Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>In what ratio salt costing Rs.13\/kg mixed with another salt costing Rs.7\/ kg to get a mixture costing Rs.9\/kg?<\/p>\n a)\u00a04 : 7<\/p>\n b)\u00a03 : 5<\/p>\n c)\u00a02 : 3<\/p>\n d)\u00a01 : 2<\/p>\n Question 2:\u00a0<\/b>$1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{16}+\\frac{1}{32}+\\frac{1}{64}$ is approximately equal to ___________ .<\/p>\n a)\u00a01.5<\/p>\n b)\u00a02<\/p>\n c)\u00a02.5<\/p>\n d)\u00a03<\/p>\n Question 3:\u00a0<\/b>The sum of the present ages of a father and his son is 126 years. 8 years ago their respective ages were in the ratio of 7 : 4. After 7 years what will be the ratio of ages of father and son?<\/p>\n a)\u00a023 : 19<\/p>\n b)\u00a017 : 11<\/p>\n c)\u00a027 : 20<\/p>\n d)\u00a034 : 23<\/p>\n Question 4:\u00a0<\/b>A bus can travel at 150 km\/hr without stoppages and at 120 km\/hr with stoppages. How many minutes per hour does the bus stop?<\/p>\n a)\u00a015 minutes<\/p>\n b)\u00a012 minutes<\/p>\n c)\u00a016 minutes<\/p>\n d)\u00a020 minutes<\/p>\n Question 5:\u00a0<\/b>A train of length 120 m crossed a platform in 36 seconds. It crosses a man standing on the platform in 24 seconds running at the same speed. Find the length of the platform.<\/p>\n a)\u00a0120 m<\/p>\n b)\u00a060 m<\/p>\n c)\u00a0104 m<\/p>\n d)\u00a072 m<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>A train crosses a pole in 12 seconds and a platform of length 160 m in 32 seconds running at the same speed. Find the length of the train.<\/p>\n a)\u00a0108 m<\/p>\n b)\u00a096 m<\/p>\n c)\u00a0120 m<\/p>\n d)\u00a084 m<\/p>\n Question 7:\u00a0<\/b>\u0394XYZ is right angled at Y. If $sinX = \\frac{4}{5}$, then what is the value of $cos Z$ ?<\/p>\n a)\u00a0$\\frac{3}{4}$<\/p>\n b)\u00a0$\\frac{5}{3}$<\/p>\n c)\u00a0$\\frac{4}{5}$<\/p>\n d)\u00a0$\\frac{4}{3}$<\/p>\n Question 8:\u00a0<\/b>What is the value of $(\\frac{1}{3} – sec 45^\\circ)$ ?<\/p>\n a)\u00a0$\\frac{(4-\\sqrt3)}{2}$<\/p>\n b)\u00a0$\\frac{(\\sqrt{6}-1)}{\\sqrt{3}}$<\/p>\n c)\u00a0$\\frac{(1-3\\sqrt2)}{3}$<\/p>\n d)\u00a0$\\frac{(\\sqrt2-1)}{\\sqrt{2}}$<\/p>\n Question 9:\u00a0<\/b>If \u03b8 is positive acute angle Find whether the equation 3 $(\\sec^{2}\\theta-\\tan^{2}\\theta)=5$ is true or not ?<\/p>\n a)\u00a0True<\/p>\n b)\u00a0False<\/p>\n c)\u00a0can not be determined<\/p>\n d)\u00a0None of these<\/p>\n Question 10:\u00a0<\/b>What is the absolute difference of the roots of the equation $2x^{2}-6x+8 = 0$ ?<\/p>\n a)\u00a0$\\sqrt{6}i$<\/p>\n b)\u00a0$\\sqrt{5}i$<\/p>\n c)\u00a0$\\sqrt{7}i$<\/p>\n d)\u00a0$\\sqrt{3}i$<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 11:\u00a0<\/b>A class has two sections A and B. The average marks of two Sections in an examination is 75. The average marks of Section A is 72 and that of Section B is 80. Find the ratio of the number of students in Section A to Section B<\/p>\n a)\u00a0$5 : 3$<\/p>\n b)\u00a0$3 : 5$<\/p>\n c)\u00a0$9 : 10$<\/p>\n d)\u00a0$10 : 9$<\/p>\n Question 12:\u00a0<\/b>If \u2018a\u2019 is a positive integer such that $a^2=2a+3$ then what is the value of $a^3$?