{"id":39929,"date":"2020-01-06T16:10:45","date_gmt":"2020-01-06T10:40:45","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=39929"},"modified":"2020-01-06T16:10:45","modified_gmt":"2020-01-06T10:40:45","slug":"surds-and-indices-questions-for-ssc-chsl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/surds-and-indices-questions-for-ssc-chsl-pdf\/","title":{"rendered":"Surds and Indices Questions for SSC CHSL PDF"},"content":{"rendered":"

Surds and Indices Questions for SSC CHSL PDF<\/strong><\/span><\/h1>\n

For Previous year Surds and Indices Questions of SSC CHSL 2018-2019 Tier I exam download PDF. Go through the video of Repeatedly asked Surds and Indices questions explanations most important for the CHSL exam.<\/p>\n

Download Surds and Indices Questions for SSC CHSL PDF<\/a><\/p>\n

Get 200 SSC mocks for just Rs. 249. Enroll here<\/a><\/p>\n

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More SSC CHSL Important Questions and Answers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>Arranging the following in descending order, we get
\n$\\sqrt[3]{4},\\sqrt{2},\\sqrt[6]{3},\\sqrt[4]{5}$<\/p>\n

a)\u00a0$\\sqrt[3]{4}>\\sqrt[4]{5}>\\sqrt{2}>\\sqrt[6]{3}$<\/p>\n

b)\u00a0$\\sqrt[4]{5}<\\sqrt[3]{4}>\\sqrt[6]{3}>\\sqrt{2}$<\/p>\n

c)\u00a0$\\sqrt{2}>\\sqrt[6]{3}>\\sqrt[3]{4}>\\sqrt[4]{5}$<\/p>\n

d)\u00a0$\\sqrt[6]{3}>\\sqrt[4]{5}>\\sqrt[3]{4}>\\sqrt{2}$<\/p>\n

Question 2:\u00a0<\/b>Number of digits in the square root of 62478078 is:<\/p>\n

a)\u00a04<\/p>\n

b)\u00a05<\/p>\n

c)\u00a06<\/p>\n

d)\u00a03<\/p>\n

Question 3:\u00a0<\/b>The approx value of $5\\frac{1}{3}+1\\frac{2}{9}\\times \\frac{1}{4}(10+\\frac{3}{1-\\frac{1}{5}})$ is<\/p>\n

a)\u00a0$10$<\/p>\n

b)\u00a0$\\frac{67}{25}$<\/p>\n

c)\u00a0$\\frac{128}{11}$<\/p>\n

d)\u00a0$\\frac{128}{99}$<\/p>\n

Question 4:\u00a0<\/b>Which is the largest of the following fractions ?
\n$\\frac{2}{5},\\frac{3}{5},\\frac{8}{11},\\frac{11}{17}$<\/p>\n

a)\u00a0$\\frac{8}{11}$<\/p>\n

b)\u00a0$\\frac{3}{5}$<\/p>\n

c)\u00a0$\\frac{11}{17}$<\/p>\n

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

Question 5:\u00a0<\/b>The simplified value of $999\\frac{1}{7}+999\\frac{2}{7}+999\\frac{3}{7}+999\\frac{4}{7}+999\\frac{5}{7}+999\\frac{6}{7}$ is<\/p>\n

a)\u00a0$10009\\frac{2}{7}$<\/p>\n

b)\u00a0$5994\\frac{6}{7}$<\/p>\n

c)\u00a0$9999\\frac{2}{7}$<\/p>\n

d)\u00a05997<\/p>\n

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Question 6:\u00a0<\/b>If $x=6+2\\sqrt{6}$, then what is the value of $\\sqrt{x-1}+\\frac{1}{\\sqrt{x-1}}$ ?<\/p>\n

a)\u00a0$2\\sqrt{3}$<\/p>\n

b)\u00a0$3\\sqrt{2}$<\/p>\n

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

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

Question 7:\u00a0<\/b>If $x=\\frac{2\\sqrt{15}}{\\sqrt{3}+\\sqrt{5}}$, then what is the value of $\\frac{x+\\sqrt{5}}{x-\\sqrt{5}}+\\frac{x+\\sqrt{3}}{x-\\sqrt{3}}$<\/p>\n

a)\u00a0$\\sqrt{5}$<\/p>\n

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

c)\u00a0$\\sqrt{15}$<\/p>\n

d)\u00a0$2$<\/p>\n

Question 8:\u00a0<\/b>Which value among $\\sqrt[3]{5},\\sqrt[4]{6},\\sqrt[6]{12},\\sqrt[12]{276}$ is the largest ?<\/p>\n

a)\u00a0$\\sqrt[3]{5}$<\/p>\n

b)\u00a0$\\sqrt[4]{6}$<\/p>\n

c)\u00a0$\\sqrt[6]{12}$<\/p>\n

d)\u00a0$\\sqrt[2]{12}$<\/p>\n

Question 9:\u00a0<\/b>If $x-y-\\sqrt{18}=-1$ and $x + y – 3\\sqrt{2} = 1$, then what is the value of $12xy(x^{2} – y^{2})$ ?<\/p>\n

