Author name: Prasanna

In this Page, you will learn how to solve questions of Ex 3.1 Class 10 Maths NCERT Solutions recommended in CBSE 10th Board Maths Exam syllabus.

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting)? Represent this situation algebraically and graphically.
Solution:
Let Aftab’s present age be x and his daughter’s present age by y.
∴ Seven years ago,
Aftab’s age was x – 7
and his daughter’s age was y – 7
According to question
(x – 7) = 7 (y – 7)
or, x – 7 = 7y – 49
or, x – 7y + 42 = 0 … (i)
In case II
Three vears hence
Aftab’s age will be = x + 3
and his daughter’s age will be = y + 3
Again, according to question,
(x + 3) = 3(y + 3)
or, x + 3 = 3y + 9
or, x – 3y – 6 = 0 … (ii)
From equation (i)
x – 7y + 42 = 0
x = 7y – 42

Now from equation (ii)
x – 3y – 6 = 0
x = 3y + 6

x – 7y + 42=0
x – 3y – 6 = 0
Subtracting both equations
4y = 48
y = 12
Putting this value in equation (ii)
x – 3(12) – 6 = 0
x – 36 – 6 = 0
x = 2
Hence the age of Aftab is 42 years and his daughter is 12 years old.

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let the cost of each bat = x
and the cost of each ball = y
Then the algebric representation is given by the following equation.

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Solution:
Let cost of one kg of apples = ₹ x
and the cost of one kg of grapes = ₹ y
Then the algebric representation is given by the following equations :

In this Page, you will learn how to solve questions of Ex 4.1 Class 10 Maths NCERT Solutions recommended in CBSE 10th Board Maths Exam syllabus.

These NCERT Solutions for Class 10 Maths Chapter Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x+ 1)² = 2(x – 3)
(ii) x – 2x = (- 2) (3 – x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3) (2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x² + 3x + 1 = (x – 2)²
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ -4x² – x + 1 = (x – 2)³
Solution:
(i) (x+ 1)² = 2(x – 3)
⇒ x² + 2x +1 = 2x – 6
⇒ x² + 2x – 2x+1 + 6 = 0
⇒ x² + 7 = 0
Now, it is in the form of ax² + bx + c = 0, where b = 0.
Therefore, the given equation is a quadratic equation.

(ii) x² – 2x = (- 2) (3 – x)
⇒ x² – 2x = – 6 + 2x
⇒ x² – 4x + 6 = 0
Which is of the form
ax² + bx + c = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(iii) We have,
(x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x² + x – 2x – 2 = x² + 3x – x – 3
⇒ x² + x – 2x – 2 = x² – 3x + x + 3 = 0
⇒ – 3x + 1 = 0
It is not in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(iv) We have
(x-3) (2x+ 1) = x (x + 5)
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² + x – 6x – 3 – x² – 5x = 0
⇒ x² – 10x – 3 = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(v) We have
(2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x² – 6x-x + 3 = x² -x + 5x – 5
2x² – 6x-x + 3 = x² + x – 5x + 5 = 0
⇒ x² – 11x + 8 = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(vi) We have
x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² + 4 – 4x
⇒ x² + 3x + 1 = x²– 4 + 4c = 0
⇒ 7x – 3 = 0
It is not in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(vii) We have
(x + 2)³ = 2x (x² – 1)
⇒ x³ + 8 + 3.x.2 (x + 2) = 2x³ – 2x
⇒ x³ + 8 + 6x² + 12x = 2x³ – 2x
⇒ x³ – 6x² – 14x – 8 = 0
It is not in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(viii) We have
x³ – 4x² – x+1 = (x-2)³
⇒ x³ – 4x² – x + 1 = x³-8 + 3x(-2)(x – 2)
⇒ x³ – 4x² -x + 1 = x³ – 6x² + 12x – 8
⇒ 2x² – 13x + 9 = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
(i) Let breadth of the rectangular plot = x m
Then, length of the plot = (2x + 1)m
Area of a rectangular plot = l x b ,
⇒ 528 (2x + 1)x
⇒ 528 = 2x² +x
⇒ 2x² + x – 528 = 0
Which is the required quadratic equation.
Therefore, area of rectangle, satisfies the quadratic equation 2x² + x – 528 = 0, where x is breadth (in metres) of the plot.

