{"id":28777,"date":"2019-05-09T18:55:03","date_gmt":"2019-05-09T13:25:03","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=28777"},"modified":"2019-05-09T18:55:03","modified_gmt":"2019-05-09T13:25:03","slug":"rrb-je-mathematics-questions-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/rrb-je-mathematics-questions-set-2-pdf\/","title":{"rendered":"RRB JE Mathematics Questions Set-2 PDF"},"content":{"rendered":"

RRB JE Mathematics Questions Set-2 PDF<\/strong><\/span><\/h1>\n

Download Top 15 RRB JE Mathematics Set-2 Questions and Answers PDF. RRB JE Mathematcis Questions based on asked questions in previous exam papers very important for the Railway JE exam.<\/p>\n

Download RRB JE Mathematics Questions Set-2 PDF<\/a><\/p>\n\n

Download RRB JE Previous Papers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>If tan(\u03b8\/2) = 5\/12, then find sin\u03b8 = ?<\/p>\n

a)\u00a05\/13<\/p>\n

b)\u00a0119\/120<\/p>\n

c)\u00a0120\/169<\/p>\n

d)\u00a012\/13<\/p>\n

Question 2:\u00a0<\/b>Find the value of $x$ (from the given options), if $x^2-16x+64 = 4$ ?<\/p>\n

a)\u00a08<\/p>\n

b)\u00a010<\/p>\n

c)\u00a012<\/p>\n

d)\u00a014<\/p>\n

Question 3:\u00a0<\/b>Sum of a natural number and 5 times its reciprocal is 14\/3, then find that number ?<\/p>\n

a)\u00a06<\/p>\n

b)\u00a05<\/p>\n

c)\u00a04<\/p>\n

d)\u00a03<\/p>\n

FREE RRB JE YOUTUBE VIDEOS<\/a><\/p>\n

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

Question 4:\u00a0<\/b>If p, q are the sum and product of the roots of $3x^3-4x^2+5x-6=0$, then find p\/q ?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a01\/3<\/p>\n

c)\u00a02\/3<\/p>\n

d)\u00a03\/2<\/p>\n

Question 5:\u00a0<\/b>Given a = 3, b =4 and c = 6, which of the following statements is true?
\n3a+ 4b -c = 20
\nab + bc – ca = 18
\nabc – $c^2$ + ca – b = 52
\n$a^2c > b^2a$<\/p>\n

a)\u00a0Only 2<\/p>\n

b)\u00a0Only 4<\/p>\n

c)\u00a0Only 1 and 3<\/p>\n

d)\u00a0Only 2 and 4<\/p>\n

Question 6:\u00a0<\/b>Find the value of the expression:$\\frac{114^3+7^3}{114^2+7^2-(114*7)}$<\/p>\n

a)\u00a0118<\/p>\n

b)\u00a0124<\/p>\n

c)\u00a0125<\/p>\n

d)\u00a0121<\/p>\n

RRB JE Free Mock Test<\/a><\/p>\n

RRB JE Free Mock Test (latest pattern)<\/a><\/p>\n\n

Question 7:\u00a0<\/b>Three numbers A, B and C are such that A is 50% more than B which is twice of C. Then find the ratio between A and C.<\/p>\n

a)\u00a02 : 3<\/p>\n

b)\u00a03 : 2<\/p>\n

c)\u00a03 : 1<\/p>\n

d)\u00a01 : 2<\/p>\n

Question 8:\u00a0<\/b>The ratio of two numbers is 2:5. If the difference between the squares of the numbers is 1029, then find the sum of the two numbers.<\/p>\n

a)\u00a035<\/p>\n

b)\u00a056<\/p>\n

c)\u00a049<\/p>\n

d)\u00a028<\/p>\n

Question 9:\u00a0<\/b>The ratio of boys to girls in a class is 10:7. After a few days, 60 boys were added to the class. Now, the ratio of boys to girls changed to 5:2. What was the total number of student in the class initially?<\/p>\n

