Bhagya

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 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) \(\sqrt{2} x^{2}+7 x+5 \sqrt{2}\) = 0
(iv) 2x² – x + \(\frac{1}{8}\) = 0
(v) 100x² – 20x + 1 =0
Solution:
(i) We have, x² – 3x – 10 = 0
⇒ x² – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0 (x – 5)(x + 2) = 0
Either x – 5 = 0 or x + 2 = 0
x = 5 or x = – 2
Thus, the required roots are 5 and -2.

(ii) We have, 2x² + x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2)(2x – 3) = 0
Either x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or x = \(\frac{3}{2}\)
Thus, the required roots are -2 and \(\frac{3}{2}\)

(iii)

(iv) We have, 2x² – x + \(\frac{1}{8}\) = 0
⇒ 16x² – 8x + 1 = 0
⇒ 16x² – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) -1(4x – 1) = 0
⇒ 4x – 1 = 0
⇒ x = \(\frac{1}{4}\)

(v) We have 100x² – 20x + 1 = 0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) -1(10x – 1) = 0
⇒ (10x – 1)(10x – 1) = 0
⇒ (10x – 1) = 0
⇒ x = \(\frac{1}{10}\)
Thus the required roots are \(\frac{1}{10}, \frac{1}{10}\)

Question 2.
Solve the problems:
(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 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 John had x marbles and Jivanti had (45 – x) marbles.
When both of them lost 5 marbles then equation becomes (x – 5) × (45 – x – 5) = 124
⇒ (x – 5) × (40 – x) = 124
⇒ x² – 45x + 324 = 0
⇒ x² – 9x – 36x + 324 = 0
⇒ x(x – 9) – 36(x – 9) = 0
⇒ (x – 9)(x – 36) = 0
Either x – 9 = 0 or x – 36 = 0
Thus, x = 9 or x = 36
∴ If John had 9 marbles, then Jivanti had 45 – 9 = 36 marbles.
If John had 36 marbles, then Jivanti had 45 – 36 = 9 marbles.

(ii) Let the number of toys produced in a day be x.
Then cost of 1 toy = \(\frac{750}{x}\)
⇒ \(\frac{750}{x}\) = 55 – x
⇒ 750 = 55x – x²
⇒ x² – 55x + 750 = 0
⇒ x² – 30x – 25x + 750 = 0
⇒ x(x – 30) – 25(x – 30) = 0
⇒ (x – 30)(x – 25) = 0
Either x – 30 = 0 or x – 25 = 0
x = 30 or x = 25

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one of the numbers be x.
∴ Other number = 27 – x
According to the condition,
x(27 – x) = 182
⇒ 21x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 13x – 14x +182 = 0
⇒ x(x – 13) – 14(x – 13) = 0
⇒ (x – 13)(x – 14) = 0
Either x – 13 = 0 or x -14 = 0
⇒ x = 13 or x = 14
Thus, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the two consecutive positive integers be x and (x + 1).
Since, the sum of the square of the numbers is 365.
∴ x² + (x + 1)² = 365
⇒ x² + (x² + 2x + 1) = 365
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 – 365 = 0
⇒ 2x² + 2x – 364 = 0
⇒ x² + x – 182 = 0
⇒ x² + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x -13) = 0
Either x + 14 = 0 or x – 13 = 0
⇒ x = -14 or x = 13
Since, x has to be a positive integer
⇒ x = -14 is rejected.
∴ x = 13 ⇒ x + 1 = 13 + 1 = 14
Thus, the required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of the given right triangle be x cm.
∴ Its height = (x – 7) cm

Squaring both sides, we get
169 = x² + (x – 7)²
⇒ 169 = x² + x² – 14x + 49
⇒ 2x² – 14x + 49 – 169 = 0
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x- 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
Either x – 12 = 0 or x + 5 = 0
⇒ x = 12 or x = -5
But the sides of a triangle can never be negative
⇒ x = -5 is rejected.
∴ x = 12
∴ Length of base = 12 cm
⇒ Length of altitude = (12 – 7)cm = 5 cm
Thus, the required base = 12 cm and altitude = 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 number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day = x
∴ Cost of production of each article = ₹ (2x + 3)
According to the condition,
Total cost = ₹ 90
⇒ x × (2x + 3) = 90
⇒ 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² – 12x + 15x – 90 = 0
⇒ 2x(x – 6) + 15(x – 6) = 0
⇒ (x – 6)(2x + 15) = 0
Either x – 6 = 0 or 2x + 15 = 0
⇒ x= 6 or x = \(\frac{-15}{2}\)
But the number of articles produced can never be negative.
⇒ x = \(\frac{-15}{2}\) is rejected
∴ Cost of production of each article = ₹ (2 × 6 + 3) = ₹ 15
Thus, the required number of articles produced is 6 and the cost of each article is ₹ 15.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the AP : 121,117,113, ……., is its first negative term? [Hint: Find n for a_n < 0]
Solution:
We have the A.P. having a = 121 and d = 117 – 121 = -4

Thus, the first negative term is 32nd term.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP
Solution:

Question 3.
A ladder has rungs 25 cm apart (see figure).The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2 \frac{1}{2}\) m apart, what is the length of the wood required for the rungs? 250
[Hint: Number of rungs = \(\frac{250}{25}+1\)]

Solution:

Length of the 1^st rung (bottom rung) = 45 cm
Length of the 11^th rung (top rung) = 25 cm
Let the length of each successive rung decreases by x cm.
∴ Total length of the rungs = 45 cm + (45 – x) cm + (45 – 2x) cm + ….. + 25 cm
Here, the number 45, (45 – x), (45 – 2x), …., 25 are in an AP such that
First term (a) = 45
and last term (l) = 25
Number of terms, n = 11

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: S_x-1 = S₄₉ – S_x]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see fig.) Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the 1st step \(\frac{1}{4} \times \frac{1}{2} \times 50 m^{3}\)]

Solution:


\(d=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\)
Here, total number of steps n = 15
Total volume of concrete required to build 15 steps is given by the sum of their individual volumes.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

In each of the following, give the justification of the construction also.

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction :
I. Draw a line segment AB = 7.6 cm.
II. Draw a ray AX making an acute angle with AB.
III. Mark 13 = (8 + 5) equal points on AX, such that AX₁ = X₁X₂ = ……….. X₁₂X₁₃.
IV. Join points X₁₃ and B.
V. From point X₅, draw X₅C || X₁₃B, which meets AB at C.
Thus, C divides AB in the ratio 5 : 8 On measuring the two parts, we get AC = 2.9 cm and CB = 4.7 cm.

Justification:
In ∆ABX₁₃ and ∆ACX₅, we have
CX₅ || BX₁₃
∴ \(\frac{A C}{C B}=\frac{A X_{5}}{X_{5} X_{13}}=\frac{5}{8}\) [By Thales theorem]
⇒ AC : CB = 5 : 8.

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Draw a ∆ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
II. Draw a ray BX making an acute angle ∠CBX.
III. Mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂, X₃ on BX₁ such that BXj = X₁X₂ = X₂X₃.

IV. Join X₃C.
V. Draw a line through X₂ such that it is parallel to X₃C and meets BC at C’.
VI. Draw a line through C parallel to CA which intersect BA at A’.
Thus, ∆A’BC’ is the required similar triangle.
Justification :
By construction, we have X₃C || X₂C’

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Construct a ∆ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.

II. Draw a ray BX such that ∠CBX is an acute angle.
III. Mark 7 points of X₁, X₂, X₃, X₄, X₅, X₆ and X₇ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ – X₄X₅ = X₅X₆ = X₆X₇
IV Join X₅ to C.
V. Draw a line through X₇ intersecting BC (produced) at C’ such that X₅C || X₇C’
VI. Draw a line through C’ parallel to CA to intersect BA (produced) at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ Using AA similarity, ∆ABC ~ ∆A’BC’

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :
I. Draw BC = 8 cm
II. Draw the perpendicular bisector of BC which intersects BC at D.

III. Mark a point A on the above perpendicular such that DA = 4 cm.
IV. Join AB and AC.
Thus, ∆ABC is the required isosceles triangle.
V. Now, draw a ray BX such that ∠CBX is an acute angle.
VI. On BX, mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂ and X₃ such that BX₁ = X₁X₂ = X₂X₃
VII. Join X₂C.
VIII. Draw a line through X₃ parallel to X₂C and intersecting BC (extended) to C’.
IX. Draw a line through C’ parallel to CA intersecting BA (extended) to A’, thus, ∆A’BC’ is the required triangle.
Justification:
We have C’A’ || CA [By construction]
∴ Using AA similarity, ∆A’BC’ ~ ∆ABC

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Steps of construction :
I. Construct a ∆ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
II. Draw a ray BX such that ∠CBX is an acute angle.

