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Page No 11:
Question 1:
The exponent of 2 in the prime factorisation of 144, is
(a) 4
(b) 5
(c) 6
(d) 3
Answer:
Using the factor tree for prime factorization, we have:
Therefore,
Thus, the exponent of 2 in 144 is 4.
Hence the correct choice is (a).
Page No 11:
Question 2:
The LCM of two numbers is 1200. Which of the following cannot be their HCF?
(a) 600
(b) 500
(c) 400
(d) 200
Answer:
It is given that the LCM of two numbers is 1200.
We know that the HCF of two numbers is always the factor of LCM
Checking all the options:
(a) 600 is the factor of 1200.
So this can be the HCF.
(b) 500 is not the factor of 1200.
So this cannot be the HCF.
(c) 400 is the factor of 1200.
So this can be the HCF.
(d) 200 is the factor of 1200.
So this can be the HCF.
Hence the correct choice is (b).
Page No 11:
Question 3:
If n = 23 ✕ 34 ✕ 54 ✕ 7, then the number of consecutive zeros in n, where n is a natural number, is
(a) 2
(b) 3
(c) 4
(d) 7
Answer:
Since, it is given that
n =
So, this means the given number n will end with 3 consecutive zeroes.
Page No 11:
Question 4:
The sum of the exponents of the prime factors in the prime factorisation of 196, is
(a) 1
(b) 2
(c) 4
(d) 6
Answer:
Using the factor tree for prime factorization, we have:
Therefore,
The exponents of 2 and 7 are 2 and 2 respectively.
Thus the sum of the exponents is
Hence the correct choice is (c).
Page No 11:
Question 5:
The number of decimal place after which the decimal expansion of the rational number will terminate, is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
We have,
Theorem states:
Let be a rational number, such that the prime factorization of q is of the form, where m and n are non-negative integers.
Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.
This is given that the prime factorization of the denominator is of the form.
Hence, it has terminating decimal expansion which terminates after places of decimal.
Hence, the correct choice is (b).
Page No 11:
Question 6:
If two positive ingeters a and b are expressible in the form a = pq2 and b = p3q; p, q being prime number, then LCM (a, b) is
(a) pq
(b) p3q3
(c) p3q2
(d) p2q2
Answer:
Two positive integers are expressed as follows:
p and q are prime numbers.
Then, taking the highest powers of p and q in the values for a and b we get:
LCM
Hence the correct choice is (c).
Page No 11:
Question 7:
In Q.No. 7, HCF (a, b) is
(a) pq
(b) p3q3
(c) p3q2
(d) p2q2
Answer:
Two positive integers are expressed as follows:
p and q are prime numbers.
Then, taking the smallest powers of p and q in the values for a and b we get
HCF
Hence the correct choice is (a).
Page No 11:
Question 8:
If two positive integers m and n are expressible in the form m = pq3 and n = p3q2, where p, q are prime numbers, then HCF (m, n) =
(a) pq
(b) pq2
(c) p3q2
(d) p2q2
Answer:
Two positive integers are expressed as follows:
p and q are prime numbers.
Then, taking the smallest powers of p and q in the values for m and n we get
HCF
Hence the correct choice is (b).
Page No 11:
Question 9:
The HCF of 95 and 152, is
(a) 57
(b) 1
(c) 19
(d) 38
Answer:
Using the factor tree for 95, we have:
Using the factor tree for 152, we have:
Therefore,
HCF
Hence the correct choice is (c).
Page No 11:
Question 10:
If HCF (26, 169) = 13, then LCM (26, 169) =
(a) 26
(b) 52
(c) 338
(d) 13
Answer:
HCF
We have to find the value for LCM
We know that the product of numbers is equal to the product of their HCF and LCM.
Therefore,
Hence the correct choice is (c).
Page No 11:
Question 11:
Which of the following rational numbers have terminating decimal?
(i)
(ii)
(iii)
(iv)
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i) and (iv)
Answer:
(i) We have,
Theorem states:
Let be a rational number, such that the prime factorization of q is not of the form, where m and n are non-negative integers.
Then, x has a decimal expression which does not have terminating decimal.
(ii) We have,
Theorem states:
Let be a rational number, such that the prime factorization of q is not of the form, where m and n are non-negative integers.
Then, x has a decimal expression which does not have terminating decimal.
(iii) We have,
Theorem states:
Let be a rational number, such that the prime factorization of q is not of the form, where m and n are non-negative integers.
Then, x has a decimal expression which does not have terminating decimal.
(iv) We have,
Theorem states:
Let be a rational number, such that the prime factorization of q is of the form, where m and n are non-negative integers.
Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.
Then, x has a decimal expression which will have terminating decimal after 3 places of decimal.
Hence the (iv) option will have terminating decimal expansion.
There is no correct option.
Page No 11:
Question 12:
The decimal expansion of the rational number will terminate after
(a) one decimal place
(b) five decimal places
(c) three decimal places
(d) two decimal places
Answer:
We have, .
