{"id":40201,"date":"2020-01-28T17:32:55","date_gmt":"2020-01-28T12:02:55","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40201"},"modified":"2020-01-28T17:32:55","modified_gmt":"2020-01-28T12:02:55","slug":"ssc-cgl-maths-questions-and-answers-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/ssc-cgl-maths-questions-and-answers-pdf\/","title":{"rendered":"SSC-CGL Maths Questions and Answers PDF"},"content":{"rendered":"

SSC-CGL Maths Questions and Answers PDF<\/strong><\/span><\/h1>\n

Download SSC CGL Maths Questions with answers PDF based on previous papers very useful for SSC CGL exams. Top-15 Very Important Questions for SSC Exams.<\/p>\n

Download SSC CGL Maths Questions and Answers PDF<\/a><\/p>\n

Take a free SSC CGL mock test<\/a><\/p>\n

Get 200 SSC mocks for just Rs. 249. Enroll here<\/a><\/p>\n

Take a free mock test for SSC CGL<\/a><\/p>\n

Download SSC CGL Previous Papers PDF<\/a><\/p>\n

More SSC CGL Important Questions and Answers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is<\/p>\n

a)\u00a01120<\/p>\n

b)\u00a04714<\/p>\n

c)\u00a05200<\/p>\n

d)\u00a05600<\/p>\n

Question 2:\u00a0<\/b>A number, when divided by 114, leaves remainder 21. If the same number is divided by 19, then the remainder will be<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a07<\/p>\n

d)\u00a017<\/p>\n

Question 3:\u00a0<\/b>A manufacturer marked an article at Rs. 50 and sold it allowing 20% discount. If his profit was 25%, then the cost price of the article was<\/p>\n

a)\u00a0Rs. 40<\/p>\n

b)\u00a0Rs. 35<\/p>\n

c)\u00a0Rs. 32<\/p>\n

d)\u00a0Rs. 30<\/p>\n

Question 4:\u00a0<\/b>At what rate percent per annum will a sum of Rs. 1,000 amount to Rs. 1,102.50 in 2 years at compound interest?<\/p>\n

a)\u00a05<\/p>\n

b)\u00a05.5<\/p>\n

c)\u00a06<\/p>\n

d)\u00a06.5<\/p>\n

Question 5:\u00a0<\/b>In how many years will a sum of Rs. 800 at 10% per annum compound interest, compounded semiannually becomes Rs. 926.10 ?<\/p>\n

a)\u00a01$\\frac{1}{2}$<\/p>\n

b)\u00a01$\\frac{2}{3}$<\/p>\n

c)\u00a02$\\frac{1}{3}$<\/p>\n

d)\u00a02$\\frac{1}{2}$<\/p>\n

SSC CGL Previous Papers Download PDF<\/a><\/p>\n

SSC CGL Free Mock Test<\/a><\/p>\n

Question 6:\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 7:\u00a0<\/b>$\\sqrt{8 + \\sqrt{57 + \\sqrt{38 + \\sqrt{108 + \\sqrt{169}}}}}$<\/p>\n

a)\u00a04<\/p>\n

b)\u00a06<\/p>\n

c)\u00a08<\/p>\n

d)\u00a010<\/p>\n

Question 8:\u00a0<\/b>The unit digit in the product $122^{173}$\u00a0is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a04<\/p>\n

c)\u00a06<\/p>\n

d)\u00a08<\/p>\n

Question 9:\u00a0<\/b>A copper wire is bent in the form of an equilateral triangle, and has an area $121\\sqrt{3}$ cm2\u00a0<\/sup>. If the same wire is bent into the form of a circle, the area(in cm2<\/sup>) enclosed by the wire in(Take $\\pi = \\frac{22}{7}$)<\/p>\n

a)\u00a0364.5<\/p>\n

b)\u00a0693.5<\/p>\n

c)\u00a0346.5<\/p>\n

d)\u00a0639.5<\/p>\n

Question 10:\u00a0<\/b>At present, the ratio of the ages of Maya and Chhaya is 6:5 and fifteen years from now, the ratio will get changed to 9:8. Maya’s present age is<\/p>\n

a)\u00a021 years<\/p>\n

b)\u00a024 years<\/p>\n

c)\u00a030 years<\/p>\n

d)\u00a040 years<\/p>\n

Question 11:\u00a0<\/b>Which one of the following will completely divide 571<\/sup> + 572<\/sup> + 573<\/sup> ?<\/p>\n

