Category: Class 7

MP Board Class 7th Maths Solutions Chapter 5 Lines and Angles Ex 5.2

Question 1.
State the property that is used in each of the following statements?
(i) If a||b, then ∠1 = ∠5.
(ii) If ∠4 = ∠6, then a||b.
(iii) If ∠4 + ∠5 = 180°,then a||b.

Solution:
(i) Corresponding angles property
(ii) Alternate interior angles property
(iii) Interior angles on the same side of the transversal are supplementary.

Question 2.
In the adjoining figure, identify

(i) the pairs of corresponding angles.
(ii) the pairs of alternate interior angles.
(iii) the pairs of interior angles on the same side of the transversal.
(iv) the vertically opposite angles.
Solution:
(i) ∠1 and ∠5; ∠2 and ∠6;
∠3 and ∠7; ∠4 and ∠8
(ii) ∠2 and ∠8; ∠3 and ∠5
(iii) ∠2 and ∠5; ∠3 and ∠8
(iv) ∠1 and ∠3; ∠2 and ∠4;
∠5 and ∠7; ∠6 and ∠8

Question 3.
In the adjoining figure, p||q. Find the unknown angles.

Solution:
∠d = 125° (Corresponding angles)
∠e = 180° – 125° = 55° (Linear pair)
∠f = ∠e = 55° (Vertically opposite angles)
∠c = ∠f = 55° (Corresponding angles)
∠a = ∠e – 55° (Corresponding angles)
∠b = ∠d = 125° (Vertically opposite angles)

Question 4.
Find the value of x in each of the following figures l ||m

Solution:
(i) We have,

y = 110° (Corresponding angles)
x + y = 180° (Linear pair)
⇒ x = 180° -110° = 70°

(ii) x = 100° (Corresponding angles)

Question 5.
In the given figure, the arms of two angles are parallel. If ∠ABC = 70°, then find
(i) ∠DGC
(ii) ∠DEF

Solution:
(i) Consider that AB||DG and a transversal line BC is intersecting them.
∴ ∠DGC = ∠ABC (Corresponding angles)
⇒ ∠DGC = 70°

(ii) Consider that BC 11 EF and a transversal line DE is intersecting them.
∴ ∠DEF = ∠DGC (Corresponding angles)
⇒ ∠DEF = 70°

Question 6.
In the given figures below, decide whether l is parallel to m.


Solution:
(i) Consider two lines l and m and a transversal line n which is intersecting them. Sum of the interior angles on the same side of transversal = 126° + 44° = 170°.
As the sum of interior angles on the same side of the transversal is not 180°, therefore, 1 is not parallel to m.

(ii) We have,

x + 75° = 180°(Linear pair on line l)
⇒ x = 180° – 75° = 105°
For l and m to be parallel to each other, corresponding angles (∠ABC and x) should be equal. However, here their measures are 75° and 105°. Hence, the lines l and m are not parallel to each other.

(iii) We have,

x + 123° = 180° (Linear pair on line m)
⇒ x = 180° – 123° = 57°
For l and m to be parallel to each other, corresponding angles (∠ABC and x) should be equal. However their measures are equal to 57°. Hence, l||m.

(iv) We have,

98° + x = 180°(Linear pair on line l)
⇒ x = 180° – 98° = 82°
For l and m to be parallel to each other, corresponding angles (∠ABC and x) should be equal. However, their measures are 72° and 82°. Hence, the lines l and m are not parallel to each other.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 14 Symmetry Ex 14.3

Question 1.
Name any two figures have both line symmetry and rotational symmetry.
Solution:
An equilateral triangle and regular hexagon have both line symmetry and rotational symmetry.

Question 2.
Draw, wherever possible, a rough sketch of
(i) a triangle with both line and rotational symmetries of order more than 1.
(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
(iii) a quadrilateral with a rotational symmetry of order more than 1 but not a line symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Solution:
(i) An equilateral triangle has 3 lines of symmetry and rotational symmetry of order 3.

(ii) An isosceles triangle has only 1 line of symmetry and no rotational symmetry of order more than 1.

(iii) A parallelogram is a quadrilateral which has no line symmetry but a rotational symmetry of order 2.

(iv) A kite is a quadrilateral which has only 1 line of symmetry and no rotational- symmetry of order more than 1.

Question 3.
If a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 1 ?
Solution:
Yes, if a figure has two or more lines of symmetry, then it will definitely have its rotational symmetry of order more than 1.

