The decimal expansion of the rational number will terminate after how many places of decimals? (2013)
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Write the decimal form of
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Nonterminating nonrepeating.
Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.
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Algorithm
398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527
HCF of 391, 425, 527 = 17
Express 98 as a product of its primes.
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2 × 7^{2}
If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.
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HCF of 408 and 1032 is 24.
1032 × 2 + 408 × (p) = 24
408p = 24 – 2064
p = 5
HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number. (2012)
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We know,
1st number × 2nd number = HCF × LCM
⇒ 27 × 2nd number = 9 × 459
⇒ 2nd number = = 153
Find HCF and LCM of 13 and 17 by prime factorisation method. (2013)
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13 = 1 × 13; 17 = 1 × 17
HCF = 1 and LCM = 13 × 17 = 221
Find LCM of numbers whose prime factorisation are expressible as 3 × 5^{2 }and 3^{2 }× 7^{2 }(2014)
Answer :
LCM (3 × 5^{2}, 3^{2} × 7^{2}) = 3^{2} × 5^{2} × 7^{2} = 9 × 25 × 49 = 11025
Find the LCM of 96 and 360 by using fundamental theorem of arithmetic. (2012)
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96 = 2^{5} × 3
360 = 2^{3} × 3^{2} × 5
LCM = 2^{5} × 3^{2} × 5 = 32 × 9 × 5 = 1440
Find the HCF (865, 255) using Euclid’s division lemma. (2013)
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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
Prove that √5 is irrational and hence show that 3 + √5 is also irrational. (2012)
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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.
Prove that 3 + 2√3 is an irrational number. (2014)
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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.
Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together? (2013)
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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.
Two tankers contain 850 liters and 680 liters of petrol. Find the maximum capacity of a container which can measure the petrol of each tanker in the exact number of times. (2012)
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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.
The length, breadth, and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly. (2015)
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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
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 
Answer :
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… 
Answer :
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 
Answer :
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 
Answer :
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 
Answer :
1B, 2A, 3D, 4C
The chapter discusses the real numbers and their applications. The divisibility of integers using Euclid’s division algorithm says that any positive integer a can be divided by another positive integer b such that the remainder will be smaller than b. On the other hand, The Fundamental Theorem of Arithmetic works on multiplication of positive integers. 10th Maths solution for CBSE makes you prepare for the examination questions and answers.
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.
For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.