{"id":31502,"date":"2019-07-12T18:56:50","date_gmt":"2019-07-12T13:26:50","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=31502"},"modified":"2019-07-16T17:08:51","modified_gmt":"2019-07-16T11:38:51","slug":"elementary-statistics-questions-for-rrb-group-d-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/elementary-statistics-questions-for-rrb-group-d-pdf\/","title":{"rendered":"Elementary Statistics Questions for RRB Group-D PDF"},"content":{"rendered":"

Elementary Statistics Questions for RRB Group-D PDF<\/strong><\/span><\/h2>\n

Download RRB NTPC Elementary Statistics Questions and Answers PDF. Top 10 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n

Download Elementary Statistics Questions for RRB Group-D PDF<\/a><\/p>\n\n

Download RRB Group-D Previous Papers PDF<\/a><\/p>\n

Take a RRB Group-D free mock test<\/a><\/p>\n

Question 1:\u00a0<\/b>Four integers w,x,y,z are selected at random from 0 to 1000 numbers both inclusive. The probability that wz-xy is even is<\/p>\n

a)\u00a0$\\frac{1}{16}$<\/p>\n

b)\u00a0$\\frac{5}{8}$<\/p>\n

c)\u00a0$\\frac{9}{16}$<\/p>\n

d)\u00a0$\\frac{5}{16}$<\/p>\n

Question 2:\u00a0<\/b>$x$ and $y$ are integers such that $\\dfrac{3}{x} + \\dfrac{2}{y} = \\dfrac{1}{12} $. What is the sum of all such integral values of $y$?<\/p>\n

Question 3:\u00a0<\/b>If $f(x + 2) = f(x) + f(x + 1)$ for all positive integers x, and $f(11) = 91, f(15) = 617$, then $f(10)$ equals<\/p>\n

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Question 4:\u00a0<\/b>$|7x-5|<16$ and $|16y+11|<37$. If $x$ and $y$ are integers then find the maximum value of $|x+7|*|y-5|$<\/p>\n

a)\u00a063<\/p>\n

b)\u00a042<\/p>\n

c)\u00a081<\/p>\n

d)\u00a0None of These<\/p>\n

Question 5:\u00a0<\/b>The arithmetic mean of 9 distinct integers is 87. If none of the numbers is more than 100 and the average of the smallest five numbers is 78, find the minimum value of the sixth number.<\/p>\n

Question 6:\u00a0<\/b>If $f$, $g$, $h$ are all positive integers and $f+g+h = 24$, then what is the maximum value of $(f-4)(g-2)(h-6)$ provided (f-4), (g-2), and (h-6) are positive integers?<\/p>\n

RRB Group D previous year papers<\/a><\/p>\n

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Question 7:\u00a0<\/b>The sum of five integers is 5. What is the minimum value of the sum of reciprocal of the five numbers?<\/p>\n

a)\u00a0– 45\/7<\/p>\n

b)\u00a0-35\/9<\/p>\n

c)\u00a05<\/p>\n

d)\u00a0-2<\/p>\n

e)\u00a0none of these<\/p>\n

Question 8:\u00a0<\/b>For two positive integers a and b define the function h(a,b):as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the GCF of the elements of set A is computed by repeatedly using the function h.
\nThe minimum number of times h is required to be used to compute G is:<\/p>\n

a)\u00a01\/2 n<\/p>\n

b)\u00a0(n – 1)<\/p>\n

c)\u00a0n<\/p>\n

d)\u00a0None of these<\/p>\n

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

Question 9:\u00a0<\/b>Find b-a if the arithmetic mean of ordered set of integers S={5, a, 13, 16, b, 24} is 14?<\/p>\n

a)\u00a08<\/p>\n

b)\u00a010<\/p>\n

c)\u00a012<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

Question 10:\u00a0<\/b>Find b-a if the arithmetic mean of ordered set of non-repeating integers S={8, a, 11, 14, 17, 21, b, 24} is 16?<\/p>\n