Question 4.
Fill in the blanks:

Solution:
The given table can be completed as follows:

Question 5.
Name the quadrilaterals which have both line and rotational symmetry of order more than 1.
Solution:
Square, rectangle, and rhombus are the quadrilaterals which have both line and rotational symmetry of order more than 1. A square has 4 lines of symmetry and rotational symmetry of order 4. A rectangle has 2 lines of symmetry and rotational symmetry of order 2. A rhombus has 2 lines of symmetry and rotational symmetry of order 2.

Question 6.
After rotating by 60° about a centre, a figure looks exactly the same as its original position. At what other angles will this happen for the figure?
Solution:
It can be observed that if a figure looks symmetrical on rotating by 60°, then it will also look symmetrical on rotating by 120°, 180°, 240°, 300°, 360° i.e., further multiples of 60°.

Question 7.
Can we have a rotational symmetry of order more than 1 whose angle of rotation is
(i) 45°?
(ii) 17°?
Solution:
It can be observed that if the angle of rotation of a figure is a factor of 360°, then it will have a rotational symmetry of order more than 1.
(i) It can be checked that 45° is a factor of 360°. Therefore, the figure having its angle of rotation as 45° will have its rotational symmetry of order more than 1.
(ii) 17° is not a factor of 360°. Therefore, the figure having its angle of rotation as 17° will not be having its rotational symmetry of order more than 1.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 5 Lines and Angles Ex 5.1

Question 1.
Find the complement of eachof the following angles:

Solution:
The sum of the measures of complementary angles is 90°.
(i) Complement of 20° = 90°- 20° = 70°
(ii) Complement of 63° = 90° – 63° = 27°
(iii) Complement of 57° = 90° – 57° = 33°

Question 2.
Find the supplement of each of the following angles:

Solution:
The sum of measures of supplementary angles is 180°.
(i) Supplement of 105° = 180° – 105° = 75°
(ii) Supplement of 87° = 180° – 87° = 93°
(iii) Supplement of 154° = 180° – 154° = 26°

Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary.
(i) 65°, 115°
(ii) 63°, 27°
(iii) 112°, 68°
(iv) 130°, 50°
(v) 45°, 45°
(vi) 80°, 10°
Solution:
The sum of the measures of complementary angles is 90° and that of supplementary angles is 180°.
(i) Sum of the measures of angles = 65° +115° = 180°
∴ These angles are supplementary angles.
(ii) Sum of the measures of angles = 63° + 27° = 90°
∴ These angles are complementary angles.
(iii) Sum of the measures of angles = 112° + 68° = 180°
∴ These angles are supplementary angles.
(iv) Sum of the measures of angles = 130° + 50° = 180°
∴ These angles are supplementary angles.
(v) Sum of the measures of angles = 45° + 45° = 90°
∴ These angles are complementary angles.
(vi) Sum of the measures of angles = 80°+ 10° =90°
∴ These angles are complementary angles.

Question 4.
Find the angle which is equal to its complement.
Solution:
Let the angle be x.
Complement of this angle is also x.
∵ The sum of the measures of a pair of complementary angles is 90°.
∴ x + x = 90° ⇒ 2x = 90°

Question 5.
Find the angle which is equal to its supplement.
Solution:
Let the angle be x.
Supplement of this angle is also x.
∵ The sum of the measures of a pair of supplementary angles is 180°.
∴ x + x = 180°
⇒ 2x = 180° ⇒ x = 90°

Question 6.
In the given figure, ∠1 and ∠2 are supplementary angles. If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary.

Solution:
∠1 and ∠2 are supplementary angles.
If ∠1 is decreased, then ∠2 should be increased by the same measure so that the given pair of angles remains supplementary.

Question 7.
Can two angles be supplementary if both of them are:
(i) acute?
(ii) obtuse?
(iii) right?
Solution:
(i) No. Acute angle is always less than 90°. It can be observed that two angles, even of 89°, cannot add up to 180°. Therefore, two acute angles cannot form a supplementary.
(ii) No. Obtuse angle is always greater than 90°. It can be observed that two angles, even of 91°, will always add up to more than 180°. Therefore, two obtuse angles cannot form a supplementary.
(iii) Yes. Right angles are of 90° and 90° + 90° = 180°.
Therefore, two right angles form a supplementary angle together.

Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?
Solution:
Let A and B are two angles making a complementary angle pair and A is greater than 45°.
A + B = 90°
⇒ B = 90° – A
Therefore, B will be less than 45°.

