Get 125 SSC CGL Mocks – Just Rs. 199<\/a><\/p>\n Take a free SSC CGL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>A point in the 4th quadrant is 6 unit away from x-axis and 7 unit away from y-axis. The point is at<\/p>\n a)\u00a0(7, -6)<\/p>\n b)\u00a0(-7, 6)<\/p>\n c)\u00a0(-6, -7)<\/p>\n d)\u00a0(-6, 7)<\/p>\n Question 2:\u00a0<\/b>If $\\frac{\\sqrt{a+2b}+\\sqrt{a-2b}}{\\sqrt{a+2b} – \\sqrt{a-2b}}=\\frac{\\sqrt{3}}{1}$, find the value of $\\frac{a}{b}$<\/p>\n a)\u00a0$2:\\sqrt{3}$<\/p>\n b)\u00a0$\\sqrt{3}:4$<\/p>\n c)\u00a0$\\sqrt{3}:2$<\/p>\n d)\u00a0$4:\\sqrt{3}$<\/p>\n Question 3:\u00a0<\/b>The sum of four numbers is 48. When 5 and 1 are added to the first two; and 3 and 7 are subtracted from the 3rd and 4th, all the four numbers will be equal. The numbers are<\/p>\n a)\u00a09, 7, 15, 17<\/p>\n b)\u00a04, 12, 12, 20<\/p>\n c)\u00a05, 11, 13, 19<\/p>\n d)\u00a06, 10, 14, 18<\/p>\n Question 4:\u00a0<\/b>The length of the portion of the straight line 3x + 4y = 12 intercepted between the axes is<\/p>\n a)\u00a05<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a07<\/p>\n Question 5:\u00a0<\/b>If x = 332, y = 332, z = 332, then the value of $x^3 + y^3 + z^3 – 3xyz$ is<\/p>\n a)\u00a010000<\/p>\n b)\u00a00<\/p>\n c)\u00a08000<\/p>\n d)\u00a09000<\/p>\n Question 6:\u00a0<\/b>2x- ky + 7 = 0 and 6x- 12y+ 15 = 0 has no solution for<\/p>\n a)\u00a0k=- 1<\/p>\n b)\u00a0k=- 4<\/p>\n c)\u00a0k = 4<\/p>\n d)\u00a0k = 1<\/p>\n Question 7:\u00a0<\/b>A number exceeds its two fifth by 75. The number is<\/p>\n a)\u00a0125<\/p>\n b)\u00a0112<\/p>\n c)\u00a0100<\/p>\n d)\u00a0150<\/p>\n Question 8:\u00a0<\/b>Given that $x^{3} + y^{3} = 72$ and $xy = 8$ with $x > y$. Then the value of $(x – y)$ is<\/p>\n a)\u00a04<\/p>\n b)\u00a0-4<\/p>\n c)\u00a02<\/p>\n d)\u00a0-2<\/p>\n Question 9:\u00a0<\/b>If the sum of two numbers, one of which is $\\frac{2}{5}$ times the other, is 50, then the numbers are<\/p>\n a)\u00a0$\\frac{115}{7}$ and $\\frac{235}{7}$<\/p>\n b)\u00a0$\\frac{150}{7}$ and $\\frac{200}{7}$<\/p>\n c)\u00a0$\\frac{240}{7}$ and $\\frac{110}{7}$<\/p>\n d)\u00a0$\\frac{250}{7}$ and $\\frac{100}{7}$<\/p>\n Question 10:\u00a0<\/b>If $\\frac{3}{4}$ of a number is 7 more then $\\frac{1}{6}$ of the number, then $\\frac{5}{3}$ of the number is<\/p>\n a)\u00a012<\/p>\n b)\u00a020<\/p>\n c)\u00a015<\/p>\n d)\u00a018<\/p>\n Join Exam Preparation Telegram Group<\/a><\/p>\n Question 11:\u00a0<\/b>The area of the triangle formed by the graphs of the equations x= 0, 2x+ 3y= 6 and x+ y= 3 is :<\/p>\n a)\u00a03 sq. unit<\/p>\n b)\u00a0$4\\frac{1}{2}$ sq. unit<\/p>\n c)\u00a0$1\\frac{1}{2}$ sq. unit<\/p>\n d)\u00a01 sq. unit<\/p>\n Question 12:\u00a0<\/b>The graphs of x = a and y = b intersect at<\/p>\n a)\u00a0(a, b)<\/p>\n b)\u00a0(b, a)<\/p>\n c)\u00a0(-a, b)<\/p>\n d)\u00a0(a, -b)<\/p>\n Question 13:\u00a0<\/b>The area