@scrabbler said:Fraction is 1900/27% = 19/27...so remaining is 8/27 of vol => 2/3 of height. regardsscrabbler
bhai ye wala part me confusion vol is dirctly proportional to height, why cube root..
@mailtoankit said:bhai approach batao ?
Log bhai ne de diya hai hai...I guess sabne waise hi kiya hai...
no different approach bhai..:)
@swapnil4ever2u said:bhai ye wala part me confusion vol is dirctly proportional to height, why cube root..
Similar figs...r and h are in proportion...so r^2*h is prop to cube...draw and see.
regards
scrabbler
regards
scrabbler
@saurav205 said:Log bhai ne de diya hai hai...I guess sabne waise hi kiya hai...no different approach bhai..
abey yaar..10 baar kiya hua hai yeh question.....but fir wohi baat ho jaati hai na.....sahi time par dimaag mein click karna chahiye.(or may be needs more practice)..wohi to nahi hota..
@mailtoankit said:abey yaar..10 baar kiya hua hai yeh question.....but fir wohi baat ho jaati hai na.....sahi time par dimaag mein click karna chahiye.(or may be needs more practice)..wohi to nahi hota..
bhai relax...we have not even entered March yet..sab kar lenge yaar mil ke...my geometry is pathetic...bas yahan jo question hote hai wo dekh leta hun...abhi bohot time hai...relax karo... 😃
@Logrhythm said:bhai relax...we have not even entered March yet..sab kar lenge yaar mil ke...my geometry is pathetic...bas yahan jo question hote hai wo dekh leta hun...abhi bohot time hai...relax karo...
O i m relax yaar....chadi thodi na hai mennu..
...prep mil kar hi karenge...
...prep mil kar hi karenge...If f(x)f(y) =f(x) +f(y) +f(xy) -2 ;f(2)=5
find f(5) ??
@abhijit90 said:oye yaar plz help..............is single sitting degree valid for cat
plz do not spam in this thread...
ask here.
http://www.pagalguy.com/discussions/official-cat-2013-aspirants-25088203/3610422
ask here.
http://www.pagalguy.com/discussions/official-cat-2013-aspirants-25088203/3610422
@amresh_maverick If f(x)f(y) =f(x) +f(y) +f(xy) -2 ;f(2)=5
find f(5)
Put x = 1, y = 2, we get f(1)f(2) = f(1) + f(2) + f(2) - 2
i.e. 5f(1) = f(1) + 10 - 2
i.e. f(1) = 2
Put x = y = 2, we get {f(2)}^2 = 2f(2) + f(2^2) - 2
i.e. 25 = 10 + f(4) - 2
i.e. f(4) = 17
You can safely observe that f(x) = x^2 + 1
So f(5) = 26
find f(5)
Put x = 1, y = 2, we get f(1)f(2) = f(1) + f(2) + f(2) - 2
i.e. 5f(1) = f(1) + 10 - 2
i.e. f(1) = 2
Put x = y = 2, we get {f(2)}^2 = 2f(2) + f(2^2) - 2
i.e. 25 = 10 + f(4) - 2
i.e. f(4) = 17
You can safely observe that f(x) = x^2 + 1
So f(5) = 26
@pathetic said:Rohan writes all three-digit numbers of base 6, one below the other in an order. Find the number of times the digit '1' is used by Rohan.
1 at Left most place - other two places can be filled in 6*6 = 36 ways.
1 at middle place - other two places can be filled in 5*6 = 30 ways.
1 at Right most place - other two places can be filled in 5*6 = 30 ways.
So total '1' has been used 36 + 30 + 30 = 96 times.
Alternatively, while writing all numbers from 000 to 555 in base-6, each digit has been used equally at all places. As total numbers from 000 to 555 are 6^3 = 216 and each uses 3-digits, so total 216*3 digits are used where each of the six digits (0 - 5) has been used equally i.e. 216*3/6 = 108 number of times.
We just need to subtract the number of '1' which have been counted for 0xx numbers i.e. from 00 to 55. With same reasoning, as above, this number will be 2*6^2/6 = 12.
Thus required number is = 108 - 12 = 96.
1 at middle place - other two places can be filled in 5*6 = 30 ways.
1 at Right most place - other two places can be filled in 5*6 = 30 ways.
So total '1' has been used 36 + 30 + 30 = 96 times.
Alternatively, while writing all numbers from 000 to 555 in base-6, each digit has been used equally at all places. As total numbers from 000 to 555 are 6^3 = 216 and each uses 3-digits, so total 216*3 digits are used where each of the six digits (0 - 5) has been used equally i.e. 216*3/6 = 108 number of times.
We just need to subtract the number of '1' which have been counted for 0xx numbers i.e. from 00 to 55. With same reasoning, as above, this number will be 2*6^2/6 = 12.
Thus required number is = 108 - 12 = 96.