Mark four points [greater of 3 and 4 in \(\frac{3}{4}\)] X₁, X₂, X₃, X₄ on BX such that 4
BX₁ = X₁X₂ = X₂X₃ = X₃X₄
IV. Join X₄C and draw a line through X₃ parallel to X₄C to intersect BC at C’.
V. Also draw another line through C’ and parallel to CA to intersect BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
In ∆BX₄C we have
X₄C || X₃C’ [By construction]

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°.Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
Solution:
Steps of Construction :
I. Construct a AABC such that BC = 7 cm, ∠B = 45°, ∠A = 105° and ∠C = 30°
II. Draw a ray BX making an acute angle ∠CBX with BC.

III. On BX, mark four points [greater of 4 and 3 in \(\frac{4}{3}\) ] X₁, X₂, X₃ and X₄ such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄.
IV. Join X₃C.
V. Draw a line through X₄ parallel to X₃C intersecting BC(extended) at C’.
VI. Draw a line through C parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle. Justification:
By construction, we have
C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
I. Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.
II. Draw a ray BX such that an acute angle ∠CBX is formed.
III. Mark 5 points X₁, X₂, X₃, X₄ and X₅ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅.
IV. Join X₃C.
V. Draw a line through X₅ parallel to X₃C, intersecting the extended line segment BC at C’.
VI. Draw another line through C’ parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle by the rope with the ground level is 30° (see figure).

Solution:
In right ∆ABC,

Thus, the required height of the pole is 10 m.

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Let the tree is broken at A and its top is touching the ground at B.
Now, in right ∆AOB, we have
\(\frac{A O}{O B}\) = tan 30°

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
In the figure, DE is the slide for younger children, whereas AC is the slide for elder children.

In right ∆ABC, AB = 3m

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution:
In right ∆ABC, AB = height of the tower and point C is 30 m away from the foot of the tower,

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Let OB = Length of the string
AB = 60 m = Height of the kite.
In the right ∆AOB,
∴ \(\frac{O B}{A B}\) = cosec 60° = \(\frac{2}{\sqrt{3}}\)

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building.The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
Here, OA is the building.

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
Let the height of the building be BC
∴ BC = 20 m and height of the tower be CD.
Let the point A be at a distance y metres from the foot of the building.
Now, in right ∆ABC,

Thus, the height of the tower is 14.64 m.

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution:
In the figure, DC represents the statue and BC represents the pedestal.
Now, in right ∆ABC, we have
\(\frac{A B}{B C}\) = cot 45° = 1
⇒ \(\frac{A B}{h}\) = 1 ⇒ AB = h metres.
Now in right ∆ABD,

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
In the figure, let height of the building = AB = h m

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Solution:
Let AB = h metres = CD

and AP = x m
∴ CP = (80 – x) m
Now, in right ∆APB,
we have

Thus, the required point is 20 m away from the first pole and 60 m away from the second pole.
Height of each pole = 34.64 m.

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of tower is 30° (see figure). Find the height of the tower and the width of the canal.

Solution:
Let the TV tower be AB = h m.
Let the point ‘C be such that BC = x m and CD = 20 m.
Now, in right ∆ABC, we have

Thus, the height of the tower = 17.32 m.
Also width of the canal = 10 m.

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
In the figure, let AB be the height of the building.
∴ AB = 7 metres.
Let BC = X m = AE

Let CD be the height of the cable tower and DE = h m
∴ In right ∆DAE, we have

Question 13.
As observed from the top of a 75 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships.
Solution:
In the figure, let AB represent the light house.
∴ AB = 75 m.
Let the positions of two ships be C and D such that angle of depression from A are 45° and 30° respectively.
Now, in right ∆ABC,
we have

Thus, the required distance between the ships is 54.9 m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.

Solution:
In the figure, let C be the position of the observer (the girl).
A and P are two positions of the balloon.
CD is the horizontal line from the eyes of the observer (girl).
Here PD = AB = 88.2 m – 1.2 m = 87 m
In right ∆ABC, we have
\(\frac{A B}{B C}\) = tan 60°
⇒ \(\frac{87}{B C}=\sqrt{3}\) ⇒ BC = \(\frac{87}{\sqrt{3}}\) m
In right ∆PDC, we have

Thus, the required distance between the two positions of the balloon = \(58 \sqrt{3}\) m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
In the figure, let AB be the height of the tower and C, D be the two positions of the car.

⇒ AC = \(\sqrt{3} \times \sqrt{3}\) × AD = 3 AD
Now, CD = AC – AD = BAD – AD = 2AD
Since the distance 2 AD is covered in 6 seconds,
∴ The distance AD will be covered in \(\frac{6}{2}\) i.e., 3 seconds,
Thus, the time taken by the car to reach the tower from D is 3 seconds.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Let the height of the tower be represented by AB in the figure.

⇒ h = ± 6 m
∴ h = 6 m [∵ Height is can be only positive]
Thus, the height of the tower is 6 m.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4\(\sqrt{3} x\) + 4 = 0
(iii) 2x² – 6x + 3 = 0
Solution:
(i) Comparing the given quadratic equation with ax² + bx + c = 0, we get a = 2, b = -3, c = 5
b² – 4 ac = (-3)² – 4(2)(5) = 9 – 40 = -31 < 0
Since b² – 4ac is negative.
∴ The given quadratic equation has no real roots.

(ii) Comparing the given quadratic equation with ax² + bx + c = 0, we get

Since, b² – 4ac is zero.
∴ The given quadratic equation has two real roots which are equal. Hence, the roots are

(iii) Comparing the given quadratic equation with ax² + bx + c = 0, we get a = 2,b = -6,c = 3
∴ b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24 = 12 > 0
Since, b² – 4ac is positive.
∴ The given quadratic equation has two real and distinct roots, which are given by

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(k – 2) + 6 = 0
Solution:
(i) 2x² + kx + 3 = 0
Here, a = 2, b = k, c = 3
b² – 4ac = (k)² – 4(2) (3)
= k² – 24
∵ For a quadratic equation to have equal root, b² – 4ac = 0

(ii) kx(k – 2) + 6 = 0
Comparing kx² – 2kx + 6 = 0 with ax² + bx + c, we get
a = k, b = -2k, c = 6
∴  b² – 4ac = (-2k)² – 4(k)(6) = 4k² – 24k
Since, the roots are equal
∴  b² – 4ac ⇒ 4k² – 24k = 0.
⇒ 4k(k – 6) = 0
⇒ 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m² ? If so, find its length and breadth.
Solution:
Let the breadth be x m.
∴ Length = 2x m
Now, Area = Length × Breadth = 2x × x m² = 2x² m²
According to the given condition,
2x² = 800

Therefore, x = 20 or x = -20
But x = -20 is not possible
[ ∵ Breadth cannot be negative]
∴ x = 20 ⇒ 2x = 2 × 20 = 40
Thus, length = 40 m and breadth = 20 m

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the age of one of the friends be ‘x’.
Age of another friend is (20 – x).
4 years back age of 1st friend is (x – 4)
4 years back age of 2nd friend = (20 – x – 4) = (16 – x)
Product of their ages is 48.
∴ (x – 4) (16 – x) = 48
16x – x² – 64 + 4x = 48
-x² + 20x – 64 = 48
-x² + 20x – 64 – 48 = 0
-x² + 20x – 112 = 0
x² – 20x + 112 = 0
Here, a = 1, b = -20, c = 112
b² – 4ac = (-20)² – 4( 1)( 112) = 400 – 448 = – 48
Here, b² – 4ac = -48 < 0.
∴ It has no real roots.
∴ This situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution:
Let the breadth of the rectangle be x m.
Since, the perimeter of the rectangle = 80 m
∴ 2(Length + breadth) = 80
2(Length + x) = 80

⇒ Length = (40 – x) m
∴ Area = (40 – x) × x m² = (40x – x²) m²
According to the given condition,
Area of the rectangle = 400 m²
⇒ 40x – x² = 400
⇒ – x² + 40x – 400 = 0
⇒ x² – 40x +400 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get
a = 1, b = -40, c = 400

