SSC CHSL Algebra Previous Year Questions download PDF based on previous year question paper of SSC exams. 20 Very important Algebra questions for SSC CHSL Exam.<\/p>\n
Download SSC CHSL Previous Year Algebra Questions<\/a><\/p>\n Download SSC CHSL Biology Previous Year Questions<\/a><\/p>\n\n Download Biology Previous Questions Set-1<\/a><\/p>\n Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Which of the following has to be added to $36x^{2}-87x+52$ to make it a perfect square ?<\/p>\n a)\u00a03x+3<\/p>\n b)\u00a0-3x+3<\/p>\n c)\u00a03x-3<\/p>\n d)\u00a0-3x-3<\/p>\n Question 2:\u00a0<\/b>Which of the following has to be added to $16x^{2}-38x+23$ to make it a perfect square ?<\/p>\n a)\u00a0-2x+2<\/p>\n b)\u00a0-x+3<\/p>\n c)\u00a0-2x-2<\/p>\n d)\u00a0-x-3<\/p>\n Question 3:\u00a0<\/b>If a:b=6:5 and b:c=12:14 then what is the value of $a^{2}:b^{2}:bc$<\/p>\n a)\u00a0216:150:225<\/p>\n b)\u00a0108:150:175<\/p>\n c)\u00a0216:125:175<\/p>\n d)\u00a0216:150:175<\/p>\n Question 4:\u00a0<\/b>If a:b=7:3 and b:c=9:11 then what is the value of $a^{2}:b^{2}:bc$<\/p>\n a)\u00a06:49:11<\/p>\n b)\u00a09:49:11<\/p>\n c)\u00a049:9:11<\/p>\n d)\u00a049:9:13<\/p>\n Question 5:\u00a0<\/b>What is the least number that has to be added to $9x^{2}-30x+42$ to make it a perfect square ?<\/p>\n a)\u00a017<\/p>\n b)\u00a0-17<\/p>\n c)\u00a07<\/p>\n d)\u00a0-7<\/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 least number that has to be added to $4x^{2}-12x+3$ to make it a perfect square ?<\/p>\n a)\u00a0-3<\/p>\n b)\u00a0+3<\/p>\n c)\u00a0-6<\/p>\n d)\u00a06<\/p>\n Question 7:\u00a0<\/b>What is the least number that has to be added to $49x^{2}-28x+9$ to make it a perfect square ?<\/p>\n a)\u00a0-5<\/p>\n b)\u00a05<\/p>\n c)\u00a0-8<\/p>\n d)\u00a08<\/p>\n Question 8:\u00a0<\/b>What is the least number that has to be added to $9x^{2}-42x+36$ to make it a perfect square ?<\/p>\n a)\u00a0-11<\/p>\n b)\u00a013<\/p>\n c)\u00a0-13<\/p>\n d)\u00a011<\/p>\n Question 9:\u00a0<\/b>If a:b=1:4 and b:c=8:7 then what is the value of $a^{2}:b^{2}:ac$<\/p>\n a)\u00a02:32:7<\/p>\n b)\u00a01:16:9<\/p>\n c)\u00a02:32:13<\/p>\n d)\u00a01:16:11<\/p>\n Question 10:\u00a0<\/b>If $x^{4}+(\\frac{1}{x})^{4}$=34 then what is the value of $x-1\/x$<\/p>\n a)\u00a02<\/p>\n b)\u00a06<\/p>\n c)\u00a08<\/p>\n d)\u00a04<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 11:\u00a0<\/b>If $x^{4}+(\\frac{1}{x})^{4}$=727 then what is the value of $x^{3}-(\\frac{1}{x})^{3}$<\/p>\n a)\u00a0140<\/p>\n b)\u00a0138<\/p>\n c)\u00a0134<\/p>\n d)\u00a0136<\/p>\n Question 12:\u00a0<\/b>If $x^{4}+(\\frac{1}{x})^{4}$=119 then what is the value of $x^{3}-(\\frac{1}{x})^{3}$<\/p>\n a)\u00a036<\/p>\n b)\u00a024<\/p>\n c)\u00a042<\/p>\n d)\u00a030<\/p>\n Question 13:\u00a0<\/b>If a and b are positive integers and $a^{2}-b^{2}$=11 then find the value of b+(1\/b)<\/p>\n a)\u00a021\/5<\/p>\n b)\u00a026\/5<\/p>\n c)\u00a013\/5<\/p>\n d)\u00a029\/5<\/p>\n Question 14:\u00a0<\/b>If p and q are positive integers and $p^{2}-q^{2}$=59 then find the value of (1\/p)+(1\/q).<\/p>\n a)\u00a059\/870<\/p>\n b)\u00a029\/870<\/p>\n c)\u00a03\/87<\/p>\n d)\u00a0133\/870<\/p>\n Question 15:\u00a0<\/b>If x and y are positive integers and $x^{2}-y^{2}$=31 then find the value of y.