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Given:

x + 1/x = 23

Formula Used:

(a – b)2 = a2 + b2 – 2ab

Calculations:

x + 1/x = 23

Subtracting 2 on both sides

x + 1/x – 2 = 23 – 2

⇒ (√x)² + 1/(√x)² – 2 = 21

⇒ (√x – 1/√x)² = 21

∴ (√x – 1/√x) is √21

Given:

√x = √3 – √ 5

Calculation:

Squaring both sides

(√x)² = (√3 – √5)²

⇒ x = 3 + 5 – 2√15 

⇒ x – 8 = -2√15

Again, squaring both sides

(x – 8)² = (-2√15)²

⇒ x² + 64 – 16x = 4 × 15

⇒ x² + 4 – 16x = 0

⇒ x² – 16x + 6 = 2  

∴ The value is 2.

I. x2 + 3√ 3 x - 120 = 0

⇒ x² + 8√3x - 5√ 3x - 120 = 0

⇒ x (x + 8√ 3x) - 5√ 3(x + 8√ 3) = 0

⇒ (x - 5√ 3)(x + 8√ 3 ) = 0

⇒ x = 5√ 3, -8√ 3

II 6y2 + 13√ 3y + 20  = 0

⇒ 6y2 + 8√3x + 5√ 3x +20 = 0

⇒2√3 y (√ 3 y + 4) - 5(√ y + 4) = 0

⇒ (2√ 3 y - 5)(√ 3 y + 4 ) = 0

⇒ y = -5//2√ 3, -4/√ 3

∴ There is no relation between x and y

Given-

Total number of soldiers = 784

Concept Used-

Concept of square root

Calculation-

Let the total number of rows and number of soldiers in each row be x

Total number of soldiers = x × x

According to the question -

x² = 784

x = √784

x = 28

∴ Number of soldiers in each row is 28

Given:

[(x + a)/a] + [(x + b)/b] + [(x + c)/c] = 3

Calculation:

∵ [(x + a)/a] + [(x + b)/b] + [(x + c)/c] = 3

⇒ x/a + a/a + x/b + b/b + x/c + c/c = 3

⇒ x/a + 1 + x/b + 1 + x/c + 1 = 3

⇒ x/a + x/b + x/c + 3 = 3

⇒ (bcx + acx + abx)/abc = 3 – 3

⇒ (bcx + acx + abx)/abc = 0

⇒ [x(ab + bc + ca)]/abc = 0

As from the question x ≥ 0

∴ (ab + bc + ca) = 0

Given, (a⁶⁸ + 1)/a³⁴ = 34

⇒ a³⁴  + a^-34 = 34      ---- (1)

⇒ a³⁴  + a^-34 + 2 = 36      {adding 2 both the sides}   

⇒ (a¹⁷  + a^-17)² = 6²      {using, (a + b)² = a² + b² + 2ab}   

⇒ (a¹⁷  + a^-17) = 6      ---- (2)

Now, (a⁵¹  + a^-51) = (a¹⁷  + a^-17)( a³⁴  + a^-34 - a¹⁷ × a^-17)      {using, a³ + b³ = (a + b) (a² + b² – ab)}   

⇒ (a¹⁰² + 1) / a⁵¹ = 6(34 – 1) = 198      {from eq. (1) and (2)}

Let the cost of one cycle tyre be x, and the cost of one tube be y.

Here, The cost of four-cycle tyres and Three tubes is Rs. 720, and The cost of three cycle tyres and four tubes is Rs. 610.

So, 4x + 3y = 720      -----(1)

3x + 4y = 610      ----(2)

Multiplying equation (1) by 3 and equation (2) by 4,

12x + 9y = 2160     ----(3)

12x + 16y = 2440     ----(4)

Subtracting equation (3) from equation (4),

7y = 280

⇒ y = 40

Here, y is the cost of one tube.

Hence, the cost of a tube is "Rs. 40".

Given:

[(127)³ + (173)³]/[(1.27)² – 2.1971 + (1.73)²]

Concept:

Compare the given values with the formula of a³ + b³ and solve to get the desired answer.

