MCQ Questions for Class 10 Maths Real Numbers with Answers

MCQ Questions for Class 10 Maths Real Numbers with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 1 Real Numbers Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Real Numbers MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 1 Real Numbers

1. The decimal form of \frac{129}{2^{2} 5^{7} 7^{5}} is
(a) terminating
(b) non-termining
(c) non-terminating non-repeating
(d) none of the above

Answer

Answer: c


2. HCF of 8, 9, 25 is
(a) 8
(b) 9
(c) 25
(d) 1

Answer

Answer: d


3. Which of the following is not irrational?
(a) (2 - √ 3)2
(b) (√2 + √3)2
(c) (√2 -√3)(√2 + √3)
(d)\frac{2 \sqrt{7}}{7}

Answer

Answer: c


4. The product of a rational and irrational number is
(a) rational
(b) irrational
(c) both of above
(d) none of above

Answer

Answer: b


5. The sum of a rational and irrational number is
(a) rational
(b) irrational
(c) both of above
(d) none of above

Answer

Answer: b


6. The product of two different irrational numbers is always
(a) rational
(b) irrational
(c) both of above
(d) none of above

Answer

Answer: b


7. The sum of two irrational numbers is always
(a) irrational
(b) rational
(c) rational or irrational
(d) one

Answer

Answer: a


8. If b = 3, then any integer can be expressed as a =
(a) 3q, 3q+ 1, 3q + 2
(b) 3q
(c) none of the above
(d) 3q+ 1

Answer

Answer: a


9. The product of three consecutive positive integers is divisible by
(a) 4
(b) 6
(c) no common factor
(d) only 1

Answer

Answer: b


10. The set A = {0,1, 2, 3, 4, …} represents the set of
(a) whole numbers
(b) integers
(c) natural numbers
(d) even numbers

<strongAnswer

Answer: a


11. Which number is divisible by 11?
(a) 1516
(b) 1452
(c) 1011
(d) 1121

Answer

Answer: b


13. The largest number that will divide 398,436 and 542 leaving remainders 7,11 and 15 respectively is
(a) 17
(b) 11
(c) 34
(d) 45

Answer/ Explanation

Answer: a
Explaination:(a); [Hint. Algorithm 398 - 7 - 391; 436 - 11 = 425; 542 - 15 = 527; HCF of 391, 425, 527 = 17]


14. There are 312, 260 and 156 students in class X, XI and XII respectively. Buses are to be hired to take these students to a picnic. Find the maximum number of students who can sit in a bus if each bus takes equal number of students
(a) 52
(b) 56
(c) 48
(d) 63

Answer/ Explanation

Answer: a
Explaination:(a); [Hint. HCF of 312, 260, 156 = 52]


15. There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field. Harish takes 12 minutes. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet ?
(a) 36 minutes
(b) 18 minutes
(c) 6 minutes
(d) They will not meet

Answer/ Explanation

Answer: a
Explaination:(a); [Hint. LCM of 18 and 12 = 36]


16. Express 98 as a product of its primes
(a) 22x7
(b) 22x72
(c) 2x72
(d) 23x7

Answer

Answer: c


17. Three farmers have 490 kg, 588 kg and 882 kg of wheat respectively. Find the maximum capacity of a bag so that the wheat can be packed in exact number of bags.
(a) 98 kg
(b) 290 kg
(c) 200 kg
(d) 350 kg

Answer/ Explanation

Answer: a
Explaination:(a); [Hint. HCF of 490, 588, 882 = 98 kg]


18. For some integer p, every even integer is of the form
(a) 2p + 1
(b) 2p
(c) p + 1
(d) p

Answer

Answer: b


19. For some integer p, every odd integer is of the form
(a) 2p + 1
(b) 2p
(c) p + 1
(d) p

Answer

Answer: a


20. m² - 1 is divisible by 8, if m is
(a) an even integer
(b) an odd integer
(c) a natural number
(d) a whole number

Answer

Answer: b


21. If two positive integers A and B can be ex-pressed as A = xy3 and B = xiy2z; x, y being prime numbers, the LCM (A, B) is
(a) xy2
(b) x4y2z
(c) x4y3
(d) x4y3z

Answer

Answer: d


22. The product of a non-zero rational and an irrational number is
(a) always rational
(b) rational or irrational
(c) always irrational
(d) zero

Answer

Answer: c


23. If two positive integers A and B can be expressed as A = xy3 and B = x4y2z; x, y being prime numbers then HCF (A, B) is
(a) xy2
(b) x4y2z
(c) x4y3
(d) x4y3z

Answer

Answer: a


24. The largest number which divides 60 and 75, leaving remainders 8 and 10 respectively, is
(a) 260
(b) 75
(c) 65
(d) 13

