જવાબ :
જવાબ : Nonterminating nonrepeating.
જવાબ : Algorithm
398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527
HCF of 391, 425, 527 = 17
જવાબ : 2 × 7^{2}
જવાબ : HCF of 408 and 1032 is 24.
1032 × 2 + 408 × (p) = 24
408p = 24 – 2064
p = 5
જવાબ : We know,
1st number × 2nd number = HCF × LCM
⇒ 27 × 2nd number = 9 × 459
⇒ 2nd number = = 153
જવાબ : 13 = 1 × 13; 17 = 1 × 17
HCF = 1 and LCM = 13 × 17 = 221
જવાબ : LCM (3 × 5^{2}, 3^{2} × 7^{2}) = 3^{2} × 5^{2} × 7^{2} = 9 × 25 × 49 = 11025
જવાબ : It is given that on dividing 70 by the required number, there is a remainder 5.
This means that 70 – 5 = 65 is exactly divisible by the required number.
Similarly, 125 – 8 = 117 is also exactly divisible by the required number.
65 = 5 × 13
117 = 3^{2} × 13
HCF = 13
Required number = 13
જવાબ : 96 = 2^{5} × 3
360 = 2^{3} × 3^{2} × 5
LCM = 2^{5} × 3^{2} × 5 = 32 × 9 × 5 = 1440
જવાબ : 865 > 255
865 = 255 × 3 + 100
255 = 100 × 2 + 55
100 = 55 × 1 + 45
55 = 45 × 1 + 10
45 = 10 × 4 + 5
10 = 5 × 2 + 0
The remainder is 0.
HCF = 5
જવાબ : Let x = …(i)
100x = 612. …(ii)
…[Multiplying both sides by 100]
Subtracting (i) from (ii),
99x = 606
x = =
Denominator = 33
Prime factorisation = 3 × 11
જવાબ : y = 5 × 13 = 65
x = 3 × 195 = 585
જવાબ : Let us assume, to the contrary, that 2 + 3√5 is rational.
So that we can find integers a and b (b ≠ 0).
Such that 2 + 3√5 = , where a and b are coprime.
Rearranging the above equation, we get
Since a and b are integers, we get is rational and so √5 is rational.
But this contradicts the fact that √5 is irrational.
So, we conclude that 2 + 3√5 is irrational.
જવાબ : Let us assume, to the contrary, that 3√7 is rational.
That is, we can find coprime a and b (b ≠ 0) such that 3√7 =
Rearranging, we get √7 =
Since 3, a and b are integers, is rational, and so √7 is rational.
But this contradicts the fact that √7 is irrational.
So, we conclude that 3√7 is irrational.
જવાબ : 17 × 5 × 11 × 3 × 2 + 2 × 11 …(i)
= 2 × 11 × (17 × 5 × 3 + 1)
= 2 × 11 × (255 + 1)
= 2 × 11 × 256
Number (i) is divisible by 2, 11 and 256, it has more than 2 prime factors.
Therefore (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.
જવાબ : 4^{n} = (2^{2})n = 2^{2n}
The only prime in the factorization of 4^{n} is 2.
There is no other prime in the factorization of 4^{n} = 2^{2n}
(By uniqueness of the Fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of 4^{n} for any n.
Therefore, 4^{n} does not end with the digit zero for any natural number n.
જવાબ : No, LCM = Product of the highest power of each factor involved in the numbers.
HCF = Product of the smallest power of each common factor.
We can conclude that LCM is always a multiple of HCF, i.e., LCM = k × HCF
We are given that,
LCM = 175 and HCF = 15
175 = k × 15
⇒ 11.67 = k
But in this case, LCM ≠ k × HCF
Therefore, two numbers cannot have LCM as 175 and HCF as 15.
જવાબ : Irrational No.
જવાબ : Neither Rational nor Irrational no.
જવાબ : Irrational No.
જવાબ : Rational No.
જવાબ : Irrational No.
જવાબ : Rational No.
જવાબ : Irrational No.
જવાબ : RationalNo.
જવાબ : Rational
જવાબ : quotient
જવાબ : Euclid
જવાબ : remainder
જવાબ : Composite
જવાબ : Prime No.
જવાબ : Terminating
જવાબ : (√2−√3)(√3+√2)(2−3)(3+2)
=(√2−√3)(√2+√3)=(2−3)(2+3)
=23=1
So, it is rational number
જવાબ : 29029=29×13×11×729029=29×13×11×7
580=29×5×4580=29×5×4
HCF=29
LCM=29×13×11×7×4×5=58058029×13×11×7×4×5=580580
જવાબ : 120=2×2×3×2×5120=2×2×3×2×5
144=2×2×3×2×2×3144=2×2×3×2×2×3
HCF=23×3=24HCF=23×3=24
LCM=720
જવાબ : 27=3×3×327=3×3×3
81=3×3×3×381=3×3×3×3
HCF=27
LCM=81
જવાબ : 2 & 5
જવાબ : axb
જવાબ : 54
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : False
જવાબ : True
જવાબ : True