<\/p>\n a)\u00a01<\/p>\n b)\u00a030<\/p>\n c)\u00a015<\/p>\n d)\u00a016<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 16:\u00a0<\/b>If 1<\/p>\n a)\u00a01<\/p>\n b)\u00a04<\/p>\n c)\u00a0-1<\/p>\n d)\u00a0-5<\/p>\n Question 17:\u00a0<\/b>If 2<x<3 then what is the value of $\\sqrt{(x-2)^{2}}+\\sqrt{(x-3)^{2}}$<\/p>\n a)\u00a01<\/p>\n b)\u00a05<\/p>\n c)\u00a0-1<\/p>\n d)\u00a0-5<\/p>\n Question 18:\u00a0<\/b>If 5<x<6 then what is the value of $\\sqrt{(x-5)^{2}}+\\sqrt{(x-6)^{2}}$<\/p>\n a)\u00a02<\/p>\n b)\u00a011<\/p>\n c)\u00a0-1<\/p>\n d)\u00a01<\/p>\n Question 19:\u00a0<\/b>If the sum of the interior angles of a regular polygon is $1260^\\circ$ then how many sides does it have?<\/p>\n a)\u00a010<\/p>\n b)\u00a08<\/p>\n c)\u00a09<\/p>\n d)\u00a07<\/p>\n Question 20:\u00a0<\/b>Find the volume(approx) $(in\u00a0 cm^3)$ of a cube of side 6.5 cm.<\/p>\n a)\u00a0274.625<\/p>\n b)\u00a0264.625<\/p>\n c)\u00a0224.625<\/p>\n d)\u00a0314.625<\/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 let us write the equation in the form of $a^{2}-2ab+b^{2}$ 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n let us write the equation in the form of $a^{2}-2ab+b^{2}$ 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n let a=6x,b=5x,b=12y,c=14y 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n let a=7x,b=3x,b=9y,c=11y 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n let us write the equation in the form of $a^{2}-2ab+b^{2}$ 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n let us write the equation in the form of $a^{2}-2ab+b^{2}$ 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n let us write the equation in the form of $a^{2}-2ab+b^{2}$ 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n let us write the equation in the form of $a^{2}-2ab+b^{2}$ 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n let a=x,b=4x,b=8y,c=7y 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $x^{4}+(\\frac{1}{x})^{4}$=34 ($x-\\frac{1}{x})^{2}$=4 <\/p>\n GK Questions And Answers PDF<\/a><\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $x^{4}+(\\frac{1}{x})^{4}$=727 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $x^{4}+(\\frac{1}{x})^{4}$=119 $(x^{2}+\\frac{1}{x^{2}})^{2}$=121<\/p>\n $x^{2}+(\\frac{1}{x})^{2}$=11 $x^{3}-(\\frac{1}{x})^{3}$=36<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n As 11 is a prime number we have 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n As 59 is a prime number we have the other case i.e p+q=1 and p-q=31 is not considered because it gives q value as negative.<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n As 31 is a prime number we have The other case i.e x+y=1 and x-y=31 is not considered because it gives y value as negative.<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n value of $\\sqrt{(z+1)^{2}}$ for 1Value of $(z-3)^{2}$ for 1So required sum=z+1-z+3 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n value of $\\sqrt{(x-2)^{2}}$ for 2<x<3 is $+$. Hence, +(x-2)<\/p>\n Value of $\\sqrt(x-3)^2$ for 2<x<3 is $-$. Hence, -(x-3)<\/p>\n So required sum=x-2-x+3 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n value of $\\sqrt{(x-5)^{2}}$ for 5<x<6 is always positive and so we take +(x-5)<\/p>\n Value