Answer

Answer: d


25. The least number that is divisible by all the numbers from 1 to 5 (both inclusive) is
(a) 5
(b) 60
(c) 20
(d) 100

Answer

Answer: b
Explaination:(b); [Hint. LCM of 2, 3, 4, 5 = 60


26. The least number that is divisible by all the numbers from 1 to 8 (both inclusive) is
(a) 840
(b) 2520
(c) 8
(d) 420

Answer

Answer: a


27. The decimal expansion of the rational number \frac{14587}{250} will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places

Answer

Answer: c


28. The decimal expansion of the rational number \frac{97}{2 \times 5^{4}} will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places

Answer

Answer: d


29. The product of two consecutive natural numbers is always:
(a) prime number
(b) even number
(c) odd number
(d) even or odd

Answer

Answer: b


30. If the HCF of 408 and 1032 is expressible in the form 1032 x 2 + 408xp, then the value of p is
(a) 5
(b) -5
(c) 4
(d) -4

Answer/ Explanation

Answer: b
Explaination:(b); [Hint. HCF of 408 and 1032 is 24, .-. 1032 x 2 + 408 x (-5)]


31. The number in the form of 4p + 3, where p is a whole number, will always be
(a) even
(b) odd
(c) even or odd
(d) multiple of 3

Answer

Answer: b


32. When a number is divided by 7, its remainder is always:
(a) greater than 7
(b) at least 7
(c) less than 7
(d) at most 7

Answer

Answer: c


33. (6 + 5 √3) - (4 - 3 √3) is
(a) a rational number
(b) an irrational number
(c) a natural number
(d) an integer

Answer

Answer: b


34. If HCF (16, y) = 8 and LCM (16, y) = 48, then the value of y is
(a) 24
(b) 16
(c) 8
(d) 48

Answer

Answer: a


35. According to the fundamental theorem of arith-metic, if T (a prime number) divides b2, b > 0, then
(a) T divides b
(b) b divides T
(c) T2 divides b2
(d) b2 divides T2

span style=”color: #ff0000; cursor: pointer;”>Answer

Answer: a


36. The number ‘π' is
(a) natural number
(b) rational number
(c) irrational number
(d) rational or irrational

Answer

Answer: c


37. If LCM (77, 99) = 693, then HCF (77, 99) is
(a) 11
(b) 7
(c) 9
(d) 22

Answer

Answer: a


38. Euclid's division lemma states that for two positive integers a and b, there exist unique integer q and r such that a = bq + r, where r must satisfy
(a) a < r < b
(b) 0 < r ≤ b
(c) 1 < r < b
(d) 0 ≤ r < b

Answer

Answer: d


39. For positive integers 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

Answer: a


40. Find the greatest number of 5 digits, that will give us remainder of 5 when divided by 8 and 9 respectively.
(a) 99921
(b) 99931
(c) 99941
(d) 99951

Answer/ Explanation

Answer: c
Explaination: The greatest number will be multiple of LCM (8, 9)
LCM of 8 and 9 = 72
On verification we find that 99941 when divided by 72 leaves remainder 5.


41. For some integersp and 5, there exist unique integers q and r such that p - 5q + r. Possible values of r are
(a) 0 or 1
(b) 0, 1 or 2
(c) 0, 1, 2 or 3
(d) 0, 1, 2, 3 or 4

Answer/ Explanation

Answer: d
Explaination: According to Euclids division lemma, p - 5q + r, where 0 < r < 5
⇒ r = 0,1,2, 3, 4
So, possible values of r are 0, 1,2, 3 or 4.


42. If two positive integers a and b are written as a = x'y1 and b = xyi, where x, y are prime numbers, then HCF(a, b) is
Also, find LCM of (a, b). [NCERT Exemplar Problems; Delhi 2019]
(a) xy
(b) xy²
(c) x3y3
(d) x²y²

Answer/ Explanation

Answer: b
Explaination: Here, a = x3y² and b = xy3
⇒ a=xxxxxxyxy
and b = xyxyxy
∴ HCF(a, b) = xxyxy
= xxy² = xy²
LCM = x3y3


43. If two positive integers p and q can be expressed as p = ab² and q = c3b; where a, b being prime numbers, then LCM (p, q) is equal to [NCERT Exemplar Problems]
(a) ab
(b) crb²
(c) a3b²
(d) c²b3

Answer/ Explanation

Answer: c
Explaination:LCM (p, q) = a3b²


44. The ratio between the LCM and HCF of 5, 15, 20 is:
(a) 9 : 1
(b) 4:3
(c) 11:1
(d) 12:1

Answer/ Explanation

Answer: d
Explaination:
5, 15 = 5x3, 20 = 2x2x5
LCM(5, 15, 20) = 5x3x2x2 = 60
HCF(5, 15, 20) = 5
MCQ Questions for Class 10 Maths Real Numbers with Answers 1


45. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time ?
(a) 12.20 pm
(b) 12.12 pm
(c) 12.11pm
(d) none of these

Answer/ Explanation

Answer: d
Explaination:
LCM of 50 and 48 = 1200
∴ 1200 sec = 20 min
Hence at 12.20 pm they will beep again for the first time.


