{"id":40374,"date":"2020-02-03T10:46:25","date_gmt":"2020-02-03T05:16:25","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40374"},"modified":"2020-02-03T10:46:25","modified_gmt":"2020-02-03T05:16:25","slug":"maths-questions-for-ssc-chsl-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/maths-questions-for-ssc-chsl-set-2-pdf\/","title":{"rendered":"Maths Questions for SSC CHSL Set-2 PDF"},"content":{"rendered":"

Maths Questions for SSC CHSL Set-2 PDF<\/strong><\/span><\/h2>\n

Download SSC CHSL Maths Questions with answers Set-2\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Questions for SSC Exam<\/p>\n

Download Maths Questions Set-2 PDF<\/a><\/p>\n

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Question 1:\u00a0<\/b>What is the mean of the range, median and mode of the data:
\n$2, 3, 1, 1, 3, 4, 2, 1, 1, 4, 5, 2, 4, 2, 2, 1, 3, 3, 2, 5$<\/p>\n

a)\u00a0$3\\frac{1}{3}$<\/p>\n

b)\u00a0$4\\frac{1}{2}$<\/p>\n

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

d)\u00a0$4$<\/p>\n

Question 2:\u00a0<\/b>The value of $\\left(\\frac{7}{8}\\div5\\frac{1}{4}\\times7\\frac{1}{5}-\\frac{3}{20}of\\frac{2}{3}\\right)\\div\\frac{1}{2}+\\left(\\frac{3}{5}\\times7\\frac{1}{2}+\\frac{2}{3}\\div\\frac{8}{15}\\right)$\u00a0is equal to 7 + k, where k =<\/p>\n

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

b)\u00a0$\\frac{1}{5}$<\/p>\n

c)\u00a0$\\frac{1}{20}$<\/p>\n

d)\u00a0$\\frac{19}{20}$<\/p>\n

Question 3:\u00a0<\/b>The value of $2 \\times 3 \\div 2\u00a0 of\u00a0 3 \\times 2 \\div (4 + 4 \\times 4 \\div 4\u00a0 of\u00a0 4 – 4 \\div 4 \\times 4)$ is:<\/p>\n

a)\u00a08<\/p>\n

b)\u00a01<\/p>\n

c)\u00a04<\/p>\n

d)\u00a02<\/p>\n

Question 4:\u00a0<\/b>The value of
\n$\\frac{9}{15}\u00a0 of\u00a0 \\left(\\frac{2}{3} \\div \\frac{2}{3}\u00a0 of\u00a0 \\frac{3}{2}\\right) \\div \\left(\\frac{3}{4} \\times \\frac{3}{4} \\div \\frac{3}{4}\u00a0 of\u00a0 \\frac{4} {3}\\right)\u00a0 of\u00a0 \\left(\\frac{5}{4} \\div \\frac{5}{2} \\times \\frac{2}{5}\u00a0 of\u00a0 \\frac{4}{5}\\right)$ is:<\/p>\n

a)\u00a0$\\frac{20}{9}$<\/p>\n

b)\u00a0$\\frac{4}{25}$<\/p>\n

c)\u00a0$\\frac{18}{125}$<\/p>\n

d)\u00a0$\\frac{40}{9}$<\/p>\n

Question 5:\u00a0<\/b>The value of\u00a0$\\frac{8}{9}\u00a0 of\u00a0 \\left(5\\frac{1}{4} \\div 2\\frac{1}{3}\u00a0 of\u00a0 4\\right) \\div \\left(8 \\div \\frac{2}{3}\u00a0 of\u00a0 \\frac{4}{5}\\right)\u00a0 of\u00a0 \\left(8 \\times \\frac{2}{3} \\div \\frac{4}{5}\\right)$ is:<\/p>\n

a)\u00a0$\\frac{1}{200}$<\/p>\n

b)\u00a0$1\\frac{1}{8}$<\/p>\n

c)\u00a0$\\frac{4}{15}$<\/p>\n

d)\u00a0$\\frac{1}{100}$<\/p>\n

SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n\n

SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n

Question 6:\u00a0<\/b>The simplified value of $\\frac{\\left(3\\frac{1}{5} + \\frac{3}{5}\\right) \\div \\frac{8}{5}}{1\\frac{1}{7} \\div \\left\\{\\frac{5}{7} + \\left(\\frac{1}{7} \\div \\frac{1}{3}\\right)\\right\\}}$ is:<\/p>\n

