Vikram bhai, answer is nt 2.5.
Your earlier approach was right. Even dis one is rite bt i think u have done mistake in solving eqn.
The eqn will b x^2-20*x+50=0
On solving we will get 10(+-)(5* sqrt 2)
We will consider negative value of the 2 value
Ans - 10-5*sqrt 2

1. 8
2. 1744...............

bro explain it ???

Q1:-A man borrows Rs 6000 at 5% interest, on reducing balance, at the start of the year.if he repays Rs 1200 at the end of each year, find the amount of loan outstanding in Rs , at the beginning of the third year.

1. 3162.75
2. 4125.00
3. 4155.00
4. 5100.00
5. 5355.00

My take is 4155

Amount at the end of first year = 6000 + 5% *6000 = 6300
Less amount paid = 6300-1200 =5100

Amount at the end of second year = 5100 + 5% *5100 = 5100+255+5355
Less amount paid = 5355-1200 = 4155

Hence, amount at the beginning of third year = 4155

P.S.: @ NITISH NARANG. Brother, answer to that should be 2.5. Have edited my post. Made a mistake earlier.

ramya , based in shanpur, took her car for a 400km trip to rampur. She maintained a log of the odometer readings and the amount of petrol she purchased at different petrol pumps at different prices(given below). Her car already had 10 litres of petrol at the start of the journey, and she first purchased petrol at the start of the journey, as given in table below, and she had 5 litres remaining at the end of the journey.

Odometer reading - petrol purchased - rate of petrol
(km) (litre) (rs/litre)

start of journey- 400 - 20 - 30
600 - 15 - 35
650 - 10 - 40
end of journey- 800


q1:- what has been the mileage (in km/l) of her car over the entire trip?

1. 8.00
2. 8.50
3. 9.00
4. 9.50
5. None of the above

q2:- her car's tank- capacity is 35 litres. Petrol costs rs 45/litre in rampur. What is the minimum amount of money she would need for purchasing petrol for the return trip from rampur to shanpur, using the same route? Assume that the mileage of the car remains unchanged throughout the route, and she did not use her car to travel around in rampur.

1. 1714
2. 1724
3. 1734
4. 1744
5. Data insufficient to answer.

1. 8
2. 1744...............

Q1:- A computer program was tested 300 times before its release . The testing was done in three stages of 100 tests each. The software failed 15 times in stage 1, 12 times in stage 2, 8 times in stage 3 , 6 times in both stage 1 and stage 2, 7 times in both stage 2 and stage 3, 4 times in both stage 1 and stage 3, and 4 times in all three stages. how many times the software failed in a single stage only?

1. 10
2. 13
3. 15
4. 17
5. 35

Q2:- Suresh who runs a bakery , uses a conical shaped equipment to write decorative labels(e.g happy birthday etc.) using cream. The height of this equipment is 7cm and the diameter of the base is 5mm. A full charge of the equipment will write 330 words on an average . How many words can be written using three fifth of a litre of cream?

1. 45090
2. 45100
3. 46000
4. 43200
5. none of the above

nitish narang Says

There is 10 lit. container containing orange juice. A jug of orange juice is drawn out and replaced with pineapple juice.One more time this operation is done and then there is equal ratio of orange and pineapple juice in container.The volume of jug in litres is ?

10 orange, 0 pine; Let the jug be of x liters.
In the 1st draw out x liters of orange is taken out.
In the 2nd draw out (x/10)*(10-x) liters of orange is drawn out

x+(x/10)*(10-x) = 5
10x+10x-x2=50
x2-20x+50=0

x comes as 5*(2-root(2)) liters!!!!!!!!!!!

Ramya , based in shanpur, took her car for a 400km trip to Rampur. she maintained a log of the odometer readings and the amount of petrol she purchased at different petrol pumps at different prices(given below). her car already had 10 litres of petrol at the start of the journey, and she first purchased petrol at the start of the journey, as given in table below, and she had 5 litres remaining at the end of the journey.

Odometer Reading - Petrol Purchased - Rate of Petrol
(Km) (Litre) (Rs/litre)

Start of journey- 400 - 20 - 30
600 - 15 - 35
650 - 10 - 40
End of journey- 800


Q1:- what has been the mileage (in km/l) of her car over the entire trip?

