Monday, February 14, 2011

Problem on Permutation (Interesting One)

Using the digits 1,5,2,8 four digit numbers are formed and the sum of all possible such numbers?

Answer
24 numbers can be formed. 1528 1582 1258 1285 1825 1852 2158 2185 2581 2518 2851 2815 5128 5182 5218 5281 5812 5821 8152 8125 8512 8521 8215 8251.

each number comes 6 times in unit 10's 100's and 1000th places.

The sum is 106656
1*6+5*6+2*6+8*6 96

96+96(10)+96(100)+96(1000) 106656

 

Friday, February 11, 2011

Problems on Time and Work

1) Problem: If 9 men working 6 hours a day can do a work in 88 days. Then 6 men working 8 hours a day
can do it in how many days?
Solution: From the above formula i.e (m1*t1/w1) = (m2*t2/w2)
 so (9*6*88/1) = (6*8*d/1)
 on solving, d = 99 days.
2) Problem: If 34 men completed 2/5th of a work in 8 days working 9 hours a day. How many more man
should be engaged to finish the rest of the work in 6 days working 9 hours a day?
Solution: From the above formula i.e (m1*t1/w1) = (m2*t2/w2)
 so, (34*8*9/(2/5)) = (x*6*9/(3/5))
so x = 136 men
 number of men to be added to finish the work = 136-34 = 102 men
3) Problem: If 5 women or 8 girls can do a work in 84 days. In how many days can 10 women and 5 girls can
do the same work?
Solution: Given that 5 women is equal to 8 girls to complete a work. So, 10 women = 16 girls. Therefore 10
women + 5 girls = 16 girls + 5 girls = 21 girls. 8 girls can do a work in 84 days then 21 girls can do a work in
(8*84/21) = 32 days. Therefore 10 women and 5 girls can a work in 32 days
4) Problem: Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long it
take both A & B, working together but independently, to do the same job?
Solution: A's one hour work = 1/8. B's one hour work = 1/10. (A+B)'s one hour work = 1/8+1/10 = 9/40. Both
A & B can finish the work in 40/9 days
5) Problem: A can finish a work in 18 days and B can do the same work in half the time taken by A. Then,
working together, what part of the same work they can finish in a day?
Solution: Given that B alone can complete the same work in days = half the time taken by A = 9 days
 A's one day work = 1/18
 B's one day work = 1/9
 (A+B)'s one day work = 1/18+1/9 = 1/6
6) Problem: A is twice as good a workman as B and together they finish a piece of work in 18 days.In how
many days will A alone finish the work.
Solution: if A takes x days to do a work then B takes 2x days to do the same work
 = > 1/x+1/2x = 1/18
 = > 3/2x = 1/18
 = > x = 27 days.
 Hence, A alone can finish the work in 27 days.

Problems on Ages with solution

PROBLEMS ON AGES
1. Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
          a. 24                     b. 27                     c. 40            d. None of these
Basic Formula:
If the present age of A is ‘x’ years, the age of A, n years ago was (x-n) years, and the age of a after n years will be (x+n) years.

Answer with Explanation:

Given : The ratio of the present ages of Sameer and Anand is 5:4
Let the present ages of Sameer be 5x years
Let the present ages of Anand be 4x years
Given: 3years hence (after), the ratio is 11 : 9
5x + 3                   :         4x + 3
11               :         9
                   =       11/9
(5x+3) 9      =       11 (4x + 3)
45x +27      =       44x +33
45x -44x     =       33-27
X                =       6
To find: Present age of Anand
The present age of Anand is 4(6) years = 24 years  
2. The ratio of the present ages of two brothers is 1 : 2 and 5 years back, the ratio was 1 : 3. What will be the ratio of their ages after 5 years?
          a. 1 : 4                            b. 2 : 3                           c. 3 : 5                       d. 5 : 6

Answer with Explanation:

