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..:)

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

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

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

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

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

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

Rajiv is a student in a business school. After every test he calculates his cumulative average.QT and OB were his last two tests. 83 marks in QT increased his average by 2. 75 marks in OB further increased his average by 1. reasoning is the next test, if he gets 51 in reasoning , his average will be-

1. 63
2. 62
3. 61
4. 60
5. 59

It should be 63

Say so far 'n' exams have taken place and his avg. is 'x'.
So, score = nx

Now, when he gets 83; his avg. of (n+1) tests becomes x+2.

=> nx + 83 = (n+1)(x+2)
=> nx + 83 = nx + 2n + x + 2
=> 2n + x = 81 .................... Eqn. 1

Similarly, nx + 83 + 75 = (n+2)(x+2+1)
=> nx + 158 = nx + 3n + 2x + 6
=> 3n + 2x = 152 ................. Eqn. 2

Solving two equations; n = 10 and x = 61

So, his new score = 10*61 + 83 + 75 + 51 = 819

And new avg. = 819 / 13 = 63

guys. pls solve dis question..!

The equation px2 + (p + 1)x + p = 0 has real and distinct roots. Also, exactly one of its roots
lies between 2 and 3. Which of the following is the range the values of 'p'?
(a)-2/7

<3/13 (b) p<-2/7 or p>3/13 (c) p > 1 (d) p < -/7 or p>=1 e) p<-1/3 or p>1

(p + 1)^2 - 4p^2 > 0
3p^2 - 2p - 1 < 0
=> (p - 1)(3p + 1) < 0

=> -1/3 < p < 1

Also, exactly one root lies between 2 and 3
=> f(2)*f(3) < 0
=> (7p + 2)(13p + 3) < 0
=> -2/7 < p < -3/13

So, a should be the answer (but I think that it should be -3/13 instead of 3/13)

sol:
(p+1)^2-4p^2>=0
=> (3p+1)(p-1)<=0
=> p <=-1/3 & p>=1
here 1 lies between 2 & 3
is it option e)

dude answer is a) also pls use the condition dat root loes between 2 and 3

i guess the OA =63.

guys. pls solve dis question..!

The equation px2 + (p + 1)x + p = 0 has real and distinct roots. Also, exactly one of its roots
lies between 2 and 3. Which of the following is the range the values of 'p'?
(a)-2/7

<3/13 (b) p<-2/7 or p>3/13 (c) p > 1 (d) p < -/7 or p>=1 e) p<-1/3 or p>1

sol:
(p+1)^2-4p^2>=0
=> (3p+1)(p-1)<=0
=> p <=-1/3 & p>=1
here 1 lies between 2 & 3
is it option e)

Rajiv is a student in a business school. After every test he calculates his cumulative average.QT and OB were his last two tests. 83 marks in QT increased his average by 2. 75 marks in OB further increased his average by 1. reasoning is the next test, if he gets 51 in reasoning , his average will be-

1. 63
2. 62
3. 61
4. 60
5. 59

guys. pls solve dis question..!

The equation px2 + (p + 1)x + p = 0 has real and distinct roots. Also, exactly one of its roots
lies between 2 and 3. Which of the following is the range the values of 'p'?
(a)-2/7

<3/13 (b) p<-2/7 or p>3/13 (c) p > 1 (d) p < -/7 or p>=1 e) p<-1/3 or p>1

q1 120
q2 1/2
q3 45

can you plz post the solns as well:)

q1.abc is a triangle with

q1 120
q2 1/2
q3 45

Q1.ABC is a triangle with

coefficient of x^11 in ((x+2x^2+x^3)^20??
Need full approach and also easy path if it is there?:biggrin:

My take is 0

(x + 2x^2 + x^3)^20 = x^20 * (1 + 2x + x^2)^20
Hence, coefficient of x^11 will be 0.

coefficient of x^11 in ((x+2x^2+x^3)^20??
Need full approach and also easy path if it is there?:biggrin:

A cow is tethered at pt. A Neither cow nor rope can enter triangle ABC
angle BAC =30
m(AB)=m(AC)=10
m(BC)=6

WHAT is the area cow can graze if rope length =
i> 8
ii> 12

Ans. i> (176pi)/3
ii> (403pi)/3

ist case take length of rope 8
area wud b a circle of radius 8 and segment angle 330 degree.
solving (330/360 )64pi=176pi
second case it wud be again a circle of radius 12 and segment angle 330+
2 circlesof segment angle 105 degree and radius 2 so area
(330/360)144pi+((2*105/360))*4pi=403pi/3

HARIV Says

The avg of 5 two-digit numbers is 60. X and Y are two of the five numbers. If the digits of each of X and Y are interchanged, then the average of the five numbers will be 51. Instead of this, if the other numbers are multiplied by -1, then the average of the five numbers will be -18. Find the sum of the tens digit of X and Y.

let the numbers be y,x,c,d,e. x+y+c+d+e = 300;

now x= 10p+q and y=10a+b on reversing them we get sum of nums = 255

diff = 45 ==> 9(p-q) +9(a-b) = 45

p-q + a-b = 5 ..... (1)

x+y -c-d-e = -90 ==> c+d+e = 90 +x+y
so 90 + 2(x+y) = 300

x+y = 105

10p +q + 10a+b=105

we need p+a value

10p+10a-5+p+a = 105
11(p+a) = 110 ==> p+a = 10

A cow is tethered at pt. A Neither cow nor rope can enter triangle ABC
angle BAC =30
m(AB)=m(AC)=10
m(BC)=6

WHAT is the area cow can graze if rope length =
i> 8
ii> 12

Ans. i> (176pi)/3
ii> (403pi)/3

in 2nd year , students at a business school can opt for systems,operations or hr electives only. The no. Of girls opting for operations and the no. Of boys opting for systems elective is 37. Twenty two students opt for operations elective. Twenty girls opt for systems and operations electives. The number of students opting for systems elective and the no. Of boys opting for operations electives is 37. Twenty-five students opt for hr electives.

Q1:- the no.of students in the 2nd year is-

1. 73
2. 74
3. 75
4. 76
5. 77

q2:- if 20% of the girls opt for hr electives , then the total no. Of boys in the 2nd year is

1. 54
2. 53
3. 52
4. 51
5. 50

1 4)76
2. 4)51

AM>=GM

(l+m+n)/3>=(lmn)^1/3
18/3>=(lmn)^1/3

(lmn)<=216

Options theek hain?

Wrong approach! :banghead:



It won't be 216. Adding them up will violate the three given conditions.

got it apply A.m. >= G.m. on individual equations
7/2 >= (mn)^0.5
8 >= (nl)^0.5
13/2 >= (ml) ^0.5

and multiply all 3

AM>=GM

(l+m+n)/3>=(lmn)^1/3
18/3>=(lmn)^1/3

(lmn)<=216

Options theek hain?

but the ans. given is ii mnl<=182