of \u00a0$(x-6)^{2}$ for 5<x<6 is always $(-)^{2}$ an so we take -(x-6)<\/p>\n So required sum=x-5-x+6<\/p>\n =6-5<\/p>\n =1<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, sum of the interior angles of a regular polygon is $1260^\\circ$<\/p>\n Sum of all interior angles of a polygon with $’n’$ sides = $(n-2)\\times180^\\circ$<\/p>\n Let the number of sides be $n$<\/p>\n => Sum of interior angles =\u00a0$(n-2)\\times180^\\circ=720^\\circ$<\/p>\n =\u00a0$(n-2)=\\large\\frac{1260^\\circ}{180^\\circ}=7$<\/p>\n => $n=7+2=9$<\/p>\n => Ans – (C)<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Side of cube =\u00a0$a=6.5$ cm<\/p>\n => Volume of cube = $a^3$<\/p>\n = $(6.5)^3=274.625$ $cm^3$<\/p>\n => Ans – (A)<\/p>\n\n
\n$(6x-7)^{2}$=$36x^{2}-2*6x*7+7^{2}$
\nBut in the question it is given as -87x+52 so we should add 3x-3 to it in order to make it a perfect square.<\/p>\n
\n$(4x-5)^{2}$=$16x^{2}-2*4x*5+5^{2}$
\nBut in the question it is given as 38x+23 so we should add -2x+2 to it in order to make it a perfect square.<\/p>\n
\n5x=12y,x=12\/5y
\n$a^{2}$=$36*x^{2}$
\n$b^{2}$=$25*x^{2}$
\nbc=$5x*5x*14\/12$
\nbc=$175\/6*x^{2}$
\n$a^{2}:b^{2}:bc$=216:150:175<\/p>\n
\n3x=9y,x=3y,a=7x
\nb=3x,c=11x\/3
\n$a^{2}$=$49*x^{2}$
\n$b^{2}$=$9*x^{2}$
\nbc=$3x*11x\/3$
\nbc=$11*x^{2}$
\n$a^{2}:b^{2}:bc$=49:9:11<\/p>\n
\n$(3x-5)^{2}$=$3x^{2}-2*3x*5+5^{2}$
\nBut in the question it is given as 42 so we should add -17 to it in order to make it a perfect square.<\/p>\n
\n$(2x-3)^{2}$=$4x^{2}-2*2x*3+3^{2}$
\nBut in the question it is given as 3 so we should add 6 to it in order to make it a perfect square.<\/p>\n
\n$(7x-2)^{2}$=$49x^{2}-2*7x*2+2^{2}$
\nBut in the question it is given as 9 so we should add -5 to it in order to make it a perfect square.<\/p>\n
\n$(3x-7)^{2}$=$9x^{2}-2\\times3x\\times7+7^{2}$
\nBut in the question it is given as 36 so we should add 13 to it in order to make it a perfect square.<\/p>\n
\n4x=8y,x=2y,
\na=x, b=4x, c=7x\/2
\n$a^{2}$=$x^{2}$
\n$b^{2}$=$16\\times x^{2}$
\nac=$x\\times7x\/2$
\nac=$(7\/2)\\times x^{2}$
\n$a^{2}:b^{2}:ac$=2:32:7<\/p>\n
\nAdd 2 on both sides
\n$x^{4}+2+(\\frac{1}{x})^{4}$=34+2
\n($x^{2}+\\frac{1}{x^{2}})^{2}$=36
\n$x^{2}+(\\frac{1}{x})^{2}$=6
\nSubtracting 2 on both sides
\n$x^{2}+(\\frac{1}{x})^{2}-2$=4<\/p>\n
\n$x-1\/x$=2<\/p>\n
\nAdd 2 on both sides
\n$x^{4}+2+(\\frac{1}{x})^{4}$=727+2
\n($x^{2}+\\frac{1}{x^{2}})^{2}$=729
\n$x^{2}+(\\frac{1}{x})^{2}$=27
\nSubtracting 2 on both sides
\n$x^{2}+(\\frac{1}{x})^{2}-2$=27-2
\nx-1\/x=5
\nCubing on both sides we have
\n$x^{3}-(\\frac{1}{x})^{3}-15$=125
\n$x^{3}-(\\frac{1}{x})^{3}$=125+15=140<\/p>\n
\nAdd 2 on both sides
\n$x^{4}+2+(\\frac{1}{x})^{4}$=119+2<\/p>\n
\nSubtracting 2 on both sides
\n$x^{2}+(\\frac{1}{x})^{2}-2$=11-2
\nx-1\/x=3
\nCubing on both sides we have
\n$x^{3}-(\\frac{1}{x})^{3}-9$=27<\/p>\n
\n(a+b)(a-b)=11*1
\na+b=11
\na-b=1
\n2a=12
\na=6
\nb=5
\n(b)+(1\/b)=(1\/5)+(5)
\n=(26\/5)
\nThe other case i.e a+b=1 and a-b=11 is not taken because it will give negative values for a and b.<\/p>\n
\n(p+q)(p-q)=59*1
\np+q=59
\np-q=1
\n2p=60
\np=30
\nq=29
\n(1\/p)+(1\/q)=((1\/30)+(1\/29).
\n=(59\/870)<\/p>\n
\n(x+y)(x-y)=31*1
\nx+y=31
\nx-y=1
\n2x=32
\nx=16
\ny=15<\/p>\n
\n=3+1
\n=4<\/p>\n
\n=3-2
\n=1<\/p>\n