Download Top-20 SSC MTS Expected Maths Questions PDF. Most Expected Maths questions based on asked questions in previous year exam papers very important for the SSC MTS exam.<\/p>\n
Download Expected Maths Questions For SSC MTS PDF<\/a><\/p>\n Last Day – 10 SSC MTS Mocks for just Rs. 117<\/a><\/p>\n Take a free mock test for SSC MTS<\/a><\/p>\n SSC MTS Previous Papers<\/a> (Download PDF)<\/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 SSC MTS Previous Papers PDF<\/a><\/p>\n Take a free mock test for SSC MTS<\/a><\/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 SSC Study Material (18000 Solved Questions)<\/a><\/p>\n General Science Notes for SSC Exams<\/a><\/p>\n Question 11:\u00a0<\/b>Find equation of the perpendicular bisector of segment joining the points (2,-5) and (0,7)?<\/p>\n a)\u00a0x – 6y = 5<\/p>\n b)\u00a0x + 6y = -5<\/p>\n c)\u00a0x – 6y = -5<\/p>\n d)\u00a0x + 6y = 5<\/p>\n Question 12:\u00a0<\/b>Find equation of the perpendicular to segment joining the points A(0,4) and B(-5,9) and passing through the point P. Point P divides segment AB in the ratio 2:3.<\/p>\n a)\u00a0x – y = 8<\/p>\n b)\u00a0x – y = -8<\/p>\n c)\u00a0x + y = -8<\/p>\n d)\u00a0x + y = 8<\/p>\n Question 13:\u00a0<\/b>The co-ordinates of the centroid of a triangle ABC are (-1,-2) what are the co-ordinates of vertex C, if co-ordinates of A and B are (6,-4)\u00a0 and (-2,2) respectively?<\/p>\n a)\u00a0(-7,-4)<\/p>\n b)\u00a0(7,4)<\/p>\n c)\u00a0(7,-4)<\/p>\n d)\u00a0(-7,4)<\/p>\n Question 14:\u00a0<\/b>What is the slope of the line parallel to the line passing through the points (4,-2) and (-3,5)?<\/p>\n a)\u00a03\/7<\/p>\n b)\u00a01<\/p>\n c)\u00a0-3\/7<\/p>\n d)\u00a0-1<\/p>\n Question 15:\u00a0<\/b>The line passing through (-2,5) and (6,b) is perpendicular to the line 20x + 5y = 3. Find b?<\/p>\n a)\u00a0-7<\/p>\n b)\u00a04<\/p>\n c)\u00a07<\/p>\n d)\u00a0-4<\/p>\n Question 16:\u00a0<\/b>Find k, if the line 4x+y = 1 is perpendicular to the line 5x+ky = 2?<\/p>\n a)\u00a020<\/p>\n b)\u00a0-20<\/p>\n c)\u00a04<\/p>\n d)\u00a0-4<\/p>\n Question 17:\u00a0<\/b>The square root of 0.09 is<\/p>\n a)\u00a00.30<\/p>\n b)\u00a00.03<\/p>\n c)\u00a00.81<\/p>\n d)\u00a00.081<\/p>\n Question 18:\u00a0<\/b>How many perfect squares lie between 120 and 300 ?<\/p>\n a)\u00a05<\/p>\n b)\u00a06<\/p>\n c)\u00a07<\/p>\n d)\u00a08<\/p>\n Question 19:\u00a0<\/b>A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is:<\/p>\n a)\u00a02.5<\/p>\n b)\u00a03<\/p>\n c)\u00a03.5<\/p>\n d)\u00a04<\/p>\n Question 20:\u00a0<\/b>The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is<\/p>\n a)\u00a0135<\/p>\n b)\u00a0240<\/p>\n c)\u00a073<\/p>\n d)\u00a0106<\/p>\n 200+ SSC Important Practice Sets<\/a><\/p>\n GK Q&A for Competitive Exams (Download 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 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let line $l$ perpendicularly bisects line joining \u00a0A(2,-5) and B(0,7) at C, thus C is the mid point of AB.<\/p>\n => Coordinates of C = $(\\frac{2 + 0}{2} , \\frac{-5 + 7}{2})$<\/p>\n = $(\\frac{2}{2} , \\frac{2}{2}) = (1,1)$<\/p>\n Now, slope of AB = $\\frac{y_2 – y_1}{x_2 – x_1} = \\frac{(7 + 5)}{(0 – 2)}$<\/p>\n = $\\frac{12}{-2} = -6$<\/p>\n Let slope of line $l = m$<\/p>\n Product of slopes of two perpendicular lines = -1<\/p>\n => $m \\times -6 = -1$<\/p>\n => $m = \\frac{1}{6}$<\/p>\n Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$<\/p>\n $\\therefore$ Equation of line $l$<\/p>\n => $(y – 1) = \\frac{1}{6}(x – 1)$<\/p>\n => $6y – 6 = x – 1$<\/p>\n => $x – 6y = 1 – 6 = -5$<\/p>\n => Ans – (C)<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a\u00a0: b<\/p>\n = $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n Coordinates of A(0,4) and B(-5,9). Let coordinates of\u00a0P = (x,y) which divides AB in ratio = 2\u00a0: 3<\/p>\n => $x = \\frac{(2 \\times -5) + (3 \\times 0)}{2 + 3}$<\/p>\n => $5x = -10$<\/p>\n => $x = \\frac{-10}{5} = -2$<\/p>\n Similarly, $y = \\frac{(2 \\times 9) + (3 \\times 4)}{2 + 3}$<\/p>\n => $5y = 18 + 12 = 30$<\/p>\n => $y = \\frac{30}{5} = 6$<\/p>\n => Point P = (-2,6)<\/p>\n Slope of AB = $\\frac{9 – 4}{-5 – 0} = \\frac{5}{-5} = -1$<\/p>\n Let slope of line perpendicular to AB = $m$<\/p>\n Also, product of slopes of two perpendicular lines is -1<\/p>\n => $m \\times -1 = -1$<\/p>\n => $m = 1$<\/p>\n Equation of lines having slope $m$ and passing through point P(-2,6) is<\/p>\n => $(y – 6) = 1(x + 2)$<\/p>\n => $y – 6 = x + 2$<\/p>\n => $x – y = -8$<\/p>\n => Ans – (B)<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Coordinates of centroid of triangle with vertices $(x_1 , y_1)$ , $(x_2 , y_2)$ and $(x_3 , y_3)$ is $(\\frac{x_1 + x_2 + x_3}{3} , \\frac{y_1 + y_2 + y_3}{3})$<\/p>\n Let coordinates of vertex C = $(x , y)$<\/p>\n Vertex A(6,-4) and Vertex B(-2,2) and Centroid = (-1,-2)<\/p>\n => $-1 = \\frac{-2 + 6 + x}{3}$<\/p>\n => $x + 4 = -1 \\times 3 = -3$<\/p>\n => $x = -3 – 4 = -7$<\/p>\n Similarly, => $-2 = \\frac{-4 + 2 + y}{3}$<\/p>\n => $y – 2 = -2 \\times 3 = -6$<\/p>\n => $y = -6 + 2 = -4$<\/p>\n $\\therefore$ Coordinates of vertex C = (-7,-4)<\/p>\n => Ans – (A)<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Slope of line passing through points\u00a0(4,-2) and (-3,5)<\/p>\n = $\\frac{5 + 2}{-3 – 4} = \\frac{7}{-7} = -1$<\/p>\n Slope of two parallel lines is always equal.<\/p>\n => Slope of the line parallel to the line having slope -1 = $-1$<\/p>\n => Ans – (D)<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Slope of line having equation : $ax + by + c = 0$ is $\\frac{-a}{b}$<\/p>\n => Slope of line $20x + 5y = 3$ is $\\frac{-20}{5} = -4$<\/p>\n Slope\u00a0line passing through (-2,5) and (6,b) = $\\frac{b – 5}{6 + 2} = \\frac{(b – 5)}{8}$<\/p>\n Also, product of slopes of two perpendicular lines is -1<\/p>\n => $\\frac{(b – 5)}{8} \\times -4 = -1$<\/p>\n => $b – 5 = \\frac{8}{4} = 2$<\/p>\n => $b = 2 + 5 = 7$<\/p>\n => Ans – (C)<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Slope of line having equation : $ax + by + c = 0$ is $\\frac{-a}{b}$<\/p>\n Thus, slope of line $4x + y = 1$ is $\\frac{-4}{1} = -4$<\/p>\n Similarly, slope of line $5x + ky = 2$ is $\\frac{-5}{k}$<\/p>\n Also, product of slopes of two perpendicular lines is -1<\/p>\n => $\\frac{-5}{k} \\times -4 = -1$<\/p>\n => $\\frac{20}{k} = -1$<\/p>\n => $k = -20$<\/p>\n => Ans – (B)<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 0.09 = $\\frac{9}{100}$ 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n let’s say n be a number whose perfect square lie between 120 and 300 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n since we know volume will remain same while melting 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n As we know $(a-b)^{2}$ = $a^{2} + b^{2} – 2ab$
\nHence, the line passes through origin.<\/p>\n
\n$\\sqrt{\\frac{9}{100}} = \\frac{3}{10}$ = 0.3<\/p>\n
\nhence 120<$n^{2}$<300
\nor $121\\leq n^{2} \\leq289$
\nor $11\\leq n^{2} \\leq17$<\/p>\n
\n$\\pi r_{1}^{2}h= \\frac{4}{3}\\pi r_{2}^{3}$
\nwhere $r_{1}$ is radius of cylinderical wire and $r_{2}$ is radius of sphere and h is length of wire
\nputting values we will get $r_{2}$ = 3 cm.<\/p>\n
\nWe assume that first number is a and second number is b hence ab = 45
\nand a – b = 4
\nafter putiing values we will get\u00a0$a^{2} + b^{2}$ = 106<\/p>\n