Prasanna

These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Exercise 1.4

Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or non-terminating repeating decimal expansion:
(i) \(\frac { 13 }{ 3125 }\)
(ii) \(\frac { 17 }{ 8 }\)
(iii) \(\frac { 64 }{ 455 }\)
(iv) \(\frac{15}{1600}\)
(v) \(\frac { 29 }{ 343 }\)
(vi) \(\frac{23}{2^{3} 5^{2}}\)
(vii) \(\frac{129}{2^{2} 5^{7} 7^{5}}\)
(viii) \(\frac { 6 }{ 15 }\)
(ix) \(\frac { 35 }{ 50 }\)
(x) \(\frac { 77 }{ 210 }\)
Solution:
(i) \(\frac { 13 }{ 3125 }\) = \(\frac{17}{2 \times 2 \times 2}\) = \(\frac{17}{2^{3}}\)
Because the denominator can be in the form 2ⁿ 5ⁿ, hence it will have terminating decimal expansion.

(ii) \(\frac { 17 }{ 8 }\) = \(\frac { 17 }{ 8 }\)
It will have terminating decimal expansion.

(iii) \(\frac { 64 }{ 455 }\) = \(\frac{64}{5 \times 7 \times 13}\)
Non terminating repeating decimal expansion.

(iv) \(\frac{15}{1600}\) = \(\frac{15}{2^{2} \times 5^{2}}\)
It will have terminating decimal expansion.

(v) \(\frac { 29 }{ 343 }\) = \(\frac{29}{7^{3}}\)
Non terminating repeating decimal expansion.

(vi) \(\frac{23}{2^{3} 5^{2}}\)
It will have terminating decimal expansion.

(vii) \(\frac{129}{2^{2} 5^{7} 7^{5}}\)
Non terminating repeating decimal expansion.

(viii) \(\frac { 6 }{ 15 }\) = \(\frac{6}{3 \times 5}\)
It will have terminating decimal expansion.

(ix) \(\frac { 35 }{ 50 }\) = \(\frac{35}{2 \times 5^{2}}\)
It will have terminating decimal expansion.

(x) \(\frac { 77 }{ 210 }\) = \(\frac{17}{2 \times 3 \times 5 \times 7}\)
Non terminating repeating decimal expansion.

Question 2.
Write down the decimal expansion of those rational numbers in Question 1 above which terminating decimal expansions.
Solution:
(i) \(\frac { 13 }{ 3125 }\) = 0.00146
(ii) \(\frac { 17 }{ 8 }\) = 2.125
(iii) \(\frac { 15 }{ 1600 }\) = 0.009375
(iv) \(\frac{23}{2^{3} 5^{2}}\) = \(\frac { 23 }{ 200 }\) = 0.115
(v) \(\frac { 6 }{ 15 }\) = 0.4
(vi) \(\frac { 35 }{ 50 }\) = 0.7

Question 3.
The following real numbers have decimal expansions as given below. In each case decide whether they are rational or not. If they are rational and of the form \(\frac { p }{ q }\), what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000
(iii) \(43 . \overline{123456789}\)
Solution:
(i) 43.123456789
\(=\frac{43123456789}{1000000000}=\frac{43123456789}{2^{9} 5^{9}}\)
it is a rational, number The prime factors of q are 2⁹5⁹

(ii) 0.120120012000120000 …….
\(\begin{aligned}
&=\frac{120120012000012}{100000000000000 \ldots \ldots . .} \\
&=\frac{120120012000012}{\left(2^{1} \times 2^{2} \times 2^{3} \ldots . .\right) \times\left(5^{1} \times 5^{2} \times 5^{3} . \ldots .\right)}
\end{aligned}\)
it is a rational number The prime factors of q are (2¹ x 2² x 2³ ….) x (5¹ x 5² x 5³ ….)

(iii) \(43 . \overline{123456789}\)
It is not-terminating decimal expansion.
But it is a rational number, whose prime factors of q will also have a factor other than 2 or 5.

