SSC CGL 2019 objective Ques (6 results)
SSC CGL 20191)The graphs of the equations 3x + y - 5 = 0 and 2x - y - 5 = 0 intersect at the point P(\({\alpha,\beta}\)). What is the value of (\({3\alpha+\beta}\)) ?
5
When graphs of the equations intersect at the point
\( P(\alpha, \beta) then,; 3\alpha + \beta - 5 = 0; 2\alpha - \beta - 5 = 0; \alpha = 2; \beta = -1; (3\alpha + \beta) = 3 \times 2 - 1 = 6 - 1 = 5 \)
SSC CGL 20192)The graph of the equation x - 7y = -42, intersects the y-axis at \(P\left(\alpha,\beta\right)\) and the graph of 6x + y - 15 = 0, intersects the x-axis at \( Q\left(\gamma,\delta\right)\) , What is the value of \(\alpha+\beta+\gamma+\delta\)?
\(17\over2\)
The
graph of the equation x — 7y = —42, intersects the y-axis at \(P\left(\alpha,\beta\right)\);
So, x = 0;
0 - 7y = -42;
y = 6;
\(\alpha\) = 0;
\(\beta\) = 6;
graph of 6x + y - 15 = 0, intersects the x-axis at\( Q\left(\gamma,\delta\right)\);
So, y = 0;
6x - 15 = 0;
x = 5/2;
\(\gamma\) = 5/2;
\(\delta \)= 0;
Now,
\(\alpha+\beta+\gamma+\delta = 0 + 6 + 5/2 + 0 = \frac{17}{2}\)
SSC CGL 20193)What is the area (in square units) of the triangular region enclosed by the graphs of the equations x + y = 3, 2x + 5y= 12 and the x-axis?
3
x + y = 3;
2x + 2y = 6 ---(1);
2x + 5y = 12 ---(2);
From eq (1) and eq (2), 3y = 6; y=2; so height = 2; put the value of y in eq(1) and (2), x = 3; and x = 6; Area = \({1\over2}\times base\times height\)= \({1\over2}\times(6-3)\times2=3\) square units
SSC CGL 20194)The graph of the equations 5x - 2y + 1 = 0 and 4y - 3x +5 = 0 , intersect at the point P (\(\alpha, \beta\) ). What is the value of \((2\alpha - 3\beta)\)?
4
SSC CGL 20195)The point of intersection of the graphs of the equations 3x - 5y = 19 and 3y - 7x + 1 =0 is P\((\alpha,\beta)\) . What is the value of \((3\alpha-\beta)\)?
-1
SSC CGL 20196)The graphs of the equations 2x+3y=11 and x-2y+12=0 intersects at P (\(x_1,y_1\)) and the graph of the equations x-2y+12=0 intersects the x-axis at \(Q(x_2,y_2)\). What is the value of \((x_1-x_2+y_1+y_2)\)?
15
2x+3y=11 and x-2y+12=0 intersects at P (\(x_1,y_1\)) ; x-2y+12=0 intersects the x-axis at \(x_2,y_2\).
therefore \(x_1=-2,y_1=5 \)
Also, when a line intersect x-axis the y cordinate is 0.
therefore \(x_2=-12,y_2=0\)
\((x_1-x_2+y_1+y_2) = \) 15