Theorem states:
Let be a rational number such that the prime factorization of q is of the form , where m and n are non-negative integers.
Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. This is given that the prime factorization of the denominator is of the form .
Thus, it has terminating decimal expansion which terminates after 5 places of decimal.
Hence, the correct answer is option (b).
Page No 11:
Question 13:
The decimal expansion of will be
(a) terminating
(b) non-terminating
(c) non-terminating and repeating
(d) non-terminating and non-repeating
Answer:
Since the denominator is of the form , the decimal expansion of the given number is terminating.
=
= 0.155
So, the decimal expansion of is terminating.
Hence, the correct answer is option (a).
Page No 11:
Question 14:
The LCM of 23 × 32 and 22 × 33 is
(a) 23
(b) 33
(c) 23 × 33
(d) 22 × 32
Answer:
We have,
LCM of 23 × 32 and 22 × 33 = 23 × 33
Hence, the correct answer is option (c).
Page No 12:
Question 15:
The HCF of two numbers is 18 and their product is 12960. Their LCM will be
(a) 420
(b) 600
(c) 720
(d) 800
Answer:
Given that,
HCF of two numbers = 18
Product of numbers = 12960
Now,
Product of two numbers = HCF × LCM
⇒ 12960 = 18 × LCM
Hence, the correct answer is option (c).
Page No 12:
Question 16:
The decimal expansion of will terminate after how many places?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
We have,
The power of 2 in the denominator gives the number of decimal places after which the expansion terminates.
Thus, the places of decimal in the decimal expansion is 3.
Hence, the correct answer is option (c).
Page No 12:
Question 17:
The total number of factors of a prime numbers is
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
We know that the factors of any prime number are 1 and the prime number itself.
Hence, the correct answer is option (c).
Page No 12:
Question 18:
The HCF and LCM of 12, 21, 15 respectively are
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Answer:
Here,
Therefore, HCF(12, 21, 15) = 3 and
LCM(12, 21, 15) =
Hence, the correct answer is option C.
Page No 12:
Question 19:
The product of a non-zero rational number and an irrational number is
(a) always rational
(b) always irrational
(c) rational or irrational
(d) none of these
Answer:
The product of a non-zero rational number and an irrational number is always an irrational number.
Hence, the correct option is (a).
Page No 12:
Question 20:
If two positive integers a and b are written as and are prime numbers , then HCF(a,b) is
(a) xy (b) xy2 (c) x3y3 (d) x2y2
Answer:
It is given that,
Hence, the correct answer is option B.
Page No 12:
Question 21:
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(a) 10
(b) 100
(c) 504
(d) 2520
Answer:
To find the least number that is divisible by all the numbers between 1 and 10 (both inclusive), find the L.C.M of all the numbers between 1 and 10.
LCM = Product of the highest power of the prime factors
= 23 × 32 × 5 × 7
= 2520
Hence, the correct answer is option (d).
Page No 12:
Question 22:
The largest number which divides 70 and 125, leaving remainders 5 and 8 , respectively,is
(a) 13 (b) 65 (c) 875 (d) 1750
Answer:
It is given that on dividing 70 by the required number, there is a remainder of 5. This means that 70 − 5 = 65 is exactly divisible by the required number.
Similarly, 125 − 8 = 117 is exactly divisible by the required number.
The required number is HCF of 65 and 117.
By Euclid's division algorithm,
Here, the remainder is zero. Therefore, the HCF of 65 and 117 is 13.
So, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively.
Hence, the correct answer is option A.
Page No 12:
Question 23:
If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n =
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
LCM ……(I)
We have to find the value for n
Also
We know that the while evaluating LCM, we take greater exponent of the prime numbers in the factorization of the number.
Therefore, by applying this rule and taking we get the LCM as
LCM ……(II)
On comparing (I) and (II) sides, we get:
Hence the correct choice is (b).
Page No 12:
Question 24:
The decimal expansion of the rational number will terminate after
(a) one decimal place
(b) two decimal place
(c) three decimal place
(d) four decimal place
Answer:
We have,
Theorem states:
Let be a rational number, such that the prime factorization of q is of the form, where m and n are non-negative integers.
Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.
This is given that the prime factorization of the denominator is of the form.
Hence, it has terminating decimal expansion which terminates after places of decimal.
Hence, the correct choice is (d).
Page No 12:
Question 25:
is
(a) an integer
(b) a rational number
(c) a natural number
(d) an irrational number
Answer:
We have,
Let
Then,
Subtract these to get
Thus, we can also conclude that all infinite repeating decimals are rational numbers.
Hence, the correct choice is (b).
Page No 12:
Question 26:
The LCM and HCF of two rational numbers are equal, then the numbers must be
(a) prime
(b) co-prime
(c) composite
(d) equal
Answer:
Let the two numbers be a and b.
(a) If we assume that the a and b are prime.
Then,
HCF
LCM
(b) If we assume that a and b are co-prime.
Then,
HCF
LCM
(c) If we assume that a and b are composite.