These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Exercise 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² -15
(vi) 3x² – x – 4
Solution:
(i) x² – 2x – 8
⇒ x² – 4x + 2x – 8
⇒ x(x – 4) + 2(x – 4)
⇒ (x – 4) (x + 2)
⇒ x – 4 = 0 or x + 2 = 0
⇒ x = 4 or x= -2
Verification:

(ii) 4s² – 4s + 1
⇒ 4s² – 2s – 2s + 1
⇒ 2s(2s – 1) – 1(2s – 1)
⇒ (2s – 1) (2s – 1)
2s – 1 = 0 or 2s – 1 = 0
⇒ s = \(\frac { 1 }{ 2 }\) or s = \(\frac { 1 }{ 2 }\)
Verification:

(iii) 6x² – 3 – 7x
⇒ 6x² – 7x + 2x – 3
⇒ 3x(2x – 3) + 1(2x – 3)
⇒ (2x – 3) (3x + 1)
2x – 3 = 0 or 3x + 1 = 0
⇒ x = \(\frac { 3 }{ 2 }\) or x = \(\frac { -1 }{ 3 }\)
Verification:

(iv) 4u² + 8u
⇒ 4u(u + 2)
⇒ u – 4 = 0 or u + 2 = 0
⇒ u = 4 or u= – 2
Verification:

(v) t² -15
⇒ t² = 15
⇒ t = ± \(\sqrt{15}\)
t = \(\sqrt{15}\) or t = – \(\sqrt{15}\)
Verification:

(vi) 3x² – x – 4
⇒ 3x² – 4x + 3x – 4
⇒ x(3x – 4) + 1(3x – 4)
⇒ (3x – 4) (x + 1)
3x – 4 = 0 or x + 1 = 0
⇒ x = \(\frac { 4 }{ 3 }\) or x = – 1
Verification:

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac { 1 }{ 4 }\), – 1
(ii) \(\sqrt{2}\), \(\frac { 1 }{ 3 }\)
(iii) 0, \(\sqrt{5}\)
(iv) 1, 1
(v) \(\frac { -1 }{ 4 }\), \(\frac { 1 }{ 4 }\)
(vi) 4, 1
Solution:
(i) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = \(\frac { 1 }{ 4 }\) = \(\frac { -b }{ a }\)
and α.β = – 1 = \(\frac { c }{ a }\)
If a = 4, then b = – 1 and c = – 4
So, one quadratic polynomial which fits the given condition is 4x² – x – 4.

(ii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = \(\sqrt{2}\) = \(\frac { -b }{ a }\)
and α.β = \(\frac { 1 }{ 3 }\) = \(\frac { c }{ a }\)
If a = 4, then b = – \(\sqrt{2}\) and c = \(\frac { 1 }{ 3 }\)
So, one quadratic polynomial which fits the given condition is x² – \(\sqrt{2x}\) + \(\frac { 1 }{ 3 }\) or 3x² – 3\(\sqrt{2x}\) + 1

(iii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 0 = \(\frac { -b }{ a }\)
and α.β = \(\sqrt{5}\) = \(\frac { c }{ a }\)
If a = 1, then b = 0 and c = \(\sqrt{5}\)
So, one quadratic polynomial which fits the given condition is x² – \(\sqrt{5}\)

(iv) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 1 = \(\frac { -b }{ a }\)
and α.β = 1 = \(\frac { c }{ a }\)
If a = 1, then b = – 1 and c = 1
Now, put the values of a, b and c in equation ax² + bx + c
So, one quadratic polynomial which fits the given condition is x² – \(\sqrt{5}\)
1x² – 1x 1 = 0
or x² – x + 1 = 0

(v) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = –\(\frac { 1 }{ 4 }\) = \(\frac { -b }{ a }\)
and α.β = \(\frac { 1 }{ 4 }\) = \(\frac { c }{ a }\)
If a = 1, then b = 1 and c = 1
Now, put the values of a, b and c in equation ax² + bx + c
So, one quadratic polynomial which fits the given condition is 4x² + x + 1 = 0

(vi) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 4 = \(\frac { -b }{ a }\)
and α.β = 1 = \(\frac { c }{ a }\)
If a = 1, then b = – 4 and c = 1
So, one quadratic polynomial which fits the given condition is x² – 4x + 1 = 0

