CBSE Class 10 Mathematics Real Numbers Worksheet Set 03

OBJECTIVE QUESTIONS

Question. If \( X = 28 + (1 \times 2 \times 3 \times 4 \times \dots \times 16 \times 28) \) and \( Y = 17 + (1 \times 2 \times 3 \times \dots \times 17) \), then which of the following is/are true?
1. \( X \) is a composite number
2. \( Y \) is a prime number
3. \( X - Y \) is a prime number
4. \( X + Y \) is a composite number.
(a) Both (1) and (4)
(b) Both (2) and (3)
(c) Both (2) and (4)
(d) Both (1) and (2)
Answer: (a) Both (1) and (4)
We have, \( X = 28 + (1 \times 2 \times 3 \times \dots \times 16 \times 28) \)
\( X = 28 [1 + (1 \times 2 \times 3 \times \dots \times 16)] \)
Hence, \( X \) is a composite number.
Also, we have
\( Y = 17 + (1 \times 2 \times 3 \times \dots \times 17) \)
\( = 17 [1 + (1 \times 2 \times 3 \times \dots \times 16)] \)
Hence, \( Y \) is a composite number.
Now, \( X - Y = [1 + (1 \times 2 \times \dots \times 16)](28 - 17) \)
\( = [1 + (1 \times 2 \times 3 \times \dots \times 16)](11) \)
Hence, \( X - Y \) is a composite number.
and, \( X + Y = [1 + (1 \times 2 \times 3 \times \dots \times 16)](28 + 17) \)
\( = [1 + (1 \times 2 \times 3 \times \dots \times 16)] \times 45 \)
Hence, \( X + Y \) is a composite number.

 

Question. Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers, is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (a) 2
Let the numbers be \( 12x \) and \( 12y \) where \( x \) and \( y \) are co-primes.
Product of these numbers = \( 144xy \)
Hence, \( 144xy = 6336 \Rightarrow xy = 44 \)
Since, 44 can be written as the product of two factors in three ways. i.e. \( 1 \times 44 \), \( 2 \times 22 \), \( 4 \times 11 \). As \( x \) and \( y \) are coprime, so \( (x, y) \), can be \( (1, 44) \) or \( (4, 11) \) but not \( (2, 22) \).
Hence, two possible pairs exist.

 

Question. The value of \( (12)^{3x} + (18)^{3x} \), \( x \in N \), end with the digit.
(a) 2
(b) 8
(c) 0
(d) Cannot be determined
Answer: (c) 0
For all \( x \in N \), \( (12)^{3x} \) ends with either 8 or 2 and \( (18)^{3x} \) ends with either 2 or 8.
If \( (12)^{3x} \) ends with 8, then \( (18)^{3x} \) ends with 2.
If \( (12)^{3x} \) ends with 2, then \( (18)^{3x} \) ends with 8.
Thus, \( (12)^{3x} + (18)^{3x} \) ends with 0 only.

 

Question. If \( n \) is an even natural number, then the largest natural number by which \( n(n + 1)(n + 2) \) is divisible, is
(a) 6
(b) 8
(c) 12
(d) 24
Answer: (d) 24
Since \( n \) is divisible by 2 therefore \( (n + 2) \) is divisible by 4, and hence \( n(n + 2) \) is divisible by 8.
Also, \( n, n+1, n+2 \) are three consecutive numbers.
So, one of them is divisible by 3.
Hence, \( n(n + 1)(n + 2) \) must be divisible by 24.

 

Question. If \( p_1 \) and \( p_2 \) are two odd prime numbers such that \( p_1 > p_2 \), then \( p_1^2 - p_2^2 \) is
(a) an even number
(b) an odd number
(c) an odd prime number
(d) a prime number
Answer: (a) an even number
\( p_1^2 - p_2^2 \) is an even number.
Let us take \( p_1 = 5 \) and \( p_2 = 3 \)
Then, \( p_1^2 - p_2^2 = 25 - 9 = 16 \)
16 is an even number.