Since, Length = Breadth
⇒ This rectangle is a square.
Thus, it is not possible to design a rectangular park of given perimeter and area.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.
Solution:
Since, diameter of the cylinder = 10 cm 10
∴ Radius of the cylinder (r) = \(\frac{10}{2}\) cm = 5cm
⇒ Length of wire in one round = 2πr
= 2 × 3.14 × 5 cm = 31.4 cm
∵ Diameter of wire = 3 mm = \(\frac{3}{10}\) cm
∴ The thickness of cylinder covered in one round = \(\frac{3}{10}\) cm
⇒ Number of rounds (turns) of the wire to cover 12 cm = \(\frac{12}{3 / 10}=12 \times \frac{10}{3}\) = 40
∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds
= l = 40 × 31.4 cm = 1256 cm
Now, radius of the wire = \(\frac{3}{2}\) mm = \(\frac{3}{20}\) cm
∴ Volume of wire = πr²l
= 3.14 × \(\frac{3}{20} \times \frac{3}{20}\) × 1256 cm³
∵ Density of wire = 8.88 g/cm³
∴ Mass of the wire = [Volume of the wire] × density

Question 2.
A right triangle, whose sides are 3cm and 4cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).
Solution:
Let us consider the right ABAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse BC = \(\sqrt{3^{2}+4^{2}}\) = 5cm
Obviously, we have obtained two cones on the same base AA’ such that radius = DA or DA’.
Now,

Question 3.
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Dimensions of the cistern are 150 cm, 120 cm and 100 cm.
∴ Volume of the cistern = 150 × 120 × 110 cm³ = 1980000 cm³
Volume of water contained in the cistern = 129600 cm³
∴ Free space (volume) which is not filled with water = (1980000 – 129600) cm³
= 1850400 cm³
Now, volume of one brick
= (22.5 × 7.5 × 6.5) cm³ = 1096.875 cm³
∴ Volume of water absorbed by one brick
= \(\frac{1}{17}\) × 1096.875 cm³
Let n bricks can be put in the cistern.
∴ Volume of water absorbed by n bricks
= \(\frac{n}{17}\) × 1096.875 cm³
∴ Volume occupied by n bricks = [free space in the cistern + volume of water absorbed by n bricks]
⇒ [n × 1096.875] = [1850400 + \(\frac{n}{17}\)(1096.875)]
⇒ 1096.875 n – \(\frac{n}{17}\)(1096.875) = 1850400
⇒ (n – \(\frac{n}{17}\)) × 1096.875 = 1850400
⇒ \(\frac{16}{17} n=\frac{1850400}{1096.875} \Rightarrow n=\frac{1850400}{1096.875} \times \frac{17}{16}\)
= 1792.4102 ≈ 1792
Thus, 1792 bricks can be put in the cistern.

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the norma water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of three rivers = 3 {(Surface area of a river) × Depth}

Since 0.7236 km³ ≠ 9.728 km³
∴ The additional water in the three rivers is not equivalent to the rainfall.

Question 5.
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see figure).

Solution:
We have,
For the cylindrical part:
Diameter = 8 cm ⇒ Radius (r) = 4 cm
Height = 10 cm

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone.
Solution:

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

Now, the total surface area of the frustum = (curved surface area) + (base surface area) + (top surface area)
= πl(r₁ + r₂) + πr₂² + πr₁² = π [(r₁ + r₂)l + r₁² + r₂²]

Question 7.
Derive the formula for the volume of the frustum of a cone.
Solution:
We have,
[Volume of the frustum RPQS] = [Volume of right circular cone OPQ] – [Volume of right circular cone ORS]

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

From (1) and (2), we have
Volume of the frustum RPQS

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.1

Question 1.
Fill in the blanks using the correct word given in brackets:
(i) All circles are ……….. (congruent, similar)
(ii) All squares are ………….. (similar, congruent)
(iii) All ………….. triangles are similar (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are ………… and
(b) their corresponding sides are …………. (equal, proportional).
Solution:
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar.
(a) Their corresponding angles are equal and
(b) Their corresponding sides are proportional.

Question 2.
Give two different examples of pair of
i) similar figures
Solution:
coin, wheel of a cart.

ii) non-similar figures.
Solution:
A square Rhombus

Question 3.
State whether the following quadrilaterals are similar or not:

Solution:
On observing the given figures, we find that
Their corresponding sides are proportional but their corresponding angles are not equal.
∴ The given figures are not similar.

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals

MP Board Class 10th Science Chapter 3 Intext Questions

Intext Questions Page No. 40

Question 1.
Give an example of a metal which:

1. Is a liquid at room temperature.
2. can be easily cut with a knife.
3. Is the best conductor of heat.
4. Is a poor conductor of heat.

Answer:

1. Metal that exists in a liquid state at room temperature → Mercury.
2. Metal that can be easily cut with a knife → Sodium.
3. Metal that is the best conductor of heat → Silver.
4. Metals that are poor conductors of heat → Lead.

Question 2.
Explain the meanings of malleable and ductile.
Answer:

1. Malleable: Materials that can be beaten into thin sheets are called malleable.
2. Ductile: Materials that can be drawn into thin wires are called ductile.

Metals can be hammered into thin sheets. This property of a metal is called malleability and the metals showing this property are called malleable. Gold, Silver, Copper, aluminium etc are malleable metals. Metals can be drawn into wires. The ability of metals to be drawn into thin wires is called ductility. Gold is the most ductile metal. It is interesting to know that a wire of about 2 km length can be drawn from one gram of gold.

Intext Questions Page No. 46

Question 1.
Why is sodium kept immersed in kerosene oil?
Answer:
Sodium is highly reactive metal. It catches fire if kept in the open. Hence, to protect this to prevent accidental fires, it is kept immersed in kerosene oil.

Question 2.
Write equations for the reactions of:
Iron with steam.
Calcium and potassium with water.
Answer:

1. 3Fe(s) + 4H₂O(g) ➝ Fe₃O₄(aq) + 4H₂(g)
2. Ca(s) + 2H₂O(l) ➝ Ca(OH)₂(aq)+ H₂(g)+ Heat
   2K(s) + 2H₂O(l) ➝ 2KOH(aq) + H₂(g) + Heat

Question 3.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows.

Use the table above to answer the following questions about metals A, B, C and D.

1. Which is the most reactive metal?
2. What would you observe if B is added to a solution of Copper(II) sulphate?
3. Arrange the metals A, B, C and D in the order of decreasing reactivity.

Answer:

1. B is most reactive metal.
2. If B is added to a solution of copper sulphate it displaces copper from copper sulphate.
3. If metals are written in the order of decreasing reactivity it is B > A > C > D.

Question 4.
Which gas is produced when diluting hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H₂SO₄.
Answer:
Hydrogen gas is evolved when diluting hydrochloric acid is added to a reactive metal. When iron reacts with dilute H₂SO₄, Iron(II) sulphate with the evolution of hydrogen gas is formed.
Fe(s) + H₂SO₄aq) → FeSO₄(aq) + H₂(g)

Question 5.
What would you observe when zinc is added to a solution of iron(II) sulphate? Write the chemical reaction that takes place.
Answer:
Zinc is more reactive than iron. When zinc is added to iron (II) sulphate, then it will displace the iron from iron sulphate solution as shown in the following chemical reaction,
Zn(s) + FeSO₄(aq) → ZnSO₄(aq) + Fe(s)

Intext Questions Page No. 49

Question 1.
(i) Write the electron-dot structures for sodium, oxygen and magnesium.
(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds?
Answer:

(iii) The ions present in Na₂O are Na⁺ and O^2- ions and MgO are Mg²⁺
and O^2- ions.

Question 2.
Why do ionic compounds have high melting points?
Answer:
Ionic compounds have high melting points because there is electrostic forces of attraction between their charges.

Intext Questions Page No. 53

Question 1.
Define the following terms:

1. Mineral
2. Ore
3. Gangue

Answer:

1. Mineral: Compounds which occur Naturally are called minerals.
2. Ore: Metals can be obtained from minerals. These are called ores.
3. Gangue: Ores mined from the earth are usually contaminated with large amounts of impurities such as soil, sand etc., called gangue.

Question 2.
Name two metals which are found in nature in the free state.
Answer:
The metals at the bottom of the reactivity series are mostly found in a free state. For example gold, silver, and platinum.

Question 3.
What chemical process is used for obtaining a metal from its oxide?
Answer:
Metal can be extracted from its oxide by the process of reduction.

Intext Questions Page No. 55

Question 1.
Metallic oxides of zinc, magnesium and copper were heated with the following metals.

In which cases will you find displacement reactions taking place?
Answer:

Question 2.
Which metals do not corrode easily?
Answer:
Silver and Gold are not corrode easily.

Question 3.
What are alloys?
Answer:
An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal. For example, brass is an alloy of copper and zinc.

