Download Top-20 Arithmetic questions for SSC CGL exam. Most important arithmetic questions based on asked questions in previous exam papers for SSC CGL.<\/p>\n
Download SSC CGL Arithmetic questions <\/a><\/p>\n Get 125 SSC CGL Mocks – Just Rs. 199<\/a><\/p>\n Take a free SSC CGL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n <\/p>\n Question 1:\u00a0<\/b>If X = 0.3 $\\times$ 0.3, the value of X is<\/p>\n a)\u00a00.009<\/p>\n b)\u00a00.03<\/p>\n c)\u00a00.09<\/p>\n d)\u00a00.08<\/p>\n Question 2:\u00a0<\/b>An equation of the form ax + by + c = 0. Where, a\u00a0\u2260 0, b\u00a0\u2260 0 and c = 0 represents a straight line which passes through<\/p>\n a)\u00a0(2, 4)<\/p>\n b)\u00a0(0, 0)<\/p>\n c)\u00a0(3, 2)<\/p>\n d)\u00a0None of these<\/p>\n Question 3:\u00a0<\/b>The fifth term of the sequence for which $t_{1}=1$, $t_{2}=2$ and $t_{n+2}$ = $t_{n}+t_{n+1}$, is<\/p>\n a)\u00a05<\/p>\n b)\u00a010<\/p>\n c)\u00a06<\/p>\n d)\u00a08<\/p>\n Question 4:\u00a0<\/b>Reduce 3596 \/ 4292 to lowest terms.<\/p>\n a)\u00a029\/37<\/p>\n b)\u00a017\/43<\/p>\n c)\u00a031\/37<\/p>\n d)\u00a019\/23<\/p>\n Question 5:\u00a0<\/b>Reduce 2530\/1430 to lowest terms.<\/p>\n a)\u00a047\/17<\/p>\n b)\u00a023\/13<\/p>\n c)\u00a047\/19<\/p>\n d)\u00a029\/17<\/p>\n Question 6:\u00a0<\/b>The first and last terms of an arithmetic progression are -32 and \u00ad43. If the sum of the series is \u00ad88, then it has how many terms?<\/p>\n a)\u00a016<\/p>\n b)\u00a015<\/p>\n c)\u00a017<\/p>\n d)\u00a014<\/p>\n Question 7:\u00a0<\/b>29 is 0.8% of?<\/p>\n a)\u00a03625<\/p>\n b)\u00a01450<\/p>\n c)\u00a07250<\/p>\n d)\u00a010875<\/p>\n Question 8:\u00a0<\/b>5*[-0.6 (2.8 + 1.2)] of 0.3 is equal to<\/p>\n a)\u00a0-1.44<\/p>\n b)\u00a0-1.08<\/p>\n c)\u00a0-1.2<\/p>\n d)\u00a0-3.6<\/p>\n Question 9:\u00a0<\/b>Find the value of p if 3x + p, x – 10 and -x + 16 are in arithmetic progression.<\/p>\n a)\u00a016<\/p>\n b)\u00a036<\/p>\n c)\u00a0-16<\/p>\n d)\u00a0-36<\/p>\n Question 10:\u00a0<\/b>If 9\/4th of 7\/2 of a number is 126, then 7\/2th of that number is …………..<\/p>\n a)\u00a056<\/p>\n b)\u00a0284<\/p>\n c)\u00a072<\/p>\n d)\u00a026<\/p>\n Question 11:\u00a0<\/b>The 4th term of an arithmetic progression is 15, 15th term is -29, \ufb01nd the 10th term?<\/p>\n a)\u00a0-5<\/p>\n b)\u00a0-13<\/p>\n c)\u00a0-17<\/p>\n d)\u00a0-9<\/p>\n Question 12:\u00a0<\/b>(91 + 92 + 93 + \u2026\u2026\u2026 +110) is equal to<\/p>\n a)\u00a04020<\/p>\n b)\u00a02010<\/p>\n c)\u00a06030<\/p>\n d)\u00a08040<\/p>\n Question 13:\u00a0<\/b>40.36 – (9.347 – x ) – 29.02 = 3.68. Find x.<\/p>\n a)\u00a0-56.353<\/p>\n b)\u00a01.687<\/p>\n c)\u00a0-17.007<\/p>\n d)\u00a082.407<\/p>\n Question 14:\u00a0<\/b>What is the value of (81 + 82 + 83 + \u2026\u2026\u2026 +130)?<\/p>\n a)\u00a05275<\/p>\n b)\u00a010550<\/p>\n c)\u00a015825<\/p>\n d)\u00a021100<\/p>\n Question 15:\u00a0<\/b>In an arithmetic progression if 13 is the 3rd term, \u00ad47 is the 13th term, then \u00ad30 is which term?