These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Exercise 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 -x²
Solution:
(i)

Therefore,
quotient = x – 3 and remainder = 7x – 9

(ii) First we arrange the terms of the dividend and the divisor in the decreasing order of their degrees.
∴ p(x) = x⁴– 3x² + 4x + 5 and g(x) = x² – x + 1

Therefore,
quotient = x2 +x-3 and remainder = 8

(iii) First we arrange the terms of the dividend and the divisior in the decreasing order of their degrees.
∴ p(x) = x² + x – 3 and g(x) = – x² + 2

Therefore,
quotient = – x² – 2 and remainder = – 5x + 10

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 + t³ – 2t² – 9t – 12
(ii) x² + 3x + 1, 3x⁴ + 5x³ -7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + l
Solution:
We have,
P(t) = 2t⁴ + 3t³ – 2t² – 9t – 12
q(x) = t² – 3
By actual division, we have

Here, remainder is zero.
Therefore, q(x) = t² – 3 is the factor of p(x) = 2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) We have,
p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2
and q(x) = x² + 3x + 1
By actual division, we have

Here, remainder is not zero.
Therefore, q(x) = x² – 3x + 1 is not a factor of p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) We have,
p(x) = x⁵ – x³ + x² + 3x + 1
and q(x) = x³ – 3x + 1
By actual division, we have

Here, remainder is not zero.
Therefore, q(x) = x² – 3x + 1 is not a factor of p(x) = x⁵ – x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt { \frac { 5 }{ 3 } }\) and – \(\sqrt { \frac { 5 }{ 3 } }\)
Solution:
We have given that two zeroes of polynomial p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5 are \(\sqrt { \frac { 5 }{ 3 } }\) and – \(\sqrt { \frac { 5 }{ 3 } }\)
∴ (x – \(\sqrt { \frac { 5 }{ 3 } }\))(x + \(\sqrt { \frac { 5 }{ 3 } }\)) = x² – \(\frac { 5 }{ 3 }\) is a factor is given polynomial p(x).
Now apply the division algorithm to the given polynomial and x² – \(\frac { 5 }{ 3 }\)

So, 3x⁴ + 6x³ – 2x² – 10x – 5

If 3x⁴ + 6x³ – 2x² – 10x – 5 = 0
Then, (x + \(\sqrt { \frac { 5 }{ 3 } }\))(x – \(\sqrt { \frac { 5 }{ 3 } }\))(x + 1)(3x + 3) = 0
∴ x = \(\sqrt { \frac { 5 }{ 3 } }\) or x = \(\sqrt { \frac { 5 }{ 3 } }\) Therefore, the zeroes of polynomial p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5 are – \(\sqrt { \frac { 5 }{ 3 } }\), \(\sqrt { \frac { 5 }{ 3 } }\), -1 and -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x).
Solution:
We know that
Dividend = Divisor x Quotiet + Remainder
∴ x² – 3x² + x + 2 = g(x) x (x – 2) + (- 2x + 4)
or x³– 3x² + x + 2 = g(x) (x – 2) + (-2x + 4)
or x³ – 3x² + x + 2 + 2 x- 4 = g(x) x (x-2)
or x³ – 3x² + 3x – 2 = g(x) x (x – 2)

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:
(i) deg p(x) = deg q(x)
Polynomial p(x) = 2x²– 2x + 14; g(x) = 2
q(x) = x² – x + 7 r(x) = 0
Here, deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)
Polynomial p(x) = x³+ x² + x + 1; g(x) = x² – 1,
q(x) = x + 1, r(x) = 2x + 2

(iii) deg r(x) is 0.
Polynomial p(x) = x²+ 2x² – x + 2; g(x) = x² – 1, q(x) = x + 1, r(x) = 4

These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Exercise 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;  \(\frac { 1 }{ 2 }\), 1, – 2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Solution:
(i) Comparing the given polynomial with ax³ + bx² + cx + d, we get:
a = 2, b – 1, c = -5 and d = 2.
Now, we get the zeroes