1. 8.00
2. 8.50
3. 9.00
4. 9.50
5. none of the above

Q2:- Her car's tank- capacity is 35 litres. Petrol costs Rs 45/litre in Rampur. what is the minimum amount of money she would need for purchasing petrol for the return trip from Rampur to Shanpur, using the same route? Assume that the mileage of the car remains unchanged throughout the route, and she did not use her car to travel around in Rampur.

1. 1714
2. 1724
3. 1734
4. 1744
5. data insufficient to answer.

nitish narang Says

There is 10 lit. container containing orange juice. A jug of orange juice is drawn out and replaced with pineapple juice.One more time this operation is done and then there is equal ratio of orange and pineapple juice in container.The volume of jug in litres is ?

My take is 2.5

Let capacity of mug = x

After first process:

Orange juice = 10-x
Pineapple juice = x

Ratio = 10-x:x
After 2nd process :

Orange juice = (10-x) - x(10-x)/10 ---- (1)
Pineapple juice = x-x(x)/10 +x ----- (2)

Now, (1) and (2) are equal

So, (10-x)-x+(x^2/10) = 2x+(x^2/10)
=>10-x-x=2x
=>10 =4x
=>x=2.5

Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of
donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21 (B) 42 (C) 120 (D) 504 (E) 5040

A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees
cross pollinated. The number of his trees that are pure Fuji plus the crosspollinated ones totals 187, while 3/4 of
all his trees are pure Fuji. How many of his trees are pure Gala?
22 33 55 77 88

A certain established organization has exactly 4096 members. A certain new organization has exactly 4 members.
Every 5 months the membership of the established organization increases by 100 percent. Every 10 months the
membership of the new organization increases by 700 percent. New members join the organizations only on the
last day of each 5 or 10month period. Assuming that no member leaves the organizations, after how many
months will the two groups have exactly the same number of members?
(A) 20 (B) 40 (C) 50 (D) 80 (E) 100

120(nt sure)
33
40

Q1:-A man borrows Rs 6000 at 5% interest, on reducing balance, at the start of the year.if he repays Rs 1200 at the end of each year, find the amount of loan outstanding in Rs , at the beginning of the third year.

1. 3162.75
2. 4125.00
3. 4155.00
4. 5100.00
5. 5355.00

Q2:- A city has a park shaped as a right angled triangle. the length of the longest side of this park is 80m. the mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments. determine the sum of the squares of the lengths of the three paths.

1. 4000m
2. 4800m
3. 5600m
4. 6400m
5. 7200m

Abhiup87 Says

Is it 2.5 ltr?

I dnt know the answer...
approach bata do apni..

upudit0 Says

ace bhai ans 3024 hai

I think they took 4 teams out of 2 men and 2 women.
Like each man can go with 2 women and hence, 2*2 = 4.

But, with (M1, M2) and (W1, W2) there can be 2 possible matches.
Those are:
(M1, W1) Vs (M2, W2) and (M1, W2) Vs (M2, W1).

If you have original solution, please post the same. I will see if I have something

Question..
The football league of a certain country is played according to the following rules :
Each team plays exactly one game against each of the other team.
The winning team of each game is awarded 2 points and lossing team gets 0 point.
If a match ends a draw, both the teams get 1 point.
After the league was over, the team were ranked according to the points that they earned at the end of the tournament. Analysis of the points table revealed the following :
Exactly half of the points earned by each team were earned in games against the ten teams finished at the bottom of the table.
Each of bottom ten teams earned half of their total points against the other nine teams in the bottom ten.
How many teams participated in the league ?

