Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Find the coordinates of the point which divides the line joining (3,4) and (-1,8) internally in the ratio 1:3.<\/p>\n a)\u00a0(5 , 2)<\/p>\n b)\u00a0(4 , 2.6)<\/p>\n c)\u00a0(2 , 5)<\/p>\n d)\u00a0(4 , 2)<\/p>\n Question 2:\u00a0<\/b>Find the coordinates of the point which divides the line joining the points (4,2) and (4,6) internally in the ratio 2:3.<\/p>\n a)\u00a0(4 , 3.6)<\/p>\n b)\u00a0(3 , 4)<\/p>\n c)\u00a0(3.6 , 4.2)<\/p>\n d)\u00a0(2.8 , 3.6)<\/p>\n Question 3:\u00a0<\/b>Find the equation of the line which passes through the midpoint of (2,3) and (-4,1) having slope=-2.<\/p>\n a)\u00a0-2x+y=0<\/p>\n b)\u00a02x+y=0<\/p>\n c)\u00a02x+2y-2=0<\/p>\n d)\u00a02x-2y+2=0<\/p>\n Question 4:\u00a0<\/b>Find the equation of the line which is parallel to the line 4x+3y-7=0 and passes through the point (4,-3).<\/p>\n a)\u00a03x-4y=0<\/p>\n b)\u00a04x+3y-7=0<\/p>\n c)\u00a04x+3y-10=0<\/p>\n d)\u00a03x+4y=0<\/p>\n Question 5:\u00a0<\/b>Find the equation of the line which is parallel to the line 8x+7y-3=0 and passes through the point (3,-2).<\/p>\n a)\u00a05x+7y-1=0<\/p>\n b)\u00a08x+7y+10=0<\/p>\n c)\u00a08x+7y-10=0<\/p>\n d)\u00a05x+7y+1=0<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>What is the equation of the line which is perpendicular to 2x+3y-5=0 and passing through the point of intersection of the lines x+2y-7=0 and 5x-3y+4=0<\/p>\n a)\u00a03x+2y-3=0<\/p>\n b)\u00a03x-2y+3=0<\/p>\n c)\u00a03x-2y-3=0<\/p>\n d)\u00a03x+2y+3=0<\/p>\n Question 7:\u00a0<\/b>What is the area of the triangle formed with coordinate axis by the line having slope equal to the line x+2y-5=0 and passing through (2,1) ?<\/p>\n a)\u00a04<\/p>\n b)\u00a06<\/p>\n c)\u00a07<\/p>\n d)\u00a03<\/p>\n Question 8:\u00a0<\/b>What is the area of the triangle formed with coordinate axis by the line having slope equal to the line 2x+3y-5=0 and passing through (1,2) ?<\/p>\n a)\u00a014\/3<\/p>\n b)\u00a016\/3<\/p>\n c)\u00a017\/3<\/p>\n d)\u00a013\/3<\/p>\n Question 9:\u00a0<\/b>What is the equation of a line which passes through the point of intersection of lines A:x-3y=7 and B:2x+y=-7 and the required line is also perpendicular to line x+4y=10 ?<\/p>\n a)\u00a04x-y+4=0<\/p>\n b)\u00a04x-y+5=0<\/p>\n c)\u00a04x-3y-1=0<\/p>\n d)\u00a0x-4y-10=0<\/p>\n Question 10:\u00a0<\/b>Equation of one of the diagonals of a rhombus is 2x+3y=10. what is the equation of the other diagonal passing through the point (1,3) ?<\/p>\n a)\u00a0$3x+2y-9=0$<\/p>\n b)\u00a0$6x-4y+5=0$<\/p>\n c)\u00a0$3x-2y+3=0$<\/p>\n d)\u00a0$2x+3y-11=0$<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 11:\u00a0<\/b>The straight line y=3x must pass through the point:<\/p>\n a)\u00a0(0,1)<\/p>\n b)\u00a0(0,0)<\/p>\n c)\u00a0(2,0)<\/p>\n d)\u00a0(1,2)<\/p>\n Question 12:\u00a0<\/b>A is a point on the x-axis with abscissa – 8 and B is the pint on the y-axis with ordinate 15. The length of AB is<\/p>\n a)\u00a019 units<\/p>\n b)\u00a021 units<\/p>\n c)\u00a017 units<\/p>\n d)\u00a023 units<\/p>\n Question 13:\u00a0<\/b>What is the reflection of the point (-5, -2) in the line y = 1?