{"id":28770,"date":"2019-05-09T13:26:06","date_gmt":"2019-05-09T07:56:06","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=28770"},"modified":"2019-05-09T13:26:06","modified_gmt":"2019-05-09T07:56:06","slug":"number-system-questions-for-rrb-ntpc-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/number-system-questions-for-rrb-ntpc-set-2-pdf\/","title":{"rendered":"Number System Questions for RRB NTPC Set-2 PDF"},"content":{"rendered":"

Number System Questions for RRB NTPC Set-2 PDF<\/strong><\/span><\/h2>\n

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

Download Number System Questions for RRB NTPC Set-2 PDF<\/a><\/p>\n\n

Take a free mock test for RRB NTPC<\/a><\/p>\n

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

Question 1:\u00a0<\/b>During the school assembly, if the teachers make the students stand in a row of 14 each, 10 students are left. Instead, if they decide to make the students stand in a row of 24 each, 20 students will be left. If they decide to make them stand in rows of 25 each then 21 students will be left. What is the minimum number of students present in the school?<\/p>\n

a)\u00a03742<\/p>\n

b)\u00a04196<\/p>\n

c)\u00a03192<\/p>\n

d)\u00a04852<\/p>\n

Question 2:\u00a0<\/b>What is the largest possible number which will leave the same remainder on dividing 4048, 3096 and 2024?<\/p>\n

a)\u00a08<\/p>\n

b)\u00a016<\/p>\n

c)\u00a032<\/p>\n

d)\u00a024<\/p>\n

Question 3:\u00a0<\/b>Three numbers A, B and C are such that B is 40 percent less than A and C is 60 percent more than B. Another number D is twice as big as the difference between A and C. Find the value of D as a percentage of C.<\/p>\n

a)\u00a012.5 %<\/p>\n

b)\u00a020 %<\/p>\n

c)\u00a016.67 %<\/p>\n

d)\u00a08.33 %<\/p>\n

RRB NTPC Previous Papers [Download PDF]<\/a><\/p>\n

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

Question 4:\u00a0<\/b>Ravi gives equal number of chocolates to some children. If he gives 4 chocolates to each children then he is left with 9 chocolates in the end. Instead if he decides to give 5 chocolates to each children then the number of chocolates fall short by 8. Find the sum of the number of chocolates with Ravi and the total number of children to whom he distributed the chocolates.<\/p>\n

a)\u00a069<\/p>\n

b)\u00a084<\/p>\n

c)\u00a094<\/p>\n

d)\u00a077<\/p>\n

Question 5:\u00a0<\/b>When the ten\u2019s and units digit of a two digit number are interchanged, the resulting number is 9 more than the original number. How many such two digit numbers exist?<\/p>\n

a)\u00a07<\/p>\n

b)\u00a08<\/p>\n

c)\u00a09<\/p>\n

d)\u00a010<\/p>\n

Question 6:\u00a0<\/b>There are three bells which are named as A, B and C. Bell A rings after every 7 minutes. Bell B rings after every 9 minutes and bell C rings after every 10 minutes. The bells rang together at 12 pm on Friday. How many times will the three bells have rung together(exclude the one at 12 pm) by 6 pm on the next day (Saturday)?<\/p>\n

a)\u00a09<\/p>\n

b)\u00a07<\/p>\n

c)\u00a02<\/p>\n

d)\u00a05<\/p>\n

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

Question 7:\u00a0<\/b>Find the smallest number which leaves remainders 3, 5 and 7 when divided by 11, 13 and 15?<\/p>\n

a)\u00a0135<\/p>\n

b)\u00a0187<\/p>\n

c)\u00a02137<\/p>\n

d)\u00a02797<\/p>\n

Question 8:\u00a0<\/b>The sum of two numbers x and y is multiplied with x and y separately. The product obtained in each of these cases will be 1080 and 520. Find the difference between the given numbers.<\/p>\n

a)\u00a017<\/p>\n

b)\u00a014<\/p>\n

c)\u00a011<\/p>\n

d)\u00a013<\/p>\n

Question 9:\u00a0<\/b>The ratio of the sum of two numbers and the difference of the two numbers is 17:4. Which of the following cannot be the two numbers?<\/p>\n