(ii) Let first consecutive positive integer = x
∴ Second consecutive positive integer = x + 1
According to question,
x (x + 1) = 306
or x² + x = 306
or x² + x – 306 = 0
Therefore, the two consecutive positive integers whose product is 306, satisfies the quadratic equation.
x² + x – 306 = 0, where x is the smallest integer.

(iii) Let the present age of Rohan = x years
∴ Rohan’s mother’s present age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
After 3 years, Rohan’s mother’s age = (x + 26 + 3) years
According to question,
(x + 3) (x + 29) = 360
⇒ x² + 29x + 3x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Therefore, product or Rohan’s and his mother’s age three years now satisfies the quadratic equation
x² + 32x – 273 = 0, where x (in years) is the present age of Rohan.

(iv) Case I :
Let the uniform speed of the train = u km/h
Total distance covered by the train = 480 km.
distance
speed
Case: II
Speed of train = (u – 8) km/h
and total distance covered by the train = 480 km
∴ Time taken = \(\frac { distance }{ speed }\) = \(\frac { 480 }{ u }\)h
Case: II
Speed of train = (u – 8) km/h
and total distance covered by the train = 480 km
∴ Time taken = \(\frac { distance }{ speed }\) = \(\frac { 480 }{ u – 8 }\)h
According to question,

Therefore, speed of train satisfies the quadratic equation 3u2 – 24K – 1280, where K (in km/h) is the speed of the train.

In this Page, you will learn how to solve questions of Ex 2.1 Class 10 Maths NCERT Solutions recommended in CBSE 10 Board Maths Exam syllabus.

These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1

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

(i) No. of zeroes of p(x) is zero because the graph does not intersect x-axis point.
(ii) No. of zeroes of p(x) is 1 as the graph intersects the x-axis at one point only.
(iii) No. of zeroes p(x) is 3 as the graph intersects x-axis at three points.
(iv) No. of zeroes p(x) is as the graphs intersects the x-axis at two points.
(v) No. of zeroes p(x) is 4 as the graph intersects x-axis at four points.
(vi) No. of zeroes p(x) is 3 as the graph intersects the x-axis at three points.

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
Since, ∆ABC ~ ∆DEF
The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Solution:
ABCD is a trapezium with AB || DC and AB = 2 CD

Question 3.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Solution:

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given : Areas of two similar triangles are equal.
To Prove : Triangles are congruent. Ratio in the areas of two similar triangles is equal to the ratio of their respective sides.
Proof: Let ∆ABC and ∆PQR be two triangles.

Hence, by SSS congruence theorem
∆ ABC ≅ ∆PQR (Proved)

Question 5.
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution:
ABC is a triangle and D, E, F are the mid¬points of the sides AB, BC and CA respectively

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Given ∆ ABC ~ ∆DEF, and AP and DQ are their medians.

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Given A square ABCD. Equilateral ABCE and AACF have been described on side BC diagonal AC respectively.

Question 8.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is _________.
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4
Solution:
(c) 4 : 1

Question 9.
Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio __________.
(a) 2 : 3
(b) 4 : 9
(c) 81 : 16
(d) 16 : 81
Solution:
(d) 16 : 81

These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3

Question 1.
Find the sum of the following APs:
(i) 2, 7, 12,…… to 10 terms.
(ii) -37, -33, -29, …… to 12 terms.
(iii) 0.6, 1.7, 2.8, ……, to 100 terms.
(iv) \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), …….., to 11 terms.
Solution:
(i) We have
2, 7, 12, … to 10 terms
Here, a = 2, d = 7 – 2 = 5 and n = 10

Therefore, the sum of 10 terms of the AP 2, 7, 12, ….. is 245

(ii) We have,
– 37, – 33, – 29, … to 12 terms
Here, a = – 37, d = – 33 – (- 37) = 4 and n = 12

Therefore, the sum of 12 terms of the AP – 37, – 33, – 29,… is – 180.

(iii) We have,
0.6,1.7, 2.8,… to 100 terms
Here, a = 0.6, d = 1.7 – 0.6 = 1.1 and n = 100

Therefore, the sum of 100 terms of the AP 0.6, 1.7,2.8,. .. is 5505.

(iv) We have,

Therefore, the sum of 11 terms of the AP \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), ….. is \(\frac { 33 }{ 20 }\).