<\/p>\n a)\u00a01<\/p>\n b)\u00a08<\/p>\n c)\u00a027<\/p>\n d)\u00a064<\/p>\n Question 13:\u00a0<\/b>If $\\log{2} = 0.301$ and $log{3}=0.477$, find the value of $log{72}$.<\/p>\n a)\u00a01.424<\/p>\n b)\u00a01.857<\/p>\n c)\u00a01.987<\/p>\n d)\u00a02.357<\/p>\n Question 14:\u00a0<\/b>If $\\log{16}=1.2$ then what is the value of $\\log{64}$?<\/p>\n a)\u00a01.6<\/p>\n b)\u00a01.7<\/p>\n c)\u00a01.8<\/p>\n d)\u00a01.9<\/p>\n Question 15:\u00a0<\/b>If $log_{x}y = 5 $ and $log_{x}729= 6$ then find the value of y.<\/p>\n a)\u00a0625<\/p>\n b)\u00a0243<\/p>\n c)\u00a081<\/p>\n d)\u00a0343<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let quantity of Rs. 13\/kg and Rs. 7\/kg salt mixed be $x$ and $y$ kg respectively.<\/p>\n According to ques,<\/p>\n => $13x+7y=9(x+y)$<\/p>\n => $13x+7y=9x+9y$<\/p>\n => $13x-9x=9y-7y$<\/p>\n => $4x=2y$<\/p>\n => $\\frac{x}{y}=\\frac{2}{4}=\\frac{1}{2}$<\/p>\n $\\therefore$ Required ratio = $1:2$<\/p>\n => Ans –\u00a0(D)<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression =\u00a0$1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{16}+\\frac{1}{32}+\\frac{1}{64}$<\/p>\n =\u00a0$(1+\\frac{1}{2}+\\frac{1}{2^2}+\\frac{1}{2^3}+\\frac{1}{2^4}+\\frac{1}{2^5}+\\frac{1}{2^6})-(\\frac{1}{8})$<\/p>\n The first term above is a geometric progression with first term, $a=1$ and common ratio, $r=\\frac{1}{2}$<\/p>\n Number of terms, $n=7$<\/p>\n Sum of $n$ terms of G.P. (when $r<1$) = $\\frac{a(1-r^n)}{1-r}$<\/p>\n = $[\\frac{1(1-\\frac{1}{2}^7)}{1-\\frac{1}{2}}]-(\\frac{1}{8})$<\/p>\n = $[2\\times(1-\\frac{1}{128})]-(\\frac{1}{8})$<\/p>\n = $\\frac{127}{64}-\\frac{8}{64}$<\/p>\n = $\\frac{119}{64}=1.85\\approx2$<\/p>\n => Ans – (B)<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Sum of the present ages of a father and his son\u00a0= 126 years<\/p>\n Sum of their ages 8 years ago = $126-8(2)=110$ years<\/p>\n => Father’s age at that time = $\\frac{7}{(7+4)}\\times110=70$ years<\/p>\n => Son’s age at that time = $110-70=40$ years<\/p>\n Thus, present age of father = $70+8=78$ years<\/p>\n => Son’s present age = $40+8=48$ years<\/p>\n $\\therefore$ Ratio of ages of father and son after 7 years = $\\frac{78+7}{48+7}=\\frac{85}{55}$<\/p>\n = $17:11$<\/p>\n => Ans – (B)<\/p>\n DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n SSC CHSL Important Q&A PDF<\/a><\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Due to stoppages, bus travelled 30 km less in an hour. 