a)\u00a080<\/p>\n

b)\u00a0120<\/p>\n

c)\u00a0136<\/p>\n

d)\u00a0196<\/p>\n

Question 10:\u00a0<\/b>A beaker contains acid solution of concentration 60%. What fraction of the acid solution should be replaced with water, so that the resultant solution has a concentration of 50%?<\/p>\n

a)\u00a01\/3<\/p>\n

b)\u00a01\/4<\/p>\n

c)\u00a01\/6<\/p>\n

d)\u00a01\/5<\/p>\n

Daily Free Online Tests for RRB Exams<\/a><\/p>\n

Download RRB GK Material PDF<\/a><\/p>\n\n

Question 11:\u00a0<\/b>The difference of two positive numbers is 10. 2 and 48 are the HCF and LCM of those two numbers, then find the smaller number ?<\/p>\n

a)\u00a012<\/p>\n

b)\u00a016<\/p>\n

c)\u00a06<\/p>\n

d)\u00a08<\/p>\n

Question 12:\u00a0<\/b>In an exam, Amit is supposed to calculate the product of 34.1 * 4.67. He instead calculates the product of 0.0341 * 46.7 as 1.59. What is the actual answer in the exam?<\/p>\n

a)\u00a01.59<\/p>\n

b)\u00a015.90<\/p>\n

c)\u00a00.159<\/p>\n

d)\u00a0159.0<\/p>\n

Question 13:\u00a0<\/b>Given that log2 = 0.3 approx, one billion would be approximately<\/p>\n

a)\u00a0$2^{9}$<\/p>\n

b)\u00a0$2^{10}$<\/p>\n

c)\u00a0$2^{20}$<\/p>\n

d)\u00a0$2^{30}$<\/p>\n

Question 14:\u00a0<\/b>If the hypotenuse of an isosceles right angled triangle is $6 \\times \\sqrt 2$m, then find the area of that triangle ?<\/p>\n

a)\u00a036 sq.m<\/p>\n

b)\u00a018 sq.m<\/p>\n

c)\u00a024 sq.m<\/p>\n

d)\u00a048 sq.m<\/p>\n

Question 15:\u00a0<\/b>A third angle orthographic projection of an object is given below. What is the object ?\"\"<\/p>\n

a)\u00a0Triangle<\/p>\n

b)\u00a0Trapezium<\/p>\n

c)\u00a0Cone<\/p>\n

d)\u00a0Frustrum of a cone<\/p>\n

General Science Notes for RRB Exams (PDF)<\/a><\/p>\n\n

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

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

$Sin\u03b8$ = $\\frac{2tan(\u03b8\/2)}{1+tan^2(\u03b8\/2)}$ = $\\frac{2(5\/12)}{1+(5\/12)^2}$ = $120\/169$<\/p>\n

So the answer is option C.<\/p>\n

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

$x^2-16x+64 = 4$<\/p>\n

$x^2-16x+60 = 0$<\/p>\n

$x^2-10x-6x+60 = 0$<\/p>\n

$x(x-10)-6(x-10) = 0$<\/p>\n

$(x-10)(x-6) = 0$<\/p>\n

$x = 10$ or $x = 6$<\/p>\n

So the answer is option B.<\/p>\n

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

Let the number be x<\/p>\n

$x+5(1\/x) = 14\/3$<\/p>\n

$x^2+5 = 14x\/3$<\/p>\n

$3x^2+15 = 14x$<\/p>\n

$3x^2-14x+15 = 0$<\/p>\n

$3x^2-9x-5x+15 = 0$<\/p>\n

$(3x-5)(x-3) = 0$<\/p>\n

$x = 5\/3$ or $x = 3$<\/p>\n

So the answer is option D.<\/p>\n

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

Sum of roots = p = -b\/a = 4\/3<\/p>\n

Product of roots = -d\/a = 6\/3\u00a0 = 2<\/p>\n

p\/q = (4\/3)\/2 = 2\/3<\/p>\n

So the answer is option C.<\/p>\n

 <\/p>\n

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

Substituting the values in (1), we get 3 * 3 + 4 * 4 – 6 = 19
\nSo, (1) is false.<\/p>\n