A. 16
B. 18
C. 19
D. 25
E. 30 plz solve

b1,b2,b3,....b10 => bottom 10 teams
t1,t2,t3,...tn => top n teams
(b1+b2+b3+...b10)/2 => scored from b's and same amount scored from t's
(t1+t2+t3+...tn)/2=>half scored from b's and remaining from t's
Now lets take the bottom 10 teams alone.. number of matches between them = 10C2 = 45(45*2=90 total points). These 90 points are distributed among the ten teams and it constitutes half of their score.
(b1+b2+b3+...b10)/2 = 90
(b1+b2+b3+...b10) = 180
Each team in the top group will play 10 matches with teams from bottom group = n*10 matches = 20n points. Among this 20 n points 90 points are scored by bottom teams and remaining by top teams..(t->20n-90, b->90)
Each team in top will play n(n-1)/2 matches with other teams from top grp = n(n-1) points
n^2-n = 20n-90
n^2-21n+90=0
n=6 or 15
total points of bottom teams = 2*90 = 180
avg points of bottom teams = 18
for n=6 -> 20n-90 = 30
total points of top teams = 2*30 = 60
avg points of top teams = 10
This is not possible..
for n=15 -> 20n-90 = 210
total points of top teams = 2*210 = 420
avg points of top teams = 28
This is possible.
so ans = 10+15 = 25 total teams..
I think better to solve it using options, if possible...

(1) My take is 11


(2) My take is 1512

Two men can be selected in 9C2 ways.
Now, we can select 2 women from 7 women as wives of those 2 men cannot be selected.

And then each man can make a team with any of two women. So, total two possibilities of games between 4 selected players.

So, total ways = 9C2 * 7C2 * 2 = 1512

ace bhai ans 3024 hai

Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of
donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21 (B) 42 (C) 120 (D) 504 (E) 5040

A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees
cross pollinated. The number of his trees that are pure Fuji plus the crosspollinated ones totals 187, while 3/4 of
all his trees are pure Fuji. How many of his trees are pure Gala?
22 33 55 77 88

A certain established organization has exactly 4096 members. A certain new organization has exactly 4 members.
Every 5 months the membership of the established organization increases by 100 percent. Every 10 months the
membership of the new organization increases by 700 percent. New members join the organizations only on the
last day of each 5 or 10month period. Assuming that no member leaves the organizations, after how many
months will the two groups have exactly the same number of members?
(A) 20 (B) 40 (C) 50 (D) 80 (E) 100

Is it 2.5 ltr?

There is 10 lit. container containing orange juice. A jug of orange juice is drawn out and replaced with pineapple juice.One more time this operation is done and then there is equal ratio of orange and pineapple juice in container.The volume of jug in litres is ?

some PnC plzzzzzzz puys help

1.in a certain test there are n question and atleast 2^(n-k) students give wrong answers to atleast k questions where k=1,2,3,4............,n now if total wrong answers are 2047 then then whats is the value of n ? options
10/11/12/13

2.no of ways in which a mixed double game can be arranged from amongst 9 couples such that no wife and husband play in same game ???options
756/1512/3024/none of these

3. given that n is odd the number of ways in which three numbers in AP can be selected in 1,2,3,4.............,n is
>(n-1)^2/4
>(n+1)^2/4
>(n+1)^2/2
>(n-1)^2/4

4.

(1) My take is 11

We have total number of wrong answers = 2^(n-n) + 2^(n-(n-1) + ...
= 2^0 + 2^1 + 2^2 + 2^3 + ..... + 2^(n-1)
= 2047

=> n = 12 as 2^0 + 2^1 + 2^2 + 2^3 + ...... + 2^10 = 2^11 - 1

(2) My take is 1512

Two men can be selected in 9C2 ways.
Now, we can select 2 women from 7 women as wives of those 2 men cannot be selected.

And then each man can make a team with any of two women. So, total two possibilities of games between 4 selected players.

So, total ways = 9C2 * 7C2 * 2 = 1512

(3) My take is (n-1)^2 / 4

We have numbers like 1, 2, 3, 4, ....., n
Now, we can AP with d = 1 as (1, 2, 3), (2, 3, 4), ....., (n-2, n-1, n)
=> (n-2) ways