Formula used:

a³ + b³ = (a + b)(a² + b² – ab)

⇒ (a³ + b³)/(a² + b² – ab) = a + b

Calculation:

∵ [(127)³ + (173)³]/[(1.27)² – 2.1971 + (1.73)²]

⇒ {[(127)³ + (173)³]/[(127)² – 21971 + (173)²]} × 10000

Now comparing the given values with the formula of a³ + b³:

⇒ {[(127)³ + (173)³]/[(127)² – 21971 + (173)²]} × 10000 = (127 + 173) × 10000

⇒ 300 × 10000

⇒ 30 × 10⁵

∴ x = 5

Given:

x² + xy + y² = 8

x + √(xy) + y = 4

Formula used:

(x + y)² = x² + y² + 2xy

(x – y)² = x² + y² – 2xy

Calculation:

∵ x + √(xy) + y = 4

⇒ x + y = 4 – (√xy)

Squaring both sides;

⇒ x² + y² + 2xy = 16 + xy – 8√xy

⇒ x² + y² + 2xy – xy = 16 – 8√xy

⇒ x² + y² + xy = 16 – 8√xy

⇒ 8 = 16 – 8√xy

⇒ 8√xy = 16 – 8

⇒ 8√xy = 8

⇒ √xy = 8/8 = 1

Given:

a(a + 3) = 4

⇒ a² + 3a = 4

Concept:

Cube both side of the given equation and then proceed.

Formula used:

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculation:

∵ a² + 3a = 4

On cubing both sides, we get;

(a² + 3a)³ = 4³

⇒ a⁶ + 27a³ + 3 × a² × 3a(a² + 3a) = 64

⇒ a⁶ + 27a³ + 3 × a² × 3a(4) = 64

⇒ a⁶ + 27a³ + 36a³ = 64

⇒ a⁶ + 63a³ = 64      -----(1)

Putting the value of (1) in (a⁶ + 63a³ – 60)/4;

⇒ (64 – 60)/4

⇒ 4/4 = 1

Concept:

If, (x + 1) or (x – 1) are factors of a given equation, then on putting x = 1 or x = (-1) in place of ‘x’ in the given equation will give 0. SO, by using option, put the value ‘y’ in the equation and test if the result comes 0 or not. If not, then it is not the factor of the equation, if yes, then it a factor of the given equation.

Calculation:

If (y – 1) is a factor,

Then, putting y = 1 in the given equation will give 0.

⇒ 1³⁰ – 1²⁷ + 1²⁴ – 1

⇒ 1 – 1 + 1 – 1 = 0

∴ (y – 1) is a factor of the given equation.

Now with (y + 1);

Put y = (-1) in the given equation;

⇒ (-1)³⁰ – (-1)²⁷ + (-1)²⁴ – 1

⇒ 1 + 1 + 1 – 1 = 2

∴ (y + 1) is not a factor of given equation.

Given:

a = (√5 + 2)/(√5 – 2)

b = (√5 – 2)/( √5 + 2)

Concept:

First rationalize the values of ‘a’ and ‘b’ and then proceed.

Formula used:

(a + b)² = a² + b² + 2ab

(a + b)(a – b) = a² – b²

Calculation:

∵ a = (√5 + 2)/(√5 – 2)

⇒ a = [(√5 + 2)( (√5 + 2)]/[(√5 – 2)(√5 + 2)]

⇒ a = [(√5 + 2)²]/[(√5)² – (2)²]

⇒ a = [5 + 4 + 4√5]/(5 – 4)

⇒ a = (9 + 4√5)

Similarly;

b = (√5 – 2)/( √5 + 2)

⇒ b = [(√5 – 2)(√5 - 2)]/[(√5 + 2)(√5 – 2)]

⇒ b = [(√5 – 2)²]/[(√5)² – (2)²]

⇒ b = [5 + 4 – 4√5]/(5 – 4)

⇒ b = (9 – 4√5)

Now, a² + b² + ab = (a + b)² – ab

⇒ (9 + 4√5 + 9 – 4√5)² – [(9 + 4√5)( 9 – 4√5)]

⇒ (18)² – [(9)² – (4√5)²]

⇒ 324 – (81 – 80)

⇒ 324 – 1

⇒ 323

∴ The value of the given expression is 323.