Then,
HCFor any other highest common integer
LCM
(d) If we assume that a and b are equal and consider a=b=k.
Then,
HCF
LCM
Hence the correct choice is (d).
Page No 12:
Question 27:
If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is
(a) 203400
(b) 194400
(c) 198400
(d) 205400
Answer:
Let the HCF be x and the LCM of the two numbers be y.
It is given that the sum of the HCF and LCM is 1260
…… (i)
And, LCM is 900 more than HCF.
…… (ii)
Substituting (ii) in (i), we get:
Substituting in (ii), we get:
We also know that the product the two numbers is equal to the product of their LCM and HCF
Thus the product of the numbers
Hence, the correct choice is (b).
Page No 12:
Question 28:
The ratio of LCM and HCF of the least composite number and the least prime number is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 1 : 3
Answer:
We have,
Least composite number = 4
Least Prime Number = 2
Hence, the correct answer is option (b).
Page No 12:
Question 29:
Prime factors of the denominator of a rational number with decimal expansion 44.123 are
(a) 2, 3
(b) 2, 3, 5
(c) 2, 5
(d) 3, 5
Answer:
Since 44.123 has a terminating decimal expansion, the prime factors of the denominator will be 2 and 5.
Hence, the correct answer is option (c).
Page No 12:
Question 30:
If than a is
(a) rational
(b) irrational
(c) whole number
(d) integer
Answer:
Given that,
which is an irrational number.
Hence, the correct answer is option (b).
Page No 12:
Question 31:
If LCM (x, 18) = 36 and HCF (x, 18) = 2, then x is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
We have,
Thus, the value of x is 4.
Hence, the correct answer is option (c).
Page No 13:
Question 32:
If the sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such number is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
Given that, HCF of two numbers = 81
Let the two numbers are 81x and 81y.
Thus, the four co-prime numbers are (1, 14), (2, 13), (4, 11), (7, 8).
So,
For x = 1, y = 14, the numbers are 1 × 81 + 14 × 81 = 81 + 1134 = 1215
For x = 7, y = 8, the numbers are 7 × 81 + 8 × 81 = 567 + 648 = 1215
For x = 2, y = 13, the numbers are 2 × 81 + 13 × 81 = 162 + 1053 = 1215
For x = 4, y = 11, the numbers are 4 × 81 + 11 × 81 = 324 + 891 = 1215
Therefore, the numbers of such pairs are 4.
Hence, the correct answer is option (c).
Page No 13:
Question 33:
If the LCM of two prime numbers p and q(p > q) is 221 then the value of 3p – q is
(a) 4
(b) 28
(c) 38
(d) 48
Answer:
We have,
LCM of two prime numbers = Product of two numbers
= 13 × 17
Therefore, p = 17 and q = 13.
Hence, the correct answer is option (c).
Page No 13:
Question 34:
The smallest number by which should be multiplied so that its decimal expansion terminates after two decimal places
(a)
(b)
(c)
(d)
Answer:
For terminating decimal expansion after one decimal place, the highest power of m and n in should be 1.
Let .
We will get:
Thus, it is evident that we multiplied it by so that its decimal expansion terminates after two decimal places.
Hence, the correct answer is option (a).
Page No 13:
Question 35:
Any one of numbers a, a + 2 and a + 4 is a multiple of
(a) 2
(b) 3
(c) 5
(d) 7
Answer:
Consider three numbers a, a + 2 and a + 4.
Case 1: If , then a, a + 2 and a + 4 are even.
Let a = 2q.
So, a + 2 = 2q + 2 = 2(q + 1) and
a + 4 = 2q + 4 = 2(q + 2)
Out of three consecutive numbers q, q + 1 and q + 2, one of them will always be divisible by 3. Therefore, one out of 2q, 2q + 2 and 2q + 4, one of them will always be divisible by 3.
For a = 2, the three numbers are 2, 4, 6 and only 6 is divisible by 3.
For a = 4, the three numbers are 4, 6, 8 and only 6 is divisible by 3.
For a = 6, the three numbers are 6, 8, 10 and only 6 is divisible by 3 and so on.
Case 2: If , then a, a + 2 and a + 4 are odd.
Then, out of these consecutive odd numbers, any one of them will always be divisible by 3 as every third natural number is a multiple of 3.
For a = 1, the three numbers are 1, 3, 5 and only 3 is divisible by 3.
For a = 3, the three numbers are 3, 5, 7 and only 3 is divisible by 3.
For a = 5, the three numbers are 5, 7, 9 and only 9 is divisible by 3 and so on.
Thus, in both the cases, any one of numbers a, a + 2 and a + 4 is a multiple of 3.
Hence, the correct answer is option (b).
Page No 13:
Question 36:
If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
(a) 2
(b) 3
(c) 4
(d) 1
Answer:
LCM
HCF
We know that the product of numbers is equal to the product of their HCF and LCM.
Therefore,
Hence the correct choice is (c).