With d = 2:
(1, 3, 5), (2, 4, 6), ....., (n-4, n-2, n)
=> (n-4) ways

With d = 3:
(1, 4, 7), (2, 5, 8 ), ........, (n-6, n-3, n)
=> (n-6) ways

Total = (n-2) + (n-4) + (n-6) + ..... + 1 ways
=> Sum of odd numbers from 1 till (n-2)
=> Sum of first (n-1)/2 odd number
=> ^2 ........ Sum of first 'x' odd numbers = x^2
=> (n-1)^2 / 4

some PnC plzzzzzzz puys help

1.in a certain test there are n question and atleast 2^(n-k) students give wrong answers to atleast k questions where k=1,2,3,4............,n now if total wrong answers are 2047 then then whats is the value of n ? options
10/11/12/13

2.no of ways in which a mixed double game can be arranged from amongst 9 couples such that no wife and husband play in same game ???options
756/1512/3024/none of these

3. given that n is odd the number of ways in which three numbers in AP can be selected in 1,2,3,4.............,n is
>(n-1)^2/4
>(n+1)^2/4
>(n+1)^2/2
>(n-1)^2/4

4.

1)in how many ways can the letters of the word rainbow be arranged such that only two vowels always remain together

2)vowels are never together .
a)720 b)1440 c)360 d)3600

plz explain in detail..:)

is it 1440

1)in how many ways can the letters of the word rainbow be arranged such that only two vowels always remain together

2)vowels are never together .
a)720 b)1440 c)360 d)3600

plz explain in detail..:)

Hi;
there can be 3 cases:::: (a,i) OR (i,o) OR (a,o) together
Let us consider one case when (a,i) together in 2! ways and for each such (a,i), r,n,b,w,o in 6! ways .But o can't seat before/after (a,i) ;so we have to eliminate this case::: (6!)(2!) - (5!)(2!)
1200 & for 3 such sub cases 1200x3 = 3600
For never together All-Always together= But it's not matching any options
But there is another way::::::

_r _ n _ b_ w_ :::: rnbw in 4! ways & a,i,o in 5C3(3!) ways =1440 ways

Regards,
Never Back Down

1)in how many ways can the letters of the word rainbow be arranged such that only two vowels always remain together

2)vowels are never together .
a)720 b)1440 c)360 d)3600

plz explain in detail..:)

sol:
let a's expenditure are Ea and savings 2s where Ea+2s=15000
let b's expenditure are 15000 and savings are s

1)the combined income of a and b is more than rs 45000
30000+s >45000 for this s>15000 can't possible since
Ea+2s=15000 => s<15000/2

2) the combined expenditure of a and b is not less than 20000
means Ea+15000 well it depends on s<15000/2 s can take value
7000 Ea will be 8000 so Ea+15000 always >20000

3) A's expenditure added to twice of b's income is equal to rs 45000
Ea+2(15000+s)=30000+Ea+2S =45000 since Ea+2s =15000

4) b's expenditure added to twice of a 's income is equal to rs 30000
15000+2(15000)=45000 not equal to 30000
so ithink ans is option 2) & 3)

2) check the question

i checked the question i was getting ans as 46 but no option matches..:)

upudit0 Says

big bro this form only take 44 form whats about 40,50,.........,90

ya saar those numbers are obvious one .. i was saying about 44 which people find it difficult to track

Make it a Thumb rule

numbers of form 50k +/- 12 will have this property

big bro this form only take 44 form whats about 40,50,.........,90

very very urgent puys

Q) how many for digit perfect squares have the property that their last two digits are same??

sol:
only square of 2 digit result in 4 digit number
a 2 digit number ab =10a+b
square it (10a+b)^2=100a^2+b^2+20ab
last 2 digits comes from (b^2+20ab)
now for odd b unit digit same not possible since 2a is even
i will denote U as unt digit
b=0 => a=4,5,6,7,8,9 -------------> 6 numbers
b=2 => 4+40a this means U(4a)=4 =>a=6 number is 62
b=4 => 16+80a this means U(8a+1)=6 not possible since U(8a+1) always odd
b=6 => 36+120a means U(2a+3)=6 not possible
b=8 => 64+160a means U(6a+6)=4 for a=3 numberis 34

total 8 numbers

sol:
apply am>=gm
M=4^(1/2) + 4^(1/3) + 4^(1/4)>= 3*(4)^(1/2+1/3+1/4)*1/3
>=3*(4)^((13/12)*1/3)
>=3*(2)^(13/18 )

(2)^1/18 >1
(2)^13/18 >2
3*(2)^(13/18 ) >=6
option e)

bhai 1 baat btao 4^1/2=2
4^1/4=1.414
4^1/3<=1,6
adding them gives 5.1 almost
then how can this be greater than 6....

esotericecstasy Says

Q) how many 4 digit perfect squares have the property that their last two digits are same??

