CBSE Class 10 Real Numbers Sure Shot Questions Set 02

Assertion-Reason Questions

Direction: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assetion (A) is false but Reason (R) is true.

Question. Assertion: \( \frac{13}{3125} \) is a terminating decimal fraction.
Reason: If \( q = 2^n \cdot 5^m \) where \( n \) and \( m \) are non-negative integers, then \( \frac{p}{q} \) is a terminating decimal fraction.
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation: Since the factors of the denominator 3125 is of the form \( 2^0 \times 5^5 \), \( \frac{13}{3125} \) is a terminating decimal fraction.

Question. Assertion: A number N when divided by 15 gives the remainder 2. Then the remainder is same when N is divided by 5.
Reason: \( \sqrt{3} \) is an irrational number.
Answer: (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Explanation: Let we take three numbers which are divisible by 5 and 15 both, are 30, 45, 60. Now, we add the remainder 2, we get 32, 47, 62. Therefore, we can see that as one numbers are divisible by 5 & 15 but remainder is same as 2.

Question. Assertion: Denominator of 34.12345. When expressed in the form \( \frac{p}{q}, q \neq 0 \), is of the form \( 2^m \times 5^n \), where m and n are non-negative integers.
Reason: 34.12345 is a terminating decimal fraction.
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation: Reason is clearly true. Here, \( 34.12345 = \frac{3412345}{100000} = \frac{682469}{20000} = \frac{682469}{2^5 \times 5^4} \). Its denominator is of the form \( 2^m \times 5^n \).

Question. Assertion: When a positive integer a is divided by 3, the values of remainder can be 0, 1 or 2.
Reason: According to Euclid’s Division Lemma \( a = bq + r \), where \( 0 \le r < b \) and r is an integer.
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation: Given positive integers A and B, there exists unique integers Q and R satisfying \( a = bq + r \), where \( 0 \le r < b \). This is known as Euclid’s Division Lemma.

Question. Assertion: The H.C.F. of two numbers is 16 and their product is 3072. Then their L.C.M. = 162.
Reason: If a and b are two positive integers, then \( H.C.F. \times L.C.M. = a \times b \).
Answer: (d) Assetion (A) is false but Reason (R) is true.
Explanation: Since \( HCF \times LCM = a \times b \), \( 3072 = 16 \times 162 \) would imply \( 3072 = 2592 \), which is false. \( \therefore 3072 \neq 2592 \).

Case Based Questions

I. To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections—section A and Section B of grade X. There are 32 students in Section A and 36 students in Section B.

Question. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?
(a) 144 (b) 128 (c) 288 (d) 272
Answer: (c) 288
Explanation: LCM of 32 and 36; \( 32 = 2^5 \); \( 36 = 2^2 \times 3^2 \). \( \therefore LCM = 2^5 \times 3^2 = 288 \)

Question. If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32, 36) is
(a) 2 (b) 4 (c) 6 (d) 8
Answer: (b) 4
Explanation: \( HCF \times LCM = 32 \times 36 \). \( HCF = \frac{32 \times 36}{288} = \frac{1152}{288} = 4 \). \( \therefore HCF = 4 \)

Question. 36 can be expressed as a product of its primes as
(a) \( 2^2 \times 3^2 \) (b) \( 2^1 \times 3^3 \) (c) \( 2^3 \times 3^1 \) (d) \( 2^0 \times 3^0 \)
Answer: (a) \( 2^2 \times 3^2 \)

Question. \( 7 \times 11 \times 13 \times 15 + 15 \) is a
(a) Prime number (b) Composite number (c) Neither prime nor composite (d) None of the above
Answer: (b) Composite number

Question. If p and q are positive integers such that \( p = ab^2 \) and \( q = a^2b \), where a and b are prime numbers, then the LCM (p, q) is
(a) \( ab \) (b) \( a^2b^2 \) (c) \( a^3b^2 \) (d) \( a^3b^3 \)
Answer: (b) \( a^2b^2 \)

II. A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.

Question. In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number of participants that can accommodated in each room are
(a) 14 (b) 12 (c) 16 (d) 18
Answer: (b) 12
Explanation: H=60; E=84; M=108. HCF of 60, 84 and 108: \( 60 = 2^2 \times 3 \times 5 \); \( 84 = 2^2 \times 3 \times 7 \); \( 108 = 2^2 \times 3^3 \). \( HCF = 2^2 \times 3 = 12 \)

Question. What is the minimum number of rooms required during the event?
(a) 11 (b) 31 (c) 41 (d) 21
Answer: (d) 21
Explanation: \( \frac{60}{12} + \frac{84}{12} + \frac{108}{12} = 5 + 7 + 9 = 21 \)