a)\u00a0$\\frac{19}{7}$<\/p>\n

b)\u00a0$\\frac{19}{16}$<\/p>\n

c)\u00a0$\\frac{19}{8}$<\/p>\n

d)\u00a0$\\frac{19}{64}$<\/p>\n

Question 7:\u00a0<\/b>The value of $\\frac{3 \\div \\left\\{5 – 5 \\div (6 – 7) \\times 8 + 9\\right\\}}{4 + 4 \\times 4 \\div 4\u00a0 of\u00a0 4}$ is:<\/p>\n

a)\u00a0$\\frac{1}{3}$<\/p>\n

b)\u00a0$\\frac{1}{45}$<\/p>\n

c)\u00a0$\\frac{1}{90}$<\/p>\n

d)\u00a0$\\frac{1}{18}$<\/p>\n

Question 8:\u00a0<\/b>The value of $\\frac{18.43 \\times 18.43 – 6.57 \\times 6.57}{11.86}$ is:<\/p>\n

a)\u00a023.62<\/p>\n

b)\u00a025<\/p>\n

c)\u00a026<\/p>\n

d)\u00a024.12<\/p>\n

Question 9:\u00a0<\/b>The value of $\u00a0\\frac{(0.67 \\times 0.67 \\times0.67)\u00a0 \\times (0.33 \\times 0.33 \\times 0.33)}{(0.67 \\times 0.67) \\div (0.67 \\times 0.33) \\div (0.33 \\times 0.33)} $<\/p>\n

a)\u00a011<\/p>\n

b)\u00a00.34<\/p>\n

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

d)\u00a03.4<\/p>\n

Question 10:\u00a0<\/b>If $\\left(1 + \\frac{1}{2}\\right)\\left(1 + \\frac{1}{4}\\right)\\left(1 + \\frac{1}{6}\\right)\\left(1 + \\frac{1}{8}\\right)\\left(1 – \\frac{1}{3}\\right)\\left(1 – \\frac{1}{5}\\right)\\left(1 – \\frac{1}{7}\\right) = 1 + \\frac{1}{x}$, then what is the value of x?<\/p>\n

a)\u00a06<\/p>\n

b)\u00a08<\/p>\n

c)\u00a05<\/p>\n

d)\u00a07<\/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(C)<\/strong><\/p>\n

Arranging them in the ascending order we have
\n1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,4,4,4,5,5
\nMedian=middle value i.e average of 2 and 2 i. e 2
\nMode=highest frequency number i.e 2
\nRange=highest-least value=5-1=4
\nAverage=(2+2+4)\/3
\n=$2\\frac{2}{3}$<\/p>\n

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

Given\u00a0:\u00a0$\\left(\\frac{7}{8}\\div5\\frac{1}{4}\\times7\\frac{1}{5}-\\frac{3}{20}of\\frac{2}{3}\\right)\\div\\frac{1}{2}+\\left(\\frac{3}{5}\\times7\\frac{1}{2}+\\frac{2}{3}\\div\\frac{8}{15}\\right)=7+k$<\/p>\n

L.H.S. =\u00a0$\\left(\\frac{7}{8}\\div5\\frac{1}{4}\\times7\\frac{1}{5}-\\frac{3}{20}of\\frac{2}{3}\\right)\\div\\frac{1}{2}+\\left(\\frac{3}{5}\\times7\\frac{1}{2}+\\frac{2}{3}\\div\\frac{8}{15}\\right)$<\/p>\n

=\u00a0$\\left(\\frac{7}{8}\\div\\frac{21}{4}\\times\\frac{36}{5}-\\frac{1}{10}\\right)\\div\\frac{1}{2}+\\left(\\frac{3}{5}\\times\\frac{15}{2}+\\frac{2}{3}\\div\\frac{8}{15}\\right)$<\/p>\n

= $\\left(\\frac{7}{8}\\times\\frac{4}{21}\\times\\frac{36}{5}-\\frac{1}{10}\\right)\\div\\frac{1}{2}+\\left(\\frac{3}{5}\\times\\frac{15}{2}+\\frac{2}{3}\\times\\frac{15}{8}\\right)$<\/p>\n

= $[(\\frac{6}{5}-\\frac{1}{10})\\times2]+\\frac{9}{2}+\\frac{5}{4}$<\/p>\n

= $\\frac{11}{5}+\\frac{9}{2}+\\frac{5}{4}$<\/p>\n

= $\\frac{44+90+25}{20}=\\frac{159}{20}=7+\\frac{19}{20}$<\/p>\n

Comparing both sides, we get\u00a0: $k=\\frac{19}{20}$<\/p>\n

=> Ans – (D)<\/p>\n

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

Following BODMAS ,<\/p>\n

$2 \\times 3 \\div 2 of 3 \\times 2 \\div (4 + 4 \\times 4 \\div 4 of 4 – 4 \\div 4 \\times 4) $<\/p>\n