Make it a Thumb rule

numbers of form 50k +/- 12 will have this property

very very urgent puys

Q) how many for digit perfect squares have the property that their last two digits are same??

sol:
only square of 2 digit result in 4 digit number
a 2 digit number ab =10a+b
square it (10a+b)^2=100a^2+b^2+20ab
last 2 digits comes from (b^2+20ab)=b(b+20a)
try for b and a

Q) how many 4 digit perfect squares have the property that their last two digits are same??

very very urgent puys

Q) how many 4 digit perfect squares have the property that their last two digits are same??

Hello, Could some one please give the answers along with the explanations.

1)If M = 4^(1/2) + 4^(1/3) + 4^(1/4), the the value of M is
a)less than 3
b)equal to 3
c)between 3 and 4
d)equal to 4
e)greater than 4

sol:
apply am>=gm
M=4^(1/2) + 4^(1/3) + 4^(1/4)>= 3*(4)^(1/2+1/3+1/4)*1/3
>=3*(4)^((13/12)*1/3)
>=3*(2)^(13/18 )

(2)^1/18 >1
(2)^13/18 >2
3*(2)^(13/18 ) >=6
option e)

Hello, Could some one please give the answers along with the explanations.

1)If M = 4^(1/2) + 4^(1/3) + 4^(1/4), the the value of M is
a)less than 3
b)equal to 3
c)between 3 and 4
d)equal to 4
e)greater than 4


2) Last month 15 homes were sold in Town X. The average(arithmentic mean) sale price of the homes was $150,000 and median sale price was $130,000.Which of the following statements must be true?

i)At least one of the homes was sold for more than $165,000
ii)At least one of the homes was sold for more than $130,000 and less than $150,000
iii)At least one of the homes was sold for less than $130,000

a)i only
b)ii only
c)iii only
d)i and ii
e)i and iii

3) At least 100 students at a certain high school study Japanese.If 4 percent of the students at school who study French also study Japanese,do more students at school study French than Japanese?
(1) 16 students at the school study both French and Japanese
(2) 10 percent of the students at the school who study Japanese also study French

a) Statement (1) ALONE is sufficent, but statement (2) alone is not sufficient
b)Statement (2) ALONE is sufficent, but statement (1) alone is not sufficient
c)Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
d)Each statement ALONE is sufficient
e)Statements (1) and (2) TOGETHER are NOT sufficient

Five three digits no including N, were to be added. While adding the reverse of N was added by mistake instead of N. Hence the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit.
a)4 b)6 c)8 d)2

let numbers be abc and bac...make equations and solve for a and c...
it comes out to be...a=1 c=2...
2b1-1b2=99 =11*(a+b+c)
=>9=1+b+2
so b=6...

Five three digits no including N, were to be added. While adding the reverse of N was added by mistake instead of N. Hence the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit.
a)4 b)6 c)8 d)2

Ans = 6? Equations:
number be x=abc
reversed number y=cba

8(c-a) = 6+2c......................................I
p+q+y = p+q+x+11(a+b+c)
y = x + 11(a+b+c)
100c+10b+a=100a+100b+100c + 11(a+b+c)......II

Solve both equations value sets of (a,c) = {(1,2);(5,7)}
for both value sets, b=6 !
Cheerio

Five three digits no including N, were to be added. While adding the reverse of N was added by mistake instead of N. Hence the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit.
a)4 b)6 c)8 d)2

bhai koi xat dekar aya hai to uske question post karo

The income of a is rs 15000 and is equal to the expenditure of b,if the ratio of the savings of a to the savings of b is 2:1 ,which of the following statements is definitely true?
1)the combined income of a and b is more than rs 45000
2)the combined expenditure of a and b is not less than 20000
3)A's expenditure added to twice of b's income is equal to rs 45000
4)b's expenditure added to twice of a 's income is equal to rs 30000

p and q are the centre of two circles intersecting at r and s .if pr=26cm ,qr=30cm and pq=28 cm
find the length (in cms) of the common chord of the two circles
1)40 2)42 3)48 4)50

sol:
let a's expenditure are Ea and savings 2s where Ea+2s=15000
let b's expenditure are 15000 and savings are s