 

Question. The rational form of \( 0.25\overline{4} \) is in the form of \( \frac{p}{q} \) then \( (p + q) \) is
(a) 14
(b) 55
(c) 69
(d) 79
Answer: (c) 69
Let, \( x = 0.2\overline{54} \)
\( x = 0.2545454 \dots \dots \dots \) ...(1)
Multiplying Eq. (1) by 100, we get
\( 100x = 25.4545 \dots \dots \dots \) ...(2)
Subtracting Eq. (1) from Eq. (2), we get
\( 99x = 25.2 \Rightarrow x = \frac{252}{990} = \frac{14}{55} \)
Comparing with \( \frac{p}{q} \), we get
\( p = 14 \) and \( q = 55 \)
Hence, \( p + q = 14 + 55 = 69 \)

 

Question. If \( a = 2^3 \times 3 \), \( b = 2 \times 3 \times 5 \), \( c = 3^n \times 5 \) and \( \text{LCM} (a, b, c) = 2^3 \times 3^2 \times 5 \), then \( n = \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
Value of \( n = 2 \)

 

Question. There sets of Mathematics, Science and Biology books have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. The number of Mathematics books is 240, the number of Science books is 960 and the number of Biology books is 1024. The number of stack of Mathematics, Science and Biology books, assuming that the books are of the same thickness are respectively.
(a) 15, 60, 64
(b) 60, 15, 64
(c) 64, 15, 60
(d) None of the options
Answer: (a) 15, 60, 64
The prime factorisation of 240, 960 and 1024 is given below:
\( 240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 2^4 \times 3 \times 5 \)
\( 960 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 2^6 \times 3 \times 5 \)
and \( 1024 = 2^{10} \)
\( \text{HCF of } 240, 960 \text{ and } 1024 = 2^4 = 16 \)
Hence, there must be 16 books in each stack.
Now, number of stacks of Mathematics books \( = \frac{240}{16} = 15 \)
Number of stacks of Science books \( = \frac{960}{16} = 60 \)
and, number of stacks of Biology books \( = \frac{1024}{16} = 64 \)

 

Question. If \( a + bp^{1/3} + cp^{2/3} = 0 \), where \( a, b, c, p \) are rational numbers and \( p \) is not perfect cube, then
(a) \( a \neq b = c \)
(b) \( a = b \neq c \)
(c) \( a \neq b \neq c \)
(d) \( a = b = c = 0 \)
Answer: (d) a = b = c = 0
We have, \( a + bp^{1/3} + cp^{2/3} = 0 \) ...(1)
On multiplying both sides by \( p^{1/3} \), we get
\( ap^{1/3} + bp^{1/3} \times p^{1/3} + cp^{2/3} \times p^{1/3} = 0 \)
\( ap^{1/3} + bp^{2/3} + cp = 0 \)
\( ap^{1/3} + bp^{2/3} + cp = 0 \) ...(2)
On multiplying Eq. (1) by \( b \) and Eq. (2) by \( c \), then subtracting Eq. (2) from Eq. (1), we get
\( (ab + b^2 p^{1/3} + bcp^{2/3}) - (acp^{1/3} + bcp^{2/3} + c^2 p) = 0 \)
\( (b^2 - ac)p^{1/3} + ab - c^2 p = 0 \)
[Here, \( p^{1/3} \) is an irrational number and \( ab - c^2 p \) is a rational number]
Note that, sum of rational and irrational numbers cannot be zero.
So, \( (b^2 - ac)p^{1/3} + ab - c^2 p = 0 \), only if \( b^2 - ac = 0 \) and \( ab - c^2 p = 0 \)
\( b^2 = ac \) ...(3)
and \( ab = c^2 p \)
\( a^2 b^2 = c^4 p^2 \) ...(4) [squaring both sides]
From Eq. (4), \( a^2(ac) = c^4 p^2 \) [since, \( b^2 = ac \)]
\( a^3 c - c^4 p^2 = 0 \Rightarrow c(a^3 - c^3 p^2) = 0 \)
\( c = 0 \) or \( a^3 - c^3 p^2 = 0 \Rightarrow c = 0 \) or \( p^2 = \frac{a^3}{c^3} \)
On taking, \( p^2 = \frac{a^3}{c^3} \)
We get, \( (p^2)^{1/3} = \frac{a}{c} \) which is not possible as \( p^{2/3} \) is an irrational number and \( \frac{a}{c} \) is a rational number.
Hence, \( c = 0 \)
On putting \( c = 0 \) in Eq. (3), we get \( b = 0 \)
On putting \( b = 0 \), \( c = 0 \) in Eq. (1), we get \( a = 0 \)
Hence, \( a = b = c = 0 \)

 

Question. The rational number of the form \( \frac{p}{q}, q \neq 0, p \) and \( q \) are positive integers, which represents \( 0.13\overline{4} \) i.e., \( (0.1343434 \dots \dots \dots) \) is
(a) \( \frac{134}{999} \)
(b) \( \frac{134}{990} \)
(c) \( \frac{133}{999} \)
(d) \( \frac{133}{990} \)
Answer: (d) \frac{133}{990}
\( 0.1\overline{34} = \frac{134 - 1}{990} = \frac{133}{990} \)

