a^2+b^2+c^2+d^2+e^2=16.
Find the maximum value of e
a+b+c+d+e=8
@ishu1991 said:Cauchy-Schwarz's inequality: (a1^2 + a2^2 + a3^3 +...)*(b1^2 + b2^2 + b3^3 +...) >= (a1*b1 + a2*b2 + a3*b3 + ...)^2(a^2 + b^2 + c^2 + d^2)*(2^2 + 3^2 + 6^2 + 24^2) >= (2a + 3b + 6c + 24d)^2100*625 >= (2a + 3b + 6c + 24d)^2250
thanx:)
e=3?
a=1 b=1 c=1 d=2 .. @Logrhythm said:
a+b+c+d+e=8a^2+b^2+c^2+d^2+e^2=16.Find the maximum value of e
How many natural numbers less than 100 are there such that n^2 does not divide n!?(a) 24 (b) 25 (c) 26 (d) none of the foregoing
@ishu1991 said:3
@ChirpiBird said:e=3?a=1 b=1 c=1 d=2 ..
no, the answer is 16/5....i posted this question as it uses cauchy's inequality
a+b+c+d = (8-e)
a^2+b^2+c^2+d^2 = (16-e^2)
we know, (a^2+b^2+c^2+d^2)(1+1+1+1) >= (a+b+c+d)^2
4(16-e^2) >= (8-e)^2
64-4e^2 >= 64 - 16e + e^2
16 >= 5e
e =
How many 4-digit integers have the sum of their two leftmost digits equal to the sum of their two rightmost digits? (a) 615 (b) 485 (c) 545 (d) none of the foregoing
@ishu1991 said:How many natural numbers less than 100 are there such that n^2 does not divide n!?(a) 24 (b) 25 (c) 26 (d) none of the foregoing
26??
All primes + 4... hence 26...
@ishu1991 said:How many natural numbers less than 100 are there such that n^2 does not divide n!?(a) 24 (b) 25 (c) 26 (d) none of the foregoing
2,3,4,5,7, and all the prime numbers which follow... 11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97... 26 numbers?
@ishu1991 said:How many natural numbers less than 100 are there such that n^2 does not divide n!?(a) 24 (b) 25 (c) 26 (d) none of the foregoing
all prime no. and 4 --> total = 26?
@ishu1991 said:How many 4-digit integers have the sum of their two leftmost digits equal to the sum of their two rightmost digits? (a) 615 (b) 485 (c) 545 (d) none of the foregoing
1*2 + 2*3 + 3*4 + 4*5 + 5*6 + 6*7 + 7*8 + 8*9 + 9*10 + 9^2 + 8^2 + 7^2 + 6^2 + 5^2 + 2^2 + 3^2 + 1^2 = 615...
@Logrhythm said:1*2 + 2*3 + 3*4 + 4*5 + 5*6 + 6*7 + 7*8 + 8*9 + 9*10 + 9^2 + 8^2 + 7^2 + 6^2 + 5^2 + 2^2 + 3^2 + 1^2 = 615...
yes crct please explain
@ishu1991 said:yes crct please explain
abcd number maan liya...
a+b = c+d hona chahiye....
when sum is 1...
(a,b) = 1,0
(c,d) = (1,0);(0,1)
when sum is 2...
(a,b) = (1,1);(2,0)
(c,d) = (1,1);(2,0);(0,2)
so we see a pattern...for sum 1 ->1*2 = 2 numbers..
for sum 2 -> 2*3 numbers...
likewise till 9... (9*10 numbers)
now when sum is 10 (>9)
(a,b) = (c,d) = (9,1);(8,2);(7,3);(6,4);(5,5);(4,6);(3,7);(2,8);(1,9) -> so 9*9 or 9^2 numbers...
and so on for all sum combinations till 18...
i hope samajh aa gaya hoga..
Find the maximum value of 3x + 4y if x^2 + y^2 = 6x - 4y - 4
(a) 8 (b) 9 (c) 10 (d) none of the foregoing
1001,1010. .... 1*2 where sum is 1.. such numbers are 2.
1102,1120,1111 ... 2*3... where sum is 2 ..such numbers are 3.
aise 8*9 tk same.
then for 10 onwards
like 1991, 1982,1973,1964,1955,1946,1937,1928,1919.... 9*9.
similary for sum 9, 1881,1872,1863,1854,1845,1836,1827,1818... 8*8
last number, 9999 hoga. sum coming to 18.
toh sum k liye.. add up.. 1*2+2*3+... 8*9 + 9*10 + 9*9+8*8+ ... 1*1... =615.
@saurav.kgp said:@DeAdLy -haa bhai...correct hain tera ans ... plz post the solution bhai..
assume values in succession. then calculate total. Divide by 4. Thats the share of vodka they drink.
Now find out how much each of the three contributed to the fourth. Given second receives half of what first receives. Solve it for the equation.
@Logrhythm said:Find the maximum value of 3x + 4y if x^2 + y^2 = 6x - 4y - 4(a) 8 (b) 9 (c) 10 (d) none of the foregoing
none its 16
Repost from Bank PO thread:
plz explain me how to solve Q16, 17 QUICKLY
@Logrhythm said:Find the maximum value of 3x + 4y if x^2 + y^2 = 6x - 4y - 4(a) 8 (b) 9 (c) 10 (d) none of the foregoing
d) the second equation is the equ of a circle with centre at (3,-2) and radius 3......at one of the points on the circle(3,1) ( at angle- 90 degrees) the expression 3x+4y gives 13.......so option d.....looking for a better soln
@adityaknsit said:d) the second equation is the equ of a circle with centre at (3,-2) and radius 3......at one of the points on the circle(3,1) ( at angle- 90 degrees) the expression 3x+4y gives 13.......so option d.....looking for a better soln
okay
second equation is the equ of a circle with centre at (3,-2 )
shifting the coordinates of center to (0,0)
we get x^2 + y^2 =9
and 3x + 4y becomes 3x + 4y + 1
now x=((9-y^2))^1/2
put this in 3x + 4y + 1 and diff wrt y to get max of y
y=12/5
and the exp gives 16
second equation is the equ of a circle with centre at (3,-2 )
shifting the coordinates of center to (0,0)
we get x^2 + y^2 =9
and 3x + 4y becomes 3x + 4y + 1
now x=((9-y^2))^1/2
put this in 3x + 4y + 1 and diff wrt y to get max of y
y=12/5
and the exp gives 16