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These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
(ii) x = -1
(iii) x = 2
Solution:
(i) Let P(x) = 5x – 4x² + 3
Therefore, P(0) = 5(0) – 4(0)² + 3
= 0 – 0 + 3
= 3
So, the value of P(x) at x = 0 is 3.

(ii) Let P(x) = 5x – 4x² + 3
Therefore, P(-1) = 5(-1) – 4(-1)² + 3
= -5 – 4 + 3
= -6
So, the value of P(x) at x = -1 is -6.

(iii) Let P(x) = 5x – 4x² + 3
Therefore, P(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3
= -3
So, the value of P(x) at x = 2 is -3.

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² – y + 1
(ii) p(t) = 2 + t + 2t² – t³
(iii) p(x) = x³
(iv) p(x) = (x – 1) (x + 1)
Solution:
(i) We have given
p(y) = y² – y + 1
Therefore, the value of polynomial p(y) at y = 0 is
p(0) = 0² – 0 + 1 = 1
Again, the value of polynomial p(y) at y = 1 is
p(1) = 1² – 1 + 1 = 1
Again, the value of polynomial p(y) at y = 2 is
p(2) = 2² – 2 + 1 = 3

(ii) We have given,
p(t) = 2 + t + 2t² – t³
Therefore, the value of polynomial p(t) at t = 0 is
p(0) = 2 + 0 + 2(0)² – (0)³ = 2
Again, the value of polynomial p(t) at t = 1 is
p(1) = 2 + 1 + 2(1)² – (1)³ = 4
Again, the value of polynomial p(t) at t = 2 is
p(2) = 2 + 2 + 2(2)² – (2)³ = 4

(iii) We have given,
P(x) = x³,
Therefore, the value of polynomial p(x) at x = 0 is
p(0) = (0)³ = 0
Again the value of polynomial p(x) at x = 1 is
p(1) = (1)³ = 1
Again, the value of polynomial p(x) at x = 2 is
P(2) = (2)³ = 8

(iv) We have given
p(x) = (x – 1)(x + 1)
Therefore, the value of polynomial p(x) at x = 0 is
P(0) = (0 – 1)(0 + 1)
= (-1) × (+1)
= -1
Again, the value of polynomial p(x) at x = 1 is
p(1) = (1 – 1) × (1 + 1)
= 0 × 2
= 0
Again, the value of polynomial p(x), at x = 2 is
p(2) = (2 – 1)(2 + 1)
= 1 × 3
= 3

Question 3.
Verify whether the following are zeros of the polynomial indicated against them:
(i) p(x) = 3x + 1, x = \(-\frac{1}{3}\)
(ii) p(x) = 5x – p, x = \(\frac{4}{5}\)
(iii) p(x) = x² – 1, x = 1, -1
(iv) p(x) = (x + 1) (x – 2), x = -1, 2
(v) p(x) = x², x = 0
(vi) p(x) = 1x + m, x = \(-\frac{m}{1}\)
(vii) p(x) = 3x² – 1, x = \(-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
(viii) p(x) = 2x + 1, x = \(\frac{1}{2}\)
Solution:
(i) We have given that, p(x) = 3x + 1.
Therefore, the value of polynomial p(x) at x = \(-\frac{1}{3}\) is
\(P\left(-\frac{1}{3}\right)=3 \times \frac{(-1)}{3}+1\) = 0
yes x = \(-\frac{1}{3}\) is the zero of polynomial p(x).

(ii) We have given that p(x) = 5x – p
Therefore, the value of polynomial p(x) at x = \(\frac{4}{5}\) is
\(P\left(\frac{4}{5}\right)=5 \times \frac{4}{5}-\frac{22}{7}\) (∵ π = \(\frac {22}{7}\))
= 4 – \(\frac{22}{7}\)
\(P\left(\frac{4}{5}\right)=\frac{6}{7}\)
No, x = \(\frac{4}{5}\) is not the zero of p(x) = 5x – π.

