Free PDF Download of CBSE Class 10 Maths Chapter 2 Polynomials 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 Polynomials MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 2 Polynomials

1. If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) -5

2. Given that two of the zeroes of the cubic poly-nomial ax³ + bx² + cx + d are 0, the third zero is

3. If one of the zeroes of the quadratic polynomial (k – 1) x² + kx + 1 is – 3, then the value of k is

4. A quadratic polynomial, whose zeroes are -3 and 4, is
(a) x²- x + 12
(b) x² + x + 12
(c) (frac{x^{2}}{2}-frac{x}{2}-6)
(d) 2x² + 2x – 24

5. If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a – 0, b = -6

6. The number of polynomials having zeroes as -2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3

7. Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is

8. If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is -1, then the product of the
other two zeroes is
(a) b – a + 1
(b) b – a – 1
(c) a – b + 1
(d) a – b – 1

9. The zeroes of the quadratic polynomial x2 + 99x + 127 are
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal

10. The zeroes of the quadratic polynomial x² + kx + k, k? 0,
(a) cannot both be positive
(b) cannot both be negative
(c) are always unequal
(d) are always equal

11. If the zeroes of the quadratic polynomial ax² + bx + c, c # 0 are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have the same sign
(d) c and b have the same sign

12. If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.

13. Which of the following is not the graph of quadratic polynomial?

14. The number of polynomials having zeroes as 4 and 7 is
(a) 2
(b) 3
(c) 4
(d) more than 4

15. A quadratic polynomial, whose zeores are -4 and -5, is
(a) x²-9x + 20
(b) x² + 9x + 20
(c) x²-9x- 20
(d) x² + 9x- 20

16. The zeroes of the quadratic polynomial x² + 1750x + 175000 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

17. The zeroes of the quadratic polynomial x² – 15x + 50 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

18. The zeroes of the quadratic polynomial 3x² – 48 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

19. The zeroes of the quadratic polynomial x² – 18x + 81 are
(a) both negative
(b) one positive and one negative
(c) both positive and unequal
(d) both equal and positive

20. The zeroes of the quadratic polynomial x² + px + p, p ≠ 0 are
(a) both equal
(b) both cannot be positive
(c) both unequal
(d) both cannot be negative

21. If one of the zeroes of the quadratic polynomial (p – l)x² + px + 1 is -3, then the value of p is

22. If the zeroes of the quadratic polynomial Ax² + Bx + C, C # 0 are equal, then
(a) A and B have the same sign
(b) A and C have the same sign
(c) B and C have the same sign
(d) A and C have opposite signs

23. If x³ + 1 is divided by x² + 5, then the possible degree of quotient is
(a) 0
(b) 1
(c) 2
(d) 3

24. If x³ + 11 is divided by x² – 3, then the possible degree of remainder is
(a) 0
(b) 1
(c) 2
(d) less than 2

25. If x⁴ + 3x² + 7 is divided by 3x + 5, then the possible degrees of quotient and remainder are:
(a) 3, 0
(b) 4, 1
(c) 3, 1
(d) 4, 0

26. If x⁵ + 2x⁴ + x + 6 is divided by g(x), and quotient is x² + 5x + 7, then the possible degree of g(x) is:
(a) 4
(b) 2
(c) 3
(d) 5

27. If x⁵ + 2x⁴ + x + 6 is divided by g(x) and quo-tient is x² + 5x + 7, then the possible degree of remainder is:
(a) less than 1
(b) less than 2
(c) less than 3
(d) less than 4

28. What is the number of zeroes that a linear poly-nomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

29. What is the number(s) of zeroes that a quadratic polynomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

30. What is the number(s) of zeores that a cubic polynomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

31. If one of the zeroes of the cubic polynomial x³ + px² + qx + r is -1, then the product of the other two zeroes is
(a) p + q + 1
(b) p-q- 1
(c) q – p + 1
(d) q – p – 1