Therefore, \(\frac { 1 }{ 2 }\), 1, – 2 are the zeroes of 2x³ + x² – 5x + 2.
So we take α, ß, γ are the co-efficient of cubic polynomial.
Sum of zeroes

(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Compearing the given polynomial with ax³ + bx² + cx + d, we get:
a = 1, b = -4, c = 5 and d = – 2.
∴ p (x) = x³ – 4x² + 5x – 2
⇒  p(2) = (2)³ – 4(2)² + 5 x 2 – 2
= 8 – 16+ 10 – 2 = 0
p(1) = (1)³ – 4(1)² + 5 x 1- 2
= 1 – 4 + 1 – 2
= 6 – 6 = 0
Therefore, 2, 1 and 1 are the zeroes of x³ – 4x² + 5x – 2.
So, take α, ß, γ are the coefficient of cubic polynomial.
Comparing x² – 4x² + 5x – 2 = 0
Sum of zeroes

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:
We know that a cubic equation is ax³ + bx² + cx + d = 0
But given, α + β + γ = 2
αβ + βγ + γα = -7 and αβγ = – 14
Also, we know that a + b + g = \(\frac { -b }{ a }\) = 2, hence, b = – 2
αβ + βγ + γα = \(\frac { c }{ a }\) = – 7, hence c = – 7 a
αβγ = \(\frac { -d }{ a }\) = – 14, hence d = – 14, and a = 1
Now, put the value of a, b, c, d in equation (1), we get
x³ – 2x² – 7x + 14 = 0

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

⇒ a³ – b²a = 1
Put a = 1, in a² – b²a = – 1
(1)³ – b² = – 1 ⇒ b² = 2
b= ± \(\sqrt{2}\)
values of a and b are 1, ± \(\sqrt{2}\)

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt{3}\), finnd other zeroes.
Solution:
Let two zeroes are 2 + \(\sqrt{3}\) and 2 – \(\sqrt{3}\),

(x² – 4x +1) is a factor of the given polynomial.
Now, we divide the given polynomial by

So, x⁴ – 6x³ – 26x² + 138x – 35 = (x² – 4x + 1) (x² – 2x – 35)
Now, by spilitting – 2x, we factorise x² – 2x – 35
= x³ – 7x + 5x – 35
= x(x – 7) (x + 5)
= (x – 7) (x + 5) px = 7, x = – 5
So, zeroes of the given polynomials are
2 + \(\sqrt{3}\), 2 – \(\sqrt{3}\), 7 and – 5.

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:
If p(x) and g(x) are any two polynomials of the form x² – 2x and x² + bx + c. We know by formula.
Dividend = Divisor x Quotient + Remainder
x⁴ – 6x² + 16x³ – 25x + 10 = (x² – 2x + k) (x² + bx + c) + (x + a)
x⁴ – 6x³ + 16x² – 25x + 10 = x⁴ + bx³ + x²c – 2x³ – 2bx² – 2cx + kx² + kbx + kc + x + a
x⁴ – 6x³ + 16x² – 25x + 10 = x⁴ + (b – 2)x³ + (c – 2b + k) x² + (- 2x + kb + 1)x + kc + a
Now comparing the co-efficients of both sides b – 2 = – 6 … (1)
[Comparing the co-efficient of x³]
c – 2b + k = 16 … (2)
[Comparing the co-efficient of x²]
– 2c + kb + 1 = – 25 … (3)
[Comparing the co-efficient of x]
kc + a = 10 … (4)
[Comparing the constant term]
From (1), b = – 4
Now, putting the vlaue of b in equation (2) and (3) we get,
⇒ c – 2(- 4) + k = 16
or c + k = 8 … (5)
and -2c – 4k + 1 = – 25
or – 2c – 4k = – 26
or – c – 2k = – 13 … (6)
Adding equation (5) and (6), we get, c + k = 8
– c – 2k = – 13 – k = – 5
So, k = 5,
From (5), c = 8 – 5 ⇒ c = 3,
Now put the value of k and c in equation (4), we get,
kc + a = 10
5(3) + a = 10
a = – 5
∴ Value of k and a is 5 and – 5 respectively.

