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Parin Sir > Quantitative Aptitude (Maths) > Linear Equations > Examples 4 to 10 > Examples 4 to 10

Examples 4 to 10

July 19, 2020
Category: Examples 4 to 10

(4) Solve : $\frac{7}{x+y}$ – $\frac{15}{x-y}$ = 4, $\frac{4}{x+y}$ + $\frac{5}{x-y}$ = 5.

Soln. Here,

$\frac{7}{x+y}$ – $\frac{15}{x-y}$ – 4 = 0

$\frac{4}{x+y}$ + $\frac{5}{x-y}$ – 5 = 0

( x + y, x – y ) = $\left (\frac{(7)(5)-(4)(-15)}{(-15)(-5)-(5)(-4)},- \frac{(7)(5)-(4)(-15)}{(7)(-5)-(4)(-4)} \right )$ = $\left (\frac{95}{95}, - \frac{95}{- 19} \right )$ = ( 1, 5)

i.e. x + y = 1 and x – y = 5

$\therefore$ Solution is ( 3, – 2 )

(5) Find k for which the system 15x – 6y = 9, k x – y = 7 has a unique solution.

Soln. The given system will have a unique solution, if

i.e., $\frac{15}{k} \ne \frac{-6}{-1}$ or k $\ne \frac{5}{2}$ . i.e., k $\in$ R – $\left \{ \frac{5}{2} \right \}$

(6) Find k for which the system x + 3y = 3, 4x + ky = 11 is inconsistent.

Soln. The given system will be inconsistent if $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$ $\ne \frac{c_{1}}{c_{2}}$ , i.e., if

$\frac{1}{4}= \frac{3}{k}$ $\ne \frac{3}{11}$ i.e., if k = 12.

(7) Find k for which the system 2x + 5y – 1 = 0, 4x + ky – 2 = 0 has an infinite number of solutions.

Soln. $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$ $= \frac{c_{1}}{c_{2}}$ , i.e., $\frac{2}{4}= \frac{5}{k}$ $= \frac{-1}{-2}$ or k = 10.

(8) Find whether the system 2x + 3y = 1, 8x + 12y = 2 has (i) a unique solution or (ii) an infinite number of solutions or (iii) no solution.

Soln. We have : $\frac{a_{1}}{a_{2}}= \frac{3}{6}= \frac{1}{2}$ , $\frac{b_{1}}{b_{2}}= \frac{-2}{-4}= \frac{1}{2}$ & $\frac{c_{1}}{c_{2}}= \frac{1}{2}$

$\therefore \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$ . So, the given system has no solution.

(9) Find k such that the system 5x + 2y = 0, kx + 4y = 0 has a non – zero solution.

Soln. The given system has non – zero solution, if $\frac{a_1}{a_2} = \frac{b_{1}}{b_{2}}$ or $\frac{5}{k} = \frac{2}{4}$ or k = 10.

(10) Samkit purchased 5 apples, 2 oranges and 7 bananas for Rs. 102.5. If each apples costs thrice as much as ( or twice as much again as ) orange and each orange is twice as expensive as a banana, how money will Samkit need to buy 2 apples, 5 oranges and 8 bananas?

Soln. Let the cost of banana be Rs. x.

$\therefore$ Cost of orange is Rs. 2x.

$\therefore$ Cost of an apple is Rs. 6x

Now, 5 ( 6x ) + 2 ( 2x ) + 7 ( x ) = 102.5

$\therefore$ 41x = 102.5 $\therefore$ x = 2.5

$\therefore$ 2 apples, 5 oranges and 8 bananas cost 2 ( 6x ) + 5 ( 2x ) + 8 ( x ) = 30x

= 30 $\times$ 2.5 = Rs. 75

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Examples 4 to 10

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