32. If one zero of the quadratic polynomial x² + 3x + b is 2, then the value of b is
(a) 10
(b) -8
(c) 9
(d) -10

33. If 1 is one of the zeroes of the polynomial x² + x + k, then the value of k is:
(a) 2
(b) -2
(c) 4
(d) -4

We hope the given MCQ Questions for Class 10 Maths Polynomials with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 2 Polynomials Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

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Get Free NCERT Solutions for Class 10 Maths Chapter 3 Ex 3.2 PDF. Pair of Linear Equations in Two Variables Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 3.2 Class 10 Maths NCERT Solutions were prepared by Experienced Teachers. Detailed answers of all the questions in Chapter 3 Maths Class 10 Pair of Linear Equations in Two Variables Exercise 3.2 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables:

• Pair Of Linear Equations In Two Variables Class 10 Ex 3.1
• प्रश्नावली 3.1 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.2
• प्रश्नावली 3.2 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.3
• प्रश्नावली 3.3 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.4
• प्रश्नावली 3.4 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.5
• प्रश्नावली 3.5 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.6
• प्रश्नावली 3.6 का हल हिंदी में
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.7
• प्रश्नावली 3.7 का हल हिंदी में
• Extra Questions for Class 10 Maths Linear Equations in Two Variables

You can also download the free PDF of  Ex 3.2 Class 10 Pair of Linear Equations in Two Variables NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Ex 3.2 Class 10 Maths Question 1.
 Form the pair of linear equations of the following problems and find their solutions graphically:
(i) 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Solution:



Worksheets for Class 10 Maths

Ex 3.2 Class 10 Maths Question 2.
On comparing the ratios (frac { { a }_{ 1 } }{ { a }_{ 2 } }), (frac { { b }_{ 1 } }{ { b }_{ 2 } })
and (frac { { c }_{ 1 } }{ { c }_{ 2 } }) , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0, 2x -y + 9 = 0
Solution:

Ex 3.2 Class 10 Maths Question 3.
On comparing the ratios (frac { { a }_{ 1 } }{ { a }_{ 2 } }), (frac { { b }_{ 1 } }{ { b }_{ 2 } })
and (frac { { c }_{ 1 } }{ { c }_{ 2 } }), find out whether the following pairs of linear equations are consistent, or inconsistent:
~
Solution:

Ex 3.2 Class 10 Maths Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x-y – 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Solution:

Ex 3.2 Class 10 Maths Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden graphically.
Solution:

Ex 3.2 Class 10 Maths Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:

Ex 3.2 Class 10 Maths Question 7.
Draw the, graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:

NCERT Solutions for Class 10 Maths Chapter 3 Pairs of Linear Equations in Two Variables (Hindi Medium) Ex 3.2

We hope the NCERT Solutions for Class 10 Maths Chapter Pair of Linear Equations in Two Variables Ex 3.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2, drop a comment below and we will get back to you at the earliest.

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Important Questions for Class 10 Maths Chapter 1 Real Numbers

Real Numbers Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
The decimal expansion of the rational number (frac { 43 }{ { 2 }^{ 4 }{ 5 }^{ 3 } }) will terminate after how many places of decimals? (2013)
Solution:

Question 2.
Write the decimal form of (frac { 129 }{ { 2 }^{ 7 }{ 5 }^{ 7 }{ 7 }^{ 5 } })
Solution:
Non-terminating non-repeating.

Question 3.
Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.
Solution:
Algorithm
398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527
HCF of 391, 425, 527 = 17

Question 4.
Express 98 as a product of its primes.
Solution:
2 × 7²

Question 5.
If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.
Solution:
HCF of 408 and 1032 is 24.
1032 × 2 + 408 × (p) = 24
408p = 24 – 2064
p = -5

Real Numbers Class 10 Important Questions Short Answer-I (2 Marks)

Question 6.
HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number. (2012)
Solution:
We know,
1st number × 2nd number = HCF × LCM
⇒ 27 × 2nd number = 9 × 459
⇒ 2nd number = (frac { 9times 459 }{ 27 }) = 153