(iii) We have given that p(x) = x² – 1
Therefore, the value of polynomial p(x) at x = 1 is
p(1) = 1² – 1 = 0
Again, the value of polynomial p(x) at x = -1 is
p(-1) = (-1)² – 1 = 0
yes x = 1, -1 are the zero of polynomial p(x) = x² – 1

(iv) We have given that
p(x) = (x + 1)(x – 2)
Therefore, the value of polynomial p(x) at x = -1 is
p(-1) = (-1 + 1)(-1 – 2)
= 0 × (-3)
= 0
Again, the value of polynomial p(x) at x = 2 is
P(2) = (2 + 1)(2 – 2)
= 3 × 0
= 0
yes x = -1, 2 are the zero of polynomial p(x) = (x + 1)(x – 2)

(v) We have given that p(x) = x²
Therefore, the value of polynomial p(x) at x = 0 is
p(0) = 0² = 0
yes x = 0 is the zero of polynomial p(x) = x².

(vi) We have given that, p(x) = lx + m
Therefore, the value of polynomial p(x) at x = \(-\frac{m}{\ell}\) is
\(P\left(-\frac{m}{\ell}\right)=\ell\left(-\frac{m}{\ell}\right)+m\) = 0
yes x = \(-\frac{m}{\ell}\) is the zero of polynomial p(x) = lx + m.

(vii) We have given that, p(x) = 3x² – 1
Therefore, the value of polynomial p(x) at x = \(-\frac{1}{\sqrt{3}}\) is
\(P\left(-\frac{1}{\sqrt{3}}\right)=3 \times\left(-\frac{1}{\sqrt{3}}\right)^{2}-1\)
= \(3 \times \frac{1}{3}-1\)
= 0
Again, the value of polynomial p(x) at x = \(\frac{2}{\sqrt{3}}\) is
\(P\left(\frac{2}{\sqrt{3}}\right)=3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}-1\)
= \(3 \times \frac{4}{3}-1\)
= 4 – 1
= 3
Therefore, x = \(-\frac{1}{\sqrt{3}}\) is the zero of polynomial p(x) and x = \(\frac{2}{\sqrt{3}}\) is not the zero of polynomial p(x).

(viii) We have given that p(x) = 2x + 1
Therefore, the value of polynomial p(x) at x = \(\frac{1}{2}\) is
\(\mathrm{P}\left(\frac{1}{2}\right)=2 \times \frac{1}{2}+1\) = 2
Therefore, x = \(\frac{1}{2}\) is not the zero of polynomial p(x).

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0; c, d are real numbers.
Solution:
(i) We have given that
p(x) = x + 5 ……..(i)
To find the zero of polynomial p(x), we can take
P(x) = 0 ………(ii)
from equation (i) and (ii)
x + 5 = 0
∴ x = -5
Therefore, x = -5 is the zero of polynomial p(x) = x + 5.

(ii) We have given that,
p(x) = x – 5 ……..(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ……..(ii)
from equation (i) and (ii)
x – 5 = 0
∴ x = 5
Therefore x = 5 is the zero of polynomial p(x) = x – 5.

(iii) We have given that
p(x) = 2x + 5 ………(i)
To find the zero of polynomial p(x), we can take
p(x) = 0 …….(ii)
from equation (i) and (ii)
2x + 5 = 0
or x = \(-\frac{5}{2}\)
Therefore x = \(-\frac{5}{2}\) is the zero of polynomial p(x) = 2x + 5.

(iv) We have given that
p(x) = 3x – 2 …….(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ……..(ii)
3x – 2 = 0
x = \(\frac{2}{3}\)
Therefore, x = \(\frac{2}{3}\) is the zero of polynomial p(x) = 3x – 2.

(v) We have given that
p(x) = 3x ……..(i)
To find the zero of polynomial p(x) we can take
P(x) = 0 ……..(ii)
from equation (i) and (ii)
3x = 0
x = 0
Therefore, x = 0 is the zero of polynomial p(x) = 3x.