જવાબ : False
જવાબ : Let us assume, to the contrary, that √5 is rational.
So, we can find integers p and q (q ≠ 0), such that
√5 = , where p and q are coprime.
Squaring both sides, we get
5 =
⇒ 5q^{2} = p^{2} …(i)
⇒ 5 divides p^{2}
5 divides p
So, let p = 5r
Putting the value of p in (i), we get
5q^{2} = (5r)^{2}
⇒ 5q^{2 }= 25r^{2}
⇒ q^{2} = 5r^{2}
⇒ 5 divides q^{2}
5 divides q
So, p and q have atleast 5 as a common factor.
But this contradicts the fact that p and q have no common factor.
So, our assumption is wrong, is irrational.
√5 is irrational, 3 is a rational number.
So, we conclude that 3 + √5 is irrational.
જવાબ : Let us assume to the contrary, that 3 + 2√3 is rational.
So that we can find integers a and b (b ≠ 0).
Such that 3 + 2√3 = , where a and b are coprime.
Rearranging the equations, we get
Since a and b are integers, we get is rational and so √3 is rational.
But this contradicts the fact that √3 is irrational.
So we conclude that 3 + 2√3 is irrational.
જવાબ : 9 = 3^{2}, 12 = 2^{2} × 3, 15 = 3 × 5
LCM = 2^{2} × 3^{2} × 5 = 4 × 9 × 5 = 180 minutes or 3 hours
They will next toll together after 3 hours.
જવાબ : To find the maximum capacity of a container which can measure the petrol of each tanker in the exact number of times, we find the HCF of 850 and 680.
850 = 2 × 5^{2} × 17
680 = 2^{3} × 5 × 17
HCF = 2 × 5 × 17 = 170
Maximum capacity of the container = 170 liters.
જવાબ : To find the length of the longest rod that can measure the dimensions of the room exactly, we have to find HCF.
L, Length = 8 m 50 cm = 850 cm = 2^{1} × 5^{2} × 17
B, Breadth = 6 m 25 cm = 625 cm = 5^{4}
H, Height = 4 m 75 cm = 475 cm = 5^{2} × 19
HCF of L, B and H is 5^{2} = 25 cm
Length of the longest rod = 25 cm
જવાબ : To find the time when the clocks will next ring together,
we have to find LCM of 4, 12 and 20 minutes.
4 = 2^{2}
12 = 2^{2} × 3
20 = 2^{2} × 5
LCM of 4, 12 and 20 = 2^{2} × 3 × 5 = 60 minutes.
So, the clocks will ring together again after 60 minutes or one hour.
જવાબ : Since the books are to be distributed equally among the students of Section A and Section B. therefore, the number of books must be a multiple of 48 as well as 60.
Hence, required num¬ber of books is the LCM of 48 and 60.
48 = 2^{4} × 3
60 = 2^{2} × 3 × 5
LCM = 2^{4} × 3 × 5 = 16 × 15 = 240
Hence, required number of books is 240.
જવાબ :
જવાબ : Given numbers are 650 and 1170.
1170 > 650
1170 = 650 × 1 + 520
650 = 520 × 1 + 130
520 = 130 × 4 + 0
HCF = 130
The required largest number is 130.
જવાબ : 867 is greater than 255. We apply the division lemma to 867 and 255, to get
867 = 255 × 3 + 102
We continue the process till the remainder is zero
255 = 102 × 2 + 51
102 = 51 × 2 + 0, the remainder is zero.
HCF = 51
જવાબ : Let us assume, to the contrary, that 3 + 2√5 is rational
So that we can find integers a and b (b ≠ 0), such that
3 + 2 √5 = , where a and b are coprime.
Rearranging this equation, we get
Since a and b are integers, we get that – is rational and so √5 is rational.
But this contradicts the fact that √5 is irrational.
So we conclude that 3 + 2√5 is irrational.
જવાબ : (a) Find the maximum number of parallel rows of each class for the seating arrange¬ment.
(b) Also, find the number of students of class IX and also of class X in a row.
(c) What is the objective of the school administration behind such an arrangement?
Solution:
104 = 2^{3} × 13
96 = 2^{5} × 3
HCF = 2^{3} = 8
(a) Number of rows of students of class X = = 13
Number maximum of rows class IX = = 12
Total number of rows = 13 + 12 = 25
(b) No. of students of class IX in a row = 8
No. of students of class X in a row = 8
