SSC CHSL Maths Previous Year Questions download PDF based on previous year question paper of SSC exams. 20 Very important Maths questions for SSC CHSL Exam.<\/p>\n
Download SSC CHSL Previous Year Maths 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>What is the value of $462.767-241\\div7\\times3+134.568 $<\/p>\n a)\u00a0497.345<\/p>\n b)\u00a0494.050<\/p>\n c)\u00a0498.389<\/p>\n d)\u00a0499.987<\/p>\n Question 2:\u00a0<\/b>What is the value of $1225\\div5+400\\times3.2-312 $<\/p>\n a)\u00a01213<\/p>\n b)\u00a01211<\/p>\n c)\u00a01209<\/p>\n d)\u00a01208<\/p>\n Question 3:\u00a0<\/b>$77 + 78 + 79 + \u2026\u2026\u2026 +113$ is equal to<\/p>\n a)\u00a03215<\/p>\n b)\u00a03615<\/p>\n c)\u00a03415<\/p>\n d)\u00a03515<\/p>\n Question 4:\u00a0<\/b>Find the 13th term, if the 3rd term of an arithmetic progression is 1, 9th term is 25?<\/p>\n a)\u00a049<\/p>\n b)\u00a037<\/p>\n c)\u00a045<\/p>\n d)\u00a041<\/p>\n Question 5:\u00a0<\/b>Reduce fraction $\\large\\frac{13167}{19019}$ into lowest terms.<\/p>\n a)\u00a0$\\large\\frac{7}{13}$<\/p>\n b)\u00a0$\\large\\frac{9}{13}$<\/p>\n c)\u00a0$\\large\\frac{19}{13}$<\/p>\n d)\u00a0$\\large\\frac{11}{13}$<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>45 is 1.8% of?<\/p>\n a)\u00a02500<\/p>\n b)\u00a02900<\/p>\n c)\u00a02800<\/p>\n d)\u00a02400<\/p>\n Question 7:\u00a0<\/b>Reduce fraction $\\large\\frac{2275}{2457}$ into lowest terms.<\/p>\n a)\u00a0$\\large\\frac{25}{29}$<\/p>\n b)\u00a0$\\large\\frac{21}{27}$<\/p>\n c)\u00a0$\\large\\frac{24}{27}$<\/p>\n d)\u00a0$\\large\\frac{25}{27}$<\/p>\n Question 8:\u00a0<\/b>Find the fourth term of the sequence for which $t_{1}=2$, $t_{2}=3$ and $t_{n+2}$ = $t_{n}+2t_{n+1}$, is<\/p>\n a)\u00a017<\/p>\n b)\u00a019<\/p>\n c)\u00a012<\/p>\n d)\u00a013<\/p>\n Question 9:\u00a0<\/b>If $\\ p^{2}+16q^{2}\\ = 8pq$, then $p: q$ is<\/p>\n a)\u00a04: 1<\/p>\n b)\u00a04: 3<\/p>\n c)\u00a04: 5<\/p>\n d)\u00a02: 3<\/p>\n Question 10:\u00a0<\/b>$7\\Large\\frac{1}{3}$ $+$ $5\\Large\\frac{1}{4}$ $-$ $8\\Large\\frac{3}{5}$ $+$ $15\\Large\\frac{2}{7}$ $=$?<\/p>\n a)\u00a0$19\\large\\frac{321}{420}$<\/p>\n b)\u00a0$19\\large\\frac{231}{420}$<\/p>\n c)\u00a0$19\\large\\frac{113}{420}$<\/p>\n d)\u00a0$17\\large\\frac{221}{420}$<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 11:\u00a0<\/b>$7\\Large\\frac{1}{5}$ $+$ $2\\Large\\frac{1}{3}$ $-$ $4\\Large\\frac{2}{15}$ $-$ $3\\Large\\frac{5}{18}$ $+$ $5\\Large\\frac{2}{9}$ $=$?<\/p>\n a)\u00a0$5\\large\\frac{35}{78}$<\/p>\n b)\u00a0$7\\large\\frac{31}{90}$<\/p>\n c)\u00a0$9\\large\\frac{36}{73}$<\/p>\n d)\u00a0$4\\large\\frac{37}{79}$<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>$\\Large\\frac{628}{60}$ $-$ $\\Large\\frac{44}{45}\\div\\frac{6}{9}$ $+$ $\\Large\\frac{575}{624}\\div\\frac{25}{104}$ $+$ $\\Large\\frac{13}{15}\\div\\frac{2}{5}$ $=$ $\\large?