Question 9.
In the adjoining figure:

(i) Is ∠1 adjacent to ∠2?
(ii) Is ∠AOC adjacent to ∠AOE?
(iii) Do ∠COE and ∠EOD form a linear pair?
(iv) Are ∠BOD and ∠DOA supplementary?
(v) Is ∠1 vertically opposite to ∠4?
(vi) What is the vertically opposite angle of ∠5?
Solution:
(i) Yes. Since, ∠1 and ∠2 have a common vertex O and a common arm OC. Also, their non-common arms OA and OE are on the opposite side of the common arm.
(ii) No. ∠AOC and ∠AOE have a common vertex O and also a common arm OA. But, their non-common arms OC and OE are on the same side of the common arm. Therefore, they are not adjacent to each other.
(iii) Yes. Since, ∠COE and ∠EOD have a common vertex O and a common arm OE. Also, their non-common arms, OC and OD, are opposite rays.
(iv) Yes. Since, ∠BOD and ∠DOA have a common vertex O and their non-common arms are opposite to each other.
(v) Yes. Since, ∠1 and ∠4 are formed due to the intersection of two straight lines, AB and CD.
(vi) ∠COB is the vertically opposite angle of ∠5 as they are formed due to the intersection of two straight lines, AB and CD.

Question 10.
Indicate which pairs of angles are:
(i) Vertically opposite angles.
(ii) Linear pairs.

Solution:
(i) ∠1 and ∠4, ∠5 and ∠2 + ∠3 are vertically opposite angles as they are formed due to the intersection of two straight lines.
(ii) ∠1 and ∠5, ∠5 and ∠4 are forming linear pairs as they have a common vertex and also, have non-common arms opposite to each other.

Question 11.
In the following figure, is ∠1 adjacent to ∠2? Give reasons.

Solution:
∠1 and ∠2 are not adjacent angles because their vertex is not common.

Question 12.
Find the values of the angles x, y and z in each of the following:

Solution:
(i) Since, x and 55° are vertically opposite angles,
x = 55° .
x + y = 180° (Linear pair)
⇒ 55° + y = 180°
⇒ y = 180° – 55° = 125°
y = z (Vertically opposite angles)
∴ z = 125°

(ii) z = 40° (Vertically opposite angles)
y + z = 180° (Linear pair)
⇒ y = 180° – 40° = 140°
∵ 40° + x + 25° = 180°
(Angles on a straight line)
⇒ 65° + x = 180°
⇒ x = 180° – 65° = 115°

Question 13.
Fill in the blanks:
(i) If two angles are complementary, then the sum of their measures is ___ .
(ii) If two angles are supplementary, then the sum of their measures is ___.
(iii) Two angles forming a linear pair are ___.
(iv) If two adjacent angles. are supplementary, they form a ___.
(v) If two lines intersect at a point, then the vertically opposite angles are always ___.
(vi) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are ___.
solution:
(i) 90°
(ii) 180°
(iii) supplementary
(iv) linear pair
(v) equal
(vi) obtuse angles

Question 14.
In the adjoining figure, name the following pairs of angles.

(i) Obtuse vertically opposite angles
(ii) Adjacent complementary angles
(iii) Equal supplementary angles
(iv) Unequal supplementary angles
(v) Adjacent angles that do not form a linear pair
Solution:
(i) ∠AOD, ∠BOC
(ii) ∠EOA, ∠AOB
(iii ) ∠EOB, ∠EOD
(iv) ∠EOA, ∠EOC
(v) ∠AOB and ∠AOE; ∠AOE and ∠EOD; ∠EOD and ∠DOC

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 14 Symmetry Ex 14.2

Question 1.
Which of the following figures have rotational symmetry of order more than 1 :

Solution:
(a) The given figure has its rotational symmetry of order 4.

(b) The given figure has its rotational symmetry of order 3.

(c) The given figure has its rotational symmetry of order 1.

(d) The given figure has its rotational symmetry of order 2.

(e) The given figure has its rotational symmetry of order 3.

(f) The given figure has its rotational symmetry of order 4.

Hence, figures (a), (b), (d), (e) and (f) have rotational symmetry of order more than 1.

Question 2.
Give the order of rotational symmetry for each figure:

Solution:
(a) The given figure has rotational symmetry of order 1.

(b) The given figure has rotational symmetry of order 2.