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Speed of the train to cross the man = 120\/24 = 5 m\/sec 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the length of the train be \u2018T\u2019 metres. \u21d2 $8T = 3T + 480$ 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given\u00a0:\u00a0$\\sin X$ = $\\frac{4}{5}$<\/p>\n Also, $\\sin X=\\frac{YZ}{XZ}=\\frac{4}{5}$<\/p>\n Let YZ = 4 cm and XZ = 5 cm<\/p>\n To find\u00a0:\u00a0$\\cos Z=\\frac{YZ}{XZ}$<\/p>\n = $\\frac{4}{5}$<\/p>\n => Ans – (C)<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0:\u00a0$(\\frac{1}{3} – sec 45^\\circ)$<\/p>\n = $\\frac{1}{3}-\\sqrt2$<\/p>\n =\u00a0$\\frac{(1-3\\sqrt2)}{3}$<\/p>\n => Ans – (C)<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given that : 3 $(\\sec^{2}\\theta-\\tan^{2}\\theta)=5$<\/p>\n we know that 1 + $tan^2 \\theta = sec^2 \\theta$<\/p>\n using the above identity ,<\/p>\n L.H.S :: 3 $(\\sec^{2}\\theta-\\tan^{2}\\theta)$ = 3 $(1 + tan^2 \\theta – tan^2 \\theta)$ = 3 <\/span> R.H.S :: 5 <\/span><\/span><\/p>\n As , L.H.S =\/= R.H.S the given equation is false <\/span><\/span><\/p>\n \u00a0<\/span><\/span><\/p>\n \u00a0<\/span><\/span><\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let a and b be roots of the equation. 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number of students in Section A and Section B be $x$ and $y$ respectively 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The given equation can be written as, 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $72 = 2^3 \\times 3^2$ 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It is given that, 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $log_{x}y = 5 $ and $log_{x}729= 6$
\nTime taken to travel 30 km $= \\dfrac{30}{150}\\times60 = 12$ minutes<\/p>\n
\nSpeed of the train to cross the platform = (120+P)\/36
\n$\\dfrac{120+P}{36} = 5$
\n\u21d2 $120+P = 180$
\n\u21d2 $P = 60$<\/p>\n
\nSpeed of the train to cross the pole in 12 seconds = T\/12 m\/sec
\nSpeed of the train to cross the platform in 32 seconds = (T+160)\/32 m\/sec
\nHere, Speeds are equal.
\n\u21d2 $\\dfrac{T}{12} = \\dfrac{T+160}{32}$<\/p>\n
\n\u21d2 $5T = 480$
\n\u21d2 $T = 96 m$
\nHence, The Length of the train = 96 m<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/28221_EKKXK7P.PNG\")
<\/figure>\n
\n<\/span><\/p>\n
\na+b = $\\frac{6}{2}$ = $3$
\nab = $\\frac{8}{2}$ = $4$
\n$(a-b)^{2}$ = $(a+b)^{2}$ -$4ab$ = 9 -4(4) = -7
\n$a-b$ = $\\sqrt{7} * \\sqrt {-1}$ = $\\sqrt{7}i$<\/p>\n
\nTotal marks of the students from Section A = $72x$
\nTotal marks of the students from Section B = $80y$
\n$\\therefore 72x + 80y = 75(x+y)$
\n$\\implies 5y = 3x$
\n$ \\implies \\frac{x}{y}=\\frac{5}{3}$
\nHence , the correct option is A<\/p>\n
\n$a^2-2a-3=0$
\n=>$(a-3)(a+1)=0$
\nAs it is given that \u2018a\u2019 is a positive integer we get, a = 3.
\nSo we get, $a^3=3^3=27$<\/p>\n
\n=> $log{72} = log{2^3 \\times 3^2} = log{2^3} + log{3^2} = 3log{2} + 2log{3} = 3\\times 0.301 + 2 \\times 0.477 = 1.857$<\/p>\n
\n$\\log{16}=1.2$
\n$=>log{2^4}=1.2$
\n$=>4log{2}=1.2$
\n$=>log{2}=0.3$
\nNow let us find out the value of $\\log{64}$
\n$\\log{64}=log{2^6}=6log{2}=6*0.3=1.8$<\/p>\n
\n=> $x^5 = y$ and $x^6 = 729$ => $x = 729^{1\/6} = (3^6)^{1\/6} = 3$
\nHence
\n$ y = 3^5 = 243$<\/p>\n\n