{"id":43858,"date":"2020-12-31T17:41:25","date_gmt":"2020-12-31T12:11:25","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=43858"},"modified":"2020-12-31T17:41:25","modified_gmt":"2020-12-31T12:11:25","slug":"top-20-ssc-cgl-algebra-questions","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/top-20-ssc-cgl-algebra-questions\/","title":{"rendered":"Top-20 SSC CGL Algebra questions"},"content":{"rendered":"

SSC CGL Algebra Questions<\/strong><\/h1>\n

Download Top-20 Algebra questions for SSC CGL exam. Most important Algebra questions based on asked questions in previous exam papers for SSC CGL.<\/p>\n

Download Top – 20 SSC CGL Algebra questions <\/a><\/p>\n

Get 125 SSC CGL Mocks – Just Rs. 199<\/a><\/p>\n

 <\/p>\n

Take a free SSC CGL Tier-1 mock test<\/a><\/p>\n

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Question 1:\u00a0<\/b>If a * b = 2a + 3b – ab, then the value of (3 * 5 + 5 * 3) is<\/p>\n

a)\u00a010<\/p>\n

b)\u00a06<\/p>\n

c)\u00a04<\/p>\n

d)\u00a02<\/p>\n

Question 2:\u00a0<\/b>If a * b\u00a0= $a^{b}$, then the value of 5 * 3 is<\/p>\n

a)\u00a0125<\/p>\n

b)\u00a0243<\/p>\n

c)\u00a053<\/p>\n

d)\u00a015<\/p>\n

Question 3:\u00a0<\/b>If $x = 1 + \\sqrt{2} + \\sqrt{3}$ , then the value of $(2x^4 – 8x^3 – 5x^2 + 26x- 28)$ is __?<\/p>\n

a)\u00a0$6\\sqrt{6}$<\/p>\n

b)\u00a0$0$<\/p>\n

c)\u00a0$3\\sqrt{6}$<\/p>\n

d)\u00a0$2\\sqrt{6}$<\/p>\n

Question 4:\u00a0<\/b>If $a^2+b^2+c^2=2(a-2b-c-3)$ then the value of a+b+c is<\/p>\n

a)\u00a03<\/p>\n

b)\u00a00<\/p>\n

c)\u00a02<\/p>\n

d)\u00a04<\/p>\n

Question 5:\u00a0<\/b>Find the simplest value of $2\\sqrt{50} + \\sqrt{18} – \\sqrt{72}$ is __? $(\\sqrt{2} = 1.414)$.<\/p>\n

a)\u00a09.898<\/p>\n

b)\u00a010.312<\/p>\n

c)\u00a08.484<\/p>\n

d)\u00a04.242<\/p>\n

Question 6:\u00a0<\/b>If $a^{3}-b^{3}-c^{3}=0$ then the value of $a^{9}-b^{9}-c^{9}-3a^{3} b^{3} c^{3}$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a00<\/p>\n

d)\u00a0-1<\/p>\n

Question 7:\u00a0<\/b>If $a^{3}-b^{3}-c^{3}=0$ then the value of $a^{9}-b^{9}-c^{9}-3a^{3} b^{3} c^{3}$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a00<\/p>\n

d)\u00a0-1<\/p>\n

Question 8:\u00a0<\/b>If x + y + z = 6 and $x^{2}+y^{2}+z^{2}$=20 then the value of $x^{3}+y^{3}+z^{3}$-3xyz is<\/p>\n

a)\u00a064<\/p>\n

b)\u00a070<\/p>\n

c)\u00a072<\/p>\n

d)\u00a076<\/p>\n

Question 9:\u00a0<\/b>If $\\frac{p^2}{q^2}+\\frac{q^2}{p^2}$=1 then the value of $(p^{6}+q^{6})$ is <\/span><\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a03<\/p>\n

