Arithmetic Progressions

Exercise 5.1
ANS: 1
(i) It can be observed that
Taxi fare for 1^st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39
Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.
(ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes of air remaining in the cylinder at a time. In other words, after every stroke, only part of air will remain.
Therefore, volumes will be
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
(iii) Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.
(iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be aftern years.
Therefore, after every year, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
 

Quadratic Equations

Exercise 4.1
ANS:1

It is of the form .
Hence, the given equation is a quadratic equation.

It is of the form .
Hence, the given equation is a quadratic equation.

It is not of the form .
Hence, the given equation is not a quadratic equation.

It is of the form .
Hence, the given equation is a quadratic equation.
It is of the form .
Hence, the given equation is a quadratic equation.

It is not of the form .
Hence, the given equation is not a quadratic equation.
It is not of the form .
Hence, the given equation is not a quadratic equation.
It is of the form .
Hence, the given equation is a quadratic equation.
 

Pair of Linear Equations in Two Variables

Exercise 3.1
ANS: 1
Let the present age of Aftab be x.
And, present age of his daughter = y
Seven years ago,
Age of Aftab = x − 7
Age of his daughter = y − 7
According to the question,

Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the question,

Therefore, the algebraic representation is

For ,

The solution table is

x	− 7	0	7
y	5	6	7

For ,

The solution table is

x	6	3	0
y	0	− 1	− 2

The graphical representation is as follows.

Polynomials

 Exercise 2.1
ANS: 1
(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Real Numbers

Exercise 1.1

1. (i) 135 and 225
   Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
   225 = 135 × 1 + 90
   Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
   135 = 90 × 1 + 45
   We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
   90 = 2 × 45 + 0
   Since the remainder is zero, the process stops.
   Since the divisor at this stage is 45,
   Therefore, the HCF of 135 and 225 is 45.
   (ii)196 and 38220
   Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
   38220 = 196 × 195 + 0
   Since the remainder is zero, the process stops.
   Since the divisor at this stage is 196,
   Therefore, HCF of 196 and 38220 is 196.
   (iii)867 and 255
   Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
   867 = 255 × 3 + 102
   Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
   255 = 102 × 2 + 51
   We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
   102 = 51 × 2 + 0
   Since the remainder is zero, the process stops.
   Since the divisor at this stage is 51,
   Therefore, HCF of 867 and 255 is 51.