Can anyone tell me the approach to this problem
In an examination,the maximum marks for each of the three papers is 50 each.THe maximum marks for the fourth paper is 100.Find the number of ways with which a student can score 60% marks in aggregate.
pls it would be very helpful if someone tells me the solution quickly
hi puys i dont know if this question has been posted before or not,,this thread is so big cant search the whole thread ,so posting this question:
find the 28383rd term of the series:123456789101112........
(a)3
(b)4
(c)9
(d)7
hi puys i dont know if this question has been posted before or not,,this thread is so big cant search the whole thread ,so posting this question:
find the 28383rd term of the series:123456789101112........
(a)3
(b)4
(c)9
(d)7
I think this question has been replied to umpteen number of times, after every few pages. I think you need not even look for this in the whole thread, just a glance at the last few pages would do.
two joggers left delfi for noida simultaniously.First Jogger stopped after 42 mins later when he was 1 km short of noida.The second Jogger stopped after 52 mins when he was 2 km short of noida.if both of them travelled same distance.The frist jogger would take 17 mins less than second jogger.Find the distance between the delhi and noida.
Options 1>5km 2>15 3>25 d>35
CAn someone help me with solution
can anyone pls help me with soln of following prblms:
1) x-6> x-5x+9
2) x-6
two joggers left delfi for noida simultaniously.First Jogger stopped after 42 mins later when he was 1 km short of noida.The second Jogger stopped after 52 mins when he was 2 km short of noida.if both of them travelled same distance.The frist jogger would take 17 mins less than second jogger.Find the distance between the delhi and noida.
Options 1>5km 2>15 3>25 d>35
CAn someone help me with solution
are you sure the first jogger took 17 mins less??its shud be 15..
madnikhil Saysare you sure the first jogger took 17 mins less??its shud be 15..
As per the question it is 17
two joggers left delfi for noida simultaniously.First Jogger stopped after 42 mins later when he was 1 km short of noida.The second Jogger stopped after 52 mins when he was 2 km short of noida.if both of them travelled same distance.The frist jogger would take 17 mins less than second jogger.Find the distance between the delhi and noida.
Options 1>5km 2>15 3>25 d>35
CAn someone help me with solution
madnikhil Saysare you sure the first jogger took 17 mins less??its shud be 15..
bhupi3149 SaysAs per the question it is 17
gone through options is it 15??
madnikhil Saysgone through options is it 15??
Dont remember the answer can you post the approach you followed
bhupi3149 SaysDont remember the answer can you post the approach you followed
option a) 5km, he had covered 4km in 42mins..10min 30sec each km..so the total distance will be traveled in 52mins 30secs..
and 2nd person will take around 28 mins more..
similarly option b is somewhere satisfying the givin conditions..
speed of 1st = 3min/km so total distance 45 mins
speed of 2nd = 4min/km so total distance travelled in 60mins..
option 3 and 4 both were also not satisfying the condition..
that's y gone with option b..
hope it helps..
Is the question worded correctly? What does "if both of them travelled same distance.The frist jogger would take 17 mins less than second jogger." What is the distance here?
ShortCourage SaysIs the question worded correctly? What does "if both of them travelled same distance.The frist jogger would take 17 mins less than second jogger." What is the distance here?
done it by very foolish method.....but tried and interpreted it in some other way also which i think is the correct way but not able to get the answer..
it is something like this: that 1st person have traveled for another 42 mins after 1km, and similarly with the 2nd person traveled for another 52 mins after the time when he was 2km apart...but not getting any answer..
puys need ur help...
the latest registration number issued by delhi authority is DL-5S 2234. If all the no.s and alphabets before this have been used then find how many vehicles have no. starting with DL-5?
a)192234 b)192225 c)172227 d)182216 e)none of these
thank you really for this thread, i am new to the forum and need to clear some concepts and quant problems.
:o
the latest registration number issued by delhi authority is DL-5S 2234. If all the no.s and alphabets before this have been used then find how many vehicles have no. starting with DL-5?
a)192234 b)192225 c)172227 d)182216 e)none of these
thank you really for this thread, i am new to the forum and need to clear some concepts and quant problems.:o
my take:
option d (182216)..
if correct will post the approach..
the latest registration number issued by delhi authority is DL-5S 2234. If all the no.s and alphabets before this have been used then find how many vehicles have no. starting with DL-5?
a)192234 b)192225 c)172227 d)182216 e)none of these
thank you really for this thread, i am new to the forum and need to clear some concepts and quant problems.
As we have to calculate the total No of vehicles till DL-5S 2234 So we will take all the alphabets from DL-5A to DL-5R and in one alphabet there are 9999 Nos so total till R is 179982 and then add 2234 for S series and the total will come out to be 182216...:o
Hope this helps. ATB
my take:
option d (182216)..
if correct will post the approach..
ya its the right answer
thanks for the method!!!!
As we have to calculate the total No of vehicles till DL-5S 2234 So we will take all the alphabets from DL-5A to DL-5R and in one alphabet there are 9999 Nos so total till R is 179982 and then add 2234 for S series and the total will come out to be 182216...:o
Hope this helps. ATB
thanks for the explaination, A.G Rox
i have a problem in (Q)53 LOD III NUMBER SYSTEM 
anushreebhatt7 Saysi have a problem in
---
128^1000 mod 153
=128^40 mod 153 (euler no 96)
by chinese remainder theorm:
128^40 mod 3=1
128^40 mod 17=1
so 3x+1=17y+1
min number that satisfies the eq is 52