Question 7.
Find HCF and LCM of 13 and 17 by prime factorisation method. (2013)
Solution:
13 = 1 × 13; 17 = 1 × 17
HCF = 1 and LCM = 13 × 17 = 221

Question 8.
Find LCM of numbers whose prime factorisation are expressible as 3 × 5² and 3² × 7². (2014)
Solution:
LCM (3 × 5², 3² × 7²) = 3² × 5² × 7² = 9 × 25 × 49 = 11025

Question 9.
Find the LCM of 96 and 360 by using fundamental theorem of arithmetic. (2012)
Solution:
96 = 2⁵ × 3
360 = 2³ × 3² × 5
LCM = 2⁵ × 3² × 5 = 32 × 9 × 5 = 1440

Question 10.
Find the HCF (865, 255) using Euclid’s division lemma. (2013)
Solution:
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

Question 11.
Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively. (2015)
Solution:
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 = 3² × 13
HCF = 13
Required number = 13

Question 12.
Find the prime factorisation of the denominator of rational number expressed as (6.bar { 12 }) in simplest form. (2014)
Solution:
Let x = (6.bar { 12 }) …(i)
100x = 612.(bar { 12 }) …(ii)
…[Multiplying both sides by 100]
Subtracting (i) from (ii),
99x = 606
x = (frac { 606 }{ 99 }) = (frac { 202 }{ 33 })
Denominator = 33
Prime factorisation = 3 × 11

Question 13.
Complete the following factor tree and find the composite number x. (2014)

Solution:
y = 5 × 13 = 65
x = 3 × 195 = 585

Question 14.
Prove that 2 + 3√5 is an irrational number. (2014)
Solution:
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 = (frac { a }{ b }), where a and b are coprime.
Rearranging the above equation, we get

Since a and b are integers, we get (frac { a }{ 3b } -frac { 2 }{ 3 }) 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.

Question 15.
Show that 3√7 is an irrational number. (2016)
Solution:
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 = (frac { a }{ b })
Rearranging, we get √7 = (frac { a }{ 3b })
Since 3, a and b are integers, (frac { a }{ 3b }) is rational, and so √7 is rational.
But this contradicts the fact that √7 is irrational.
So, we conclude that 3√7 is irrational.

Question 16.
Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number? (2015)
Solution:
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.

Question 17.
Check whether 4n can end with the digit 0 for any natural number n. (2015)
Solution:
4ⁿ = (2²)ⁿ = 2²ⁿ
The only prime in the factorization of 4ⁿ is 2.
There is no other prime in the factorization of 4ⁿ = 2²ⁿ
(By uniqueness of the Fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of 4ⁿ for any n.
Therefore, 4ⁿ does not end with the digit zero for any natural number n.

Question 18.
Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons. (2017 OD)
Solution:
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.

Real Numbers Class 10 Important Questions Short Answer-II (3 Marks)

Question 19.
Prove that √5 is irrational and hence show that 3 + √5 is also irrational. (2012)
Solution:
Let us assume, to the contrary, that √5 is rational.
So, we can find integers p and q (q ≠ 0), such that
√5 = (frac { p }{ q }), where p and q are coprime.
Squaring both sides, we get
5 = (frac { { p }^{ 2 } }{ { q }^{ 2 } })
⇒ 5q² = p² …(i)
⇒ 5 divides p²
5 divides p
So, let p = 5r
Putting the value of p in (i), we get
5q² = (5r)²
⇒ 5q² = 25r²
⇒ q² = 5r²
⇒ 5 divides q²
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.

Question 20.
Prove that 3 + 2√3 is an irrational number. (2014)
Solution:
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 = (frac { a }{ b }), where a and b are coprime.
Rearranging the equations, we get

Since a and b are integers, we get (frac { a }{ 2b } -frac { 3 }{ 2 }) 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.

