{"id":27235,"date":"2019-04-08T15:06:49","date_gmt":"2019-04-08T09:36:49","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=27235"},"modified":"2019-04-08T17:53:44","modified_gmt":"2019-04-08T12:23:44","slug":"rrb-je-hcf-lcm-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/rrb-je-hcf-lcm-questions-pdf\/","title":{"rendered":"RRB JE HCF & LCM Questions PDF"},"content":{"rendered":"

RRB JE HCF & LCM Questions PDF<\/strong><\/h1>\n

Download Top 20 RRB JE HCF & LCM Questions and Answers PDF. RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam<\/p>\n

Download RRB JE HCF & LCM Questions PDF<\/a><\/p>\n\n

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

Question 1:\u00a0<\/b>What is the product of LCM and HCF of two numbers having their product as 120 and sum as 22?<\/p>\n

a)\u00a0120<\/p>\n

b)\u00a0240<\/p>\n

c)\u00a0300<\/p>\n

d)\u00a0360<\/p>\n

Question 2:\u00a0<\/b>The HCF of two numbers 24 and their LCM is 216. If one of the number is 72, then the other number is<\/p>\n

a)\u00a027<\/p>\n

b)\u00a072<\/p>\n

c)\u00a08<\/p>\n

d)\u00a024<\/p>\n

Question 3:\u00a0<\/b>The product of two numbers is 2160 and their HCF is 12. Numbers of such possible pairs is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a04<\/p>\n

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

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

Question 4:\u00a0<\/b>A number between 1000 and 2000 which when divided by 30, 36 & 80 gives a remainder 11 in each case is<\/p>\n

a)\u00a01451<\/p>\n

b)\u00a01641<\/p>\n

c)\u00a01712<\/p>\n

d)\u00a01523<\/p>\n

Question 5:\u00a0<\/b>The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is<\/p>\n

a)\u00a04302<\/p>\n

b)\u00a04032<\/p>\n

c)\u00a04023<\/p>\n

d)\u00a04203<\/p>\n

Question 6:\u00a0<\/b>The difference between two numbers is 1146. When we divide the larger number by smaller we get 4 as quotient and 6 as remainder. Find the larger number.<\/p>\n

a)\u00a01526<\/p>\n

b)\u00a01431<\/p>\n

c)\u00a01485<\/p>\n

d)\u00a01234<\/p>\n

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Question 7:\u00a0<\/b>The sum of two numbers is 7 and the sum their squares is 23, their product is equal to:<\/p>\n

a)\u00a010<\/p>\n

b)\u00a011<\/p>\n

c)\u00a012<\/p>\n

d)\u00a013<\/p>\n

Question 8:\u00a0<\/b>Find the HCF of the following fractions.
\n$\\frac{5}{12}$, $\\frac{15}{14}$, $ \\frac{25}{21}$<\/p>\n