<\/p>\n a)\u00a0(-5, 1)<\/p>\n b)\u00a0(-5, 4)<\/p>\n c)\u00a0(-5, 0)<\/p>\n d)\u00a0(-5, 5)<\/p>\n Question 14:\u00a0<\/b>If (1, 3) is the centroid of triangle ABC, with co-ordinates A(1, 6), B(3 , a) & C(b, 1), what is the value of a + b?<\/p>\n a)\u00a05<\/p>\n b)\u00a01<\/p>\n c)\u00a06<\/p>\n d)\u00a04<\/p>\n Question 15:\u00a0<\/b>What is the reflection of the point ( 6, -3) about the line y = -3?<\/p>\n a)\u00a0(6, 3)<\/p>\n b)\u00a0(6, 0)<\/p>\n c)\u00a0(6, -6)<\/p>\n d)\u00a0(6, -3)<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 16:\u00a0<\/b>Find the area (in sq units) enclosed by y = 0, x+6=0 and 2x-3y=6.<\/p>\n a)\u00a09<\/p>\n b)\u00a016<\/p>\n c)\u00a020<\/p>\n d)\u00a027<\/p>\n Question 17:\u00a0<\/b>What is the area of the triangle made by line 3x+4y = 24 with the coordinate axes?<\/p>\n a)\u00a06 square units<\/p>\n b)\u00a012 square units<\/p>\n c)\u00a024 square units<\/p>\n d)\u00a048 square units<\/p>\n Question 18:\u00a0<\/b>A cube is cut into 8 equal smaller cubes.The percentage change in the surface area is<\/p>\n a)\u00a080%<\/p>\n b)\u00a050%<\/p>\n c)\u00a0150%<\/p>\n d)\u00a0100%<\/p>\n Question 19:\u00a0<\/b>Find the centroid of the triangle formed by the vertices, P(-4,-2) Q(-3, -5) and R(1, 1) ?<\/p>\n a)\u00a0(2, -2)<\/p>\n b)\u00a0(-2, 2)<\/p>\n c)\u00a0(-2,-2)<\/p>\n d)\u00a0(2, 2)<\/p>\n Question 20:\u00a0<\/b>Which of the following represents hyperbola ?<\/p>\n a)\u00a0$x^2+y^2 = a^2$<\/p>\n b)\u00a0$\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$<\/p>\n c)\u00a0$\\frac{x^2}{a^2}-\\frac{y^2}{b^2} = 1$<\/p>\n d)\u00a0None of the above<\/p>\n DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n SSC CHSL Important Q&A PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Here, $x_1 = 3, x_2 = -1$ Required coordinates of the point = $(\\dfrac{mx_2+nx_1}{m+n} , \\dfrac{my_2+ny_1}{m+n})$<\/p>\n $= (\\dfrac{-1+9}{4} , \\dfrac{8+12}{4})$<\/p>\n $= (\\dfrac{8}{4} , \\dfrac{20}{4})$ 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Here, $x_1 = 4, x_2 = 4$ Required coordinates of the point = $(\\dfrac{mx_2+nx_1}{m+n} , \\dfrac{my_2+ny_1}{m+n})$<\/p>\n $= (\\dfrac{8+12}{5} , \\dfrac{12+6}{5})$<\/p>\n $= (\\dfrac{20}{5} , \\dfrac{18}{5})$ 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n midpoint of (2,3) and (-4,1) is ((2-4)\/2,(3+1)\/2) 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n slope of the line ax+by+c=0 is -a\/b 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n slope of the line ax+by+c=0 is -a\/b 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given line 2x+3y-5=0 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n x+2y-5=0 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 2x+3y-5=0 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The point of intersection of the lines A and B is (-2,-3) 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Diagonals of a rhombus are perpendicular to each other. GK Questions And Answers PDF<\/a><\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Equation of line = $y=3x$<\/p>\n (A) : (0,1)<\/p>\n => L.H.S. = $1\\neq$ R.H.S. = $3(0)=0$<\/p>\n (B) : (0,0)<\/p>\n =>\u00a0L.H.S. = $0=$ R.H.S. = $3(0)=0$<\/p>\n => Ans – (B)<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Now, OA = 8 units and OB = 15 units<\/p>\n => $(AB)^2=(OA)^2+(OB)^2$<\/p>\n => $(AB)^2=(8)^2+(15)^2$<\/p>\n => $(AB)^2=64+225=289$<\/p>\n => $AB=\\sqrt{289}=17$ units<\/p>\n => Ans – (C)<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We can see that (-5, -2) is 3 units away from y = 1 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Centroid = $(\\frac{x_{1}+x_{2}+x_{3}}{3} , \\frac{y_{1}+y_{2}+y_{3}}{3})$ $\\Rightarrow$ $1 = \\frac{1 + 3 + b}{3}$ Also, we know that $3 = \\frac{6 + a + 1}{3}$ Therefore, a+b = 2 -1 = 1. 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Since the given point (6, -3) lies on the line y = -3 itself, the point and its reflection will coincide. 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The graph represented by the lines\u00a0: $y=0$, $x+6=0$ and $2x-3y=6$ is<\/p>\n Now, area of $\\triangle$ ABC with base $BC=9$ units and height $AB=6$ units<\/p>\n = $\\frac{1}{2}bh$<\/p>\n = $\\frac{1}{2}\\times9\\times6=27$ sq. units<\/p>\n => Ans – (D)<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The given line will cut the x-axis at a point where the y co-ordinate will be equal to zero.<\/p>\n $\\Rightarrow$ 3x + 4*0 = 24.<\/p>\n $\\Rightarrow$ x = 8.<\/p>\n So, the point of intersection on the x axis will be (8, 0).<\/p>\n Similarly, the line will cut y-axis at a point where the x co-ordinate will be equal to zero.<\/p>\n $\\Rightarrow$ 3*0 + 4y = 24.<\/p>\n $\\Rightarrow$ y = 6.<\/p>\n So, the point of interaction on the y axis will be (0, 6)<\/p>\n We can see that $\\triangleOAB$ is right-angled at O.<\/p>\n So, the area of $\\triangleOAB$ = $\\frac{1}{2} * OA * OB$<\/p>\n = $\\frac{1}{2} * 6 * 8$ 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let us assume the side of the original cube to be 2a units. 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Centroid = ($\\frac{x_1+x_2+x_3}{3}, \\frac{y_1+y_2+y_3}{3}$) = ($\\frac{-4-3+1}{3}, \\frac{-2-5+1}{3}$) = (-2, -2)<\/p>\n So the answer is option C.<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Option A & B, represent Ellipse. ($\\because$ general form of Ellipse is $\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$)<\/p>\n Option C represents Hyperbola.\u00a0($\\because$ general form of Hyperbola is $\\frac{x^2}{a^2}-\\frac{y^2}{b^2} = 1$)<\/p>\n So the answer is option C.<\/p>\n\n
\n$y_1 = 4$ and $y_2 = 8$
\nm = 1 and n = 3<\/p>\n
\n$= (2 , 5)$<\/p>\n
\n$y_1 = 2$ and $y_2 = 6$
\nm = 2 and n = 3<\/p>\n
\n$= (4 , 3.6)$<\/p>\n
\n=((-2\/2),(4\/2))
\n=(-1,2)
\nGiven that line has a slope of -2
\nRequires equation is (-2)(x+1)=y-2
\n-2x-2=y-2
\n-2x-y=0
\n2x+y=0<\/p>\n
\nSlope of the line 4x+3y-7=0 is -4\/3
\nTwo parallel lines has same slope
\n(-4\/3)(x-4)=(y+3)
\n-4x+16=3y+9
\n4x+3y-7=0<\/p>\n
\nSlope of the line 8x+7y-3=0 is -8\/7
\nTwo parallel lines has same slope
\n(-8\/7)(x-3)=(y+2)
\n-8x+24=7y+14
\n8x+7y-10=0<\/p>\n
\ny=-(2\/3)x+5\/3
\nslope=-2\/3
\nProduct of slopes of 2 perpendicular lines=-1
\n(-2\/3) *m =-1
\nm=3\/2
\nGiven lines are x+2y-7=0 and 5x-3y+4=0
\n5x+10y-35=0
\n5x-3y+4=0
\n13y=39
\ny=3
\nx=1
\nRequired equation is (3\/2)(x-1)=y-3
\n3x-3=2y-6
\n3x-2y+3=0<\/p>\n
\ny=-x\/2 +5\/2
\nIt is in the form of y=mx+c
\nm=-1\/2
\nSlope of two parallel lines is equal.