Let the two brothers be X and Y
Given: The ratio of the present ages of X and Y is 1:2
Let the present age of X be x years
Let the present age of Y be 2x years
   X : Y
 x : 2x
Given : 5 years back (before), the ratio was 1:3
x-5    =       1        :         2x-5  =       3
                   =       1/3
3 (x-5)         =       2x-5
3x-15          =       2x-5  3x-2x      =       -5 +15  x = 10

To find : The ratio of their ages after 5 years
X+5   = ?     :         2x + 5 =      ?
10 + 5 = 15 :         2(10)+5 = 25
 15 : 25    3 : 5


3. The present ages of three persons are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
 a. 8, 20, 28          b. 16, 28, 36                  c. 20, 35, 45                   d. None of these

Answer with Explanation:
Let the 3 persons be A, B and C
Given : The ratio of the present ages of 3 persons is 4:7:9
Let the present age of A be 4x years
Let the present age of B be 7x years
Let the present age of C be 9x years
A : B : C  4x : 7x : 9x
Given : 8 years ago, the sum of their ages was 56.
                   8 years ago, the ratio of the ages is
A : B : C  4x – 8 : 7x – 8 : 9x – 8
Given : The sum of the above ages was 56
                   4x -8 + 7x-8 + 9x-8       =       56
                   20x – 24 = 56        20x = 56+24              20x=80
                             x        =       4
To find : A, B and C’s present age.
 A : B : C
4x : 7x : 9x  4 (4) : 7 (4) : 9 (4)

 A’s Age = 16 ; B’s Age = 28 ; C’s Age = 36

 4. The ratio between the present ages of A and B is 5 : 3 respectively. The ratio between A’s age 4 years ago and B’s age 4 years hence is 1 : 1. What is the ratio between A’s age 4 years hence and B’s age 4 years ago?
          a. 1 : 3                            b. 2 : 1                           c. 3 : 1                           d. 4 :
Answer with Explanation:
Given : The ratio between the present ages of A and B is 5 : 3
          Let the present age of A is 5x years
          Let the present age of B is 3x years
Given : The ratio between A’s age 4 years ago and B’s age 4 years hence is 1:1
A’s age 4 years ago = 5x - 4
B’s age 4 years hence    = 3x + 4               
5x-4  = 1 ; 3x +4  = 1
                   =       1/1
5x-4  = 3x +4  5x – 3x        =       4 + 4  2x =8      x = 4
To find :
5x+4 = ?     :        3x - 4  = ?
20 + 4         =       24 : 12 – 4
24 : 8                3 : 1


 

FOR MORE PROBLEMS CLICK HERE

Sunday, February 6, 2011

Important Concepts on Permutation

Permutation

Number of permutations of ‘n’ different things taken ‘r’ at a time is given by:-
nPr       =            n!/(n-r)!
Proof:     Say we have ‘n’ different things a1, a2……, an.
Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = n-1
So the second-place can be filled-up in (n-1) ways. Now number of things left after filling-up the first and second places = n - 2
Now the third place can be filled-up in (n-2) ways.
Thus number of ways of filling-up first-place = n
Number of ways of filling-up second-place = n-1
Number of ways of filling-up third-place = n-2
Number of ways of filling-up r-th place = n – (r-1) = n-r+1
By multiplication – rule of counting, total no. of ways of filling up, first, second --  rth-place together :-
n (n-1) (n-2) ------------ (n-r+1)
Hence:
nPr       = n (n-1)(n-2) --------------(n-r+1)
= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-1)] ----3.2.1
nPr = n!/(n-r)!
Number of permutations of ‘n’ different things taken all at a time is given by:-
nPn               =         n!
Proof   :
Now we have ‘n’ objects, and n-places.
Number of ways of filling-up first-place  = n
Number of ways of filling-up second-place = n-1
Number of ways of filling-up third-place  = n-2
Number of ways of filling-up r-th place, i.e. last place =1
Number of ways of filling-up first, second, --- n th place
= n (n-1) (n-2) ------ 2.1.
nPn =  n!
Concept. 
We have   nPr  =     n!/n-r
Putting r = n, we have :-
nPr  =   n! / (n-r)
But   nP=  n!
Clearly it is possible, only when  n!  = 1
Hence it is proof that     0! = 1

Note : Factorial of negative-number is not defined. The expression  –3! has no meaning.

