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Correct option is D. none of these

The given system of equations is :-

x– 3y +2z = 1

x + 4y – 3z = 5

As there are three variables x, y, z and we have only two equations so it is impossible to find the solution. Thus, no solution exists for this system of equations.

\(\frac{3(y\,-\,5)}{4}-4y=3-\frac{(y\,-\,3)}{2}\)

Taking LCM of 4 and 1 on LHS = 4 and 1 and 2 on RHS = 2

\(\frac{3(y\,-\,5)-16y}{4}=\frac{6\,-\,y\,+\,3}{2}\)

By cross multiplication 

⇒ 3y - 15 - 16y = 2(9 - y) 

⇒ - 13y + 2y = 18 + 15 

⇒ - 11y = 33 

⇒ y = - 3

\(\frac{9x}{7\,-\,6x}=15\)

By cross multiplication 

9x = 15(7 - 6x) 

⇒ 9x + 90x = 105 

⇒ 99x = 105

⇒ \(\text{x}=\frac{35}{33}\)

Using the identity: a² – b² = (a + b)(a-b) 

{(83)² - (17)²} = (83 - 17)(83 + 17) 

= 66 × 100= 6600

(i) True 

It has two terms so binomial. 

(ii) True 

It has single term so monomial. 

(iii) False 

(5a - 9b) - ( - 6a + 2b) = 5a + 6a - 9b - 2b = 11a - 11b 

(iv) True

\(\frac{-8}{7}x^3y^4=\frac{-8}{7}\times8\times1=\frac{-64}{7}\)

(v) True 

Taking the LCM of 4,6 and 2 = 12

\(\frac{3\text{x}\,+\,2\text{x}-\,6\text{x}}{12}=\frac{3}{4}\)

\(\Rightarrow\) -4x = 36

\(\Rightarrow\) x = -9

(vi) False

2x - 5 = 0

\(\Rightarrow \text{x}=\frac{5}{2}\)

(i) \(\text{x}-\frac{1}{\text{x}}=4\)

Squaring both the sides,

\((\text{x}-\frac{1}{\text{x}})^2=4^2\)

Using the identity, (a – b)² = a² - 2ab + b²

\(\text{x}^2-2+\frac{1}{\text{x}^{2}}=4^2\)

\(\Rightarrow \text{x}^2+\frac{1}{\text{x}^{2}}=16+2=18\) -----------(1)

(ii) Squaring equation (1) using the identities, (a + b)² = a² + 2ab + b²

\(\Rightarrow\text{x}^4+2+\frac{1}{\text{x}^{4}}=324\)

\(\Rightarrow \text{x}^4+\frac{1}{\text{x}^4}=324-2=322\)

(i) (4a + 5b) x (5a - 6b) 

= 4a (5a - 6b) + 5b (5a - 6b) 

= 20a² - 24ab + 25ab - 30b² 

= 20a² + ab - 30b² 

(ii) (6x² - x + 8) × (x² - 3) 

(6x² - x + 8) × (x² - 3) 

= x² (6x² - x + 8) - 3 (6x² - x + 8) 

= 6x⁴ - x³ + 8x² - 18 x² + 3x – 24 

6x⁴ - x³ – 10x² + 3x - 24

Using the identity : a² – b² = (a + b)(a-b) 

(i) x² - 18x + 81 = x² - (9x) + 81 = (x - 9)(x - 9) = (x - 9)² 

(ii) (4 - 36 x²) = 4(1 - 9x²) = 4(1 - 3x)(1 + 3x) 

(iii) x² - 14x + 13 = x²–(13 + 1)x + 13 = x(x - 13) - 1(x - 13) = (x - 13)(x - 1) 

(iv) 9z² - x² - 4y² + 4xy = 9z² - (x - 2y)² = (3z - x + 2y)(3z + x - 2y) 

(v) abc - ab - c + 1 = ab(c - 1) - (c - 1) = (ab - 1)(c - 1)

10p² + 11p + 3 

By using Splitting the middle term 

10 p² + 11p + 3 

= 10 p² + (5 + 6)p + 3 

= 5p(2p + 1) + 3(2p + 1) 

= (2p + 1)(5p + 3)

12x² + 60x + 75 

By using Splitting the middle term 

12 x² + 60x + 75 

= 3(4 x² + (10 + 10)x + 25) 

= 3(2x(2x + 5) + 5(2x + 5)) 

= 3(2x + 5)(2x + 5)

8x³ - 2x 

Using the identity: a² - b² = (a + b)(a - b) 

