Real Numbers Class 10 MCQ with Solution Maths

In this section provides Mathematics Real Number MCQ with Solutions for CBSE Class 10 to prepare the student higher level competitive exams like NTSE, NSO, KVPY & NSEJS.

Real Numbers Class 10 MCQ with Solution

1. Which one is divisible by 3 in the following (for n $\in$ N) :
(A) n
(B) n + 2
(C) n + 4
(D) n or (n + 2) or (n + 4)
Solution

Ans. (D)
Sol. If n = 3q, 3q + 1, 3q + 2
then n or (n + 2) or (n + 4) are divisible by 3.

2. A number in the form 25q + 8 can be written as :
(A) 3m + 3
(B) 5m+2
(C) 5m + 3
(D) 3m+2
Solution

Ans. (C)
Sol. 25q + 8 = 25q + 5 + 3
= 5(5q + 1) + 3
= 5m + 3

3. The product of three consecutive numbers is not always divisible by :
(A) 6
(B) 2
(C) 3
(D) 4
Solution

Ans. (D)
Sol. Least group will be 1.2.3 = 6
i.e. It is divisible by 6, 2 & 3 but not by ‘4’ always.

4. When n is divided by 4, the remainder is 3. What is the remainder when 2n is divided by 4 :
(A) 1
(B) 2
(C) 3
(D) 6
Solution

Ans. (B)
Sol. n = 4q + 3
2n = 4q₁ + r
2(4q + 3) = 4q₁ + r
8q + 6 = 4q₁ + r
4(2q + 1) + 2 = 4q₁ + r
on comparsion r = 2

5. Which positive odd integer is of the form of (q is any integer) :
(A) 4q + 1
(B) 4q + 3
(C) both
(D) none
Solution

Ans. (C)
Sol. 4q + 1

\to

odd integer
4q + 3

\to

odd integer

6. If n² – 1 is divisible by 8 then n is an :
(A) Odd positive integer
(B) Odd negative integer
(C) Even Positive integer
(D) Both (A) & (B)
Solution

Ans. (D)
Sol. Let n = 2m + 1
(n² – 1) = (n – 1)(n + 1)
= (2m + 1 – 1) (2m + 1 + 1)
= 2m × 2(m + 1)
= 4m (m + 1)
Case I
If m = odd
then (m + 1) is even,& m(m + 1) is multiple of 2
So, 4(m) × (m + 1) is a multiple of 8
Case II
If m = even
then m(m + 1) again a multiple of 2
So, 4(m) × (m + 1) is multiple of 8.

7. If x and y are even positive integer then x² + y² is :
(A) Even and divisible by 4
(B) Odd and divisible by 4
(C) Even but not divisible by 4
(D) Odd but not divisible by 4
Solution

Ans. (A)
Sol. x = 2m, y = 2n
x² + y² = (2m)² + (2n)² = 4m² + 4n²
= 4(m² + n²)
It is an even and divisible by 4.

8. A number when divided by 169 leaves 91 as remainder. If the same number is divided by 13, what will be the remainder?
(A) 0
(B) 1
(C) 6
(D) 9
Solution

Ans. (A)
Sol. Let number is N
Then, according to the question
N = 169 K + 91 (Here K is whole number)
When N is divided by 13

\frac{{169K + 91}}{{13}}

\frac{{13 \times 13K + 13 \times 7}}{{13}}

\frac{{13(13K + 7)}}{{13}}

So, here remainder will be zero.

9. A positive integer of the form 24q + 9 can also be written as [m, q are integers] :
(A) 8m
(B) 8m + 2
(C) 8m + 1
(D) 8m + 7
Solution

Ans. (C)
Sol. 24q + 9 = 8 × 3q + 8 + 1
= 8(3q + 1) + 1
= 8m + 1

10. Rational number between $\sqrt 2$ and $\sqrt 3$ is :

(A) $\frac{{\sqrt 2 + \sqrt 3 }}{2}$

(B) $\frac{{\sqrt 2 \times \sqrt 3 }}{2}$

(D) 1.8

Solution

Ans. (C)

Sol. $\sqrt 2$ = 1.414…..
$\sqrt 3$ = 1.732…..
So, 1.5 lies between $\sqrt 2 \,\& \,\sqrt 3$ Hence C is correct option.

