# 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) #### What's trending in BankExamsToday

Smart Prep Kit for Banking Exams by Ramandeep Singh - Download here 1. 2017 aane do.....punjab mein e-badal bhi nahi milega.....SAD will be wiped up...time for AAP !!

2. Thank you :)

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

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

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

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

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

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

10. 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

11. ha ha ha karenge bhai jarur lekin kb ye pta nhi

12. Sir pls post rbi assistant previous year model question paper

13. 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

14. 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...

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

16. 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..

17. 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 :)

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

19. 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..!!

20. 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..

21. 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...

22. Hello ramandeep sir

23. 24. 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??

25. 26. Thanks a lot @Prateek for saving my time :)

27. what can I do for you ?

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

29. 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

30. 31. 