GIVEN:

Expression = x²/(2x + 3)

CONCEPT:

Maximum value of the expression

CALCULATION:

Let k = x²/(2x + 3)

⇒ k (2x + 3) = x²

⇒ 2kx + 3k = x²

⇒ x² – 2kx – 3k = 0

For all real value of ‘x’, D ≥ 0

D = b² – 4ac ≥ = 0

After comparing above equation with ax² + bx + c = 0

a = 1, b = -2k and c = -3k

⇒ (-2k)² – 4 × 1 × (-3k) ≥ 0

⇒ 4k² + 12k ≥ 0

⇒ 4k (k + 3) ≥ 0

⇒ k ≥ 0 and k ≤ -3

⇒ k = (-∞, -3] ∪ [0, ∞)

Given:

x – (2√5/x) = 7

⇒ x² – 2√5 = 7x

Concept:

Arrange the asked equation and put the value of the given equation in it to get the answer.

Calculation:

∵ x² – 2√5 = 7x      ------(1)

⇒ x² – 7x = 2√5

On arranging the equation = (x² – 5x – 2√5)/[x²(x – 7)]

We get;

[(x² – 2√5) – 5x]/[x(x² – 7x)]

⇒ (7x – 5x)/(x × 2√5)

⇒ 2x/(2√5x)

⇒ 1/√5 = √5/5

Given:

Equation 1 = 2(5y + 7)

Equation 2 = 11(y -3)

Equation 1 = Equation 2

Concept:

Just equate the equations and reach the answer.

Calculation:

∵ 2(5y + 7) = 11(y – 3)

⇒ 10y + 14 = 11y – 33

⇒ 11y – 10y = 14 + 33

⇒ y = 47

Given:

x + (1/x) = 3

Calculation:

∵ x + (1/x) = 3

⇒ (x² + 1)/x = 3

⇒ x² + 1 = 3x     -----(1)

Now, rearranging the given equation;

⇒ [(x² + 1) – 2x]/[(x² + 1) + 3x]

⇒ (3x – 2x)/(3x + 3x)     -----(From 1)

⇒ x/6x

⇒ 1/6

Given:

x + 16/x = 8

Concept used:

Put the value and get the answer

Calculation:

Let the value of x be 4, then

4 + 16/4 = 8 (satisfy)

so 4² + 32/4² = 16 + 2 = 18

∴ The value of given identities is 18.

Given:

x + y = 10

x² + y² = 68

Formula Used:

(a + b)² = a² + b² + 2ab

Calculation:

(x + y)2 = x2 + y2 + 2xy

⇒ (10)² = 68 + 2xy

⇒ 100 - 68 = 2xy

⇒ xy = 32/2 = 16

∴ The value of xy is 16.

Let price per earphone be Rs. x and price per charging chord be Rs. y

According to question,

4x + 12y = 920      ----(1)

6x + 10y = 980      ----(2)

By equation (1) × 3 – equation (2) × 2,

⇒ 12x + 36y – 12x – 20y = 2760 – 1960

⇒ 16y = 800

⇒ y = 50

Put the value of y in eq. (1)

∴ 4x + 12 × 50 = 920

⇒ 4x = 920 – 600

x = 320/4 = 80

Now, Cost of 5 earphones and 11 charging chords = 5 × 80 + 11 × 50 = Rs. 950

∴ Cost of 5 earphones and 11 charging chords is Rs. 950

Short Trick: 

Let price per earphone be Rs. x and price per charging chord be Rs. y

According to question,

4x + 12y = 920      ----(1)

6x + 10y = 980      ----(2)

Add equation (1) and (2) 

4x + 12y + 6x + 10y = 1900

⇒ 10x + 22y = 1900

⇒ 2(5x + 11y) = 1900

⇒ 5x + 11y = 950

∴ Cost of 5 earphones and 11 charging chords is Rs. 950

Given:

4x2 + 1/3x2 = 6

Formula used:

(a + b)2 = a2 + b2 + 2ab

Calculation:

As, 4x2 + 1/3x2 = 6

Multiplying both sides by 3/4, we’ll get

(3/4) × (4x2 + 1/3x2) = 6 × (3/4)