a)\u00a0$\\frac{10}{21}$<\/p>\n

b)\u00a0$\\frac{5}{84}$<\/p>\n

c)\u00a0$\\frac{5}{42}$<\/p>\n

d)\u00a0$\\frac{5}{21}$<\/p>\n

Question 9:\u00a0<\/b>Find the HCF of 3\/8, 17\/19 and 21\/23.<\/p>\n

a)\u00a01\/2392<\/p>\n

b)\u00a01\/3192<\/p>\n

c)\u00a01\/3128<\/p>\n

d)\u00a01\/3496<\/p>\n

Question 10:\u00a0<\/b>Find the HCF of 0.8, 0.125, 0.625 and 0.5.<\/p>\n

a)\u00a00.1<\/p>\n

b)\u00a01\/40<\/p>\n

c)\u00a01\/20<\/p>\n

d)\u00a01\/80<\/p>\n

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

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

Question 11:\u00a0<\/b>Find the LCM of 3\/4, 5\/8 and 9\/27.<\/p>\n

a)\u00a015<\/p>\n

b)\u00a045<\/p>\n

c)\u00a01\/24<\/p>\n

d)\u00a01\/48<\/p>\n

Question 12:\u00a0<\/b>Find the HCF of 8\/3, 12\/7 and 13\/12.<\/p>\n

a)\u00a01\/504<\/p>\n

b)\u00a01\/126<\/p>\n

c)\u00a01\/84<\/p>\n

d)\u00a01\/1008<\/p>\n

Question 13:\u00a0<\/b>Find the HCF of 2\/3 , 6\/8, 7\/12 and 2\/5<\/p>\n

a)\u00a01\/120<\/p>\n

b)\u00a01\/60<\/p>\n

c)\u00a01\/30<\/p>\n

d)\u00a01\/240<\/p>\n

Question 14:\u00a0<\/b>The LCM of two numbers is 4 times their HCF. The sum of LCM and HCF is 125. If one of the numbers is 100, then the other number is<\/p>\n

a)\u00a05<\/p>\n

b)\u00a025<\/p>\n

c)\u00a0100<\/p>\n

d)\u00a0125<\/p>\n

Question 15:\u00a0<\/b>What is the HCF of 7\/9, 2\/3, 5\/8 and 7\/12?<\/p>\n

a)\u00a01\/18<\/p>\n

b)\u00a01\/36<\/p>\n

c)\u00a01\/144<\/p>\n

d)\u00a01\/72<\/p>\n

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 <\/p>\n

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Question 16:\u00a0<\/b>What is the LCM of 64 and 56?<\/p>\n

a)\u00a0448<\/p>\n

b)\u00a0488<\/p>\n

c)\u00a0484<\/p>\n

d)\u00a0408<\/p>\n

Question 17:\u00a0<\/b>Find the HCF of $\\frac{2}{3}, \\frac{1}{2}, \\frac{3}{5}$<\/p>\n

a)\u00a06<\/p>\n

b)\u00a0$\\frac{1}{30}$<\/p>\n

c)\u00a0$\\frac{1}{15}$<\/p>\n

d)\u00a0$\\frac{1}{10}$<\/p>\n

Question 18:\u00a0<\/b>The 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

a)\u00a048<\/p>\n

b)\u00a036<\/p>\n

c)\u00a024<\/p>\n

d)\u00a016<\/p>\n

Question 19:\u00a0<\/b>Two numbers are in the ratio 3:4. Their L.C.M. is 84. The greater number is<\/p>\n

a)\u00a021<\/p>\n

b)\u00a024<\/p>\n

c)\u00a028<\/p>\n

d)\u00a084<\/p>\n

Question 20:\u00a0<\/b>The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is<\/p>\n

a)\u00a02\/35<\/p>\n

b)\u00a03\/25<\/p>\n

c)\u00a04\/35<\/p>\n

d)\u00a02\/25<\/p>\n

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

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

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

Given $a\\times b$=120
\na+b=22
\nSubstituting a=$\\frac{120}{b}$
\nWe have $\\frac{120}{b}$ + b=22
\n$b^{2}-22b+120$=0
\nSolving we get b=10,12
\nLCM of 10 and 12 =60
\nHCF of 10 and 12 =2
\nProduct of LCM and HCF=120.<\/p>\n

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

Let the number be $a$ and other number = $b=72$<\/p>\n

We know that\u00a0: $H.C.F(a,b)\\times L.C.M.(a,b)=a\\times b$<\/p>\n

=> $a\\times72=24\\times216$<\/p>\n

=> $a=\\frac{24\\times216}{72}$<\/p>\n

=> $a=24\\times3=72$<\/p>\n

=> Ans – (B)<\/p>\n

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

H.C.F. of the two numbers is 12, let the numbers be $12x$ and $12y$, where $x$ and $y$ are co-prime<\/p>\n