Question 2.
Find the sums given below:
(i) 7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84
(ii) 34 + 32 + 30 + … + 10
(iii) -5 + (-8) + (-11) + ….. + (-230)
Solution:
(i) We have
7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84

Therefore 7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84 = 1046\(\frac { 1 }{ 2 }\)

(ii) We have
34 + 32 + 30 + … + 10

Therefore, 34 + 32 + 30 + … + 10 = 286

(iii) We have
-5 + (-8) + (-11) + ….. + (-230)

Therefore, -5 + (-8) + (-11) + ….. + (-230) is – 8390.

Question 3.
In an AP:
(i) given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) given a₃ = -15, S₁₀ = 125, find d and a₁₀.
(v) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) given a_n = 4, d = 2, S_n = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
(i) We have
a = 5, d = 3, a_n = 50
But, We know that

Therefore, the value of n = 16 and S_n = 440.

(ii) We have a = 7, a₁₃ = 35
But, We know that
a₁₃ = a + 12d
or 35 = 7 + 12 d
or d = \(\frac { 28 }{ 12 }\) = \(\frac { 7 }{ 3 }\)

Therefore, the value of d = \(\frac { 7 }{ 3 }\) and S₁₃ = 273.

(iii) We have a₁₂ = 37, d = 3
But, We know that
a₁₂ = a + 11d
or 37 = a + 11 x 3
or 37 = a ÷ 33
or a = 37 – 33 = 4
Again we know that

Therefore, the value of a = 4 and S₁₂ = 246.

(iv) We have a₃ = -15, S₁₀ = 125
But, We know that

Multiplying equation (i) by 10 then subtracting from equation (ii), we get,

Putting the value of d = – 1 in equation, (i) we get,
a + 2(-1) = 15
a – 2 = 15 ⇒ a = 17
Now, a = a+ (n – 1)d
a₁₀ = 17 + (10 – 1) (- 1)
a₁₀ = 17 – 9 = 8
Therefore, d = – 1 and a₁₀ = 8

(v) We have d = 5, S₉ = 75
But, We know that

Therefore, the value of a = \(\frac { -35 }{ 3 }\) and a₉ = \(\frac { 85 }{ 3 }\).

(vi) We have a = 2, d = 8, S_n = 90
But, We know that

But number of terms cannot be in negative
∴ n = \(\frac { -9 }{ 2 }\) is neglected.
So, n = 5
Again we know that
a_n = a + (n – 1)d
∴ a₅ = 2 + (5 – 1) x 8
= 2 + 4 x 8
or a₅ = 2 + 32 = 34
Therefore, the value of n = 5 and a₅ = 34.

(vii) We have a = 8, a_n = 62, S_n = 210
But, We know that
a_n = a + (n – 1)d
62 = 8 + (n – 1) d
∴ (n – 1)d = 62 – 8 = 54
or a₅ = 2 + 32 = 34 … (i)
Again

Therefore, the value of n = 6 and d = \(\frac { 54 }{ 5 }\).

(viii) We have a_n = 4, d = 2, S_n = – 14
But, We know that
a_n = a + (n – 1)d
or 4 = a+ (n – 1)2
or 4 = a + 2n – 2
∴ a + 2n = 6
∴ a = 6 – 2n … (i)
Again, we know that
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
or – 14 = \(\frac { n }{ 2 }\)[2 x a + (n – 1) x 2]
or – 14 x 2 = n (2a + 2n – 2)
or – 28 = n[2(6 – 2n) + 2n – 2]
[Putting the value of a from equation (i)]
or – 28 = n (12 – 4n + 2n – 2)
or – 28 = n (- 2n +10)
or – 28 = – 2n² + 10n
or 2n² – 10n – 28 = 0
or n² – 5n – 14 = 0
or n² – 7n + 2n – 14 = 0
or n (n – 7) + 2 (n – 7) = 0
or (n + 2) (n – 7) = 0
∴ n = 7 and n= -2
But, number of terms cannot be in negative
∴ n = – 2 is neglected.
So, n = 7
Putting the value of n in equation (i), we get,
a = 6 – 2n = 6 – 2 x 7
∴ a = 6 – 14 = – 8
Therefore, the value of a = – 8 and n = 7.

(ix) We have a = 3, n = 8, S = 192
But, We know that

Therefore, the value of d = 6.