(vi) We have given that,
p(x) = ax, a ≠ 0 ……(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ………(ii)
from equation (i) and (ii)
ax = 0
x = \(\frac{0}{a}\) = 0
Therefore, x = 0 is the zero of polynomial p(x) = ax, a ≠ 0.

(vii) We have given
p(x) = cx + d …….(i)
c ≠ 0, c, d are real numbers.
To find the zero of polynomial p(x) we can take
p(x) = 0 ………(ii)
from equation (i) and (ii)
cx + d = 0
x = \(-\frac{d}{c}\)
where c ≠ 0 and c, d are real numbers.
Therefore, x = \(\frac{d}{c}\), where c ≠ 0 and c, d are real numbers is the solution of polynomial p(x) = cx + d.

These NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.3

Question 1.
Estimate each of the following using general rule:
(a) 730 + 998
(b) 796-314
(c) 12,904 + 2,888
(d) 28,292-21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Answer:
(a) 730 round off to 700
998 round off to 1000
Estimated sum = 1700

(b) 796 round off to 800
314 round off to 300
Estimated sum = 500

(c) 12904 round off to 13000
2888 round off to 3000
Estimated sum = 16000

(d) 28292 round off to 28000
21496 round off to 21000
Estimated difference = 7000

Question 2.
Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(a) 439 + 334 + 4317
(b) 1,08,737-47,599
(c) 8325 – 491
(d) 4,89,348-48,365
Make four more such examples
Answer:
(a) Rounding off to nearest hundreds
439 round off to 400
334 round off to 300
4,317 round off to 4,300
Estimated sum = 5,000
Again, Rounding off to nearest tens
439 → 440
334 → 330
4,317 → 4,320
∴ Closer estimate = 440 + 330 + 4,320 = 5,090

(b) Rounding off to Areas tens
= 440 + 330 + 4320 = 5090
108737 round off to 108700
47599 round off to 47600
Estimated difference = 61100
Again, Rounding off to nearest tens
1,08,734 → 1,08,730
47,599 → 47,600
∴ Closer estimate=1,08,730 – 47,600 = 61,130

(c) Rounding off to nearest tens
= 108730-47600 = 61130
8325 round off to 8300
491 round off to 500
Estimated difference = 7800
Again, Rounding off to nearest tens
8,325 → 8,330
491 → 490
∴ Closer estimate = 8,330 – 490 = 7,800

(d) Rounding off to nearest tens
= 8330 – 490 = 7840
489348 round off to 489300
48365 round off to 48400
Estimated difference = 440900
Again, Rounding off to nearest tens
4,89,348→ 4,89,300
48,365 → 48,370
Closer estimate=4,89,350-48,370=4,40,980

Question 3.
Estimate the following products using general rule:
(a) 578 x 161
(b) 5281 x 3491
(c) 1291 x 592
(d) 9250 x 29
Make four more such examples
Answer:
(a) 578 x 161
578 round off to 600
161 round off to 200
The estimated product
= 600 x 200 = 1,20,000

(b) 5281 x 3491
5281 round of to 5,000
3491 round off to 3,500
The estimated product
= 5,000 x 3,500 = 1,75,00,000

(c) 1291 x 592
1291 round off to 1300
592 round off to 600
The estimated product
= 1300 x 600 = 7,80,000

(d) 9250 x 29
9250 round off to 10,000
29 round off to 30
The estimated product
= 10,000 x 30 = 3,00,000

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x² – 3x + 7
(ii) y² + √2
(iii) 3√t + t√2
(iv) y + \(\frac{2}{y}\)
(v) x¹⁰ + y³ + t⁵⁰
Solution:
(i) Yes, 4x² – 3x + 7 is polynomial of one variable. As x has degree 2 in it and the only x is one variable.

(ii) Yes, y is only one variable.

