RBSE Solutions Class 10 Maths Chapter 1 Vedic Mathematics Exercise 1.4

Using the Sutra Paravartya Yojayet, Solve Orally the Following Questions

Question 1. \( 13x - 14 = 9x + 10 \)
Answer: This equation is in the form \( ax + b = cx + d \). The Vedic math rule (Sutra Paravartya Yojayet) gives the solution for x as \( x = \frac{d-b}{a-c} \).
Here, \( a = 13, b = -14, c = 9, d = 10 \).
Now, we substitute these values into the formula:
\( x = \frac{10 - (-14)}{13 - 9} \)
\( \implies x = \frac{10 + 14}{4} \)
\( \implies x = \frac{24}{4} \)
\( \implies x = 6 \)
In simple words: When you have an equation like this, you can find 'x' by subtracting the constant terms and dividing by the difference of the 'x' coefficients. The number 6 is our answer.

🎯 Exam Tip: Remember to correctly identify the values for a, b, c, and d from the equation, paying close attention to their signs, especially with negative numbers.

Question 2. \( 3y + 4 = 5y - 4 \)
Answer: This equation also fits the form \( ay + b = cy + d \). We use the same Vedic math rule as before: \( y = \frac{d-b}{a-c} \).
Here, \( a = 3, b = 4, c = 5, d = -4 \).
Let's put these values into the formula:
\( y = \frac{-4 - 4}{3 - 5} \)
\( \implies y = \frac{-8}{-2} \)
\( \implies y = 4 \)
In simple words: Like the last problem, we use the special math trick. We take the last number minus the second number, then divide it by the first number minus the third number. This gives us 4.

🎯 Exam Tip: Be careful with negative signs when performing subtraction in both the numerator and denominator to avoid calculation errors.

Question 3. \( \frac {2x+1}{3x+4} = \frac {1}{3} \)
Answer: This equation is in the form \( \frac{ax+b}{cx+d} = \frac{p}{q} \). A specific Vedic math rule provides the solution for x as \( x = \frac{dp - bq}{aq - cp} \).
Here, we identify the values: \( a=2, b=1, c=3, d=4, p=1, q=3 \).
Now, we substitute these values into the formula:
\( x = \frac{(4 \times 1) - (1 \times 3)}{(2 \times 3) - (3 \times 1)} \)
\( \implies x = \frac{4 - 3}{6 - 3} \)
\( \implies x = \frac{1}{3} \)
In simple words: For fractions with 'x' like this, we use a special cross-multiplication formula. It helps us find 'x' quickly by combining the numbers in a specific way. The answer is one-third.

🎯 Exam Tip: When using this formula, ensure you correctly match 'a', 'b', 'c', 'd', 'p', and 'q' to their respective positions in the equation.

Question 4. \( \frac {5x-3}{2} = \frac {2x+1}{5} \)
Answer: This equation matches the form \( \frac{ax+b}{p} = \frac{cx+d}{q} \). We use the Vedic math formula \( x = \frac{dp - bq}{aq - cp} \).
From the given equation, we have: \( a=5, b=-3, p=2, c=2, d=1, q=5 \).
Substitute these values into the formula:
\( x = \frac{(1 \times 2) - (-3 \times 5)}{(5 \times 5) - (2 \times 2)} \)
\( \implies x = \frac{2 - (-15)}{25 - 4} \)
\( \implies x = \frac{2 + 15}{21} \)
\( \implies x = \frac{17}{21} \)
In simple words: This equation has fractions with 'x'. We use a special formula that helps us solve for 'x' by combining the numbers in a specific way. The answer is seventeen twenty-firsts.

🎯 Exam Tip: Be very careful with the negative signs, especially when multiplying negative numbers, to ensure the correct result.