Page No 13:
Question 37:
If p and q are co-prime numbers, then p2 and q2 are
(a) coprime
(b) not coprime
(c) even
(d) odd
Answer:
We know that the co-prime numbers have no factor in common, or, their HCF is 1.
Thus,and have the same factors with twice of the exponents of p and q respectively, which again will not have any common factor.
Thus we can conclude that and are co-prime numbers.
Hence, the correct choice is (a).
Page No 13:
Question 38:
If 3 is the least prime factor of number a and 7 is the least prime factor of number b, then the least prime factor of a + b, is
(a) 2
(b) 3
(c) 5
(d) 10
Answer:
Since
The least prime factor of has to be 2; unless is a prime number greater than 2.
Suppose is a prime number greater than 2. Then must be an odd number
o one of a or b must be an even number.
Suppose then that a is even. Then the least prime factor of a is 2; which is not 3 or 7. So a can not be an even number nor can b be an even number. Hence can not be a prime number greater than 2 if the least prime factor of a is 3 and b is 7.
Thus the answer is 2.
Hence the correct choice is (a).
Page No 13:
Question 39:
The smallest number by which should be multiplied so as to get a rational number is
(a)
(b)
(c)
(d) 3
Answer:
Out of the given choices is the only smallest number by which if we multiply we get a rational number.
Hence, the correct choice is (c).
Page No 13:
Question 40:
The smallest rational number by which should be multiplied so that its decimal expansion terminates after one place of decimal, is
(a)
(b)
(c) 3
(d)
Answer:
For terminating the decimal expansion after one place of decimal, the highest power of m and n in should be 1.
Let
We will get:
Thus, it is evident that we multiplied it by
Hence, the correct choice is (a).
Page No 13:
Question 41:
If n is a natural number, then 92n − 42n is always divisible by
(a) 5
(b) 13
(c) both 5 and 13
(d) None of these
[Hint : 92n − 42n is of the form a2n − b2n which is divisible by both a − b and a + b. So, 92n − 42n is divisible by both 9 − 4 = 5 and 9 + 4 = 13.]
Answer:
We know that is always divisible by both and .
So, is always divisible by both and.
Hence, the correct choice is (c).
Page No 13:
Question 42:
If n is any natural number, then 6n − 5n always ends with
(a) 1
(b) 3
(c) 5
(d) 7
[Hint: For any n ∈ N, 6n and 5n end with 6 and 5 respectively. Therefore, 6n − 5n always ends with 6 − 5 = 1.]
Answer:
We know that will end in 6
And will end in 5.
Now, always end with
Hence the correct choice in (a).
Page No 13:
Question 43:
The remainder when the square of any prime number greater than 3 is divided by 6, is
(a) 1
(b) 3
(c) 2
(d) 4
[Hint: Any prime number greater than 3 is of the from 6k ± 1, where k is a natural number and (6k ± 1)2 = 36k2 ± 12k + 1 = 6k(6k ± 2) + 1]
Answer:
Any prime number greater than 3 is of the form, where k is a natural number.
Thus,
When, is divided by 6, we get, and remainder as 1.
Hence, the correct choice is (a).
Page No 13:
Question 44:
For some integer m, every integer is of the form
(a) m (b) m + 1 (c) 2m (d) 2m+ 1
Answer:
It is known that, even integers are ...,−4, −2, 0, 2, 4,...
Observe that, all of the even numbers are the multiple of 2.
So, even numbers can be written as 2m, where, m is an integer.
m can be ...,−2, −1, 0, 1, 2,...
∴ 2m can be ...,−4, −2, 0, 2, 4,...
Hence, the correct answer is option C.
Page No 13:
Question 45:
For some integer q , every odd integer is of the form
(a) q (b) q + 1 (c) 2q (d) 2q + 1
Answer:
We know that, all numbers that are not the multiple of 2 are odd numbers.
Odd integers are ...,−3, −1, 1, 3, 5,...
So, odd numbers can be written as 2m + 1, where m is an integer.
m can be ..., −2, −1, 0, 1, 2,...
∴ 2m + 1 can be ..., −3, −1, 1, 3,...
Hence, the correct answer is option D.
Page No 13:
Question 46:
For any positive integer a and 3, there exist unique integers q and r such that a = 3q + r, where r must satisfy
(a) 0 ≤ r < 3
(b) 1 < r < 3
(c) 0 < r < 3
(d) 0 < r ≤ 3
Answer:
From Euclid's lemma, we have
a = 3q + r
Hence, the correct answer is option (a).
Page No 14:
Question 47:
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:
From Euclid's Lemma, for two positive integers a and b, there exist unique integers q and r such that a = bq + r,
Where 0 ≤ r < b.
Hence, the correct answer is option (c).
Page No 14:
Question 48:
The product of a non-zero rational number and an irrational number is
(a) always irrational
(b) always rational
(c) rational or irrational
(d) one
Answer:
The product of a non-zero rational number and an irrational number is always an irrational number.
Hence, the correct option is (a).