a)\u00a013<\/p>\n

b)\u00a014<\/p>\n

c)\u00a016<\/p>\n

d)\u00a0Can’t be determined<\/p>\n

RRB Group-D Important Questions (download PDF)<\/a><\/p>\n

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

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

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

wz-xy will be even only when both wz and xy are even or odd
\nThe probability of wz being odd is $\\frac{1}{2}$*$\\frac{1}{2}$
\nProbability for wz -xy being odd is $(\\frac{1}{4}^)2$
\nSimilarly, the probability of wz to be even is 1-probability of being odd=$\\frac{3}{4}$
\nProbability for wz -xy being even is $(\\frac{3}{4}^)2$
\nTotal probability=$\\frac{1}{16}$+$\\frac{9}{16}$
\n=$\\frac{5}{8}$
\nB is the correct answer.<\/p>\n

2)\u00a0Answer:\u00a01152<\/b><\/p>\n

$\\dfrac{3}{x} + \\dfrac{2}{y} = \\dfrac{1}{12} $<\/p>\n

$ \\Rightarrow \\dfrac{2x+3y}{xy} = \\dfrac{1}{12} $<\/p>\n

$ \\Rightarrow 24x+36y = xy $<\/p>\n

$ \\Rightarrow 24x-xy+36y = 0 $<\/p>\n

This can be expressed as<\/p>\n

$ \\Rightarrow\u00a024x-xy+36y+(36 \\times 24)-(36 \\times 24) = 0 $<\/p>\n

$ \\Rightarrow x(24-y)-36(24-y)+(36 \\times 24)=0 $<\/p>\n

$ \\Rightarrow (x-36)(24-y)= -864 $<\/p>\n

For the integral values of $y$, $(24-y)$ must be an integer as well.<\/p>\n

This means that $24-y$ will be one of the factors of 864.<\/p>\n

Now, let us consider a case.<\/p>\n

2 is a factor of 864. For 24-y to be equal to 2, y = 22<\/p>\n

-2 is also a factor of 864. For 24-y to be equal to -2, y = 26<\/p>\n

We know that $864 = 2^5 \\times 3^3 $<\/p>\n

Number of factors of 864 = 6 x 4 = 24<\/p>\n

We need to consider that 864 has 24 positive and 24 negative factors.<\/p>\n

Thus, the sum of negative and non-negative factors of 864 = 0<\/p>\n

Number of negative and non-negative factors = 24+24 = 48<\/p>\n

Thus, the sum of values of $y$ = $24 \\times 48 = 1152$<\/p>\n

3)\u00a0Answer:\u00a054<\/b><\/p>\n

$f(x + 2) = f(x) + f(x + 1)$
\nAs we can see, the value of a term is the sum of the 2 terms preceding it.<\/p>\n

It has been given that\u00a0$f(11) = 91$ and $f(15) = 617$.
\nWe have to find the value of\u00a0$f(10)$.<\/p>\n

Let\u00a0$f(10)$ = b
\n$f(12)$ = b + 91
\n$f(13)$ = 91 + b +\u00a091 = 182 + b
\n$f(14)$ = 182+b+91+b = 273+2b
\n$f(15)$ = 273+2b+182+b = 455+3b
\nIt has been given that 455+3b = 617
\n3b = 162
\n=> b = 54<\/p>\n

Therefore, 54 is the correct answer.<\/p>\n

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

$|7x-5|<16$
\nTherefore, $-16<7x-5<16$
\n=> $-11<7x<21$
\n=>\u00a0$\\frac{-11}{7}<x<3$<\/p>\n

Thus, the maximum value of $|x+7|$ is when $x = 2$.
\nThus, the maximum value of $|x+7|$ is $9$
\n$|16y+11|<37$
\nHence, $-37<16y+11<37$
\n=> $-48<16y<37$
\n=> $-3<y<\\frac{37}{16}$<\/p>\n

