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Given:

Two expressions 19 – 5x and 19x + 5

Concept Used:

We have to equate both the given expressions.

Calculation:

19 – 5x = 19x + 5

⇒ 24x = 14

⇒ x = 14/24 = 7/12

∴ The Value of x for which both expressions become equal is 7/12

Given

Sum of two numbers is 25

Calculation

Let the one number be x

And the second number be y

According to question

x + y = 25

⇒ x = 25 - y    ---- (1)

x = (1/2)y + 7     ---- (2)

From (1) and (2)

⇒ 25 - y = (1/2)y + 7

⇒ 25 - 7 = (1/2)y + y

⇒ 18 = 3y/2

⇒ y = (18 × 2)/3

⇒ y = 12

⇒ x = 25 - 12

⇒ x = 13

Now, required difference = 2(larger number) - 2(smaller number)

⇒ 2(13) - 2(12)

⇒ 26 - 24 = 2

∴ The required difference is 2.

Formula:

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

Calculation:

(2x + y)³ = 8x³ + 12x²y + 6xy² + y³

(2x - y) = 8x³ - 12x²y + 6xy² - y³

( 2x + y )³ - ( 2x - y)³ = 8x³ + 12x²y + 6xy² + y³ - 8x³ + 12x²y - 6xy² + y³

( 2x + y )³ - ( 2x - y)³ = 24x²y + 2y³

∴ The coefficient of x²y is 24.

i.  11x² + 18x + 7 = 0

⇒ 11x² + 11x + 7x + 7 = 0

⇒ 11x(x + 1) + 7(x + 1) = 0

⇒ (x + 1)(11x + 7) = 0

⇒ x = -1, -7/11

ii. 3y² + 2y - 1 = 0

⇒ 3y² + 3y - y - 1 = 0

⇒ 3y(y + 1) - 1(y + 1) = 0

⇒ (y + 1) (3y - 1) = 0

⇒ y = -1, 1/3

Hence, the relation cannot be determined​.

Calculation:

I. 3X² – 42X + 135 = 0

⇒ X² – 14X + 45 = 0      ----(∵ divide by 3 )

⇒ X² – 9X – 5X + 45 = 0

⇒ X(X – 9) – 5(X – 9) = 0

⇒ (X – 5) (X – 9) = 0

∴ The value of X is 5, 9.

II. 2Y² – 42y + 220 = 0

⇒ Y² – 11Y – 10Y + 110 = 0      ----(∵ taking 2 as common)

⇒ Y(Y – 11) – 10Y(Y – 11) = 0

⇒ (Y – 10) (Y – 11) = 0

∴ The value of Y is 10, 11.

∴ X < Y.

I. 2X² – 18X + 40 = 0

⇒ X² – 4X – 5X + 20 = 0 (∵ taking 2 as common)

⇒ (X – 4) (X – 5)

⇒ X = 4, 5

II. 2Y² – 15Y + 28 = 0

⇒ 2Y² – 7Y – 8Y + 28 = 0

⇒ Y(2Y – 7) – 4(2Y – 7) = 0

⇒ (Y – 4) (2Y – 7) = 0

⇒ Y = 4, 7/2

∴ X ≥ Y

Given :

Cost price of 4 apples and 4 oranges = 76 rupees 

Cost price of 6 apples and 2 oranges = 78 rupees 

Calculations :

Let the cost price of apple = A 

Let the cost price of orange = O

According to the question 

4A + 4O = 76       ....(1)

6A + 2O = 78       ....(2)

multiply equ (1) with 3 and equ (2) with 2 and solving, we get 

A = 10 and O = 9 

Now 

A - O = 10 - 9 = 1 

∴ Diffrence of cost price of an Apple and an orange is 1 rupees

Given:

x : y = 2 : 3

Calculation:

As, x : y = 2 : 3

Then, x = 2k and  y = 3k

⇒ Put this ratio in this equation  \(\frac{3x + 2y}{9x + 5y}\)

⇒ \(\frac{3(2k) + 2(3k)}{9(2k) + 5(3k)}\)

⇒ 12/33 = 4/11

Given:

\({x^2} - 3√ 2 x + 1 = 0\)

Formula used:

(a + b)3 = a3 + b3 + 3ab(a + b)

Calculation:

\({x^2} - 3√ 2 x + 1 = 0\)