(x) We have l = 28, S = 144, and n = 9
But, We know that
a_n = l
or a + (n – 1)d = 28
or a + (9 – 1)d = 28
or a + 8d = 28 … (i)
Again we know that

Multiply equation (i) by 2 then subtract equation (ii) from equation (i) we get,

Putting the value of d in equation (i)
a + 8 d = 28
or a + 8 x 3 = 28
∴ a = 28 – 24 = 4
Therefore, the value of a = 4.

Question 4.
How many terms of AP: 9, 17, 25, ….. must be taken to give a sum of 636?
Solution:
We have, 9,17,25 Here,
a = 9 and d = 17 – 9 = 8
Let n terms of this AP must be taken to give a sum of 636.
We know that

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:

Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
We have
S₇ and S₁₇ = 289
But, we know that

Again, we know that

Subtracting equation (ii) from equation (i), we get,
3d – 8d = 7 – 17
or – 5d = – 10
Putting the value of d in equation (i), we get,
a + 3d = 7
or a + 3 x 2 = 7
∴ a = 7 – 6 = 1
Therefore, sum of n terms of this AP is

Question 10.
Show that a₁, a₂, ……. a_n,…… form an AP where an is defined as below:
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
(i) We have
a = 3 + 4n
Put n = 1 in this equation
∴ a₁ = 3 + 4 x 1 = 7
Now, put n = 2 in this equation
a₂ = 3 + 4 x 2 = 11
Again, put n = 3 in this equation
a₃ = 3 + 4 x 3 = 15
Therefore, the required AP is 7, 11, 15, ….
Here, a = 7, d = 11 – 7 = 4
We know that

Therefore, sum of 15 terms of the sequence a_n = 3 + 4n is 525.

(ii) We have,
∴ a_n = 9 – 5n
a₁ = 9 – 5 x 1 = 4
a₂ = 9 – 5 x 2 = – 1
and a₃= 9 – 5 x 3 = – 6
∴ AP is 4,-1,-6,…
∴ Common difference (d) = – 1 – 4 = – 5

Therefore, sum of 15 terms of the sequence a_n = 9 – 5n is – 465.

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
S_n = 4n – n²
Put n = 1, 2, 3, …
S₁ = 4(1) – 1
S₁ = 3,
S₂=4,
S₃=3,
The second term is
a₂ = S₂ – S₁
a₂ = 4 – 3 ⇒ a₂ = 1;
Third term,
a₂= S₃ – S₂
a₃ = 3 – 4 ⇒ a₂ = – 1
Third term
a₂ = S₃ – S₂
a₃ = 3 – 4 ⇒ a₃ = – 1;
⇒ – 60 + 45 = – 15
and nth term is
a_n = (4n – n²) – [(4n – 4) – (n – 1)²]
= 4n – n² – 4n + 4 + n² + 1 – 2n
= 5 – 2n
nth term is (5 – 2n).

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The positive integers divisible by 6 are
6,12,18, …….
Here, a = 6 and d = 12 – 6 = 6
∴ S_n = \(\frac { 40 }{ 2 }\)[2 x 6 + (40 – 1) x 6]
= 20 (12 + 39 x 6) = 20 (12 + 234)
∴ S₄₀ = 20 x 246 = 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are 8,16, 24…
Here, a = 8,d = 16-8 = 8
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
⇒ S₁₅ = \(\frac {15 }{ 2 }\)[2 x 8 + (15 – 1)8]
⇒ S₁₅ = \(\frac { 15 }{ 2 }\)[16 + 112]
⇒ S₁₅ = \(\frac { 15 }{ 2 }\) x 128
⇒ S₁₅ = 15 x 64 = 960
∴ The sum of first 15 multiples of 8 is 960.

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Let odd numbers between 0 and 50 be 1, 3, 5, 7,…….., 49.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:
₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
We have given that penalty for delay of completion beyond a certain date are 200, 250, 300, ….
It is the form of an AP, where a = 200 d = 50
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
S₃₀= \(\frac { 30 }{ 2 }\) [2 x 200 + (30 – 1)50]
= 15 [400 + 1450]
= (15 x 1850) = 27750
The contractor has to pay ₹ 27750 as penalty.