Hence, the maximum value of $|y-5|$ will be when $y=-2$
\nThus, the maximum value of $|y-5|$ is $7$
\nTherefore, the maximum value of $|x+7|*|y-5|$ is $9*7 = 63$
\nHence, option A is the correct answer.<\/p>\n

5)\u00a0Answer:\u00a096<\/b><\/p>\n

If the average of 9 numbers is 87, then the sum of these 9 distinct numbers will be 9 X 87 = 783
\nLet the numbers be a1, a2, a3, \u2026..a9 where a9> a8> \u2026a1.
\nSo, a1+a2+a3+a4+a5 = 78 X 5 = 390
\nSmallest value of a5 can be 80 when a1, a2, a3, a4, and a5 are 76, 77, 78, 79, 80.
\nThis means that a6 > 80.
\nNow, the sum of the rest of the four numbers is 738-390 = 393
\nFor a6 to be min, a7, a8, a9 must be max.
\n=> a6 + 98 + 99 + 100 = 393 or a6 = 96.
\nThus, 96 is the correct answer.<\/p>\n

6)\u00a0Answer:\u00a064<\/b><\/p>\n

If $f+g+h = 24$
\nThen $(f-4)+(g-2)+(h-6) = 12$
\nMaximum value of $(f-4)(g-2)(h-6)$ is obtained when each of the three terms are equal
\n$f-4 = g-2 = h-6 =4$
\nTherefore, the required maximum value is $4*4*4 = 64$.<\/p>\n

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

Let the five integers be a, b, c, d and e. Given, a + b + c + d + e = 5
\nWe have to find the minimum value of $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{1}{e}$
\nThe above equation would be minimum if its value is negative and magnitude is maximum.
\nThis occurs when a = b = c = d = -1 and e = 9
\nThus, the minimum value is -1 + -1 + -1 + -1 + 1\/9 = -35\/9<\/p>\n

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

Let p and q be any two elements of the set A.
\nFor the computation of the GCF of elements of the set A, we can replace both p and q by just the GCF(p,q)\u00a0and the result is unchanged.<\/p>\n

So, for every application of the function h, we are reducing the number of elements of the set A by 1. (In this case two numbers p and q are replaced by one number GCF(p,q)).
\nExpanding this concept further, the minimum number of times the function h should be called is n-1<\/p>\n

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

The arithmetic mean of the numbers is the average of the numbers.
\nHence, from the given information, we know that the sum of all the numbers is 14*6 = 84.
\nSo, a+b=14*6-58=26.
\nFrom the set ordering, ‘b’ can be from 17-21 with ‘a’ correspondingly being 9-5. Hence, answer cannot be determined.<\/p>\n

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

As the mean of S is 16, (8 + a + 11 + 14 + 17 + 21 + b + 24)\/8 = 16<\/p>\n

=> (a + b + 95)\/8 = 16<\/p>\n

Thus, a+b=16*8-95=33.
\nFrom the set ordering, b can be 22 or 23 and a can be 9 or 10.
\nAs only one pair 23+10 exists which equals to 33, b-a=23-10=13<\/p>\n\n

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

We hope this Elementary Questions\u00a0 PDF for RRB Group-D Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Elementary Statistics Questions for RRB Group-D PDF Download RRB NTPC Elementary Statistics Questions and Answers PDF. Top 10 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Download RRB Group-D Previous Papers PDF Take a RRB Group-D free mock test Question 1:\u00a0Four integers w,x,y,z are […]<\/p>\n","protected":false},"author":41,"featured_media":31552,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[1],"tags":[],"class_list":{"0":"post-31502","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-uncategorized"},"better_featured_image":{"id":31552,"alt_text":"Elementary Statistics Questions for RRB Group-D PDF","caption":"Elementary Statistics Questions for RRB Group-D PDF","description":"Elementary Statistics Questions for RRB Group-D 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