8x³ - 2x 

= 2x(4x² - 1) 

= 2x(2x - 1)(2x - 1)

Let the number of boys = number of girls = x

2 (x – 8) = x

2x – 16 = x, 2x – x = 16

x – 16 = 0

∴ x = 16

∴ The number of boys = No. of girls = 16

Let the numbers be x and 95 - x 

⇒ 95 - x - x = 15 

By cross multiplication 

⇒ - 2x = - 80 

⇒ x = 40 

So, the numbers are 40 and 95 – 40 = 55

Given : 

Volume of the existing solution = 1125 liters 

Amount of acid in the existing solution = 45% of 1125 …(i) 

And the rest 55% of 1125 liters is the amount of water in it, which need not be computed. 

Let the water added (in liters) be x in 1125 liters of solution. 

According to the question,

x liters of water has to be added to 1125 liters of the 45% solution. 

We can say that,

Even if x liters of water is added to the 1125 liters of solution, acid content will not change. 

Only water content and the whole volume of the solution will get affected. 

So, the resulted solution will have acid content as follows : 

The acid content in the solution after adding x liters of water = 45% of 1125 …(ii) 

[∵ We know that the amount of acid content will not change after adding water to the whole solution. So, from equation (i), we have this conclusion] 

Also,

According to the question,

This resulting mixture will contain more than 25% acid content. 

So,

We have,

Acid content in the solution after adding x litres of water > 25% of new mixture 

⇒ 45% of 1125 > 25% of (1125 + x) 

[∵ from equation (ii)]

⇒ \(\frac{45}{100}\) \(\times\) 1125 > \(\frac{25}{100}\) \(\times\) (1125 + x)

⇒ 45 × 1125 > 25(1125 + x) 

⇒ 9 × 1125 > 5(1125 + x) 

⇒ 9 × 225 > 1125 + x 

⇒ 2025 > 1125 + x 

⇒ x < 2025 – 1125 

⇒ x < 900

Also, 

This resulting mixture will contain less than 30% acid content. 

So, 

We have Acid content in the solution after adding x litres of water < 30% of new mixture 

⇒ 45% of 1125 < 30% of (1125 + x) 

[∵ from equation (ii)]

⇒ \(\frac{45}{100}\)\(\times\)1125 >\(\frac{30}{100}\) \(\times\)(1125 + x) 

⇒ 45 × 1125 < 30(1125 + x) 

⇒ 9 × 1125 < 6(1125 + x) 

⇒ 3 × 1125 < 2(1125 + x) 

⇒ 3375 < 2250 + 2x 

⇒ 2x + 2250 > 3375 

⇒ 2x > 3375 – 2250 

⇒ 2x > 1125

⇒ x > \(\frac{1125}{2}\)

⇒ x > 562.5 

Thus, 

We have got,

x < 900 and x > 562.5. 

⇒ 562.5 < x < 900 

Hence,

The required liters of water to be added to 1125 liters of solution is between 562.5 liters and 900 liters.

Let the cost of a pen = x

cost of a book = x + 4

Cost of a pencil = x – 2

ie.5(x + 4) + 2x + 3(x – 2) = 74

5x + 20 + 2x + 3x – 6 = 74

10 x + 14 = 74, 10x = 74 – 14

x = \(\frac{74-14}{10}\) = 6

Cost of pen = Rs. 6

Cost of book = Rs. 10

Cost of pencil = Rs. 4

Let the speed of motorcyclists be x km/h and y km/h 

According to the question 

x + 7 = y (A) 

And 2y + 2x + 34 = 300 

Putting (A) we get 

⇒ 2(x + 7) + 2x + 34 = 300 

⇒ 2x + 14 + 2x = 300 - 34 

⇒ 4x = 266 - 14

\(\Rightarrow\text{x}=\frac{252}{4}=63\) km/h

⇒ y = \(\text{x}+7\) = 63 \(\frac{km}{h}\) + 7 = 70 km/h

Checking the answer:

2(70) + 2(63) + 34 = 140 + 126 + 34 = 300 = Distance between them

Hence, verified .