 

Question. If \( x \) and \( y \) are odd positive integers, then \( x^2 + y^2 \) is
(a) even and divisible by 4
(b) even and not divisible by 4
(c) odd and divisible by 4
(d) odd and not divisible by 4
Answer: (b) even and not divisible by 4
We know that, any odd positive integer is of the form \( 2q + 1 \), where \( q \) is any integer. So, \( x = 2m + 1 \) and \( y = 2n + 1 \) for some integers \( m \) and \( n \).
Now, \( x^2 + y^2 = (2m + 1)^2 + (2n + 1)^2 \)
\( = 4m^2 + 1 + 4m + 4n^2 + 1 + 4n \)
\( = 4(m^2 + n^2) + 4(m + n) + 2 \)
\( = 4[(m^2 + n^2) + (m + n)] + 2 \)
\( = 4r + 2 \),
Where, \( r = m^2 + n^2 + m + n \)
Clearly, \( 4r + 2 \) is an even number and not divisible by 4.
Hence, \( x^2 + y^2 \) is even but not divisible by 4.

 

Question. The least number which is a perfect square and is divisible by each of 16, 20 and 24 is
(a) 240
(b) 1600
(c) 2400
(d) 3600
Answer: (d) 3600
The L.C.M. of 16, 20 and 24 is 240. The least multiple of 240 that is a perfect square is 3600 and also we can easily eliminate choices (a) and (c) since they are not perfect number.

 

Question. Which of the following rational number have non-terminating repeating decimal expansion?
(a) \( \frac{31}{3125} \)
(b) \( \frac{71}{512} \)
(c) \( \frac{23}{200} \)
(d) None of the options
Answer: (d) None of the options
3125, 512 and 200 has factorization of the form \( 2^m \times 5^n \) (where \( m \) and \( n \) are whole numbers). So given fractions has terminating decimal expansion.

 

Question. When \( 2^{256} \) is divided by 17 the remanider would be
(a) 1
(b) 16
(c) 14
(d) None of the options
Answer: (a) 1
When \( 2^{256} \) is divided by 17 then,
\( \frac{2^{256}}{2^4 + 1} = \frac{(2^4)^{64}}{2^4 + 1} \)
By remainder theorem when \( f(x) \) is divided by \( x + a \) the remainder \( = f(-a) \)
Here, \( f(x) = (x)^{64} \) and \( x = 2^4 \) and \( a = 1 \)
Hence, \( \text{Remainder} = f(-1) = (-1)^{64} = 1 \)

 

Question. The least number which when divided by 15, leaves a remainder of 5, when divided by 25, leaves a remainder of 15 and when divided by 35 leaves a remainder of 25, is
(a) 515
(b) 525
(c) 1040
(d) 1050
Answer: (a) 515
The number is short by 10 for complete division by 15, 25 or 35.

 

Question. Without Actually performing the long division, the terminating decimal expansion of \( \frac{51}{1500} \) is in the form of \( \frac{17}{2^n \times 5^m} \), then \( (m + n) \) is equal to
(a) 2
(b) 3
(c) 5
(d) 8
Answer: (c) 5
We have, \( \frac{51}{1500} = \frac{17}{500} \)
Prime factorization of 500
2 | 500
2 | 250
5 | 125
5 | 25
5 | 5
| 1
\( = 2 \times 2 \times 5 \times 5 \times 5 = 2^2 \times 5^3 \)
which is in the form \( 2^n \times 5^m \)
So, it has a terminating decimal expansion.
Now, \( \frac{51}{1500} = \frac{17}{2^2 \times 5^3} \)
By comparing, we get \( n = 2 \) and \( m = 3 \)
\( m + n = 2 + 3 = 5 \)

 

Question. The sum of three non-zero prime numbers is 100. One of them exceeds the other by 36. Then the largest number is
(a) 73
(b) 91
(c) 67
(d) 57
Answer: (c) 67
Let \( X, X + 36 \) and \( y \) are the three prime numbers.
According to the given condition,
\( x + x + 36 + y = 100 \)
\( 2x + y = 64 \) ...(1)
Since \( X \) is a prime number, then \( 2x \) will be an even number
Also addition of two even numbers results in an even number,
Hence, from equation (1), we can conclude that \( y \) must be an even prime number.
\( y = 2 \) as 2 is the only even prime number.
Put \( y = 2 \) in equation (1), we get
\( 2x + 2 = 64 \)
\( x = 31 \)
So the required prime numbers are 31, \( 31+36 \), 2 or 31, 67 and 2
Hence, Largest number is 67.