Question 10:\u00a0<\/b>If $x=\\frac{a-b}{a+b},y=\\frac{b-c}{b+c},z=\\frac{c-a}{c+a}$ then $\\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$ is equal to<\/p>\n

a)\u00a01<\/p>\n

b)\u00a00<\/p>\n

c)\u00a02<\/p>\n

d)\u00a0$\\frac{1}{2}$<\/p>\n

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Question 11:\u00a0<\/b>If $\\frac{\\sqrt{7}-1}{\\sqrt{7}+1}-\\frac{\\sqrt{7}+1}{\\sqrt{7}-1}=a+\\sqrt{7} b$ the values of a and b are respectively<\/p>\n

a)\u00a0$\\sqrt{7},-1$<\/p>\n

b)\u00a0$\\sqrt{7}, 1$<\/p>\n

c)\u00a0$0, -\\frac{2}{3}$<\/p>\n

d)\u00a0$-\\frac{2}{3}, 0$<\/p>\n

Question 12:\u00a0<\/b>If $x=\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}}$ then $x^{3} + \\frac{1}{x^{3}}$ is equal to<\/p>\n

a)\u00a098<\/p>\n

b)\u00a01000<\/p>\n

c)\u00a05<\/p>\n

d)\u00a0970<\/p>\n

Question 13:\u00a0<\/b>$\\sqrt{\\sqrt{\\sqrt{0.00000256}}}$<\/p>\n

a)\u00a00.4<\/p>\n

b)\u00a00.02<\/p>\n

c)\u00a00.04<\/p>\n

d)\u00a00.2<\/p>\n

Question 14:\u00a0<\/b>If m = – 4, n = – 2, then the value of $m^3 – 3m^2 + 3m + 3n + 3n^2 + n^3$ is<\/p>\n

a)\u00a0– 126<\/p>\n

b)\u00a0124<\/p>\n

c)\u00a0– 124<\/p>\n

d)\u00a0126<\/p>\n

Question 15:\u00a0<\/b>If $x+\\frac{1}{x}=1$ then the value of $\\frac{x^2+3x+1}{x^2+7x+1}$<\/p>\n

a)\u00a0$1$<\/p>\n

b)\u00a0$\\frac{3}{7}$<\/p>\n

c)\u00a0$\\frac{1}{2}$<\/p>\n

d)\u00a02<\/p>\n

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Question 16:\u00a0<\/b>If the cube root of 79507 is 43, then the value of $\\sqrt[3]{79.507}+\\sqrt[3]{0.079507}+\\sqrt[3]{0.000079507}$
\nis<\/p>\n

a)\u00a00.4773<\/p>\n

b)\u00a0477.3<\/p>\n

c)\u00a047.73<\/p>\n

d)\u00a04.773<\/p>\n

Question 17:\u00a0<\/b>If $\\frac{x}{y}$=$\\frac{3}{4}$ the ratio of $(2x+3y)$ and $(3y-2x)$ is<\/p>\n

a)\u00a02 : 1<\/p>\n

b)\u00a03 : 2<\/p>\n

c)\u00a01 : 1<\/p>\n

d)\u00a03 : 1<\/p>\n

Question 18:\u00a0<\/b>If m – 5n = 2, then the vlaue of $(m^{3} – 125n^{3}$ – 30 mn) is<\/p>\n

a)\u00a06<\/p>\n

b)\u00a07<\/p>\n

c)\u00a08<\/p>\n

d)\u00a09<\/p>\n

Question 19:\u00a0<\/b>If $x+\\frac{1}{x}=2$ then the value of $x^{12}+\\frac{1}{x^{12}}$ is <\/span><\/p>\n

a)\u00a02<\/p>\n

b)\u00a0-4<\/p>\n

c)\u00a00<\/p>\n

d)\u00a04<\/p>\n

Question 20:\u00a0<\/b>If 5x + 9y = 5 and $125x^{3}$ + $729y^{3}$ = 120 then the value of the product of x and y is<\/p>\n

a)\u00a0$\\frac{1}{9}$<\/p>\n

b)\u00a0$\\frac{1}{135}$<\/p>\n

c)\u00a0$45$<\/p>\n

d)\u00a0$135$<\/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

For 3*5 put a=3 and b=5 in given equation
\nand for 5*3 put a=5 and b=3 in equation
\nnow add both values<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Put a=5 and b=3 in given equation
\nhence it will be $5^{3}$ = 125<\/p>\n

3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

x = 1+ $\\sqrt {2} + \\sqrt {3} $
\n$(x-1)^{2}$ = $(\\sqrt {2} + \\sqrt {3}) ^ {2} $
\n$x^{2} +1 – 2x = 5 + 2 \\sqrt {6}$
\n$x^{2} – 2x = 4 + 2 \\sqrt {6}$ ( eq. (1) )
\n$(x^{2} – 2x)^{2} = x^{4} + 4x^{2} – 4x^{3} = 40 + 16\\sqrt{6} $ eq (2)
\nNow in $2x^{4} – 8x^{3} – 5x^{2} + 26x – 28 $
\nor $2(x^{4} – 4x^{3}) – 5x^{2} + 26x – 28 $ ( putting values from eq (1) and eq (2) )
\nAfter solving we will get it reduced to $6\\sqrt{6}$<\/p>\n

