HCF & LCM - Practice Set 2

Q1. 468 can be expressed as as a product of prime as :a) 2×2×13×7×2×3 b) 2×2×13×7 c) 2×2×13×3×3 d) 2×2×3××7 e) None of these

Q2. A number n is said to be perfect if the sum of all its divisor (excluding n itself ) is equal to n. An example of perfect number is:
a) 27 b) 35 c) 21 d) 6 e) None of these


Q3. 70105/21035 when expressed in simplest form is
a) 203/601 b) 2003/603 c) 2003/601 d) 2001/603 e) None of these

Q4. H.C.F. of 22×33×55, 23×32×52×7 and 24×34×72×5×11 is :
a)22×32×5
b)22×32×5×7×11
c)24×34×55
d)24×34×55×7×11
e) None of these

Q5. 52×3×24×22×32×7 Find the L.C.M.
a) 12300 b) 12600 c) 24600 d) 25200

Q6. H.C.F. of 4×27×3125, 8×9×25×27 & 16×81×5×11×49 is:
a) 180 b) 360 c) 540 d) 1260 e) None of these

Q7. The greatest 5-digit number that is exactly divisible by 100 is:
a) 99899 b) 99800 c) 99900 d) 99889 e) None of these

Q8. What will be the remainder when (29)36 is divided by 28 ?
a) 0 b) 1 c) 29 d) 5 e) Cannot be determined

Q9. A number when divided by 627 leaves a remainder 43. By dividing the same number by 19, the remainder will be
a)19 b) 24 c) 43 d) 5 e) 13

Q10. The numbers 1, 3, 5 ... 25 are multiplied together. The number of zeroes at the right end of the product is :
a) 22 b) 8 c) 13 d) 6 e) 0

Q11. When a certain number is multiplied by 21, the product consist oof only fours. The smallest such number is:
a) 21164 b) 4444 c) 444444 d) 444 e) None of these

Q12. In a question, divisor is 2/3 of the dividend and twice the remainder. If the remainder is 5, then the dividend is
a) 85 b) 145 c) 225 d) 65 e) None of these

Solution

(1)
Sol : Option (c)
221333 = 468

(2)

Sol : Option (d)

n Divisors excluding n Sum of divisor
27
3✘9✘1
13
35
5✘7✘1
13
21
3✘7✘1
11
6
3✘2✘1
6

(3)

Sol : Option (c)
(4)
Sol: Option (a)

(5)
Sol:Option (d)
50+75+150+250+350
=875

(6)
Sol: Option (a)
4✘27✘3125=23✘33✘55
8✘9✘25✘7=23✘32✘52✘7 
16✘81✘5✘11*49=24✘34✘72✘5✘11
 H.C.F. = 22✘32✘7 = 180

(7)
Sol: Option (c)

(8)
Sol: Option (b)

(9)
Sol: Option (d)

(10)
Sol: Option (e)

(11)
Sol: Option (a)

(12)
Sol: Option (e)



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31 comments:

  1. 2017 aane do.....punjab mein e-badal bhi nahi milega.....SAD will be wiped up...time for AAP !!

    ReplyDelete
  2. Thank you :)

    ReplyDelete
  3. Post with full solution step by step it will be helpful for those who is novice,please...sir.

    ReplyDelete
  4. hello friends suggest me, rbi assistant material is effective ? yet to buy plz

    ReplyDelete
  5. Practice set paper 1 with explanation
    http://www.bankexamstoday.com/2015/07/hcf-lcm-quiz-part-1.html

    ReplyDelete
  6. khattar sir g kuch krna mat ..bus announcemnt krte rhna aap...

    ReplyDelete
  7. plz provide the solu of que no-8,.step by step

    ReplyDelete
  8. Q6. ans sud be option (c) 540.
    Q.wht will be the remainder when 2222^5555 + 5555^2222 is divided by 7?