⇒ x(x – 3√2 + 1/x) = 0

⇒ (x – 3√2 + 1/x) = 0

⇒ x + 1/x = 3√2      ----(1)

Cubing on both side

⇒ (x + 1/x)³ = (3√2)³

(a + b)³ = a³ + b³ + 3ab(a + b)

⇒ x³ + 1/x³ + 3(x + 1/x) = 54√2

⇒ x3 + 1/x3 + 3 × 3√2  = 54√2  (from equation 1)

⇒ x3 + 1/x3 = 54√2 – 9√2

⇒ x3 + 1/x3 = 45√2

∴ The value of x3 + 1/x3 is 45√2

Given:

The total number of participants = 110

The number of females who participated = 7/15 times of the number of males

Calculations:

Let the number of males be 'x'

The total number of female = (7/15) × x

The total number of participants  = x + 7x/15

⇒ 22x/15 = 110

⇒ 22x/15 = 110

⇒ x = 110 × 15/22

⇒ x = 75

∴ The total number of male participated in event is 75

I. 14x³ = 378

⇒ x³ = 27

⇒ x = 3

II. 27y² = 243

⇒ y² = 9

⇒ y = +3 or -3

Comparison between x and y (via Tabulation):

∴ x ≥ y

Calculation:

l. x² – 43x – 188 = 0

⇒ x² – (47 – 4)x – 188 = 0

⇒ x² – 47x + 4x -188 = 0

⇒ x(x – 47) + 4(x – 47) = 0

⇒ (x – 47) (x + 4) = 0

⇒ x = 47, -4

II. y² + 21y – 196 = 0

⇒ y² + (28 – 7)y – 196 = 0

⇒ y² + 28y – 7y – 196 = 0

⇒ y(y + 28) – 7(y + 28) = 0

⇒ (y + 28) (y – 7) = 0

⇒ y = -28, 7

Comparison between x and y (via Tabulation):

∴ Relation between x and y can not be determined.

Given:

If the total number of children are increased by 5, then each children will get 2 book less.

If total number of children are decreased by 10, then each children will get 10 books more.

Calculation:

Let the total number of children be x and each children get books be y

Total books = x × y

According to question

If children = x + 5, and books per child = y – 2

Then (x + 5) × (y – 2) = x × y

⇒ x × y – 2x + 5y – 10 = x × y

⇒ 2x – 5y = -10      ---- (1)

If children = x – 10, and books per child = y + 10

Then (x – 10) × (y + 10) = x × y

⇒ x × y + 10x – 10y – 100 = x × y

⇒ 10x – 10y = 100

⇒ x – y = 10      ---- (2)

Subtracting eq. (1) from 5 × eq. (2)

⇒ 5 × (x – y) – (2x – 5y) = 5 × 10 – (-10)

⇒ 5x – 5y – 2x + 5y = 50 + 10

⇒ 3x = 60

⇒ x = 20

Put the value of x in eq. (2)

⇒ 20 – y = 10

⇒ y = 10

Total book = x × y

⇒ 20 × 10

⇒ 200

∴ The total number of books distributed in the orphanage is 200.

Given:

√3x² – 8x + 5√3 = 0  

Concept used:

By using factorization method (Middle term split)

Calculation:

√3x² – 8x + 5√3 = 0  

⇒ √3x² – 5x – 3x + 5√3 = 0

⇒ x(√3x – 5) – √3(√3x – 5) = 0

⇒ (√3x – 5)(x – √3) = 0

⇒ (√3x – 5) = 0 or (x – √3) = 0

⇒ x = 5/√3 or x = √3

∴ The two roots of equation is √3 and 5/√3.

Let the number be x

The sum of one‐half, one‐third, and one‐fourth of that number = x/2 + x/3 + x/4

Given that it exceeds that number by 12

⇒ x/2 + x/3 + x/4 - x = 12

⇒ (6x + 4x + 3x - 12x)/12 = 12

⇒ x/12 = 12

⇒ x = 144

∴ The number is 144

Given:

(a + 2b)² - (a - 2b)²

Concept used:

(a + b)² = a² + 2ab + b²

Calculation:

⇒ (a² + 4ab + 4b²) - (a² - 4ab + 4b²)

⇒ a² + 4ab + 4b² - a² + 4ab - 4b²

⇒ 4ab + 4ab

∴ 8ab.