Let the given two angles of a triangle be 4x and 5x 

According to the question 

3^rd angle = 4x + 5x = 9x 

Using angle sum property of a triangle 

4x + 5x + 9x = 180° 

⇒ 18x = 180° 

⇒ x = 10 

So, the angles of the given triangle are: 

4x = 40° , 5x = 50° and 9x = 90°

Let the first part and second part be x and y respectively

According to the question

\(\frac{5}{100}\text{x}=\frac{10}{100}y\)

\(\Rightarrow\) y = \(\frac{5}{10}\text{x}=\frac{1}{2}\text{x}\)

Adding them

\(\text{x}+\frac{1}{2}\text{x}=4500\)

\(\Rightarrow \frac{3\text{x}}{2}=4500\)

⇒ 3x = 4500 × 2

\(\Rightarrow \text{x}=\frac{9000}{3}=3000\)

Second part = \(\frac{1}{2}\text{x}=1500\)

Let the first part be x of 150

According to the question second part is \(\frac{5}{6}\text{x}\)

And the third part is \(\frac{4}{5}(\frac{5}{6}\text{x})\)

Adding all of them

\(\text{x}+\frac{5}{6}\text{x}+\frac{20}{30}\text{x}=150\)

Taking LCM of 6 and 30 = 30

\(\Rightarrow \frac{30\text{x+25x+20x}}{30}=150\)

⇒ 75x = 150 × 30

\(\Rightarrow \text{x}=\frac{4500}{75}=60\)

Second part = \(\frac{5}{6}\text{x}=50\)

Third part = \(\frac{4}{5}(\frac{5}{6}\text{x})=40\)

Let the ages of cousins be x years and x - 10 years

According to the question

x - 15 = 2(x - 10 - 15)

By cross multiplication

⇒ x - 15 = 2x - 50 

⇒ x = 35 

So, cousins are 35 Years and 25 years in age

Let the age of Rakhi and Rakhi’s mother be x and 4x respectively 

According to the question

(4x + 5) = 3(x + 5)

⇒ 4x - 3x = 15 - 5

So, Rakhi' age = x = 10 Years

and Rakhi' s mother is 4x = 40 years

\(\frac{4}{5}\text{x}-\frac{3}{4}\text{x}=4\)

\(\Rightarrow \frac{16\text{x - 15x}}{20}=4\)

⇒ x = 80

Let the ages of A and B be 5x and 7x

\(\Rightarrow \frac{5\text{x + 4}}{7\text{x} \,+ \,4}=\frac{3}{4}\)

By cross multiplication 

⇒ 4(5x + 4) = 3(7x + 4) 

⇒ 21x - 20x = 16 - 12 

⇒ x = 4 

Age of B = 7x = 28 years

Let the length of equal sides be x cm. 

We know that, Perimeter = 16 cm 

⇒ x + x + 6 = 16 

⇒ 2x = 10 

⇒ x = 5cm

If x be the breadth, Then length = 2x + 1

2 (x + 2x + 1) = 80

2(3x + 1) = 80

6x + 2 = 80

6x = 80 – 2

x = \(\frac{80-2}{6}\) = 13.

breadth = 13 metre

Length = 2 × 13 + 1 = 27 metre

\(5x-\frac{1}{3}(x+1)=6(x+\frac{1}{30})\)

Taking LCM on both the sides

\(\frac{15x-(x+1)}{3}=\frac{6(30x+1)}{30}\)

By cross multiplication 

⇒ 10(14x - 1) = 6(30x + 1) 

⇒ 140x - 180x = 6 + 10 

⇒ - 40x = 16

⇒ \(\text{x}=-\frac{2}{5}\)

Let the second side = x cm

First side = x + 7 cm

Third side = x + 12 cm

Perimeter = x + x + 7 + x + 12 = 49

= 3x = 49 – 19 = 30

3x = 20, x = 10 cm

The sides are = 10 cm, 17 cm, 22 cm

5x + 7 = 2x - 8

By transposition, 

⇒ 5x−2x = - 8 - 7 

⇒ 3x = - 15 

⇒ x = - 5

ab - a - b + 1Taking 'a' as common from first two terms of the above polynomial. 

= a(b - 1) - (b - 1) 

Taking (b - 1) as common, in the above equation 

= (b - 1)(a - 1)= (a - 1)(b - 1)

7x² - 19x - 6 

By using splitting the middle term 

= 7 x² - 19x - 6 

= 7x² + ( - 21 + 2)x - 6 

= 7x(x - 3) + 2(x - 3) 

= (x - 3)(7x + 2)

\(\frac{x-5}{2}-\frac{x-3}{5}=\frac{1}{2}\)

Taking LCM of 2 and 5 = 10 on LHS

\(\Rightarrow \frac{5(x-5)-2(x-3)}{10}=\frac{1}{2}\)

By cross multiplication

⇒ 5x - 25 - 2x + 6 = 10/2

⇒ 3x = 5 + 19

\(\Rightarrow \text{x}=\frac{24}{3}=8\)