Page No 14:
Question 49:
For any natural numbers, 252n – 92n is always divisible by
(a) 16
(b) 34
(c) both 16 and 34
(d) none of these
Answer:
For n = 1,
Thus, 252n – 92n is always divisible by 16 and 34 both for n = 1.
For n = 2,
Thus, 252n – 92n is always divisible by 16 and 34 both for n = 2.
So, for any value of n, 252n – 92n is always divisible by 16 and 34 both.
Hence, the correct answer is option (c).
Page No 14:
Question 50:
HCF of two positive integers is always
(a) a multiple of their LCM
(b) a factor of their LCM
(c) divisible by their LCM
(d) none of these
Answer:
HCF of two numbers is always a factor of their LCM.
Hence, the correct answer is option (b).
Page No 14:
Question 51:
If the decimal expansion of the rational number terminates after k decimal places, then
(a) k = 1
(b) k = 2
(c) k = 3
(d) k > 3
Answer:
Given that,
The number of terminating decimal places is decided by the number of 2's.
Thus, the given rational number will terminate after 3 decimal places.
Hence, the correct answer is option (c).
Page No 14:
Question 52:
If (a × 5)n ends with the digit zero for every natural number n, then a is
(a) any natural number
(b) an even number
(c) an odd number
(d) none of these
Answer:
Given that, (a × 5)n ends with the digit zero for every natural number n.
For every natural number n, (a × 5)n can end with zero only if (a × 5) ends with 0. This happens only for even values of a.
Hence, the correct answer is option (b).
Page No 14:
Question 53:
All decimal numbers are
(a) rational numbers
(b) irrational numbers
(c) real numbers
(d) integers
Answer:
All decimal numbers are rational numbers.
Hence, the correct answer is option (a).
Page No 14:
Question 54:
Three bells ring at intervals of 4, 7 and 14 minutes. All the three rang at 6 AM. When will they ring together again?
(a) 6 : 07 AM
(b) 6 : 14 AM
(c) 6 : 28 AM
(d) 6 : 25 AM
Answer:
LCM of 4, 7 and 14 = 28
Thus, all the bells rang together again at 6 : 28 AM.
Hence, the correct answer is option (c).
Page No 14:
Question 55:
How many prime numbers are of form 10n + 1, where n is a natural number such that 1 ≤ n < 10?
(a) 5
(b) 6
(c) 4
(d) 3
Answer:
Given that,
Numbers of the form 10n +1, where n is a Natural number such that 1 ≤ n <10
Now, put n = 1
The number is 10(1) + 1 = 11
Again, putting n = 2
The number is 10(2) + 1 = 21
and so on till n = 9 as n < 10
The number is 10(9) + 1 = 91
Numbers are 11, 21, 31, 41, 51, 61, 71, 81, 91
Prime numbers are numbers that have exactly 2 factors, 1 and the number itself.
Out of these, the prime numbers are 11, 31, 41, 61, 71.
Thus, 5 prime numbers are of form 10n + 1, where n is a whole number such that 1 ≤ n < 10.
Hence, the correct answer is option (a).
Page No 14:
Question 56:
1192 – 112 is a
(a) prime number
(b) composite number
(c) an odd prime number
(d) an odd composite number
Answer:
We have,
1192 – 112 = (119 + 11)(119 − 11)
= 130 × 108
Since 130 and 108 both are even numbers.
Thus 1192 – 112 is a composite number.
Hence, the correct answer is option (b).
Page No 14:
Question 57:
π is
(a) a rational number
(b) an irrational number
(c) a prime number
(d) an odd number
Answer:
π is an irrational number as it has non-terminating, non-repeating decimal expansion given as follows:
π = 3.1415926535897932...
Hence, the correct answer is option (b).
Page No 14:
Question 58:
is
(a) a natural number
(b) an integer
(c) a rational number
(d) an irrational number
Answer:
The sum of irrational numbers is an irrational number. Also, the sum of a rational and an irrational number is irrational.
Thus, is an irrational number because are irrationals.
Hence, the correct answer is option (d).
Page No 14:
Question 59:
The smallest irrational number by which should be multiplied so as to get a rational number is
(a)
(b)
(c)
(d) 2
Answer:
We have,
Thus, the smallest number by which should be multiplied is .
which is a rational number.
Hence, the correct answer is option (c).
Page No 14:
Question 60:
The decimal expansion of π is
(a) terminating
(b) non-terminating non-repeating
(c) non-terminating
(d) doesn't exist
Answer:
The decimal expansion of π is 3.14159...
Since it is non-terminating and non-repeating.
Thus, π is irrational.
Hence, the correct answer is option (b).
Page No 15:
Question 61:
The decimal expansion of numbers has
(a) terminating decimal
(b) non-terminating but repeating
(c) non-terminating non-repeating
(d) terminating after two places of decimal
Answer:
We have, .
Since the denominator of this rational number is in the form of 2m5n, so it will have a terminating decimal representation.
Hence, the correct answer is option (a).