Substituting the values in (2), we get 3 * 4 + 4 * 6 – 6 *3 = 18
\nSo, (2) is true<\/p>\n

Substituting the values in (3), we get 3 * 4 * 6 – 4 * 4 + 6 * 3 – 4 = 50
\nSo, (3) is false.<\/p>\n

Substituting the values in (4), we get 3 * 3 * 6 > 4 * 4 * 3
\nSo, (4) is true
\nHence the correct option is (d)<\/p>\n

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

$\\frac{114^3+7^3}{114^2+7^2-(114*7)} = \\frac{(114+7)(114^2+7^2-(114*7))}{114^2+7^2-(114*7)}$ = 121<\/p>\n

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

Let C be x.
\nThen, B = 2*x = 2x
\nA = 150% of 2x = 3x
\nThen, A : C = 3x : x = 3 : 1.<\/p>\n

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

Let the two numbers be 2x and 5x.
\nGiven, $(5x)^2 – (2x)^2 = 1029
\n\u21d2 $25x^2 – 4x^2 = 1029
\n\u21d2 $21x^2 = 1029
\n\u21d2 x^2 = 49
\n\u21d2 x = 7.
\nThen, the two numbers will be,
\n2x = 2*7 = 14
\n5x = 5*7 = 35
\nThen, Sum of two numbers = 14+35 = 49.<\/p>\n

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

Boys : Girls = 10 : 7
\nLet the number of boys and girls initially be 10x and 7x respectively.
\nNow, 60 boys were added and ratio changed to 5 : 2.
\nTotal number of boys = 10x + 60
\nTotal number of girls = 7x
\n=> $\\frac{10x+60}{7x}$ = $\\frac{5}{2}$
\n=> 20x + 120 = 35x
\n=> 15x = 120
\n=> x = 8
\n=> Total number of boys initially = 80 and Total number of girls initially = 56
\n=> Total number of students initially = 80 + 56 = 136<\/p>\n

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

Pure water => concentration is 0%
\nSo, the situation can be represented as follows:<\/p>\n

\"\"<\/p>\n

The required ratio of acid solution to water in the resultant solution is 50 : 10 = 5 : 1
\nSo, the fraction of acid solution that has to be replaced with water is 1\/(5+1) = 1\/6<\/p>\n

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

Let one number is X, then another number is 10+X<\/p>\n

Product of two numbers\u00a0 = (LCM)*(HCF)<\/p>\n

(X)(X+10) = (2)(48)<\/p>\n

X^2+10X = 96<\/p>\n

X^2+10X-96 = 0<\/p>\n

(X+16)(X-6) = 0<\/p>\n

X = -16 or 6<\/p>\n

Since X is positive, X = 6 = smaller number<\/p>\n

So the answer is option C.<\/p>\n

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

0.0341 * 46.7 = 1.59
\nSo, 34.1 * 46.7 = 1.59 * 1000
\n34.1 * 4.67 = 1.59 * 100
\nHence the answer is 159<\/p>\n

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

1 billion = $10^9$<\/p>\n

log (1 billion) = 9<\/p>\n

log 2 = 0.3<\/p>\n

$log_2$1billion = 30<\/p>\n

=> 1 billion = $2^{30}$<\/p>\n

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

Let a, a are the adjacent sides of the isosceles triangle,<\/p>\n

$a^2+a^2 = (6\\sqrt2)^2$<\/p>\n

$2a^2 = 72$<\/p>\n

$a = 6$<\/p>\n

Area of triangle = $\\frac{1}{2}\\times 6 \\times 6$ = $18sq.m$<\/p>\n

So the answer is option B.<\/p>\n

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

By cutting a portion of the cone parallel to its base at the top, a frustum is formed.<\/p>\n