Given:

3x² + 5x - 2 = 0

Concept used:

Factorisation 

Calculation:

⇒ 3x² + 5x - 2

⇒ 3x² + 6x - x - 2

⇒ 3x(x + 2) -1(x + 2)

⇒ (3x - 1)(x + 2)

⇒ 3x - 1 = 0 so x = 1/3

⇒ x + 2 = 0 so x = -2

∴ x = -2 and 1/3

Calculations:

From I,

x² = 49

⇒ x = –7, 7 

From II,

y² – 16y + 63 = 0

⇒ y² – 9y – 7y + 63 = 0

⇒ y(y – 9) – 7(y – 9) = 0

⇒ (y – 9)(y – 7) = 0

Taking,

⇒ y – 9 = 0 or y – 7 = 0

⇒ y = 9 or y = 7

Comparison between x and y (via Tabulation):

∴ x ≤ y.

Calculations:

From I,

x² + 7x + 12 = 0

⇒ x² + 4x + 3x + 12 = 0

⇒ x(x + 4) + 3(x + 4) = 0

⇒ (x + 4)(x + 3) = 0

Taking,

⇒ x + 4 = 0 or x + 3 = 0

⇒ x = –4 or x = –3

From II,

y² + 8y + 16 = 0

⇒ y² + 4y  + 4y + 16 = 0

⇒ y(y + 4) + 4(y + 4) = 0

⇒ (y + 4)(y + 4) = 0

Taking,

⇒ y + 4 = 0 or y + 4 = 0

⇒ y = –4 or y = –4

Comparison between x and y (via Tabulation):

∴ x ≥ y.

Calculation:

According to the question,

3x + 2 = 23

⇒ 3x = 23 – 2

⇒ 3x = 21

⇒ x = 7

∴ The value of x is 7

Given:

One of the roots of equation x² + mx + n = 0 is the square of the other.

Formula Used:

If ax² + bx + c = 0, then

Sum of the roots = -(b/a)

Product of the roots = (c/a)

(a + b)³ = a³ + b³ + 3 × a × b × (a + b)

Calculation:

Let α and α² be the roots of equation x² + mx + n = 0

Product of the roots = (c/a)

⇒ α × α² = n

⇒ α³ = n

⇒ α = (n)^1/3      ---- (1)

Sum of the roots = -(b/a)

⇒ α + α² = -m

⇒ α + α² = -m      ---- (2)

Put the value of eq. (1) in the eq. (2)

⇒ (n)^1/3 + (n)^2/3 = -m

On cubing both the sides,

⇒ n + n² + 3 × n^1/3 × n^2/3 × (n^1/3 + n^2/3) = (-m)³

⇒ n + n² + 3 × n × (-m) = -m³

⇒ n + n² + m³ = 3mn

Given:

The polynomial is x² – 13x + 36

Formula Used:

If polynomial ax² + bx + c, then

Min/Max value of polynomial = c – b²/4a

Calculation:

For polynomial x² – 13x + 36

a = 1, b = -13, and c = 36

Minimum value of the polynomial = c – b²/4a

⇒ 36 – (-13)²/4 × 1

⇒ (36 × 4 – 169)/4

⇒ (144 – 169)/4

⇒ -25/4

∴ The minimum value of x² – 13x + 36 is -25/4.

Concept Used:

The last digit obtained for any power of a number ending with digit 6 is always 6

i.e. (6)¹ = 6

(6)2 = 6

(6)3 = 6 and so on...

Calculation:

⇒ last digit of the number (986)986  = 6  

∴ Required last digit = 6          

Calculation:

l. 24x² + 38x + 15 = 0

⇒ 24x² + 18x + 20x + 15 = 0

⇒ 6x(4x + 3) + 5(4x + 3) = 0

⇒ (6x + 5) (4x + 3) = 0

⇒ x = -5/6, -3/4

ll. 54y² + 123y + 65 = 0

⇒ 54y² + 78y + 45y + 65 = 0

⇒ 6y(9y + 13) + 5(9y + 13) = 0

⇒ (6y + 5) (9y + 13) = 0

⇒ y = -5/6, -13/9

Comparison between x and y (via Tabulation):

∴ The result will be x ≥ y.

