{"id":44127,"date":"2021-01-19T12:00:18","date_gmt":"2021-01-19T06:30:18","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=44127"},"modified":"2021-01-19T12:00:18","modified_gmt":"2021-01-19T06:30:18","slug":"top-20-rrb-ntpc-lcm-and-hcf-questions","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/top-20-rrb-ntpc-lcm-and-hcf-questions\/","title":{"rendered":"Top-20 RRB NTPC LCM AND HCF Questions"},"content":{"rendered":"

RRB NTPC LCM AND HCF Questions<\/strong><\/h1>\n

Download Top-20 LCM and HCF questions for RRB NTPC exam 2020. Most important\u00a0 LCM and HCF Questions based on asked questions in previous exam papers for RRB NTPC. The following questions are helpful to crack the RRB NTPC Exam 2020.<\/p>\n

Download Top-20 RRB NTPC LCM AND HCF Questions <\/a><\/p>\n

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Question 1:\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 2:\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 3:\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

Question 4:\u00a0<\/b>L.C.M. of two numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers?<\/p>\n

a)\u00a0140<\/p>\n

b)\u00a080<\/p>\n

c)\u00a060<\/p>\n

d)\u00a070<\/p>\n

Question 5:\u00a0<\/b>Product of two coprime numbers is 117. Then their LCM is<\/p>\n

a)\u00a09<\/p>\n

b)\u00a013<\/p>\n

c)\u00a039<\/p>\n

d)\u00a0117<\/p>\n

Question 6:\u00a0<\/b>HCF and LCM of two numbers are 11 and 825 respectively. If one number is 275 find the other number.<\/p>\n

a)\u00a053<\/p>\n

b)\u00a045<\/p>\n

c)\u00a033<\/p>\n

d)\u00a043<\/p>\n

Question 7:\u00a0<\/b>What is the LCM (least common multiple) of 57 and 93?<\/p>\n

a)\u00a01767<\/p>\n

b)\u00a01567<\/p>\n

c)\u00a01576<\/p>\n

d)\u00a01919<\/p>\n

Question 8:\u00a0<\/b>What is the HCF (highest common factor) of 57 and 513?<\/p>\n

a)\u00a010<\/p>\n

b)\u00a057<\/p>\n

c)\u00a03<\/p>\n

d)\u00a027<\/p>\n

Question 9:\u00a0<\/b>The two numbers are 63 and 77, HCF is 7, Find the LCM.<\/p>\n

a)\u00a0668<\/p>\n

b)\u00a0693<\/p>\n

c)\u00a0674<\/p>\n

d)\u00a0680<\/p>\n

Question 10:\u00a0<\/b>What is the HCF (highest common factor) of 77 and 275?<\/p>\n

a)\u00a012<\/p>\n

b)\u00a011<\/p>\n

c)\u00a07<\/p>\n

d)\u00a025<\/p>\n

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Question 11:\u00a0<\/b>The two numbers are 55 and 99, HCF is 11, What is their LCM?<\/p>\n

a)\u00a0486<\/p>\n

b)\u00a0479<\/p>\n

c)\u00a0476<\/p>\n

d)\u00a0495<\/p>\n

Question 12:\u00a0<\/b>What is the HCF (highest common factor) of 133 and 112?<\/p>\n

a)\u00a015<\/p>\n

b)\u00a07<\/p>\n

c)\u00a019<\/p>\n

d)\u00a016<\/p>\n

Question 13:\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 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>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

Get 20 RRB NTPC Mocks – Just Rs. 149<\/a><\/p>\n

Question 16:\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

Question 17:\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 18:\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 19:\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

Question 20:\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

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

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

1)\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

2)\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

3)\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

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

We assume that numbers are $hr_1$ and $hr_2$ (where h= H.C.F. of numbers and $r_1$\u00a0and $r_2$ are prime factors)
\nSo L.C.M. will be = $hr_1r_2$ = 120
\nor $r_1r_2$ = 12
\nSo $r_1$ =4 and $r_2$ = 3 ; numbers will be 40 and 30, sum is 70
\nor $r_1$ = 12 and $r_2$ = 1 ; numbers will be 120 and 10, sum is 130
\nHence only option D justifies.<\/p>\n