Question 21.
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)
Solution:
9 = 3², 12 = 2² × 3, 15 = 3 × 5
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes or 3 hours
They will next toll together after 3 hours.

Question 22.
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)
Solution:
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 × 5² × 17
680 = 2³ × 5 × 17
HCF = 2 × 5 × 17 = 170
Maximum capacity of the container = 170 liters.

Question 23.
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)
Solution:
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 = 2¹ × 5² × 17
B, Breadth = 6 m 25 cm = 625 cm = 5⁴
H, Height = 4 m 75 cm = 475 cm = 5² × 19
HCF of L, B and H is 5² = 25 cm
Length of the longest rod = 25 cm

Question 24.
Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they start ringing together, after how much time will they next ring together? (2015)
Solution:
To find the time when the clocks will next ring together,
we have to find LCM of 4, 12 and 20 minutes.
4 = 2²
12 = 2² × 3
20 = 2² × 5

LCM of 4, 12 and 20 = 2² × 3 × 5 = 60 minutes.
So, the clocks will ring together again after 60 minutes or one hour.

Question 25.
In a school, there are two Sections A and B of class X. There are 48 students in Section A and 60 students in Section B. Determine the least number of books required for the library of the school so that the books can be distributed equally among all students of each Section. (2017 OD)
Solution:
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 = 2⁴ × 3
60 = 2² × 3 × 5
LCM = 2⁴ × 3 × 5 = 16 × 15 = 240
Hence, required number of books is 240.

Question 26.
By using Euclid’s algorithm, find the largest number which divides 650 and 1170. (2017 OD)
Solution:
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.

Question 27.
Find the HCF of 255 and 867 by Euclid’s division algorithm. (2014)
Solution:
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

Question 28.
Using Euclid’s division algorithm, find whether the pair of numbers 847, 2160 are coprime or not.
To find out the minimum (least) time when the bells toll together next, we find the LCM of 9, 12, 15.
Solution:

Real Numbers Class 10 Important Questions Long Answer (4 Marks)

Question 29.
Prove that 3 + 2√5 is irrational. (2012, 2017 D)
Solution:
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 = (frac { a }{ b }), where a and b are coprime.
Rearranging this equation, we get

Since a and b are integers, we get that (frac { a }{ 2b }) – (frac { 3 }{ 2 }) 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.

Question 30.
There are 104 students in class X and 96 students in class IX in a school. In a house examination, the students are to be evenly seated in parallel rows such that no two adjacent rows are of the same class. (2013)
(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 = 2³ × 13
96 = 2⁵ × 3
HCF = 2³ = 8

(a) Number of rows of students of class X = (frac { 104 }{ 8 }) = 13
Number maximum of rows class IX = (frac { 96 }{ 8 }) = 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.

Question 31.
Dudhnath has two vessels containing 720 ml and 405 ml of milk respectively. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled. (2014)
Solution:
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 = 3⁴ × 5
720 = 2⁴ × 3² × 5
HCF = 3² × 5 = 45 ml = Capacity of glass
No. of glasses filled from 1st vessel = (frac { 720 }{ 45 }) = 16
No. of glasses filled from 2nd vessel = (frac { 405 }{ 45 }) = 9
Total number of glasses = 25

Question 32.
Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together? (2013)
Solution:
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 = 2⁴
20 = 2² × 5
LCM = 2⁴ × 5 = 16 × 5 = 80 minutes
They will start preparing a new card together after 80 minutes.

Question 33.
Find HCF of numbers 134791, 6341 and 6339 by Euclid’s division algorithm. (2015)
Solution:
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.

Question 34.
If two positive integers x and y are expressible in terms of primes as x = p²q³ and y = p³q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain. (2014)
Solution:
x = p²q³ and y = p³q
LCM = p³q³
HCF = p²q …..(i)
Now, LCM = p³q³
⇒ LCM = pq² (p²q)
⇒ LCM = pq² (HCF)
Yes, LCM is a multiple of HCF.
Explanation:
Let a = 12 = 2² × 3
b = 18 = 2 × 3²
HCF = 2 × 3 = 6 …(ii)
LCM = 2² × 3² = 36
LCM = 6 × 6
LCM = 6 (HCF) …[From (ii)]
Here LCM is 6 times HCF.