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

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

c)\u00a0$512\\sqrt{2}$<\/p>\n

d)\u00a0$612\\sqrt{2}$<\/p>\n

Question 10:\u00a0<\/b>If $\\frac{p}{q}=\\frac{r}{s}=\\frac{t}{u}=\\sqrt{5}$, then what is the value of $[\\frac{(3p^{2} + 4r^{2} + 5t^{2})}{(3q^{2} + 4s^{2} + 5u^{2})}]$ \u00a0?<\/p>\n

a)\u00a01\/5<\/p>\n

b)\u00a05<\/p>\n

c)\u00a025<\/p>\n

d)\u00a060<\/p>\n

Question 11:\u00a0<\/b>If $x+[\\frac{1}{(x+7)}]=0$, then what is the value of $x-[\\frac{1}{(x+7)}]$ ?<\/p>\n

a)\u00a0$3\\sqrt{5}$<\/p>\n

b)\u00a0$3\\sqrt{5}-7$<\/p>\n

c)\u00a0$3\\sqrt{5}+7$<\/p>\n

d)\u00a0$8$<\/p>\n

Question 12:\u00a0<\/b>If $\\frac{x+\\sqrt{x^2-1}}{x-\\sqrt{x^2-1}}+\\frac{x-\\sqrt{x^2-1}}{x+\\sqrt{x^2-1}}=194$, then what is the value of x?<\/p>\n

a)\u00a07\/2<\/p>\n

b)\u00a04<\/p>\n

c)\u00a07<\/p>\n

d)\u00a014<\/p>\n

Question 13:\u00a0<\/b>If $x=8+2\\sqrt{15}$, then what is the value of $\\sqrt{x}+\\frac{1}{\\sqrt{x}}$ ?<\/p>\n

a)\u00a0$2\\sqrt{5}$<\/p>\n

b)\u00a0$2\\sqrt{3}$<\/p>\n

c)\u00a0$\\frac{(3\\sqrt{5}+\\sqrt{3})}{2}$<\/p>\n

d)\u00a0$\\frac{(3\\sqrt{3}-\\sqrt{5})}{2}$<\/p>\n

Question 14:\u00a0<\/b>What is the value of $\\frac{1+a}{a^{\\frac{1}{2}}+a^{\\frac{-1}{2}}}-\\frac{a^{\\frac{1}{2}}+a^{\\frac{-1}{2}}}{1+a}+a^{\\frac{-1}{2}}$ ?<\/p>\n

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

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

c)\u00a0$\\sqrt{a}+1$<\/p>\n

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

Question 15:\u00a0<\/b>If $\\frac{\\sqrt{5+x}+\\sqrt{5-x}}{\\sqrt{5+x}-\\sqrt{5-x}}=3$, then what is the value of x?<\/p>\n

a)\u00a05\/2<\/p>\n

b)\u00a025\/3<\/p>\n

c)\u00a04<\/p>\n

d)\u00a03<\/p>\n

SSC CHSL FREE MOCK TEST<\/a><\/p>\n

FREE SSC STUDY MATERIAL<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

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

Expression : $\\sqrt[3]{4},\\sqrt{2},\\sqrt[6]{3},\\sqrt[4]{5}$<\/p>\n

= $4^{\\frac{1}{3}} , 2^{\\frac{1}{2}} , 3^{\\frac{1}{6}} , 5^{\\frac{1}{4}}$<\/p>\n

Now, L.C.M. of the powers i.e. 3,2,4,6 = 12<\/p>\n

Multiplying the powers by 12 in each of the numbers, we get :<\/p>\n

= $4^4 , 2^6 , 3^2 , 5^3$<\/p>\n

= $256 , 64 , 9 , 125$<\/p>\n

Now arranging them in descending order,<\/p>\n

=> $256 > 125 > 64 > 9$<\/p>\n

$\\equiv$ $\\sqrt[3]{4} > \\sqrt[4]{5} > \\sqrt{2} > \\sqrt[6]{3}$<\/p>\n

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

When the number of digits in a no. is 7 or 8 , then no. of digits in square root will be 4.<\/p>\n