Question. The LCM of 60, 84 and 108 is
(a) 3780 (b) 3680 (c) 4780 (d) 4680
Answer: (a) 3780
Explanation: \( LCM = 2^2 \times 3^3 \times 5 \times 7 = 3780 \)

Question. The product of HCF and LCM of 60, 84 and 108 is
(a) 55360 (b) 35360 (c) 45500 (d) 45360
Answer: (d) 45360
Explanation: \( 12 \times 3780 = 45360 \)

Question. 108 can be expressed as a product of its primes as
(a) \( 2^3 \times 3^2 \) (b) \( 2^3 \times 3^3 \) (c) \( 2^2 \times 3^2 \) (d) \( 2^2 \times 3^3 \)
Answer: (d) \( 2^2 \times 3^3 \)

III. A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience. Observe the following factor tree and answer the following:

Question. What will be the value of x?
(a) 15005 (b) 13915 (c) 56920 (d) 17429
Answer: (b) 13915
Explanation: Value of \( x = 2783 \times 5 = 13915 \)

Question. What will be the value of y?
(a) 23 (b) 22 (c) 11 (d) 19
Answer: (c) 11
Explanation: \( 253y = 2783 \Rightarrow y = \frac{2783}{253} = 11 \)

Question. What will be the value of z?
(a) 22 (b) 23 (c) 17 (d) 19
Answer: (b) 23
Explanation: \( 11z = 253 \Rightarrow z = \frac{253}{11} = 23 \). \( \therefore z = 23 \)

Question. According to Fundamental Theorem of Arithmetic 13915 is a
(a) Composite number (b) Prime number (c) Neither prime nor composite (d) Even number
Answer: (a) Composite number

Question. The prime factorisation of 13915 is
(a) \( 5 \times 11^2 \times 13^2 \) (b) \( 5 \times 11^3 \times 23^2 \) (c) \( 5 \times 11^2 \times 23 \) (d) \( 5 \times 11^3 \times 13^2 \)
Answer: (c) \( 5 \times 11^2 \times 23 \)

IV. For 71th republic day Parade on 26/01/2021 in Delhi, Captain RS Meel is planning for parade of following two groups: (1) First group of Army contingent of 624 members behind an army band of 32 members. (2) Second group of CRPF troops with 468 soldiers behind the 228 members of bikers. These two groups are to march in the same number of columns. This sequence of soldiers is followed by jhankis of difference states which are showing the culture of the respective states.

Question. What is the maximum number of columns in which the army troops can march?
(a) 8 (b) 16 (c) 4 (d) 32
Answer: (b) 16

Question. What is the maximum number of columns in which the CRPF troops can march?
(a) 4 (b) 8 (c) 12 (d) 16
Answer: (c) 12
Explanation: HCF of 624, 468 and 228

Question. What is the maximum number of columns in which total army troops and CRPF troops together can march past?
(a) 2 (b) 4 (c) 6 (d) 8
Answer: (b) 4
Explanation: HCF of 228 and 32

Question. What should be subtracted from the numbers of CRPF soldiers and the number of bikers so that their maximum number of columns is equal to the maximum number of columns of army troops?
(a) 4 Soldiers and 4 Bikers (b) 4 Soldiers and 2 Bikers (c) 2 Soldiers and 4 Bikers (d) 2 Soldiers and 2 Bikers
Answer: (a) 4 Soldiers and 4 Bikers
Explanation: Maximum no. of columns of army troops = 16 (from point (i)). But 228 and 468 are not divisible by 16. If we subract 4 from 228 and 468, both 224 and 464 are divisible by 16.

Question. What should be added with the number of CRPF soldiers and the number of bikers so that their maximum number of columns is equal to the maximum number of columns of army troops?
(a) 4 Soldiers and 4 Bikers (b) 12 Soldiers and 12 Bikers (c) 6 Soldiers and 6 Bikers (d) 12 Soldiers and 6 Bikers
Answer: (b) 12 Soldiers and 12 Bikers

V. In a classroom activity on real numbers, the students have to pick a number card from a pile and frame question on it if it is not a rational number for the rest of the class. The number cards picked up by first 5 students and their questions on the numbers for the rest of the class are as shown below. Answer them:

Question. Suraj picked up \( \sqrt{8} \) and his question was: Which of the following is not irrational?
(a) It is a natural number.
(b) It is an irrational number.
(c) It is a rational number.
(d) None of these.
Answer: (b) It is an irrational number.
Explanation: Here, \( \sqrt{8} = 2\sqrt{2} \) i.e., product of rational and irrational numbers = irrational number