(c) The given figure has rotational symmetry of order 3.

(d) The given figure has rotational symmetry of order 4.

(e) The given figure has rotational symmetry of order 4.

(f) The given figure has rotational symmetry of order 5.

(g) The given figure has rotational symmetry of order 6.

(h) The given figure has rotational symmetry of order 3.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 4 Simple Equations Ex 4.4

Question 1.
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the result is 23.
Solution:
(a) Let the number be x.
8 times of this number = 8x
According to question;
8x + 4 = 60 ⇒ 8x = 60 – 4
(Transposing 4 to R.H.S.)
⇒ 8x = 56
Dividing both sides by 8,

(b) Let the number be x
One-fifth of this number \(\frac{x}{5}\)
According to question;

(Transporting -4 to R.H.S.)
Mutiplying both sides by 5,

(c) Let the number be x.
Three-fourth of this number = \(\frac{3 x}{4}\)
According to question;

Multiplying both sides by 4,

Dividing both sides by 3,

(d) Let the number be x.
Twice of this number = 2x
According to question;
2x – 11 = 15 ⇒ 2x = 15 + 11
(Transposing -11 to R.H.S.)
⇒ 2x = 26
Dividing both sides by 2,

(e) Let the number of notebooks be x.
Thrice the number of notebooks = 3x
According to question;
⇒ 50 – 3x = 8 ⇒ – 3x = 8 – 50
(Transposing 50 to R.H.S.)
⇒ – 3x = – 42
Dividing both sides by – 3,

(f) Let the number be x.
According to question;

Multiplying both sides by 5,

⇒ x = 40 – 19 (Transposing 19 to R.H.S.)
⇒ x = 21

(g) Let the number be x.
\(\frac{5}{2}\) of this number = \(\frac{5x}{2}\)
According to question;

Question 2.
Solve the following;
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:
(a) Let the lowest score be l.
According to question;
Highest marks = 2 × Lowest marks + 7
⇒ 87 = (2 × l) + 7 × 2l + 7 = 87
⇒ 2l = 87 – 7
(Transposing 7 to R.H.S.)
⇒ 2l = 80
Dividing both sides by 2,

Therefore, the lowest score is 40.

(b) Let the base angles be equal to b.
Vertex angle = 40°
The sum of all interior angles of a triangle is 180°.
b + b + 40° = 180° ⇒ 2b + 40° = 180°
⇒ 2b = 180° – 40° = 140°
(Transposing 40° to R.H.S.)
Dividing both sides by 2,

Therefore, the base angles of the triangle are of 70°.

(c) Let Rahul’s score be x.
Therefore, Sachin’s score = 2x
According to question;
Rahul’s score + Sachin’s score = 200 – 2
2x + x = 198 ⇒ 3x = 198
Dividing both sides by 3,

∴ Rahul’s score = 66
Sachin’s score = 2 × 66 = 132

Question 3.
Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Solution:
(i) Let Parmit has x marbles.
5 times the number of marbles Parmit has = 5x
According to question;
5x + 7 = 37 ⇒ 5x = 37 – 7 = 30
(Transposing 7 to R.H.S.)
Dividing both sides by 5,

Therefore, Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years.
According to question;
Her father’s age = 3 × Laxmi’s age + 4
⇒ 49 = (3 × x) + 4
⇒ 3x + 4 = 49 ⇒ 3x = 49 – 4
(Transposing 4 to R.H.S.)
⇒ 3x = 45
Dividing both sides by 3,

Therefore, Laxmi’s age is 15 years.

(iii) Let the number of fruit trees be x. According to question;
Number of non-fruit trees = 3 × Number of fruit trees + 2
77 = (3 × x) + 2
⇒ 3x + 2 = 77 ⇒ 3x = 77 – 2
(Transposing 2 to R.H.S.)
⇒ 3x = 75
Dividing both sides by 3,

Therefore, the number of fruit trees planted were 25.

Question 4.
Solve the following riddle:

Solution:
Let the number be x.
Seven times x = 7x
According to question; .
(7x + 50) + 40 = 300 ⇒ 7x + 90 = 300
⇒ 7x = 300 – 90 (Transposing 90 to R.H.S.)
⇒ 7x = 210
Dividing both sides by 7,

Therefore, the number is 30.

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Chapter 17 The Governor and State Council of Ministers

MP Board Class 7th Social Science Chapter 17 Text Book Questions

Fill in the blanks:

1. …………. is the head of the State.
2. The Chief Minister is appointed by the ………….
3. The leader of die Council of Minister is the ………….
4. The minimum age for the post of die Governor is …………..