62478078 has 8 digits, => Its square root has 4 digits.<\/p>\n

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

$5\\frac{1}{3}+1\\frac{2}{9}\\times \\frac{1}{4}(10+\\frac{3}{1-\\frac{1}{5}})$<\/p>\n

$5\\frac{1}{3}+1\\frac{2}{9}\\times \\frac{1}{4}(10+\\frac{15}{4})$
\n<\/span><\/p>\n

$\\frac{16}{3}+\\frac{11}{9}\\times \\frac{55}{16}$ = 9.9 ~ 10
\n<\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

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

Fractions : $\\frac{2}{5},\\frac{3}{5},\\frac{8}{11},\\frac{11}{17}$<\/p>\n

L.C.M. of 5,11,17 = 935<\/p>\n

Now, multiplying each fraction by 935, we get :<\/p>\n

=> 374 , 561 , 680 , 605<\/p>\n

Since, among these numbers, 680 is the largest $\\equiv \\frac{8}{11}$<\/p>\n

=> $\\frac{8}{11}$ is the largest.<\/p>\n

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

we need to find value of $999\\frac{1}{7}+999\\frac{2}{7}+999\\frac{3}{7}+999\\frac{4}{7}+999\\frac{5}{7}+999\\frac{6}{7}$<\/p>\n

$6000 – \\frac{1+2+3+4+5+6}{7}$<\/p>\n

= 6000 – $\\frac{21}{7}$<\/p>\n

= 6000 – 3 = 5997<\/p>\n

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

We need to calculate $\\sqrt{x-1}+\\frac{1}{\\sqrt{x-1}}$
\nThis equals $\\frac{x-1 + 1}{\\sqrt{x-1}} = \\frac{x}{\\sqrt{x-1}}$
\n$x-1 = 5+2\\sqrt{6} = (\\sqrt{3} + \\sqrt{2})^2$
\nTherefore, $\\sqrt{x-1} = \\sqrt{3} + \\sqrt{2}$<\/p>\n

Hence, the required expression becomes $\\frac{6+2\\sqrt{6}}{\\sqrt{2}+\\sqrt{3}}$
\nThis equals $2*\\frac{3+\\sqrt{6}}{\\sqrt{2}+\\sqrt{3}}=2\\sqrt{3}$<\/p>\n

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

$x=\\frac{2\\sqrt{3}\\sqrt{5}}{\\sqrt{3}+\\sqrt{5}}$<\/p>\n

$\\frac{x}{\\sqrt{5}}=\\frac{2\\sqrt{3}}{\\sqrt{3}+\\sqrt{5}}$<\/p>\n

By C-D rule,<\/p>\n

$\\frac{x+\\sqrt{5}}{x-\\sqrt{5}}=\\frac{3\\sqrt{3}+\\sqrt{5}}{\\sqrt{3}-\\sqrt{5}}$——-(1)<\/p>\n

and<\/p>\n

$\\frac{x}{\\sqrt{3}}=\\frac{2\\sqrt{5}}{\\sqrt{3}+\\sqrt{5}}$<\/p>\n

$\\frac{x+\\sqrt{3}}{x-\\sqrt{3}}=\\frac{3\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}}$——-(2)<\/p>\n

add (1) & (2)<\/p>\n

$\\frac{x+\\sqrt{5}}{x-\\sqrt{5}}+\\frac{x+\\sqrt{3}}{x-\\sqrt{3}}$<\/p>\n

= $\\frac{3\\sqrt{3}+\\sqrt{5}}{\\sqrt{3}-\\sqrt{5}}+\\frac{3\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}}$<\/p>\n