⇒ 3x2 + 1/4x2 = 9/2

Now squaring both sides, we’ll get

(3x2 + 1/4x2)2 = (9/2)2

⇒ 9x4 + 1/16x4 + 2 × 3x2 × (1/4x2) = 81/4

⇒ 9x4 + 1/16x4 + 3/2 = 81/4

⇒ 9x4 + 1/16x4 = 81/4 – 3/2

⇒ 9x4 + 1/16x4 = 18.75

∴ The value of 9x4 + 1/16x4 is 18.75

Given :-

Difference between two positive number = 5

sum of their number of their square = 73

Concept :- 

(a - b)² = a² + b² - 2ab

Calculation :-

Let first number be x 

⇒ Second number = y 

From question

⇒ x - y = 5    .....(1)

Again from question 

⇒ x² + y² = 123    ....(2) 

Now squaring both side of equation (1)

⇒ (x - y)² = 5²

⇒ x2 + y² - 2xy = 25    ....(3)

Now put the value of equation (2) in equation (3)

⇒ 123 - 2xy = 25

⇒ 2xy = 123 – 25 

⇒ 2xy = 98 

⇒ xy = (98/2)

⇒ xy = 49

Now 

⇒ Square root of xy = √49 

⇒ Square root of xy = 7

∴ Square root of xy is 7

(x - y)² - 7(x² - y²) + 12(x + y)²

= (x - y)² – 7(x – y) (x + y) + 12(x + y)² [As x² – y² = (x +y) (x + y)]

Let (x – y) = p & (x + y) = q, then

(x - y)² - 7(x² - y²) + 12(x + y)²

= p² - 7pq + 12q²

= p² - 4pq - 3pq + 12q²

= p(p - 4q) - 3q(p - 4q)

= (p - 3q) (p - 4q)

= [(x - y) - 3(x + y)] [(x - y) - 4(x + y)] [As (x - y) = p and (x + y) = q]

= (- 2x - 4y) (- 3x - 5y)

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

Calculation∶

I. x² - 35x + 306 = 0

⇒ x² - 17x - 18x + 306 = 0

⇒ x (x - 17) - 18 (x - 17) = 0

⇒ (x - 17) (x - 18) = 0

⇒ x = 17, 18

II. y² - 25y + 144 = 0

⇒ y² - 9y - 16y + 144 = 0

⇒ y (y - 9) - 16(y - 9) = 0

⇒ (y - 9) (y - 16) = 0

⇒ y = 9, 16

Calculation∶

I. x² + 31x + 240 = 0

⇒ x² + 15x + 16x + 240 = 0

⇒ x (x + 15) + 16(x + 15) = 0

⇒ (x + 15) (x + 16) = 0

⇒ x = -15, -16

II. 5y² + 60y + 135 = 0

⇒ 5y² + 45y + 15y + 135 = 0

⇒ 5y (y + 9) + 15 (y + 9) = 0

⇒ (y + 9) (5y + 15) = 0

⇒ y = -9, -3

∴ x < y

Calculation∶

I. x² + 22x + 120 = 0

⇒ x² + 12x + 10x + 120 = 0

⇒ x(x + 12) + 10(x + 12) = 0

⇒ (x + 12)(x + 10) = 0

⇒ x = -12, -10

II. 10y² + 23y + 12 = 0

⇒ 10y² + 15y + 8y + 12 = 0

⇒ 5y(2y + 3) + 4(2y + 3) = 0

⇒ (2y + 3)(5y + 4) = 0

⇒ y = -3/2, -4/5

Given-

Price of 14 mangoes = Rs. 39.2

Concept Used-

Solving linear equations.

Calculation-

Let price of one mango be Rs x 

According to condition- 

14x = 39.2

⇒ x = 2.8

⇒ 12x = 33.6

∴ The price of 12 mangoes is Rs. 33.6.

Calculations:

(21)³ – (39)³ + (18)³ = (a - b) (a² + b² + ab) + c³

⇒ (21-39) (21² + 39² + 21 × 39) + (18)³

⇒ -18 × (441 + 1521 + 819) + (18)³

⇒ 18(-2781 + (18)²)

⇒ 18(-2781 + 324)

⇒ 18 × (-2457)

⇒ -44,226.

∴ The value of expresssion - (21)3 – (39)3 + (18)3 is -44,226.