Answer:

1. Governor
2. Governor
3. Chief Minister
4. 35 yrs.

Match the column ‘A ’ with Column ‘B’

Answer:

1. (d) President of Ministers.
2. (a) five + years by
3. (c) four of Ministers
4. (b) to make laws

MP Board Class 7th Social Science Chapter 17 Short Answer Type Questions

Question 1.
In whose name are the functions of the state government carried out?
Answer:
The functions of the State Government are carried out in the name of the Governor.

Question 2.
On whose advice are the members of the Council of Ministers appointed?
Answer:
On the advice of the Chief Ministers, the members of the Council of Ministers are appointed.

Question 3.
Who summons the meeting of the State Legislative Assembly?
Answer:
The Governor summons the meeting of the state Legislative Assembly.

MP Board Class 7th Social Science Chapter 17 Long Answer Type Questions

Question 1.
Describe the functions of the State Council of Ministers.
Answer:
The council of Ministers executes die laws, made by the Vidhan Sabha. It decides the polices of die government and advice the Governor in the matter of administration. It is the real executive of the state.

The main functions of the State Council of Ministers:

• Being the executive of the state, the council of Ministers formulates the policies of die state and gives advice to the Governor in administrative matters.
• All the bills are introduced by the Council of Ministers and play an important role in the making of laws.
• It formulates the economic and taxation policy of the state.
• It makes plans for the welfare of the state.

Question 2.
What are the qualifications for the post of Governor?
Answer:
Qualifications for Governor:

• A person has to be citizen of India.
• He must have completed the age of 35 years.
• He must not have held the office of profit, under the Central or State Government.
• He must not be a members of Parliament or State Assembly.

Question 3.
When and why does the Governor issue ordinance?
Answer:
The bills passed by the Assembly become laws only after the assent of the Governor. When Vidhan Sabha is not in session, and a law is urgently required the Governor himself pass some orders which are known as ordinances. It has the effect of law.

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 4 Simple Equations Ex 4.3

Question 1.
Solve the following equations:
(a) \(2 y+\frac{5}{2}=\frac{37}{2}\)
(b) 5t + 28 = 10
(c) \(\frac{a}{5}+3\) = 2
(d) \(\frac{q}{4}+7\) = 5
(e) \(\frac{5}{2} x\) = -10
(f) \(\frac{5}{2} x=\frac{25}{4}\)
(g) \(7 m+\frac{19}{2}\) = 13
(h) 6z + 10 = -2
(i) \(\frac{3 l}{2}=\frac{2}{3}\)
(j) \(\frac{2 b}{3}-5\) = 3
Solution:

Question 2.
Solve the following equations.
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = -21
(d) -4(2 + x) = 8
(e) 4(2 – x) = 8
Solution:
(a) 2(x + 4) = 12
Dividing both sides by 2,

(Transposing 4 to R.H.S.)

(b) 3(n – 5) = 21
Dividing both sides by 3,

(Transposing – 5 to R.H.S.)

(c) 3(n – 5) = -21
Dividing both sides by 3,

(Transposing – 5 to R.H.S.)

(d) – 4(2 + x) = 8
Dividing both sides by – 4,

(Transposing 2 to R.H.S.)

(e) 4(2 – x) = 8
Dividing both sides by 4,

(Transposing 2 to R.H.S.)
⇒ x = 0

Question 3.
Solve the following equations:
(a) 4 = 5(p – 2)
(b) -4 = 5(p – 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p – 1) = 34
(e) 0 = 16 + 4(m – 6)
Solution:
(a) 4 = 5(p – 2)
Dividing both sides by 5,

Question 4.
(a) Construct 3 equations starting with x= 2.
(b) Construct 3 equations starting with x = -2.
Solution:
(a) x = 2
Multiplying both sides by 5,
5x = 10 … (i)
(b) Let the number be x.
Subtracting 3 from both sides,
5x – 3 = 10 – 3
5x – 3 = 7 …(ii)
Dividing both sides by 2,

(b) x= -2
Subtracting 2 from both sides,
x – 2 = -2 – 2 ⇒ x – 2 = -4 … (i)
Again, take x = – 2
Multiplying both sides by 6,
6 × x = -2 × 6 ⇒ 6x = -12
Subtracting 12 from both sides,
6x – 12 = – 12 – 12 ⇒ 6x – 12 = – 24 …(ii)
Adding 24 to both sides,
6x – 12 + 24 = – 24 + 24
⇒ 6x + 12 = 0 …(iii)