in sq. unit. of the triangle formed by the graphs ofx= 4, y= 3 and 3x+ 4y= 12 is<\/p>\n a)\u00a012<\/p>\n b)\u00a08<\/p>\n c)\u00a010<\/p>\n d)\u00a06<\/p>\n Question 14:\u00a0<\/b>The equations 3x+ 4y = 10 -x+ 2y = 0 have the solution (a, b). The value of a + b is<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n Question 15:\u00a0<\/b>If $2x+3y=\\frac{11}{2}$ and $xy=\\frac{5}{6}$ then the value of $8x^{3}+27y^{3}$ is<\/p>\n a)\u00a0$583$<\/p>\n b)\u00a0$ –\u00a0\\frac{583}{4}$<\/p>\n c)\u00a0$187$<\/p>\n d)\u00a0$- \\frac{429}{8}$<\/p>\n Get 125 SSC CGL Mocks – Just Rs. 199<\/a><\/p>\n Question 16:\u00a0<\/b>If $x^{2} – y^{2} = 80$ and x- y= 8, then the average of x and y is<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n Question 17:\u00a0<\/b>If $x,y,z \\neq 0$ and $\\frac{1}{x^{2}}+\\frac{1}{y^{2}}+\\frac{1}{z^{2}}$ = $\\frac{1}{xy}+\\frac{1}{yz}+\\frac{1}{zx}$ then the relation among x, y, z is<\/p>\n a)\u00a0x+ y+ z= 0<\/p>\n b)\u00a0x+y=z<\/p>\n c)\u00a0x-y=0<\/p>\n d)\u00a0x=y = z<\/p>\n Question 18:\u00a0<\/b>Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is<\/p>\n a)\u00a015<\/p>\n b)\u00a020<\/p>\n c)\u00a025<\/p>\n d)\u00a010<\/p>\n Question 19:\u00a0<\/b>If a* b= a+ b+ a\/b, then the value of 12 * 4 is :<\/p>\n a)\u00a020<\/p>\n b)\u00a021<\/p>\n c)\u00a043<\/p>\n d)\u00a019<\/p>\n Question 20:\u00a0<\/b>If $x^{2}+y^{2}+z^{2}=2(x-y-z)-3$, then the value of $2x-3y+4z$ is [Assume that x, y, z are all real numbers) :<\/p>\n a)\u00a09<\/p>\n b)\u00a01<\/p>\n c)\u00a03<\/p>\n d)\u00a00<\/p>\n More SSC CGL Important Questions and Answers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n A point in the 4th quadrant will be in the form of $(x,-y)$<\/p>\n Since, the point is 6 units away from x axis, => y coordinate = 6<\/p>\n and the point is 7 units away from y axis, => x coordinate = 7<\/p>\n => Point = (7,-6)<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n here in this question $\\frac{\\sqrt{a+2b}+\\sqrt{a-2b}}{\\sqrt{a+2b-}\\sqrt{a-2b}}=\\frac{\\sqrt{3}}{1}$<\/p>\n using componendo and dividendo, we will get<\/p>\n $\\frac{\\sqrt(a+2b)}{\\sqrt(a-2b)} = \\frac{\\sqrt3 + 1 }{\\sqrt3 – 1}$<\/p>\n now on squaring both side and solving, we will get<\/p>\n 16 b = 4a$\\surd3$<\/p>\n $\\frac{a}{b}$ = $\\frac{4}{\\surd3}$<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the numbers be $a,b,c,d$<\/p>\n => $a+b+c+d$ = 48 ——–Eqn(1)<\/p>\n When 5 & 1 are added to first two, => $(a+5) and (b+1)$<\/p>\n and when 3 & 7 are subtracted from last two, => $(c-3) and (d-7)$<\/p>\n According to question :<\/p>\n => $a+5 = b+1 = c-3 = d-7 = k$ (let)<\/p>\n Now, in eqn(1)<\/p>\n $(a+5) + (b+1) + (c-3) + (d-7)$ = 48 + (5+1-3-7)<\/p>\n => $k+k+k+k$ = 48-4<\/p>\n => $k$ = 11<\/p>\n => Numbers are : $a = k-5 = 11-5 = 6$<\/p>\n Similarly, $b$ = 10<\/p>\n $c$ = 14<\/p>\n $d$ = 