Last two digits of numbers ending in 3, 7 or 9
For all such numbers, it is easy to get the last 2 digits since now we know about a number ending in 1. Any number with last two digits ending in 3 and 7 can be converted into a number ending in 1 easily by raising it to power of 4.
eg 73 ^ 2 = 5329 which again squared will end in 1.
Number ending in 9 will have to be squared once.
Eg : to find last two digits
55557^987654
-->take the last two digits of the number as we are concerned with that only
57 in this case raised to the power 987654.
--> to make it into a number ending in 1.
(57) ^ 4 * 246913 * 57 ^ 2
--> 10556001 ^ 246913 * 3249
considering only the last 2 digits.
--> 1 ^ 246913 * 49 = 49 which would be the last two digits.
Similarly for 33 ^ 288
(33)^4*72 . 33^4 ends in 21. So finally we are interested in 21^72
It will definitely end in 1 and for tens digit, multiply the tens digit 21 with the unit digit of power that is 72 to give 4 which would be the tens digit. hence last two digits would be 41.
Hope it helps.
10. Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:
Solution
CP of each Article=5/6 Re
SP of each Article=6/5 Re
Profit=(6/5)-(5/6)=9/30=3/10
CP=5/6
Profit %age= (Profiit/CP)*100=*100=36%
Hope u all get it pls do tell me if there is confusion in any of the problems.
ATB
Rgrds
Myth Less
i feel there s a small correction here, Profit=(6/5)-(5/6)=11/30 ----> profit = (11/25)*100 = 44%
If a jar that fills up in 8 seconds is used to fill up a tank that fills up in 40 minutes, the time taken to cover the distance between the jar and the tank being 10 secs (person has to fill the jar walk upto the tank and pourin from the jar walk back and fill up the jar again).How long will it take a person to fill up the tank using the jar if it had a leak that empties the jar in 42 seconds?
subscribed
Sol:-
OR
form triplets from RHS of numbers .. add alternate triplets and subtract the sum hence obtained ..
321 + 987 + 345= 1653
654 + 678 + 12 = 1344
1653 - 1344 = 309
hence remainder is 309
Hey..this method is easier ..but will it apply to all such lengthy numbers ? i mean can we use the same method to all other Qs ?
Also, i guess topic wise approach that was being followed was much better than the one now .Please retain the previous one only .Some answers are being left unanswered because of this .
ajayjr SaysIf a jar that fills up in 8 seconds is used to fill up a tank that fills up in 40 minutes, the time taken to cover the distance between the jar and the tank being 10 secs (person has to fill the jar walk upto the tank and pourin from the jar walk back and fill up the jar again).How long will it take a person to fill up the tank using the jar if it had a leak that empties the jar in 42 seconds?
it takes total 18 sec for the process.
18n+ 8 =2400 (For the last process there is no need of 10 sec of going back and fro)
so n= 133 =>133 +1 =134 jar
it means the tank volume is equal to 134 jar.
Now the jar can be emptied in 48 sec. Considering 18 sec(not
to fill as in 18 sec, leakage will be continue to empty the jar.1*134 = (1-18/42) *x (included last 10 second to avoid calc)
x =235
so time will be 235*18+8 => 70 min (approximately)
@PS> please check.
plz post some number system problems...
i weak in this topic.. 
:
ajayjr SaysIf a jar that fills up in 8 seconds is used to fill up a tank that fills up in 40 minutes, the time taken to cover the distance between the jar and the tank being 10 secs (person has to fill the jar walk upto the tank and pourin from the jar walk back and fill up the jar again).How long will it take a person to fill up the tank using the jar if it had a leak that empties the jar in 42 seconds?
let total capacity of JAR = 336 ltr
Filling efficiency = 42 ltr/sec
leaking efficiency = 8 ltr/sec
In 18 sec tank fills up with 336 ltr
so , total capacity of tank = 44800 ltr
after leaking , In 18 sec tank fills up with 336 - 80 = 256 ltr
44,800/256 = 175
so , total time taken = 175*18 = 3150 sec or 52.5 minutes
Hello everyone. I am looking for few puys to form a study group who are seriously interested in CAT 2012; preferably who have taken CAT shot last year and are above 80+s %tilers, based in Delhi.
Any of you dedicated to a serious preparation, may pm me.
Cheerio!
Hello everyone. I am looking for few puys to form a study group who are seriously interested in CAT 2012; preferably who have taken CAT shot last year and are above 80+s %tilers, based in Delhi.
Any of you dedicated to a serious preparation, may pm me.
Cheerio!
This thread has some other purpose...!!! Please stop Spamming out here...!!! If you have any doubts in Quant get them cleared here or help others by solving their doubts.....!!!! Kindly refrain from doing this again....!!!!
OK This one for u
For what value of 'n' will the remainder of 351n and 352n be the same when divided by 7?
a.2
b.3
c.6
d.4
the answer is 3
How many keystrokes are needed to type numbers from 1 to 1000?
3001
2893
2704
2890
its 2893 .correct me if i am wrong......
How many keystrokes are needed to type numbers from 1 to 1000?
3001
2893
2704
2890
A quesion which can be solved by good funda of number patterns
1 to 1000 is formed of numbers from 0 to 9.. Totally 10 digits
From 1- 999 - digits from 1 to 9 occur 300 times - so 300 * 9 = 2700
For digit 0 ,
1-100 - occurs 11 times
101 - 900 - 20 *8 = 160 times
901 - 1000 = 21 times...So 0 occurs 192 times between 1 to 1000
We counted all digits from 1 to 999 and all zeroes from 1 to 1000
So we have to add the '1' of 1000 too..
So total
2700 +192 +1 = 2893
poojachhoker Saysthe answer is 3
hi,
can you please explain the solution for this....
hi,
can you please explain the solution for this....
when n=2
then 351^2/7 then r is 1.
and 352^2/7 then r is 4
n=3
351^3/7... r=1
352^3/7 r=8,as 8 is greater than 7...so dividing 8 by 7 ,remainder is 1..
hence ans is 3...
How many keystrokes are needed to type numbers from 1 to 1000?
3001
2893
2704
2890
My take
2893--single digit numbers - 1 to 9 = 9*1 = 9 key strokes
--two digit numbers - 10 to 99 = 99-9 = 90 numbers =90*2= 180 key strokes
--three digit numbers = 100 - 999 = 999-99 = 900 numbers = 900*3 = 2700 keystrokes
--four digit numbers = 1 number (1000) = 4 keystrokes
total number of keystrokes = 9+180+2700+4 = 2893
OK This one for u
For what value of 'n' will the remainder of 351n and 352n be the same when divided by 7?
a.2
b.3
c.6
d.4
here '351n' and '352n' are two four digit numbers right..
In that case none of the options would satisfy.. Correct me if i am wrong. I feel there is an error in the question
sanu8080 SaysFind the remainder when 1212..1212(300 digits) is divided by 999 ?
My take
666Take three digits at a time as a single number. Add them up
121+212+121+....212
we will have 121 repeated 50 times and 212 repeated 50 times
sum = 121*50+212*50 = 16650
Now find the remainder when 16650 is divided by 999 = 666
Hey I am very weak in Algebra concepts...went through the CL material..but still i am having problems. Please suggest some good sources for concepts and solved problems
My take666
Take three digits at a time as a single number. Add them up
121+212+121+....212
we will have 121 repeated 50 times and 212 repeated 50 times
sum = 121*50+212*50 = 16650
Now find the remainder when 16650 is divided by 999 = 666
why will they be repeated 50 times ?