Question 5. \( (x + 7)(x + 9) = (x - 8)(x - 11) \)
Answer: This equation is in the form \( (x+a)(x+b) = (x+c)(x+d) \). There is a specific Vedic math rule that solves for x directly using the constants: \( x = \frac{cd-ab}{a+b-c-d} \).
Here, we identify the values: \( a=7, b=9, c=-8, d=-11 \).
Now, substitute these values into the formula:
\( x = \frac{(-8)(-11) - (7)(9)}{7+9 - (-8) - (-11)} \)
\( \implies x = \frac{88 - 63}{7+9+8+11} \)
\( \implies x = \frac{25}{35} \)
\( \implies x = \frac{5}{7} \)
In simple words: When two sets of numbers multiplied by 'x' equal each other, we use a special formula with the constant numbers. We multiply and add them in a certain way to get the final 'x' value, which is five-sevenths.

🎯 Exam Tip: Pay close attention to the signs of 'c' and 'd' when substituting them into the formula, as this often leads to errors.

Question 6. \( (x + 5)(x + 1) = (x + 3)(x + 2) \)
Answer: This equation follows the form \( (x+a)(x+b) = (x+c)(x+d) \). We can use the Vedic math formula for x: \( x = \frac{cd-ab}{a+b-c-d} \).
Here, the values are: \( a=5, b=1, c=3, d=2 \).
Let's plug these values into the formula:
\( x = \frac{(3 \times 2) - (5 \times 1)}{5+1 - 3 - 2} \)
\( \implies x = \frac{6 - 5}{6 - 5} \)
\( \implies x = \frac{1}{1} \)
\( \implies x = 1 \)
In simple words: For this type of equation, we use the same formula as before. By correctly putting in the numbers and doing the math, we find that 'x' is equal to 1.

🎯 Exam Tip: Double-check both the numerator and the denominator calculations, as a small error in one will affect the final answer.

Question 7. \( \frac {1}{x+1} - \frac {2}{x+1} = 0 \)
Answer: This equation can be written as \( \frac{1}{x+1} + \frac{(-2)}{x+1} = 0 \). Using a specific algebraic formula for such cases, \( x = \frac{(1 \times 1) + (-2 \times 1)}{1 + (-2)} \). This formula helps find the value of x when terms share a common denominator.
\( x = \frac{1 - 2}{1 - 2} \)
\( \implies x = \frac{-1}{-1} \)
\( \implies x = 1 \)
In simple words: Since both parts of the equation share the same bottom number (\( x+1 \)), we can add the top numbers. Then, using a special rule, we find that 'x' is 1.

🎯 Exam Tip: When the denominators are identical, combine the numerators first. If a Vedic Sutra is specified, apply it carefully with the correct parameters.

Question 8. \( \frac {5}{2x-1} - \frac {9}{3x-2} = 0 \)
Answer: We can rewrite the equation as \( \frac{5}{2x-1} = \frac{9}{3x-2} \). To solve this, we can first convert it to the form \( \frac{ax+b}{p} = \frac{cx+d}{q} \). We can rearrange this to \( \frac{2x-1}{5} = \frac{3x-2}{9} \).
Now, we use the formula \( x = \frac{dp - bq}{aq - cp} \).
Here, \( a=2, b=-1, p=5, c=3, d=-2, q=9 \).
Substitute these values into the formula:
\( x = \frac{(-2 \times 5) - (-1 \times 9)}{(2 \times 9) - (3 \times 5)} \)
\( \implies x = \frac{-10 - (-9)}{18 - 15} \)
\( \implies x = \frac{-10 + 9}{3} \)
\( \implies x = \frac{-1}{3} \)
In simple words: We first move the second fraction to the other side to make both sides equal. Then we change the equation to a special form and use a formula that helps us find 'x' by multiplying and subtracting numbers in a particular order. The answer is negative one-third.

🎯 Exam Tip: For equations with fractions on both sides, cross-multiplication or converting to a standard formula type (like the one used here) are effective methods. Always check your signs.