Product = $(12x)\\times(12y)=2160$<\/p>\n

=\u00a0$xy=\\frac{2160}{144}$<\/p>\n

=> $xy=15$<\/p>\n

Now, factors of 15 = 1, 3, 5, 15<\/p>\n

Thus, possible values of $(x,y)=(1,15),(3,5)$<\/p>\n

$\\therefore$ 2<\/strong> such pairs are possible.<\/p>\n

=> Ans – (B)<\/p>\n

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

LCM of given 3 numbers (30, 36, 80) = 720<\/p>\n

Multiple of 720 between 1000 and 2000 is 1440.<\/p>\n

$\\therefore$ Number which gives a remainder 11 in each case (1440 + 11) = 1451<\/p>\n

Hence, option A\u00a0is the correct answer.<\/p>\n

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

LCM of 12,18,21,32 is 252<\/p>\n

Multiples of 252 between 4000 and 5000 are 4032, 4284, 4536, 4788.<\/p>\n

4032 is present in the options.<\/p>\n

Hence, option B is the correct answer.<\/p>\n

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

Let the smaller number be $x$ and the larger number = $(x+1146)$<\/p>\n

According to ques, on dividing the larger term by smaller one,<\/p>\n

=> $(x+1146)=4x+6$<\/p>\n

=> $4x-x=1146-6$<\/p>\n

=> $3x=1140$<\/p>\n

=> $x=\\frac{1140}{3}=380$<\/p>\n

$\\therefore$ Larger number = $380+1146=1526$<\/p>\n

=> Ans – (A)<\/p>\n

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

Let the numbers be $x$ and $y$<\/p>\n

It is given that $x^2+y^2=23$ ———–(i)<\/p>\n

Also, $x+y=7$<\/p>\n

Squaring both sides, we get\u00a0:<\/p>\n

=> $x^2+y^2+2xy=49$<\/p>\n

=> $23+2xy=49$<\/p>\n

=> $2xy=49-23=26$<\/p>\n

=> $xy=\\frac{26}{2}=13$<\/p>\n

$\\therefore$ Product of the numbers =\u00a013<\/strong><\/p>\n

=> Ans – (D)<\/p>\n

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

$\\frac{5}{12}$, $\\frac{15}{14}$, $ \\frac{25}{21}$
\nWe know that HCF of fractions is given by
\n$\\frac{\\textrm{HCF of fractions}}{\\textrm{LCM of fractions}}$
\nSo applying this we can calculate the required fraction.
\nHCF of 5, 15 and 25 is 5.
\nLCM of 12, 14 and 21 is 4*3*7 = 84
\nHence, the required HCF will be $\\frac{5}{84}$<\/p>\n

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

HCF of fractions = HCF of the numerators \/ LCM of the denominators.
\nAll the fractions must be in their empirical form to apply this formula.<\/p>\n

The HCF of (3,17,21) is 1.
\nThe LCM of (8,19,23) is 3496.
\nTherefore, the HCF of the given fractions is 1\/3496.
\nHence, option D is the right answer.<\/p>\n

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

HCF of fractions = HCF of the numerators \/ LCM of the denominators.
\nAll the fractions must be in their empirical form to apply this formula.
\n0.8 = 4\/5
\n0.125 = 1\/8
\n0.625 = 5\/8
\n0.5 = 1\/2<\/p>\n

The given numbers can be written as 4\/5, 1\/8, 5\/8, 1\/2.
\nHCF of the numerators = 1.
\nLCM of the denominators (5,8,2) = 40.
\nTherefore, the HCF of the given numbers will be 1\/40 or 0.025.
\nTherefore, option B is the right answer.<\/p>\n

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

LCM of fractions = LCM of the numerators \/ HCF of the denominators.
\nAll the fractions must be in their empirical form to apply this formula.
\n9\/27 can be reduced to 1\/3.
\nTherefore, the given fractions are 3\/4, 5\/8 and 1\/3.
\nLCM of the numerators (3,5,1) is 15.
\nHCF of the denominators (4,8,3) is 1.
\nTherefore, the LCM is 15\/1 = 15.
\nHence, option A is the right answer.<\/p>\n