4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given\u00a0$a^2+b^2+ c^2=2(a-2b-c-3)$,<\/p>\n

So, $(a-1)^2+(b+2)^2+(c-1)^2=0$<\/p>\n

Hence, a=1, b=-2 and c=1<\/p>\n

So, the sum of the equation is<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Given equation can be reduced in the form of $10\\sqrt2 + 3\\sqrt2 – 6\\sqrt2 = 7\\sqrt2$
\nHence\u00a0\u00a0$7\\sqrt2$ will be around 9.898<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

shortcut :<\/p>\n

put c = 0 in\u00a0\u00a0$a^{3}-b^{3}-c^{3}=0$ $\\Rightarrow$\u00a0$a^{3}=b^{3}$<\/p>\n

$a^{9}-b^{9}-(0)^{9}-3a^{3} b^{3} (0)^{3}$ =\u00a0$a^{9}-b^{9}$ =\u00a0$(a^{3})^{3}-(b^{3})^{3}$ =\u00a0\u00a0$(a)^{3}-(a)^{3}$ = 0\u00a0 ( $\\because$ $a^{3}=b^{3}$ )<\/p>\n

so the answer is option C.<\/p>\n

normal method :<\/p>\n

$a^{3}-b^{3}-c^{3}=0$<\/p>\n

$a^{3}=b^{3}+c^{3}$<\/p>\n

cubing on both sides,<\/p>\n

$(a^{3})^{3}=(b^{3}+c^{3})^{3}$<\/p>\n

$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(b^{3}+c^{3})$<\/p>\n

$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(a^{3})$<\/p>\n

$a^{9}-b^{9}-c^{9}-3a^{3}b^{3} c^{3}=0$<\/p>\n

so the answer is option C.<\/p>\n

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

shortcut :<\/p>\n

put c = 0 in\u00a0\u00a0$a^{3}-b^{3}-c^{3}=0$ $\\Rightarrow$\u00a0$a^{3}=b^{3}$<\/p>\n

$a^{9}-b^{9}-(0)^{9}-3a^{3} b^{3} (0)^{3}$ =\u00a0$a^{9}-b^{9}$ =\u00a0$(a^{3})^{3}-(b^{3})^{3}$ =\u00a0\u00a0$(a)^{3}-(a)^{3}$ = 0\u00a0 ( $\\because$ $a^{3}=b^{3}$ )<\/p>\n

so the answer is option C.<\/p>\n

normal method :<\/p>\n

$a^{3}-b^{3}-c^{3}=0$<\/p>\n

$a^{3}=b^{3}+c^{3}$<\/p>\n

cubing on both sides,<\/p>\n

$(a^{3})^{3}=(b^{3}+c^{3})^{3}$<\/p>\n

$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(b^{3}+c^{3})$<\/p>\n

$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(a^{3})$<\/p>\n

$a^{9}-b^{9}-c^{9}-3a^{3}b^{3} c^{3}=0$<\/p>\n

so the answer is option C.<\/p>\n

8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

We know that $x^{3}+y^{3}+z^{3}-3xyz = (x + y + z)(x^2 +y^2 + z^2 -xy-yz-xz)$
\n$x^{3}+y^{3}+z^{3}-3xyz = (6)(20 -xy-yz-xz)$
\nHence the solution must be a multiple of 6.
\nOut of the given options only Option C is a multiple of 6.
\nHence Option C is the correct answer.<\/span><\/span><\/span>
\n<\/span><\/span><\/span><\/span><\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Expression : $\\frac{p^2}{q^2}+\\frac{q^2}{p^2}$ = 1<\/p>\n