11. If a and b be any two positive integers then :
(A) HCF (a, b) + LCM (a, b) = a + b
(B) HCF (a, b) + LCM (a, b) = a × b
(C) HCF (a, b) × LCM (a, b) = a × b
(D) HCF (a, b) = LCM (a, b)
Solution

Ans. (C)
Sol. If ‘a’ and ‘b’ are two positive integers then
HCF (a, b) × LCM (a, b) = a × b

12. The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm. respectively, then the longest rod which can measure the three dimensions of the room exactly :
(A) 65 cm
(B) 70 cm
(C) 75 cm
(D) 80 cm
Solution

Ans. (C)
Sol. If Length, breadth and height of room are 8 m 25 cm, 6 m 75 cm and 5 m 50 cm.
If it will be converted in to cm then dimension will be 825 cm, 675 cm and 450 cm.
If we will take H.C.F. of above dimension then we will get the desired Dimension.
825 = 3 × 5² × 11
675 = 33 × 5²
450 = 2 × 3² × 5²
HCF of 825, 675 and 450 = 5² × 3 = 75 cm.

13. Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolate of both kinds, what is the least number of boxes of each kind. I would need to buy
(A) 8 of first kind, 5 of second kind
(B) 5 of first kind, 8 of second kind
(C) 5 of first kind, 7 of second kind
(D) None of these
Solution

Ans. (B)
Sol. Let first brand have number of the toffee = 24
Second brand have number of toffee = 15
Minimum number of chocolates = Take LCM of 24 and 15 = 120
First Brand have minimum =

\frac{{120}}{{24}}\, = \,5

Second Brand have minimum chocolates =

\frac{{120}}{{15}}\, = \,8

So 5 of first kind and 8 of second kind.

14. If p is a prime number, then the L.C.M. of p and (p + 1)

(A) p²

(B) $\frac{{p(p + 1)}}{2}$

(D) p(p + 1)

Solution

Ans. (D)
Sol. Here p is a prime number then there will be no common factor among p and (p+ l)
Hence LCM of p & (p + 1) is p(p + 1)

15. The product of two numbers is 1600 and their H.C.F. is 5. The L.C.M. of the numbers is :
(A) 8000
(B) 1600
(C) 320
(D) 1605
Solution

Ans. (C)
Sol. Let numbers are ‘a’ and ‘b’
Here a × b = 1600
H.C.F. = 5
Here H.C.F. × L.C.M. = Product of numbers
5 × L.C.M. = 1600
L.C.M. =

\frac{{1600}}{5}\, = \,320

16. The G.C.D. of 3556 and 3444 is :
(A) 25
(B) 28
(C) 3
(D) 26
Solution

Ans. (B)
Sol. G.C.D. of 3556 and 3444
3556 = 2² × 7 × 127
3444 = 2² × 3 × 7 × 41
G.C.D. of 3556 and 3444
= 2² × 7 = 28

17. H.C.F. of 2³ × 3² × 5, 2² × 3³ × 5² and 2⁴ × 3 × 5³ × 7 is :
(A) 30
(B) 48
(C) 60
(D) 105
Solution

Ans. (C)
Sol. H.C.F. of 2³ × 3² × 5, 2² × 3³ × 5² and 2⁴ × 3 × 5³ × 7
= 2² × 3 × 5 = 60.

18. LCM of two prime numbers x and y (x > y) is 161. The value of 3y – x is ?
(A) –2
(B) – 1
(C) 1
(D) 2
Solution

Ans. (A)
Sol. LCM of x and y = 161
xy = 161
For x = 2, y will not have integral solution
x = 3, y will not have integral solution
x = 5, y will not have integral solution
x = 7, y =

\frac{{161}}{7} = 23

Here x > y So, x = 23 and y = 7
So 3y – x = (3 × 7 – 23) = (21 – 23) = –2

19. The G.C.D. of two numbers is 38 and their L.C.M. is 98154. If one of the number is 1558, the other number is :
(A) 1260
(B) 3450
(C) 2394
(D) 2395
Solution

Ans. (C)
Sol. Let Numbers are ‘a’ and ‘b’
G.C.D. of two numbers = 38
L.C.M. of two numbers = 98154
G.C.D. × L.C.M. = Product of numbers
Here a = 1558 then b = ?
b =

\frac{{38 \times \,98154}}{{1558}}

= 2394

20. If the sum of two numbers is 55 and the HCF and LCM of these numbers are 5 and 120 respectively then the sum of the reciprocals of the numbers is equal to