(iii) No as 3√t + t√2 can be written as \(3 t^{\frac{1}{2}}+t \sqrt{2}\), Here the exponent of t in \(3 t^{\frac{1}{2}}\) is \(\frac {1}{2}\) which is not a whole number.

(iv) No, as y + \(\frac{2}{y}\) can be written as y + 2y^-1 where exponent of y in \(\frac{2}{y}\) is -1, which is not a whole number.

(v) Yes, it is a polynomial in three variables x, y, and 1.

Question 2.
Write the co-efficients of x² in each of the following:
(i) 2 + x² + x
(ii) 2 – x² + x³
(iii) \(\frac{\pi}{2} x^{2}+x\)
(iv) \(\sqrt{2 x}-1\)
Solution:
(i) We have given that the equation 2 + x² + x
we can also write 1x² + 1x + 2
Therefore, the coefficient of x² in this equation is 1.

(ii) we have given that the equation:
2 – x² + x³
we can also write
x³ – 1x² + 2
Therefore, the coefficient of x² in this equation is -1.

(iii) We have given that the equation
\(\frac{\pi}{2} x^{2}+x\)
or, \(\frac{\frac{22}{7} x^{2}+1 x}{2}\) (since π = \(\frac {22}{7}\))
or, \(\frac{22}{7 \times 2} x^{2}+1 x\)
or, \(\frac{11}{7} x^{2}+1 x\)
Therefore, the coefficient of x² in this equation is \(\frac{11}{7}\).

(iv) We have given that the equation √2x – 1.
We can also write 0x² + √2x – 1
Therefore, the coefficient of x² in this equation is 0.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
We know that polynomials having only two terms is called binomial.
Therefore, the example of a binomial of degree 35 is ax³⁵ + b where a and b are any real number.
Again, the example of a monomial of degree 100 is ax¹⁰⁰ where a is any real number.

Question 4.
Write the degree of each of the following polynomials:
(i) 5x³ + 4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3
Solution:
(i) We know that the highest power of the variable in a polynomial is called the degree of the polynomial.
In polynomial 5x³ + 4x² + 7x.
The highest power of variable x is 3.
Therefore, the degree of polynomial 5x³ + 4x² + 7x is 3.

(ii) In polynomial 4 – y², the highest power of the variable y is 2.
Therefore, the degree of the polynomial 4 – y² is 2.

(iii) In polynomial 5t – 5, the highest power of the variable t is 1.
Therefore, the degree of polynomial 5t – 5 is 1.

(iv) The only term here is 3 which can be written as 3x⁰: So the highest power of the variable x is 0.
Therefore, the degree of the polynomial 3 is 0.

Question 5.
Classify the following as linear, quadratic and cubic polynomials
(i) x² + x
(ii) x – x³
(iii) y + y² + 4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x³
Solution:
(i) In polynomial x² + x the highest power of the variable x is 2. So the degree of the polynomial is 2.
We know that the polynomial of degree 2 is called a quadratic polynomial.
Therefore, polynomial x² + x is a quadratic polynomial.

(ii) In polynomial x – x³, the highest power of the variable x is 3. So the degree of the polynomial is 3.
We know that the polynomial of degree 3 is called a cubic polynomial.
Therefore, polynomial x – x³ is a cubic polynomial.

(iii) In polynomial y + y² + 4, the highest power of the variable y is 2. So the degree of the polynomial is 2.
We know that the polynomial of degree 2 is called a quadratic polynomial.
Therefore, polynomial y + y² + 4 is a quadratic polynomial.

(iv) In polynomial 1 + x, the highest power of the variable x is 1. So the degree of the polynomial is 1.
We know that the polynomial of degree 1 is called a linear polynomial.
Therefore, polynomial 1 + x is a linear polynomial.

(v) In polynomial 3t, the highest power of the variable t is 1. So, the degree of the polynomial is 1.
We know that the polynomial of degree 1 is called a linear polynomial.
Therefore, polynomial 3t is a linear polynomial.