Question. Shreya picked up ‘BONUS’ and her question was. Which of the following is not irrational?
(a) \( 3 - 4\sqrt{5} \)
(b) \( \sqrt{7} - 6 \)
(c) \( 2\sqrt{2} + \sqrt{9} \)
(d) \( 4\sqrt{11} - 6 \)
Answer: (c) \( 2\sqrt{2} + \sqrt{9} \)
Explanation: Here, \( \sqrt{9} = 3 \). So, \( 2\sqrt{2} + 9 = 2\sqrt{2} + 6 + 8 \), which is not irrational. (Note: Possible typo in original source explanation, as \( 2\sqrt{2} + 3 \) remains irrational unless added to something else. Based on source Ans (c).)

Question. Ananya picked up \( \sqrt{5} - \sqrt{10} \) and her question was \( \sqrt{5} - \sqrt{10} \) is ______ number.
(a) a natural (b) an irrational (c) a whole (d) a rational
Answer: (b) an irrational
Explanation: Here \( \sqrt{5} \) and \( \sqrt{10} \) are both irrational and difference of two irrational numbers is also irrational.

Question. Suman picked up \( \frac{1}{\sqrt{5}} \) and her question was: \( \frac{1}{\sqrt{5}} \) is ________ number.
(a) a whole (b) a rational (c) an irrational (d) a natural
Answer: (c) an irrational
Explanation: As \( \sqrt{5} \) is irrational, so its reciprocal is also irrational.

Question. Preethi picked up \( \sqrt{6} \) and her question was. Which of the following is not irrational?
(a) \( 15 + 3\sqrt{6} \)
(b) \( \sqrt{24} - 9 \)
(c) \( 5\sqrt{150} \)
(d) None of these
Answer: (d) None of these
Explanation: We know that \( \sqrt{6} \) is irrational. So, \( 15 + 3\sqrt{6} \) is irrational. Similarly, \( \sqrt{24} - 9 = 2\sqrt{6} - 9 \) is irrational and \( 5\sqrt{150} = 5 \times 5\sqrt{6} = 25\sqrt{6} \) is irrational.

Very Short Answer

Question. The decimal expansion of the rational number \( \frac{43}{2^4 \cdot 5^3} \) will terminate after how many places of decimals?
Answer: \( \frac{43}{2^4 \cdot 5^3} = \frac{43 \times 5^1}{2^4 \cdot 5^3 \cdot 5^1} = \frac{43 \times 5}{2^4 \cdot 5^4} = \frac{215}{10^4} \)
After 4 decimal places.

Question. Write the decimal form of \( \frac{129}{2^2 \cdot 5^7 \cdot 7^5} \).
Answer: Non-terminating repeating.

Question. Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.
Answer: Algorithm
\( 398 - 7 = 391 \), \( 436 - 11 = 425 \), \( 542 - 15 = 527 \)
HCF of 391, 425, 527 = 17

Question. Express 98 as a product of its primes.
Answer: \( 2 \times 7^2 \)

Question. If the HCF of 408 and 1032 is expressible in the form \( 1032 \times 2 + 408 \times p \), then find the value of \( p \).
Answer: HCF of 408 and 1032 is 24.
\( \therefore 1032 \times 2 + 408 \times (p) = 24 \)
\( 408p = 24 - 2064 \)
\( p = \frac{-2040}{408} = -5 \)

Short Answer-I 

Question. HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number.
Answer: We know,
\( 1^{st} \text{ number} \times 2^{nd} \text{ number} = HCF \times LCM \)
\( \Rightarrow 27 \times 2^{nd} \text{ number} = 9 \times 459 \)
\( \therefore 2^{nd} \text{ number} = \frac{9 \times 459}{27} = 153 \)

Question. Find HCF and LCM of 13 and 17 by prime factorisation method.
Answer: \( 13 = 1 \times 13 \); \( 17 = 1 \times 17 \)
\( \therefore HCF = 1 \) and \( LCM = 13 \times 17 = 221 \)

Question. Find LCM of numbers whose prime factorisation are expressible as \( 3 \times 5^2 \) and \( 3^2 \times 7^2 \).
Answer: \( LCM(3 \times 5^2, 3^2 \times 7^2) = 3^2 \times 5^2 \times 7^2 \)
\( = 9 \times 25 \times 49 = 11025 \)

Question. Find the LCM of 96 and 360 by using fundamental theorem of arithmetic.
Answer: \( 96 = 2^5 \times 3 \)
\( 360 = 2^3 \times 3^2 \times 5 \)
\( \therefore LCM = 2^5 \times 3^2 \times 5 = 32 \times 9 \times 5 = 1440 \)