These NCERT Solutions for Class 10 Maths Chapter Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x² – 3x – 10 = 0
(ii) 2x² + x – 6 = 0
(iii) √2x² + 7x + 5√2 = 0
(iv) 2x² – x + \(\frac { 1 }{ 8 }\) = 0 8
(v) 100x² – 20 x + 1 = 0
Solution:
(i) We have,
x² – 3x – 10 = 0
or x² – 5x + 2x – 10 = 0
or x(x – 5) – 2(x – 5) = 0
or, (x – 5) (x + 2) = 0
∴ x – 5 = 0 or, x + 2 = 0
⇒ x = 5 or, x + 2 = 0
Therefore, the roots x² – 2x -10 = 0 are – 5 and

(ii) We have,
2x² + x – 6 = 0
or, 2x² + 4x – 3x – 6 = 0
or, 2x(x + 2) – 3(x + 2) = 0
or, (x + 2) (2x – 3) = 0
∴ (x + 2) = 0 or, (2x – 3) = 0
⇒ x = – 2 or, x = \(\frac { 3 }{ 2 }\)
Therefore, the roots of 2x² + x – 6 = 0 are – 2 and \(\frac { 3 }{ 2 }\)

(iii) We have,

Therefore, the root of \(\sqrt{2}\)2x² + 7x + 5\(\sqrt{2}\) = o are \(\frac{-5}{\sqrt{2}}\) and \(\sqrt{2}\)

(iv) We have,

Therefore, the roots of 2x² – x + \(\frac { 1 }{ 8 }\) = 0 is \(\frac { 1 }{ 4 }\) ; \(\frac { 1 }{ 4 }\)

(v) We have,

Therefore, the roots of 100x² – 20x + 1 = 0 is \(\frac { 1 }{ 10 }\), \(\frac { 1 }{ 10 }\).

Question 2.
Solve the problems given in Example 1.
(i) John and Jivanti together have 45 marbles Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy in (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let number of marbles John have be x
∴ Number of marbles Jivanti have = 45 – x
The number of marbles left with John when he lost 5 marbles = x – 5
The number of marbles left with Jivanti when she lost 5 marbles = 45x – x – 5 = 40 – x
According to question,
(x – 5) (40 – x) = 124
or 40x – x² – 200 + 5x = 124
or 45x – x² – 200 = 124
or x² – 45x + 324 = 0
or x² – 36x – 9x + 324 = 0
or x(x – 36) – 9(x – 36) = 0
or (x – 36) (x – 9) = 0
∴ (x – 36) =0 or (x – 9) = 0
⇒ x = 36 or x = 9
Therefore, if number of marbles John have be 36
Then number of marbles Jivanti have = 45 – x = 45 – 36 = 9
And, if number of marbles John have be 9
Then, number of marbles Jivanti have = 45 – x = 45 – 9 = 36

(ii) Let the number of toys produced on that day be x
Therefore; the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x(55 – x)
According to question,
x(55 – x) = 750
or 55x – x² = 750
or x² – 55x + 750 = 0
or x² – 30x – 25x + 750 = 0
or x (x – 30) – 25 (x – 30) = 0
(x – 30) (x – 25) = 0
∴ (x – 30) = 0 or (x – 25) = 0
⇒ x = 30 or x = 25
Therefore, if the number of toys produced on that day is x = 30, then, the cost of production (in rupees) or each toy = 55 – x = 55 – 30 = ₹ 25.
And, if the number of toys produced is x = 25,
then, the cost of production = 55 – x = 55 – 25 = ₹ 30.