Page No 15:
Question 62:
The decimal expansion of will terminate after how many places of decimals?
(a) 1
(b) 2
(c) 3
(d) will not terminate
Answer:
We have,
Since the denominator of the rational number is of the form 23 , therefore it will terminate after 3 decimal places.
Hence, the correct answer is option (c).
Page No 15:
Question 63:
The decimal expansion of is
(a) terminating
(b) non-terminating
(c) non-terminating and repeating
(d) none of these
Answer:
We have, .
Since the denominator of this rational number is in the form of 2m × 5n, so it will have a terminating decimal representation.
Hence, the correct answer is option (a).
Page No 15:
Question 64:
Ajay wants to host a party on his 50th birthday in a large banquet hall having a certain number of chairs. He wants that guests should sit in different groups like in pairs, triplets, quadruplets, fives and sixes etc. When the banquet hall manager arranges chairs in such pattern like in 2's, 3's, 4's, 5's and 6's then 1, 2, 3, 4 and 5 chairs are left respectively. But, when he arranges in groups of 11's no chair is left.
(i) How many chairs are in the banquet hall?
(b) 209
(c) 539
(d) 149
(ii) If three chairs are removed, then the remaining chairs can be arranged in groups of
(b) 3's
(c) 4's
(d) 5's
(iii) If one chair is added, then the total number of chairs can be arranged in groups of
(b) 3's
(c) 4's
(d) 11's
(iv) If one chair is added to the total number of chairs, how many chairs will be left when arranged in groups of 11's
(b) 2
(c) 3
(d) 4
(v) How many chairs will be left in the original arrangement if some number of chairs is arranged in groups of 9's
(b) 1
(c) 6
(d) 3
Answer:
(i) By dividing all the options by 2, 3, 4, 5, 6, and 11, we will get that 539 is the only option which leaves the remainder of 1, 2, 3, 4, 5, and 0 respectively.
Dividing by 2, it leaves remainder 1. So, the number should be odd.
Dividing by 5, it leaves remainder 4. So, the number should end with 4 or 9. Option (a) is eliminated.
Dividing by 4, it leaves remainder 1. So, options (b) and (d) are eliminated.
Hence, the correct answer is option (c).
(ii) After removing 3 chairs, the number of chairs left are 536 chairs. Since it is an even number and divisible completely by both 2 and 4, the chairs can be arranged in pairs of 2's or 4's.
Hence, the correct answers are options (a) and (c).
Disclaimer: There are two correct answers for the question and the explanation is given above.
(iii) If one chair is added to 536 chairs, the number of chairs become 537.
Now, 537 is completely divisible by 3. So, the chairs can be arranged in groups of 3's.
Hence, the correct answer is option (b).
Disclaimer: The correct answer for the question is option (b) and the explanation for the same is given above.
(iv) If one chair is added to 539 chairs, the total number of chairs become 540. It is already given in the question that the initial number of chairs is completely divisible by 11.
Therefore, when one more chair is added, the number of chairs left after they are arranged in groups of 11 is 1.
Hence, the correct answer is option (a).
(v) When 539 is divided by 9, the remainder will be 8. Thus, arranging chairs in pairs of 9's, 8 chairs will be left.
Hence, the correct answer is option (a).
Page No 15:
Question 65:
Mira is very health conscious and avoids fast food, cakes, ice creams, etc. On her birthday she decided to serve fruits to her friend's guests. She had 60 bananas and 36 apples which are to be distributed equally among all.
(i) How many maximum guests Mira can invite?
(b) 96
(c) 12
(d) 180
(ii) How many apples will each guest get?
(b) 6
(c) 4
(d) 5
(iii) How many bananas will each guest get?
(b) 6
(c) 4
(d) 5
(iv) If Mira also decides to distribute 42 mangoes, how many maximum guests she can invite?
(b) 6
(c) 8
(d) 180
(v) How many total fruits will each guest get?
(b) 25
(c) 17
(d) 18
Answer:
(i) To find the maximum number of guests Mira can invite, find the HCF of the quantities of two fruits which is the highest common factor between the two quantities.
Now, factors of 60 bananas and 36 apples
So, the highest common factor (HCF) between them is 2 × 2 × 3 = 12.
Thus, the maximum number of guests Mira can invite is 12.
Hence, the correct answer is option (c).
(ii) As Mira wants to distribute the apples equally among all the guests and the number of guests is 12. So,
Number of apples each guest will get =
Hence, the correct answer is option (a).
(iii) As Mira wants to distribute the bananas equally among all the guests and the number of guests is 12. So,
Number of bananas each guest will get =
Hence, the correct answer is option (d).
(iv) As Mira decides to add 42 mangoes, so we need to find out the HCF of quantities of all three fruits.
Factors 36, 60, and 42 are-
So, the highest common factor (HCF) between them is 2 × 3 = 6.
Therefore, she can invite a maximum of 6 guests in that case.
Hence, the correct answer is option (b).