<\/p>\n a)\u00a09<\/p>\n b)\u00a010<\/p>\n c)\u00a07<\/p>\n d)\u00a08<\/p>\n Get 125 SSC CGL Mocks – Just Rs. 199<\/a><\/p>\n Question 16:\u00a0<\/b>The first and last terms of an arithmetic progression are 37 and \u00ad-18. If the sum of the series is 114, then it has how many terms?<\/p>\n a)\u00a013<\/p>\n b)\u00a012<\/p>\n c)\u00a014<\/p>\n d)\u00a015<\/p>\n Question 17:\u00a0<\/b>If 4\/5th of 6\/7th of a number is 216, then 8\/9th of that number will be<\/p>\n a)\u00a0179<\/p>\n b)\u00a0280<\/p>\n c)\u00a0160<\/p>\n d)\u00a0269<\/p>\n Question 18:\u00a0<\/b>199994 x 200006 = ?<\/p>\n a)\u00a039999799964<\/p>\n b)\u00a039999999864<\/p>\n c)\u00a039999999954<\/p>\n d)\u00a039999999964<\/p>\n Question 19:\u00a0<\/b>In an arithmetic progression, if 17 is the 3rd term, -25 is the 17th term, then -1 is which term?<\/p>\n a)\u00a010<\/p>\n b)\u00a011<\/p>\n c)\u00a09<\/p>\n d)\u00a012<\/p>\n Question 20:\u00a0<\/b>In an arithmetic progression, if 9 is the 5th term, -26 is the 12th term, then -6 is which term?<\/p>\n a)\u00a011<\/p>\n b)\u00a08<\/p>\n c)\u00a010<\/p>\n d)\u00a07<\/p>\n Join Exam Preparation Telegram Group<\/a><\/p>\n Free SSC Preparation Videos<\/a><\/p>\n More SSC CGL Important Questions and Answers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0: $X=0.3\\times0.3$<\/p>\n => $X=0.09$<\/p>\n => Ans – (C)<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n As c=0, and substituting the point (0,0) in the equation, we get ax+by+c = 0 at the point (0,0). 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $t_{1}=1$, $t_{2}=2$<\/p>\n $t_{n+2}$ = $t_{n}+t_{n+1}$<\/p>\n put n=3, then\u00a0\u00a0$t_{5}$ = $t_{3}+t_{4}$<\/p>\n $t_{3}$ = $t_{1}+t_{2}$ = 1+2 = 3<\/p>\n $t_{4}$ = $t_{2}+t_{3}$ = 2+3 = 5<\/p>\n $t_{5}$ = $t_{3}+t_{4}$ = 3+5 = 8<\/p>\n so the answer is option D.<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0: $\\frac{3596}{4292}$<\/p>\n Dividing both numerator and denominator by 4, = $\\frac{899}{1073}$<\/p>\n Similarly, dividing by 29, we get\u00a0:<\/p>\n = $\\frac{31}{37}$<\/p>\n => Ans – (C)<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression\u00a0: $\\frac{2530}{1430}$<\/p>\n Dividing both numerator and denominator by 10, we get\u00a0= $\\frac{253}{143}$<\/p>\n Similarly, dividing by 11, we get\u00a0:<\/p>\n = $\\frac{23}{13}$<\/p>\n => Ans – (B)<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n First term of AP, $a=-32$ and last term,\u00a0$l=43$<\/p>\n Let there be $n$ terms<\/p>\n Sum of AP = $\\frac{n}{2}(a+l) = 88$<\/p>\n => $\\frac{n}{2}(-32+43)=88$<\/p>\n => $\\frac{11n}{2}=88$<\/p>\n => $n=88 \\times \\frac{2}{11}$<\/p>\n => $n=8 \\times 2=16$<\/p>\n => Ans – (A)<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number be $x$<\/p>\n According to ques, 0.8% of $x$ = 29<\/p>\n => $\\frac{0.8}{100} \\times x = 29$<\/p>\n => $\\frac{x}{125} = 29$<\/p>\n => $x = 29 \\times 125 = 3625$<\/p>\n => Ans – (A)<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression\u00a0:\u00a05*[-0.6 (2.8 + 1.2)] of 0.3<\/p>\n = $5 [(-0.6) \\times (4)] \\times 0.3$<\/p>\n = $5 \\times (-2.4) \\times 0.3$<\/p>\n = $(-12) \\times 0.3 = -3.6$<\/p>\n => Ans – (D)<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Terms in arithmetic progression\u00a0: $(3x + p) , (x – 10) , (-x + 16)$<\/p>\n => Difference between first two terms is equal to the difference between last two terms<\/p>\n => $(x – 10) – (3x + p) = (-x + 16) – (x – 10)$<\/p>\n => $-2x -10 – p = -2x + 16 + 10$<\/p>\n => $-p = 26 + 10 = 36$<\/p>\n => $p = -36$<\/p>\n => Ans – (D)<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number be $x$<\/p>\n According to ques,<\/p>\n => $\\frac{9}{4} \\times \\frac{7}{2} \\times x = 126$<\/p>\n => $\\frac{63}{8} x = 126$<\/p>\n => $x = \\frac{126}{63} \\times 8$<\/p>\n => $x = 2 \\times 8 = 16$<\/p>\n $\\therefore (\\frac{7}{2})^{th}$ of the number = $\\frac{7}{2} \\times 16$<\/p>\n = $7 \\times 8 = 56$<\/p>\n => Ans – (A)<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n 11)\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 4th term, $A_4 = a + (4 – 1) d = 15$<\/p>\n => $a + 3d = 15$ —————–(i)<\/p>\n Similarly, 15th term, $A_{15} = a + 14d = -29$ ——————(ii)<\/p>\n Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n => $(14d – 3d) = -29 – 15$<\/p>\n => $d = \\frac{-44}{11} = -4$<\/p>\n Substituting it in equation (i), => $a – 12 = 15$<\/p>\n => $a = 15 + 12 = 27$<\/p>\n $\\therefore$ 10th term, $A_{10} = a + (10 – 1)d$<\/p>\n = $27 + (9 \\times -4) = 27 – 36 = -9$<\/p>\n => Ans – (D)<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression\u00a0:\u00a0(91 + 92 + 93 + \u2026\u2026\u2026 +110)<\/p>\n This is an arithmetic progression with first term, $a = 91$ , last term, $l = 110$ and common difference, $d = 1$<\/p>\n Let number of terms = $n$<\/p>\n Last term in an A.P. = $a + (n – 1)d = 110$<\/p>\n => $91 + (n – 1)(1) = 110$<\/p>\n => $n – 1 = 110 – 91 = 19$<\/p>\n => $n = 19 + 1 = 20$<\/p>\n $\\therefore$ Sum of A.P. = $\\frac{n}{2} (a + l)$<\/p>\n = $\\frac{20}{2} (91 + 110)$<\/p>\n = $10 \\times 201 = 2010$<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression\u00a0:\u00a040.36 – (9.347 – x ) – 29.02 = 3.68<\/p>\n => 40.36 – 9.347 + x = 3.68 + 29.02<\/p>\n =>\u00a031.013 + x = 32.7<\/p>\n => x = 32.7 – 31.013<\/p>\n => x = 1.687<\/p>\n => Ans – (B)<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression\u00a0:\u00a0(81 + 82 + 83 + \u2026\u2026\u2026 +130)<\/p>\n This is an arithmetic progression with first term, $a = 81$ , last term, $l = 130$ and common difference, $d = 1$<\/p>\n Let number of terms = $n$<\/p>\n Last term in an A.P. = $a + (n – 1)d = 130$<\/p>\n => $81 + (n – 1)(1) = 130$<\/p>\n => $n – 1 = 130 – 81 = 49$<\/p>\n => $n = 49 + 1 = 50$<\/p>\n $\\therefore$ Sum of A.P. = $\\frac{n}{2} (a + l)$<\/p>\n = $\\frac{50}{2} (81 + 130)$<\/p>\n = $25 \\times 211 = 5275$<\/p>\n 