Question 10. \( a(x - 1) + b(x - 1) = c(x - 1) + d(x - 1) \)
Answer: In this equation, the term \( (x-1) \) is a common factor on both sides of the equation. According to a Vedic math principle, if a common factor exists on both sides of an equation like this, setting that common factor to zero will give a solution for x. This means if \( (x-1) \) is a common factor, then `\( x-1 = 0 \)` must be true for a solution.
\( x - 1 = 0 \)
\( \implies x = 1 \)
In simple words: Since the part \( (x-1) \) is in every term on both sides, we can just say \( x-1 \) equals zero to find 'x'. So, 'x' is 1.

🎯 Exam Tip: When a common factor appears in every term of an equation, setting that factor to zero is a quick way to find one of the solutions.

Question 11. \( (x + 1)(x + 9) = (x + 3)(x + 3) \)
Answer: This equation is in the form \( (x+a)(x+b) = (x+c)(x+d) \). A Vedic math rule states that if the product of the constant terms on both sides is equal (i.e., \( ab = cd \)), then \( x=0 \) is a solution.
On the left side, the product of constants is \( 1 \times 9 = 9 \).
On the right side, the product of constants is \( 3 \times 3 = 9 \).
Since both products are equal to 9, according to the formula, \( x = 0 \).
Let's verify this by expanding both sides: \( x^2 + 10x + 9 = x^2 + 6x + 9 \). Subtracting \( x^2+9 \) from both sides gives \( 10x = 6x \), which simplifies to \( 4x = 0 \), so \( x=0 \).
In simple words: If the last numbers multiplied together are the same on both sides of this kind of equation, then 'x' will always be 0.

🎯 Exam Tip: For equations of the form \( (x+a)(x+b) = (x+c)(x+d) \), if the constant products (ab and cd) are equal, then x=0 is a solution.

Question 12. \( \frac {x}{2} + \frac {x}{3} = \frac {x}{4} + \frac {x}{1} \)
Answer: This equation has 'x' as the numerator in all terms. According to a Vedic math rule, if we have fractions with the same numerator 'x' and the sum of the denominators on the left side equals the sum of the denominators on the right side, then \( x=0 \) is the solution.
Sum of denominators on the Left Hand Side (L.H.S.): \( 2 + 3 = 5 \).
Sum of denominators on the Right Hand Side (R.H.S.): \( 4 + 1 = 5 \).
Since \( 5 = 5 \), the condition for the rule is met, and thus \( x = 0 \).
If we solve by common algebra: \( x(\frac{1}{2}+\frac{1}{3}) = x(\frac{1}{4}+\frac{1}{1}) \implies x(\frac{3+2}{6}) = x(\frac{1+4}{4}) \implies x(\frac{5}{6}) = x(\frac{5}{4}) \). This implies \( \frac{5}{6}x - \frac{5}{4}x = 0 \implies x(\frac{10-15}{12}) = 0 \implies x(-\frac{5}{12}) = 0 \implies x=0 \).
In simple words: When 'x' is on top of all the fractions, and the bottom numbers add up to the same total on both sides, then 'x' is simply 0.

🎯 Exam Tip: This Vedic math rule simplifies equations where 'x' is a common numerator. Always confirm that 'x' is the common numerator in all terms and that the sums of denominators are equal.

Question 13. \( \frac {1}{x+4} + \frac {1}{x-6} = 0 \)
Answer: For an equation of the form \( \frac{A}{x+a} + \frac{B}{x+b} = 0 \), if A and B are constants, a Vedic math rule states that we can set the sum of the denominators (if they were constant terms) equal to zero, provided the numerators are also constant. In this specific case, where the numerators are both 1, we can simply add the expressions in the denominators and set them to zero.
Thus, \( (x+4) + (x-6) = 0 \).
\( \implies 2x - 2 = 0 \)
\( \implies 2x = 2 \)
\( \implies x = 1 \)
In simple words: When you add two fractions with '1' on top and the equation equals zero, you can just add the bottom parts and set that sum to zero to find 'x'. So, 'x' is 1.

🎯 Exam Tip: When fractions sum to zero and have constant numerators, remember to set the sum of the denominators equal to zero to find the value of x.