18<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Equation : $3x + 4y = 12$<\/p>\n To find $x$-intercept, put $y$=0<\/p>\n => $3x + 0 = 12$<\/p>\n => $x$ = 4<\/p>\n Similarly to find $y$-intercept, we need to put $x$ = 0<\/p>\n => $0 + 4y = 12$<\/p>\n => $y$ = 3<\/p>\n Thus, the line passes through (4,0) in x-axis and (0,3) in y-axis<\/p>\n Using, $d = \\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$<\/p>\n => $x_1 = 0 , x_2 = 4 , y_1 = 0 , y_2 = 3$<\/p>\n => $d = \\sqrt{4^2 + 3^2}$<\/p>\n => $d = \\sqrt{25} = 5$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n when x = y = z= 332 , then $(x^2 + y^2 + z^2 – xy – yz – xz)$ = 0<\/p>\n and hence $x^3 + y^3 + z^3 – 3xyz$ = 0 as $x^3 + y^3 + z^3 – 3xyz$ = (x+y+z) $(x^2 + y^2 + z^2 – xy – yz – xz)$<\/span><\/p>\n and hence the answer for this question is = 0<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n NOTE : – For the pair of equations : $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$<\/p>\n The equations have no solution only if : $\\frac{a_1}{a_2} = \\frac{b_1}{b_2} \\neq \\frac{c_1}{c_2}$<\/p>\n Equations : $2x- ky + 7 = 0$ and $6x- 12y+ 15 = 0$<\/p>\n Comparing with above formula, we get :<\/p>\n => $\\frac{2}{6} = \\frac{-k}{-12}$<\/p>\n => $\\frac{k}{12} = \\frac{1}{3}$<\/p>\n => $k = 4$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number be $x$<\/p>\n Acc to ques :<\/p>\n => $x – \\frac{2}{5}x = 75$<\/p>\n => $\\frac{3x}{5} = 75$<\/p>\n => $x = 125$<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x^{3} + y^{3} = 72$ and $xy = 8$<\/p>\n Solution : $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$<\/p>\n => $(x+y)^3 = 72 + 3.8(x+y)$<\/p>\n => $(x+y)^3 – 24(x+y) – 72 = 0$<\/p>\n This is a cubic equation in terms of $(x+y)$ which has one real root = 6<\/p>\n => $x+y = 6$<\/p>\n Now, $(x-y)^2 = (x+y)^2 – 4xy$<\/p>\n => $(x-y) = \\sqrt{6^2 – 4*8} = \\sqrt{4}$<\/p>\n => $(x-y) = 2$<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the number be $x$<\/p>\n => Other number will be $\\frac{2x}{5}$<\/p>\n Acc to ques :<\/p>\n => $x$ + $\\frac{2x}{5}$ = 50<\/p>\n => $7x$ = 250<\/p>\n => $x$ = $\\frac{250}{7}$<\/p>\n and second number = $\\frac{2}{5} * \\frac{250}{7} = \\frac{100}{7}$<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the number be $x$<\/p>\n Acc to ques :<\/p>\n => $\\frac{3x}{4} = \\frac{x}{6} + 7$<\/p>\n => $\\frac{14x}{24} = 7$<\/p>\n => $x = 12$<\/p>\n => $\\frac{5}{3}$ of the number = $\\frac{5}{3}$ * 12 = 20<\/p>\n Free SSC Preparation Videos<\/a><\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n AC represents $x+y=3$<\/p>\n BC represents $2x+3y=6$<\/p>\n AB represents $x=0$<\/p>\n => ABC is the required triangle.