a)\u00a0150<\/p>\n

b)\u00a0160<\/p>\n

c)\u00a0155<\/p>\n

d)\u00a030<\/p>\n

Question 12:\u00a0<\/b>A student was asked to divide a number by 6 and add 12 to the quotient. He, however, first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been<\/p>\n

a)\u00a0124<\/p>\n

b)\u00a0122<\/p>\n

c)\u00a0118<\/p>\n

d)\u00a0114<\/p>\n

Question 13:\u00a0<\/b>The difference between the compound interest and simple interest for the amount Rs. 5,000 in 2 years is Rs.32. The rate of interest is<\/p>\n

a)\u00a05%<\/p>\n

b)\u00a08%<\/p>\n

c)\u00a010%<\/p>\n

d)\u00a012%<\/p>\n

Instructions<\/b>Directions: The following graph shows the production of cotton bales of 100 kg each in lakhs by different states A, B, C, D and E over the years. Study the graph and answer the following Questions.
\n\"\"<\/p>\n

Question 14:\u00a0<\/b>In which State(s) is there a steady increase in the production of cotton during the given period?<\/p>\n

a)\u00a0A and B<\/p>\n

b)\u00a0B and D<\/p>\n

c)\u00a0A and C<\/p>\n

d)\u00a0D and E<\/p>\n

Question 15:\u00a0<\/b>If $x = 1 + \\sqrt{2} + \\sqrt{3}$ , then the value of $(2x^4 – 8x^3 – 5x^2 + 26x- 28)$ is __?<\/p>\n

a)\u00a0$6\\sqrt{6}$<\/p>\n

b)\u00a0$0$<\/p>\n

c)\u00a0$3\\sqrt{6}$<\/p>\n

d)\u00a0$2\\sqrt{6}$<\/p>\n

FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n

18000+ Questions – Free SSC Study Material<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given: Numbers- First = 5834
\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

 <\/p>\n

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the given number be x<\/em>
\nLet a be the quotient when x<\/em>\u00a0is divided by 114
\nSo $\\frac{x}{114}$ = a$\\frac{21}{114}$
\nso x<\/em>\u00a0= 114a + 21
\nwhen x<\/em>\u00a0is divided by 19 it can be written as
\n$\\frac{x}{19} = \\frac{114a + 21}{19}$
\n114 is divisible by 19 and 21 leaves a remainder of 2.<\/p>\n

3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Given: Marked Price = 50
\nDiscount = 20%of 50 = $50\\times0.5$ = 10
\nHence sold price = 50 – 10 = 40
\nLet’s say cost price is $x$
\nProfit = 25% of x (Always remember profit and loss applicable only on cost price) = $\\frac{x}{4}$
\nHence sold price will be $x$\u00a0+ $\\frac{x}{4}$ = $\\frac{5x}{4}$
\nor\u00a0$\\frac{5x}{4}$ = 40
\n$x$ = 32<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let’s say rate is r
\nhence $1000\\times(1+\\frac{r}{100})^{2}$ = 1102.5
\nnow on solving the we will get r = 5<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

When we are compounding it semiannually its rate becomes 5% and number of years will 2n
\nso for compound interest:
\n926.10=$800\\times(1+\\frac{10}{100})^{2n}$
\nsolve for n.<\/p>\n

6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

since we know volume will remain same while melting
\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

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4<\/p>\n

8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

As we know a number with unit digit 2 have repeating cycle of 2,4,8,6 after every fourth power
\nas power is 173 or (172+1) where till 172 , 43rd cycle will get complete and next unit digit will be 2.<\/p>\n

9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Area of equilateral triangle is $\\frac{\\sqrt{3}}{4} a^{2}$ where a is side of triangle
\nwhich is equals to $121{\\sqrt{3}}$
\nor a = 22 and whole length of wire will be 66
\nfrom here when it is bend to make a circle, circumference will be $2\\pi r$ = 66
\nr = 10.5
\nhence area of circle will be $\\pi r^{2}$ = 346.5<\/p>\n

10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let’s say maya’s age is $6x$ and chaya’s age is $5x$.
\nafter 15 years ages will be $6x+15$ and $5x+15$.
\nNew ratio will be $\\frac{6x+15}{5x+15} = \\frac{9}{8}$.
\nAfter solving above equation we will get $x$ equals to 5
\nSo maya’s age will be 30.<\/p>\n

11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Among all options only option C has unit digit 5, and in given equation unit digit will also be 5.
\nSo only 155 can divide the given equation completely.<\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let’s say number is N
\nSo according to student result is $112= \\frac{N+12}{6} $
\nor N = 660
\nCorrect answer will be = $\\frac{660}{6} +12 = 110+12 = 122$<\/p>\n