(vi) In polynomial r², the highest power of the variable r is 2. So, the degree of the polynomial is 2.
We know that the polynomial of degree 2 is called a quadratic polynomial.
Therefore, polynomial r is a quadratic polynomial.

(vii) In polynomial 7x³, the highest power of the variable x is 3. So the degree of the polynomial is 3.
We know that the polynomial of degree 3 is called a cubic polynomial.
Therefore polynomial 7x³ is a cubic polynomial.

These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.6

Question 1.
Find
(i) \(64^{\frac{1}{2}}\)
(ii) \(32^{\frac{1}{5}}\)
(iii) \(125^{\frac{1}{3}}\)
Solution:

Question 2.
Find
(i) \(9^{\frac{3}{2}}\)
(ii) \(32^{\frac{2}{5}}\)
(iii) \(16^{\frac{3}{4}}\)
(iv) \((125)^{-\frac{1}{3}}\)
Solution:
(i) We have given
\(9^{\frac{3}{2}}=\left[(9)^{\frac{1}{2}}\right]^{3}\)
= (3)³
= 27
Therefore, \((9)^{\frac{3}{2}}\) = 27

(ii) We have given,
\(32^{\frac{2}{5}}=\left[32^{\frac{1}{5}}\right]^{2}\)
= (2)²
= 4
Therefore, \((32)^{\frac{2}{5}}\) = 4

(iii) We have given,
\((16)^{\frac{3}{4}}=\left[(16)^{\frac{1}{4}}\right]^{3}\)
= (2)³
= 8
Therefore, \((16)^{3 / 4}\) = 8

(iv) We have given,
\((125)^{-\frac{1}{3}}=\left[(125)^{\frac{1}{3}}\right]^{-1}\)
= (5)^-1
= \(\frac{1}{5}\)
Therefore, \((125)^{-\frac{1}{3}}=\frac{1}{5}\)

Question 3.
Simplify
(i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)
(ii) \(\left(\frac{1}{3^{3}}\right)^{7}\)
(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
(iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\)
Solution:
(i) We have given \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)

These NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.2

Question 1.
A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094,1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Answer:
Number of tickets sold on first day = 1,094 Number of tickets sold on second day = 1,812
Number of tickets sold on third day = 2,050
Number of tickets sold on fourth day = 2,751
Total tickets sold = 7,707
Therefore, 7,707 tickets were sold on all the four days.

Question 2.
Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Answer:
Runs to achieve = 10,000
Runs scored = 6,980
Runs required = 10000 – 6980 = 3,020
Therefore, he needs 3,020 more runs.

Question 3.
In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Answer:
Number of votes secured by successful candidates = 5,77,500
Number of votes secured by his nearest rival = 3,48,700
Margin between them = 2,28,800
Therefore, the successful candidate won by a margin of 2,28,800 votes.

Question 4.
Kirti bookstore sold books worth ? 2,85,891 in the first week of June and books worth ? 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Answer:
Books sold in first week = 2,85,891
Books sold in second week = 4,00,768
Total books sold = 6,86,659
Since, 4,00,768 – 2,85,891
Therefore sale of second week is greater than that of first week.
More books sold in second week = 1,14,877
Therefore, 1,14,877 more books were sold in second week.

Question 5.
Find the difference between the greatest and the least 5-digits number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Answer:
Greatest five-digit number using digits 6, 2, 7, 4, 3 = 76,432
Smallest five-digit number using digits 6, 2, 7, 4, 3 = 23,467
Difference = 76,432 – 23,467 Therefore the difference is 52,965.

Question 6.
A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Answer:
Number of screws manufactured in one day = 2,825
Number of days in the month of January (31 days) = 2,825 x 31 = 87,575
Therefore, the machine produced 87,575 screws in the month of January.

