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Age problems, number problems, Area problems etc. can be solved by linear equations.

(a) Transposition

The shifting of a number from one side of an equation to other is called Transposition.

(c) a rational number

Given, 8x – 3 = 25 + 17x

Transposing -3 to RHS and it becomes 3 and 17x to LHS it becomes -17x.

8x – 17x = 25 + 3

-9x = 28

X = -28/9

Therefore x is a rational number.

Correct option is  B) 180°

(c) x = -b/a

Given, ax + b = 0

Transposing b to RHS and it becomes -b

Then,

ax = -b

x = -b/a

Let Leena’s present age be ‘x’ years 

16 years later she will be (x + 16) years 

Given x + 16 = 3x 

16 = 2x 

x = \(\frac{16}{2}\)

x = 8 years 

Leena’ s present age = 8 years

Correct option is  C) -37/4

Correct option is (C) -37/4

\(\frac{7}{4}\) – p = 11

\(\Rightarrow\) p = \(\frac{7}{4}\) - 11 \(=\frac{7-4\times11}{4}=\frac{7-44}{4}=\frac{-37}{4}.\)

Let Bindu’s present age be ‘x’ years. 

Mrs. Joseph’s present age = x + 27 

years. After 8 years Bindu’s age = x + 8 and Mrs. 

Josephs age = x + 27 + 8 

= x + 35 years. 

Given that x + 35 = 2(x + 8) 

x + 35 = 2x + 16 

35 – 16 = 2x – x 

19 = x 

x= 19 

Bindu’s present age = 19 years 

Mrs.Joseph’s age = x + 27 = 19 + 27 = 46 years

Correct option is   A) -c/a

Correct option is (A) -c/a

ax + c = 0

\(\Rightarrow\) ax = -c

\(\Rightarrow\) x = \(\frac{-c}a\).

Correct option is  D) none

Correct option is (D)

\(\frac{x}{5}+11=\frac{1}{15}\)

\(\Rightarrow\) \(\frac{x}{5}=\frac{1}{15}-11\) \(=\frac{1-11\times15}{15}=\frac{1-165}{15}=\frac{-164}{15}.\)

\(\Rightarrow\) \(x=\frac{-164}{15}\times5=\frac{-164}3.\)

Let Deep’s age be ‘x’, Viji’s age is 2x 

Difference of their age = 11 

2x – x = 11 

x = 11 

∴ Deepu’s age = x = 11 years 

Viji’s age = 2x = 2 x 11 = 22 years

D) 2x – x = 9, x = 9

Correct option is (D) 2x – x = 9, x = 9

(A) 3x = 10 \(\Rightarrow\) x = \(\frac{10}3.\) Thus, (A) is false.

(B) 2m = 1 \(\Rightarrow\) m = \(\frac12.\) Thus, (B) is false.

(C) \(\frac{2}{3}x^2=1\) \(\Rightarrow\) \(x^2=\frac1{\frac23}=\frac{3}{2}\) \(\Rightarrow\) \(x=\pm\sqrt{\frac32}.\)

(D) 2x – x = 9 \(\Rightarrow\) x = 9.

Let the present age of Shikha be x years.

Therefore, the present age of Shikha’s brother Ravish = (x + 3) years.

So, sum of their ages = x + (x+ 3)

⇒ x +(x + 3) = 37

⇒ 2x + 3 = 37

Subtracting 3 from both sides, we get

⇒ 2x+ 3 – 3 = 37 – 3

⇒ 2x = 34

Dividing both sides by 2, we get

⇒ 2x/2 = 34/2

⇒ x = 17

Therefore, the present age of Shikha = 17 years,

And the present age of Ravish = x + 3 = 17 + 3 = 20 years.

Let Nishu’s age be ‘x’ 

Sanju’s age = x + 6 

Sum of their ages = 28x + (x + 6) = 28 

2x + 6 = 28 

2x = 28 - 6 

2x = 22 

x = \(\frac{22}{11}\)

x = 11 

Nishu’sage = x = 11 years 

Sanju’ s age = x + 6 = 11 + 6 = 17 years

Correct option is   A) 35

Correct option is (A) 35

Let the number be x.