Given:

a + b = 7/3

a² + b² = 31/9

Formula used:

(a + b)² = a² + b² + 2ab

a³ + b³ = (a + b)( a² + b² – ab)

Calculation:

According to the question,

⇒ a + b = 7/3

Squaring both sides,

⇒ (a + b)² = (7/3)²

⇒ a² + b² + 2ab = 49/9

⇒ 31/9 + 2ab = 49/9

⇒ 2ab = 49/9 – 31/9

⇒ 2ab = 18/9

⇒ 2ab = 2

⇒ ab = 1

According to the formula,

⇒ a³ + b³ = (a + b)( a² + b² – ab)

Multiply 27 on both sides,

⇒ 27(a³ + b³) = 27(a + b)( a² + b² – ab)

⇒ 27(a³ + b³) = 27(7/3)( 31/9 – 1)

⇒ 27(a³ + b³) = 27(7/3)( 22/9)

⇒ 27(a³ + b³) = 22 × 7

⇒ 27(a³ + b³) = 154

∴ The value of 27(a³ + b³) is 154.

Formula Used:

(a + b)3 = a3 + b3 + 3a2b + 3ab2

(a + b)2 = a2 + 2ab + b2

Calculation:

(x + 2)³ = x³ + (2)³ + 3(x)²(2) + 3(x)(2)²

(x + 6)² = (x)² + 2(x)(6) + (6)²

According to question 

(x + 2)3 – (x + 6)2

⇒ x³ + 8 + 6x² + 12x – (x² + 12x + 36)

⇒ x³ + 8 + 6x² + 12x – x² - 12x – 36

∴ x³ + 5x² – 28

The correct option is 1 i.e. x3 + 5x2 – 28 

Given:

a2 - 2a = 1

Formula Used:

(a - b)³ = a³ - b³ - 3a²b + 3ab² = a3 - b3 - 3(ab)(a - b)

If (a - 1/a) = p

Then, (a³ - 1/a³) = p³ + 3p

Calculation:

a2 - 2a = 1

Divide by 'a'

⇒ a - 2 = 1/a

⇒ a - 1/a = 2

Now,

if (a - 1/a) = 2

∴ Then, (a3 - 1/a3) = (2)3 + 3(2) = 14

Given:

Girls ∶ Boys = 3 ∶ 2

Calculation:

Let assume that number of girls be 3x

And, boys are be 2x

According to the question: 

⇒ (3x - 5) = (2x + 3)

⇒ x = 8

Total strength of the class in the beginning of the year = 2x + 3x = 5x

Total strength of the class in the beginning of the year = 5x = 5 × 8 = 40

∴ The total strength of the class at the beginning of the year was 40

Given:

\(\frac{{10 × 10 × 10 + 9 × 9 × 9}}{{10 × 10 - 10 × 9 + 9 × 9}}\)

Concept used:

(a³ + b³) = (a + b) × (a² – ab + b²)

Calculations:

a = 10

b = 9 

⇒ (103 + 93) = (10 + 9) × (102 – 10 × 9 + 92)

Putting the values,

⇒ [(10 + 9) × (102 – 10 × 9 + 92)]/(102 – 10 × 9 + 92)

⇒ 10 + 9

⇒ 19

∴ The result of the given equation is 19

GIVEN:

 f(x) = x2 - 14x + 45

FORMULA USED:

Sum of roots = (-b/a) and Product of roots = c/a for f(x) = ax² + bx + c 

CALCULATION:

f(x) = x2 - 14x + 45

⇒ a = 1, b = -14, c = 45

⇒ Sum of roots(P + Q) = (-b/a)

⇒ 14

⇒ Product of roots (PQ) = c/a

⇒ 45

⇒ (1/P + 1/Q) = (P + Q)/PQ

⇒ 14/45

∴  (1/P + 1/Q) = 14/45

Given:

Let constant expenses and other expenses for each traveler be Rs. a and Rs. b respectively.

Calculation:

⇒ 46000 = a + 30 × b

⇒ 19000 = a + 12 × b

Solving,

⇒ b = 1500 and a = Rs. 1000

∴ Required expenses = 1000 + 5 × 1500 = Rs. 8500

Given:

356 ×  7693

Calculations:

356

3⁰ = 1    

3¹ = 3    

3² = 9    

3³  = 27 

3⁴ = 81

For every 4 powers the unit digit is same 

Here, power of 3 is 56 

56 ÷ 4 = 14 

So the unit digit  = 1 for  356

 7693.

