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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Polynomials Class 10 Maths NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials are part of NCERT Solutions for Class 10 Maths. Here we have given Maths NCERT Solutions Class 10 Chapter 2 Polynomials.

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.1

Question 1:
The graphs of y = p(x) are given below for some polynomials p(x). Find the number of zeroes of p(x) in each case.



Solution:

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4
Solution:

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively:

Solution:

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p(x) = x⁴– 5x + 6, g(x) = 2 – x^2
Solution:

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i) t² – 3, 2t⁴ + 3t³ – 2t²– 9t – 12
(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x² + 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Solution:

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are  and (sqrt { frac { 5 }{ 3 } }) and –(sqrt { frac { 5 }{ 3 } })
Solution:

Question 4.
On dividing x³ – 3x² + x + 2bya polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and:
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² – 5x + 2;  (frac { 1 }{ 4 }), 1, -2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Solution:

Question 2.
Find a cubic polynomial with the sum, some of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a-b, a, a + b, find a and b.
Solution:

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± √3, find other zeroes.
Solution:

Question 5.
If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:

Class 10 Maths Polynomials Mind Map

Polynomial

An algebraic expression f(x) of the form
f(x) = a₀ + a₁x + a₂x² +…. + a_nxⁿ,
where a₀, a₁, ……., a_n are real numbers and all indices of variables are non-negative integers is called polynomial in variable x.
(i) The highest power of x is called degree of the polynomial.
(ii) a₀, a₁x,….,a_nxⁿ are terms of the polynomial.
(iii) a₀, a₁ ,….a_n are co-efficients of the polynomial.

Standard Forms of Linear, Quadratic and Cubic Polynomials

(i) Linear Polynomial:
ax + b, where a, b are real numbers and a ≠ 0.
(ii) Quadratic Polynomial:
ax² + bx + c, where a, b, c are real numbers & a ≠ 0.
(iii) Cubic Polynomials:
ax³ + bx² + cx + d, where a, b, c, d are real numbers and a ≠ 0.

Value of a Polynomial

The value of a polynomial f(x) at x = a is obtained by substituting x = a in the given polynomial and is denoted by/(a).

Zero(es)/Root(s) of Polynomial

x = r is a zero of a polynomial p(x) if p(r) = 0.

Geometrical Meaning of Zeroes of a Polynomial

Zero(es) of a polynomial is/are the x-coordinate of the point(s) where graph y = fix) intersects the x-axis.
(i) Linear polynomial: Graph of linear polynomial is a straight line and has exactly one zero.
(ii) Quadratic polynomial: Graph of quadratic polynomial is always a parabola and this polynomial can have atmost two zeroes.
(iii) Cubic polynomial: Cubic polynomial can have atmost three zeroes.

Cases of Quadratic Polynomial

Case-I : If a quadratic polynomial P(x) = ax² + bx + c has two zeroes, then its graph will intersect the x-axis at two distinct points A and B as shown in the figure.

Case-II: If a quadratic polynomial P(x) = ax² + bx + c has only one zero, then its graph will touch the x-axis at only one point A as shown in the figure.

Case-III : If a quadratic polynomial P(x) = ax² + bx + c has no zero, then its graph will not intersect /touch the x-axis at any point as shown in the figure.

Relationship between Zeroes and Coefficients of a Polynomial
(i) Zero of a linear polynomial ax + b is x = (-frac{b}{a})
(ii) If α and β are the zeroes of the quadratic polynomial ax² + bx + c, then
α + β = (-frac{b}{a})

(iii) If α, β and γ are zeroes of the cubic polynomial ax³ + bx² + cx + d then

Division Algorithm

If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x)
where either r(x) = 0 or degree of r(x) < degree of g(x)

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Free PDF Download of CBSE Class 10 Maths Chapter 5 Arithmetic Progressions 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 Arithmetic ProgressionsMCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 5 Arithmetic Progressions

1. The n^th term of an A.P. is given by a_n = 3 + 4n. The common difference is
(a) 7
(b) 3
(c) 4
(d) 1

2. If p, q, r and s are in A.P. then r – q is
(a) s – p
(b) s – q
(c) s – r
(d) none of these

3. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
(a) 2, 4, 6
(b) 1, 5, 3
(c) 2, 8, 4
(d) 2, 3, 4

4. The (n – 1)^th term of an A.P. is given by 7,12,17, 22,… is
(a) 5n + 2
(b) 5n + 3
(c) 5n – 5
(d) 5n – 3

5. The n^th term of an A.P. 5, 2, -1, -4, -7 … is
(a) 2n + 5
(b) 2n – 5
(c) 8 – 3n
(d) 3n – 8

6. The 10^th term from the end of the A.P. -5, -10, -15,…, -1000 is
(a) -955
(b) -945
(c) -950
(d) -965

7. Find the sum of 12 terms of an A.P. whose nth term is given by a_n = 3n + 4
(a) 262
(b) 272
(c) 282
(d) 292

8. The sum of all two digit odd numbers is
(a) 2575
(b) 2475
(c) 2524
(d) 2425

9. The sum of first n odd natural numbers is
(a) 2n²
(b) 2n + 1
(c) 2n – 1
(d) n²

10. If (p + q)^th term of an A.P. is m and (p – q)^tn term is n, then pth term is

11. If a, b, c are in A.P. then (frac{a-b}{b-c}) is equal to

12. The number of multiples lie between n and n² which are divisible by n is
(a) n + 1
(b) n
(c) n – 1
(d) n – 2

13. If a, b, c, d, e are in A.P., then the value of a – 4b + 6c – 4d + e is
(a) 0
(b) 1
(c) -1.
(d) 2

14. The next term of the sequence

15. n^th term of the sequence a, a + d, a + 2d,… is
(a) a + nd
(b) a – (n – 1)d
(c) a + (n – 1)d
(d) n + nd