Question 3.
Find two numbers whose sum is 27, and product is 182.
Solution:
Let first number be x Second number be = 27 – x According to question,
x(27 – x) = 182 ‘
or 27x – x² = 182
or x² – 27x + 182 = 0
or x² – 14x – 13 + 182 = 0
or x (x – 14) – 13 (x – 14) = 0
or (x – 14) (x – 13) = 0
∴ (x – 14) = 0 or, (x – 13) = 0
⇒ x = 14 or x = 13
Therefore, if first number is 14 then second number is
27 – x = 27 – 14 = 13
And, if first number is 13 then second number is
27 – x = 27 – 13 = 14

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let first consecutive positive integer be x
∴Second consecutive positive integer be x + 1
According to question,
x² + (x + 1)² = 365
or x² + x² + 2x + 1 = 365
or 2x² + 2x + 1 – 365 = 0
or 2x² + 2x – 364 = 0
or 2(x² + x – 182) = 0
or x² + x – 182 = 0
or x² + 14x – 13x – 182 = 0
or x(x + 14) – 13 (x + 14) = 0
or (x + 14) (x- 13) = 0
∴ (x + 14) = 0 or (x – 13) = 0
⇒ x = – 14 or x = 13
But, we has given that numbers are positive
∴ x = – 14 is neglected.
Therefore, x = 13
∴ First consecutive positive integer = x = 13
and second consecutive positive integer = 27 – x = 14

Question 5.
The altitude of a right traingle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of right angle triangle = x
∴ The height of right angle triangle = x – 7
And hypotenuse of right angle triangle = 13 cm
By Pythagoras theorem, we have,
(height)² + (base)² = (hypotenuse)²
∴ (x – 7)² + x² = (13)²
or x² – 14x + 49 + x² = 169
or 2x² – 14x + 49 – 169 = 0
or 2x² -14x – 120 = 0
or 2(x² – 7x – 60) = 0
or x² – 7x – 60 = 0
or x² – 12x + 5x – 60 = 0
or x (x – 12) + 5 (x – 12) = 0
or (x -12) (x + 5) = 0
∴ x – 12 = 0 or x + 5=0
⇒ x = 12 or x = – 5
But, length cannot be in negative
So, x = – 5 is neglected.
∴ x = 12
Therefore, base of the right angle triangle = x
∴ x = 12 cm
and height of the right angle triangle = x – 7 = 12 – 7 = 5 cm

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the numbr of articles produced on that day. It the total cost of production on that was ₹ 90. Find the number of articles produced and the cost of each article.
Solution:
Let total number of articles produced = x
Cost of production = 2x + 3
According to question,
x (2x + 3) = 90
or 2x² + 3x = 90
or 2x² + 3x – 90 = 0
or 2x² + 15x – 12x – 90 = 0
or x(2x + 15) – 6 (2x + 15) = 0
or (2x +15) (x – 6) = 0
∴(2x + 15) = 0 or (x – 6) = 0
x = \(\frac { -15 }{ 2 }\) or x = 6
But, number of articles cannot be in negative.
So, x = \(\frac { -15 }{ 2 }\) is neglected.
∴ Total number of articles produced = x = 6
and cost of production = 2x + 3 = 2 x 6 + 3 = ₹ 15.

NCERT Solutions for Class 10 English

Footprints without Feet NCERT Solutions for Class 10 English Footprints Without Feet Chapter 5

Footprints without Feet NCERT Text Book Questions and Answers

Footprints without Feet Read and Find out

Question 1.
How did the invisible man first become visible?
Answer:
The invisible man became visible because he wanted to save himself from bitter cold of winter. He stole warm clothes from a store and put them on. He dressed himself up with an overcoat, shoes and wide brimmed hat.

Question 2.
Why was he wandering in the streets?
Answer:
He was wandering in the streets because he was homeless after being ejected by the landlord. He set the house
on fire.

Question 3.
Why does Mrs Hall find the scientist eccentric?
Answer:
Mrs Hall found the scientist eccentric because he had paid the rent in advance. He never allowed Mrs Hall to talk to him and he had an uncommon appearance.

Question 4.
What curious episode occurs in the study?
Answer:
The curious episode was that money was being taken from the clergyman’s desk but there was no one in the room.

Question 5.
What other extraordinary things happen in the inn?
Answer:
The other extraordinary things were that Mrs Hall felt a sniff very close to her car. The chair started moving and it hit Mrs Hall-the hat leapt up and dashed into her face.