3 + 23x - 8x² 

By using Splitting the middle term 

= 3 + 23x - 8 x² 

= 3 + (24 - 1)x - 8 x² 

= 3(1 + 8x) - x(1 + 8x) 

= (1 + 8x)(3 - x)

8x + 3 = 27 + 2x 

By transposition, 

⇒ 8x – 2x = 27 − 3 

⇒ 6x = 24 

⇒ x = 4

15(y - 4) - 2(y - 9) + 5(y + 6) = 0 

Opening the brackets and multiplying, we get, 

⇒ 15y - 60 - 2y + 18 + 5y + 30 = 0 

⇒ 15y - 2y + 5y - 60 + 18 + 30 = 0 

⇒ 18y = 12

⇒ y = \(\frac{12}{18}=\frac{2}{3}\)

\(\frac{8x\,-\,3}{3x}=2\)

By cross multiplication

8x - 3 = 6x 

⇒ 2x = 3

⇒ \(\text{x}=\frac{3}{2}\)

3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17 

Multiplying we get, 

⇒ 15x - 21 - 18x + 22 = 32x - 52 – 17 

Solving, we get 

(15x - 18x) + (22 - 21) = 32x - (52 + 17) 

⇒ - 3x + 1 = 32x - 69 

⇒ 35x = 70 

⇒ x = 2

Total amount spent = Rs. 5200

Share of one = Rs.1300

∴ Members in the group = \(\frac{5200}{1300}\) = 4

Money Lissy got from mother = Rs. 60

Amount she returned = Rs. 13

Amount Lissy used to buy books = 60 – 13 = Rs. 47

Amount spent by Vimala for shopping = Rs. 163

Amount left with her after shopping Rs. 217

Total amount = 163 + 217 = Rs. 380

Number = 452 – 264 = 188

Number = 198 + 163 = 361

Let the number be x 

According to the question 

10 + 4x = 5x - 5 [ 10 is added to 4 times the number, 5 less than 5 times the number] 

By transposing 

⇒ 5x - 4x = 10 + 5 

⇒ x = 15 

So the number is 15

Let the two parts be x and (24 - x) 

According to the question 

7x + 5(24 - x) = 146 

By cross multiplication 

⇒ 2x = 146 - 120 

⇒ 2x = 26 

⇒ x = 13 

So the parts are 13 and (24 - 13) = 11

Let the numbers be x 

According to the question

\(\frac{1}{5}\text{x}+5=\frac{1}{4}\text{x}-5\)

Taking LCM of 5 and 1 on LHS = 5 and 4 and 1 on RHS = 1

\(\Rightarrow \frac{1}{5}\text{x}-\frac{1}{4}\text{x}=-10\)

\(\Rightarrow \frac{4\text{x-5x}}{20}=-10\)

\(\Rightarrow \text{x}=200\)

So the number 200

Let the number of correct answers = x

Total number of questions = 25

No. of wrong answers = 25 – x 

Marks for correct answers = 2x

marks losses for wrong answers = 25

ie. 2x – (25 – x) = 25

2x – 25 + x = 35

3x = 60

x = \(\frac{60}{3}\) = 20,

∴ number of correct answers = 20.

Let the cost of chair = x

Cost of table = 4x

x + 4 x = 1500

5x = 1500

x = 300

cost of chair = Rs. 300

Cost of table = Rs. 1200

\(\frac{6x\,+\,1}{3}+1\) = \(\frac{x\,-\,3}{6}\)

Taking LCM of 1 and 3 = 3,

\(\frac{6\text{x}\,+\,1\,+\,3}{3}=\frac{\text{x}-3}{6}\)

⇒ 2(6x + 4) = (x - 3) 

⇒ 12x - x = - 3 - 8 

⇒ x = - 1

\(\frac{x\,+\,1}{2x\,+\,3}=\frac{3}{8}\)

By cross multiplication, 8(x + 1) = 3(2x + 3)

⇒ 8x - 6x = 9 - 8 

⇒ 2x = 1

⇒ \(\text{x}=\frac{1}{2}\)

\(\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21\)

Taking LCM of 2, 4, 6 = 12

\(\frac{6n\,-\,9n\,+\,10n}{12}=21\)

⇒ 7n = 21 × 12 

⇒ n = 36

\(\frac{4x\,+\,7}{9\,-\,3x}=\frac{1}{4}\)

By cross multiplication

4(4x + 7) = (9 - 3x) 

⇒ 16x + 28 = 9 - 3x 

⇒ 19x = - 19 

⇒ x = - 1