CBSE Class 10 Mathematics Real Numbers Worksheet Set 06

VERY SHORT ANSWER TYPE QUESTIONS

Question. If n is a natural number, then 25²ⁿ – 9²ⁿ is always divisible by :
(i) 16
(ii) 34
(iii) both 16 or 34
(iv) None of these
Answer. (i) 16

Question. All decimal numbers are
(a) rational number
(b) irrational numbers
(c) real numbers
(d) integers
Answer. (c) real numbers

Question. Which of these numbers always end with the digits 6.
(a) 4ⁿ
(b) 2ⁿ
(c) 6ⁿ
(d) 8ⁿ
Answer. (c) 6ⁿ

Question. Euclid’s division Lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy.
(a) 1 < r < b
(b) 0 < r ≤ b
(c) 0 ≤ r < b
(d) 0 < r < b
Answer. (c) 0 ≤ r < b

Question. A number N when divided by 16 gives the remainder 5 ______ is the remainder when the same number is divided by 8.
Answer. 5

Question. Write the form in which every odd integer can be written taking t as variable.
Answer. 2t + 1

Question. What would be the value of n for which n²–1 is divisible by 8.
Answer. An odd integer

Question. HCF of 3³ × 5⁴ and 3⁴ × 5² is ________ .
Answer. 3³ × 5²

Question. If a = xy² and b = x³y⁵ where x and y are prime numbers then LCM of (a, b) is _____ .
Answer. x³ × y⁵

SHORT ANSWER TYPE QUESTIONS-I

Question. What is the smallest number by which √5 – √2 is to be multiplied to make it a rational number? Also find the number so obtained?
Answer. √5 + √2 , 3

Question. Find one rational and one irrational no between √3 and √5.
Answer. Rational number between √3 and √5 is √4, and irrational number between √3 and √5 is (√3+√5)/2.

Question. A rational number in its decimal expansion is 327. 7081. What can you say about the prime factors of q, when this number is expressed in the form of p/q ? Give reasons.
Answer. Denominator is the multiple of 2’s and 5’s.

Question. If HCF of 144 and 180 is expressed in the form 13m – 3, find the value of m.
Answer. By Euclid’s division lemma
180 = 144 × 1 + 36
144 = 36 × 4 + 0

HCF of 180 and 144 is 36.
13m – 3 = 36
13 m = 39
m = 3

Question. Use Euclid’s division algorithm to find the HCF of 16 and 28.
Answer. 4

Question. If n is an odd integer then show that n² – 1 is divisible by 8.
Answer. Any +ve odd integer is of the form 4q + 1 or 4q + 3 for some integer q so if n = 4q + 1.
n2 – 1 = (4q + 1)²–1 = 16q² + 8q = 8q(2q + 1) ⇒ n² – 1 is divisible by 8.
If n = 4q + 3
n² – 1 =(4q + 3)² – 1 = 16q² + 24q + 8 = 8 (2q² + 3q + 1) ⇒ n² – 1 is divisible by 8.

SHORT ANSWER TYPE QUESTIONS-II

Question. If 105/(2³×5²×3^p×7^q) is a terminating decimal, what are the least possible values of p and q.
Answer. 0 and 1

Question. Three bells ring at intervals of 9, 12, 15 minutes respectively. If they start ringing together at a time, after how much time will they next ring together?
Answer. LCM of (9, 12, 15) = 180 minutes.

Question. Explain why: 7 × 11 × 13 × 15 + 15 is a composite number
Answer. 15 × (7 × 11 × 13 + 1) as it has more than two factors so it is composite number.

Question. On a morning walk, three persons steps off together and their steps measure 40 cm, 42 cm, and 45 cm respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps?
Answer. LCM of 40, 42, 45 = 2520
Minimum distance each should walk 2520 cm.

Question. During a sale, colour pencils were being sold in the pack of 24 each and crayons in the pack of 32 each. If you want full packs of both and the same number of pencils and crayons, how many packets of each would you need to buy?
Answer. LCM of 24 and 32 is 96
96 crayons or 96/32 = 3 packs of crayons
96 pencils or 96/24 = 4 packs of pencils.