(A) $\frac{{55}}{{601}}$

(B) $\frac{{601}}{{55}}$

(D) $\frac{{120}}{{11}}$

Solution

Ans. (C)

Sol. Let numbers are A and B
A + B = 55 . . .(1)

$\because$ (HCF × LCM) = Product of numbers
So A × B = 5 × 120
A × B = 600 . . .(2)
then $\frac{1}{A} + \frac{1}{B}$ = $\frac{{A + B}}{{AB}}$ = $\frac{{55}}{{600}} = \frac{{11}}{{120}}$

21. Which one of the following is not an irrational number :

(A) $2 + \sqrt 3$

(B) $2 + \sqrt 5$

(D) $2 + \sqrt 4$ Solution

Ans. (D)

Sol. As we know $2 + \sqrt 4 = 2 + 2 = 4\,\, \Rightarrow \,\,\frac{p}{q} = \frac{4}{1}$

22. π is a/an :

(A) Rational number

(B) Irrational number

(D) Can’not say

Solution

Ans. (B)

Sol. We know that π is non-terminating, non-repeating numbers and its value $\frac{{22}}{7}$ is not the exact value. It is just an approximate value. Hence p is irrational.

23. The product of two irrational numbers is :

(A) Always an irrational number

(B) Always a rational number

(D) None of these

Solution

Ans. (C)

Sol. Let us consider

$2 + \sqrt 3 \,\,{\bf{and}}\,\,2 - \sqrt 3$ two irrational number

Now $\left( {2 + \sqrt 3 } \right)\,\,\left( {2 - \sqrt 3 } \right)$ = 4 – 3 = 1 which is rational

Also Let us take $\sqrt 3$ & $\sqrt 2$ two irrational number

Now $\sqrt 3 \,\sqrt 2 = \sqrt 6$ which is irrational.

Hence (C)

24. The number 1.12112111211112… is a/an :

(A) Irrational number

(B) Rational number

(D) None of these

Solution

Ans. (A)

Sol. We can see that this number is non-repeating and non-terminating numbers.

Hence irrational.

25. Which one is not correct in the following :

(A) There are infinitely many irrational numbers

(B) The sum of two irrational numbers is always irrational

(D) Zero is a rational number

Solution

Ans. (B)

Sol. The sum of two irrational numbers is always irrational

26. $\sqrt n$ in an irrational number, if n is :

(A) A natural number

(B) A prime number

(D) An odd number

Solution

Ans. (B)

Sol. $\sqrt n$ = Irrational number if n is a prime number.

27. A rational number $\frac{a}{b}$ will have a terminal decimal expansion, if b is of the form :

(A) 2^m × 3ⁿ

(B) 2^m × 5ⁿ

(D) 2^m + 5ⁿ

Solution

Ans. (B)

Sol. Product of ‘2’ and ‘5’ gives ‘0’ at unit place.

28. ${(7 + \sqrt 3 )^2} + {(7 - \sqrt 3 )^2}$ is a/an :

(A) Rational number

(B) Irrational number

(D) None

Solution

Ans. (A)

Sol. ${(7 + \sqrt 3 )^2} + {(7 - \sqrt 3 )^2}$

$= 49 + 14\sqrt 3 + 49 - 14\sqrt 3 + 3 + 3$

= 98 + 6 = 104 = $\frac{{104}}{1}\left( {\frac{p}{q}\,\,{\bf{form}}} \right)$

Hence rational number (A)

29. Which one of the following rational numbers have the non terminating decimal representation :

(A) $\frac{7}{{25}}$

(B) $\frac{{15}}{{16}}$

(D) $\frac{{17}}{{45}}$ Solution

Ans. (D)

Sol. In $\frac{{17}}{{45}} = \frac{{17 \times 20}}{{9 \times 5 \times 20}} = \frac{{340}}{{9 \times 100}} = 0.37777\,.\,\,.\,\,.\,\,.$ Non-terminating

Hence (D)

30. Which one of the following rational numbers has the terminating decimal representation :

(A) $\frac{{29}}{{343}}$

(B) $\frac{{129}}{{{2^2} \times {5^7} \times {7^5}}}$

(D) $\frac{{13}}{{3125}}$ Solution

Ans. (D)

Sol. In $\frac{{13}}{{3125}}$ there is multiple of 25.

31. If n is a natural number, then $\sqrt n$ is :

(A) Always a natural number

(B) Always a rational number

(D) Sometimes a natural number or rational number and sometimes an irrational number

Solution

Ans. (D)

Sol. n = natural number

then $\sqrt n \, = \,?$

If n will be perfect square than it will be rational number and if n will be any number except perfect square than it will be irrational number.