15)\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, $A_3 = a + (3 – 1) d = 13$<\/p>\n => $a + 2d = 13$ —————–(i)<\/p>\n Similarly, 13th term, $A_{13} = a + 12d = 47$ ——————(ii)<\/p>\n Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n => $(12d – 2d) = 47 – 13 = 34$<\/p>\n => $d = \\frac{34}{10} = 3.4$<\/p>\n Substituting it in equation (i), => $a + 2 \\times 3.4 = 13$<\/p>\n => $a = 13 – 6.8 = 6.2$<\/p>\n Let $n^{th}$ term = 30<\/p>\n => $a + (n – 1) d = 30$<\/p>\n => $6.2 + (n – 1) (3.4) = 30$<\/p>\n => $(n – 1) (3.4) = 30 – 6.2 = 23.8$<\/p>\n => $(n – 1) = \\frac{23.8}{3.4} = 7$<\/p>\n => $n = 7 + 1 = 8$<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n In an arithmetic progression with first term, $a = 37$ , last term, $l = -18$<\/p>\n Let number of terms = $n$<\/p>\n $\\therefore$ Sum of A.P. = $\\frac{n}{2} (a + l) = 114$<\/p>\n => $\\frac{n}{2} (37 – 18) = 114$<\/p>\n => $19n = 114 \\times 2 = 228$<\/p>\n => $n = \\frac{228}{19} = 12$<\/p>\n => Ans – (B)<\/p>\n Get 125 SSC CGL Mocks – Just Rs. 199<\/a><\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the number be $x$<\/p>\n According to ques,<\/p>\n => $\\frac{4}{5} \\times \\frac{6}{7} \\times x = 216$<\/p>\n => $x = 216 \\times \\frac{35}{24} = 9 \\times 35$<\/p>\n $\\therefore$\u00a08\/9th of the number = $\\frac{8}{9} \\times (35 \\times 9)$<\/p>\n = $8 \\times 35 = 280$<\/p>\n => Ans – (B)<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression\u00a0:\u00a0\u00a0199994 x 200006<\/p>\n = (200000 – 6)\u00a0x (200000 + 6)<\/p>\n = $(200000)^2 – (6)^2$<\/p>\n = 40000000000 – 36 = 39999999964<\/p>\n => Ans – (D)<\/p>\n 19)\u00a0Answer\u00a0(C)<\/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, $A_3 = a + (3 – 1) d = 17$<\/p>\n => $a + 2d = 17$ —————–(i)<\/p>\n Similarly, 17th term, $A_{17} = a + 16d = -25$ ——————(ii)<\/p>\n Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n => $(16d – 2d) = -25 – 17$<\/p>\n => $d = \\frac{-42}{14} = -3$<\/p>\n Substituting it in equation (i), => $a – 6 = 17$<\/p>\n => $a = 17 + 6 = 23$<\/p>\n Let $n^{th}$ term = -1<\/p>\n => $a + (n – 1) d = -1$<\/p>\n => $23 + (n – 1) (-3) = -1$<\/p>\n => $(n – 1) (-3) = -1 – 23 = -24$<\/p>\n => $(n – 1) = \\frac{-24}{-3} = 8$<\/p>\n => $n = 8 + 1 = 9$<\/p>\n 20)\u00a0Answer\u00a0(B)<\/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 5th term, $A_5 = a + (5 – 1) d = 9$<\/p>\n => $a + 4d = 9$ —————–(i)<\/p>\n Similarly, 12th term, $A_{12} = a + 11d = -26$ ——————(ii)<\/p>\n Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n => $(11d – 4d) = -26 – 9$<\/p>\n => $d = \\frac{-35}{7} = -5$<\/p>\n Substituting it in equation (i), => $a – 20 = 9$<\/p>\n => $a = 9 + 20 = 29$<\/p>\n Let $n^{th}$ term = -6<\/p>\n => $a + (n – 1) d = -6$<\/p>\n => $29 + (n – 1) (-5) = -6$<\/p>\n => $(n – 1) (-5) = -6 – 29 = -35$<\/p>\n => $(n – 1) = \\frac{-35}{-5} = 7$<\/p>\n => $n = 7 + 1 = 8$<\/p>\n
\nHence, the line passes through origin.<\/p>\n