Question 14. \( \frac {5}{3x+2} + \frac {5}{2x+8} = 0 \)
Answer: This equation has identical numerators (which is 5) for both fractions. According to a Vedic math rule, if two fractions with the same non-zero numerator sum to zero, then the sum of their denominators must also be equal to zero. This simplifies the problem significantly.
So, we set the sum of the denominators to zero:
\( (3x+2) + (2x+8) = 0 \)
\( \implies 5x + 10 = 0 \)
\( \implies 5x = -10 \)
\( \implies x = \frac{-10}{5} \)
\( \implies x = -2 \)
In simple words: Because the '5' on top is the same for both fractions, and they add up to zero, we just add the bottom parts together and make that sum zero. This tells us 'x' is -2.

🎯 Exam Tip: When solving equations where fractions with identical numerators sum to zero, remember that the sum of their denominators must also be zero, which makes finding 'x' very straightforward.

Question 15. \( \frac {2x+4}{2x+1} = \frac {2x+1}{2x+4} \)
Answer: This equation is in the form \( \frac{N_1}{D_1} = \frac{N_2}{D_2} \). A Vedic math rule applies here: if the sum of the numerator and denominator on one side is equal to the sum of the numerator and denominator on the other side, then this sum must be equal to zero.
Sum of numerator and denominator on L.H.S.: \( (2x+4) + (2x+1) = 4x + 5 \).
Sum of numerator and denominator on R.H.S.: \( (2x+1) + (2x+4) = 4x + 5 \).
Since both sums are equal (\( 4x+5 \)), we can set this sum to zero:
\( 4x + 5 = 0 \)
\( \implies 4x = -5 \)
\( \implies x = -\frac{5}{4} \)
In simple words: When the top number plus the bottom number equals the same total on both sides of the equal sign, that total must be zero. So, we set \( 4x+5 \) to zero and solve for 'x', which gives us negative five-fourths.

🎯 Exam Tip: For reciprocal-like fractions equated, check if the sum of the numerator and denominator is equal on both sides; if so, setting that sum to zero provides the solution.

Question 16. \( \frac {3x+2}{5x+7} = \frac {x+1}{3x-1} \) where the ratio of (sum of numerators) to (sum of denominators) is 1 : 2.
Answer: This equation follows a Vedic math principle where if the ratio of the sum of the numerators to the sum of the denominators is a constant value (in this case, 1:2), then the sum of the numerators must be zero.
Sum of numerators: \( (3x+2) + (x+1) = 4x + 3 \).
Sum of denominators: \( (5x+7) + (3x-1) = 8x + 6 \).
The ratio of these sums is \( \frac{4x+3}{8x+6} = \frac{4x+3}{2(4x+3)} = \frac{1}{2} \). This matches the given condition.
Therefore, according to the rule, we set the sum of the numerators to zero:
\( 4x + 3 = 0 \)
\( \implies 4x = -3 \)
\( \implies x = -\frac{3}{4} \)
In simple words: We add up the top numbers and the bottom numbers separately. If the total of the top numbers is half of the total of the bottom numbers, then the sum of the top numbers must be zero. This helps us find 'x' as negative three-fourths.

🎯 Exam Tip: When working with fractions equated, always check for relationships between the sum of numerators and sum of denominators, as specific Vedic math rules might apply.

Question 17. \( \frac {5x+7}{2x+1} = \frac {x+1}{3x+5} \)
Answer: This equation is in the form \( \frac{N_1}{D_1} = \frac{N_2}{D_2} \). A Vedic math rule states that if the difference between the numerator and denominator on one side is proportional to the difference between the numerator and denominator on the other side, then that difference can be set to zero.
Difference for L.H.S.: \( (5x+7) - (2x+1) = 3x + 6 \). This can be written as \( 3(x+2) \).
Difference for R.H.S.: \( (x+1) - (3x+5) = -2x - 4 \). This can be written as \( -2(x+2) \).
The differences are \( 3(x+2) \) and \( -2(x+2) \), which have a common factor of \( (x+2) \). So, we can set \( (x+2) \) to zero to find x.
\( x + 2 = 0 \)
\( \implies x = -2 \)
In simple words: We find the difference between the top and bottom numbers for both sides. If these differences share a common part, we set that common part to zero to get 'x'. Here, 'x' is -2.

