2008} - p - p^2 - \frac{p}{2008} = \frac{1}{2008} \)
\( \frac{1}{2008} \)
In simple words: Name the long sum p; everything cancels except 1/2008.

Teacher's Note:
1. Substitution turns a long expression into simple algebra.
2. Expand carefully and cancel p and p².

Question 18. 4 bells ring together at 9.00 a.m. They ring after 7, 8, 11 and 12 seconds respectively. How many times will they ring together again in the next 3 hours?
Answer:
LCM of 7, 8, 11 and 12: \( 7 \times 8 \times 11 \times 3 = 1848 \) seconds.
In 3 hours there are 3 × 3600 = 10800 seconds. 10800 ÷ 1848 = 5.84…, so the whole number of times is 5.
5 times (at 1848 s, 3696 s, 5544 s, 7392 s and 9240 s).
In simple words: They ring together every 1848 seconds.

Teacher's Note:
1. Count only complete multiples of 1848 inside 10800.
2. The first ring at 9.00 a.m. is not counted.

Question 19. Find the greatest number which when divides 245 and 1029 leaves remainder 5 in each case.
Answer:
Required number = HCF of (245 − 5) and (1029 − 5) = HCF (240, 1024).
\( 240 = 2^4 \times 3 \times 5 \) and \( 1024 = 2^{10} \), so HCF = \( 2^4 = 16 \).
16
In simple words: Subtract the remainder, then take the HCF.

Teacher's Note:
1. The divisor must be greater than the remainder 5.
2. 16 > 5, so it is valid.

Question 20. Find the largest number which when divides 445, 572 and 699 leaves remainders 4, 5 and 6 respectively.
Answer:
The numbers are 445 − 4 = 441, 572 − 5 = 567 and 699 − 6 = 693.
\( 441 = 3^2 \times 7^2,\ 567 = 3^4 \times 7,\ 693 = 3^2 \times 7 \times 11 \), so HCF = \( 3^2 \times 7 = 63 \).
63
In simple words: Subtract each remainder, then find the HCF of the three results.

Teacher's Note:
1. Common prime factors with the smallest powers give the HCF.
2. 63 is greater than all three remainders.