$<\/p>\n a)\u00a015<\/p>\n b)\u00a025<\/p>\n c)\u00a010<\/p>\n d)\u00a014<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>$15.921$ $+$ $9.679$ $+$ $101.36$ $-$ $94.76$ $=$ $17.976$ $+$ $6.224$ $+$ ?<\/p>\n a)\u00a07<\/p>\n b)\u00a05<\/p>\n c)\u00a08<\/p>\n d)\u00a09<\/p>\n e)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>$\\sqrt{1936}\\div\\sqrt[3]{1331}\\times\\sqrt{50625}\\div\\sqrt{14400}\\times\\sqrt{64}$ $=$ $7920\\div132$ $+$ ?<\/p>\n a)\u00a01<\/p>\n b)\u00a0-3<\/p>\n c)\u00a00<\/p>\n d)\u00a0-1<\/p>\n e)\u00a0None of these<\/p>\n Question 15:\u00a0<\/b>$522\\div150\\times25$ $+$ $96\\div180\\times15$ $-$ $\\Large\\frac{1125}{135}$ $\\div$ $\\Large\\frac{5}{9}$ $=$ ?<\/p>\n a)\u00a080<\/p>\n b)\u00a075<\/p>\n c)\u00a0105<\/p>\n d)\u00a065<\/p>\n e)\u00a0None of these<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 16:\u00a0<\/b>What would come in place of ($) mark in the following equation ? a)\u00a06<\/p>\n b)\u00a02<\/p>\n c)\u00a04<\/p>\n d)\u00a00<\/p>\n Question 17:\u00a0<\/b>If $\\ x^{2}+9y^{2}\\ $= 6xy, then x: y is<\/p>\n a)\u00a01 : 3<\/p>\n b)\u00a03 : 2<\/p>\n c)\u00a03 : 1<\/p>\n d)\u00a02 : 3<\/p>\n Question 18:\u00a0<\/b>Which value among \u00a0$3^{200},\\ 2^{300}\\ and\\ 7^{100}$ is the largest?<\/p>\n a)\u00a0$3^{200}$<\/p>\n b)\u00a0$2^{300}$<\/p>\n c)\u00a0$7^{100}$<\/p>\n d)\u00a0All are equal<\/p>\n Question 19:\u00a0<\/b>In a class, there are 40 students. Some of them passed the examination and others failed. Raman\u2019s rank among the student who have passed is 13 th from top and 17 th from bottom. How many students have failed?<\/p>\n a)\u00a011<\/p>\n b)\u00a010<\/p>\n c)\u00a09<\/p>\n d)\u00a0Cannot be determined<\/p>\n Question 20:\u00a0<\/b>By interchanging which two signs the equation will be correct? a)\u00a0+ and \u00f7<\/p>\n b)\u00a0+ and –<\/p>\n c)\u00a0\u00f7 and –<\/p>\n d)\u00a0None of these<\/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(B)<\/strong><\/p>\n by applying BODMAS we have 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n by applying BODMAS we have 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression\u00a0:\u00a0$77 + 78 + 79 + \u2026\u2026\u2026 +113$<\/p>\n This is an arithmetic progression with first term, $a = 77$ , last term, $l = 113$ and common difference, $d = 1$<\/p>\n Let number of terms = $n$<\/p>\n Last term in an A.P. = $a + (n – 1)d = 113$<\/p>\n => $77 + (n – 1)(1) = 113$<\/p>\n => $n – 1 = 113 – 77 = 36$<\/p>\n => $n = 36 + 1 = 37$<\/p>\n $\\therefore$ Sum of A.P. = $\\frac{37}{2} (a + l)$<\/p>\n = $\\frac{37}{2} (77 + 113)$<\/p>\n = $95 \\times 37 = 3515$<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.