Short trick:

The algebraic identity states

If a – b + c = 0, then a3 + b3 + c3 = -3abc

As 21 – 39 + 18 = 0

⇒ (21)3 – (39)3 + (18)3 

⇒ -3 ×  21 × 39 × 18

⇒ -44,226. 

∴ The value of expresssion (21)3 – (39)3 + (18)3 is -44,226.

Given:

a and b are two positive real numbers such that a + b = 20 and ab = 4. We have to find the value of a³ + b³

Formula Used:

a3 + b³ = (a + b)³ – 3ab(a + b)

Calculation:

a3 + b3 = (a + b)3 – 3ab(a + b)

⇒ a3 + b³ = (20)³ – 3 × 4 × 20       [∵ Given a + b = 20 and ab = 4]

⇒ a3 + b³ = 20 × (20² – 12)

⇒ a3 + b³ = 20 × (400 – 12)

⇒ a3 + b³ = 20 × 388

⇒ a3 + b³ = 7760

∴ Value of a3 + b³ is 7760

We will solve both the equations separately.

Equation I:

x2 + 9x + 18 = 0

⇒ x2 + 6x + 3x + 18 = 0

⇒ x(x + 6) + 3(x + 6) = 0

⇒ x = -6 or x = -3

Equation II:

y2 + 8y + 12 = 0

⇒ y2 + 6y + 2y + 12 = 0

⇒ y(y + 6) + 2(y + 6) = 0

⇒ y = -6 or y = -2

Comparing the values of x and y we get,

∴ x = y or the relationship cannot be determined

Given:

Total amount = Rs. 120

A's share = Rs. 20 + B's share = C's share - 20

Calculations:

Let the share of A be Rs. y.

B's share = y - 20

C's share = y + 20

As given, Total amount = Rs. 120

⇒ y + y - 20 + y + 20 = 120

⇒ 3y = 120

⇒ y = 40

B's share = y - 20

⇒ 40 - 20

⇒ 20

∴ B's share is Rs. 20.

Given:

x + y + z = 3 and xy + yz + zx = -11

Concept used:

Algebra

Calculation:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

Using the above values in the equation,

⇒ (3)² = x² + y² + z² + 2(-11)

⇒ x2 + y2 + z² = 22 + 9

⇒ x2 + y2 + z2 = 31

\({x^3} + {y^3} + {z^3} - 3xyz\) = (x + y + z) [(x2 + y2 + z2 - (xy + yz + zx)]

\({x^3} + {y^3} + {z^3} - 3xyz\) = (3) [(31 + 11)]

\({x^3} + {y^3} + {z^3} - 3xyz\) = 126

Given:

A positive number which when increased by 17 is equal to 84 times the reciprocal of the number

calculation:

Let the positive number be x 

Reciprocal of a positive number = 1/x

x + 17 = 84 × 1/x

⇒ x (x + 17) = 84

⇒ x² + 17x = 84

⇒ x2 + 17x - 84 = 0

⇒ x² + 21x - 4x - 84 = 0

⇒ x(x + 21) - 4( x + 21) = 0

⇒ (x + 4)(x - 21) = 0

⇒ x = 4, -21

The value of x is positive, So x = 4

∴ The required value is 4

Let the positive integer be 'x'

Then, 20 times integer = 20x

Square of integer = x²

Twenty times a positive integer is less than its square by 96

x²  = 20x + 96

⇒ x² - 20x - 96 = 0

⇒ x² - 24x + 4x - 96 = 0

⇒ x ( x -24) + 4( x - 24) = 0

⇒ (x + 4) × (x -24) = 0

⇒ x = -4,  x = 24

∴ The positive integer is 24

Given

xa × xb × xc = 1

Formula used

1) a^m × aⁿ = a^{m + n}

2) a⁰ = 1

3) a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)

Calculation

xa × xb × xc = 1

⇒ x^{a + b + c} = 1

⇒ x^{a + b + c} = x⁰

On comparing both sides we get,

a + b + c = 0

a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)

Putting (a + b + c) = 0 we get,

a3 + b3 + c3 − 3abc = 0

⇒ a3 + b3 + c² = 3abc

Given:

\({x^2} - √ 3 x + 1 = 0\)

Concept used:

Algebra

Calculation:

Divide the given equation by x

⇒ \(x + \frac{1}{x} = √ 3 \)