\(\therefore\) \(\frac x7\) = 5

\(\Rightarrow\) x = 5 \(\times\) 7 = 35.

Let the numbers be 3x and 5x.

According to the question we can write as

(3x+10) / (5x+10) = (5/7)

On cross multiplying we get

7(3x+10) = 5(5x+10)

21x+70=25x+50

On rearranging or transposing

70-50=25x-21x

4x=20

x=20/4=5

So the numbers are 15 and 25

Let the required number be x 

4 times the number = 4x 

12 less than 4 times the number = 4x – 12 

According to the statement 

4x – 12 = 28 

=> 4x = 28 + 12 

=> 4x = 40 

x = 10 

Required number = 10

Let the number be x.

According to the question we can write as

10+4x=5x-5

On rearranging

4x-5x=-5-10

-x=-15

So x=15

Let the bigger number be ‘x’. 

The sum of two numbers = 58 

∴ Smaller number = 58 – x 

The difference of two numbers = 28 

∴ x – (58 – x) = 28 

x – 58 + x = 28 

2x = 28 + 58 = 86 

∴ x = \(\frac{86}2\) = 43 

∴ Bigger number or one number = 43 

Smaller number or second number = 58 – 43 = 15

Let the bigger number be x. 

The difference between two numbers 8 

∴ Smaller number = x – 8

If 2 is added to the bigger number the result will be three times the smaller number. 

So x + 2 = 3(x – 8) 

x + 2 = 3x – 24

x – 3x = -24 – 2 

– 2x = -26

∴ x = \(\frac{26}2\) = 13

∴ Bigger number = 13 

Smaller number = 13 – 8 = 5

Let the required number be x.

So, according to question:

⇒ x/3 + 5 = 2x

Transposing x/3 to RHS, we get

⇒ 5 = 2x – (x/3)

⇒ 5 = (6x – x)/3

⇒ 5 = (5x/3)

Multiplying both sides by 3 we get,

⇒ 5 × 3 = (5x/3) × 3

⇒ 15 = 5x

Dividing both sides by 5 we get

⇒ 15/5 = 5x/5

⇒ 3 = x

Thus the number thought of by the man is 3.

Let first number be x.

According to the question second number is x + 1 and the third is x + 2

Sum of first and second numbers = (x) + (x + 1).

According to question:

⇒ (x) + (x + 1) = 15 + (x + 2)

⇒ 2x + 1 = 17 + x

Transposing x to LHS and 1 to RHS, we get

⇒2x – x = 17 – 1

⇒ x = 16

So, first number = x = 16,

Second number = x + 1 = 16 + 1 = 17

And third number = x + 2 = 16 + 2 = 18

Thus, the required consecutive natural numbers are 16, 17 and 18.

Let the smaller number be ‘x’.

So, the larger number = x + 7.

According to question:

⇒ 6x + (x + 7) = 77

⇒ 6x + x + 7 = 77

On simplifying we get

⇒ 7x + 7 = 77

Subtracting 7 from both sides, we get

⇒ 7x + 7 – 7 = 77 – 7

⇒ 7x = 70

Dividing both sides by 7, we get

⇒ 7x/7 = 70/7

⇒ x = 10

Thus, the smaller number = x = 10

And the larger number = x + 7 = 10 + 7 = 17.

The two required numbers are 10 and 17.

x⁹ – 5x⁸ – 14x⁷ = 0 

P(x) = x⁹ – 5x⁸ – 14x⁷. 

The number of sign changes is P(x) is 1. 

The number of positive roots is 1. P (-x) = -x⁹ – 5x⁸ + 14x⁷

The number of sign changes is P(-x) is one. The number of negative zero of P(-x) is 1. It is clear that 0 is a root of the equation. 

∴ The number of the imaginary roots is at least 6.