=> $\\frac{p^{4}+q^{4}}{p^2q^2}$ = 1<\/p>\n

=> $p^4+q^4 = p^2q^2$ ————–Eqn(1)<\/p>\n

Now, to find : $(p^{6}+q^{6})$<\/p>\n

=> $(p^2)^3 + (q^2)^3$<\/p>\n

Using the formula, $a^3 + b^3 = (a+b)(a^2+b^2-ab)$<\/p>\n

=> $(p^2+q^2)(p^4+q^4-p^2q^2)$<\/p>\n

From eqn (1), we get :<\/p>\n

=> $(p^2+q^2)(p^2q^2-p^2q^2)$<\/p>\n

=> $(p^2+q^2)*0$<\/span><\/p>\n

= 0<\/span><\/span><\/p>\n

10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

If $x=\\frac{a-b}{a+b}$<\/p>\n

=> $(1-x) = 1- (\\frac{a-b}{a+b})$<\/p>\n

=> $(1-x) = \\frac{2b}{a+b}$<\/span><\/p>\n

Similarly, $(1+x) = \\frac{2a}{a+b}$<\/span><\/p>\n

Applying the same method, we get :<\/span><\/p>\n

=> $(1-y) = \\frac{2c}{b+c}$ and => $(1+y) = \\frac{2b}{b+c}$<\/span><\/span><\/p>\n

=> $(1-z) = \\frac{2a}{c+a}$ and => $(1+z) = \\frac{2c}{c+a}$<\/span><\/span><\/span><\/span><\/p>\n

Putting above values in the equation : $\\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$<\/span><\/span><\/span>
\n<\/span><\/span><\/p>\n

=> $\\frac{(\\frac{2b}{a+b})(\\frac{2c}{b+c})(\\frac{2a}{c+a})}{(\\frac{2a}{a+b})(\\frac{2b}{b+c})(\\frac{2c}{c+a})}$<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

=> $\\frac{2a*2b*2c}{2a*2b*2c}$<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

= 1<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

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11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$\\frac{\\sqrt{7}-1}{\\sqrt{7}+1}-\\frac{\\sqrt{7}+1}{\\sqrt{7}-1}=a+\\sqrt{7} b$<\/p>\n

L.H.S. = $\\frac{\\sqrt{7}-1}{\\sqrt{7}+1}-\\frac{\\sqrt{7}+1}{\\sqrt{7}-1}$<\/p>\n

= $\\frac{(\\sqrt{7}-1)^2 – (\\sqrt{7}+1)^2}{(\\sqrt{7}-1)(\\sqrt{7}+1)}$<\/span><\/p>\n

= $\\frac{(7+1-2\\sqrt{7})-(7+1+2\\sqrt{7})}{7-1}$<\/span><\/p>\n

= $\\frac{-4\\sqrt{7}}{6}$<\/span><\/p>\n

= $\\frac{-2\\sqrt{7}}{3}$<\/span><\/p>\n

Now, comparing with R.H.S. $a+\\sqrt{7} b$<\/span><\/p>\n

we get, <\/span><\/span><\/p>\n

$a=0$ and $b=\\frac{-2}{3}$<\/span><\/span><\/p>\n

 <\/p>\n

12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$x=\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}}$<\/p>\n

=> $x=\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}} * \\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}$<\/p>\n

=> $x= 5+2\\sqrt{6}$ —————Eqn(1)<\/span><\/p>\n

Now, $\\frac{1}{x}=\\frac{1}{5+2\\sqrt{6}}$<\/span><\/p>\n

=> $\\frac{1}{x} = \\frac{1}{5+2\\sqrt{6}} * \\frac{5-2\\sqrt{6}}{5-2\\sqrt{6}}$<\/span><\/p>\n

=> $\\frac{1}{x}= 5-2\\sqrt{6}$ —————Eqn(2)<\/span><\/span><\/p>\n

Now, cubing eqns (1)&(2), we get :<\/span><\/p>\n

=> $x^3 = 125+72\\sqrt{6}+150\\sqrt{6}+360 = 485+222\\sqrt{6}$<\/span><\/p>\n

and $\\frac{1}{x^3} = 125-72\\sqrt{6}-150\\sqrt{6}+360 = 485-222\\sqrt{6}$<\/span><\/p>\n

To find : $x^{3} + \\frac{1}{x^{3}}$<\/span><\/span><\/p>\n

= $485+222\\sqrt{6} + 485-222\\sqrt{6}$<\/span><\/span><\/span><\/p>\n

= 970<\/span><\/span><\/span><\/span><\/p>\n

13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Expression : $\\sqrt{\\sqrt{\\sqrt{0.00000256}}}$<\/p>\n