1)the combined income of a and b is more than rs 45000
30000+s >45000 for this s>15000 can't possible since
Ea+2s=15000 => s<15000/2

2) the combined expenditure of a and b is not less than 20000
means Ea+15000 well it depends on s<15000/2 s can take value
7000 Ea will be 8000 so Ea+15000 always >20000

3) A's expenditure added to twice of b's income is equal to rs 45000
Ea+2(15000+s)=30000+Ea+2S =45000 since Ea+2s =15000

4) b's expenditure added to twice of a 's income is equal to rs 30000
15000+2(15000)=45000 not equal to 30000
so ithink ans is option 2) & 3)

2) check the question

The income of a is rs 15000 and is equal to the expenditure of b,if the ratio of the savings of a to the savings of b is 2:1 ,which of the following statements is definitely true?
1)the combined income of a and b is more than rs 45000
2)the combined expenditure of a and b is not less than 20000
3)A's expenditure added to twice of b's income is equal to rs 45000
4)b's expenditure added to twice of a 's income is equal to rs 30000

p and q are the centre of two circles intersecting at r and s .if pr=26cm ,qr=30cm and pq=28 cm
find the length (in cms) of the common chord of the two circles
1)40 2)42 3)48 4)50

1. the ant, decided to take a stroll in a square shaped field. He reached a point and decided that he had walked enough. This point is located 13 meters from one of the field's corners, 17 meters from the corner that is diagonally opposite that one, and 20 meters from a third corner post. Using this data, deduce the area of the field. (Assumption: The land is flat.)

i want detailed approach @varun.tyagi plzz solve this

Q x and y are positive integers. How many pairs of (x, y) satisfy the following relation? 1/(x)^1/2 +1/(y)^1/2=1/(20)^1/2


OPTIONS
1)8 2)11 3)3 4)1 5)None of these

1.If the positive real numbers x and y satisfy the equation x + y = 1, what is the maximum possible value of S = x^4 + y^4 (x^5 + y^5)?
OPTIONS 1)1/2 2)1/3 3)1/4 4)1/12 5)1/16

2. What is the value of J73? If for a sequence Jk is defined by the relation
2 = (J(k + 1) + k) - 2Jk andJ1 = 4
OPTIONS
1)2^74 + 74
2)2^73 + 73
3)2^72 + 73
4)2^74 + 72
5)2^72 + 74

ans is option 4
3.Q If Z is a set of three digit numbers of base 10 which are divisible by 3 and end with 1 when written in base 5 and base 7 notations, then how many elements does Z has?
OPTIONS
1)18 2)9 3)12 4)15 5)6

ans is option 2

Ashok has only one 5 rupee coin - 65 possible, 21 possible
Ashok has only two 5 rupee coins - 65 possible, 21 possible(naturally as 2>1)
Ashok has only five 2 rupee coins - 65 possible, 21 posible(3*2+3*5)
Ashok has only one 2 rupee coin - 65 possible(13*5), 21 not possible
Ashok has only seven 5 rupee coins - 65 possible, 21 possible..

4th choice

ans ekdum ryt koi question post karo bhai

Shekhar has 9 colleagues in his office, 5 of them are ladies and 4 gentlemen, his wife Priya also has 9 colleagues in her office, 4 of them are ladies and 5 gentlemen. In how many ways, they can invite a dinner party of 4 ladies and 4 gentlemen so that there are 4 of Shekhars colleagues and 4 of Priyas colleagues?
OPTIONS
1)15876
2)4286
3)7626
4)5626
5)None of these

L1 - 5, G1-4
L2-4, G2 - 5
Lets select the ladies first.. from 1st and 2nd group and get the gents based on that.. it can be
L(0,4) => 5C0*4C4 = 1 G(4,0) =>4C4*5C0 = 1
L(1,3) => 5C1*4C3 = 20 G(3,1) => 4C3*5C1 = 20
L(2,2) => 5C2*4C2 = 60 G(2,2) => 4C2*5C2 = 60
L(3,1) => 5C3*4C1 = 40 G(1,3) => 4C1*5C3 = 40
L(4,0) => 5C4*4C0 = 5 G(0,4) => 4C0*5C1 = 5

Ans = 1^2 + 20^2 + 60^2 + 40^2 + 5^2 = 1+400+3600+1600+25 = 5626

4 kadam ke bad .. ?

