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Given

In certain city the Taxi charges comprise a fixed charge and the charge of the distance travelled. A person paid Rs. 156 for a journey of 16 km and another person paid Rs. 204 for the journey of 24 km.

Calculation

Let the fixed charge be Rs. 'x'

And let the charge of 1 km travelled be Rs. 'y'

x + 16y = 156      _____(1)

And x + 24y = 204      _____(2)

Subtracting equation (1) from equation (2), we get

8y = 48

⇒ y = 6

By putting the value of y in equation (1) we get,

x = 156 - 16(6) = 60

∴ The amount paid by a passenger who has travelled 30 km = x + 30y

⇒ 60 + 30(6) = 60 + 180 = Rs. 240

Given:

a – b = 8

ab = 3

Formula used:

(a – b)³ = a³ – b³ – 3ab(a –  b)

Calculation:

(a – b)³ = a3 – b3 – 3ab(a –  b)

⇒ 8³ = a³ – b³ – 3 × 3 × (8)

⇒ 512 = a3 – b3 – 72

⇒ a3 – b3 = 512 + 72 = 584 

∴ The correct answer is 584

Given:

Number = 27

5 times the first part and 11 times the second part both together = 195

Calculation:

Let, the first part = x

And second part = y

According to the question,

5x + 11y = 195      ---- (1) × 1

x + y = 27      ---- (2) × 5

From (1) – (2) we get,

⇒ 6y = 60

⇒ y = 10

From (1) we get,

⇒ 5x = 85

⇒ x = 17

∴ Required ratio of x : y is 17 : 10

Given:

a - b  = 5

a² + b² = 31

Formula used:

(a - b)² = a² + b² - 2ab

Calculation:

(a - b)2 = a2 + b2 - 2ab

⇒ 5² = 31 - 2ab

⇒ 25 = 31 - 2ab

⇒ 2ab = 31 - 25

⇒ 2ab = 6

⇒ ab = 3

∴ The value of ab is 3.

Given:

2(a2 + b2) = (a + b)2

Formula used:

(a + b)2 = (a2 + b2 + 2ab)

Calculation:

According to the question:

2(a2 + b2) = (a + b)2

⇒ 2(a2 + b2) = (a2 + b2 + 2ab)

⇒ 2a² + 2b2 = (a2 + b2 + 2ab)

⇒ 2a2 + 2b2 – a2 – b2 – 2ab = 0

⇒ a2 + b2 – 2ab = 0

⇒ (a – b)² = 0

⇒ a – b = 0

⇒ a = b

∴ The answer is a = b.

Given:

Donated by  Raj + Rohit = Rs. 195    

Donated by  Rohit + Raksha = Rs. 512      

Rohit donated = 1/7 of Raksha            

Calculation:

Donated by  Raj + Rohit = Rs. 195      ……………….(1)

Donated by  Rohit + Raksha = Rs. 512           ……………………(2)

Rohit donated = 1/7 of Raksha                     …………………….(3)

⇒ Let assume that  amount donated by Raksha = Rs. x

From equation (3) and (2)

⇒ 1/7 of x + x = Rs. 512

⇒ 8/7 x = 512

⇒ x = (512 × 7)/8

⇒ x = Rs. 448

⇒ Amount donated by Raksha = Rs. 448

According to equation (3)

⇒ Amount donated by Rohit =  448 × 1/7 = Rs. 64                …………….(4)

According to equation (1) and (4)

⇒ Amount donated by Raj = Rs. 195 – Rs.64 = Rs. 131

⇒ Amount more donated by Raj = Rs. 131 – Rs. 64 = Rs. 67

∴ Rs. 67 more donated by Raj than Rohit.                        

The correct option is 1 i.e. Rs. 67

Given:

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

Calculation:

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

⇒ 10x - 10 + 2 = 6x + 12

⇒ 10x - 6x = 12 + 8

⇒ 4x = 20

⇒ x = 20/4

⇒ x = 5

∴ The value of x is 5.

Given:

(x/3) + (3/x) = 1

Calculation:

∵ (x/3) + (3/x) = 1

⇒ (x² + 9)/3x = 1

⇒ x² + 9 = 3x      ------(1)

Now, rearranging the given equation;

⇒ [(x² + 9) – 3x]/(x² - 3x)

⇒ (3x – 3x)/(x² – x² - 9)      ------(From 1)

Given:

7x + 3y = 4 and 2x + y = 2

Calculation:

7x + 3y = 4      ----(1)

2x + y = 2     ----(2)

By solving,

equation (1) - 3 × equation(2)

⇒ 7x + 3y - 3(2x + y) = 4 - (3 × 2)

⇒ x = -2

Putting the value of x in equation (1)

7x + 3y = 4

⇒ 7 × (-2) + 3y = 4

⇒ y = 6

∴ The value of x and y is -2 and 6.