a)\u00a063, 39<\/p>\n

b)\u00a0210, 130<\/p>\n

c)\u00a0189, 104<\/p>\n

d)\u00a0147, 91<\/p>\n

Download General Science Notes PDF<\/a><\/p>\n\n

Question 10:\u00a0<\/b>A number when successively divided by 7, 8, 9 leave remainders 3, 2 and 1 respectively. What will be the smallest such number?<\/p>\n

a)\u00a0219<\/p>\n

b)\u00a052<\/p>\n

c)\u00a0480<\/p>\n

d)\u00a073<\/p>\n

Download General Science Notes PDF<\/a><\/p>\n

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

Question 11:\u00a0<\/b>A church has 3 bells. All of them start ringing together.The first bell rings every 5 seconds and the second bell rings every 7 seconds. The third bell rings every 15 seconds. Ram noticed that all the 3 bells rang together 210 seconds after he came home. When will the 3 bells ring together again?<\/p>\n

a)\u00a0420 seconds after Ram came home<\/p>\n

b)\u00a0305 seconds\u00a0after Ram came home<\/p>\n

c)\u00a0415 seconds\u00a0after Ram came home<\/p>\n

d)\u00a0315 seconds\u00a0after Ram came home<\/p>\n

Question 12:\u00a0<\/b>Geetha thinks of a number X. She notices that 2X has an odd number of factors. Which of the following statements about X is definitely true?<\/p>\n

a)\u00a0X is a perfect cube.<\/p>\n

b)\u00a0X is an odd number.<\/p>\n

c)\u00a0X is not a perfect square.<\/p>\n

d)\u00a0X can be expressed as the square of a prime number.<\/p>\n

Question 13:\u00a0<\/b>Sheela has a certain number of marbles. She notices that she can divide the marbles in groups of 4 but not in groups of 8. She also notices that she can divide the marbles in groups of 5 but not in groups of 15. Which of the following cannot be the number of marbles with Sheela?<\/p>\n

a)\u00a0100<\/p>\n

b)\u00a0820<\/p>\n

c)\u00a0340<\/p>\n

d)\u00a0720<\/p>\n

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

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

Question 14:\u00a0<\/b>The ratio of the product of 2 numbers to their sum is 8:3. The difference between the 2 numbers is 4. What is the difference between the squares of the 2 numbers if it is known that both the numbers are whole numbers?<\/p>\n

a)\u00a056<\/p>\n

b)\u00a048<\/p>\n

c)\u00a064<\/p>\n

d)\u00a036<\/p>\n

Question 15:\u00a0<\/b>The remainder when a number is divided by 289 is same as the remainder when the number is divided by 17. How many 3 digit numbers satisfy this condition?<\/p>\n

a)\u00a042<\/p>\n

b)\u00a048<\/p>\n

c)\u00a051<\/p>\n

d)\u00a068<\/p>\n\n

Download Current Affairs Questions & Answers PDF<\/a><\/p>\n

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

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

We can observe that 14 – 10 = 4
\n24 – 20 = 4
\nand
\n25 – 21 = 4
\nHence, the required number of students will be nothing but LCM ( 14, 24, 25) – 4
\nWe have
\n14 = 2*7
\n24 = 8*3
\n25 = 5*5
\nHence, the required LCM will be 8*25*3*7 = 600*7 = 4200
\nHence, the required number of students will be 4200 – 4 = 4196<\/p>\n

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

We know that the largest number which leaves same remainder on dividing A, B and C is given by HCF ( B-A, C-B)
\nThus, here we have A = 4048, B = 3096 and C =2024
\nSo, the required number will be
\nHCF (952, 1072)
\n952 = 8*7*17
\n1072 = 16*67
\nThus, we can see that the HCF is 8
\nHence, option A is the correct answer.<\/p>\n

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

Let A be 100x. So B will be 60x. C in this case will be 60x*1.6 = 96x
\nHence, the difference between A and C is 4x. Thus, D must be 8x. Hence, D as a percentage of C will be 8x*100\/96x = 8.33 %
\nThus, option D is the correct answer.<\/p>\n