जनसंख्या बढ़ने के कारण वातावरण प्रदूषित हुआ है। वातावरणीय प्रदूषण के कारण ऐतिहासिक इमारतों की बाहरी परत खराब होने लगी है। उदाहरण के लिए वायु प्रदूषण के कारण आगरा स्थित ताजमहल का रंग पीला पड़ने लगा है इसके अतिरिक्त ऐसे स्थानों पर बढ़ती भीड़ के कारण वहाँ गंदगी भी फैलने लगी है।

अधिक जनसंख्या के कारण आज बसों, रेलगाड़ियों, हवाई अड्डों आदि सभी जगह अपार भीड़ लगी रहती है। बहुत से लोगों को बसों तथा गाड़ियों में बैठने तक की जगह नहीं मिल पाती है और उन्हें खड़े रहकर यात्रा करनी पड़ती है। इसके अलावा सड़कों पर बढ़ती वाहनों की संख्या के कारण कई-कई घंटों का जाम भी लग जाता है तथा दुर्घटनाएँ हो जाती हैं।

माल्थस के अनुसार खाद्य सामग्री में वृद्धि सदैव अंकगणितीय क्रम (जैसे १, २, ३, ४, ५…) में होती है, जबकि जनसंख्या वृद्धि ज्यामितीय क्रम (जैसे २, ४, ८, १६, …) में होती है।

लोगों की संख्या बढ़ने के कारण जब हमारे प्राकृतिक संसाधन कम पड़ने लगते हैं तो हम इसे जनसंख्या वृद्धि कहते हैं। जनसंख्या वृद्धि का पर्यावरण पर बुरा प्रभाव पड़ता है। आज का पर्यावरण प्रदूषण जनसंख्या वृद्धि की ही देन है।

Given:

Slant height of conical mountain = 2.5 km

Area of its base = 1.54 km²

Let the radius of base be ‘r’ km, height of the mountain is ‘h’ km and slant height be ‘l’ km

Area of base = πr²

1.54 = πr²

1.54 = 22/7 r²

or r = 0.7 km

We know that, l² = r² + h²

(2.5)² = (0.7)² + h²

6.25 – 0.49 = h²

or h = 2.4 km

Let radius = r and

Height = h

The sum of the radius of the base and the height of a solid cylinder is 37 metres.

⇨ r + h = 37 m …(1)

Total surface area of the cylinder = 2πr(r + h)

r = 7

From (1): 7 + h = 37

h = 30

Again,

Volume of the cylinder = πr² h

= 22/7 x 7 x 7 x 30

= 4620 cm³

Let the area of the circles be A₁ and A₂ and their circumference be c₁ and c₂ respectively.

According to the question it is clear that A₁ = 100 A₂

⇒ π (r₁)² = 100 × π (r₂)²

⇒ r₁ = 10 r₂

⇒ r₁/r₂ = 10/1 … (i)

Finding the ratios of the circumference;

C₁: C₂ = 2πr₁:  2πr₂

C₁/C₂ = (2πr₁)/ (2πr₂)

C₁/C₂ = r₁/r₂

Putting the value of r₁/r₂ from equation (i)

C₁/C₂ = 10/1

C₁: C₂ = 10: 1

Hence, the ratio of their circumferences is 10: 1.

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.
⇒ Number of elements in set B = 3
Number of elements in (A × B)
= (Number of elements in A) × (Number of elements in B)
= 3 × 3 = 9
Thus, the number of elements in (A × B) is 9.

Given: Circumference of the circle = 39.6 cm.

We know, Circumference of circle = 2πr

Where, r = Radius of the circle

⇨ 2πr = 39.6

r = 39.6/2π

r = 6.3

(put value of π = 22/7)

Area of the circle = πr²

Where, r = radius of the circle

⇨ Area of the circle = π(6.3)²

= 124.74

So, area of circle is 124.74 cm².