Given: 

(a × b)/c = (a2 + ab - b2)/ (c + a)

Calculation:

According to question

(4² + 4 × 5 - 5²)/ (7 + 4)

⇒ (16 + 20 – 25)/11

⇒ (36 – 25)/11

⇒ 11/11

⇒ 1

∴ (4 × 5)/7 =  (42 + 4 × 5 - 52)/ (7 + 4) = 1

The correct option is 2 i.e. 1.

Given:

a = 2

b = 7

c = 5

Calculation:

[(2)² + (2 × 7) – (7) ²]/ (5 - 2)

⇒ (4 + 14 – 49)/3

⇒ (18 – 49)/3

⇒ (-31/3)

∴ (2 × 7)/(5) = [(2)2 + (2 × 7) – (7) 2]/ (5 - 2) = - 31/3

The correct option is 2 i.e. - 31/3

Concept

A Prime number is a number divisible by 1 and itself only. 

a² - b² = (a - b)(a + b)

Calculation

a2 - b2 = (a - b)(a + b) is a prime

So, it is divisible by 1 and itself 

So, either (a - b) = 1 or (a + b) = 1

But (a + b) ≠ 1 {a and b are natural numbers}

∴ (a - b) = 1

⇒ a2 - b2 = (1)(a + b) = a + b

Given :

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaGioaiabgUcaRiaaikdacqGHAiI1caaIZaaa % paqaa8qacaaIZaWaaOaaa8aabaWdbiaaiodaaSqabaGccqGHRaWkca % aI1aGaaiiOaaaaaaa!3FA0! \frac{{8 + 2\surd 3}}{{3√ 3 + 5\;}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaGioaiabgUcaRiaaikdacqGHAiI1caaIZaaa % paqaa8qacaaIZaWaaOaaa8aabaWdbiaaiodaaSqabaGccqGHRaWkca % aI1aGaaiiOaaaaaaa!3FA0! \frac{{8 + 2\surd 3}}{{3\sqrt 3 + 5\;}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaGioaiabgUcaRiaaikdacqGHAiI1caaIZaaa % paqaa8qacaaIZaWaaOaaa8aabaWdbiaaiodaaSqabaGccqGHRaWkca % aI1aGaaiiOaaaaaaa!3FA0! \frac{{8 + 2\surd 3}}{{3\sqrt 3 + 5\;}}\) = a√3 - b 

Calculations :

\(\frac {8 + 2\sqrt 3}{3\sqrt 3 + 5}\) multiply numerator and denominator by \(\frac {3\sqrt 3 - 5}{3\sqrt 3 - 5} \)

We get ⇒ 7√3 - 11 = a√3 - b 

Now compare both sides of the equation 

we will get, a = 7 and b = 11 

a + b = 7 +11 

⇒ 18 

∴ the value of a + b will be 18 

Let the tens and units digit of number are 'x', 'y' respectively

First condition:

10x + y = 4(x + y)

⇒ 10x + y = 4x + 4y

⇒ 10x - 4x = 4y - y

⇒ 6x = 3y

⇒ 2x = y        ----(1)

Second condition:

Reversed Number = (10y + x)

(10x + y) + 27 = 10y + x

⇒ 10x - x + y - 10y = -27

⇒ 9x - 9y = -27

⇒  9(x - y) = -27

⇒ (x - y) = -27/3

⇒ (x - y) = -3      ----(2)

Substitute the value of eq.1 in equation.2, we can get the value of x

x - 2x = -3

⇒ x = 3

Substitute the value of x in in eq.1, we can get the value of y

2 × 3 = y

⇒ y = 6

According to first condition the original number = 10 × 3 + 6

⇒ 36

Adding 27 to the first condition = 36 + 27

⇒ 63

According to the second condition the reversed number = 10 × 6 + 3

⇒ 6

∴ The original number is 36

Calculation

x2 - 6x + 1 = 0

Dividing the whole equation by x,

x - 6 + 1/x = 0

\(\Rightarrow x + {1 \over x} = 6\)