MP Board Class 10th Science Chapter 3 Ncert Textbook Exercises

Question 1.
Which of the following pairs will give displacement reactions?
(a) NaCl solution and copper metal.
(b) MgCl₂ solution and aluminium metal.
(c) FeSO₄ solution and silver metal.
(d) AgNO₃ solution and copper metal.
Answer:
(d) AgNO₃ solution and copper metal.

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting?
(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) all of the above.
Answer:
(c) Applying a coating of zinc

Question 3.
An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be:
(a) Calcium
(b) Carbon
(c) Silicon
(d) Iron
Answer:
(a) Calcium

Question 4.
Food cans are coated with tin and not with zinc because:
(a) Zinc is costlier than tin.
(b) Zinc has a higher melting point than tin.
(c) Zinc is more reactive than tin.
(d) Zinc is less reactive than tin.
Answer:
(c) Zinc is more reactive than tin.

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch.

1. How could you use them to distinguish between samples of metals and non-metals?
2. Assess the usefulness of these tests in distinguishing between metals and non-metals.

Answer:

1. Metals can be spread into sheets with the help of a hammer while non metals give powder. When metals are connected into circuit using battery, bulb, wires and a switch current passes through the circuit and the bulb glows.
2. Hammer is a reliable method because no non metal can be spread into sheet.

Question 6.
What are amphoteric oxides? Give two examples of amphoteric oxides.
Answer:
Metal oxides which react with both acids as well as bases to produce salt and water are known as amphoteric oxides.
Eg: Al₂O₃ – Aluminium oxide
ZnO – Zinc Oxide.

Question 7.
Name two metals which will displace hydrogen from dilute acids, and two metals which will not.
Answer:
Iron and Aluminium are more reactive than Hydrogen and displace hydrogen from dilute acids. Mercury and copper re less reactive and these do not displace hydrogen from dilute acids.

Question 8.
In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?
Answer:
In the electrolytic refining of a metal, M:

1. Anode → Impure metal, M.
2. Cathode → Thin strip of pure metal, M.
3. Electrolyte → Aqueous solution of a salt of the metal, M.

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in the figure below.
(a) What will be the action of gas on:

1. Dry litmus paper?
2. Moist litmus paper?

(b) Write a balanced chemical equation for the reaction taking place.

Answer:
(a)

1. There will be no action on dry litmus paper.
2. The colour of litmus paper will turn red because sulphur is a non-metal and the oxides of non-metal are acidic in nature.

(b)

Question 10.
State two ways to prevent the rusting of iron.
Answer:
Two ways to prevent the rusting or iron are
(a) Applying oil, paint and grease we can prevent rusting.
(b) By coating with zinc to iron, we can prevent using. This is called Galvanisation.

Question 11.
What type of oxides are formed when non-metals combine with oxygen?
Answer:
When non-metals are combined with oxygen then neutral or acidic oxides are formed. Examples of acidic oxides are NO₂, SO₂ and examples of neutral oxides are NO, CO etc.

Question 12.
Give reasons:

1. Platinum, gold and silver are used to make jewellery.
2. Sodium, potassium and lithium are stored under oil.
3. Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.
4. Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

Answer:
1. Platinum, gold and silver are used to make jewellery because they are very lustrous. Also, they are very less reactive, ductile and do not corrode easily.

2. Sodium, potassium, and lithium are very reactive metals and react very vigorously with air and water. Therefore, they are kept immersed in oil.

3. Though aluminium is a highly reactive metal, it is resistant to corrosion. This is because aluminium reacts with oxygen present in the air to form a thin layer of aluminium oxide. This oxide layer is very stable and prevents further reaction of aluminium with oxygen. Also, it is light in weight and a good conductor of heat.
Hence, it is used to make cooking utensils.

4. Carbonate and sulphide ores are usually converted into oxides during the process of extraction because metals can be easily extracted from their oxides rather than from their carbonates and sulphides.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.
Answer:
Copper reacts with moist carbon dioxide in the air and slowly tosses its shiny brown surface and gains a green coat. This green substance is basic copper carbonate. Citric acid and tartaric acid neutralise copper carbonate. Hence citric or tartaric acid are effective in cleaning the vessel.

Question 14.
Differentiate between metal and non-metal on the basis of their chemical properties.
Answer:

Question 15.
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument, the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?
Answer:
The solution he had used was Aqua regia. Aqua regia is a Latin word which means ‘Royal Water’. It is the mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3:1. It is capable of dissolving metals like Gold and Platinum. Since the outer layer of the gold bangles is dissolved in aqua regia, so their weight was reduced drastically.

Question 16.
Give reasons why copper is used to making hot water tanks and not steel (an alloy of iron).
Answer:
Iron do not react with hot water, but reacts with steam and forms metallic oxide and Hydrogen. But copper do not reacts with water. Hence copper is used to make hot water tanks and not steel (an alloy of iron).

MP Board Class 10th Science Chapter 3 Additional Questions

MP Board Class 10th Science Chapter 3 Multiple Choice Questions

Question 1.
What kind of element is carbon?
(a) Metal
(b) Metalloid
(c) Non-metal
(d) An alloy
Answer:
(c) Non-metal

Question 2.
A metal of daily use which does not get rusted is:
(a) Steel
(b) Iron
(c) Gold
(d) Silver
Answer:
(a) Steel

Question 3.
Liquid non-metal at room temperature:
(a) Oxygen
(b) Nitrogen
(c) Mercury
(d) Bromine
Answer:
(c) Mercury

Question 4.
Steel is primarily made up of:
(a) Fe and C
(b) Cu and C
(c) Fe and S
(d) Zn and C
Answer:
(a) Fe and C

Question 5.
Silver is coated over iron in electroplating, it represents:
(a) Silver is more reactive than Iron.
(b) Iron is more reactive than silver.
(c) Both are equally reactive
(d) None of above.
Answer:
(b) Iron is more reactive than silver.

Question 6.
Brass is an alloy of:
(a) Cu and Mn
(b) Cu and Zn
(c) Cu and Fe
(d) Cu, Fe and Zn.
Answer:
(b) Cu and Zn

Question 7.
During electrolytic refining of a metal, metal gets deposited at:
(a) Anode
(b) Cathode
(c) Solution
(d) None
Answer:
(b) Cathode

Question 8.
What basic physical property can differentiate metal and non-metal?
(a) Hardness
(b) Lustre
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 9.
We can cut ………. metal with an ordinary knife:
(a) Sodium
(b) Carbon
(c) Gold
(d) Silver
Answer:
(a) Sodium

Question 10.
Hardest metal present in our nature is:
(a) Gold
(b) Diamond
(c) Tungsten
(d) Copper
Answer:
(c) Tungsten

Question 11.
Lustrous non-metal is:
(a) Oxygen
(b) Nitrogen
(c) Iodine
(d) Gold
Answer:
(c) Iodine

Question 12.
Iron pyrites contain which constituent other than Fe:
(a) Co
(b) Cl
(c) S
(d) Pt
Answer:
(c) S

Question 13.
When a metal reacts with water, it forms:
(i) Metal
(ii) H₂
(iii) Metal hydroxide
Choose the best combination:
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) All
Answer:
(c) (ii) and (iii)

Question 14.
Metal forms salts when it reacts with:
(a) Acid
(b) Water
(c) Another Metal
(d) Oxygen
Answer:
(a) Acid

Question 15.
What kinds of metals are coin metals (Cu, Ag, Au)?
(a) Most reactive
(b) Non-reactive
(c) Least reactive
(d) None
Answer:
(c) Least reactive

Question 16.
What kinds of metals are noble metals?
(a) Most reactive
(b) Non-reactive
(c) Least reactive
(d) None
Answer:
(b) Non-reactive

Question 17.
Arrange the following metals in descending order of reactivity – Na, Al, Au, H:
(a) Na > Al > Au>H.
(b) H < Au < Al < Na.
(c) Au > Al > Na > H.
(d) Na > Al > H > Au.
Answer:
(b) H < Au < Al < Na.

