Friday, 20 September 2013

RIVERS SIDE CITIES ( IMPORTANT FOR SSC 10 + 2)

TownRiver
Kabul (Afghanistan)Kabul
AllahabadConfluence of Ganga,
Yamuna,
Saraswati (invisible)
VaranasiGanga
NasikGodawari
KolkataHooghly
CuttackMahanadi
PatnaGanga
Chittagong (Bangladesh)Maiyani
LucknowGomati
JamshedpurSubarnarekha
HaridwarGanga
DelhiYamuna
KanpurGanga
SuratTapti
SrinagarJhelum
FerozepurSutlej
LudhianaSutlej
Karachi (Pak)Indus
Yangon (Myanmar)Irawady
Akyab (Myanmar)Irawady
VijaywadaKrishna
Lahore (Pak)Ravi
Paris (France)Seine
Hamburg (Germany)Elbe
Budapest (Hungary)Danube
Rome (Italy)Tiber
Warsaw (Poland)Vistula
Bristol (U.K.)Avon
London (U.K.)Thames
New Castle (U.K.)Tyre
China
ShanghaiYang-tse-Kiang
NankingYang-tse-Kiang
ChungkingYang-tse-Kiang
CantonSi-Kiang
Middle East and Africa
Cairo (Egypt)Nile
Basra (Iraq)Tigris and Euphrates
Ankara (Turkey)Kizil
Baghdad (Iraq)Tigris
Khartoum (Sudan)Blue and While Nile
Europe
Berlin (Germany)Spree
BelgradeDunube
Cologne (Germany)Rhine
Lisbon (Portugal)Tangus
Glasgow (Scotland)Clyde
USA
New YorkHudson
PhiladelphiaDelaware
New OrleansMississippi
Monetreal (Canada)Ottawa
Quebec (Canada)St. Lawrence

12 comments:

S r i N u said...

any one give solution and formula for this

what is the sum of all positive integers lying between 200 and 400 that are multiples of 7 ?

Anonymous said...

sir plz provide daily stock gk

RAHUL said...

there will be 28 multiples of 7 between 200 and 400,,,among them 1st one is 203 and last is 399..so the average is 301.......there by sum of all those is 28*301
that is 8428.....

Anonymous said...

What is the sum of all positive integers that are multiples of 7 from 200 to 400?
8729
8700
8428
8278

Correct Choice is (1) and Correct Answer is 8729

Explanatory Answer

Because 4 is the remainder when we divide 200 by 7 the least number greater than 200 divisible by 7 is 203.
When we divide 400 by 7, we get a remainder of 1.

This implies that the greatest number less than 400, which is divisible by 7 is 399.

Therefore, this is an Arithmetic Progression in which the first term, t1 = a = 203, the common difference 'd' = 7, and the last term 'l' = 399

Let 'n' be the total number of terms in this series.

The nth term of an Arithmetic Sequence is an = a + (n - 1)d

Then 399 = 203 + (n – 1) (7)

Simplifying we get, 7n = 399 – 203 + 7

Or 406 – 203 = 203 or n = 29.

Hence, the required sum is n/2[a + l]

Or 29/2[203 + 399] = (29) * (301) = 8729

bharath said...

thanq

bharath said...

thanq

Anonymous said...

ssc ka admit card kab aayega

Anonymous said...

Its an ap series....a=203,d=7,l=399.....putting the formula n/2(a+l) u'll get ans=8729.

Anonymous said...

Rahul, 301 is d avg of 203, 204, 205...399 and nt of 203,210...399. I hd also made d same mistake in hurry.

Anonymous said...

u r doing great job admin
pls upload gk capsule for ibps po as early as
thank you so much for ur help

Anonymous said...

8700

Anonymous said...

8700 hi hoga pta nhi is chutiye ne itna explain karke bhi galat bataya....
200/7 quotient will be 28 ,,,,,400/7 quotient will be 57......then 57-28 = 29
hence the value of n is 29 then apply the formula n(n+1) ....29(29+1) = 8700
okkkkkkkkkkkkk bosss

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