🎯 Exam Tip: Look for common factors after calculating the differences between numerators and denominators; these often lead to direct solutions for x.

Question 18. \( \frac {3x+6}{6x+3} = \frac {5x+4}{2x+7} \)
Answer: This equation is in the form \( \frac{N_1}{D_1} = \frac{N_2}{D_2} \). A Vedic math rule applies when the difference between the denominator and numerator is the same on both sides (or equal to zero).
Difference \( D_1 - N_1 \) for L.H.S.: \( (6x+3) - (3x+6) = 3x - 3 \).
Difference \( D_2 - N_2 \) for R.H.S.: \( (2x+7) - (5x+4) = -3x + 3 \).
If we assume the rule implies \( D_1 - N_1 = 0 \), then:
\( 3x - 3 = 0 \)
\( \implies 3x = 3 \)
\( \implies x = 1 \)
Alternatively, if \( D_1 - N_1 = -(D_2 - N_2) \), then \( 3x - 3 = -(-3x+3) \implies 3x-3 = 3x-3 \), which is an identity, indicating the condition holds true and this value of 'x' is a solution. If we set \( 3x-3=0 \) then \( x=1 \).
In simple words: We subtract the top number from the bottom number for both sides. If these results are related (like being opposites or equal), we can set that result to zero to solve for 'x'. In this case, 'x' is 1.

🎯 Exam Tip: For fractions equated, comparing the difference between denominator and numerator on both sides can reveal common factors or a direct solution for x.

Question 19. \( \frac {1}{x+2} + \frac {1}{x+6} = \frac {1}{x+1} + \frac {1}{x+7} \)
Answer: This equation is in the form \( \frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{x+c} + \frac{1}{x+d} \). A Vedic math rule states that if the sum of the constant terms in the denominators on the left side equals the sum of the constant terms in the denominators on the right side (i.e., \( a+b = c+d \)), then the solution can be found by setting \( 2x + a + b = 0 \).
Sum of constants in L.H.S. denominators: \( 2 + 6 = 8 \).
Sum of constants in R.H.S. denominators: \( 1 + 7 = 8 \).
Since \( 8 = 8 \), the condition is met. So, we set \( 2x + (\text{sum of constants}) = 0 \).
\( 2x + 8 = 0 \)
\( \implies 2x = -8 \)
\( \implies x = -4 \)
In simple words: When you have fractions like these, and the sum of the last numbers on the bottom is the same for both sides, you add '2x' to that sum and make it zero. This helps us find 'x' as -4.

🎯 Exam Tip: For equations with inverse terms, always check if the sum of the constant denominators is equal on both sides, as this directly leads to a simplified equation for x.

Question 20. \( \frac {1}{x-4} + \frac {1}{x-6} = \frac {1}{x-2} + \frac {1}{x-8} \)
Answer: This equation has the same structure as the previous one, \( \frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{x+c} + \frac{1}{x+d} \). The Vedic math rule states that if the sum of the constant terms in the denominators on both sides is equal, then \( 2x + a + b = 0 \).
Sum of constants in L.H.S. denominators: \( -4 + (-6) = -10 \).
Sum of constants in R.H.S. denominators: \( -2 + (-8) = -10 \).
Since both sums are equal to -10, we apply the rule:
\( 2x + (-10) = 0 \)
\( \implies 2x - 10 = 0 \)
\( \implies 2x = 10 \)
\( \implies x = 5 \)
In simple words: Like the last problem, we add the constant numbers at the bottom of the fractions. If the totals are the same, we add '2x' to that total and set it to zero. This helps us find 'x' as 5.

🎯 Exam Tip: Be cautious with negative constant terms when summing denominators; ensure accurate addition or subtraction for the rule to apply correctly.