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

Let the number of children be x and number of chocolates be y. So we have
\n4x + 9 = y
\nand
\n5x – 8 = y
\nSolving both equations, we get
\n4x + 9 = 5x – 8
\n=> x = 17
\nHence, there are 17 children. Hence, the number of chocolates must be 17*4 + 9 = 77
\nThus, the required sum is 77 + 17 = 94<\/p>\n

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

Let the given two digit number be 10x + y
\nSo after interchanging the digits, the new number will be 10y + x
\nWe have been given that
\n10x + y + 9 = 10y + x
\n9y – 9x = 9
\n=> y – x = 1
\nSince both x and y are single digit natural numbers so y can take values from 2 to 9. Consequently, x will vary from 1 to 8. Hence, there are 8 possible numbers which satisfy the given conditions.<\/p>\n

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

Bell A rings after every 7 minutes, bell B rings after every 9 minutes and bell C rings after every 10 minutes. Hence, the three bells will ring together after an interval of LCM(7, 9, 10) minutes. Hence, the three bells will ring together after every 630 minutes. Thus they will ring together after every 10 hours and 30 minutes. Hence, they will ring together 2 times in the next 30 hours. Thus, option C is correct.<\/p>\n

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

We can observe that in each of the cases the difference between the divisor and quotient is constant.
\n11 – 3 = 8
\n13 – 5 = 8
\n15 – 7 = 8
\nHence, the smallest number which satisfies the given conditions will be of the form
\nLCM(11, 13, 15) – 8
\nThus, the required number will be
\n11*13*15 – 8 = 2145 – 8 = 2137<\/p>\n

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

We have been given that (x + y)*x = 1080 and (x + y)*y = 520. The HCF of 1080 and 520 will be the value of x + y. 1080 = 40*27 and 520 = 40*13
\nHence, the HCF will be 40. Thus, x = 27 and y = 13
\nHence, x – y = 14<\/p>\n

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

Let the two numbers be x and y.
\nWe have been given that
\n$\\frac{x+y}{x-y} = \\frac{17}{4}$
\n=> $4x + 4y = 17x – 17y$
\n=> 13x = 21y
\n=> x\/y = 21\/13
\nHence, the numbers must be in this ratio.
\nNumbers in option C are not in this ratio. Hence, it is our answer.<\/p>\n

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

The number on division with 9 leaves are remainder of 1. Since we are looking for smallest number so the number must be 1. This must have been the quotient on division by 8. Hence, the number which would have been divided by 8 would have been 8*1 + 2 = 10.
\nThus, the number which must have been divided by 7 will be 10*7 + 3 = 73
\nHence, option D is the correct answer.<\/p>\n

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

We know that the three bells ring together at intervals of their LCMs.
\nThe LCM of 5, 7 and 15 seconds is 105 seconds.
\nIf the 3 bells rang together at 210 seconds, they will ring together again at 210+105 = 315 seconds.
\nHence, option D is the right answer.<\/p>\n

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

2X has an odd number of factors. Therefore, in 2X, the power of 2 must be even. This means that the power of 2 in X must have been an odd number. Therefore, X\u00a0is definitely an even number. We can eliminate option B.<\/p>\n

X must have been an even number. We know that 2X has an odd number of factors. Also, we know that only perfect squares have an odd number of factors. If 2X is a perfect square, X could not have been a perfect square for sure. Therefore, option C is the right answer.<\/p>\n

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

The number of marbles with Sheela is a multiple of 4 and 5, but are not multiples of 8 and 15. Therefore, the number of marbles must be of the form 20k, where k must neither be even nor must be a multiple of 3.<\/p>\n

720 is a multiple of 3. Therefore, the number of marbles with Sheela cannot be 720. Hence, option D is the right answer.<\/p>\n

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

The difference between the 2 numbers is 4.
\nLet the numbers be $x$ and $(x+4)$.
\nWe know that $\\frac{x(x+4)}{x+x+4} = \\frac{8}{3}$
\n$\\frac{x^2+4x}{2x+4} = \\frac{8}{3}$
\n$3x^2+12x = 16x+32$
\n$3x^2-4x-32 = 0$
\nApplying $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$, we get,
\nRoots = $\\frac{4\\pm\\sqrt{4^2+4*3*32}}{6}$
\n= $\\frac{4\\pm\\sqrt{400}}{6}$
\nSince we know that $x$ is a whole number, $x$ must be $4$.
\n=> The other number is $x+4$ = $8$.
\nDifference between the squares = $8^2 – 4^2$ = $64-16$ = $48$.
\nTherefore, option B is the right answer.<\/p>\n

