Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers

Ans: (d)
(a)
But  is not the smallest among all options. 
(b)  is irrational 
(c)  is irrational 
(d)

Hence, option (d) is correct.

Ans: (c)
The numbers p and q are prime numbers, 
∴ HCF (p, q) = 1 
Here, LCM(p, q) = 221 
∴ As, p > q
p = 17, q = 13 
(As p × q = 221) 
Now, 3p – q = 3 × 17 – 13 
= 51 – 13
= 38

Ans:(c)
Here √3 and √27 both are irrational numbers.
The product of

∴ 9 is a rational number.

Ans:(a)
HCF(2520, 6600) = 40
LCM(2520, 6600) = 252 × k 
∴ HCF × LCM = Ist No. × IInd No. 
∴ 40 × 252 × k = 2520 × 6600
⇒
⇒ k = 1650

Ans:
(A)

So least prime no. used by students = 3(because 2 is announced by the teacher, so the least number used by the students is 3)
(B)As the last student got  173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11
there are 7 factors other than 2, which is announced by teacher. So, Number of student = 7
OR
Highest prime number used by student = 11
(C)Prime number 5 is used maximum times i.e., 3 times.

Sol:Least composite number = 4
Least prime number = 2
∴ HCF = 2, LCM = 4
∴ Required ratio = HCF / LCM = 2/4
i.e. 1 : 2

Ans:55

Given, least number which when divided by 12, 16 and 24 leaves remainder 7 in each case
∴  Least number = LCM(12, 16, 24) + 7
= 48 + 7
= 55

Ans:30

Let the two numbers be 2x and 3x
LCM of 2x and 3x = 6x, HCF(2x, 3x) = x
Now, 6x = 180
⇒ x = 180/6
x = 30

Ans: Let us assume that √3 is a rational number.

Then √3 = a/b; where a and b ( ≠ 0) are co-prime positive integers.
Squaring on both sides, we get
3 = a²/b²
⇒ a² = 3b² 
⇒ 3 divides a²
⇒ 3 divides a   _________(i)
= a = 3c, where c is an integer
Again, squaring on both sides, we get
a² = 9c²
⇒3b² = 9c²
⇒b² = 3c²
⇒ 3 divides b²
⇒ 3 divides b   _________(ii)
From (i) and (ii), we get 3 divides both a and b.
⇒ a and b are not co- prime integers.
This contradicts the fact that a and b are co-primes.
Hence,  √3 is an irrational number.

Ans:(a)
Sol:Given, HCF = 12
Let two numbers be 12a and 12b
So. 12a x 12b = 6336 
⇒ ab = 44
We can write 44 as product of two numbers in these ways:
ab = 1 x 44 = 2 x 22 = 4x  11
Here, we will take a = 1 and b = 44 ; a = 4 and b = 11.
We do not take ab = 2  x 22 because 2 and 22 are not co-prime to each other.

For a = 1 and b = 44, 1^st no. = 12a = 12, 2^nd no. = 12b = 528
For a = 4 and b = 11, 1^st no. = 12a = 48, 2^nd no. = 12b = 132
Hence, we get two pairs of numbers, (12, 528) and (48, 132).

Ans: (d)
Sol: for n = 1, 2,  3, 4...
(12)ⁿ cannot end with 0.

Ans: (b)
Sol:We have,

∴ Prime factorisation of 385 = 5 x 7 x 11

Ans: We have, 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5.
We can write these numbers as:
2 x 3 x 5 + 5 = 5(2 x 3 + 1)
= 1 x 5 x 7
and 5 x 7 x 11 + 7 x 5 = 5 x 7(11 + 1)
= 5 x 7 x 12 = 1 x 5 x 7 x 12
Since, on simplifying. we find that both the numbers have more than two factors. 
So. these are composite numbers.

Ans: (c)
Sol:We have, 
12 = 2 x 2 x 3 = 2² x 3
21= 3 x 7
15 = 3 x 5
∴ HCF (12, 21, 15) = 3
and LCM (12, 21 ,15 ) = 2² x 3 x 5 x 7
= 420

Ans: 91
Let the other number be x
As, HCF (a, b) x  LCM (a, b) = a  x  b
⇒ 13  x  182= 26x
⇒ x = 13 x 182 / 26
= 91
Hence, other number is 91.

Ans: We know that
LCM × HCF = Product of two numbers
∴ LCM (135, 225) = Product of 135 and 225 / HCF(135, 225)
= 135 x 225 / 45
= 675
So, LCM (135, 225) = 675

Ans: 3024
Since. HCF (a, b) x LCM (a, b) = a x b
∴ HCF (336, 54) x  LCM (336, 54) = 336 x 54
⇒ 6 x LCM(336, 54) = 18144
⇒ LCM (336, 54) = 18144 / 6
= 3024

Ans: 1000
We know that HCF (a, b) x  LCM (a, b)=a x b
⇒ 5 x 200 = ab
⇒  ab = 1000

Ans: 2

Since,  HCF (65 ,117) = 13
Given HCF ( 65, 117 ) = 65n - 117
13 = 65n - 117
⇒  65n = 13 +117
⇒  n = 2

Ans:18

Prime factorisation of 612 and 1314 are
612 = 2 x 2 x 3 x 3 x 17
1314 = 2 x 3 x 3 x 73
∴ HCF (612, 1314) = 2 x 3 x 3
= 18

Ans: Let us assume that √5 is a rational number.
Then √5 = a/b where a and b (≠ 0} are co-prime integers,
if Squaring on both sides, we get

⇒ 5 divides a²
⇒ 5 divides a  ----------(i)
⇒ a = 5c, where c is an integer
Again, squaring on both sides, we get
a² = 25c²
⇒ 5b² = 25c²  
⇒ b² = 5c²
⇒ 5 divides b² ----------(ii)
⇒ 5 divides b
From (i) and {ii), we get 5 divides both a and b.
⇒  a and b are not co-prime integers.
Hence, our supposition is wrong.
Thus, √5 is an irrational number.

Ans:Let us assume √2 be a rational number.
Then, √2 = p/q where p, q (q ≠ 0) are integers and co-prime. ;
On squaring both sides. we get
------------(i)
⇒ 2 divides p²
⇒ 2 divides p -----------(ii)
So, p = 2a, where a is some integer.
Again squaring on both sides, we get
p² = 4a²
⇒ 2q² = 4a² (using (i))
⇒ q² = 2a²
⇒ 2 divides q²
⇒ 2 divides q       -----------(iii)
From (ii) and (iii), we get
2 divides both p and q.
∴ p and q are not co-prime integers.
Hence, our assumption is wrong.
Thus √2 is an irrational number.

Ans:Suppose 2 + 5√3 is a rational number.
We can find two integers a, b (b ≠ 0) such that
2 + 5√3 = a/b, where a and b are co -prime integers.

⇒ √3 is a rational number.

[ ∵ a, b are integers, so  is a rational number]
But this contradicts the fact that √3 is an irrational number.
Hence, our assumption is wrong.
Thus, 2 + 5√3 is an irrational number.

Ans:Given numbers are 306 and 657. 
The smallest number divisible by 306 and 657 = LCM(306, 657) 
Prime factors of 306 = 2 × 3 × 3 × 17 
Prime factors of 657 = 3 × 3 × 73 
LCM of (306, 657) = 2 × 3 × 3 × 17 × 73 
= 22338 
Hence, the smallest number divisible by 306 and 657 is 22,338.