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

Let the two numbers be a,b.
\nHence a * b = L.C.M(a,b) * G.C.D(a,b)
\nIt is given that a,b are co-primes, implies G.C.D(a,b) = 1
\nHence from the above equation we get L.C.M(a,b) = a*b = 117<\/p>\n

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

Let the number = $x$<\/p>\n

HCF = 11 and LCM = 825<\/p>\n

Product of HCF and LCM = Product of the two numbers<\/p>\n

=> $x \\times 275 = 11 \\times 825$<\/p>\n

=> $x = \\frac{11 \\times 825}{275}$<\/p>\n

=> $x = \\frac{825}{25} = 33$<\/p>\n

=> Ans – (C)<\/p>\n

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

Prime factorization of 57 = 3 $\\times$ 19<\/p>\n

Prime factorization of 93 = 3 $\\times$ 31<\/p>\n

=> L.C.M. of 57 and 93 = 3\u00a0$\\times$\u00a019\u00a0$\\times$\u00a031<\/p>\n

= 57\u00a0$\\times$\u00a031 = 1767<\/p>\n

=> Ans – (A)<\/p>\n

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

Factors of 57 = 1 , 3 , 19 , 57<\/p>\n

Factors of 513 = 1 , 3 , 9 , 19 , 27 , 57 , 171 , 513<\/p>\n

The common factors are = 1 , 3 , 19 , 57<\/p>\n

=> Highest common factor = 57<\/p>\n

=> Ans – (B)<\/p>\n

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

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

The numbers a = 63 and b = 77 and HCF = 7<\/p>\n

=> L.C.M. = $\\frac{a \\times b}{HCF}$<\/p>\n

= $\\frac{63 \\times 77}{7} = 63 \\times 11$<\/p>\n

= 693<\/p>\n

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

Factors of : 77 = 1 , 7 , 11 , 77<\/p>\n

275 = 1 , 5 , 11 , 25 , 55 , 275<\/p>\n

The common factors are 1 and 11<\/p>\n

and HCF = 11<\/p>\n

=> Ans – (B)<\/p>\n

Free RRB Preparation Videos<\/a><\/p>\n

 <\/p>\n

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

Let the LCM = $x$<\/p>\n

Numbers are = 55 , 99<\/p>\n

Also, product of numbers = HCF $\\times$ LCM<\/p>\n

=> $55 \\times 99 = 11 \\times x$<\/p>\n

=> $x = \\frac{55 \\times 99}{11} = 5 \\times 99$<\/p>\n

=> $x = 495$<\/p>\n

=> Ans – (D)<\/p>\n

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

Prime factorization of<\/p>\n

133 = 7\u00a0$\\times$ 19<\/p>\n

112 = $2^4$\u00a0$\\times$\u00a07<\/p>\n

There is only 1 common factor, and thus\u00a0the HCF (highest common factor)\u00a0= 7<\/p>\n

=> Ans – (B)<\/p>\n

13)\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

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

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

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16)\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

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

LCM of 12,18,21,32 is 2016<\/p>\n

Multiples of 2016 between 4000 and 5000 are 4032.<\/p>\n

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

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

18)\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

19)\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

20)\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

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 <\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

RRB NTPC LCM AND HCF Questions Download Top-20 LCM and HCF questions for RRB NTPC exam 2020. Most important\u00a0 LCM and HCF Questions based on asked questions in previous exam papers for RRB NTPC. The following questions are helpful to crack the RRB NTPC Exam 2020. Take a free RRB NTPC mock test Download RRB […]<\/p>\n","protected":false},"author":53,"featured_media":44129,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3171,31,1603],"tags":[166,1593],"class_list":{"0":"post-44127","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-railways-en","8":"category-railways","9":"category-rrb-ntpc","10":"tag-railways","11":"tag-rrb-ntpc"},"better_featured_image":{"id":44129,"alt_text":"Top 20 RRB NTPC LCM And HCF Questions","caption":"Top-20 RRB NTPC LCM And HCF Questions ","description":"Top-20 RRB NTPC LCM And HCF Questions 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