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

When a number is divided by 17, the remainder can range from 0 to 16 (17 values).
\nWhen a number is divided by 289, the remainder can range from 0 to 288 (289 values).
\nNumbers of the form 289x+k, where k is less than 17 will leave the same remainder when divided by both 17 and 289.
\nThe question asks for 3 digit numbers.
\nTherefore, x can take the values 1,2 and 3. For every value of \u2018x\u2019, 17 values of k will satisfy this condition. Therefore, the number of 3 digit numbers that satisfy the given condition is 3*17 = 51.
\nHence, option C is the right answer.<\/p>\n\n

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

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

Number System Questions for RRB NTPC Set-2 PDF Download RRB NTPC Number Syatem Set-2 Questions and Answers PDF. Top 15 RRB NTPC Number System Set-2 questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Take a free mock test for RRB NTPC Download RRB NTPC Previous Papers PDF […]<\/p>\n","protected":false},"author":41,"featured_media":28772,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603],"tags":[489,491,1647,1619],"class_list":{"0":"post-28770","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-ntpc","11":"tag-railway-exam","12":"tag-rrb","13":"tag-rrb-mocks","14":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":28772,"alt_text":"Number System Questions for RRB NTPC Set-2 PDF","caption":"Number System Questions for RRB NTPC Set-2 PDF","description":"Number System Questions for RRB NTPC Set-2 PDF","media_type":"image","media_details":{"width":1200,"height":630,"file":"2019\/05\/fig-09-05-2019_07-40-40.jpg","sizes":{"thumbnail":{"file":"fig-09-05-2019_07-40-40-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-150x150.jpg"},"medium":{"file":"fig-09-05-2019_07-40-40-300x158.jpg","width":300,"height":158,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-300x158.jpg"},"medium_large":{"file":"fig-09-05-2019_07-40-40-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-768x403.jpg"},"large":{"file":"fig-09-05-2019_07-40-40-1024x538.jpg","width":1024,"height":538,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-1024x538.jpg"},"tiny-lazy":{"file":"fig-09-05-2019_07-40-40-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-30x16.jpg"},"td_80x60":{"file":"fig-09-05-2019_07-40-40-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-80x60.jpg"},"td_100x70":{"file":"fig-09-05-2019_07-40-40-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-100x70.jpg"},"td_218x150":{"file":"fig-09-05-2019_07-40-40-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-218x150.jpg"},"td_265x198":{"file":"fig-09-05-2019_07-40-40-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-265x198.jpg"},"td_324x160":{"file":"fig-09-05-2019_07-40-40-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-324x160.jpg"},"td_324x235":{"file":"fig-09-05-2019_07-40-40-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-324x235.jpg"},"td_324x400":{"file":"fig-09-05-2019_07-40-40-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-324x400.jpg"},"td_356x220":{"file":"fig-09-05-2019_07-40-40-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-356x220.jpg"},"td_356x364":{"file":"fig-09-05-2019_07-40-40-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-356x364.jpg"},"td_533x261":{"file":"fig-09-05-2019_07-40-40-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-533x261.jpg"},"td_534x462":{"file":"fig-09-05-2019_07-40-40-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-534x462.jpg"},"td_696x0":{"file":"fig-09-05-2019_07-40-40-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-696x365.jpg"},"td_696x385":{"file":"fig-09-05-2019_07-40-40-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-696x385.jpg"},"td_741x486":{"file":"fig-09-05-2019_07-40-40-741x486.jpg","width":741,"height":486,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-741x486.jpg"},"td_1068x580":{"file":"fig-09-05-2019_07-40-40-1068x580.jpg","width":1068,"height":580,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-1068x580.jpg"},"td_1068x0":{"file":"fig-09-05-2019_07-40-40-1068x561.jpg","width":1068,"height":561,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-1068x561.jpg"},"td_0x420":{"file":"fig-09-05-2019_07-40-40-800x420.jpg","width":800,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/05\/fig-09-05-2019_07-40-40-800x420.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_07-40-40.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n