= 138
Total number of guests she invites = 6
Thus,
Hence, the correct answer is option (a).
Page No 16:
Question 66:
Jai, Jameel and Jony decided to play a game of climbing 100 stairs. Jai climbs 5 stairs and gets down by 2 stairs in one turn, Jameel goes up by 7 stairs and comes a down down by 2 stairs in one turn, Jony goes 10 stairs up and 3 stars down each time. Each one of them stops when less number of stairs are left than the number of stairs for his forward movement.
(i) Who climbs the maximum number of stairs?
(b) Jameel
(c) Jony
(d) Jai and Jameel
(ii) How many times can they meet in between on the same stair?
(b) 4
(c) 5
(d) Never
(iii) Who takes the least number of attempts to reach near the 100th stair?
(b) Jameel
(c) Jony
(d) All take equal number of steps
(iv) Who meets for the first time on a stair?
(b) Jameel and Jony after 35 turns
(c) Jai and Jony after 21 turns
(d) Jai and Jameel after 21 turns
(v) Who meet for the second time on a stair?
(b) Jameel and Jony after 35 turns
(c) Jai and Jony after 21 turns
(d) Jai and Jony after 35 turns
Answer:
(i)
Name | Forward Steps | Backward Steps | Effective Steps |
Jai | 5 | 2 | 3 |
Jameel | 7 | 2 | 5 |
Jony | 10 | 3 | 7 |
Now, effective step of Jai is 5 and he moves 5 steps forward in one turn. Thus, Jai can climb up to 96 because the number of stairs is 100 only
Similarly, Jameel can climb up to 95 stairs and Jony can climb up to 91 stairs. Thus, Jai climbs the maximum number of stairs.
Hence, the correct answer is option (a).
(ii) Since we have to calculate the meeting point of all three of them, we have to calculate the LCM of the effective steps taken by Jai, Jameel and Jony.
∴ All will meet = LCM (3, 5, 7)
= 105 steps
But, there are only 100 steps.
Thus, Jai, Jameel, and Jony will never meet at all.
Hence, the correct answer is option (d).
(iii) Since the effective number of steps taken by Jai, Jameel and Jony are 3, 5 and 7
For Jai to climb 96 steps, he can take a group of 32 steps.
Similarly, Jameel can climb 95 stairs and he can take a group of 15 steps.
Jony can climb up to 91 stairs and he can take a group of 13 steps.
Thus, Jony will take the minimum number of steps to reach the 100th stair.
Hence, the correct answer is an option (c).
(iv) Since we have to calculate the meeting point of 2 persons then we have to calculate the LCM of the effective steps taken by Jai, Jameel, and Jony.
Jai and Jameel will meet = LCM of (3, 5) = 15 steps
Jameel and Jony will meet = LCM of (5, 7) = 35 steps
Jony and Jai will meet = LCM of (7, 3) = 21 steps
Thus, Jai and Jamaal will meet first at 15 steps.
Hence, the correct answer is option (a).
(v) From the above case, we have Jony and Jai will meet first at 21 steps.
Hence, the correct answer is option (c).
Page No 16:
Question 67:
Observe the factor tree below and answer the questions:
(i) The value of x is
(b) 3825
(c) 835
(d) 3325
(ii) The value of y is
(b) 25
(c) 17
(d) 3
(iii) The value of z is
(b) 17
(c) 5
(d) 13
(iv) The value of x + y + z is
(b) 3847
(c) 3825
(d) 3874
Answer:
(i) From the factor tree, the two factors of x are observed to be 3 and 1275.
So,
Thus, the value of x is 3825.
Hence, the correct answer is option (b).
(ii) From the factor tree, the two factors of 425 are observed to be y and 85.
So,
Thus, the value of y is 5.
Hence, the correct answer is option (a).
(iii) From the factor tree, the two factors of 85 are observed to be z and 5.
So,
Thus, the value of z is 17.
Hence, the correct answer is option (b).
(iv)
Thus, the value of x is 3847.
Hence, the correct answer is option (b).
Page No 16:
Question 68:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): HCF and LCM of two natural numbers are 25 and 815 respectively.
Statement-2 (Reason): LCM of two natural numbers is always divisible by their HCF.
Answer:
Statement-1: HCF and LCM of two natural numbers are 25 and 815.
Now,
HCF of 25 and 815 is 5
LCM of 25 and 815 is (25 × 163) = 4075
Therefore, 25 and 815 are divisible by only 5. So, HCF is 5 and LCM is 4075. Also, the LCM of the two numbers is divisible by their HCF too.
Thus, Statement-1 is true.
Statement-2: LCM of two natural numbers is always divisible by their HCF.
The LCM (least common multiple) or simply, the multiple of two natural numbers is always divisible by their HCF (highest common factor).
Thus, Statement-2 is true and is the correct explanation for Statement-1.
Hence, the correct answer is option (a).
Page No 16:
Question 69:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): HCF (234, 47) = 1.
Statement-2 (Reason): HCF of two co-primes is always 1.