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 4 Simple Equations Ex 4.2

Question 1.
Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = -7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = -4
Solution:
On adding 1 to both sides of the given equation, we obtain
x – 1 + 1 = 0 + 1 ⇒ x = 1

(b) x + 1 = 0
On subtracting 1 from both sides of the given equation, we obtain
x + 1 – 1 = 0 – 1 ⇒ x = -1

(c) x – 1 = 5
On adding 1 to both sides of the given equation, we obtain
x – 1 + 1 = 5 + 1 ⇒ x = 6

(d) x + 6 = 2
On subtracting 6 from both sides of the given equation, we obtain
x + 6 – 6 = 2 – 6 ⇒ x = -4

(e) y – 4 = – 7
On adding 4 to both sides of the given equation, we obtain
y – 4 + 4 = -7 + 4 ⇒ y = -3

(f) y – 4 = 4
On adding 4 to both sides of the given equation, we obtain
y – 4 + 4 = 4 + 4 ⇒ y = 8

(g) y + 4 = 4
On subtracting 4 from both sides of the given equation, we obtain
y + 4 – 4 = 4 – 4 ⇒ y = 0

(h) y + 4 = -4
On subtracting 4 from both sides of the given equation, we obtain
y + 4 – 4 = – 4 – 4 ⇒ y = -8

Question 2.
Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
(b) \(\frac{b}{2}\) = 6
(c) \(\frac{p}{7}\) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac{z}{3}=\frac{5}{4}\)
(g) \(\frac{a}{5}=\frac{7}{15}\)
(h) 20t = -10
Solution:
(a) 3l = 42
On dividing both sides of the given equation by 3, we obtain

(b) \(\frac{b}{2}\) = 6
On multiplying both sides of the given equation by 2, we obtain

(c) \(\frac{p}{7}\) = 4
On multiplying both sides of the given equation by 7, we obtain

(d) 4x = 25
On dividing both sides of the given equation by 4, we obtain

(e) 8y = 36
On dividing both sides of the given equation by 8, we obtain

(f) \(\frac{z}{3}=\frac{5}{4}\)
On multiplying both sides of the given equation by 3, we obtain

(g) \(\frac{a}{5}=\frac{7}{15}\)
On multiplying both sides of the given equation by 5, we obtain

(h) 20t = -10
On dividing both sides of the given equation by 20, we obtain

Question 3.
Give the steps you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) \(\frac{20 p}{3}\) = 40
(d) \(\frac{3 p}{10}\) = 6
Solution:
(a) 3n – 2 = 46
On adding 2 to both sides, we obtain
3n – 2 + 2 = 46 + 2 ⇒ 3n = 48
On dividing both sides of the given equation by 3, we obtain

(b) 5m + 7 = 17
On subtracting 7 from both sides, we obtain
5m + 7 – 7 = 17 – 7 ⇒ 5m = 10
On dividing both sides of the given equation by 5, we obtain

(c) \(\frac{20 p}{3}\) = 40
On multiplying both sides of the given equation by 3, we obtain

On dividing both sides of the given equation by 20, we obtain

(d) \(\frac{3 p}{10}\) = 6
On multiplying both sides of the given equation by 10, we obtain 3pxl0

On dividing both sides of the given equation by 3, we obtain

Question 4.
Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) \(\frac{p}{4}\) = 5
(d) \(\frac{-p}{3}\) = 5
(e) \(\frac{39}{4}\) = 6
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Solution:

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Chapter 14 The Age of Magnificence

MP Board Class 7th Social Science Chapter 14 Text Book Questions

Choose the correct alternatives from the following:

Question 1.
Just after ascending the throne Jahangir had to face the revolt of:
(a) Khusrau
(b) Khurrum
(c) Salim
(d) Shiya
Answer:
(a) Khusrau

Question 2.
Tajmahal was built by:
(a) Akbar
(b) Jahangir
(c) Shahjahan
(d) Aurangzeb
Answer:
(c) Shahjahan

Question 3.
The ninth Guru of Sikhs was:
(a) AijunDev
(b) Teg Bahadur
(c) Govind Singh
(d) Ranjeet Singh
Answer:
(b) Teg Bahadur

Fill in the blanks:

1. Sir Thomas Roe and Captain Hawkins came to India during the reign of ………………
2. ………………………. was studded in the peacock.
3. ………………………. pirated in the Bay of Bengal.
4. ………………………… was die famous warrior and able statesman.