=\u00a0$2 \\times 3 \\div 6 \\times 2 \\div (4 + 4 \\times 4 \\div 16 – 4 \\div 4 \\times 4) $<\/p>\n

= $2 \\times\u00a0 \\frac{1}{2} \\times 2 \\div (4 + 4 \\times \\frac{1}{4} – 4 \\div 4 \\times 4) $<\/p>\n

= $2 \\div (4 + 1 – 1 \\times 4) $<\/p>\n

=\u00a0$2 \\div (5 – 4) $<\/p>\n

=2<\/p>\n

So , the answer would be option d) 2.<\/p>\n

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

$\\frac{9}{15} of \\left(\\frac{2}{3} \\div \\frac{2}{3} of \\frac{3}{2}\\right) \\div \\left(\\frac{3}{4} \\times \\frac{3}{4} \\div \\frac{3}{4} of \\frac{4} {3}\\right) of \\left(\\frac{5}{4} \\div \\frac{5}{2} \\times \\frac{2}{5} of \\frac{4}{5}\\right)$<\/p>\n

We know that “of” can be replace by multiply,<\/p>\n

$\\Rightarrow \\frac{9}{15} of \\left(\\frac{2}{3} \\div 1 \\right) \\div \\left(\\frac{3}{4} \\times \\frac{3}{4} \\div 1 \\right) of \\left(\\frac{5}{4} \\div \\frac{5}{2} \\times \\frac{8}{25} \\right)$<\/p>\n

Now, solving small brakets,<\/p>\n

$\\Rightarrow \\frac{9}{15} of \\left(\\frac{2}{3} \\right) \\div \\left(\\frac{9}{16} \\right) of \\left(\\frac{10}{20} \\times \\frac{8}{25} \\right)$<\/p>\n

$\\Rightarrow \\frac{9}{15} of \\left(\\frac{2}{3} \\right) \\div \\left(\\frac{9}{16} \\right) of \\left(\\frac{4}{25} \\right)$<\/p>\n

$\\Rightarrow \\dfrac{\\frac{18}{45}}{ \\left(\\frac{9}{100} \\right) }$<\/p>\n

$\\Rightarrow \\frac{18\\times 100}{45\\times 9}$<\/p>\n

$\\Rightarrow \\frac{2\\times 20}{9}$<\/p>\n

$\\Rightarrow \\frac{40}{9}$<\/p>\n

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

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

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

Apply BODMAS rule to this type of problems<\/p>\n

According\u00a0to Bodmas<\/b>\u00a0rule, if an expression contains brackets ((), {}, [])\u00a0we<\/b>\u00a0have\u00a0to<\/b>\u00a0first solve or simplify the bracket followed by of (powers and roots etc.), then division, multiplication, addition and subtraction from left\u00a0to<\/b>\u00a0right.<\/p>\n

BODMAS full form<\/p>\n

B=brackets<\/p>\n

O=orders<\/p>\n

D=division<\/p>\n

M=multiplication<\/p>\n

A=Addition<\/p>\n

S=Subtraction<\/p>\n

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

$\\frac{18.43 \\times 18.43 – 6.57 \\times 6.57}{11.86}$<\/p>\n

$\\frac{18.43^{2}-6.57^{2}}{11.86^{2}}$<\/p>\n

$\\frac{(18.43+6.57)(18.43 – 6.57)}{11.86}$<\/p>\n

$\\frac{25\\times11.86}{11.86}=25$<\/p>\n

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

Mathematical Operators provided in the question seems to be incorrect. Please review the question again and provide correct data.<\/p>\n

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

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

SSC CHSL Important Q&A PDF<\/a><\/p>\n

We hope this Maths Questions for SSC CHSL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Maths Questions for SSC CHSL Set-2 PDF Download SSC CHSL Maths Questions with answers Set-2\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Questions for SSC Exam Take a free mock test for SSC CHSL Download SSC CHSL Previous Papers More SSC CHSL Important Questions and Answers PDF Question […]<\/p>\n","protected":false},"author":42,"featured_media":40378,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1459,1611,1741,1268,1441],"tags":[3595,358],"class_list":{"0":"post-40374","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-gd","13":"category-ssc-je","14":"category-ssc-mts","15":"category-ssc-stenographer","16":"category-stenographer","17":"tag-maths-questions-for-ssc-chsl","18":"tag-ssc-chsl"},"better_featured_image":{"id":40378,"alt_text":"Maths Questions for 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