(c) The objective of school administration behind such an arrangement is fair and clean examination, so that no student can take help from any other student of his/her class.
જવાબ : 1st vessel = 720 ml; 2nd vessel = 405 ml
We find the HCF of 720 and 405 to find the maximum quantity of milk to be filled in one glass.
405 = 3^{4} × 5
720 = 2^{4} × 3^{2} × 5
HCF = 3^{2} × 5 = 45 ml = Capacity of glass
No. of glasses filled from 1st vessel = = 16
No. of glasses filled from 2nd vessel = = 9
Total number of glasses = 25
જવાબ : To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.
10 = 2 × 5
16 = 2^{4}
20 = 2^{2} × 5
LCM = 2^{4} × 5 = 16 × 5 = 80 minutes
They will start preparing a new card together after 80 minutes.
જવાબ : First, we find HCF of 6339 and 6341 by Euclid’s division method.
6341 > 6339
6341 = 6339 × 1 + 2
6339 = 2 × 3169 + 1
2 = 1 × 2 + 0
HCF of 6341 and 6339 is 1.
Now, we find the HCF of 134791 and 1
134791 = 1 × 134791 + 0
HCF of 134791 and 1 is 1.
Hence, the HCF of the given three numbers is 1.
જવાબ : x = p^{2}q^{3} and y = p^{3}q
LCM = p^{3}q^{3}
HCF = p^{2}q …..(i)
Now, LCM = p^{3}q^{3}
⇒ LCM = pq^{2} (p^{2}q)
⇒ LCM = pq^{2} (HCF)
Yes, LCM is a multiple of HCF.
Explanation:
Let a = 12 = 2^{2} × 3
b = 18 = 2 × 3^{2}
HCF = 2 × 3 = 6 …(ii)
LCM = 2^{2} × 3^{2} = 36
LCM = 6 × 6
LCM = 6 (HCF) …[From (ii)]
Here LCM is 6 times HCF.
જવાબ : Let n, n + 1, n + 2 be three consecutive positive integers.
We know that n is of the form 3q, 3q + 1, or 3q + 2.
Case I. When n = 3q,
In this case, n is divisible by 3,
but n + 1 and n + 2 are not divisible by 3.
Case II. When n = 3q + 1,
In this case n + 2 = (3q + 1) + 2
= 3q + 3
= 3(q + 1 ), (n + 2) is divisible by 3,
but n and n + 1 are not divisible by 3.
Case III.
When n = 3q + 2, in this case,
n + 1 = (3q + 2) + 1
= 3q + 3 = 3 (q + 1 ), (n + 1) is divisible by 3,
but n and n + 2 are not divisible by 3.
Hence, one and only one out of n, n + 1 and n + 2 is divisible by 3.
જવાબ : 306 = 2 × 3^{2} × 17
657 = 32 × 73
HCF = 3^{2} = 9
LCM = 2 × 3^{2} × 17 × 73 = 22338
L.H.S. = LCM × HCF = 22338 × 9 = 201042
R.H.S. = Product of two numbers = 306 × 657 = 201042
L.H.S. = R.H.S.
જવાબ : Let a be a positive odd integer
By Euclid’s Division algorithm:
a = 4q + r …[where q, r are positive integers and 0 ≤ r < 4]
a = 4q
or 4q + 1
or 4q + 2
or 4q + 3
But 4q and 4q + 2 are both even
a is of the form 4q + 1 or 4q + 3.
જવાબ : Let x be any positive integer and y = 3.
By Euclid’s division algorithm; x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3) Therefore, x = 3q, 3q + 1 and 3q + 2 As per the given question, if we take the square on both the sides, we get; x^{2} = (3q)^{2} = 9q^{2} = 3.3q^{2} Let 3q^{2} = n Therefore, x^{2} = 3n ………………….(1) x^{2} = (3q + 1)^{2} = (3q)^{2} + 1^{2} + 2 × 3q × 1 = 9q^{2} + 1 + 6q = 3(3q^{2} + 2q) + 1 Substitute, 3q^{2}+2q = n, to get, x^{2} = 3n + 1 ……………………………. (2) x^{2} = (3q + 2)^{2} = (3q)^{2 }+ 2^{2 }+ 2 × 3q × 2 = 9q^{2} + 4 + 12q = 3(3q^{2} + 4q + 1) + 1 Again, substitute, 3q^{2 }+ 4q + 1 = m, to get, x^{2} = 3n + 1…………………………… (3) Hence, from eq. 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3n or 3n + 1 for some integer n.જવાબ : (i) 1400
140 = 2 × 2 × 2 × 5 x 5 × 7 × 1
1 
Rational No 
A 
Root of 1 
2 
Irrational No. 
B 
Rational & Irrational no. 
3 
Real No. 
C 
1/3 
4 
NonReal No. 
D 
Square root of 2 
જવાબ :
1C , 2D, 3B, 4A
1 
Whole No. 
A 
1/7 
2 
Natural No 
B 
0,1,2,3…… 
3 
Integers. 
C 
1,2,3,4…… 
4 
Fractions 
D 
…2,1,0,1,2… 
જવાબ :
1B, 2C, 3D, 4A
1 
Even No. 
A 
4,6,8,9,12 
2 
Odd No . 
B 
2,3,5,7,11 
3 
Prime No. 
C 
1,3,5,7,11 
4 
Composite No. 
D 
2,4,6,8,10 
જવાબ :
1D, 2C, 3B, 4A