=\u00a0 $\\frac{2\\sqrt{3}-2\\sqrt{5}}{\\sqrt{3}-\\sqrt{5}}$<\/p>\n

= $2$<\/p>\n

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

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

Values :\u00a0$\\sqrt[3]{5},\\sqrt[4]{6},\\sqrt[6]{12},\\sqrt[12]{276}$<\/p>\n

Taking L.C.M. of exponents, => L.C.M.(3,4,6,12) = 12<\/p>\n

Now, multiplying all the exponents by 12, we get\u00a0:<\/p>\n

Values\u00a0: $(5)^4,(6)^3,(12)^2,(276)^1$<\/p>\n

=\u00a0$625,216,144,276$<\/p>\n

Thus, $625\\equiv \\sqrt[3]{5}$ is the largest.<\/p>\n

=> Ans – (A)<\/p>\n

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

Given\u00a0:\u00a0$x-y-\\sqrt{18}=-1$<\/p>\n

=> $x-y=\\sqrt{18}-1$ ————-(i)<\/p>\n

Squaring both sides,<\/p>\n

=> $(x-y)^2=(\\sqrt{18}-1)^2$<\/p>\n

=> $x^2+y^2-2xy=18+1-2\\sqrt{18}$<\/p>\n

=> $x^2+y^2-2xy=19-2\\sqrt{18}$ ————–(ii)<\/p>\n

Also,\u00a0$x + y – 3\\sqrt{2} = 1$<\/p>\n

=> $x+y=\\sqrt{18}+1$ ————-(iii)<\/p>\n

Squaring both sides,<\/p>\n

=> $(x+y)^2=(\\sqrt{18}+1)^2$<\/p>\n

=> $x^2+y^2+2xy=18+1+2\\sqrt{18}$<\/p>\n

=> $x^2+y^2+2xy=19+2\\sqrt{18}$ ————–(iv)<\/p>\n

Subtracting equation (ii) from (iv),<\/p>\n

=> $4xy=4\\sqrt{18}$<\/p>\n

=> $12xy=12\\sqrt{18}$ ————(v)<\/p>\n

Multiplying equations (i) and (iii),<\/p>\n

=> $(x-y)(x+y)=(\\sqrt{18}-1)(\\sqrt{18}+1)$<\/p>\n

=> $x^2-y^2=18-1=17$ ———–(vi)<\/p>\n

Now, multiplying equations (v) and (vi), we get\u00a0:<\/p>\n

=>\u00a0$12xy(x^{2} – y^{2})= (12\\sqrt{18})\\times17$<\/p>\n

=\u00a0$204\\sqrt{18}=612\\sqrt2$<\/p>\n

=> Ans – (D)<\/p>\n

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

Given\u00a0:\u00a0$\\frac{p}{q}=\\frac{r}{s}=\\frac{t}{u}=\\sqrt{5}$<\/p>\n

=> $p=\\sqrt5q$ ,\u00a0$r=\\sqrt5s$ ,\u00a0$t=\\sqrt5u$<\/p>\n

To find\u00a0:\u00a0$[\\frac{(3p^{2} + 4r^{2} + 5t^{2})}{(3q^{2} + 4s^{2} + 5u^{2})}]$<\/p>\n

= $\\frac{3(\\sqrt5q)^2+4(\\sqrt5s)^2+5(\\sqrt5u)^2}{3q^2+4s^2+5u^2}$<\/p>\n

=\u00a0$\\frac{15q^2+20s^2+25u^2}{3q^2+4s^2+5u^2}$<\/p>\n

=\u00a0$\\frac{5(3q^2+4s^2+5u^2)}{3q^2+4s^2+5u^2}=5$<\/p>\n

=> Ans – (B)<\/p>\n

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

Given\u00a0:\u00a0$x+[\\frac{1}{(x+7)}]=0$ ———–(i)<\/p>\n

=> $\\frac{x^2+7x+1}{x+7}=0$<\/p>\n

=> $x^2+7x+1=0$<\/p>\n

=> $x=\\frac{-7\\pm\\sqrt{49-4}}{2}$<\/p>\n

=> $x=\\frac{3\\sqrt{5}-7}{2}$ ———–(ii)<\/p>\n

From equation (i), => $\\frac{1}{(x+7)}=-x$ —————(iii)<\/p>\n

To find\u00a0:\u00a0$x-[\\frac{1}{(x+7)}]$<\/p>\n

Substituting values from equations (ii) and (iii), we get\u00a0:<\/p>\n

= $x-(-x)=2x$<\/p>\n

= $2\\times\\frac{3\\sqrt5-7}{2}$<\/p>\n

= $3\\sqrt5-7$<\/p>\n

=> Ans – (B)<\/p>\n

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

Given\u00a0:\u00a0$\\frac{x+\\sqrt{x^2-1}}{x-\\sqrt{x^2-1}}+\\frac{x-\\sqrt{x^2-1}}{x+\\sqrt{x^2-1}}=194$<\/p>\n

=> $\\frac{(x+\\sqrt{x^2-1})^2+(x-\\sqrt{x^2-1})^2}{(x-\\sqrt{x^2-1})(x+\\sqrt{x^2-1})}=194$<\/p>\n

=> $\\frac{(x^2+x^2-1+2x\\sqrt{x^2-1})+(x^2+x^2-1-2x\\sqrt{x^2-1})}{x^2-(x^2-1)}=194$<\/p>\n