@pathetic said:If a > 0; b > 0; c > 0 and a + b + c = 1, then the maximum value of (a/3+a )+(b/3+b)+(c/3+c) is ?
If you change a by b and b by c and c by a simultaneously, the expression doesn't change. This type of expression is known as cyclic expression and achieves its extreme value when all the variables are equal.
So just put a = b = c = 1/3 and get the answer as 3/10 = 0.3.
Now to check whether this is maximum or minimum value of the expression, just put the different values of a, b, c. Say for limiting case, take a = b 0 and c = 1 And the expression becomes 0 + 0 + 1/4 = 0.25
That confirms 0.3 is not the minimum rather it is the Maximum value of the expression.
Also we can infer from this that the minimum value of the expression is going to be always greater than 0.25 (for positive a, b, c)
So just put a = b = c = 1/3 and get the answer as 3/10 = 0.3.
Now to check whether this is maximum or minimum value of the expression, just put the different values of a, b, c. Say for limiting case, take a = b 0 and c = 1 And the expression becomes 0 + 0 + 1/4 = 0.25
That confirms 0.3 is not the minimum rather it is the Maximum value of the expression.
Also we can infer from this that the minimum value of the expression is going to be always greater than 0.25 (for positive a, b, c)
@maroof10 said:Volume of a right circular cylinder is 16 cubic meter. Find the dimension(r and h) that uses min material.
pi(r^2)h = 16
i.e. (r^2)h = 16/pi
i.e. (r^4)(h^2) = 256/pi^2
i.e. (2r^2)(rh)(rh) = 512/pi^2
and we want to minimise 2(pi)r(r + h)
i.e. pi(2r^2 + rh + rh)
As product of (2r^2), rh and rh is constant i.e. 512/pi^2, there sum will be least when they are equal to each other i.e. cuberoot of 512/pi^2.
So 2r^2 = rh i.e. h = 2r
And by putting this above, you can easily find their respective values also.
i.e. (r^2)h = 16/pi
i.e. (r^4)(h^2) = 256/pi^2
i.e. (2r^2)(rh)(rh) = 512/pi^2
and we want to minimise 2(pi)r(r + h)
i.e. pi(2r^2 + rh + rh)
As product of (2r^2), rh and rh is constant i.e. 512/pi^2, there sum will be least when they are equal to each other i.e. cuberoot of 512/pi^2.
So 2r^2 = rh i.e. h = 2r
And by putting this above, you can easily find their respective values also.
@rkshtsurana said:Q. there are four unit spheres inside a larger sphere such that each of the unit sphere touches large sphere and 3 other unit spheres. Find the radius of larger spheres ?
1 + rt(3/2)
I hope, there is no error in my mental calculations
I hope, there is no error in my mental calculations
In how many ways can 1000 be written as a sum of 'n' consecutive natural numbers, where 'n' is
greater than 1?
(a) 0 (b) 1 (c) 2 (d) 3
greater than 1?
(a) 0 (b) 1 (c) 2 (d) 3
A circle is drawn inside a trapezium such that it touches all the four sides of the trapezium. The line
joining the midpoints of the non-parallel sides divides the trapezium in two parts with areas in the
ratio 3 : 5. If the lengths of the non-parallel sides are 6 cm and 10 cm, then what is the length (in cm)
of the longer parallel side of the trapezium?
(a) 8 (b) 10 (c) 12 (d) Cannot be determined
joining the midpoints of the non-parallel sides divides the trapezium in two parts with areas in the
ratio 3 : 5. If the lengths of the non-parallel sides are 6 cm and 10 cm, then what is the length (in cm)
of the longer parallel side of the trapezium?
(a) 8 (b) 10 (c) 12 (d) Cannot be determined
@Koushik98 said:In how many ways can 1000 be written as a sum of 'n' consecutive natural numbers, where 'n' isgreater than 1?(a) 0 (b) 1 (c) 2 (d) 3
It is simply number of odd factors of '100' - 1 i.e. 4 - 1 = 3
And the ways are 1000 = 5*200 = 198 + 199 + 200 + 201 + 202
1000 = 16*62.5 = 55 + ....+ 62 + 63 + ...+ 70
1000 = 25*40 = 28 + ....+ 40 + ..+ 52
Hope it helps :)
And the ways are 1000 = 5*200 = 198 + 199 + 200 + 201 + 202
1000 = 16*62.5 = 55 + ....+ 62 + 63 + ...+ 70
1000 = 25*40 = 28 + ....+ 40 + ..+ 52
Hope it helps :)
@Koushik98 said:A circle is drawn inside a trapezium such that it touches all the four sides of the trapezium. The linejoining the midpoints of the non-parallel sides divides the trapezium in two parts with areas in theratio 3 : 5. If the lengths of the non-parallel sides are 6 cm and 10 cm, then what is the length (in cm)of the longer parallel side of the trapezium?(a) 8 (b) 10 (c) 12 (d) Cannot be determined
should be 8...