\nRequired equation of the line is
\n-(1\/2)((x-2)=y-1
\n-x+2=2y-2
\nx+2y-4=0
\nExpressing equation of line in the form of (x\/a)+(y\/b)=1
\nArea of triangle is (1\/2)ab
\nx\/(4)+y\/(2)=1
\nArea of the triangle=(1\/2)*4*2
\n=4<\/p>\n
\ny=-2x\/3 +5\/3
\nIt is in the form of y=mx+c
\nm=-2\/3
\nSlope of two parallel lines is equal.
\nRequired equation of the line is
\n-(2\/3)(x-1)=y-2
\n-2x+2=3y-6
\n2x+3y-8=0
\nExpressing equation of line in the form of (x\/a)+(y\/b)=1
\nArea of triangle is (1\/2)ab
\nx\/(4)+y\/(8\/3)=1
\nArea of the triangle=(1\/2)*4*8\/3
\n=16\/3<\/p>\n
\nthe line is perpendicular to x+4y=10
\nSlope of x+4y=10 is ($\\frac{-1}{4}$)
\n$\\frac{-1}{4}\\times s $=-1
\ns=4
\nEquation of line is 4(x+2)=y+3
\n4x+8=y+3
\n4x-y+5=0<\/p>\n
\nLines perpendicular to each other have the product of their slopes = -1
\nSlope of the line\u00a0 2x+3y-10=0\u00a0 is\u00a0 $-(\\frac{a}{b})$
\nSlope = $\\frac{-2}{3}$
\nThen slope of the other line is $ \\frac{3}{2}$.
\n$\\therefore$ required\u00a0 equation\u00a0 is\u00a0 $\\frac{3}{2}(x-1)=(y-3)$
\n$3x-3=2y-6$
\n$ 3x-2y+3=0$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/31926.PNG\")
<\/figure>\n
\nX coordinate of the point will remain unchanged while y axis will change
\n$\\Rightarrow$ $1 = \\frac{-2 + b}{2}$
\n$\\Rightarrow$ $2 = -2 + b$
\n$\\Rightarrow$ $b = 4$
\nAlternatively from graph we can see the reflection of point as well. (Answer :B)<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Capture_302n0eD.PNG\")
<\/figure>\n
\n$(1, 3)$ = $(\\frac{1 + 3 + b}{3} , \\frac{6 + a + 1}{3})$<\/p>\n
\n$\\Rightarrow$ $b + 4 = 3$
\n$\\Rightarrow$ $b = -1$<\/p>\n
\n$\\Rightarrow$ $a + 7 = 9$
\n$\\Rightarrow$ $a = 2$<\/p>\n
\nHence, option B is the right answer.<\/p>\n
\nTherefore, the reflection of point = (6, -3).
\nHence, option\u00a0D is the right answer.<\/p>\n<\/figure>\n
![](\"https:\/\/cracku.in\/media\/uploads\/tRAINGLE.PNG\")
<\/figure>\n
\n= 24 sq units.
\nTherefore, option\u00a0C\u00a0is the right answer.<\/p>\n
\n=> The surface area of the bigger cube = $6*(2a)^{2} = 24(a)^{2}$
\nWhen we cut the original cube to get 8 smaller cubes, each side will become half of its original length.
\nSo, the side of smaller cube will be 2a\/2 = a units in length.
\nSurface area of 1 smaller cube = $6*(a)^{2} = 6(a)^{2}$
\nTotal surface area of 8 cubes combined = $8 * 6(a)^{2} = 48(a)^{2}$
\nSo, the percentage change in the surface area = $\\frac{48(a)^{2} – 24(a)^{2}}{24(a)^{2}} \\times 100$ = 100 %.
\nTherefore, option\u00a0D is the right answer.<\/p>\n