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

HCF of fractions = HCF of the numerators \/ LCM of the denominators.
\nAll the fractions must be in their empirical form to apply this formula.<\/p>\n

HCF of the numerators (8,12,13) is 1.
\nLCM of the denominators (3,7,12) is 84.<\/p>\n

Therefore, the HCF of the given fractions is 1\/84. Hence, option C is the right answer.<\/p>\n

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

HCF of fractions = HCF of the numerators \/ LCM of the denominators.
\nAll the fractions must be in their empirical form to apply this formula.<\/p>\n

6\/8 can be written as 3\/4.
\nThe given fractions are 2\/3, 3\/4, 7\/12 and 2\/5.
\nHCF of the numerators = 1.
\nLCM of (3,4,12 and 5) is 60.
\nTherefore, the HCF of the given fractions is 1\/60.
\nHence, option B is the right answer.<\/p>\n

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

Let one of the numbers = $x$ and other number = 100<\/p>\n

Let L.C.M = $L$ and H.C.F = $H$<\/p>\n

According to ques, => $L=4H$ ————–(i)<\/p>\n

and $L+H=125$<\/p>\n

Substituting value from equation (i), we get\u00a0: $4H+H=5H=125$<\/p>\n

=> $H=\\frac{125}{5}=25$<\/p>\n

=> $L=4\\times25=100$<\/p>\n

Thus, product of numbers = $L\\times H$<\/p>\n

=> $100\\times x=100\\times25$<\/p>\n

=> $x=25$<\/p>\n

=> Ans – (B)<\/p>\n

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

HCF of fractions = (HCF of the numerators)\/ (LCM of the denominators).<\/p>\n

HCF of 7,2,5 and 7 is 1.<\/p>\n

LCM of 9, 3, 8 and 12 is 72.
\nTherefore, the HCF of the given fractions is 1\/72.<\/p>\n

Option D is the right answer.<\/p>\n

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

(diagram)<\/p>\n

so LCM of 64 & 56 is = 8*8*7 = 448<\/p>\n

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

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

HCF = HCF of numerators\/ LCM of denominators<\/p>\n

HCF of 1, 2, 3 = 1<\/p>\n

LCM of 3, 2, 5 = 30<\/p>\n

Hence, $\\frac{1}{30}$<\/p>\n

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

Given:-
\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

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

Let the numbers be 3x, 4x
\nLCM of 3x and 4x is = 12x
\nSo the number 84 is divisible by 12
\n$\\frac{84}{12}$ = 7
\nThe numbers are 7×3 = 21 , 7x 4 = 28
\nThe greatest number is 28<\/p>\n

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

let’s say numbers are $x$ and $y$
\nhence sum of the reciprocals will be \u00a0$\\frac{1}{x} + \\frac{1}{y}$
\nor\u00a0$\\frac{x +y}{xy}$
\nas $x+y$ = 36 (given)
\nand $xy$ = $HCF \\times LCM$
\n= $3 \\times 105 = 315$
\nafter putting the values we will get summation of reciprocals equals to\u00a0$\\frac{4}{35}$<\/p>\n\n

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

RRB JE HCF & LCM Questions PDF Download Top 20 RRB JE HCF & LCM Questions and Answers PDF. RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam Download RRB JE Previous Papers PDF Question 1:\u00a0What is the product of LCM and HCF of two […]<\/p>\n","protected":false},"author":41,"featured_media":27237,"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":[489,491,1635,1647,1671],"class_list":{"0":"post-27235","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-je","14":"tag-rrb-mocks","15":"tag-rrb-mocks-test"},"better_featured_image":{"id":27237,"alt_text":"RRB JE HCF & LCM Questions PDF","caption":"RRB JE HCF & LCM Questions PDF","description":"RRB JE HCF & LCM 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