Sometimes Quantitative aptitude section of SSC CHSL syllabus is referred as maths.<\/p>\n
SSC CHSL Maths Questions and Answers PDF:<\/strong><\/p>\n 20 SSC CHSL (latest pattern mocks) – Rs. 117<\/a><\/p>\n You can practice SSC CHSL Previous year solved papers<\/a><\/p>\n Download SSC CHSL Reasoning questions PDF<\/a><\/p>\n Download GK PDFs for SSC CHSL 2018<\/a><\/p>\n Take Free SSC CHSL Mock test 2018<\/a><\/p>\n Question\u00a01:<\/b><\/p>\n The average of the first 100 positive integers is<\/p>\n a) 100 Question 2:<\/b><\/p>\n Two natural numbers are in the ratio 3:5 and their product is 2160. The smaller of the numbers is<\/p>\n a) 36 Question 3:<\/b><\/p>\n By selling an article, a man makes a profit of 25% of its selling price. His profit per cent is<\/p>\n a) 20 <\/p>\n Question\u00a04:<\/p>\n The ninth term of the sequence 0, 3, 8, 15, 24, 35, is<\/p>\n a) 63 Question 5:<\/b><\/p>\n If the ratio of cost price and selling price of an article be as 10:11, the percentage of profit is<\/p>\n a) 8 Question 6:<\/p>\n By what least number should 675 be multiplied so as to obtain a perfect cube number ?<\/p>\n a) 3 Question 7: a) 4\/11 SSC CHSL Expected Cutoff 2018<\/a><\/p>\n Question 8: a) 0.30 Question 9: a) 48 a) 1120 Take a Free Mock test for SSC CHSL<\/a><\/p>\n SSC CHSL 2018 Syllabus PDF<\/a><\/p>\n Solutions (1 to 10)<\/b><\/p>\n 1) Answer (c) 2) Answer (a) 3) Answer (d)<\/p>\n Given Profit on selling price is 25% SSC CHSL General intelligence pdf<\/a><\/p>\n 4) Answer (c) 5) Answer (b) 6) Answer (b) 7) Answer (c) 8) Answer (a)<\/p>\n 0.09 = $$\\frac{9}{100}$$ 9) Answer (d) 10) Answer (b)<\/p>\n Given: Numbers- First = 5834 Download GK and Current Affairs PDF for SSC CHSL<\/a><\/p>\n
\nb) 51
\nc) 50.5
\nd) 49.5<\/p>\n
\nb) 24
\nc) 18
\nd) 12<\/p>\n
\nb) 25
\nc) 30
\nd) 33.33<\/p>\n
\nb) 70
\nc) 80
\nd) 99<\/p>\n
\nb) 10
\nc) 11
\nd) 15
\n<\/b><\/p>\n
\nb) 5
\nc) 24
\nd) 40<\/p>\n
\n<\/b>
\nThe number 0. 121212…. in the P form p\/q is equal to<\/p>\n
\nb) 2\/11
\nc) 4\/33
\nd) 2\/33<\/p>\n
\n<\/b>
\nThe square root of 0.09 is<\/p>\n
\nb) 0.03
\nc) 0.81
\nd) 0.081<\/p>\n
\n<\/b>
\nThe H.C.F. and L.C.M. of, two numbers are 8 and 48 respectively. If one of the numbers is 24, then the other number is<\/p>\n
\nb) 36
\nc) 24
\nd) 16
\n
\nQuestion 10:
\n<\/b>
\nThe greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is<\/p>\n
\nb) 4714
\nc) 5200
\nd) 5600<\/p>\n
\nsummation of first n natural numbers is $$\\frac{n\\times(n+1)}{2}$$
\nnow putting n = 100
\nsummation will be 5050
\nhence average will be 50.5<\/p>\n
\nGiven:
\nRatio of two numbers = 3:5
\nHence we can say first number is $$3x$$ and second number is $$5x$$
\nProduct of two numbers is = 2160
\nor $$15x^{2}$$ = 2160
\nor $$x$$ = 12
\nhence numbers will be 30 and 60<\/p>\n
\nSuppose selling price is y
\nhence profit will be $$\\frac{y}{4}$$ and cost price will be $$\\frac{3y}{4}$$
\nNow profit percentage on cost price will be \u00a0$$\\frac{\\frac{y}{4}}{\\frac{3y}{4}}\\times100$$
\ni.e.\u00a0$$\\frac{100}{3}$$ = 33.33<\/p>\n
\nGiven: Sequence of numbers 0,3,8,15,24,35
\nTo know:
\n9th term of the sequence
\nAs we can see difference of sequence is in arthmatic progression of 3,5,7,9,11,13 and so on. 6th term of sequence is 35
\nand adding common difference for it (that is 13)\u00a0, hence 7th term will be 48
\naccordingly 9th term will be 80<\/p>\n
\nLet the cost price be 10x<\/em>\u00a0and selling price be 11x<\/em>
\nPercentage Profit = $$\\frac{11x – 10x}{10x}$$ x 100 = 10<\/p>\n
\nFactorising 675 we get
\n675 = 5 $$\\times$$ 5\u00a0$$\\times$$ 3\u00a0$$\\times$$ 3\u00a0$$\\times$$ 3
\nThere in order to make 675 a perfect cube, it has to be multiplied by 5.<\/p>\n
\nLet
\nx<\/em>\u00a0= 0.121212 \u00a0be equation 1
\n100x<\/em>\u00a0= 12.121212 be equation 2
\nSubtracting 1 from 2
\n99x =\u00a0<\/em>12
\nx =\u00a0<\/em>$$\\frac{12}{99}$$
\n= $$\\frac{4}{33}$$<\/p>\n
\n$$\\sqrt{\\frac{9}{100}} = \\frac{3}{10}$$ = 0.3<\/p>\n
\nGiven:-
\nNumbers- First = 24
\nSecond = x (suppose)
\nH.C.F. of numbers = 8
\nL.C.M. of numbers = 48
\nAs we know:
\nH.C.F.* L.C.M. = Product of numbers
\nHence
\n48*8 = 24*x
\nx = 16<\/p>\n
\nSecond = x (Suppose)
\nAnd number (5834 – x) is divisible by each of 20,28,32,35
\nLet’s say it is y
\nHence 5834 – x = y
\nor x = 5834 – y
\nNow for x to be greatest y should be least
\nhence y should be least common multiple of 20,28,32,35
\ny = 1120
\nnow x = 5834 – 1120
\nx = 4714<\/p>\n