Answer:
Statement-1: HCF (234, 47) = 1
234 = 2 × 3 × 3 × 13
47 = 1 × 47
There is no common factor between the two numbers. So, their HCF is 1.
Thus, Statement-1 (Assertion) is true.
Statement-2: HCF of two co-primes is always 1.
If there are two co-prime numbers, then their only common factor is 1. Therefore, HCF of two co-primes is always 1.
Thus, Statement-2 (Reason) is true.
Thus, Statement-1 is true and Statement-2 is a correct explanation for Statement-1.
Hence, the correct answer is option (a).
Page No 16:
Question 70:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion); is an irrational number.
Statement-2 (Reason): If p is a prime number, then is an irrational number.
Answer:
Statement-1: is an irrational number.
The given statement is true as cannot be written in the form of .
Thus, Statement-1 is true.
Statement- 2: If p is a prime number, then is an irrational number.
Let be a rational number. Then,
∴ p divides .
But when a prime number divides the product of two numbers, it must divide at least one of them. So, let p divides a.
∴ p divides . Thus, p divides b.
Thus, a and b have at least one common multiple p. But it arises the contradiction to our assumption that a and b are coprime.
Thus, our assumption is wrong and p is irrational number.
Thus, Statement-2 is true and it is the correct explanation of Statement-1 is true.
Hence, the correct answer is option (a).
Page No 16:
Question 71:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): HCF of two consecutive natural numbers is 1.
Statement-2 (Reason): HCF of two co-primes is 1.
Answer:
Statement-1: HCF of two consecutive natural numbers is 1.
The set of natural numbers is {1, 2, 3, 4, 5, ...}. It is evident that there is no common factor (except 1) between any two consecutive natural numbers. So, the HCF of two consecutive natural numbers is 1.
Thus, Statement-1 is true.
Statement-2: HCF of two co-primes is 1.
There is no common factor between two co-primes numbers other than 1. So, their HCF is 1.
Thus, Statement-2 is true and it is the correct explanation of Statement-1.
Hence, the correct answer is option (a).
Page No 16:
Question 72:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has the following four choices (a), (b), (c), and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): For any positive integer n, n3 – n is divisible by 6.
Statement-2 (Reason): The product of three consecutive natural numbers is always a multiple of 6.
Answer:
Statement I:
Let
Now, divisible by 6
Again, divisible by 6
So the given statement is true for
Let us take some integer k divisible by 6
divisible by 6
.....(1)
Also,
Now, k(k + 1) is the product of two consecutive integers, which is always divisible by 2.
From (1) and (2), we have
So, f(k+1) is divisible by 6.
So, the statement is true for if we assume it to be true for .
Statement - II
Let the three consecutive positive integers be n, n + 1 and n + 2. When a number is divided by , the remainder obtained is either or .
Then, and are divisible by .
Hence, the correct answer is option (a).
Page No 17:
Question 73:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has the following four choices (a), (b), (c), and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If HCF (a, b) = 4 and ab = 96 × 404, then LCM (a, b) = 9696.
Statement-2 (Reason): LCM of two numbers a and b = HCF (a, b) × ab.
Answer:
Statement-1: If HCF (a, b) = 4 and ab = 96 × 404, then LCM (a, b) = 9696.
We have,
Thus, Statement-1 is true.
Statement-2 (Reason): LCM of two numbers a and b = HCF (a, b) × ab.
The product of two numbers is equal to the product of their LCM and HCF.
So, Statement-2 is false.
Thus, Statement-1 is true and Statement-2 is false.
Hence, the correct answer is option (c).
Page No 17:
Question 74:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): 997 is the largest three digit prime number.
Statement-2 (Reason): A positive integer n is a prime number, if no positive integer less than or equal to divides n.
Answer:
Statement-1 (Assertion): 997 is the largest three digit prime number.
Starting from 999, we see that 999 and 998 are not prime numbers. But 997 has only 2 factors, 1 and 997 itself, satisfying all the conditions of a prime number. Therefore, 997 is the largest three-digit prime number in the number system.
Thus, Statement-1 is true.
Statement-2 (Reason): A positive integer n is a prime number, if no positive integer less than or equal to divides n.
Let n be the positive integer such that no prime less than divides n. Suppose n is not prime, then we have
Let p be a prime factor of a. Then,
This contradicts our assumption that no prime less than divides n.
Thus, statement-2 is true but statement-1 is false.
Hence, the correct answer is option (d).
Page No 17:
Question 75:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): is an irrational number.
Statement-2 (Reason): If p and q are prime positive integers, then is an irrational number.
Answer:
Statement-1 (Assertion): is an irrational number.
The sum of two irrational numbers and is an irrational number.
Thus, Statement-1 is true.
Statement-2 (Reason): If p and q are prime positive integers, then is an irrational number.
The sum of two irrational numbers is an irrational number.
Thus, Statement-2 is true and it is a correct explanation for Statement-1.
Hence, the correct answer is option (a).
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