Answer:

1. Jahangir
2. Koh-i-noor
3. Portuguese
4. Mallik Amber

MP Board Class 7th Social Science Chapter 14 Short Answer Type Questions

Question 1.
Why did Jahangir execute Guru Arjun Dev?
Answer:
Jahangir was angry with Gum Arjun Dev (Sikh Gum) because he had financially helped prince Khusrau to revolt against him. So he arrested him and tortured him to death.

Question 2.
What did Jahangir do so that the common people received justice?
Answer:
Jahangir was famous for his judicial system. He put a gold chain with bells on it outside the palace. Anybody could appeal to the king for justice by pulling the chain.

Question 3.
What effect had Nurjahan on the administration?
Answer:
In 1611 4D Nurjahan was married to Jahangir. She was very beautiful and intelligent woman. She set new codes and ideologies for the court and even new fashion began from there. She was capable and looked after the administration very effectively. Jahangir used to take her advice in all important matters. Later Nurjahan’s influence in administration increased to such an extent that the royal firmans were issued in her name.

Question 4.
Mention the important buildings made by Shahjahan.
Answer:
The important buildings made by Shahjahan are, die Red Fort of Delhi, Diwan-i- Aaam, Diwan-i-Khas, JamaMasjid, Taj MahaL

MP Board Class 7th Social Science Chapter 14 Long Answer Type Questions

Question 1.
Why is the reign of Jahangir and Shahjahan known as the age of magnificence?
Answer:
The reign of Jahangir and Shahjahan had die most peaceful region. This was so because Akbar set up a vast empire on the basis of his power and good policies and gave it stability. So during the reign of Jahangir and Shahjahan the life was quite comfortable. They encouraged art and literature. Many beautiful and magnificent buildings were built The Royal family and the Aristocracy lived a life of great luxury. It was therefore the reign of Jahangir and Shahjahan is known as the age of magnificence.

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 4 Simple Equations Ex 4.1

Question 1.
Complete the last column of the table.

Solution:
(i) x + 3 = 0
By putting x = 3
L.H.S. = 3 + 3 = 6 ≠ R.H.S.
Hence, the equation is not satisfied.

(ii) x + 3 = 0
By putting x = 0
L.H.S. = 0 + 3 = 3 ≠ R.H.S.
Hence, the equation is not satisfied.

(iii) x + 3 = 0
By putting x = – 3
L.H.S. = -3 + 3 = 0 = R.H.S.
Hence, the equation is satisfied.

(iv) x – 7 = 1
By putting x = 7
L.H.S. = 7 – 7 = 0 ≠ R.H.S.
Hence, the equation is not satisfied.

(v) x – 7 = 1
By putting x = 8 L.H.S. = 8 – 7 = 1 = R.H.S.
Hence, the equation is satisfied.

(vi) 5x = 25
By putting x = 0
L.H.S. = 5 × 0 = 0 ≠ R.H.S.
Hence, the equation is not satisfied.

(vii) 5x = 25
By putting x = 5
L.H.S. = 5 × 5 = 25 = R.H.S.
Hence, the equation is satisfied,

(viii) 5x = 25 .
By putting x = – 5
L.H.S. = 5 × (- 5) = – 25 ≠ R.H.S.
Hence, the equation is not satisfied.

(ix) \(\frac{m}{3}\) = 2
By putting m = – 6
L.H.S. = \(\frac{-6}{3}\) = -2 ≠ R.H.S.
Hence, the equation is not satisfied.

(x) \(\frac{m}{3}\) = 2
By putting m = 0
L.H.S. = \(\frac{0}{3}\) = 0 ≠ R.H.S.
Hence, the equation is not satisfied.

(xi) \(\frac{m}{3}\) = 3
By putting m = 6
L.H.S. = \(\frac{6}{3}\) = 2 = R.H.S
Hence, the equation is satisfied.

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19(n = 1)
(b) 7n + 5 = 19(n = -2)
(c) 7n + 5 = 19(n = 2)
(d) 4p – 3 = 13(p = 1)
(e) 4p – 3 = 13(p = -4)
(f) 4p – 3 = 13 (p = 0)
Solution:
(a) n + 5 = 19 (n = 1)
Putting n = 1
L.H.S. = n + 5 = 1 + 5 = 6 ≠ 19 = R.H.S.
As L.H.S. ≠ R.H.S.
Therefore, n = 1 is not a solution of the given equation, n + 5 = 19.