Thursday, February 3, 2011

Important formula for Boats and Streams

                                        BOATS AND STREAMS

1.In water ,the direction along the stream is called downstream and ,the direction against
the stream is called upstream.
 
2.If the speed of a boat in still water is u km/hr and the speed of the stream is v
km/hr,then:
 
speed downstream=(u+v)km/hr.
speed upstream=(u-v)km/hr.
 
3.If the speed downstream is a km/hr and the speed upstream is b km/hr,then :
 
speed in still water=1/2(a+b)km/hr
 
rate of stream=1/2(a-b)km/hr

Tuesday, February 1, 2011

Karnataka Post Graduate Common Entrance Test 2010 Application Form was Available at

The Karnataka PGCET 2010 application can be obtained on payment of Rs. 600/- (Rs 300/- in case of Karnataka SC/ ST and Cat-I, on production of documentary proof) in the form of DD only, drawn in favour of the Member Secretary, CPAC payable at Belgaum.
a) Karnataka PGCET 2010 Application can be obtained only in person from :
1) Gulbarga University,
Department of Management,
“Jnana Ganga”,
Gulbarga – 585106
Phone No : 08472-263283.
2) University of Mysore,
B.N. Bahadur Institute of Management Sciences,
“Manasagangothri,” MYSORE – 570 006
Phone No : 0821-2413920/2419755
3) Karnataka University,
Kousali Institute of Management Studies,
Dharwad – 580 003
Phone No : 0836-2741882
4) Karnataka State Women’s University,
Department of PG Studies & Research in Management,
Bijapur – 586 101
Phone No : 08352-277503
5) J N N College of Engineering,
Dept. of MBA, Savalanga Road, Navule,
Shimoga – 577 204
Phone No : 08182-276568
6) Srinivasa Institute of Technology,
Merlapadavu, Arkula, P.O. Parangipet, Valachil,
Mangalore – 574 143
Phone No : 0824-2274730/32
7) Bapuji Institute of Engineering and Technology,
Shamanur Road, Davanagere – 577 004
Phone No : 08192-222245
8 ) Rao Bahadur Y Mahabaleshwarappa Engineering College,
Cantonment, Bellary-583104,
Phone No : 08392-244809
9) Sahyadri Institute of Technology & Management,
NH – 48, Adyar,
Mangalore – 575009,
Phone No : 0824 – 2277222.

b) Karnataka PGCET 2010 Application can be procured in person & also by Post from :

1) Visvesvaraya Technological University,
PGCET Cell,
“Jnana Sangama”, BELGAUM – 590 018,
2) VTU Regional Office,
No. 351, II cross, I Block,
Jayanagar, Bangalore – 560 011
3) VTU Regional Office,
Sri Jayachamarajendra College of Engineering Campus,
Mysore – 570 006
4) VTU Regional Office,
Rajapura Kusnoor Road,
Gulbarga – 585 106.
The application can be obtained through post by applying on a plain paper indicating Name of the candidate & complete postal address, name of course for which admission is sought and enclose the DD (mention DD No., Date and Bank name) for application fee.
Karnataka PGCET 2010 Important Dates :
1) Last date for issue of Karnataka PGVET 2010 application form : 10-6-2010
2) Completed application forms addressed must reach on or before 10-6-2010 only to,
The Member Secretary CPAC, & Registrar,
VTU, Jnana Sangama, Belgaum – 590 018
3) The list of colleges along with Seat Matrix will be notified in the website http://www.vtu.ac.in/ & http ://pgcet.vtu.ac.in on approval of the Seat Matrix by the Government.
Contact Nos : 0831-2498126/127/129..