Question 35.
Show that one and only one out of n, (n + 1) and (n + 2) is divisible by 3, where n is any positive integer. (2015)
Solution:
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.

Question 36.
Find the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the two numbers. (2016 D)
Solution:
306 = 2 × 3² × 17
657 = 32 × 73
HCF = 3² = 9
LCM = 2 × 3² × 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.

Question 37.
Show that any positive odd integer is of the form 41 + 1 or 4q + 3 where q is a positive integer. (2016 OD)
Solution:
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.

Important Questions for Class 10 Maths

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CBSE Class 10 Maths Notes Chapter 1 Real Numbers Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 1 Real Numbers. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 1 Real Numbers

R = Real Numbers:
All rational and irrational numbers are called real numbers.

I = Integers:
All numbers from (…-3, -2, -1, 0, 1, 2, 3…) are called integers.

Q = Rational Numbers:
Real numbers of the form (frac { p }{ q }), q ≠ 0, p, q ∈ I are rational numbers.

• All integers can be expressed as rational, for example, 5 = (frac { 5 }{ 1 })
• Decimal expansion of rational numbers terminating or non-terminating recurring.

Q’ = Irrational Numbers:
Real numbers which cannot be expressed in the form (frac { p }{ q }) and whose decimal expansions are non-terminating and non-recurring.

• Roots of primes like √2, √3, √5 etc. are irrational

N = Natural Numbers:
Counting numbers are called natural numbers. N = {1, 2, 3, …}

W = Whole Numbers:
Zero along with all natural numbers are together called whole numbers. {0, 1, 2, 3,…}

Even Numbers:
Natural numbers of the form 2n are called even numbers. (2, 4, 6, …}

Odd Numbers:
Natural numbers of the form 2n -1 are called odd numbers. {1, 3, 5, …}

• Why can’t we write the form as 2n+1?

Remember this!

• All Natural Numbers are whole numbers.
• All Whole Numbers are Integers.
• All Integers are Rational Numbers.
• All Rational Numbers are Real Numbers.

Prime Numbers:
The natural numbers greater than 1 which are divisible by 1 and the number itself are called prime numbers, Prime numbers have two factors i.e., 1 and the number itself For example, 2, 3, 5, 7 & 11 etc.

• 1 is not a prime number as it has only one factor.

Composite Numbers:
The natural numbers which are divisible by 1, itself and any other number or numbers are called composite numbers. For example, 4, 6, 8, 9, 10 etc.
Note: 1 is neither prime nor a composite number.

I. Euclid’s Division lemma
Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r ≤ b.
Notice this. Each time ‘r’ is less than b. Each ‘q’ and ‘r’ is unique.

II. Application of lemma
Euclid’s Division lemma is used to find HCF of two positive integers. Example: Find HCF of 56 and 72 ?
Steps:

• Apply lemma to 56 and 72.
• Take bigger number and locate ‘b’ and ‘r’. 72 = 56 × 1 + 16
• Since 16 ≠ 0, consider 56 as the new dividend and 16 as the new divisor. 56 = 16 × 3 + 8
• Again, 8 ≠ 0, consider 16 as new dividend and 8 as new divisor. 16 = 8 × 2 + 0

Since remainder is zero, divisor (8) is HCF.
Although Euclid’s Division lemma is stated for only positive integers, it can be extended for all integers except zero, i.e., b ≠ 0.

III. Constructing a factor tree
Steps

• Write the number as a product of prime number and a composite number
  Example:
  Factorize 48
• Repeat the process till all the primes are obtained
  ∴ Prime factorization of 48 = 2⁴ x 3
  

IV. Fundamental theorem of Arithmetic
Every composite number can be expressed as a product of primes, and this expression is unique, apart from the order in which they appear.
Applications:

1. To locate HCF and LCM of two or more positive integers.
2. To prove irrationality of numbers.
3. To determine the nature of the decimal expansion of rational numbers.