=> $\\frac{4x^2-2}{1}=194$<\/p>\n

=> $4x^2=194+2=196$<\/p>\n

=> $x^2=\\frac{196}{4}=49$<\/p>\n

=> $x=\\sqrt{49}=7$<\/p>\n

=> Ans – (C)<\/p>\n

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

Given\u00a0:\u00a0$x=8+2\\sqrt{15}$<\/p>\n

=> $x=5+3+2\\sqrt{(5)(3)}$<\/p>\n

=> $x=(\\sqrt5)^2+(\\sqrt3)^2+2(\\sqrt5)(\\sqrt3)$<\/p>\n

=> $x=(\\sqrt5+\\sqrt3)^2$<\/p>\n

=> $\\sqrt{x}=\\sqrt5+\\sqrt3$ ————(i)<\/p>\n

Now,\u00a0$\\frac{1}{\\sqrt{x}}=\\frac{1}{\\sqrt5+\\sqrt3}$<\/p>\n

=> $\\frac{1}{\\sqrt{x}}=\\frac{1}{\\sqrt5+\\sqrt3}\\times\\frac{(\\sqrt5-\\sqrt3)}{(\\sqrt5-\\sqrt3)}$<\/p>\n

=> $\\frac{1}{\\sqrt{x}}=\\frac{\\sqrt5-\\sqrt3}{5-3}=\\frac{\\sqrt5-\\sqrt3}{2}$ ———–(ii)<\/p>\n

Adding equations (i) and (ii), we get\u00a0:<\/p>\n

=> $\\sqrt{x}+\\frac{1}{\\sqrt{x}}=(\\sqrt5+\\sqrt3)+(\\frac{\\sqrt5-\\sqrt3}{2})$<\/p>\n

= $\\frac{2\\sqrt5+2\\sqrt3+\\sqrt5-\\sqrt3}{2}=\\frac{3\\sqrt5+\\sqrt3}{2}$<\/p>\n

=> Ans – (C)<\/p>\n

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

Expression\u00a0:\u00a0$\\frac{1+a}{a^{\\frac{1}{2}}+a^{\\frac{-1}{2}}}-\\frac{a^{\\frac{1}{2}}+a^{\\frac{-1}{2}}}{1+a}+a^{\\frac{-1}{2}}$<\/p>\n

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

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

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

=> Ans – (A)<\/p>\n

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

Expression\u00a0:\u00a0$\\frac{\\sqrt{5+x}+\\sqrt{5-x}}{\\sqrt{5+x}-\\sqrt{5-x}}=3$<\/p>\n

Rationalizing the denominator,<\/p>\n

=>\u00a0$\\frac{\\sqrt{5+x}+\\sqrt{5-x}}{\\sqrt{5+x}-\\sqrt{5-x}}\\times \\frac{\\sqrt{5+x}+\\sqrt{5-x}}{\\sqrt{5+x}+\\sqrt{5-x}}=3$<\/p>\n

=> $\\frac{[(\\sqrt{5+x})+(\\sqrt{5-x})]^2}{(\\sqrt{5+x})^2-(\\sqrt{5-x})^2}=3$<\/p>\n

=> $\\frac{(5+x)+(5-x)+2(\\sqrt{5+x})(\\sqrt{5-x})}{(5+x)-(5-x)}=3$<\/p>\n

=> $\\frac{10+2\\sqrt{25-x^2}}{2x}=3$<\/p>\n

=> $5+\\sqrt{25-x^2}=3x$<\/p>\n

=> $3x-5=\\sqrt{25-x^2}$<\/p>\n

Squaring both sides, we get\u00a0:<\/p>\n

=> $(3x-5)^2=(\\sqrt{25-x^2})^2$<\/p>\n

=> $9x^2+25-30x=25-x^2$<\/p>\n

=> $10x^2-30x=0$<\/p>\n

=> $10x(x-3)=0$<\/p>\n

=> $x=0,3$ \u00a0 \u00a0[But $x$ can’t be zero because the denominator can’t be zero]<\/p>\n

=> Ans – (D)<\/p>\n

DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n

SSC CHSL Important Q&A PDF<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Surds and Indices Questions for SSC CHSL PDF For Previous year Surds and Indices Questions of SSC CHSL 2018-2019 Tier I exam download PDF. Go through the video of Repeatedly asked Surds and Indices questions explanations most important for the CHSL exam. Take a free mock test for SSC CHSL Download SSC CHSL Previous Papers […]<\/p>\n","protected":false},"author":49,"featured_media":39935,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1459,1611,1741,1441],"tags":[358,3447],"class_list":{"0":"post-39929","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-stenographer","17":"tag-ssc-chsl","18":"tag-surds-and-indices-questions"},"better_featured_image":{"id":39935,"alt_text":"Surds and Indices Questions for SSC CHSL PDF","caption":"Surds and Indices Questions for SSC CHSL PDF","description":"Surds and 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