    ReplyDelete
  9. Nivetha PrabhakaranJuly 3, 2015 at 10:39 PM

    sir can u plz guide me for RBI assistant plz rely for my questions which i have posted already plz sir

    ReplyDelete
  10. ha ha ha karenge bhai jarur lekin kb ye pta nhi

    ReplyDelete
  11. Sir pls post rbi assistant previous year model question paper

    ReplyDelete
  12. qn 8 :

    (29^36)=((28+1)^36)/28, obviously only the last term in the expansion, which is equal to 1 ., does not contain 28 as a factor...therefore, upon division , remainder is 1

    qn 9 :

    Let X be the number and k be the quotient when divided by 627
    We then have:

    X = 627*k + 43

    divided by 19 we get:

    X/19 = 627*k/19 + 43/19
    19 divides 627
    627 = 19*33
    43 = 2*19 + 5
    so we have:

    X/19 = 33*k + 2 + 5/19
    So we have

    X = 19*(33*k + 2) + 5

    Remainder is therefore 5



    qn 10.


    we get a zero by multiplying a 5 wid a 2 ( i.e d product of 5 and any evn no. ).. also numbers ending wid zero contribute to d same


    1 x 2 x...x 10 -> till this , we get 2 zeros ( one from 10 and oder by d multiplication of any evn no. between (1 - 10) and 5)
    simmilarly , for 11 x 12 x ...x 20 we get 2 zeros ( one from 20 and oder by d multiplication of any evn no between (11 - 20) and 15)
    for d product 21 x 22..x 25 ., see here , we hav 22 which is an evn no. and 24 which is agen an even no. and 25 is 5x5.. thus we get 2 pairs of an evn no. and 5., thus 2 zeros..
    so total 6 zeros


    shortcut : see., here it is 25!
    divide 25 by 5, u get 5 as qotient.. count this 5
    thn divide this 5 by 5.. u get 1 as qotient.. count this 1
    now total 5 +1 = 6.. so 6 zeros
    ( reason for dividing by 5 :- between 1 and 25 , we hav too many even numbers sompared to multiples of 5.. to get a zero, we must multiply one evn no wid one 5.. so its d minimum number of 5s which wil pair wid an equal no of evn nos to get a zero at d end)


    eider of d 2 methods... choose d 1 u r feeling comfortable
    ofcourse., dey hav printed d option wrong., i hav reported dem


    qn 11 :


    chk dividing wid 3 and 7
    start trying d min no. 44 - not possible.,
    444 - divisible by 3 , not divisible by 7 ( use divisiblity criteria for 7)
    4444 - not divisible by 3 and 7
    44444 - not divisible by 3 and 7
    444444 - divisible by 3 ( dont add., just see der r six 4s ., and as 6 is divisble by 3 , so dis no. is divisible by 3) and also by 7 ( use d number-sets divisiblity rule of 7 )


    acc to d qn : d no. multiplied by 21 gives 444444 => d no. is 444444/21 = 21164 ( option a)


    qn 12 :


    dividend = d
    qotient = q


    given that remainder = 5
    and divisor = 2 x 5 = 10


    also divisor = (2/3) x dividend
    => dividend = (3/2) x 10 = 15
    which is not mentioned in d options.. so none of diz :)
    okk ?? @Gurmeet D Angel

    ReplyDelete
  13. sir do chk d solutions of qn 6 and 10..

    in qn 6 :

    product 1 = 4 x 27 x 3 x 125

    product 2 = 8 x 27 x 9 x 25

    product 3 = 16 x 81 x 5x 11 x 49 = 16 x 27 x 3 x 5 x 11 x 49

    so hcf = 4 x 27 x 3 x 5 = 1620

    in ur solution , u hav tekn 4 as 2 ^3.. do corect this solution sir Vijay@BankExamsToday

    simmilarly in qn 10 :

    we get a zero by multiplying a 5 wid a 2 ( i.e d product of 5 and any evn no. ).. also numbers ending wid zero contribute to d same

    1 x 2 x...x 10 -> till this , we get 2 zeros ( one from 10 and oder by d multiplication of any evn no. between (1 - 10) and 5)
    simmilarly , for 11 x 12 x ...x 20 we get 2 zeros ( one from 20 and oder by d multiplication of any evn no between (11 - 20) and 15)
    for d product 21 x 22..x 25 ., see here , we hav 22 which is an evn no. and 24 which is agen an even no. and 25 is 5x5.. thus we get 2 pairs of an evn no. and 5., thus 2 zeros..



    so total 6 zeros.. plzz corect me if i am wrong...

