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) 2×2×3×3 ×7 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 2^{2}×3^{3}×5^{5}, 2^{3}×3^{2}×5^{2}×7 and 2^{4}×3^{4}×7^{2}×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. 5^{2}×3×2^{4}×2^{2}×3^{2}×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 5digit 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
(2)
Sol : Option (d)
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 theseQ3. ^{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 2^{2}×3^{3}×5^{5}, 2^{3}×3^{2}×5^{2}×7 and 2^{4}×3^{4}×7^{2}×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. 5^{2}×3×2^{4}×2^{2}×3^{2}×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 5digit 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)
Sol : Option (c)
2✘2✘13✘3✘3 = 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=2^{3}✘3^{3}✘5^{5}
8✘9✘25✘7=2^{3}✘3^{2}✘5^{2✘7}
16✘81✘5✘11*49=2^{4}✘3^{4}✘7^{2}✘5✘11
H.C.F. = 2^{2}✘3^{2}✘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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ReplyDeleteplz provide the solu of que no8,.step by step
ReplyDeleteQ6. ans sud be option (c) 540.
ReplyDeleteQ.wht will be the remainder when 2222^5555 + 5555^2222 is divided by 7?
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ReplyDeleteqn 8 :
ReplyDelete(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 numbersets 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
sir do chk d solutions of qn 6 and 10..
ReplyDeletein 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...
solu are long...wats d short tricks...
ReplyDeletego 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..
ReplyDeletehi... bhul gayi ?? ya stil firing at me ??
ReplyDeletenyways.. 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 :)
ye to hme pta chlta h bhai jab yd krne pdte hain ...
ReplyDeletein product 1: its "3125" not "3 x 125"
ReplyDeleten yeah.. if u remember me firing at u... den i think u sud also remember d way u were talking..!!
yeah.. 3125., i ovrlukd at it.. thanks for rectifying
ReplyDeletei 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 selfprepared 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..
in qn 10 :
ReplyDeletewe 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...
Hello ramandeep sir
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