1. Algorithm to locate HCF and LCM of two or more positive integers:

Step I:
Factorize each of the given positive integers and express them as a product of powers of primes in ascending order of magnitude of primes.
Step II:
To find HCF, identify common prime factor and find the least powers and multiply them to get HCF.
Step III:
To find LCM, find the greatest exponent and then multiply them to get the LCM.

2. To prove Irrationality of numbers:

• The sum or difference of a rational and an irrational number is irrational.
• The product or quotient of a non-zero rational number and an irrational number is irrational.

3. To determine the nature of the decimal expansion of rational numbers:

• Let x = p/q, p and q are co-primes, be a rational number whose decimal expansion terminates. Then the prime factorization of’q’ is of the form 2^m5ⁿ, m and n are non-negative integers.
• Let x = p/q be a rational number such that the prime factorization of ‘q’ is not of the form 2^m5ⁿ, ‘m’ and ‘n’ being non-negative integers, then x has a non-terminating repeating decimal expansion.

Alert!

• 2³ can be written as: 2³ = 2³5⁰
• 5² can be written as: 5² = 2⁰5²

Class 10 Maths Notes

NCERT Solutions

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CBSE Class 10 Maths Notes Chapter 2 Polynomials Pdf free download is part of Class 10 Maths Notes for Quick Revision. Here we have given NCERT Class 10 Maths Notes Chapter 2 Polynomials. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Class 10 Maths Notes Chapter 2 Polynomials

• “Polynomial” comes from the word ‘Poly’ (Meaning Many) and ‘nomial’ (in this case meaning Term)-so it means many terms.
• A polynomial is made up of terms that are only added, subtracted or multiplied.
• A quadratic polynomial in x with real coefficients is of the form ax² + bx + c, where a, b, c are real numbers with a ≠ 0.
• Degree – The highest exponent of the variable in the polynomial is called the degree of polynomial. Example: 3x³ + 4, here degree = 3.
• Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomial respectively.
• A polynomial can have terms which have Constants like 3, -20, etc., Variables like x and y and Exponents like 2 in y².
• These can be combined using addition, subtraction and multiplication but NOT DIVISION.
• The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the graph of y = p(x) intersects the x-axis.

If α and β are the zeroes of the quadratic polynomial ax² + bx + c, then
(sumquad ofquad zeros,alpha +beta =frac { -b }{ a } =frac { -coefficientquad ofquad x }{ coefficientquad ofquad { x }^{ 2 } } )
(productquad ofquad zeros,alpha beta =frac { c }{ a } =frac { constantquad term }{ coefficientquad ofquad { x }^{ 2 } } )

If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d = 0, then
(alpha +beta +gamma =frac { -b }{ a } =frac { -coefficientquad ofquad { x }^{ 2 } }{ coefficientquad ofquad { x }^{ 3 } } )
(alpha beta +beta gamma +gamma alpha =frac { c }{ a } =frac { coefficientquad ofquad { x } }{ coefficientquad ofquad { x }^{ 3 } } )
(alpha beta gamma =frac { -d }{ a } =frac { -constantquad term }{ coefficientquad ofquad { x }^{ 3 } } )

Zeroes (α, β, γ) follow the rules of algebraic identities, i.e.,
(α + β)² = α² + β² + 2αβ
∴(α² + β²) = (α + β)² – 2αβ

DIVISION ALGORITHM:
If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then
p(x) = g(x) × q(x) + r(x)
Dividend = Divisor x Quotient + Remainder

Remember this!

• If r (x) = 0, then g (x) is a factor of p (x).
• If r (x) ≠ 0, then we can subtract r (x) from p (x) and then the new polynomial formed is a factor of g(x) and q(x).

Class 10 Maths Notes

NCERT Solutions

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