RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Ex 1.4 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 1 Vedic Mathematics Exercise 1.4.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Vedic Mathematics |

Exercise |
Exercise 1.4 |

Number of Questions Solved |
20 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 1 Vedic Mathematics Ex 1.4

**Using the Sutra Paravartya Yogurt, Solve orally the following questions**

Question 1.

13x – 14 = 9x + 10

Solution

We know (RBSESolutions.com) that

Algebraic formula ax + b = cx + d then

Question 2.

3y + 4 = 5y – 4

Solution

We know that

Algebraic formula ax + b = cx + d then

Question 3.

\(\frac { 2x+1 }{ 3x+4 }\) = \(\frac { 1 }{ 3 }\)

Solution

We know (RBSESolutions.com) that

Question 4.

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

Solution

We know that

Question 5.

(x + 7)(x + 9) = (x – 8)(x – 11)

Solution

We know that (RBSESolutions.com) if equation

(x + a)(x + b) = (x + c)(x + d)

then putting values a = 7, b = 9, c = -8 and d = – 11

Question 6.

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

Solution

We know that, (RBSESolutions.com) if equation

Question 7.

Solution

We know that if

Question 8.

Solution

Using the sutra Shunyam Samyaschahye, Solve the equation:

Question 9.

(2x + 1) + (x + 3) = 5x + 4

Solution

(2x + 1) + (x + 3) = 5x + 4 independent (RBSESolutions.com) term of both sides are same (= 4)

So, x = 0.

Question 10.

a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)

Solution

Here

a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)

Thus, (x – 1) is common factor is both sides, then

x – 1 = 0

⇒ x = 1

Question 11.

(x + 1)(x + 9) = (x + 3)(x + 3)

Solution

(x + 1)(x + 9) = (x + 3)(x + 3)

Here independent term in both sides are same (= 9) then x = 0

Question 12.

Solution

Numerator (x) is (RBSESolutions.com) same in both sides

Sum of denominators in R.H.S. = 2 + 3 = 5

Sum of denominators in R.H.S. = 4 + 1 = 5

Thus, According to formula, x = 0

Question 13.

Solution

Here Numerator of two fractions are same = 1,

So According to formula :

x + 4 + x- 6 = 0

⇒ 2x – 2 = 0

⇒ 2x =

⇒ x = 1

Question 14.

Solution

Here, Numerator of two (RBSESolutions.com) fractions are same = 5,

So according to formula :

3x + 2 + 2x + 8 = 0

⇒ 3x + 2x + 2 + 8 = 0

⇒ 5x + 10 = 0

⇒ 5x = -10

⇒ x = -2

Question 15.

Solution

Sum of numerators of both sides = 2x + 4 + 2x + 1 = 4x + 5

Sum of denominators in both sides = 2x + 1 + 2x + 4 = 4x + 5

Two sums are equal, so by the formula

4x + 5 = 0

⇒ 4x = -5

⇒ x = \(\frac { -5 }{ 4 }\)

Question 16.

Solution

Sum of numerators (RBSESolutions.com) of two sides = 3x + 2 + x + 1 = 4x + 3 …..(i)

Sum of denominators of two sides = 5x + 7 + 3x – 1 = 8x + 6 …(ii)

Ratio of (i) and (ii) is 1 : 2.

So, by formula, equation any sum equal to zero,

4x + 3 = 0

⇒ 4x = – 3

⇒ x = \(\frac { -3 }{ 4 }\)

Question 17.

Solution

Difference between numerator and denominator = 5x + 7 – 2x – 1 = 3x + 6 = 3(x + 2) …(i)

Difference between numerator and denominator is R.H.S. = 3x + 5 – x – 1 = 2x + 4 = 2(x + 2) …(ii)

Ratio of (i) and (ii) is 3 : 2 THus, by formula

3(x + 2) = 0

⇒ x + 2 = 0

⇒ x = -2

Question 18.

Solution

Difference between numerator (RBSESolutions.com) and denominator in L.H.S. = 6x + 3 – 3x – 6 = 3x – 3 …(i)

Difference numerator and denominator is L.H.S. = 5x + 4 – 2x – 7 = 3x – 3 …(ii)

Difference numerator and denominator in R.H.S.

3x – 3 = 0

⇒ 3x = 3

⇒ x = 1

Thus, x = 1

(i) and (ii) are equal, so by formula

3x + 6 + 5x + 4 = 8x + 10 …(i)

Adding denominators of both sides

6x + 3 + 2x + 7 = 8x + 10 …(ii)

(i) and (ii) are equal so by formula.

8x + 10 = 0

⇒ 8x = – 10

⇒ x = \(\frac { -10 }{ 8 }\) = \(\frac { -5 }{ 4 }\)

thus, x = \(\frac { -5 }{ 4 }\) and x = 1

Question 19.

Solution

Sum of (RBSESolutions.com) denominators of L.H.S. = x + 2 + x + 6 = 2x + 8 …(i)

Sum of denominators of R.H.S. = x + 1 + x + 7 = 2x + 8 …(ii)

(i) and (ii) are equal so by formula

2x + 8 = 0

⇒ 2x = -8

⇒ x = -4

Question 20.

Solution

Sum of denominators ofL.H.S. = x – 4 + x – 6 = 2x – 10 …(i)

Sum of denominators of R.H.S. = x – 2 + x – 8 = 2x – 10 …(ii)

(i) and (ii) are same so by formula

2x – 10 = 0

⇒ 2x = 10

⇒ x = 5

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