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let 1st prize be of ₹ a
2nd prize be ₹ (a – 20) and
3rd prize be ₹ (a – 20 – 20) = ₹ (a – 40)

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, example a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Let the trees be planted 1, 2, 3, 4, 5 , …… 12
Here, a = 1, d = 1, n = 12
Total number of trees planted by each section
S₁₂ = \(\frac { 12 }{ 2 }\) [2a + (n – 1) d] = 6 [2 x 1 + (12 – 1) x 1]
= 6 [2 + 11] = 6 x 13 = 78
Total number of trees planted by 3 sections = 78 x 3 = 234

Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?
(Take π = \(\frac { 22 }{ 7 }\))
[Hint:Length of successive semicircles is l₁, l₂, l₃, l₄, … with centres at A, B, respectively.]
Solution:
Length of successive semi circles is l₁, l₂, l₃, l₄, …
Series is 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm… we have to found l_n(total length)
Here, l₁ = 0.5 π
d = 1.0 – 0.5 = 0.5π
n = 13 (consective semi circles)
Sum of l_n terms

∴ Total length of spiral of 143 cm.

Question 19.
200 logs are stacked in the following manner 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 x 5 + 2 x (5 + 3)]

Solution:
Distance covered by the girl pick up the first potato = 10m
The distance covered by the girl to pick up the second potato = 8 x 2 = 16m
and the distance covered by the girl to pick up the third potato = 11 x 2 = 22m
So, we get the following series ;
10, 16, 22, 28, … upto 10^th term.
We know that, [where a = 10, d = 6]
Total distance We know that
∴ S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
= \(\frac { 10 }{ 2 }\)[2 x 10 + (10 – 1)6]
= 5 [20 + 54]
= 5 [74] = 370 m
Hence, total distance covered by the competitor is 370.

These NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:
(i) Let P (2, 3) and Q (4, 1) be the two points. Therefore, by distance formula we know that

Therefore distance between points (2, 3) and (4, 11) is 2\(\sqrt{2}\) unit.

(ii) Let P (- 5, 7) and Q (- 1, 3) be two points.
Therefor by distance formula we know that

Therefore, distance between points (- 5, 7) and (- 1, 3) is 4\(\sqrt{2}\) unit.

(iii) Let P (a, b) and Q (-a, – b) be the points.
Therefore, by distance formula we know that

Therefore, distance between points (a. b) and (- a, – b) is 2\(\sqrt{a^{2}+b^{2}}\) unit.

Question 2.
Find the distance between the points (0, 0) and (36, 15).
Solution:
The distance between the points (0,0) and (36,15) by using distance formula, the distance of point P (x₁, y₁) from (0, 0) are

The distance between two towns is 39 km.

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let points be A (1, 5), B (2, 3) and C (-2, -11)

Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let points be A(5, -2), B (6, 4) and C (7, -2)

Therefore, points (1,5), (2, 3) and (-2, -11) are not collinear.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
Points A (3, 4), B (6, 7), C (9, 4) and D (6, 1)

By distance formula we come to know that Champa is correct. The quadrilateral ABCD is a square
Because AB = BC = CD = DA = 3\(\sqrt{2}\)

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let points be A (-1, -2), B (1, 0), C (-1, 2) and D (-3, 0) be the vertices of quadrilateral.
Therefore, by distance formula, we know that

AB = BC = CD = AD and diagonals AC = BD
So, the quadrilateral formed by points (-1, -2), (1, 0), (-1,2) and (-3, 0) is a square.

(ii) Let A (-3, 5), B (3, 1), C = (0, 3) and D = (-1, -4) be the vertices of quadrilateral.
Therefore, by distance formula, we know that

There is no quadrilateral from the given points.

(iii) Let A (4, 5), B (7, 6), C (4,3) and D (1,2) be the vertices of quadrilateral ABCD.
Therefore, by distance formula, we know that

The points (4,5), (7,6), (4,3) and (1,2) form a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
Let P (x, 0) be a point on x-axis which is equidistant from the points. A (2, -5) and B (-2, 9)
Using the distance formula, we have

Hence, the required point on the x-axis is P (x, 0) = P (-7, 0)

Question 8.
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
Solution:
By distance formula we know that

Therefore, the values of y are – 9 and 3

Question 9.
IF Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the value of x. Also find the distances QR and PR.
Solution:
By distance formula we know that.

According the question,

Now the points Q (0,1), P (5, -3) and R (4, 6) by using of distance formula.

Question 10.
Find the relation between between x and y such that the point (x, y) is equidistant from the point (3, 6) and (3, 4).
Solution:
Let the points be P (x, y), Q (3, 6) and R (- 3, 4). By using distance formula.