16. The 10th term from the end of the A.P. 4, 9,14, …, 254 is
(a) 209
(b) 205
(c) 214
(d) 213

17. If 2x, x + 10, 3x + 2 are in A.P., then x is equal to
(a) 0
(b) 2
(c) 4
(d) 6

18. The sum of all odd integers between 2 and 100 divisible by 3 is
(a) 17
(b) 867
(c) 876
(d) 786

19. If the numbers a, b, c, d, e form an A.P., then the value of a – 4b + 6c – 4d + e is
(a) 0
(b) 1
(c) -1
(d) 2

20. If 7 times the 7^th term of an A.P. is equal to 11 times its 11^th term, then 18^th term is
(a) 18
(b) 9
(c) 77
(d) 0

We hope the given MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

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Polynomials Class 10 Extra Questions Maths Chapter 2

Extra Questions for Class 10 Maths Chapter 2 Polynomials. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download Maths NCERT Solutions Class 10 to help you to revise complete syllabus and score more marks in your examinations.

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths

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Free PDF Download of CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 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 Pair of Linear Equations in Two Variables MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 3 Pair of Linear Equations in Two Variables

1. A pair of linear equations a₁x + b₁y + c₁ = 0; a₂x + b₂y + c₂ = 0 is said to be inconsistent, if

2. Graphically, the pair of equations 7x – y = 5; 21x – 3y = 10 represents two lines which are
(a) intersecting at one point
(b) parallel
(c) intersecting at two points
(d) coincident

3. The pair of equations 3x – 5y = 7 and – 6x + 10y = 7 have
(a) a unique solution
(b) infinitely many solutions
(c) no solution
(d) two solutions

4. If a pair of linear equations is consistent, then the lines will be
(a) always coincident
(b) parallel
(c) always intersecting
(d) intersecting or coincident

5. The pair of equations x = 0 and x = 5 has
(a) no solution
(b) unique/one solution
(c) two solutions
(d) infinitely many solutions

6. The pair of equation x = – 4 and y = – 5 graphically represents lines which are
(a) intersecting at (- 5, – 4)
(b) intersecting at (- 4, – 5)
(c) intersecting at (5, 4)
(d) intersecting at (4, 5)

7. For what value of k, do the equations 2x – 3y + 10 = 0 and 3x + ky + 15 = 0 represent coincident lines

8. If the lines given by 2x + ky = 1 and 3x – 5y = 7 are parallel, then the value of k is

9. One equation of a pair of dependent linear equations is 2x + 5y = 3. The second equation will be
(a) 2x + 5y = 6
(b) 3x + 5y = 3
(c) -10x – 25y + 15 = 0
(d) 10x + 25y = 15

10. If x = a, y = b is the solution of the equations x + y = 5 and 2x – 3y = 4, then the values of a and b are respectively
(a) 6, -1
(b) 2, 3
(c) 1, 4
(d) 19/5,  6/5

11. The graph of x = -2 is a line parallel to the
(a) x-axis
(b) y-axis
(c) both x- and y-axis
(d) none of these

12. The graph of y = 4x is a line
(a) parallel to x-axis
(b) parallel to y-axis
(c) perpendicular to y-axis
(d) passing through the origin

13. The graph of y = 5 is a line parallel to the
(a) x-axis
(b) y-axis
(c) both axis
(d) none of these

14. Two equations in two variables taken together are called
(a) linear equations
(b) quadratic equations
(c) simultaneous equations
(d) none of these

15. If am bl then the system of equations ax + by = c, lx + my = n, has
(a) a unique solution
(b) no solution
(c) infinitely many solutions
(d) none of these

16. If in the equation x + 2y = 10, the value of y is 6, then the value of x will be
(a) -2
(b) 2
(c) 4
(d) 5

17. The graph of the equation 2x + 3y = 5 is a
(a) vertical line
(b) straight line
(c) horizontal line
(d) none of these

18. The value of k, for which equations 3x + 5y = 0 and kx + lOy = 0 has a non-zero solution is
(a) 6
(b) 0
(c) 2
(d) 5

19. The value of k, for which the system of equations x + (k + l)y = 5 and (k + l)x + 9y = 8k – 1 has infinitely many solutions is
(a) 2
(b) 3
(c) 4
(d) 5

20. The value of k for which the equations (3k + l)x + 3y = 2; (k2 + l)x + (k – 2)y = 5 has no solution, then k is equal to
(a) 2
(b) 3
(c) 1
(d) -1

21. The pair of equations x = a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)

22. Asha has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively
(a) 35 and 15
(b) 15 and 35
(c) 35 and 20
(d) 25 and 25

23. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages of the son and the father are, respectively
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24

24. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is
(a) 27
(b) 72
(c) 45
(d) 36

We hope the given MCQ Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

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