No. 

H.C.F. 
1 
6, 12, 15 
A 
1 
2 
4, 8, 16 
B 
3 
3 
10, 25, 50 
C 
4 
4 
2, 3, 4 
D 
5 
જવાબ :
1B,. 2C, 3D, 4A

No. 

L.C.M 
1 
12, 13 
A 
150 
2 
100, 15 
B 
156 
3 
50, 60 
C 
66 
4 
2, 33 
D 
300 
જવાબ :
1B, 2A, 3D, 4C
1 
Rational No 
A 
√2 
2 
Irrational No. 
B 
√2 & √4 
3 
Real No. 
C 
√4 
4 
NonReal No. 
D 
√4 
જવાબ :
1D, 2A, 3B, 4C
1 
Even No. 
A 
32,34,35,36 
2 
Odd No . 
B 
31,37,41,43 
3 
Prime No. 
C 
32,34,36 
4 
Composite No. 
D 
31,33,35,37 
જવાબ :
1C, 2D, 3B, 4A

No. 

H.C.F. 
1 
33,100 
A 
4 
2 
44, 46 
B 
3 
3 
3, 39 
C 
2 
4 
8,12 
D 
1 
જવાબ :
1D, 2C, 3B, 4A

No. 

L.C.M. 
1 
33,100 
A 
24 
2 
44, 46 
B 
3300 
3 
3, 39 
C 
903 
4 
8,12 
D 
39 
જવાબ :
1B, 2C, 3D, 4A

No. 

H.C.F. 
1 
12, 13 
A 
5 
2 
100, 15 
B 
2 
3 
50, 60 
C 
1 
4 
2, 32 
D 
10 
જવાબ :
1C,2A,3D,4B
Math
Chapter 01 : Real Numbers
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