Question 18.
When a metal reacts with oxygen, it forms:
(a) Hydrated metals
(b) Metal oxides
(c) Non-metals
(d) Oxygen
Answer:
(b) Metal oxides

Question 19.
Which metal violently reacts with cold water?
(a) Cu and Ag
(b) Au and Ag
(c) K and Na
(d) Hg
Answer:
(c) K and Na

Question 20.
Metals that do not react with water at all are:
(a) Alkali metal
(b) Alkaline eater
(c) Lanthanides
(d) Coin metals
Answer:
(d) Coin metals

Question 21.
Aluminium develops a thin layer of oxide when exposed to air, this process is called:
(a) Anodization
(b) Amalgamation
(c) Corrosion
(d) Rancidity
Answer:
(c) Corrosion

Question 22.
When water reacts with metal, which gas is evolved in the reaction?
(a) Oxygen
(b) Hydrogen
(c) Nitrogen
(d) Sulphur dioxide
Answer:
(b) Hydrogen

Question 23.
What will be the missing product of the following reaction?
Ca(s) + 2H₂O^– → Ca(OH)₂(aq) + ….
(a) O₂
(b) CaO
(c) H₂
(d) O₃
Answer:
(c) H₂

Question 24.
Which solution or reagent can dissolve gold and platinum?
(a) Conc. H₂SO₄
(b) Conc. HCl
(c) Conc. HNO₃
(d) Aqua regia
Answer:
(d) Agua regia

Question 25.
Benchmark element of the reactivity series is –
(a) Au
(b) H
(c) Na
(d) Fe
Answer:
(b) H

Question 26.
Elements more electropositive in nature are:
(a) Metals
(b) Non-metals
(c) Metalloids
(d) None
Answer:
(a) Metals

Question 27.
Ionic solids are:
(a) Solid and hard
(b) Having high boiling and melting point
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 28.
Which one of the following represents electron dot structures of sodium?

Answer:
(d) M

Question 29.
Carefully observe the given diagram below:

Choose the best compound to show above electron-dot structure bonding:
(a) NaCl
(b) MgCl₂
(c) H₂O
(d) None
Answer:
(c) H₂O

Question 30.
Ore of metals with high reactivity can be separated by:
(a) Roasting
(b) Calcination
(c) Electrolysis
(d) Reduction process
Answer:
(c) Electrolysis

Question 31.
Thermionic reactions are:
(a) Displacement reactions of highly exothermic nature
(b) Endothermic reaction
(c) Both (a) and (b)
(d) Electrolytic reduction reactions.
Answer:
(c) Both (a) and (b)

Question 32.
Rusting is an example of:
(a) Electrolysis
(b) Calcination
(c) Corrosion
(d) None
Answer:
(c) Corrosion

Question 33.
Which compounds of metals are basic in nature?
(a) Hydrides
(b) Chlorides
(c) Cyanides
(d) Oxides
Answer:
(d) Oxides

Question 34.
When metal reacts with water which gas is liberated?
(a) Oxygen
(b) Hydrogen
(c) Metal oxide
(d) All of these
Answer:
(b) Hydrogen

Question 35.
Which of the following gas is produced when dilute H₂SO₄ reacts with iron?
(a) Oxygen gas
(b) Hydrogen gas
(c) Both
(d) None of these
Answer:
(b) Hydrogen gas

Question 36.
What happen when Zinc is added to iron (II) sulphate Zn(s) + Fe(s):
(a) ZnSO₂ + FeO₂
(b) ZnSO₂(aq) + Fe(s)
(c) ZnFe + 2Ov
(d) None
Answer:
(b) ZnSO₂(aq) + Fe(s)

Question 37.
Choose the best reason from the following for why ionic compounds have high melting points:
(a) Electrostatic forces
(b) Low magnetic forces
(c) (a) and (b) both
(d) None of these
Answer:
(a) Electrostatic forces

Question 38.
Which of the following is gangue?
(a) Water
(b) Metals
(c) Rocks
(d) CO₂
Answer:
(c) Rocks

Question 39.
Sodium, potassium and lithium are:
(a) Very less reactive
(b) More reactive and react with air
(c) Easily converted to oxides
(d) None
Answer:
(b) More reactive and react with air

Question 40.
Choose from following ways to prevent the rusting of iron:
(a) Oiling
(b) Greasing
(c) Galvanisation
(d) All (a), (b) and (c)
Answer:
(d) All (a), (b) and (c)

Fill in the blanks:

1. Metal generally give ……………Oxides when dissolved in water.
2. Li, Na and K are …………… metals.
3. K and Na catch fire when exposed to ……….
4. Ag and Au ………….. react with water at all.
5. Aluminium is positioned ……………. hydrogen in reactivity series.
6. …………………….. is called royal water.
7. Metal is displaced from their ……………… or …………….. form.
8. Metal gives salt when it reacts with…….
9. Hydrogen is not liberated when metal reacts with ……….. acid.
10. Ionic compounds are ……………………….. in water.
11. Ionic compounds are hard but …………….. in nature hence break into pieces when pressure is applied.
12. Earth’s …………….. is main source of metal.
13. Minerals containing metals are called ……………….
14. Sulphide ores contain metals with ………………. reactivity.
15. Carbonate ore is converted to their metal oxides by the process of …………
16. Electrolysis is done to the ores of ……………….. reactivity to obtain pure metal.
17. Impurities such as soil, sand etc. are called …………… in the ore.
18. Bronze is a homogeneous mixture of ……………… and …………….

Answers:

1. basic
2. soft
3. air
4. do not
5. above
6. Aqua regia
7. solution, molten
8. alkali
9. nitric acid
10. soluble
11. brittle
12. crust
13. Ores
14. low
15. Calcination
16. high
17. gangue
18. copper, tin

MP Board Class 10th Science Chapter 3 Very Short Answer Type Questions

Question 1.
Which one is the most abundant element in our earth crust?
Answer:
Oxygen.

Question 2.
Write the names of two diatomic gaseous elements.
Answer:

1. Oxygen, O₂.
2. Nitrogen, N₂.

Question 3.
Is any metal known in gaseous form, in its natural conditions?
Answer:
No.

Question 4.
Name two naturally occurring soft metals.
Answer:
Sodium, magnesium.

Question 5.
Which metal shows poor conductivity?
Answer:
Tungsten [W] or Bismuth (Bi].

Question 6.
Which non-metal conducts electricity?
Answer:
Graphite.

Question 7.
Which metal forms amphoteric oxides when reacted with oxygen?
Answer:
Aluminium, Zinc, etc.

Question 8.
Draw an electron dot structure of SO₂.
Answer:

Question 9.
Give an example of metal oxide’s reaction with water.
Answer:
All metal oxides do not react with water but some metal oxides react with water to give alkali. For example, Na and K.
Na₂O(s) + H₂O(l) → 2NaOH(aq)

Question 10.
Give an example of amphoteric oxides.
Answer:
Al₂O₃.

Question 11.
Give a term to the following:
A process in which a carbonate ore is heated at very high temperature to get metal oxide.
Answer:
Roasting.

Question 12.
How less reactive metal oxides are reduced?
Answer:
Less reactive metal oxides are reduced by reducing agents like aluminium.

Question 13.
Give a reaction in which metal hydroxide is formed.
Answer:
‘Na’ directly reacts with hydrogen gas and forms sodium hydride.
2Na + H₂ → 2NaH.

Question 14.
Give an example for each an alloy and amalgam.
Answer:

1. Alloy – Steel.
2. Amalgam – Sodium Amalgam. (Alloy of mercury and sodium).

Question 15.
What do we call the removal of gangue from the ore?
Answer:
Enrichment of ores.

Question 16.
What is the place of less reactive metals in metal reactivity series?
Answer:
Less reactive metals are arranged at the bottom of the series.

Question 17.
Name two metals which can replace iron in electroplating.
Answer:
Gold and silver.

Question 18.
What kind of compounds has the highest melting and boiling points?
Answer:
Ionic compounds.

Question 19.
What corrodes copper?
Answer:
Moist carbon-dioxide corrodes copper.

Question 20.
Iron pillar of Qutub Minar, Delhi is prevented against rusting. Why?
Answer:
The corrosion-resistant nature is due to protective film at the non-rust interface because of high phosphorus content.

Question 21.
Why every metal has a different rate of reactivity with the same chemical or reactant? (HOTS)
Answer:
Every metal has different electronic configuration and different electro-positivity or ion forming capability, so metal reactivity differs from each other.

MP Board Class 10th Science Chapter 3 Short Answer Type Questions

Question 1.
Why can we draw gold to thin wire form?
Answer:
Gold is a metal and metal has the ability to be drawn in very thin wire or sheets without being broken, this property of a metal is called ductility. Gold is the most ductile metal.

Question 2.
What is a semiconductor?
Answer:
There are some elements known to us which are neither metal nor nonmetals, they alter their properties with the different physical and chemical environment provided. Normally, they behave as non-metals and show insulation property but when provided with extra energy, they start behaving as metal and show conduction. Example: Silicon.

Question 3.
What do you know about amphoteric oxides?
Answer:
Metal oxides which produce salts in both cases, either react with acid or base are termed amphoteric oxides.

Question 4.
What is gangue? Why is it important to remove them before the extraction process?
Answer:
Impurities associated with ores such as sand, soil etc. are called gangue. The original compound becomes bulkier with their presence and extraction of pure metal consumes lots of energy and time. Hence, gangue is removed before the extraction process.

