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11). Ten chairs and six tables together cost Rs.6200, three chairs and two tables together cost Rs.1900. The cost of 4 chairs and 5 tables is
Let the cost of one chair be Rs. x and cost of one table be y. By given condition, 10x + 6y = 6200 ...(i) and 3x + 2y = 1900 = 9x + 6y = 5700 ...(ii) On subtracting Eq. (ii) from Eq. (i), we get x = Rs.500 From Eq. (i), 5000 + 6y = 6200 = 6y = 1200 Therefore, y = Rs.200 The cost of 4 chairs and 5 tables = 4x + 5y = \( \Large 4 \times 500 + 5 \times 200 \) = 2000 + 1000 = Rs.3000 | ||||

12). The system of equations 3x + y - 4 = 0 and 6x + 2y - 8 = 0 has
Given equations of system 3x + y = 4 ...(i) 6x + 2y = 8 ...(ii) Here, \( \Large a_{1}=3, b_{1}=1, c_{1}=4, \) \( \Large a_{2}=6, b_{2}=2, c_{2}=8 \) Because, \( \Large \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2} \) So, the system of equations has infinite solutions, because it represents a parallel line. | ||||

13). If x+ y - 7 = 0 and 3x + y -13 = 0, then what is \( \Large 4x^{2} + y^{2} + 4xy\) equal to?
We have x+y-7=0 | ||||

14). The solution of the equations \( \Large \frac{p}{x}+\frac{q}{y} \ and \ \frac{q}{x}+\frac{p}{y} \) n is
\( \Large \frac{p}{x}+\frac{q}{y}=m \) ...(i) | ||||

15). If \( \Large \frac{3}{x+y}+\frac{2}{x-y}=2 \) and \( \Large \frac{9}{x+y}-\frac{4}{x-y}=1 \), then what is the value of \( \Large \frac{x}{y} \)?
Given, \( \Large \frac{3}{x+y}+\frac{2}{x-y}=2 \) ...(i) and \( \Large \frac{9}{x+y}-\frac{4}{x-y}=1 \) ...(ii) Let x + y = a and x - y = b On multiplying Eq. (i) by 3 and subtracting from Eq. (ii), we get \( \Large \frac{9}{a}-\frac{4}{b}=1 \) \( \Large \frac{9}{a}+\frac{6}{b}=6 \) \( \Large \frac{-10}{b}=5 \) Therefore, b = 2 Now, on putting the value of b in Eq. (i), we get \( \Large \frac{3}{a}+\frac{2}{2}=2 \) \( \Large \frac{3}{a}=2-1 \) = a = 3 = x+y = 3 ...(iii) and x-y = 2 ...(iv) On subtracting Eq. (iv) from Eq. (iii), we get 2y = 1 Therefore, \( \Large y=\frac{1}{2} \) Now, putting \( \Large y=\frac{1}{2} \) in Eq. (iii), we get \( \Large x+\frac{1}{2}=3 \) = x = \( \Large 3-\frac{1}{2}=\frac{6-1}{2}=\frac{5}{2} \) Therefore, \( \Large \frac{x}{y}= \frac{\frac{5}{2}}{\frac{1}{2}}=\frac{5}{2} \times \frac{2}{1}=5 \) | ||||

16). If \( \Large \frac{a}{b}-\frac{b}{a}=\frac{x}{y} \) and \( \Large \frac{a}{b}+\frac{b}{a}= x - y \) , then what is the value of x?
Given equations are | ||||

17). If \( \Large \frac{x}{2}+\frac{y}{3}=4 \) and \( \Large \frac{2}{x}+\frac{3}{y}=1 \) then what is x + y equal to?
Because, \( \Large \frac{x}{2}+\frac{y}{3}=4 \) 2 *12, 3*8,4*6, 6*4,8*3,12*2 2 and 12 cannot be the values of x and y as their sum is 14 and it is not given in options. | ||||

18). For the two given equations I and II. I. \( \Large p^{2} \)+ 5p + 6 = 0 II.\( \Large q^{2} \) + 3q + 2 = 0
I. => \( \Large p^{2} \) + 3p + 2p + 6 = 0 => p ( p + 3 ) + 2 ( p + 3 ) = 0 => ( p + 3 ) ( p + 2 ) = 0 => p = - 2 or - 3 II. => \( \Large q^{2} \) + q + 2q + 2 = 0 => q ( q + 1 ) + 2 ( q + 1 ) = 0 => ( q + 1 ) ( q + 2 ) = 0 => q = - 1 or - 2 Obviously p \( \Large \leq \) q | ||||

19). For the two given equations I and II. I. \( \Large p^{2} \) = 4 II. \( \Large q^{2} \) + 4q = - 4
I. => p = \( \Large \pm \) 2 II. => \( \Large q^{2} \) + 2q + 2q + 4 = 0 => q ( q + 2 ) + 2 ( q + 2) = 0 => ( q + 2) ( q + 2 ) = 0 => q = - 2 Obviously, p \( \Large \geq \) q | ||||

20). For the two given equations I and II. I. \( \Large p^{2} \) + p = 56 II. \( \Large q^{2} \) - 17 q + 72 = 0
I. => \( \Large p^{2} \) + p - 56 = 0 => \( \Large p^{2} \) + 8p - 7p - 56 = 0 => p ( p + 8 ) - 7 ( p + 8 ) = 0 => ( p + 8 ) ( p - 7 ) = 0 => p = 7 or - 8 II. => \( \Large q^{2} \) - 8q - 9q + 72 = 0 => q ( q - 8 ) - 9 ( q - 8 ) = 0 => ( q - 8 ) ( q - 9 ) = 0 => q = 8 or 9 Obviously, p < q |