<\/p>\n 3rd term, $A_3 = a + (3 – 1) d = 1$<\/p>\n => $a + 2d = 1$ —————–(i)<\/p>\n Similarly, 9th term, $A_{9} = a + 8d = 25$ ——————(ii)<\/p>\n Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n => $(8d – 2d) = 25 – 1$<\/p>\n => $d = \\frac{24}{6} = 4$<\/p>\n Substituting it in equation (i), => $a + 8 = 1$<\/p>\n => $a = 1 – 8 = -7$<\/p>\n $\\therefore$ 13th term, $A_{13} = a + (13 – 1)d$<\/p>\n = $-7 + (12 \\times 4) = -7+48\u00a0 = 41$<\/p>\n => Ans – (D)<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given, Expression\u00a0: $\\frac{13167}{19019}$<\/p>\n We need to check whether any of the prime factors from 2, 3, 5, 7, 11, 13, 17, 19…… divides both numerator & denominator.<\/p>\n We observe that they can be divided by 7, 11 & 19.<\/p>\n Dividing both numerator and denominator by 7, we get\u00a0= $\\frac{1881}{2717}$<\/p>\n Similarly, dividing by 11, we get\u00a0:<\/p>\n = $\\large\\frac{171}{247}$<\/p>\n Similarly, dividing by 19, we get\u00a0:<\/p>\n = $\\large\\frac{9}{13}$<\/p>\n => Ans – (B)<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number be $n$<\/p>\n According to ques, 1.8% of $n$ = 45<\/p>\n => $\\frac{1.8}{100} \\times n = 45$<\/p>\n => $\\frac{1.8n}{100} = 45$<\/p>\n => $n = \\large\\frac{45 \\times 100}{1.8}$ $ = 2500$<\/p>\n => Ans – (A)<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given Expression\u00a0: $\\large\\frac{2275}{2457}$<\/p>\n We need to check whether any of the prime factors from 2, 3, 5, 7, 11, 13, 17, 19…… divides both numerator & denominator.<\/p>\n We observe that they can be divided by 7, 13.<\/p>\n Dividing both numerator and denominator by 13, we get\u00a0= $\\frac{175}{189}$<\/p>\n Similarly, dividing by 7, we get\u00a0:<\/p>\n = $\\frac{25}{27}$<\/p>\n => Ans – (D)<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given,\u00a0 $t_{1}=2$, $t_{2}=3$<\/p>\n $t_{n+2}$ = $t_{n}+2t_{n+1}$<\/p>\n putting n=2, we get the 4th term. therefore,\u00a0 $t_{4}$ = $t_{2}+2t_{3}$<\/p>\n So,we need to find,\u00a0$t_{3}$<\/p>\n $t_{3}$ = $t_{1}+2t_{2}$ = 2+2(3) = 8<\/p>\n $t_{4}$ = $t_{2}+2t_{3}$ = 3+2(8) = 19<\/p>\n so the answer is option B.<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given, $\\ p^{2}+16q^{2}\\ = 8pq$<\/p>\n $p^{2}+16q^{2}=6pq$<\/p>\n $\\Rightarrow p^{2}-8pq+16q^{2}=0$<\/p>\n $\\Rightarrow (p-4q)^{2}=0$<\/p>\n $\\Rightarrow p-4q=0$<\/p>\n $\\Rightarrow p=4q$<\/p>\n $\\Rightarrow \\frac{p}{q}$=$\\frac{4}{1}$<\/p>\n $\\therefore p: q=4: 1$<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, $7\\Large\\frac{1}{3}$ $+$ $5\\Large\\frac{1}{4}$ $-$ $8\\Large\\frac{3}{5}$ $+$ $15\\Large\\frac{2}{7}$ $=$?