Question. Find the HCF (865, 255) using Euclid’s division lemma.
Answer: \( 865 > 255 \)
\( 865 = 255 \times 3 + 100 \)
\( 255 = 100 \times 2 + 55 \)
\( 100 = 55 \times 1 + 45 \)
\( 55 = 45 \times 1 + 10 \)
\( 45 = 10 \times 4 + 5 \)
\( 10 = 5 \times 2 + 0 \)
The remainder is 0.
\( \therefore HCF = 5 \)

Question. Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively.
Answer: It is given that on dividing 70 by the required number, there is a remainder 5. This means that \( 70 - 5 = 65 \) is exactly divisible by the required number. Similarly, \( 125 - 8 = 117 \) is also exactly divisible by the required number.
\( 65 = 5 \times 13 \)
\( 117 = 3^2 \times 13 \)
\( \therefore HCF = 13 \)
\( \therefore \text{Required number} = 13 \)

Question. Find the prime factorisation of the denominator of rational number expressed as \( 6.\overline{12} \) in simplest form.
Answer: Let \( x = 6.1212... \) ...(i)
\( 100x = 612.1212... \) ...(ii)
Subtracting (i) from (ii), \( 99x = 606 \)
\( x = \frac{606}{99} = \frac{202}{33} \)
\( \therefore \text{Denominator} = 33 \)
Prime factorisation = \( 3 \times 11 \)

Question. Prove that \( 2 + 3\sqrt{5} \) is an irrational number.
Answer: Let us assume, to the contrary, that \( 2 + 3\sqrt{5} \) is rational.
So that we can find integers \( a \) and \( b \) (\( b \neq 0 \)).
Such that \( 2 + 3\sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are coprime.
Rearranging the above equation, we get
\( 3\sqrt{5} = \frac{a}{b} - 2 \)
\( 3\sqrt{5} = \frac{a - 2b}{b} \)
\( \sqrt{5} = \frac{a - 2b}{3b} = \frac{a}{3b} - \frac{2b}{3b} \)
\( \sqrt{5} = \frac{a}{3b} - \frac{2}{3} \)
Since \( a \) and \( b \) are integers, we get \( \frac{a}{3b} - \frac{2}{3} \) is rational and so \( \sqrt{5} \) is rational.
But this contradicts the fact that \( \sqrt{5} \) is irrational.
So, we conclude that \( 2 + 3\sqrt{5} \) is irrational.

Question. Show that \( 3\sqrt{7} \) is an irrational number.
Answer: Let us assume, to the contrary, that \( 3\sqrt{7} \) is rational.
That is, we can find coprime \( a \) and \( b \) (\( b \neq 0 \)) such that \( 3\sqrt{7} = \frac{a}{b} \)
Rearranging, we get \( \sqrt{7} = \frac{a}{3b} \).
Since 3, \( a \) and \( b \) are integers, \( \frac{a}{3b} \) is rational, and so \( \sqrt{7} \) is rational.
But this contradicts the fact that \( \sqrt{7} \) is irrational.
So, we conclude that \( 3\sqrt{7} \) is irrational.

Question. Explain why \( (17 \times 5 \times 11 \times 3 \times 2 + 2 \times 11) \) is a composite number?
Answer: \( 17 \times 5 \times 11 \times 3 \times 2 + 2 \times 11 \) ...(i)
\( = 2 \times 11 \times (17 \times 5 \times 3 + 1) \)
\( = 2 \times 11 \times (255 + 1) = 2 \times 11 \times 256 \)
Number (i) is divisible by 2, 11 and 256, it has more than 2 prime factors.
Therefore \( (17 \times 5 \times 11 \times 3 \times 2 + 2 \times 11) \) is a composite number.

Question. Check whether \( 4^n \) can end with the digit 0 for any natural number n.
Answer: \( 4^n = (2^2)^n = 2^{2n} \)
The only prime in the factorization of \( 4^n \) is 2.
There is no other prime in the factorization of \( 4^n = 2^{2n} \) (By uniqueness of the fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of \( 4^n \) for any n.
Therefore, \( 4^n \) does not end with the digit zero for any natural number n.

Question. Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons.
Answer: No,
LCM = Product of highest power of each factor involved in the numbers.
HCF = Product of smallest power of each common factor.
\( \therefore \) We can conclude that LCM is always a multiple of HCF, i.e., \( LCM = k \times HCF \)
We are given that, LCM = 175 and HCF = 15
\( \therefore 175 = k \times 15 \Rightarrow 11.67 = k \)
But in this case, \( LCM \neq k \times HCF \)
Therefore, two numbers cannot have LCM as 175 and HCF as 15.