= $\\sqrt{\\sqrt{0.0016}}$<\/p>\n

= $\\sqrt{0.04} = 0.2$<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

We are given that m = -4 and n = -2<\/p>\n

Expression : $m^3 – 3m^2 + 3m + 3n + 3n^2 + n^3$<\/p>\n

= $(m^3 – 3m^2 + 3m – 1) + (n^3 + 3n^2 + 3n + 1)$<\/p>\n

= $(m-1)^3 + (n+1)^3$<\/p>\n

= $(-4-1)^3 + (-2+1)^3$<\/p>\n

= $(-5)^3 + (-1)^3$<\/p>\n

= $-125 – 1 = -126$<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Expression : $x+\\frac{1}{x}=1$<\/p>\n

=> $x^2 + 1 = x$ ——Eqn(1)<\/p>\n

To find : $\\frac{x^2+3x+1}{x^2+7x+1}$<\/p>\n

= $\\frac{(x^2+1) + 3x}{(x^2+1) + 7x}$<\/p>\n

Using eqn(1),we get : <\/span><\/p>\n

= $\\frac{x + 3x}{x + 7x} = \\frac{4}{8}$<\/p>\n

= $\\frac{1}{2}$<\/p>\n

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16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Since $\\sqrt[3]{79507}$ = 43<\/p>\n

=> $\\sqrt[3]{79.507}$ = 4.3<\/p>\n

=> $\\sqrt[3]{0.079507}$ = 0.43<\/p>\n

=> $\\sqrt[3]{0.000079507}$ = 0.043<\/p>\n

=> 4.3+0.43+0.043 = 4.773<\/p>\n

17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let $x = 3k$ and $y = 4k$<\/p>\n

=> $\\frac{2x + 3y}{3y – 2x}$<\/p>\n

= $\\frac{6k + 12k}{12k – 6k}$<\/p>\n

= $\\frac{18}{6}$<\/p>\n

= $\\frac{3}{1}$ = 3 : 1<\/p>\n

18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Using the formula, $(x-y)^3 = x^3 – y^3 -3xy(x-y)$<\/p>\n

=> $(m – 5n)^3 = m^3 – 125n^3 – 15mn(m-5n)$<\/p>\n

=> $2^3 = m^3 – 125n^3 – 15mn*2$<\/p>\n

=> $m^3 – 125n^3 – 30mn = 8$<\/p>\n

19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Expression : $x+\\frac{1}{x}=2$<\/p>\n

Squaring both sides<\/p>\n

=> $x^2 + \\frac{1}{x^2} + 2 = 4$<\/p>\n

=> $x^2 + \\frac{1}{x^2} = 2$<\/p>\n

Cubing both sides<\/p>\n

=> $x^6 + \\frac{1}{x^6} + 3.x.\\frac{1}{x}(x+\\frac{1}{x}) = 8$<\/p>\n

=> $x^6 + \\frac{1}{x^6} = 8-6 = 2$<\/p>\n

Again, squaring both sides, we get :<\/p>\n

=> $x^{12} + \\frac{1}{x^{12}} + 2 = 4$<\/p>\n

=> $x^{12} + \\frac{1}{x^{12}} = 2$<\/p>\n

20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Expression : $5x + 9y = 5$<\/p>\n

Cubing both sides, we get :<\/p>\n

=> $(5x + 9y)^3 = 125$<\/p>\n

=> $125x^3 + 729y^3 + 135xy(5x+9y) = 125$<\/p>\n

=> $125x^3 + 729y^3 + 135xy*5 = 125$<\/p>\n

Since, $125x^{3}$ + $729y^{3} = 120$<\/p>\n

=> $xy = \\frac{5}{5*135} = \\frac{1}{135}$<\/p>\n

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18000+ Questions – Free SSC Study Material<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

SSC CGL Algebra Questions Download Top-20 Algebra questions for SSC CGL exam. Most important Algebra questions based on asked questions in previous exam papers for SSC CGL.   Take a free SSC CGL Tier-1 mock test Download SSC CGL Tier-1 Previous Papers PDF Question 1:\u00a0If a * b = 2a + 3b – ab, then […]<\/p>\n","protected":false},"author":53,"featured_media":43862,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[9,504],"tags":[4093,462],"class_list":{"0":"post-43858","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ssc","8":"category-ssc-cgl","9":"tag-algebra-questions","10":"tag-ssc-cgl"},"better_featured_image":{"id":43862,"alt_text":"Top 20 SSC CGL Algebra questions","caption":"Top-20 SSC CGL Algebra questions ","description":"Top -20 SSC CGL Algebra questions 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