6⁰ = 1 

6¹ =  6

61 =   36

For every power of 6 the unit digit is 6

6 × 1 = 6

∴ The unit digit of a product 356 ×  7693 is 6

Given:

4 + 2x > 3(x – 1)

x + 3(x – 2) > x + 9

Concept:

Using inequality, find the range of x from both the equations and choose the appropriate option.

Calculation:

∵ 4 + 2x > 3(x – 1)

⇒ 4 + 2x > 3x – 3

⇒ 4 + 3 > 3x – 2x

⇒ x < 7     ----(1)

∵ x + 3(x – 2) > x + 9

⇒ x + 3x – 6 > x + 9

⇒ 4x – 6 > x + 9

⇒ 4x – x > 9 + 6

⇒ 3x > 15

⇒ x > 15/3

⇒ x > 5      ----(2)

From (1) and (2);

5 < x < 7

∴ From the options, x can take the value of 6.5.

Given:

Given equation = 7(3x – 5y) = 21x – 35y

Final equation = 2(8x – 9y) = 16x – 18y

Concept:

Subtract the given equation from the final equation to get the desired answer.

Calculation:

Let the desired equation be ‘a’.

∴ 21x – 35y + a = 16x – 18y

⇒ a = 16x – 18y – 21x + 35y

⇒ a = 17y – 5x

Given:

Let the numbers be ‘x’ and ‘y’

∴ x × y = 20

x² + y² = 41

Formula used:

(x – y)² = x² + y² – 2xy

(x + y)² = x² + y² + 2xy

Calculation:

∵ (x – y)² = x² + y² – 2xy

⇒ (x – y)² = 41 – (2 × 20)

⇒ (x – y)² = 41 – 40

⇒ (x – y)² = 1

⇒ x – y = 1      ----(1)

∵ (x + y)² = x² + y² + 2xy

⇒ (x + y)² = 41 + (2 × 20)

⇒ (x + y)² = 41 + 40

⇒ (x + y)² = 81

⇒ x + y = √81 = 9      ----(2)

∴ (x + y) ∶ (x – y) = 9 ∶ 1

Given:

[5x + 1/(3x)] = (-10/3)

Calculation:

∵ [5x + 1/(3x)] = (-10/3)

Multiplying both side with 3/5;

⇒ 3/5 × [5x + 1/(3x)] = 3/5 × (-10/3)

⇒ [(3/5) × (5x)] + [(3/5) × (1/3x)] = (-2)

⇒3x + (1/5x) = (-2)

Cube both side;

⇒ (3x)³ + (1/5x)³ + 3 × (3x) × (1/5x) × [3x + (1/5x)] = (-8)

⇒ (3x)³ + (1/5x)³ + [9/5 × (-2)] = (-8)

⇒ (3x)³ + (1/5x)³ – (18/5) = (-8)

⇒ (3x)³ + (1/5x)³ = (-8) + (18/5)

⇒ (3x)³ + (1/5x)³ = [(-40) + 18]/5

⇒ (3x)³ + (1/5x)³ = (-22)/5

GIVEN:

3x - 20y - 2 = 0 and 11x - 5y + 61 = 0

CALCULATION:

3x - 20y - 2 = 0 .......(1)

⇒ (11x - 5y + 61 = 0) × 4  ......(2)

On subtracting  these two equation

⇒ 41x = - 246

⇒ x = - 6

putting the value of x = - 6 in any of the two equation, we will get 

⇒ y = -1

both the equations intersect each other at point P(a,b) = P(x. y)

⇒ a = -6 and b = -1

⇒ (a2 + b² - ab)/(a2 - b2 + ab) = (36 + 1 - 6)/(36 - 1 + 6)

⇒ 31/41

∴  (a2 + b² - ab)/(a2 - b2 + ab)  = 31/41

GIVEN:

x² + 1/x² = 34

CONCEPT:

If x² + 1/x² = a,

then x – 1/x = √(a – 2)

x + 1/x = √(a + 2)

CALCULATION:

x² + 1/x² = 34

If x² + 1/x² = a, then x + 1/x = √(a + 2)

⇒ x + 1/x = √(a + 2) = √36 = 6

If x + 1/x = a, then √x – 1/√x = √(a – 2)

⇒ √x – 1/√x = √(6 – 2)

∴ The value of √x – 1/√x is 2.