 

Question. The values of \( x \) and \( y \) is the given figure are
(a) 7, 13
(b) 13, 7
(c) 9, 12
(d) 12, 9
Answer: (a) 7, 13
Given number is 1001. Then, the factor tree of 1001 is given as below
\( 1001 = 7 \times 11 \times 13 \)
By comparing with given factor tree, we get \( x = 7, y = 13 \)

 

Question. If \( P = (2)(4)(6)...(20) \) and \( Q = (1)(3)(5)...(19) \), then the HCF of \( P \) and \( Q \) is
(a) \( (3^3)(5)(7) \)
(b) \( (3^4)(5) \)
(c) \( (3^4)(5^2)(7) \)
(d) \( (3^3)(5^2) \)
Answer: (c) (3^4)(5^2)(7)
In \( P \), the primes that occurs are 2, 3, 5, 7.
In \( Q \), there is no 2.
So, the HCF of \( P, Q \), has powers of only 3, 5 and 7.
In \( P \), 3 comes from 6, 12, 18. So, \( 3^4 \) is the greatest power of 3 in prime factorisation of \( P \). While in \( Q \), 3 comes from 3, 9, 15.
So, \( 3^4 \) is also in prime factorisation of \( Q \).
Similarly \( 5^2 \) is the greatest power of 5 for both \( P \) and \( Q \) and \( 7^1 \) is the greatest power of 7 for both \( P \) and \( Q \).
Hence, HCF of \( P \) and \( Q \) is \( (3^4)(5^2)(7) \).

 

Question. The number \( 3^{13} - 3^{10} \) is divisible by
(a) 2 and 3
(b) 3 and 10
(c) 2, 3 and 10
(d) 2, 3 and 13
Answer: (d) 2, 3 and 13
\( 3^{13} - 3^{10} = 3^{10}(3^3 - 1) = 3^{10}(26) \)
\( = 2 \times 13 \times 3^{10} \)
Hence, \( 3^{13} - 3^{10} \) is divisible by 2, 3 and 13.

 

Question. Which of the following will have a terminating decimal expansion?
(a) \( \frac{77}{210} \)
(b) \( \frac{23}{30} \)
(c) \( \frac{125}{441} \)
(d) \( \frac{23}{8} \)
Answer: (d) \frac{23}{8}
For terminating decimal expansion, denominator must have only 2 or only 5 or 2 and 5 as factor.
Here, \( \frac{23}{8} = \frac{23}{(2)^3} \)
(only 2 as factor of denominator so terminating)

 

Question. For any natural number \( n, 9^n \) cannot end with the digit.
(a) 1
(b) 2
(c) 9
(d) None of the options
Answer: (b) 2
For \( n = 1, 9^n = 9^1 = 9 \), so \( 9^n \) can end with digit 9.
For \( n = 2, 9^n = 9^2 = 81 \), so \( 9^n \) can also end with digit 1.
Now, let if possible \( 9^n \) ends with 2 for some natural numbers \( n \).
Then, \( 9^n \) is divisible by 2.
But prime factors of 9 are \( 3 \times 3 \).
\( 9^n = (3 \times 3)^n = 3^{2n} \)
Thus, prime factorisation of \( 9^n \) does not contain 2 as a factor.
By the fundamental theorem of arithmetic, there are no other primes in factorisation of \( 9^n \).
Therefore, \( 9^n \) is not divisible by 2. So, our assumption is wrong.
Hence, there is no natural number \( n \) for which \( 9^n \) ends in the digit 2.

 

Question. A number lies between 300 and 400. If the number is added to the number formed by reversing the digits, the sum is 888 and if the unit’s digit and the ten’s digit change places, the new number exceeds the original number by 9. Then the number is
(a) 339
(b) 341
(c) 378
(d) 345
Answer: (d) 345
Sum is 888 \( \Rightarrow \) unit’s digit should add up to 8. This is possible only for option (d) as “3” + “5” = “8”.