13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Difference between compound interest and simple interest for 2 years will be
\n= $(P((1+\\frac{r}{100})^2) – P) – 2P \\frac{r}{100} = 32$ (where P is principal amount 5000 and r is rate )
\nafter solving above equation we will get r = 8%<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Only in A and C there is a steady increment in production of cotton as in D and E , It is decresed and in B production is equal for two years. Hence answer will be C).<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

x = 1+ $\\sqrt {2} + \\sqrt {3} $
\n$(x-1)^{2}$ = $(\\sqrt {2} + \\sqrt {3}) ^ {2} $
\n$x^{2} +1 – 2x = 5 + 2 \\sqrt {6}$
\n$x^{2} – 2x = 4 + 2 \\sqrt {6}$ ( eq. (1) )
\n$(x^{2} – 2x)^{2} = x^{4} + 4x^{2} – 4x^{3} = 40 + 16\\sqrt{6} $ eq (2)
\nNow in $2x^{4} – 8x^{3} – 5x^{2} + 26x – 28 $
\nor $2(x^{4} – 4x^{3}) – 5x^{2} + 26x – 28 $ ( putting values from eq (1) and eq (2) )
\nAfter solving we will get it reduced to $6\\sqrt{6}$<\/p>\n

Get 200 SSC mocks for just Rs. 249. Enroll here<\/a><\/p>\n

SSC Free Previous Papers App<\/a><\/p>\n

We hope this\u00a0Maths Questions for SSC CGL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

SSC-CGL Maths Questions and Answers PDF Download SSC CGL Maths Questions with answers PDF based on previous papers very useful for SSC CGL exams. Top-15 Very Important Questions for SSC Exams. Take a free mock test for SSC CGL Download SSC CGL Previous Papers PDF More SSC CGL Important Questions and Answers PDF Question 1:\u00a0The […]<\/p>\n","protected":false},"author":49,"featured_media":40211,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1459,1611,1741,1441],"tags":[3529,3535],"class_list":{"0":"post-40201","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-stenographer","17":"tag-maths-questions","18":"tag-ssc_cgl"},"better_featured_image":{"id":40211,"alt_text":"SSC CGL Maths Questions and Answers PDF","caption":"SSC CGL Maths Questions and Answers PDF","description":"SSC CGL Maths Questions and Answers PDF","media_type":"image","media_details":{"width":900,"height":472,"file":"2020\/01\/fig-28-01-2020_11-08-25.jpg","sizes":{"thumbnail":{"file":"fig-28-01-2020_11-08-25-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-150x150.jpg"},"medium":{"file":"fig-28-01-2020_11-08-25-300x157.jpg","width":300,"height":157,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-300x157.jpg"},"medium_large":{"file":"fig-28-01-2020_11-08-25-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-768x403.jpg"},"tiny-lazy":{"file":"fig-28-01-2020_11-08-25-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-30x16.jpg"},"td_80x60":{"file":"fig-28-01-2020_11-08-25-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-80x60.jpg"},"td_100x70":{"file":"fig-28-01-2020_11-08-25-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-100x70.jpg"},"td_218x150":{"file":"fig-28-01-2020_11-08-25-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-218x150.jpg"},"td_265x198":{"file":"fig-28-01-2020_11-08-25-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-265x198.jpg"},"td_324x160":{"file":"fig-28-01-2020_11-08-25-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-324x160.jpg"},"td_324x235":{"file":"fig-28-01-2020_11-08-25-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-324x235.jpg"},"td_324x400":{"file":"fig-28-01-2020_11-08-25-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-324x400.jpg"},"td_356x220":{"file":"fig-28-01-2020_11-08-25-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-356x220.jpg"},"td_356x364":{"file":"fig-28-01-2020_11-08-25-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-356x364.jpg"},"td_533x261":{"file":"fig-28-01-2020_11-08-25-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-533x261.jpg"},"td_534x462":{"file":"fig-28-01-2020_11-08-25-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-534x462.jpg"},"td_696x0":{"file":"fig-28-01-2020_11-08-25-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-696x365.jpg"},"td_696x385":{"file":"fig-28-01-2020_11-08-25-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-696x385.jpg"},"td_741x486":{"file":"fig-28-01-2020_11-08-25-741x472.jpg","width":741,"height":472,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-741x472.jpg"},"td_0x420":{"file":"fig-28-01-2020_11-08-25-801x420.jpg","width":801,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25-801x420.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0","keywords":[]}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-28-01-2020_11-08-25.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n