32. Irrational number between 2 and 2.5 is :

(A) $\sqrt {2.2}$

(B) $\sqrt 5$

(D) $\sqrt {2.5}$ Solution

Ans. (B)

Sol. Irrational number between 2 and 2.5

4 < 5 < 9 By taking square Root.

$\sqrt 4$ < $\sqrt 5$ < $\sqrt 9$

2 < $\sqrt 5$ < $\sqrt 9$

2 < $\sqrt 5$ < 3

2 < $\sqrt 5$ < 2.5

33. $\frac{{2\sqrt 6 }}{{\sqrt 2 \, + \sqrt 3 \, + \,\sqrt 5 }}$ is equal to :

(A) $\sqrt 2$ + $\sqrt 3$ – $\sqrt 5$

(B) 4 – $\sqrt 2$ – $\sqrt 3$

(D) $\frac{1}{2}\,(\sqrt 2 \, + \sqrt 5 \, - \sqrt 3 )$ Solution

Ans. (A)

Sol. If $\frac{{2\sqrt 6 }}{{\sqrt 2 \, + \,\sqrt 3 \, + \,\sqrt 5 }}$

After simplification

$\frac{{2\sqrt 6 \,(\sqrt 2 \, + \,\sqrt 3 \, - \,\sqrt 5 \,)}}{{(\sqrt 2 \, + \,\sqrt 3 \, + \,\sqrt 5 \,)\,(\sqrt 2 \, + \,\sqrt 3 \, - \,\sqrt 5 \,)}}$ = $\frac{{2\sqrt 6 \,(\sqrt 2 \, + \,\sqrt 3 \, - \,\sqrt 5 \,)}}{{(2\, + \,3\, + \,2\sqrt 6 \,)\, - \,5}}$

= $\frac{{2\sqrt 6 \,(\sqrt 2 \, + \,\sqrt 3 \, - \sqrt 5 )}}{{\,2\sqrt 6 \,}}$ = $\sqrt 2 + \sqrt 3 - \sqrt 5$

34. If x = 7 + 4 $\sqrt 3$ , then $\sqrt x + \frac{1}{{\sqrt x }}$ equals :

(A) 6

(B) 4

(D) 4 $\sqrt 3$ Solution

Ans. (B)

Sol. x = 7 + $4\sqrt 3$

⇒ $\sqrt x$ = $\sqrt {7 + 4\sqrt 3 }$

= $\sqrt {{2^2} + 2.2.\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}}$ = $\sqrt {{{(2 + \sqrt 3 )}^2}}$

$\sqrt x$ = $2 + \sqrt 3$

and $\frac{1}{{\sqrt x }}$ = $\frac{1}{{2 + \sqrt 3 }} = 2--\sqrt 3$

$\sqrt x + \frac{1}{{\sqrt x }}$ = $2 + \sqrt 3 + 2--\sqrt 3 = 4$

35. The sum of two irrational numbers is :

(A) Rational

(B) Irrational

(D) A or B

Solution

Ans. (D)

Sol. Let $2 + \sqrt 3$ and $2 - \sqrt 3$ are two irrational numbers.

Sum = 4 → rational number.

Let $2 + \sqrt 3$ and $\sqrt 3$ are two irrational numbers

Sum = $2 + 2\sqrt 3$ → irrational number. Hence (D)

36. Which of the following is not rational :

(A) 0.14

(B) $0.14\overline {16}$

(D) 0.4014001400014…

Solution

Ans. (D)

Sol. In option (D), it is a non-terminating non-recurring i.e., it is an irrational number.

37. Which of the following is a true statement :

(A) Every real number is rational

(B) Every real number is irrational

(D) None of these

Solution

Ans. (D)

Sol. All statements regarding to real number are false, i.e. every real number is either rational or irrational.

38. Which one of the following numbers can be represented as non-terminating, repeating decimals :

(A) $\frac{{39}}{{24}}$

(B) $\frac{3}{{16}}$

(D) $\frac{{137}}{{25}}$ Solution

Ans. (C)

Sol. In $\frac{3}{{11}}$ , 11 is not in the form of 2^m × 5ⁿ. & numerate must be in the form of prime number.