Question 5.
Give an example of displacement reaction used for extraction of metal from its oxide.
Answer:
Highly reactive metals are used to reduce metal oxide to the metal in an electrolytic displacement reaction.

Example:
3MnO₂(s) + 4Al(s) → 3Mn(l) + 2Al₂O₃
Here, aluminium helps in extraction of pure Mn.

Question 6.
Explain the reactivity series of metals in brief.
Answer:
All metals are arranged in an order on the basis of reactivity. This series represents displacing ability of one metal to displace other from its compound form when it undergoes an electrolytic displacement reaction.

Question 7.
How metal and non-metal interact to form salts?
Answer:
When metal and non-metal interact, they form ionic compounds. Metal tends to lose an electron and form positive ion while non-metal forms a negative ion and a bond is formed due to strong electrostatic forces of attraction among two ions.

Question 8.
How can we protect metals corrosion?
Answer;
Prevention from corrosion can be done by painting, oiling, greasing, galvanizing and electroplating etc. on the metal. All these processes stop the interaction of outer layer of air with metal.

Question 9.
What is galvanisation?
Answer:
Galvanisation is a protection technique against metal corrosion. In this process, metal is coated with a thin layer of zinc.

Question 10.
Why pure gold is mixed with silver or copper while making jewellery?
Answer:
Pure gold metal is very soft. So, jewellery will be brittle if used in the same form. Impurity of other metal in nominal percentage makes the metal harder and ready to use and it does not appear very different than the pure one.

Question 11.
What is an amalgam?
Answer:
When one component of any alloy is mercury it is called amalgam.

Question 12.
Why alloys are not used as conductance medium of electricity?
Answer:
Alloys are not used as conductance medium of electricity because of electrical conductivity and melting point of an alloy is lesser than the pure metal.

Question 13.
Why cooking utensils are made up of metals and their handle or knobs with non-metal materials? (HOTS)
Answer:
Metals are good conductors of heat. So, when utensil made of metals are pulled over a flame, they spread heat energy evenly and food gets cooked properly. But handles or knobs are made of non-metals because they are bad conductors of heat and give us ease to work with highly heated utensils so that we can hold them easily while cooking.

Question 14.
Why ageing is a natural phenomenon? (HOTS)
Answer:
The oxygen of the atmosphere react with the outer skin of body fat and carbohydrate, hence continuously degrade it. Similarly inside the body for the energy need, carbohydrate and body fat undergo oxidation process and get continuously decomposed which causes ageing, so ageing is a natural phenomenon.

Question 15.
Neha went to market with her grandmother to purchase some cooking utensils for their family’s wedded couple. Grandmother started buying cooking utensils made of copper and iron but Neha suggested her to take utensils made of steel. (VBQ)

1. How is steel better than copper?
2. What kind of corrosion occurs in Cu?
3. What values does Neha show in this act?

Answer:

1. Steel is an alloy. It does not get corroded or spoiled with time and it is cheap too.
2. Cu forms its oxide when it comes in contact with oxygen and turns green in colour.
3. Neha shows awareness about metal of more usability and durability.

MP Board Class 10th Science Chapter 3 Long Answer Type Questions

Question 1.
Draw line diagrams for the steps involved in the extraction of metals.
Answer:

Question 2.
Write short notes on the following:

1. Roasting
2. Calcination
3. Mineral
4. Anodising

Answer:
1. Roasting:
To extract metal of medium reactivity from its sulphide ore, the ore is heated strongly in the presence of air and this extraction process is called roasting.

2. Calcination:
When medium reactivity metal is extracted from its carbonate ore by heating moderately, the process is known as calcination.

3. Mineral:
Naturally occurring compounds of metals and non-metals in various combinations are called minerals.

4. Anodising:
When aluminium is exposed to air it develops a thick layer of oxides and turns green in colour.

Question 3.
Rupam was painting the garden chair and other pots made with iron at his home, his younger brother asked him why he is doing so and wasting his time:

1. What is your view about his work?
2. What are other methods to protect the metal from corrosion?
3. What values does Rupam express in this way?

Answer:

1. Rupam was painting the garden’s metal objects to protect them from corrosion.
2. Oiling, euchre plating, anodizing etc. are other methods.
3. Rupam was doing a great job protecting the metal from degrading because it can give a long life to objects and save money.

MP Board Class 10th Science Chapter 3 Textbook Activities

Class 10 Science Activity 3.1 Page No. 37

1. Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample.
2. Clean the surface of each sample by rubbing them with sandpaper and note their appearance again.

Observations:
Iron, copper, aluminium and magnesium have lustre which clearly appears on rubbing them with sandpaper.

Class 10 Science Activity 3.2 Page No. 37

1. Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations.
2. Hold a piece of sodium metal with a pair of tongs.

Caution:

1. Always handle sodium metal with care. Dry it by pressing between the folds of a filter paper.
2. Put it on a watch-glass and try to cut it with a knife.
3. What do you observe?

Observations:
Only Na and Mg are soft metals and we can cut them with a knife but other metals are hard. de

Class 10 Science Activity 3.3 Page No. 38

1. Take pieces of iron, zinc, lead and copper.
2. Place any one metal on a block of iron and strike it four or five times with a hammer.
3. What do you observe?
4. Repeat with other metals.
5. Record the change in the shape of these metals.

Observations:
Fe, Zn, Pb and Cu are metals and show proper malleability when hammered.

Class 10 Science Activity 3.4 Page No. 38

List the metals whose wires you have seen in daily life.

Observations:
Lead cannot be drawn to a thin wire. But Fe, Cu and Al are ductile.

Class 10 Science Activity 3.5 Page No. 38

1. Take an aluminium or copper wire. Clamp this wire on a stand, as shown in the figure.
2. Fix a pin to the free end of the wire using wax.
3. Heat the wire with a spirit lamp, candle or a burner near the place where it is clamped.
4. What do you observe after some time?
5. Note your observations. Does the metal wire melt?

Observations:

1. The pin fell down because heat is being conducted.
2. The metal wire does not melt.

Class 10 Science Activity 3.6 Page No. 39

1. Set up an electric circuit as shown in the figure.
   Place the metal to be tested in the circuit between terminals A and B as shown.
2. Does the bulb glow? What does this indicate?

Observations:
Yes, the bulb glows as metals are a good conductor of heat and electricity.

Class 10 Science Activity 3.7 Page No. 39

1. Collect samples of carbon (coal or graphite), sulphur and iodine.
2. Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample.
3. Clean the surface of each sample by rubbing them with sandpaper and note their appearance again.
4. Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations.
5. Hold a piece of sodium metal with a pair of tongs.

Caution:

1. Always handle sodium metal with care. Dry it by pressing between the folds of a filter paper.
2. Put it on a watch-glass and try to cut it with a knife.
3. What do you observe?
4. Take pieces of iron, zinc, lead and copper.
5. Place any one metal on a block of iron and strike it four or five times with a hammer.
6. What do you observe?
7. Repeat with other metals.
8. Record the change in the shape of these metals.
9. Set up an electric circuit as shown in the figure.
10. Place the metal to be tested in the circuit between terminals A and B as shown.
11. Does the bulb glow? What does this indicate?

Observations:

1. Yes, the bulb glows as metals are a good conductor of heat and electricity.
2. List the metals whose wires you have seen in daily life.
3. Iron, copper, aluminium and magnesium have lustre which clearly appears on rubbing them with sandpaper.
4. Only Na and Mg are soft metals and we can cut them with a knife but other metals are hard.
5. Fe, Zn, Pb and Cu are metals and show proper malleability when hammered.

Class 10 Science Activity 3.8 Page No. 40

1. Take a magnesium ribbon and some sulphur powder.
2. Burn the magnesium ribbon. Collect the ashes formed and dissolve them in water.
3. Test the resultant solution with both red and blue litmus paper.
4. Is the product formed on burning magnesium acidic or basic?
5. Now burn sulphur powder. Place a test tube over the burning sulphur to collect the fumes produced.
6. Add some water to the above test tube and shake.
7. Test this solution with blue and red litmus paper.
8. Is the product formed on burning sulphur acidic or basic?
9. Can you write equations for these reactions?

Observations:
(i)

Result: Oxide of metal is basic.
(ii)

Result: Oxide of non-metal is acidic.