(b) 7n + 5 = 19 (n =- 2)
Putting n = – 2
L.H.S. = 7n + 5 = 7 × (-2) + 5 = -14 + 5
= -9 ≠ 19 = R.H.S.
As L.H.S. ≠ R.H.S.
Therefore, n = -2 is not a solution of the given equation, 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)
Putting n = 2
L.HS. = 7n + 5 = 7 × (2) + 5 = 14 + 5 = 19 = R.HS.
As L.H.S. = R.H.S.
Therefore, n = 2 is a solution of the given equation, 7n + 5 = 19.

(d) 4p – 3 = 13(p = 1)
Putting p = 1
L.H.S. = 4p – 3 = (4 × 1) – 3
= 1 × 13 = R.H.S.
As L.H.S. × R.H.S.
Therefore, p = 1 is not a solution of the given equation, 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)
Putting p = – 4
L.H.S. = 4p – 3 = 4 × (- 4) – 3 = -16 – 3
= -19 ≠ 13 = R.H.S.
As L.H.S. ≠ R.H.S.
Therefore, p = – 4 is not a solution of the given equation, 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)
Putting p = 0
L.HS. = 4p – 3 = (4 × 0) – 3 = -3 ≠ 13 = RH.S.
As L.H.S. ≠ R.H.S.
Therefore, p = 0 is not a solution of the given equation, 4p – 3 = 13.

Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m -14 = 4
Solution:
(i) 5p + 2 = 17 Putting p = 1
L.H.S. = (5 × 1)+ 2 = 7 ≠ R.H.S.
Putting p = 2
L.H.S. = (5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S.
Putting p = 3
L.H.S. = (5 × 3) + 2 = 17 = R.H.S.
Hence, p = 3 is a solution of the given equation.

(ii) 3m – 14 = 4
Putting m = 4,
L.H.S. = (3 × 4) – 14 = -2 ≠ R.H.S.
Putting m = 5,
L.H.S. = (3 × 5) – 14 = 1 ≠ R.H.S.
Putting m = 6,
L.H.S. = (3 × 6) – 14 = 18 -14 = 4 = R.H.S.
Hence, m = 6 is a solution of the given equation.

Question 4.
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of f is 15.
(vi) Seven times m plus 7 gives 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Solution:
(i) x + 4 = 9
(ii) y – 2 = 8
(iii) 10a = 70
(iv) \(\frac{b}{5}\) = 6
(v) \(\frac{3}{4} t\) = 15
(vi) Seven times of m is 7m.
∴ 7m + 7 = 77
(vii) One-fourth of a number x is \(\frac{x}{4}\).

(viii) Six times of y is 6y.
∴ 6y – y = 60

(ix) One-third of z is \(\frac{x}{4}\) .

Question 5.
Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) \(\frac{m}{5}\) = 3
(v) \(\frac{3 m}{5}\) = 6
(vi) 3p + 4 = 25
(viii) \(\frac{p}{2}\) + 2 = 8
Solution:
(i) The sum of p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Twice of a number m is 7.
(iv) One-fifth of a number m is 3.
(v) Three-fifth of a number m is 6.
(vi) Three times of a number p, when added to 4, gives 25.
(vii) 2 subtracted from four times of a number p is 18.
(viii) 2 added to half of a number p gives 8.

Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87, (Take the lowest score to be l.)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees.)
Solution:
(i) Let Parmit has in marbles.
Number of marbles Irfan has = 5 × Number of marbles Parmit has +7
∴ 5 × m + 7 = 37 ⇒ 5m + 7 = 37

(ii) Let Laxmi be i years old.
Laxmi’s father’s age 3 × Laxmi’s age +4
∴ 49 = 3xy + 4 ⇒ 3y + 4 = 49

(iii) Let the lowest marks be 1.
Highest marks = 2 × Lowest marks +7
∴ 87 = 2 × l + 7 ⇒ 2l + 7 = 87

(iv) An isosceles triangle has two of its angles of equal measure.
Let base angle be b.
Vertex angle =2 × Base angle = 2b
Sum of all interior angles of a triangle 180°
∴ b + b + 2b = 180° ⇒ 4b = 180°

MP Board Class 7th Maths Solutions