Why mcqs of class 10 Maths lesson 1 Real numbers is important ?
- these mcqs are important because these questions cover a wide topic of the chapter
- mcqs are easy to understand
- mcqs break the big topic in some small topic so they are easy to learn
- they give a short revision at the time of class 10 Maths exam
How to get pdf of class 10 Maths lesson 1 Real numbers mcqs
Increasing role of mcqs in board exam
MCQ’S TEST
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CLASS 10 MATHS LESSON 1 REAL NUMBERS OBJECTIVE QUESTIONS MCQ’S TEST
1. A number when divided by 61 gives 27 as quotient and 32 as the remainder, then the number is:
1967
1569
1679
1796
Answer - C
Explanation - The dividend is equal to Divisor × Quotient + Remainder
Number(dividend) = D × Q + R
∴ the number (Dividend) = 61 × 27 + 32
= 1647 + 32
= 1679
2. Every positive even integer is of the form ________ for some integer ‘q’.
2q - 1
none of these
2q
2q + 1
Answer - C
Explanation - 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
3. Every positive odd integer is of the form ________ where ‘q’ is some integer.
2q + 2
3q + 1
2q + 1
5q + 1
Answer - C
Explanation - 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 consecutive integers, therefore, 2q + 1 is an odd integer.
4. For any two positive integers a and b, such that a > b. There exist (unique) whole numbers q and r such that
a = bq + r, 0 ≤ r < b
b = aq + r, 0 ≤ r < b
a = qbr
q = ar + b, 0 ≤ r < b
Answer - A
Explanation - 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
5. For any positive integer a and 3, there exist unique integers q and r such that a = 3q + r where r must satisfy
0 < r < 3
1 < r < 3
0 < r ≤ 3
0 ≤ r < 3
Answer - D
Explanation - Since a is a positive integer, therefore, r = 0, 1, 2 only.
So, that a = 3q, 3q + 1, 3q + 2.
6. If 112 = q × 6 + r, then the possible values of r, are:
0, 1, 2, 3
1, 2, 3, 4
0, 1, 2, 3, 4, 5
2, 3, 5
Answer - C
Explanation - For the relation x = qy + r, 0 ≤ r < y
So, here r lies between 0 ≤ r < 6
Hence, r = 0, 1, 2, 3, 4, 5
7. By Euclid’s division lemma x = qy + r, x > y the value of q and r for x = 27 and y = 5 are:
q = 5, r = 3
q = 6, r = 3
cannot be determined
q = 5, r = 2
Answer - D
Explanation - x = qy + r
⇒ 27 = 5 × 5 + 2
⇒ q = 5, r = 2
8. A die is thrown once. Find the probability of getting a number between 3 and 6.
2/3
1/2
1/3
1/4
Answer - C
Explanation - Total number of outcomes = {1, 2, 3, 4, 5, 6} = 6
Favourable outcomes in this case = {4, 5} = 2
∴ P(a number between 3 and 6) = Favourable outcomes/Total outcomes = 2/6 = 1/3
9. Any ________ is of the form 4q + 1 or 4q + 3 for some integer q.
positive even integer
positive odd integer
composite number
prime number
Answer - B
Explanation - Let a be a given positive odd integer.
Applying Euclid’s Division Lemma to a and b = 4
We have, a = 4q + r where 0 ≤ r < 4
⇒ r = 0, 1, 2, 3
⇒ a = 4q or 4q + 1 or 4q + 2 or 4q + 3
But a = 4q and 4q + 2 = 2 (2q + 1) are clearly even.
Also a = 4q, 4q + 1, 4q + 2, 4q + 3 are consecutive integers,
therefore any positive odd integer is of the form 4q + 1 and 4q + 3
where q is some integer
10. The product of three consecutive positive integers is divisible by
6
5
4
10
Answer - A
Explanation - Let n be a positive integer,
then three consecutive positive integers are (n+1)(n+2)(n+3) = n(n+1)(n+2)+3(n+1)(n+2)
Here, the first term is divisible by 6 and the second term is also divisible by 6
Because it contains a factor 3 and one of the two consecutive integers (n+1) or (n+2) is even and thus is divisible by 2
Therefore, the sum of multiple of 6 is also a multiple of 6.
11. The least number n so that 5n is divisible by 3, where n is:
no natural number
a real number
a natural number
a whole number
Answer - A
Explanation - Since 5 is a prime number so it is not divisible by 3.
Therefore there is no natural number n such that 5n is divisible by 3.
12. For every positive integer n, n2 - n is divisible by
4
6
8
2
Answer - D
Explanation - n2 - n = n(n - 1). Since n and (n - 1) are consecutive integers. Therefore, one of them must be divisible by 2.
13. A card is drawn at random from a pack of 52 cards. The probability that the card is drawn is a jack, a queen or a king is
3/13
1/13
1/26
11/13
Answer - A
Explanation - Total number of outcomes = 52
Favourable outcomes in this case = 4 + 4 + 4 = 12 {4 jacks, 4 queens, 4 kings}
∴ P(a jack, a queen or a king) = Favourable outcomesTotal outcomes = 1252 = 313
14. If m2 - 1 is divisible by 8, then m is
natural number
an odd integer
an even integer
a whole number
Answer - B
Explanation - Let a = m2 - 1
Here m can be even or odd.
Case I: m = Even i.e., m = 2k, where k is an integer,
⇒ a = (2k)2 - 1
⇒ a = 4k2 - 1
At k = -1, = 4 (-1)2 - 1 = 4 -1 = 3, which is not divisible by 8.
At k = 0, a = 4 (0)2 - 1 = 0 - 1 = -1, which is not divisible by 8, which is not.
Case II: m = Odd i.e., m = 2k + 1, where k is an odd integer.
⇒ a = 2k + 1
⇒ a = (2k + 1)2 - 1
⇒ a 4k2 + 4k + 1 - 1
⇒ a = 4k2 + 4k
⇒ a = 4k(k + 1)
At k = -1, a = 4(-1)(-1 + 1) = 0 which is divisible by 8.
At k = 0, a = 4(0)(0 + 1) = 4 which is divisible by 8.
At k = 1, a = 4(1)(1 + 1) = 8 which is divisible by 8.
Hence, we can conclude from the above two cases, if m is odd, then m2 - 1 is divisible by 8.
15. If two positive integers m and n can be expressed as m = x2y5 and n = x3y2, where x and y are prime numbers, then HCF(m, n) =
x2y3
x3y3
x3y2
X2y2
Answer - D
Explanation - x2y5 = y3(x2y2)
x3y3 = x(x2y2)
Therefore HCF (m, n) is x2y2