Class 10 Science Activity 3.9 Page No. 41

Caution:

1. The following activity needs the teacher’s assistance. It would be better if students wear eye protection.
2. Hold any of the samples taken above with a pair of tongs and try burning over a flame. Repeat with the other metal samples.
3. Collect the product if formed.
4. Let the products and the metal surface cool down.
5. Which metals burn easily?
6. What flame colour did you observe when the metal burnt?
7. How does the metal surface appear afterburning?
8. Arrange the metals in the decreasing order of their reactivity towards oxygen.
9. Are the products soluble in water?

Only Zn and Al form amphoteric oxide (in nature) other metals forms basic oxides.

Class 10 Science Activity 3.10 Page No. 42

Caution:

1. This Activity needs the teacher’s assistance.
2. Collect the samples of the same metals as in Activity 3.9.
3. Put small pieces of the samples separately in beakers half-filled with cold water.
4. Which metals reacted with cold water?
5. Arrange them in the increasing order of their reactivity with cold water.
6. Did any metal produce fire on water?
7. Does any metal start floating after some time?
8. Put the metals that did not react with cold water in beakers half-filled with hot water.
9. For the metals that did not react with hot water, arrange the apparatus as shown in the figure and observe their reaction with steam.
10. Which metals did not react even with steam?
11. Arrange the metals in the decreasing order of reactivity with water.

Observations:

1. Metals which reacted with cold water → Na, K and Ca.
2. Metals which produced fire → Na and K.
3. Metals which started floating after some time → Ca and Mg.
4. Metals which reacted with hot water → Mg.
5. Metals which did not react with steam also → Pb, Cu, Ag and Au.
6. Arranging reactivity of metals in ascending order:
   K > Na > Ca > Mg > Al > Zn > Fe

Class 10 Science Activity 3.11 Page No. 44

Caution:

1. Do not take sodium and potassium as they react vigorously even with cold water.
2. Put the samples separately in test tubes containing dilute hydrochloric acid.
3. Suspend thermometers in the test tubes, so that their bulbs are dipped
   in the acid.
4. Observe the rate of formation of bubbles carefully.
5. Which metals reacted vigorously with dilute hydrochloric acid?
6. With which metal did you record the highest temperature?
7. Arrange the metals in the decreasing order of reactivity with dilute acids.
   
8. Collect all the metal samples except sodium and potassium again. If the samples are tarnished. rub them clean with sandpaper.

Class 10 Science Activity 3.12 Page No. 44-45

1. Take a clean wire of copper and an iron nail.
2. Put the copper wire in a solution of iron sulphate and the iron nail in a solution of copper sulphate taken in test tubes figure.
   
3. Record your observations after 20 minutes.
4. In which test tube did you find that a reaction has occurred? On what basis can you say that a reaction has actually taken place?
   
5. Only Zn and Al form amphoteric oxide (in nature) other metals forms basic oxides.
6. Metals which reacted with cold water → Na, K and Ca.
7. Metals which produced fire → Na and K.
8. Metals which started floating after some time → Ca and Mg.
9. Metals which reacted with hot water → Mg.
10. Metals which did not react with steam also → Pb, Cu, Ag and Au.
11. Arranging reactivity of metals in ascending order:
    K > Na > Ca > Mg > Al > Zn > Fe
    
12. Write a balanced chemical equation for the reaction that has taken place.
13. Name the type of reaction.

Observations:

1. In these above observations show iron is more reactive than copper.
   Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
2. In the given set-up (A) and (B) reactivity of metals is as follows:
   

Class 10 Science Activity 3.13 Page No. 48

1. Taķe samples of sodium chloride, potassium iodide, barium chloride or any other salt from the science laboratory.
2. What is the physical state of these salts?
3. Take a small amount of a sample on a metal spatula and heat directly on the flame figure. Repeat with other samples.

Observations:

1. What did you observe? Did the samples impart any colour to the flame? Do these compounds melt?
2. Try to dissolve the samples in water, petrol and kerosene. Are they soluble?
3. Make a circuit as shown in figure (testing the conductivity of a salt solution) and insert the electrodes into a solution of one salt. What did you observe? Test the other salt samples too in this manner.
4. What is your inference about the nature of these compounds? Sodium chloride, potassium iodide and barium chloride give the following observation:

Class 10 Science Activity 3.14 Page No. 53

1. Take three test tubes and place clean iron nails in each of them.
2. Label these test tubes A, B and C. Pour some water in test tube A and cork it.
3. Pour boiled distilled water in test tube B, add about 1 ml of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
4. Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture, if any, from the air. Leave these test tubes for a few days and then observe (Figure).

Observations:
Set of 3 test tube A, B and C are given below:

1. Test tube A: Iron nail became rusty.
   Presence of water and air in test-tube A: Air and water are both exposed.
2. Test tube B: Iron nail do not become rusty.
   Presence of water and air in test-tube B: Due to layer of oil, air does not expose.
3. Test tube C: Iron nail do not become rusty.
   Presence of water and air in test-tube C: Air and water both are not present.

Conclusion:

1. Test-tube A: Iron nail become rusty due to exposure to air and water.
2. Test-tube B: Iron nail exposed with water but not with air, so the iron nail does not rust.
3. Test-tube C: Iron nail not exposed to air and water, so iron nail do not rust.

So, air and water both are required for an iron nail to rust.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P.
Solution:

Solution:
(i) a_n = a + (n- 1)d
a₈ = 7 + (8 – 1)3 = 7 + 7 × 3 = 7 + 21
⇒ a₈ = 28

(ii) a_n = a + (n – 1)d
⇒ a₁₀ = -18 + (10 – 1)7 ⇒ 0 = -18 + 9d
⇒ 9d = 18 ⇒ d = \(\frac{18}{9}=2\)
∴ d = 2

(iii) a_n = a + (n – 1)d
⇒ -5 = a + (18 – 1) × (-3)
⇒ -5 = a + 17 × (-3)
⇒ -5 = a – 51 ⇒ a = -5 + 51 = 46
Thus, a = 46

(iv) a_n = a + (n – 1)d
⇒ 3.6 = -18.9 + (n – 1) × 2.5
⇒ (n – 1) × 2.5 = 3.6 + 18.9

⇒ n = 9 + 1 = 10
Thus, n = 10

(v) a_n = a + (n- 1)d
⇒ a_n = 3.5 + (105 – 1) × 0
⇒ a_n = 3.5 + 104 × 0 ⇒ a_n = 3.5 + 0 = 3.5
Thus, a_n = 3.5

Question 2.
Choose the correct choice in the following and justify:
(i) 30^th term of the AP: 10,7,4, , is, ….,
(A) 97
(B) 77
(C) -77
(D) -87

(ii) 11^th term of the AP: -3, \(-\frac{1}{2}\), 2, …. ,is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
(i) (C): Here, a = 10, n = 30
∵ T₁₀ = a + (n – 1)d and d = 7 – 10 = -3
∴ T₃₀ = 10 + (30 – 1) × (-3)
⇒ T₃₀ = 10 + 29 × (-3)
⇒ T₃₀ = 10 – 87 = -77

(ii) (B): Here, a = -3, n = 11 and

Question 3.
In the following APs, find the missing terms in the boxes:

Solution:
(i) Here, a = 2, T₃ = 26
Let common difference = d
∴ T_n = a + (n- 1 )d
⇒ T₃ = 2 + (3 – 1)d
⇒ 26 = 2 + 2 d
⇒ 2d = 26 – 2 = 24

(ii) Let the first term = a
and common difference = d
Here, T₂ = 13 and T₄ = 3
T₂ = a + d = 13, T₄ = a + 3d = 3
T₁ – T₂ = (a + 3d) – (a + d) = 3 – 13

(iii)

(iv) Here, a = – 4, T₆ = 6
∵ T_n = a + (n -1 )d
T₆ = – 4 + (6 – 1)d ⇒ 6 = -4 + 5d ⇒ 5d = 6 + 4 = d = 10 – 5 = 2
T₂ = a + d = -4 + 2 =-2
T₃ = a + 2d = -4 + 2(2) = 0
T₄ = a + 3d = -4 + 3(2) = 2
T₅ = a + 4d = -4 + 4(2) = 4

(v) Here, T₂ = 38 and T₆ = -22
∴ T₂ = a + d = 38, T₆ = a + 5d = -22
⇒ T₆ – T₂ = a + 5d – (a + d) = -22 – 38 -60
⇒ 4d = -60 ⇒ d = \(\frac{-60}{4}\) = -15
a + d = 38 ⇒ a + (-15) = 38
⇒ a = 38 + 15 = 53
Now,
T₃ = a + 2d = 53 + 2(-15) = 53 – 30 = 23
T₄ = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T₅ = a + 4d = 53 + 4(-15) = 53 – 60 = -7
Thus missing terms are

Question 4.
Which term of the AP: 3, 8, 13, 18, is 78?
Solution:
Let the n^th term be 78
Here, a = 3 ⇒ T₁ = 3 and T₂ = 8
∴ d = T₂ – T₁ = 8 – 3 = 5
And, T_n = a + (n- 1 )d
⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5
⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15
⇒ n = 15 + 1 = 16
Thus, 78 is the 16^th term of the given AP.