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  14. solu are long...wats d short tricks...

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  15. go thru d solns otleast once.. dey seem to b long ., but dey r actualy not if u put thm on pen and paper ., also i hav gyvn a shortcut to qn 10.. do chk dat.. oders are short ones.. explanation is long..

    ReplyDelete
  16. hi... bhul gayi ?? ya stil firing at me ??

    nyways.. in qn no 6 ,
    product 1 = 4 x 27 x 3 x 125
    product 2 = 8 x 27 x 9 x 25
    product 3 = 16 x 81 x 5x 11 x 49 = 16 x 27 x 3 x 5 x 11 x 49

    so hcf = 4 x 27 x 3 x 5 = 1620.. how r u geting 540 ??



    and d qn dat u askd :
    2222 / 7 gives 3 as remainder and 5555/7 gives 4 as remainder
    d gyvn qn now simplifies as : 3^5 + 4^2 = 259.. divide this by 7.. u get 0 as d remainder.. and thus d remainder of d sum askd in d qn is also 0.. okk ??
    dont beliv ?? try working out on 6^11 + 11^6 remainder whn divided by 7..


    anyways.. anoder gud quality qn :)

    ReplyDelete
  17. ye to hme pta chlta h bhai jab yd krne pdte hain ...

    ReplyDelete
  18. in product 1: its "3125" not "3 x 125"
    n yeah.. if u remember me firing at u... den i think u sud also remember d way u were talking..!!

    ReplyDelete
  19. yeah.. 3125., i ovrlukd at it.. thanks for rectifying


    i was just asking ur ful name, but indirectly., u took it so seriously - m sorry ., but dat was nevr my intention.. why shud i talk rude to u.. i usd to solv ur qns coz any oder tym i dint get som gud qns to solve - but whn u posted som quality qns., it was brainstorming for me.. whn u askd me 2 share vedic maths and som oder tricks., i ced dat i can do it only on fb ., coz der r many pgs of my self-prepared notes n here i can upload only 1 snapshot per coment.. dats y i askd u - tel me ur ful name , lyk d status n i wil add u but i ced it indirectly.. u tuk it oderwise.. also i told u dat i am not simply i.. my initials are P and K.. wat r ur initials - dat was my qn..

    ReplyDelete
  20. in qn 10 :

    we get a zero by multiplying a 5 wid a 2 ( i.e d product of 5 and any evn no. ).. also numbers ending wid zero contribute to d same

    1 x 2 x...x 10 -> till this , we get 2 zeros ( one from 10 and oder by d multiplication of any evn no. between (1 - 10) and 5)
    simmilarly , for 11 x 12 x ...x 20 we get 2 zeros ( one from 20 and oder by d multiplication of any evn no between (11 - 20) and 15)
    for d product 21 x 22..x 25 ., see here , we hav 22 which is an evn no. and 24 which is agen an even no. and 25 is 5x5.. thus we get 2 pairs of an evn no. and 5., thus 2 zeros..

    so total 6 zeros.. plzz corect me if i am wrong...

    ReplyDelete
  21. Hello ramandeep sir

    ReplyDelete
  22. U can ask me anything directly..., will tell u what I can... Otherwise will say no... Simple..!! Wts d use of dis indirect way...., anyways my "FULL" name is Anu Thakur... is dat ok now??

    ReplyDelete
  23. Thanks a lot @Prateek for saving my time :)

    ReplyDelete
  24. @Prateek you can share your notes with all, send here - raman@bankexamstoday.com

    ReplyDelete
  25. Sir I have completed my btech in 2011 during filling up of banking forms they want percentage but college rewarded cpi how to convert that in percentage

    ReplyDelete
  26. Sir thanks for replying

    ReplyDelete

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