Area of a circle = 98.56 cm²

We know, Area of the circle = πr²

Where, r = radius of the circle

So, πr² = 98.56

Put π = 22/7

r² = 98.56/(22/7) = 31.36

r = 5.6

Circumference if circle is 5.6 cm

We know that area of circle = πr²

Given area = 4π cm²

A = πr²

4π = πr²

r² = 4

Therefore r = 2 cm

Given diameter = 5.6 cm

Therefore radius, r = d/2 = 5.6/2 = 2.8

We know that area of circle = πr²

A = (22/7) × (2.8)²

A = 24.64 cm²

Given area of the circle = 154 cm²

A = πr²

154 = 22/7 r²

r² = (154 × 7)/22

r² = 49

r = 7 cm

Given diameter = 7 km

Therefore radius, r = d/2 = 7/2 = 3.5

We know that area of circle = πr²

A = (22/7) × (3.5)²

A = 38.5 km²

We know that area of circle = πr²

Given area = 1.54 km²

A = πr²

1.54 = πr²

r² = (1.54 × 7)/22

r² = 0.49 km

r = 0.7 km

r = 700 m

Given circumference = 3.14 m

We know that circumference of circle = 2πr

3.14 = 2 × 3.14 × r

r = 3.14/ (2 × 3.14)

r = 0.5

We know that area of circle = πr²

A = (22/7) × (0.5)²

A = 0.785 m²

Given area of a circle is 50.24 m²

We know that area of circle = πr²

50.24 = (22/7) × r²

r² = (50.24 × 7)/22

r² = 15.985

r = 3.998 m

We know that circumference of circle = 2πr

C = 2 × (22/7) × 3.998

C = 25.12 m

From the question,

Horse is tied to a pole, then the pole will be the central point and the area over which the horse will graze will be a circle. The string by which the horse is tied will be the radius of the circle.

Then,

Radius of the circle, r = length of the string = 21 m

Then, area of the circle = πr²

= (22/7) × 21 × 21

= 22 × 3 × 21

= 1386 m²

Hence, the area of the horse can graze is 1386 m².

Given the length of the string = 28 m

The area over which the horse can graze is the same as the area of circle of radius 28 m

Hence required area = πr²

A = (22/7) × (28)²

A = 2464 m²

Let the number of boxes be x

\(\frac{13.6}{68}\) = \(\frac{20.4}{x}\)

On cross multiplication, we get

13.6x = 20.4 x 68

x = \(\frac{20.4\times 68}{13.6}\) = 102

Therefore number of boxes is 102

Let the fare be Rs x

\(\frac{12}{35}\) = \(\frac{294}{x}\)

On cross multiplication, we get

12x = 35 x 294

x = \(\frac{35\times 294}{12}\) = 857.5

Therefore thickness of the cardboard is 857.5 mm

Let the number of copies be x

\(\frac{3.4}{136}\) = \(\frac{5.1}{x}\)

On cross multiplication, we get

3.4x = 5.1 x 136

x = \(\frac{5.1\times 136}{3.4}\) = 204

Therefore number of copies is 204

Let the length of the cloth be x meter

\(\frac{242.50}{97}\) = \(\frac{302.50}{x}\)

On cross multiplication, we get

242.50x = 97 x 302.50

x = \(\frac{97\times 302.50}{242.50}\)= 121

Therefore length of the cloth is 121 M

Introduction : The hatred of Shylock is the main and base theme of the play. But he has reasons for his deep-rooted grudge against Antonio… His Hatred as a Jew: Antonio was a Christian and Shylock was a Jew. Their religions were responsible for their mutual hatred. Shylock admits that he hates Antonio because he is a Christian. Even Antonio hates Jews and Shylock was very well aware of Antonio’s feelings. Shylock says :

“He hates our sacred nation.”

Trading Competition : Antonio was a kind-hearted trader. He was indulged in reducing the rates of interest by lending out money without interest. He was only.. helping needy but it pinched Shylock, the money-lender. He used to charge high interest rates for income. But Antonio caused him great loss. So that Shylock hates Antonio. 

Antonio’s Criticism : Antonio has criticised Shylock for his charging high rates of interest, and thus, making huge profit. He scolds Shylock in public. This was all insulting for Shylock. 

Thus Shylock has powerful reasons for hating Antonio. His hatred is the main reason for his feeling of revenge.;

correct option is C) 92