We need to find (x + 1/x)³

Let (x + 1/x) = k = √3

\({\left( {x + \frac{1}{x}} \right)^3} = {x^3} + \frac{1}{{{x^3}}} + 3k\)

\({x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3k\)

\({x^3} + \frac{1}{{{x^3}}} = {\left( {\sqrt 3 } \right)^3} - 3k\)

\({x^3} + \frac{1}{{{x^3}}} = 3\sqrt 3 - 3\sqrt 3 \)

∴ \({x^3} + \frac{1}{{{x^3}}} = 0\)

Formula Used:

(a – b)³ = a³ – b³ – 3a²b + 3ab²

(a – b)² = a² + b² – 2ab

Calculation:

(x – 3)3 = (2x – 4)2 – 13x2 + x3

⇒ x³ – 27 – 9x² + 27x = 4x² + 16 – 16x – 13x² + x³

⇒ x³ – 27 – 9x² + 27x = -9x² + x³ + 16 – 16x

⇒ -27 + 27x = 16 –16x

⇒ 43x = 43

∴ x = 1

The correct option is 1 i.e. 1

Given:

1 - a2 = a

Calculation:

As, 1 - a2 = a

Dividing the given equation by 'a', we get

\( {1 \ \over a}\)- a = 1      ----(i)

Cubing equation (i), we get

\( {1 \ \over a^3}\) - a³ - 3 × \( {1 \ \over a}\)× a × (\( {1 \ \over a}\)- a) = 1

\( {1 \ \over a^3}\) - a3  = 1 + 3 = 4     ----(ii)

squaring equation (ii), we get

\({1 \ \over a^6}\) + a⁶ - 2 = 16

\({1 \ \over a^6}\) + a⁶ = 16 + 2 = 18

∴ The value of \({1 \ \over a^6}\) + a⁶ is 18

Calculation:

I: 3x2 – 15x + 12 = 0

⇒ 3x2 – 12x – 3x + 12 = 0

⇒ 3x(x – 4) – 3(x – 4) = 0

⇒ (x – 4) × (3x – 3) = 0

⇒ x = 4, 1

II: 4y2 – 80y + 256 = 0

Dividing the equation by 4 we'll get

⇒ y2 – 20y + 64 = 0

⇒ y2 – 4y – 16y + 64 = 0

⇒ y × (y – 4) – 16 × (y – 4) = 0

⇒ (y – 4) × (y – 16) = 0

⇒ y = 4 and 16

Comparing values of x and y,

∴ From the table we can say, x ≤  y.

Given:

The polynomial is x2 + 6x +15

Calculation:

x2 + 6x +15

here, a = 1, b = 6 and c = 15

Now, determinant = b² - 4ac = 6² - 4 × 1 × 15 = 36 - 60 = -24 < 0

Since the determinant is negative therefore it has no zeros.

If the determinant is negative (D < 0)

It doesn’t have a zero which implies that the quadratic equation had no roots and hence it doesn’t intersect the x-axis. As the roots of the polynomial lie on the x-axis.

Given:

The number of boys in a class is three times the number of girls

Calculation:

Let number of girls be x and number of boys = 3x.

Then, total number of students = 3x + x = 4x 

Thus, to find exact value of x, the total number of students must be divisible by 4

44, 48, and 52 are divisible by 4, whereas 42 is not divisible by 4

∴ 42 cannot represent the total number of children in the class 

Given:

a - b = 9

ab = 10

Formula Used:

(a – b)² = (a + b)² – 4ab

Calculation:

substituting the values in the formula 

9² = (a + b)2 – 4 × 10

⇒ 81 + 40 = (a + b)2

⇒ (a + b)2 = 121

Taking root on both sides 

a + b = 11

∴ The value of a + b is 11

Given:

Total number = 500

One-third of first part = Second part – 72

Calculation:

Let the second part be x

According to question

(1/3) of First part = x – 72

⇒ First part = 3 × (x – 72) = 3x – 216

First part + Second part = 500

⇒ 3x – 216 + x = 500

⇒ 4x – 216 = 500

⇒ 4x = 716

⇒ x = 179

First part = 3x – 216 = 3 × 179 – 216 = 321

∴ The numbers are 321 and 179