The first and second figures are top view and side view of a frustum.<\/p>\n\n

DOWNLOAD APP FOR RRB FREE MOCKS<\/a><\/p>\n

We hope this Mathematics Set-2 Questions for RRB JE Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

RRB JE Mathematics Questions Set-2 PDF Download Top 15 RRB JE Mathematics Set-2 Questions and Answers PDF. RRB JE Mathematcis Questions based on asked questions in previous exam papers very important for the Railway JE exam. Download RRB JE Previous Papers PDF Question 1:\u00a0If tan(\u03b8\/2) = 5\/12, then find sin\u03b8 = ? a)\u00a05\/13 b)\u00a0119\/120 c)\u00a0120\/169 […]<\/p>\n","protected":false},"author":41,"featured_media":28779,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1605],"tags":[489,491,1637,1635,1647],"class_list":{"0":"post-28777","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-je","11":"tag-railway-exam","12":"tag-rrb","13":"tag-rrb-exam","14":"tag-rrb-je","15":"tag-rrb-mocks"},"better_featured_image":{"id":28779,"alt_text":"RRB JE Mathematics Questions Set-2 PDF","caption":"RRB JE Mathematics Questions Set-2 PDF","description":"RRB JE Mathematics Questions Set-2 PDF","media_type":"image","media_details":{"width":1200,"height":630,"file":"2019\/05\/fig-09-05-2019_07-54-08.jpg","sizes":{"thumbnail":{"file":"fig-09-05-2019_07-54-08-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-150x150.jpg"},"medium":{"file":"fig-09-05-2019_07-54-08-300x158.jpg","width":300,"height":158,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-300x158.jpg"},"medium_large":{"file":"fig-09-05-2019_07-54-08-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-768x403.jpg"},"large":{"file":"fig-09-05-2019_07-54-08-1024x538.jpg","width":1024,"height":538,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-1024x538.jpg"},"tiny-lazy":{"file":"fig-09-05-2019_07-54-08-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-30x16.jpg"},"td_80x60":{"file":"fig-09-05-2019_07-54-08-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-80x60.jpg"},"td_100x70":{"file":"fig-09-05-2019_07-54-08-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-100x70.jpg"},"td_218x150":{"file":"fig-09-05-2019_07-54-08-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-218x150.jpg"},"td_265x198":{"file":"fig-09-05-2019_07-54-08-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-265x198.jpg"},"td_324x160":{"file":"fig-09-05-2019_07-54-08-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-324x160.jpg"},"td_324x235":{"file":"fig-09-05-2019_07-54-08-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-324x235.jpg"},"td_324x400":{"file":"fig-09-05-2019_07-54-08-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-324x400.jpg"},"td_356x220":{"file":"fig-09-05-2019_07-54-08-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-356x220.jpg"},"td_356x364":{"file":"fig-09-05-2019_07-54-08-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-356x364.jpg"},"td_533x261":{"file":"fig-09-05-2019_07-54-08-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-533x261.jpg"},"td_534x462":{"file":"fig-09-05-2019_07-54-08-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-534x462.jpg"},"td_696x0":{"file":"fig-09-05-2019_07-54-08-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-696x365.jpg"},"td_696x385":{"file":"fig-09-05-2019_07-54-08-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-696x385.jpg"},"td_741x486":{"file":"fig-09-05-2019_07-54-08-741x486.jpg","width":741,"height":486,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-741x486.jpg"},"td_1068x580":{"file":"fig-09-05-2019_07-54-08-1068x580.jpg","width":1068,"height":580,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-1068x580.jpg"},"td_1068x0":{"file":"fig-09-05-2019_07-54-08-1068x561.jpg","width":1068,"height":561,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-1068x561.jpg"},"td_0x420":{"file":"fig-09-05-2019_07-54-08-800x420.jpg","width":800,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08-800x420.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0"}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-54-08.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n