Cool.... @coolparul007 sab sahi hai...
ok puys..........bbye....gd ngt...........
@anticipator.lad said:2) If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:A. 55 / 601B. 601 / 55C. 11 / 120D. 120 / 11
bhai isme
ek no. 5a
dusra 5b
formula lagalo.
5a*5b=5*120
ab=24 eq 1
5a+5b=55
a+b=11
only 1 value can be so.
8, 3
so the number will be 8*5 and 3*5
40 and 15
ab iske reciprocal ka sum.
11/120
ek no. 5a
dusra 5b
formula lagalo.
5a*5b=5*120
ab=24 eq 1
5a+5b=55
a+b=11
only 1 value can be so.
8, 3
so the number will be 8*5 and 3*5
40 and 15
ab iske reciprocal ka sum.
11/120
@anticipator.lad said:3) The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:A. 3 B. 13C. 23 D. 334) The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:A. 74 B. 94C. 184 D. 364
3 ka answer hai 23
2497 mein option mein add kar lete hai
2500 2510 2520 2530
isme se only 2520 hi 3 se divisible hai bhaiya.
4.
answer 4
as 364 is the only no. divisible by 7.
bhai kisi ka sbi clerk ka exam hai sun ko??????hwz prep ?????any last movement advice??????????
@ibpspo2012 said:bhai kisi ka sbi clerk ka exam hai sun ko??????hwz prep ?????any last movement advice??????????
last moment advice --> "keep your brains working"...
i mean solve practise sets of SBI clerk level till tomorrow evening...and then sharpen your pencils and let them rest for 12 hours. :)
All the best bro!
plz solve this...
sm 2:
find the last two digits of 15*37*63*51*97*17
op:35,45,55,85
@Gubluu said:plz solve this...sm 2:find the last two digits of 15*37*63*51*97*17op:35,45,55,85
ans is 35
@anticipator.lad said:Bro please explain
tis qus is done by reminder concept u have to divide 15*37*51*63*17*97 by 100 and then find reminder as 15*37= 555 reminder is 55 51*63= 3213 reminder is 13 and 97*17=1749 so reminder is 49 and again 55*13*49/100
55*13=715 reminder is 15 and 15*49= 735 so finaly reminder is 35
@Gubluu
Find last two digits of : 15*37*63*51*97*17
last digit of 37*63*51*97*17=7*3*1*7*7=9 and 9 with 15=135 so last digit=35
This works only when, 2 digit to 2 digit multiplication is there and only when one value out of whole is ending at '5'..in this case its 15
So keep the 15 aside...do the rest unit digit multiplication and then fully multiply it with 15
@anuragtomar said:tis qus is done by reminder concept u have to divide 15*37*51*63*17*97 by 100 and then find reminder as 15*37= 555 reminder is 55 51*63= 3213 reminder is 13 and 97*17=1749 so reminder is 49 and again 55*13*49/10055*13=715 reminder is 15 and 15*49= 735 so finaly reminder is 35
Bro, i am not good in arithmetic. Sorry! Itna calculation karte karte to mein thak jaunga... 😞
@anticipator.lad said:Bro, i am not good in arithmetic. Sorry! Itna calculation karte karte to mein thak jaunga...
sorry bhai muche yahi method pata tha vase gubluu bhai ka method tumhare liye sahi hai
@anuragtomar said:sorry bhai muche yahi method pata tha vase gubluu bhai ka method tumhare liye sahi hai
gubluu bhai ne kounsa method diya?? :O
@anticipator.lad said:@GubluuFind last two digits of : 15*37*63*51*97*17last digit of 37*63*51*97*17=7*3*1*7*7=9 and 9 with 15=135 so last digit=35This works only when, 2 digit to 2 digit multiplication is there and only when one value out of whole is ending at '5'..in this case its 15So keep the 15 aside...do the rest unit digit multiplication and then fully multiply it with 15
bhai mein toh fida ho gaya is answer pe...gajab :)
Results for ibps po 2012 are out check
i got thru 😃 