Chapter 1 Real Numbers- MCQ Online Test 1 Class 10 Maths
1. A number when divided by 61 gives 27 as quotient and 32 as remainder, then the number is:
a) 1796
b) 1967
c) 1679
d) 1569
2. Every positive even integer is of the form ____ for some integer ‘q’
a) 2q – 1
b) none of these
c) 2q
d) 2q + 1
3. Every positive odd integer is of the form ________ where ‘q’ is some integer.
a) 3q + 1
b) 5q + 1
c) 2q + 2
d) 2q + 1
4. For any two positive integers a and b, such that a > b. there exist (unique) whole numbers q and r such that
a) a = bq + r,0⩽r<b
b) a = qbr
c) q = ar + b, 0⩽r<b
d) b = aq + r,0⩽r<b
5. 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 b) 1 < r < 3
c) 0 < r < 3
d) 0⩽r < 3
6. If a is a non-zero rational and √b is irrational, then a√b is:
a) a natural number
b) a rational number
c) an integer
d) an irrational number
7. The number (√3+√5)2 is
a) a rational number
b) not a real number
c) an irrational number
d) an integer
8. If ‘a’ and ‘b’ are both positive rational numbers, then (√a+√b)(√a−√b) is
a) none of these
b) a rational number
c) an irrational number
d) neither rational nor rational number
9. All non-terminating and non-recurring decimal numbers are
a) natural numbers
b) rational numbers
c) integers
d) irrational numbers
10. A rational number can be expressed as a terminating decimal if the denominator has the factors
a) 2 or 5 only
b) 2 or 3 only
c) 2, 3 or 5 only
d) 2 only
Ch 1 Real Numbers- MCQ Online Test 1 Class 10
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Question 1 of 10
1. Question
A number when divided by 61 gives 27 as quotient and 32 as remainder, then the number is:
Correct
Dividend = Divisor × Quotient + Remainder
Therefore the number (Dividend) = 61×27 + 32
= 1647 + 32
= 1679Incorrect
Dividend = Divisor × Quotient + Remainder
Therefore the number (Dividend) = 61×27 + 32
= 1647 + 32
= 1679 -
Question 2 of 10
2. Question
Every positive even integer is of the form ____ for some integer ‘q’
Correct
Let a be any positive integer and b=2
Then by applying Euclid’s Division Lemma, we have,
a = 2q + r where 0⩽r<2 r = 0 or 1
Therefore, a=2q or 2q+1
Thus, it is clear that a=2q
i.e., a is an even integer in the form of 2qIncorrect
Let a be any positive integer and b=2
Then by applying Euclid’s Division Lemma, we have,
a = 2q + r where 0⩽r<2 r = 0 or 1
Therefore, a=2q or 2q+1
Thus, it is clear that a=2q
i.e., a is an even integer in the form of 2q -
Question 3 of 10
3. Question
Every positive odd integer is of the form ________ where ‘q’ is some integer.
Correct
Let a be any positive integer and b=2
Then by applying Euclid’s Division Lemma,
we have, a=2q+r
where 0⩽r <2 ⇒r = 0 or 1
∴ a=2q or 2q+1
Therefore, it is clear that a=2q i.e., a is an even integer.
Also 2q and 2q+1 are two consecutive integers, therefore, 2q+1 is an odd integer.Incorrect
Let a be any positive integer and b=2
Then by applying Euclid’s Division Lemma,
we have, a=2q+r
where 0⩽r <2 ⇒r = 0 or 1
∴ a=2q or 2q+1
Therefore, it is clear that a=2q i.e., a is an even integer.
Also 2q and 2q+1 are two consecutive integers, therefore, 2q+1 is an odd integer. -
Question 4 of 10
4. Question
For any two positive integers a and b, such that a > b. there exist (unique) whole numbers q and r such that
Correct
Euclid’s Division Lemma states that for given positive integer a and b, there exist unique integers q and r satisfying a=bq+r; 0⩽r<b.
Incorrect
Euclid’s Division Lemma states that for given positive integer a and b, there exist unique integers q and r satisfying a=bq+r; 0⩽r<b.
-
Question 5 of 10
5. Question
If a is a non-zero rational and √b is irrational, then a√b is:
Correct
If possible let a√b be rational.
Then a√b=p/q, where pp and qq are non-zero integers, having no common factor other than 1.
Now, a√b= p/q …. (i)
But, p and aq are both rational and aq≠0
∵ p/aq is rational.
Therefore, from eq. (i), it follows that √b is rational.
The contradiction arises by assuming that a√b is rational.
Hence, a√b is irrational.Incorrect
If possible let a√b be rational.
Then a√b=p/q, where pp and qq are non-zero integers, having no common factor other than 1.
Now, a√b= p/q …. (i)
But, p and aq are both rational and aq≠0
∵ p/aq is rational.
Therefore, from eq. (i), it follows that √b is rational.
The contradiction arises by assuming that a√b is rational.
Hence, a√b is irrational. -
Question 6 of 10
6. Question
The number (√3+√5)2 is
Correct
:(√3+√5)2 = (√3)2 + (√5)2 + 2×√3×√5
= 3+5+2√15 = 8+2√15
Here √15 = √3×√5
Since √3 and √5 both are irrational number
therefore (√3+√5)2 is an irrational number.Incorrect
:(√3+√5)2 = (√3)2 + (√5)2 + 2×√3×√5
= 3+5+2√15 = 8+2√15
Here √15 = √3×√5
Since √3 and √5 both are irrational number
therefore (√3+√5)2 is an irrational number. -
Question 7 of 10
7. Question
If ‘a’ and ‘b’ are both positive rational numbers, then (√a+√b)(√a−√b) is
Correct
(√a+√b)(√a−√b) = {(√a)2−(√b)2} = (a−b)
Since a and b both are positive rational numbers, therefore difference of two positive rational numbers is also rational.Incorrect
(√a+√b)(√a−√b) = {(√a)2−(√b)2} = (a−b)
Since a and b both are positive rational numbers, therefore difference of two positive rational numbers is also rational. -
Question 8 of 10
8. Question
All non-terminating and non-recurring decimal numbers are
Correct
All non-terminating and non-recurring decimal numbers are rational numbers.
A number is rational if and only if its decimal representation is repeating or terminating.Incorrect
All non-terminating and non-recurring decimal numbers are rational numbers.
A number is rational if and only if its decimal representation is repeating or terminating. -
Question 9 of 10
9. Question
A rational number can be expressed as a terminating decimal if the denominator has the factors
Correct
A rational number can be expressed as a terminating decimal if the denominator has the factors 2 or 5 or both.
If denominator has 10 only which has factors 2 and 5 only. Any other factors in the denominator yield a non-terminating decimal expansionIncorrect
A rational number can be expressed as a terminating decimal if the denominator has the factors 2 or 5 or both.
If denominator has 10 only which has factors 2 and 5 only. Any other factors in the denominator yield a non-terminating decimal expansion -
Question 10 of 10
10. Question
For any positive integer ‘a’ and 3, there exist unique integers ‘q’ and ‘r’ such that a = 3q + r where ‘r’ must satisfy
Correct
Incorrect