<\/p>\n Base AB = 1 unit and height OC = 3 units<\/p>\n => Area of $\\triangle$ABC = $\\frac{1}{2}$ * AB * OC<\/p>\n = $\\frac{1}{2}$ * 1 * 3 = 1$\\frac{1}{2}$ sq. unit<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Clearly, the lines x = a and y = b meets at a point (a,b)<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Clearly, x=4 is a vertical line passing through (4,0) and y=3 is a horizontal line passing through the point (0,3) 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 3x + 4y = 0 —> (1) 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $2x+3y=\\frac{11}{2}$<\/p>\n cubing on both sides<\/p>\n $(2x+3y)^{3}=(\\frac{11}{2})^{3}$<\/p>\n $8x^{3}+27y^{3}+3(2x)(8y)(2x+3y)=\\frac{1331}{8}$<\/p>\n $8x^{3}+27y^{3}+3(16xy)(2x+3y)=\\frac{1331}{8}$<\/p>\n $8x^{3}+27y^{3}+3(16)(\\frac{5}{6})(\\frac{11}{2})=\\frac{1331}{8}$<\/p>\n $8x^{3}+27y^{3}+220 =\\frac{1331}{8}$<\/p>\n $8x^{3}+27y^{3} =\\frac{1331}{8} –\u00a0220$<\/p>\n $8x^{3}+27y^{3} =\\frac{1331}{8} – \\frac{1760}{8} =\u00a0– \\frac{429}{8}$<\/p>\n Get 125 SSC CGL Mocks – Just Rs. 199<\/a><\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $x^{2} – y^{2} = 80$ and $x – y= 8$<\/p>\n => $x + y = \\frac{x^2 – y^2}{x – y} = \\frac{80}{8}$<\/p>\n = 10<\/p>\n $\\therefore$ Required average = 10\/2 = 5<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression : $\\frac{1}{x^{2}}+\\frac{1}{y^{2}}+\\frac{1}{z^{2}}$ = $\\frac{1}{xy}+\\frac{1}{yz}+\\frac{1}{zx}$<\/p>\n Taking L.C.M on both sides<\/p>\n = $\\frac{x^2 + y^2 + z^2}{x^2 y^2 z^2} = \\frac{xy + yz + zx}{x^2 y^2 z^2}$<\/p>\n = $x^2 + y^2 + z^2 – xy – yz – zx = 0$<\/p>\n = $\\frac{1}{2} [(x-y)^2 + (y-z)^2 + (z-x)^2] = 0$<\/p>\n => $(x = y)$ and $(y = z)$ and $(z = x)$<\/p>\n => $x = y = z$<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let no. of correct questions be $x$ and incorrect be $y$<\/p>\n => $x + y = 30$<\/p>\n Also, $3x – 2y = 40$<\/p>\n Solving above equations, we get :<\/p>\n => $x = 20$ and $y = 10$<\/p>\n => No. of correct sums = 20<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n If $a*b = a + b + \\frac{a}{b}$<\/p>\n => $12*4 = 12 + 4 + \\frac{12}{4}$<\/p>\n = $16 + 3 = 19$<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : $x^{2}+y^{2}+z^{2}=2(x-y-z)-3$<\/p>\n => $x^2 – 2x + y^2 + 2y + z^2 + 2z + 3 = 0$<\/p>\n => $(x^2 – 2x + 1) + (y^2 + 2y + 1) + (z^2 + 2z + 1) = 0$<\/p>\n => $(x-1)^2 + (y+1)^2 + (z+1)^2 = 0$<\/p>\n => $x = 1 , y = -1 , z = -1$<\/p>\n To find : $2x-3y+4z$<\/p>\n = $(2*1) – (3* -1) + (4* -1)$<\/span><\/p>\n = $2+3-4 = 1$<\/span><\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/1612.PNG\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/1647.PNG\")
<\/p>\n
\nThe line 3x+ 4y= 12 also passes through (4,0) and (0,3) .
\nHence, it is a right angled triangle with base=4 units and height=3 units.
\nThe area of the triangle is = $\\frac{1}{2}bh$
\n= $\\frac{1}{2}\\times4\\times3$
\n=6 units<\/p>\n
\nx – 2y = 10
\n$(x – 2y = 10)\\times2$
\n2x – 4y = 20 —> (2)
\n5x = 20
\nx = 4
\ny = -3
\nx + y = 1<\/p>\n