Question 5.
Find the number of terms in each of the following APs:
(i) 7,13,19, …….. ,205
(ii) 18, \(15 \frac{1}{2}\), 13, …… ,-47
Solution:
(i) Here, a = 7,d = 13 – 7 = 6
Let total number of terms be n.
∴ T_n = 205
Now, T_n = a + (n – 1) ×d
= 7 + (n – 1) × 6 = 205
⇒ (n – 1) × 6 = 205 – 7 = 198
⇒ n – 1 = \(\frac{198}{6}=33\)
∴ n = 33 + 1 = 34
Thus, the required number of terms is 34.

(ii)

Thus, the required number of terms is 27.

Question 6.
Check whether -150 is a term of the AP:
11, 8, 5, 2…
Solution:
For the given AP,
we have a = 11, d = 8 -11 = -3
Let -150 be the n^th term of the given AP
∴ T_n = a + (n – 1 )d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 – 11 = (n – 1) × (-3)
⇒ -161 = (n – 1) ⇒ (-3)

But n should be a positive integer.
Thus, -150 is not a term of the given AP

Question 7.
Find the 31^st term of an AP whose 11^th term is 38 and the 16^th term is 73.
Solution:
Here, T₁₁ = 38 and T₁₆ = 73
Let the first term = a and the common difference = d.
T_n = a + (n – 1 )d
Then, T_n = a + (11 – 1)d = 38
⇒ a + 10d = 38 …(1)
and T₁₆ = a + (16 – 1)d = 73
⇒ a + 15d = 73 …(2)
Subtracting (1) from (2), we get
(a + 15d) – (a + 10d) = 73 – 38

Question 8.
An AP consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term.
Solution:
Here, n = 50, T₃ = 12, T_n = 106
⇒ T₅₀ = 106
Let the first term = a and the common difference = d
T_n = a + (n – 1 )d
T₃ = a + 2d = 12 …(1)
T₅₀ = a + 49d = 106 …(2)
Subtracting (1) from (2), we get

Question 9.
If the 3^rd and the 9^th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Here, T₃ = 4 and T₉ = -8
T_n = a + (n – 1)d
T₃ = a + 2d = 4 …. (1)
T₉ = a + 8d = – 8 …. (2)
Subtracting (1) from (2), we get

Question 10.
The 17^th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let a be the first term and d the common difference of the given AP
Now, using _n = a + (n – 1 )d, we have

Thus, the common difference is 1.

Question 11.
Which term of the AP : 3, 15, 27, 39,… will be 132 more than its 54^th term?
Solution:
Here, a = 3, d = 15 – 3 = 12
Using T_n = a + (n – 1 )d, we get

Thus, 132 more than 54^th term is the 65^th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let for the 1^st AP, the first term = a and common difference = d

And for the 2^nd AP, the first term = a and common difference = d

According to the condition,

Question 13.
How many three-digit numbers are divisible by 7?
Solution:
The first three-digit number divisible by 7 is 105.
The last such three-digit number divisible by 7 is 994.
∴ The AP is 105,112,119, ,994
Let n be the required number of terms Here, a = 105, d = 7 and _n = 994

Thus, 128 numbers of 3-digits are divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 beyond 10 is 12.
The multiple of 4 just below 250 is 248.

Thus, the required number of terms is 60.

Question 15.
For what value of n, are the n^th terms of two APs: 63,65,67 … and 3,10,17, …. equal?
Solution:

Thus, the 13^th terms of the two given AP’s are equal.

Question 16.
Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Solution:
Let the first term = a and the common difference = d

Question 17.
Find the 20^th term from the last term of the AP : 3, 8, 13, …, 253.
Solution:
We have, the last term (l) = 253
Here, d = 8 – 3 = 5
Since, the n^th term before the last term is given by l – (n – 1 )d
We have 20^th term from the last term = l – (20 – 1) × 5 = 253 – 19 × 5 = 253 – 95 = 158

Question 18.
The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
Let the first term = a and the common difference = d

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ? 200 each year. In which year did his income reach ₹ 7000?
Solution:
Here, a = ₹ 5000 and d = ₹ 200
Let, in the nth year Subba Rao gets ₹ 7000.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the n^th week, her weekly savings become ₹ 20.75, find n.
Solution:
Here, a = ₹ 5 and d = ₹ 1.75

Thus, the required number of weeks is 10.

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.7 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the age of Ani = x years
and the age of Biju’s = y years
Case I:
y > x
According to 1^st condition : y – x = 3 …. (1)
∵ [Age of Ani’s father] = 2[Age of Ani] = 2x years


Substituting the value of x in equation (1),
we get y – 21 = 3 ⇒ y = 3 + 21 = 24
∴ Age of Ani = 21 years
Age of Biju = 24 years

Substituting the value of y in equation (1),
we get 19 – y = 3 ⇒ y = 16
∴ Age of Ani = 19 years
Age of Biju = 16 years

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]
Solution:
Let the capital of 1^st friend = ₹ x,
and the capital of 2^nd friend = ₹ y
According to the condition,
x + 100 = 2(y -100)
⇒ x + 100 – 2y + 200 = 0
⇒ x – 2y + 300 = 0 … (1)
Also, 6(x – 10) = y + 10 ⇒ 6x – y – 70 = 0 …. (2)
From (1), x = -300 + 2y (3)
Substituting the value of x in equation (2), we get
6[-300 + 2y] – y – 70 = 0
⇒ -1870 + 11y = 0

Now, Substituting the value of y in equation (3), we get, x = – 300 + 2y
= – 300 + 2(170) = – 300 + 340 = 40
Thus, 1^st friend has ₹ 40 and the 2^nd friend has ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the actual speed of the train = x km/hr
and the actual time taken = y hours
Distance = speed × time
According to 1^st condition: (x +10) × (y – 2) = xy
⇒ xy – 2x + 10y – 20 = xy
⇒ 2x – 10y + 20 = 0 … (1)
According to 2^nd condition: (x -10) × (y + 3) = xy
⇒ xy + 1ox -10y – 30 = xy
⇒ 3x – 10y – 30 = 0 ….(2)
Using cross multiplication for solving (1) and (2), we get

Thus, the distance covered by the train = 50 × 12 km = 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of students = x
and the number of rows = y
∴ Number of students in each row

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Sum of angles of a triangle = 180°
∴ ∠A + ∠B + ∠C = 180° … (1)
∵ ∠C = 3∠B = 2(∠A + ∠B) … (2)
From (1) and (2), we have ∠A + ∠B + 2 (∠A + ∠B) = 180°
⇒ ∠A + ∠B + 2∠A + 2∠B = 180°
⇒ ∠A + ∠B = 60° …. (3)
Also, ∠A + ∠B + 3∠B = 180°
⇒ ∠A + 4∠B = 180° ….(4)
Subtracting (3) from (4), we get
∠A + 4∠B – ∠A – ∠B = 180°- 60°

Substituting ∠B = 40° in (4) we get,
∠A + 4(40°) = 180°
⇒ ∠A = 180° – 160° = 20°
∴ ∠C = 3∠B = 3 × 40° = 120°
Thus, ∠A = 20°, ∠B = 40° and ∠C = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and they-axis.
Solution:
To draw the graph of 5x – y = 5, we get

and for equation 3x – y = 3, we get

Plotting the points (1, 0), (2, 5) and (0, -5), we get a straight line l₁. Plotting the points (2, 3), (3, 6) and (0, -3), we get a straight line l₂.

From the figure, obviously, the vertices of the triangle formed are A(l, 0), B(0, -5) and C(0,-3).

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q, qx – py = p + q
(ii) ax + by = c, bx + ay = 1 + c
(iii) \(\frac{x}{a}-\frac{y}{b}\)= 0, ax + by = a² + b²
(iv) (a – b)x + (a + b)y = a² – 2ab – b²,
(a + b)(x + y) = a² + b²
(v) 152x – 378y = -74, – 378x + 152y = -604
Solution:

Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution:
ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180° and
∠B + ∠D = 180°
⇒ [4y + 20] + [- 4x] = 180°
⇒ 4y – 4x + 20° -180° = 0
⇒ 4y – 4x – 160° = 0
⇒ y – x – 40° = 0 … (1)