After 1 unit of time
Nawab -LRL
WO - LR

After 2 units of time
Nawab - RLR
WO - LR

Thus, in next step, both will put left foot together.So, 4 steps.
DAYA left hota hai na?

vikram bhai while nawab sahab heading his foot for right Wo will head her left foot isn't it ??

Ashok has some 5 rupee and some 2 rupee coins. He finds that he cannot choose a set of coins out of his collection whose total value is Rs. 21. However, he can choose a set of coins whose total value is Rs. 65. Which of these statements could be true?
OPTIONS
1)Ashok has only one 5 rupee coin
2)Ashok has only two 5 rupee coins
3)Ashok has only five 2 rupee coins
4)Ashok has only one 2 rupee coin
5)Ashok has only seven 5 rupee coins

Ashok has only one 5 rupee coin - 65 possible, 21 possible
Ashok has only two 5 rupee coins - 65 possible, 21 possible(naturally as 2>1)
Ashok has only five 2 rupee coins - 65 possible, 21 posible(3*2+3*5)
Ashok has only one 2 rupee coin - 65 possible(13*5), 21 not possible
Ashok has only seven 5 rupee coins - 65 possible, 21 possible..

4th choice

Ashok has some 5 rupee and some 2 rupee coins. He finds that he cannot choose a set of coins out of his collection whose total value is Rs. 21. However, he can choose a set of coins whose total value is Rs. 65. Which of these statements could be true?
OPTIONS
1)Ashok has only one 5 rupee coin
2)Ashok has only two 5 rupee coins
3)Ashok has only five 2 rupee coins
4)Ashok has only one 2 rupee coin
5)Ashok has only seven 5 rupee coins

Question can be solved by each of the options. From the given data we can apply cosine and sine rules.
So we will get two equations.

Statement 1 : a + b = 80 and from the equation of the given data (cosine rule) we can solve.
Statement 2 : From the statement and the equation of the given data (sine rule) we can solve.

can't get ur question

Shekhar has 9 colleagues in his office, 5 of them are ladies and 4 gentlemen, his wife Priya also has 9 colleagues in her office, 4 of them are ladies and 5 gentlemen. In how many ways, they can invite a dinner party of 4 ladies and 4 gentlemen so that there are 4 of Shekhars colleagues and 4 of Priyas colleagues?
OPTIONS
1)15876
2)4286
3)7626
4)5626
5)None of these

A function f(x) is defined for all positive integers that satisfy the condition:
f(1)+ f(2)+ f(3)+...+ f(p)= (p^2)*f(p)
If f(1) = 2009, then what is the value of f(2008 )?
OPTIONS
1)1
2)2/2000!
3)2009/2008
4)1/2008
5)1/1004

Question can be solved by each of the options. From the given data we can apply cosine and sine rules.
So we will get two equations.

Statement 1 : a + b = 80 and from the equation of the given data (cosine rule) we can solve.
Statement 2 : From the statement and the equation of the given data (sine rule) we can solve.