39. Which of the following is not an irrational number :

(A) $5 + \sqrt 5$

(B) $5 + \sqrt {10}$

(D) $5 + \sqrt {25}$ Solution

Ans. (D)

Sol. $5 + \sqrt {25} = 5 + 5 = 10$

It is a rational number.

40. The product of rational and irrational numbers is always :

(A) 0

(B) Rational

(D) None of these

Solution

Ans. (C)

Sol. Product of 2 and $(2 + \sqrt 3 ) = 4 + 2\sqrt 3$ is an irrational number.

41. Which one of the following rational numbers have the non terminating decimal representation :

(A) $\frac{{17}}{{625}}$

(B) $\frac{{15}}{{32}}$

(D) $\frac{{19}}{{65}}$ Solution

Ans. (D)

Sol. Here, in $\frac{{19}}{{65}}$ , 65 is not in form of 2^m × 5ⁿ. 19 is prime number. So D is the correct option.

42. Find the remainder when 27× 43 × 46 is divided by 10 :

(A) 1

(B) 6

(D) None of these

Solution

Sol. (B)

27× 43 × 46 Here in above given number. After multiplication last digit is 6. So when it will get divided by ‘10’ its remainder will be ‘6’

43. A number in the form 15q + 7 can be written as :

(A) 3m + 3

(B) 5m+2

(D) 3m + 2

Solution

Ans. (B)

Sol. Number is (15 q + 7)

15 q + 7 = (3× 5)q + (5+2)

= 5(3q + 1) + 2

= 5m + 2 [ 3q + 1 = m]

44. The product of three consecutive natural numbers is always divisible by :

(A) 6

(B) 24

(D) 4

Solution

Ans. (A)

Sol. Let three consecutive natural number is n, (n + 1) and (n + 2)

The product of these number is ‘P’

Then P = n(n+1) (n+2) = n(n² + 3n + 2)

= n³ + 3n² + 2n

Put n = 1 then P = 6

n = 2 then P = 24

n = 3 then P = 60

………………………..

So, for any natural value of of n it will be divisible by ‘6’

45. If 25 is divided by 7 then there exists two other unique non-negative integers q and r, find the value of q and r :

(A) 3, 1

(B) 3, 4

(D) 4, 2

Solution

Ans. (B)

Sol.

So here when 25 will get divided by, 7 then there exist quotient ‘3’ and remainder is ‘4’

so, unique non negative integer (q, r) = (3, 4)

46. How many possible values of r are there if a = 7q + r, and 7 is divisor :

(A) 3

(B) 4

(D) 7

Solution

Ans. (D)

Sol. If a = (7q + r ) and 7 is divisor then number of possible value of r will be ‘7’

i.e. r = 0, 1, 2, 3, 4, 5, 6

47. Using division algorithm, if q denotes quotient and r denotes remainder, then when 432 is divided by 201, the pair (q, r) is:

(A) (2, 2)

(B) (2, 12)

(D) (2, 30)

Solution

Ans. (D)

Sol. As per questions

432 = 201 q + r ….(1)

and 432 = 201 × 2 + 30 ….(2)

By comparing equation (1) and (2)

q = 2 and r = 30

48. For n ∈ N. 6ⁿ can end with the digit :

(A) 0

(B) 3

(D) 6

Solution

Ans. (D)

Sol. If n ∈N then 6ⁿ will be always end with the digit ‘6’

49. The greatest number of four digits which is divisible by each one of the numbers 12, 18, 21 & 28 is

(A) 9848

(B) 9864

(D) 9636

Solution

Ans. (C)

Sol. The greatest Four digit number which is divisible by 12, 18, 21 and 28

L.C.M. of numbers will get divisible by all numbers.

L.C.M. of 12, 18, 21, 28 = 252

So 252 will be least divisible by all of above given numbers

Let 252 × n = 9999

n = $\frac{{9999}}{{252}}$ = 252 × 19 + 171

= remainder 171 so 171 should be subsrated from 9999

Then the greatest four digit number = 9828

50. If ${(\sqrt 2 )^x}\, + \,{(\sqrt 3 )^x}\, = \,{(\sqrt {13} )^{\frac{x}{2}}}$ then the number of values of x is :

(A) 1

(B) 2

(D) 0

Solution

Ans. (A)

Sol. Here ${(2)^{\frac{x}{2}}}\, + \,{(3)^{\frac{x}{2}}}\, = \,{(13)^{\frac{x}{4}}}$

Here for x = 4 above relation will satisfied so x = 4, i.e. only one solution.

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