46. If A = 2n + 13, B = n + 7, where n is a
natural number, then HCF of A and B is:
(a) 2
(b) 1
(c) 3
(d) 4

Answer/ Explanation

Answer: b
Explaination:
Taking different values of n we find
that A and B are coprime.
∴ HCF = 1


47. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:
(a) 22
(b) 16
(c) 36
(d) 21

Answer/ Explanation

Answer: b
Explaination:
HCF of 576 and 448 = 64
∴ Number of sections = =\frac{576}{64}+\frac{448}{64}
= 9 + 7 = 16


48. The HCF of 2472, 1284 and a third number N is 12. If their LCM is 23 x 32 x 5 x 103 x 107, then the number Nis :
(a) 22 x 32 x 7
(b) 22 x 33 x 103
(c) 22 x 32 x 5
(d) 24 x 32 x ll

Answer/ Explanation

Answer: c
Explaination:
2472 = 23x3x103
1284 = 2²x3x107
∵ LCM = 23x3²x5x103x107
∴ N = 2²x3²x5 = 180


49. Two natural numbers whose difference is 66 and the least common multiple is 360, are:
(a) 120 and 54
(b) 90 and 24
(c) 180 and 114
(d) 130 and 64

Answer/ Explanation

Answer: b
Explaination:
Difference of 90 and 24 = 66
and LCM of 90 and 24 = 360
∴ Numbers are 90 and 24.


50. HCF of 52 x 32 and 35 x 53 is:
(a) 53 x 35
(b) 5 x 33
(c) 53 x 32
(d) 52 x 32

Answer/ Explanation

Answer: d
Explaination:
HCF of 5²x3² and 35x53
= 5²x3²


51. A number 10x + y is multiplied by another number 10a + b and the result comes as 100p + 10q + r, where r = 2y, q = 2(x + y) and p = 2x; y < 5, q ≠ 0. The value of 10a + b may be _______.

Answer/ Explanation

Answer:
Explaination:
(10x + y)(10a + b) = 100p + 10q + r
⇒ (10x + y)(10a + b) = 100x2x + 10x2(x + y) + 2y
⇒ (10x + y)(lOa + b) = 200x + 20(x + y) + 2y
⇒ (10x + y)(10a + b) = 220x + 22y
⇒ (10x + y)(\0a + b) = 22(10x + y)
⇒ 10a + b = 22


52. If the HCF of 55 and 99 is expressible in the form 55m - 99, then the value of m is _______.

Answer/ Explanation

Answer:
Explaination:
55 = 5x11, 99 = 9x11
∴ HCF(55, 99) = 11
ATQ, 55m - 99 = 11
55x2-99 = 11
m = 2


53. Euclid's division lemma states for any two positive integers a and b, there exists integers q and r such that a = bq + r. If a = 5, b = 8, then write the value of q and r.

Answer/ Explanation

Answer:
Explaination:
Using Euclid's division lemma, we get
a = bq + r
5 = 8x0 + 5
q = 0 and r = 5


54. If a and b are two positive integers such that a = 14b. Find the HCF of a and b.

Answer/ Explanation

Answer:
Explaination:
We can write
a = 14b + 0
∵ remainder is 0
∴ HCF is b.


55. Find HCF of 1001 and 385.

Answer/ Explanation

Answer:
Explaination:
1001 =385x2 + 231
385 =231xl + 154
231 =154x1 + 77
154 = 77x2 + 0
∴ HCF =77


56. 4 Bells toll together at 9.00 am. They toll after 7, 8, 11 and 12 seconds respectively. How many times will they toll together again in the next 3 hours?
(a) 3
(b) 4
(c) 5
(d) 6

Answer/ Explanation

Answer: c
Explaination:
LCM of 7, 8, 11, 12 = 1848
∴ Bells will toll together after every 1848 sec.
∴ In next 3 hrs, number of times the bells will toll together
=\frac{3 \times 3600}{1848} = 5.84
⇒ 5 times.


57. Given that LCM (91, 26) = 182, then HCF (91, 26) is _______.

Answer/ Explanation

Answer:
Explaination:
LCM (91, 26)xHCF (91, 26) = 91x26
182xHCF (91, 26) = 91x26
⇒ HCF (91, 26) = =\frac{91 \times 26}{182}
⇒ HCF (91, 26) = 13


58. Find the LCM of smallest prime and smallest odd composite natural number.

Answer/ Explanation

Answer:
Explaination:
Smallest prime number = 2
Smallest composite odd number = 9
LCM of 2 and 9 = 2x9 = 18


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