Footprints without Feet Think About it

Question 1.
“Griffin was rather a lawless person.” Comment.
Answer:
Griffin was rather a lawless person. The very fact that he used his scientific discoveries not for the welfare of mankind but for his selfish evil interests shows that he had no respect for law. After a quarrel with his landlord, he set his house on fire to seek revenge. He then removed all his clothes and made himself invisible. He did not even bother to pay for his clothes.

While coming out of the shop, he hit the shopkeeper and stole his money. When his money was exhausted at Iping, he stole money from the clergyman’s desk to pay to Mrs Hall, who called him an eccentric scientist. He had a strange encounter with Mr Jaffers, the constable, who was struggling hard to arrest the headless man. In this way Griffin did not honour the law and many times he crossed his limits that enables us to label him as a lawless person.

Question 2.
How would you assess Griffin as a scientist?
Answer:
In our assessment as a scientist, Griffin scores very high. Griffin was a magnificent scientist. His scientific discovery had indeed a revolutionary impact and it was of great magnitude that showcased the mind of a pure genius.
But mankind and science should complement each other. Science becomes a sheer waste if it is not utilised for the betterment of the humanity and welfare of mankind. No scientific discovery is justified if its very target is human race. There are so many instances in the text to highlight the same. Had Griffin used his discovery in favour of mankind and humanity he could be called a matchless person instead of being called a lawless person.

Footprints without Feet Extra Questions and Answers

Footprints without Feet Short Answer Questions

Question 1.
Why were the two boys in London surprised?
Answer:
The two boys in London were surprised and fascinated because they saw the fresh muddy impressions of a pair of bare feet on the steps of a house in London. The man was not in sight (not visible anywhere).

Question 2.
What happened when Griffin didn’t wake up on time? How did he escape from the London store?
Answer:
Griffin was still sleeping when the assistants arrived and stared towards him. When Griffin saw two of them approaching, he panicked and began to run. They chased him. He escaped from the London store after taking off his clothes. He became invisible and naked once again.

Question 3.
What experiments did Griffin carry out? What was the final result of these experiments?
Answer:
Griffin was a brilliant scientist. He conducted many experiments and discovered a drug which if swallowed could make a person invisible. Griffin misused his invention for his personal gains and selfish ends. He became invisible and got involved in lawlessness.

Question 4.
Mr Griffin entered the shop of the theatrical company. What did he do there?
Answer:
He wore bandages around his forehead. Then he put on dark glasses, false nose, big bushy side whiskers and a large hat. He escaped from there after attacking the shopkeeper from behind and robbing him of all the money.

Question 5.
What happened to the people who tried to help the constable in catching the invisible man?
Answer:
The people who tried to help the constable in catching the invisible man found themselves hit by blows that seemed to come from nowhere.

Question 6.
Why did Griffin slip into a big London store? What did he do inside the shop?
Answer:
Griffin was wandering on the streets of London in mid-winter. The air outside was chilling cold and he needed clothes to save himself from this unbearable weather. Griffin decided to enter a London store. Griffin broke open the boxes and wrappers and dressed himself in warm clothes. He ate cold meat and had coffee in the restaurant followed by sweets and wine from the grocery store.

Question 7.
Why were the landlord and his wife surprised to see the scientist’s door wide open?
Answer:
Griffin always kept his door shut and locked from inside. If anybody entered the room, he was angry.
But that morning the door of his room was wide open. This surprised the landlord and his wife.

Question 8.
Three extraordinary things happened in the room. What were they?
Answer:
The following three things happened in the room

• Mrs Hall heard a sniff quite close to her ears.
• The hat on the bedpost leapt up and dashed into her face.
• The bedroom chair sprang into the air and pushed them out of the room.

Question 9.
What did the scientist do when he became furious? Why were the people in the bar horrified?
Answer:
He threw off his bandages, whiskers, spectacles, and even nose in a minute and became headless. The people in the bar were horrified to see “a headless man”.

Question 10.
Why did Griffin set the landlord’s house on fire?
Answer:
He set his landlord’s house on fire as the landlord wanted to throw him out because of his bad behaviour.

Question 11.
Why did Griffin want to escape from crowded London?
Answer:
Griffin wanted to escape from crowded London because it was becoming difficult for him to hide himself there.