10 children r standing in a line. Each children has some chocolate with him. If the first child attempted to double the no of chocolates with each of the other he would fall short by 2 chocolates. If the second child took 2 chocolate from each of the remaining. He would have three chocolates less than what the first child initially had. Find the total no of chocolates with the 3rd to 10th child.
a)18 b)19 c)23 d)21

Let the chocolates with the 1st person be 'x'
Let the chocolates with the 2nd person be 'y'
Let the chocolates with 3rd to 10th persons be 'z'

According to first condition: It is given that if first child tries to double the number of chocolates with each child he'll fell short by 2 chocolates

=> If he had two more chocolates he would have zero chocolates and all the other would have the total chocolates

=> The sum of the chocolates with 2nd to 10th children is 2 greater than the chocolates with the first child Therefore:

(x+2) = (y+z) ------1

According to second condition: It is given that the second child takes 2 chocolates from 9 children. So total chocolates with him = (y+2*9) = (y+18 ). He fell short by 3 chocolates
Therefore:
(x - 3) = (y + 18 ) ------2

From 1 and 2
=> y+21+2 = y + z
=> z = 23

If log x/(b-c)=log y/(c-a)=log z/(a-b), mark all the correct options.

(A) xyz = 1 (B) xa yb zc = 1 (C) xb+c yc+a za+b = 1 (D) xb+c yc+a za+b = 0

log (x) = k*(b-c)
log (y) = k*(c-a)
log (z) = k*(a-b)

=> log (x) + log(y) + log(z) = 0
=> log (xyz) = 0

=> xyz = 1

P.S. Bhai the other options are not very clear. Please thoda clear karo options ko

Q-1) The cost of raw material of a product increases by 30% and manufacturing cost increases by 20% and Sellin price increases by 60%.The Raw material and manufacturing cost orignally formed 40% and 60% of the total cost respectively.If the orignal profit % was 1/4 the orignal Manufacturing cost,find the approximate new profit %?
Answer is :---48.39%

Q-2)There are 20 couples in a party.Each person greets every person except his or her spouse.People of the same sex shake hands and those of the opposite sex do Namaskar.What is the total number of handshakes and namaskar in the party??????????
Answer is 1140

Q-3)How many odd integers n,where 5<=n<=103 are divisible by a number if the form 5k+1 where k is positive integer.??



Please Explain these Question with Clear Cut method...Dont Write Direct Answers

1.) Let the cost of manufacturing be 60 (as it form 60% of the total cost) and raw materials be (as it forms 40% of the total cost) 40

=> C.P. = 100
Profit = 60/4 = 15

=> S.P. = 115

Given C.P. = 60*1.2 + 40*1.3 = 72 + 52 = 124
S.P. - 115*1.6 = 184
=> Profit = (184 - 124)/124 = 60/124 = 0.4839
=> 48.39%

2.) 20 couples mean 20 men and 20 women.
Total handshakes within same sex = C(20,2)*2 = 380.....We have multiplied by two because of men and women

Each men/women can do namaskaar with 19 others. So total namaskaars = 20*19*2 = 760

Total ways = 380+760 = 1140

3.) First such number is 6 and last number is 101
Total numbers = (101 - 6)/5 + 1 = 20
Out of these 20, the number of odd nos will be 20/2 = 10

If log x/(b-c)=log y/(c-a)=log z/(a-b), mark all the correct options.

(A) xyz = 1 (B) xa yb zc = 1 (C) xb+c yc+a za+b = 1 (D) xb+c yc+a za+b = 0

10 children r standing in a line. Each children has some chocolate with him. If the first child attempted to double the no of chocolates with each of the other he would fall short by 2 chocolates. If the second child took 2 chocolate from each of the remaining. He would have three chocolates less than what the first child initially had. Find the total no of chocolates with the 3rd to 10th child.
a)18 b)19 c)23 d)21

dd017 Says

See u r calculation dude//!!1

thanx ye hamesha hota haim mere saath 20*19*2=760
and 190+190+760=1140 ab to sahi hai

Q-3)How many odd integers n,where 5<=n<=103 are divisible by a number if the form 5k+1 where k is positive integer.??
Please Explain these Question with Clear Cut method...Dont Write Direct Answers

sol:
6,11,16,21............. all numbers are of the form 5k+1 now just check for n in range 5<=n<=103
6,11,16,...............,102
total terms 16+1=17

sol2:
20 couples have 20men and 20women then
20 men can shake hands in 20c2 =190 ways and 20 women in 20c2 =190and 1 men will namaste 19 women xcept her spouse then 20 men will do this in 20*19
since while doing namste namaste from boths sides are counted so 20*19*2=380
total ways 190+190+380=11470

See u r calculation dude//!!1