Squaring both sides we get,

\(x^2 + {1\over x^2} + 2 = 36\)

\(\Rightarrow x^2 + {1\over x^2} = 36 - 2 = 34\)

Concept:

Using the formula:

x² - y² = (x + y) (x - y)

Calculation:

(8.3)2 - (1.7)2 

= (8.3 + 1.7) (8.3 - 1.7) 

= 10 × 6.6

= 66

Calculation:

(x + y)(x - y) = x² - y² 

= 9 - 4 

= 5

Formula used

(a + b)² = a² + b² + 2ab

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculation

\(x + \frac{1}{x} = 3\)

Squaring both sides of the equation

\((x + \frac{1}{x})^2 = 9\)

\(\Rightarrow x^2 + {1 \over x^2} + 2 = 9\)

\(\Rightarrow x^2 + {1 \over x^2} = 9 - 2 = 7\)

\(x + \frac{1}{x} = 3\)

Cubing both sides we get

\((x + \frac{1}{x})^3 = 27\)

\(\Rightarrow x^3 + {1 \over x^3} + 3( x + {1 \over x}) = 27\)

\(\Rightarrow x^3 + {1 \over x^3} = 27 - 9 = 18\)

Now, 

\((x^2 + {1 \over x^2 })(x^3 + {1 \over x^3}) = x^5 + {1 \over x^5} + x + {1 \over x}\)

\(\Rightarrow 7 \times 18 = x^5 + {1 \over x^5} + 3\)

\(\Rightarrow x^5 + {1 \over x^5} = 126 - 3 = 123\)

Concept

When a² + b² + c² = 0 then a = 0, b = 0 and c = 0

Calculation

(x - 1)2 + (x - 2)2 + (x - 3)2 = 0

⇒ (x - 1)2 = 0

and (x - 2)² = 0

and (x - 3)² = 0 [square of a number can't be negative] 

⇒ x - 1 = 0, x - 2 = 0 and x - 3 = 0

⇒ x = 1, 2, 3

Given:

√ x + (1/√ x) = 4 

Calculation:

Squaring both sides,

[√ x + (1/√ x)]² = 4² =16

x + 2 + (1/x) = 16

We get, x + (1/x) = 16 - 2 = 14      ----(1)

Squaring both sides of (1),

[x + (1/x)]² = 14² = 196

x² + 2 + (1/x²) = 196

x² + (1/x²) = 194

∴ The value of x2 + (1/x2) is 194.

Given:

x = 3 + √8

Calculation:

as x = 3 + √8

(1/x) = 1/(3 + √8) 

Rationalising we get, (1/x) = 3 - √8

Then, x + (1/x) = 3 + √8 + 3 - √8 = 6      ----(1)

taking cube of (1) we get,

[x + (1/x)]³ = 6³

x³ + (1/x)³ + 3 × x × (1/x) [x + (1/x)] = 216

x3 + (1/x)3 + 3 × [x + (1/x)] = 216

x3 + (1/x)3 = 216 - 3 × [x + (1/x)]

x3 + (1/x)3 = 216 - 3 × 6 = 198

∴ The value of x3 + (1/x)3 is 198. 

Given:

When 5 children join B from A, the number of children in both classes is the same

If 25 children from B join A, the number of children in A becomes double the number in B

Calculation:

Let the number of children in class A be 'a' and that of class B be 'b'

As per question

5 children from class A join class B

⇒ a - 5 = b + 5

⇒ a - b = 10 -----eq^-n (1)

Also, when 25 children from B join A, the number of children in A becomes double the number of children in B

⇒ a + 25 = 2 × (b - 25)

⇒ 2b - a = 75  ------eq^-n (2)

By adding eq^-n (1) and (2), we get

b = 85 and 

a= 95

∴ The ratio of the number of children in A to the number of children in B is 19 ∶ 17 

Calculation:

x² + 8x + 15 = 0

⇒ x² + 3x + 5x + 15 = 0

⇒ x(x + 3) + 5(x + 3) = 0

⇒ (x +5) (x + 3) = 0

∴ The required answer is (x + 5) (x + 3)

Given:

a³ + b³ = 217

(a + b) = 7

Formula Used:

(a + b)³ = a³ + b³ + 3ab (a + b)

Calculation:

Substituting the values in the formula

(7)³ = 217 + 3ab (7)

343 – 217 = 21ab

126/21 = ab

∴The value of ab is = 6