Question 12.
Why was Mr Jaffers, the constable, surprised?
Answer:
Mr Jaffers, ‘the constable’, was surprised because he was trying to arrest a man who did not have a head.

Footprints without Feet Long Answer Questions

Question 1.
“Discoveries of science can be used for welfare as well as for destruction.” How is this applicable to Griffin’s scientific discovery? How did Griffin bring a bad name to his invention? How was his character?
Answer:
A true scientist works for the good of humanity. He wants to make man’s life easier, more comfortable and enjoyable. He does not misuse his discoveries for personal gains or selfish ends. But Griffin, though a brilliant scientist, misuses his discovery. By his experiments, he has been able to make his body transparent and invisible.

He uses this discovery to puzzle other people, enters stores and shops unseen, robs people of their money or things and escapes. He sets fire to the house of his landlord who tried to eject him. He steals food, sweets and wine. Griffin brought a bad name to science by misusing his invention. He was a selfish man with a criminal bent of mind.

Question 2.
Griffin was a brilliant scientist but not a true one. What made him a bad scientist? What does it show about his character? How is the value of Welfare for all’ important in the life of a scientist?
Answer:
Griffin was a brilliant scientist but not a true one. He discovered how human body could be made transparent. It was an amazing discovery that could be used the welfare of the society. But Griffin misused it. He used it to puzzle others. He disturbed the peace of others. He robbed the innocent people. He took revenge upon his landlady.

It shows he was not a man of good character. He lacked the value of kindness, cooperation and humility’. He was an eccentrics dishonest and lawless scientist who brought bad name to science. He did not understand the value of welfare for all. If he had used his invention for a good cause, he would have become immortal and remembered forever.

Question 3.
One must be a law abiding citizen of the country. Griffin lacked this attribute in his character. How did it make him an undesirable person? How did lawlessness overshadowed his greatness as a scientist?
Answer:
One must be a law abiding person. Those who don’t follow rules become a nuisance for society. Griffin was no doubt a brilliant scientist but his lawlessness overshadowed his brilliance. He misused his discovery and disturbed the peace of society.

He robbed a storehouse, hit the landlady and the shopkeeper, had a fight with the constable. All these lawless activities of Griffin made him an undesirable person in society. He used his invention for self-interest and for taking revenge upon the people around him.

Question 4.
Would you like to become invisible? What advantages and disadvantages do you foresee, if you did?
Answer:
Certainly, I would like to be invisible if I could. I find a lot of advantages in being invisible. I could easily be a super hero (Krish) and help the authorities to check the anti-social elements. The playground would have been more funny and entertaining with this invisible power.

I could play humorous and harmless pranks on my friends and even at home had the privilege of teasing my mother by eating the food stealthily. I think advantages and disadvantages of a special power depend upon how it is used. So, positive or negative, advantages or disadvantages all depend upon its usage.

Question 5.
Are there forces around us that are invisible, for example, magnetism? Are there aspects of matter that are ‘invisible’ or not visible to the naked eye? What would the world be like if you could see such forces or such aspects of matter?
Answer:
Yes, there are forces around us that are invisible such as magnetism, gravitation, air and electric current. These aspects of matter are not visible to the naked eye because they cannot be perceived by the sense of sight but these forces can definitely be felt.

The world would have been more adventurous and interesting because we have seen such forces only on paper or in books and never with the eye as of now. The understanding of the components of matter would have been easier and there would probably be no gap and empty spaces in the world. Above all, Nature would have never been as mysterious and mighty as it is now, if we could see some hidden aspects of nature with a naked eye.

Question 6.
What makes glass or water transparent (what is the scientific explanation for this)? Do you think it would be scientifically possible for a man to become invisible, or transparent? (Keep in mind that writers of science fiction have often turned out to be prophetic in their imagination.)
Answer:
The phenomenon of ‘Transparency of Glass and Water’ involves the science of Light. Water and glass do not refract the entire light incident on their surface but refract some of it. This makes glass and water transparent and not opaque. Man has done many wonderful inventions which could not be thought of earlier. Though it doesn’t seem to be possible nowadays science is capable of doing miracles. It may be possible in coming centuries.