Question. If 7560 = 2³ × 3^p × q × 7, find p and q.
Answer. p = 3 and q = 5

Question. Find the largest number that divides 31 and 99 leaving remainder 5 and 8 respectively.
Answer. Given number = 31 and 99
31 – 5 = 26 and 99 – 8 = 91
Prime factors of 26 = 2 × 13
91 = 7 × 13
HCF of (26, 91) = 13.
∴ 13 is the largest number which divides 31 and 99 leaving remainder 5 and 8 respectively.

Question. In a seminar, the no. of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room the some the same number of participants are to be seated and all of the them being of the the same subject.
Answer. HCF of 60, 84 and 108 is 2² × 3 = 12 = No. of participants in each row.
No. of rooms required = Total number of participants/12
= (60+84+108)/12 = 21 rooms

Question. State fundamental theorem of Arithmetic. Is it possible that HCF and LCM of two numbers be 24 and 540 respectively. Justify your answer.
Answer. HCF = 24, LCM = 540
LCM/HCF = 540/24 = 22.5, not an integer.
Hence two numbers cannot have HCF and LCM as 24 and 540 respectively.

Question. Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3 respectively.
Answer. 1251 – 1 =1250, 9377 – 2 = 9375, 15628 – 3 = 15625
HCF of (15625, 9375) = 3125
HCF of (3125, 1250) = 625
⇒ HCF of (1250, 9375, 15625) = 625

Question. Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.
Answer. Let n be any psoitve integer. Then it is of the form 4q, 4q + 1, 4q + 2 and 4q + 3.
When n = 4q, n³ = 64q³ = 4 (16q³) = 4 m, where m = 16q³
When n = 4q + 1, n³ = (4q + 1)³ = 64q³ + 48q² + 12q + 1
= 4(16q³ + 12q² + 3q) + 1 = 4 m + 1.
where m = 16q³ + 12q² + 3q
Similarly discuss for n = 4q + 2 and 4q + 3.

Question. Show that square of any odd integer is of the form 4m + 1, for some integer m.
Answer. By Euclid’s division algorithm, we have a = bq + r, where 0 ≤ r < 4. On
putting b = 4 we get a = 4q + r where, r = 0, 1, 2, 3.
If r = 0, a = 4q which is even
If r = 1, a = 4q + 1 not divisible by 2
If r = 2, a = 4q + 2 = 2(2q + 1) which is even
If r = 3, a = 4q + 3 not divisible by 2.
So, for any +ve integer q, 4q + 1 and 4q + 3 are odd integers.
How, a² = (4q + 1)² = 16q² + 1 + 8q = 4(4q² + 2q) + 1 = 4m + 1
where m = 4q² + 2q similarly for 4q + 3.

Question. Show that either the number n, n + 2 and n + 4 are three consecutive odd numbers or 3 even numbers.
Answer. Let the number divisible by 3 is of the form 3k + r, r = 0, 1, 2
a = 3k, 3k + 1 or 3k + 2
(i) When a = 3k
n = 3k ⇒ n is divisible by 3.
n + 2 = 3k + 2 ⇒ n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3k + 3 + 1 = 3(k + 1) + 1 ⇒ n + 4 is not divisible by 3.
So, only one out of n, n + 2 and n + 4 is divisible by 3.
(ii) When a = 3k + 1
n = 3k + 1 ⇒ n is not divisible by 3.
n + 2 = 3k + 1 + 2 = 3k + 3 = 3(k + 1)
⇒ n + 2 is divisible by 3.
n + 4 = 3k + 1 + 4 = 3k + 5 = 3(k + 1) + 2
⇒ n + 2 is not divisible by 3.
So, only one out of n, n + 2 and n + 4 is divisible by 3.
Similarly do for a = 3k + 2.

Question. The HCF of 65 and 117 is expressible in the form 65 m – 117. Find the value of m. Also find the LCM of 65 and 117 using prime factorisation method.
Answer. HCF of 117 and 65 by Euclid division algorithm.
117 = 65 × 1 + 52
65 = 52 × 1 = 52
52 = 13 × 4 + 0
HCF (117, 52) = 13.
Given that 65 m – 117 = 13 ⇒ 65 m = 130 ⇒ m = 2.
LCM (65, 117) = 13 × 3² × 5 = 585

Question. Express the HCF of numbers 72 and 124 as a linear combination of 72 and 124.
Answer. HCF (124, 72) = 4
4 = 124 × 7 + 72 × (– 12), x = 7, y = – 12