Question 7.
A merchant had ₹ 78,592 with her. She placed an order for purchasing 40 radio sets at ₹ 1,200 each. How much money will remain with her after the purchase?
Answer:
Cost of one radio = ₹ 1200
Cost of 40 radios = 1200 x 40 = ₹ 48,000
Now, total money with merchant = ₹ 78,592
Money left with her ₹ 78,592 – ₹ 48,000 = ₹ 30,592
Therefore, ₹ 30,592 will remain with her after the purchase.

Question 8.
A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)
Answer:
Wrong answer = 7236 x 65
Correct answer = 7236 x 56

Difference in answers
= 470340 – 405216 = 65,124.

Question 9.
To stitch a shirt 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain? (Hint: convert data in cm.)
Answer:
Cloth required to stitch one shirt = 2 m
15 cm = 2 x 100 cm + 15 cm = 215 cm
Length of cloth = 40 m = 40 x 100 cm = 4000 cm
Number of shirts can be stitched

Therefore, 18 shirts can be stitched and 130 cm (1 m 30 cm) cloth will remain.

Question 10.
Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Answer:
The weight of one box = 4 kg 500 g
= 4 x 1000 g + 500 g = 4500 g
Maximum load can be loaded in van
= 800 kg = 800 x 1000 g = 800000 g
Number of boxes = \(\frac{800000}{4500}\)

Therefore, 177 boxes can be loaded.

Question 11.
The distance between the school and a student’s house is 1 km 875 m. Every day she walks both ways. Find the total distance covered by her in six days.
Answer:
Distance between school and home = 1.875 km
Distance between home and school = 1.875 km
Total distance covered in one day = 3.750 km
Distance covered in six days = 3.750 x 6 = 22.500 km.
Therefore, 22 km 500 m distance covered in six days.

Question 12.
A vessel has 4 litres and 500 mL of curd. In how many glasses each of 25 mL capacity, can it be filled?
Answer:
Capacity of curd in a vessel
= 4 litres 500 mL
= 4 x 1000 mL + 500 mL
= 4500 mL
Capacity of one glass = 25 mL
Number of glasses can be filled = \(\frac{4500}{25}\)

Therefore, 180 glasses can be filled by curd.

These NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.1

Question 1.
Fill in the blanks:
(a) 1 lakh = ……………. ten thousand.
(b) 1 million = ……………. hundred thousand.
(c) 1 crore = ……………. ten lakh.
(d) 1 crore = ……………. million.
(e) 1 million = ……………. lakh.
Answer:
(a) 10
(b) 10
(c) 10
(d) 10
(e) 10

Question 2.
Place commas correctly and write the numerals:
(a) Seventy-three lakh seventy-five thousand three hundred seven.
(b) Nine crore five lakh forty-one.
(c) Seven crore fifty-two lakh twenty- one thousand three hundred two.
(d) Fifty-eight million four hundred twenty-three thousand two hundred two.
(e) Twenty-three lakh thirty thousand ten.
Answer:
(a) 73,75,307
(b) 9,05,00,041
(c) 7,52,21,302
(d) 58,423,202
(e) 23,30,010

Question 3.
Insert commas suitably and write the names according to Indian System of Numeration:
(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701
Answer:
(a) 8,75,95,762: Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.
(b) 85,46,283: Eight-five lakh forty-six thousand two hundred eighty-three.
(c) 9,99,00,046: Nine crore ninety-nine lakh forty-six.
(d) 9,84,32,701: Nine crore eighty-four lakh thirty-two thousand seven hundred one.

Question 4.
Insert commas suitably and write the names according to International System of Numeration:
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831
Answer:
(a) 78,921,092: Seventy-eight million nine hundred twenty-one thousand ninety-two.
(b) 7,452,283: Seven million four hundred fifty-two thousand two hundred eighty-three.
(c) 99,985,102: Ninety-nine million nine hundred eighty-five thousand one hundred two.
(d) 48,049,831: Forty-eight million forty nine thousand eight hundred thirty one.