 

Question. 1. The L.C.M. of \( x \) and 18 is 36. 2. The H.C.F. of \( x \) and 18 is 2. What is the number \( x \)?
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4
L.C.M. \( \times \) H.C.F. = First number \( \times \) second number
Hence, required number \( = \frac{36 \times 2}{18} = 4 \)

 

Question. A circular field has a circumference of 360 km. Two cyclists Sumeet and John start together and can cycle at speeds of 12 km/h and 15 km/h respectively, round the circular field. They will meet again at the starting point after
(a) 40 h
(b) 30 h
(c) 180 h
(d) 120 h
Answer: (d) 120 h
Given, Total distance = 360 km
and, Speed of Sumeet = 12 km/h
Number of hours taken by Sumeet to complete 1 round.
\( = \frac{\text{Distance}}{\text{Speed}} = \frac{360}{12} = 30 \text{ h} \)
and, Speed of John = 15 km/h
Number of hours taken by John to complete 1 round.
\( = \frac{\text{Distance}}{\text{Speed}} = \frac{360}{15} = 24 \text{ h} \)
Thus, Sumeet and John complete 1 round in 30 h and 24 h, respectively.
Now, to find required hours, we find the LCM of 30 and 24.
\( 30 = 2 \times 3 \times 5 \)
\( 24 = 2 \times 2 \times 2 \times 3 \)
Then, LCM (30, 24) \( = 2 \times 2 \times 2 \times 3 \times 5 = 120 \)
Hence, Sumeet and John will meet each other again after 120 h.

 

Question. If \( n \) is an even natural number, then the largest natural number by which \( n(n + 1)(n + 2) \) is divisible is
(a) 6
(b) 8
(c) 12
(d) 24
Answer: (d) 24
Out of \( n \) and \( n + 2 \), one is divisible by 2 and the other by 4, hence \( n(n + 2) \) is divisible by 8. Also \( n, n + 1, n + 2 \) are three consecutive numbers, hence one of them is divisible by 3. Hence, \( n(n + 1)(n + 2) \) must be divisible by 24. This will be true for any even number \( n \).

 

Question. The remainder on dividing given integers \( a \) and \( b \) by 7 are respectively 5 and 4. Then the remainder when \( ab \) is divided by 7 is
(a) 5
(b) 4
(c) 0
(d) 6
Answer: (d) 6
By using Euclid’s division lemma we get
\( a = 7p + 5 \)
and \( b = 7q + 4 \)
where \( p \) and \( q \) are integers
Hence, \( ab = (7p + 5)(7q + 4) \)
\( = 49pq + 7(4p + 5q) + 20 \)
\( = 7(7pq + 4p + 5q) + 7 \times 2 + 6 \)
\( = 7(7pq + 4p + 5q + 2) + 6 \)
Hence, when \( ab \) is divided by 7, we get the remainder 6.

 

TRUE/FALSE

Question. Given positive integers \( a \) and \( b \), there exist whole numbers \( q \) and \( r \) satisfying \( a = bq + r, 0 \leq r < b \).
Answer: True

Question. HCF of two numbers is always a factor of their LCM.
Answer: True

Question. Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Answer: True

Question. Sum of two prime numbers is always a prime number.
Answer: False

Question. The number zero is irrational.
Answer: False

Question. \( \sqrt{2} \) and \( \sqrt{3} \) are irrationals numbers.
Answer: True

Question. \( \pi \) is an irrational number.
Answer: True

Question. Some irrational numbers are negative.
Answer: True

Question. If \( x = p/q \) be a rational number, such that the prime factorisation of \( q \) is not of the form \( 2^n 5^m \), where \( n, m \) are non-negative integers. Then \( x \) has a decimal expansion which is terminates.
Answer: False

Question. All real numbers are rational numbers.
Answer: False

Question. Any positive odd integer is of the form \( 6q + 1 \), or \( 6q + 3 \), or \( 6q + 5 \), where \( q \) is some integer.
Answer: True

Question. Sum of two irrational numbers is an irrational number.
Answer: False

Question. The quotient of two integers is always a rational number
Answer: False

Question. Two numbers can have 12 as their LCM and 350 as their HCF.
Answer: False

Question. 1/0 is not rational.
Answer: True

Question. The product of any three consecutive natural numbers is divisible by 6.
Answer: True

Question. If \( x = p/q \) be a rational number, such that the prime factorisation of \( q \) is of the form \( 2^n 5^m \), where \( n, m \) are non-negative integers. Then \( x \) has a decimal expansion which terminates.
Answer: True

Question. All integers are real numbers.
Answer: True

Question. The number of irrational numbers between 15 and 18 is infinite.
Answer: True

Question. Every fraction is a rational number.
Answer: True

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