<\/p>\n = 7+5-8+15+$\\large\\frac{1}{3}+\\frac{1}{4}-\\frac{3}{5}+\\frac{2}{7}$<\/p>\n =19+ $\\large\\frac{140+105-252+120}{420}$<\/p>\n = 19$\\large\\frac{113}{420}$<\/p>\n GK Questions And Answers PDF<\/a><\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 7$\\frac{1}{5}$+2$\\frac{1}{3}$-4$\\frac{2}{15}$-3$\\frac{5}{18}$+5$\\frac{2}{9}$<\/p>\n = 7+2-4-3+5+$\\frac{1}{5}$+$\\frac{1}{3}$-$\\frac{2}{15}$-$\\frac{5}{18}$+$\\frac{2}{9}$<\/p>\n =7+ $\\frac{18+30-12-25+20}{90}$<\/p>\n = 7$\\frac{31}{90}$<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\Large\\frac{628}{60}$ $-$ $\\Large\\frac{44}{45}\\div\\frac{6}{9}$ $+$ $\\Large\\frac{575}{624}\\div\\frac{25}{104}$ $+$ $\\Large\\frac{13}{15}\\div\\frac{2}{5}$<\/p>\n $=$ $\\Large\\frac{314}{30}$ $-$ $\\Large\\frac{44}{30}$ $+$ $\\Large\\frac{115}{30}$ $+$ $\\Large\\frac{65}{30}$<\/p>\n $=$ $\\Large\\frac{450}{30}$ $=15$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $15.921$ $+$ $9.679$ $+$ $101.36$ $-$ $94.76$ $-$ $17.976$ $-$ $6.224$ $=$ ?<\/p>\n Adding all Positive terms: Adding all Negative terms: $126.96$ $-$ $118.96$ $=$ $8$<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\large\\sqrt{1936}\\div\\sqrt[3]{1331}\\times\\sqrt{50625}\\div\\sqrt{14400}\\times\\sqrt{64}$ $-$ $\\large7920\\div132$ $=$ ?<\/p>\n $\\Large\\frac{44}{11}$ $\\times$ $\\Large\\frac{225}{(\\Large\\frac{120}{8})}$ $-$ $\\Large\\frac{7920}{132}$<\/p>\n $4$ $\\times$ $15$ $-$ $60$ $=$ $0$<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $522\\div150\\times25$ $=$ $87$<\/p>\n $96\\div180\\times15$ $=$ $8$<\/p>\n $\\Large\\frac{1125}{135}$ $\\div$ $\\Large\\frac{5}{9}$ $=$ $15$<\/p>\n $87$ $+$ $8$ $-$ $15$ $=$ $80$<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $x^{2}+9y^{2}$=6xy<\/p>\n $\\Rightarrow x^{2}-6xy+9y^{2}$=0<\/p>\n $\\Rightarrow (x-3y)^{2}$=0<\/p>\n $\\Rightarrow$ x-3y=0<\/p>\n $\\Rightarrow$ x=3y<\/p>\n $\\Rightarrow \\frac{x}{y}$=$\\frac{3}{1}$<\/p>\n $\\therefore$ x:y=3:1<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Terms =\u00a0$3^{200},\\ 2^{300}\\ and\\ 7^{100}$<\/p>\n Dividing all the exponents by 100, we get\u00a0:<\/p>\n $\\equiv3^2,2^3,7^1$<\/p>\n = $9,8,7$<\/p>\n Thus, the largest number = $9\\equiv3^{200}$<\/p>\n => Ans – (A)<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Total number of students = 40<\/p>\n Among the student who have passed, Raman’s rank from top = 13th<\/p>\n Raman’s rank from bottom = 17th<\/p>\n => Total students who passed = $(13+17)-1=30-1=29$<\/p>\n $\\therefore$ Number of students who have failed = $40-29=11$<\/p>\n => Ans – (A)<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression\u00a0: 25 + 18 \u00f7 2 – 4 = 20<\/p>\n (A) :\u00a0+ and \u00f7<\/p>\n $\\equiv25\\div18+2-4=20$<\/p>\n L.H.S. = $1.39-2=-0.61\\neq20$<\/p>\n (B) :\u00a0+ and –<\/p>\n $\\equiv25-18\\div2+4=20$<\/p>\n L.H.S. = $25-9+4=20$<\/p>\n => Ans – (B)<\/p>\n\n
\n* 2 $ 20 \u00f7 156 = 145<\/p>\n
\n25 + 18 \u00f7 2 – 4 = 20<\/p>\n
\n=462.767-(34.428*3)+134.568
\n=462.767-103.285+134.568
\n=494.05<\/p>\n
\n=245+1280-312
\n=1525-312
\n=1213<\/p>\n
\n$15.921$ $+$ $9.679$ $+$ $101.36$ $=$ $126.96$<\/p>\n
\n$94.76$ $+$ $17.976$ $+$ $6.224$ $=$ $118.96$<\/p>\n