@Dexian said:8
right 
A watch correctly set at 10:00 am on Sunday shows 20 minutes more than the correct time on 6:00 pm on that day. If clock shows 10:30 pm on that day. What is the correct time ?
@pankaj1988 said:A watch correctly set at 10:00 am on Sunday shows 20 minutes more than the correct time on 6:00 pm on that day. If clock shows 10:30 pm on that day. What is the correct time ?
10.05 P.M?
@yudh said:how please tell
Replace x by y and y by x to find the inverse.
b^x=(1+_/1-y2)/y
for y^-1(0) put x=0
then, y=1+_/1-y^2
(y-1)^2=1-y^2
y^2+1-2y=1-y^2
y=0,1
(Y is a function of log hence can't be 0)
hence, y=1
@pankaj1988 said:A watch correctly set at 10:00 am on Sunday shows 20 minutes more than the correct time on 6:00 pm on that day. If clock shows 10:30 pm on that day. What is the correct time ?
10.00PM
@yudh said:@chillfactor please solve this question
ax + by = 7
ax^2 + bxy = 7x
axy + by^2 = 7y
Add last two to get
=> 49 + (a + b)xy = 7(x + y) ...........(1)
ax^2 + by^2 = 49
ax^3 + bxy^2 = 49x
ax^2y + by^3 = 49y
Add last two to get
133 + 7xy = 49(x + y) ............(2)
ax^3 + by^3 = 133
ax^4 + bxy^3 = 133x
ax^3y + by^4 = 133y
Add last two to get
406 + 49xy = 133(x + y) ...............(3)
From (2) and (3) get xy and (x + y), then from (1) get (a + b) and you are done
@iLoveTorres said:@chillfactor Sir kindly help me be clear with a concept.3x + 2y + z = 15how do i find the positive integral solutions, non negative integral solutions and integral solutions of this question?
step1
first think of any of the possible solutions x=4, y =1 and z=1 is clearly on of the solutions
step2
now split the LHS in two parts such that sum of coefficients of two parts are equal
(3x) + (2y+z) = 15 [sum of co-efficients = 3 for both the parts]
step3:
replace the variables in part1 with k+specific value (found in step1)
replace the variable in part2 with -k+specific value(found in step 1)
i.e x = k+4 and y=-k+1; z = -k+1 [ this is called generalised solution]
now for various integral values of k (i.e k=.....-3,-2,-1,0,1,2,3....) you can find integral solutions
Positive integral solutions
if you are interested in only +ve solution then chose the value of k such that all variables are +ve.
In the given case is we choose k>0 then y and z become non-positive and if we choose k
ATDH.
@iLoveTorres said:@chillfactor Sir kindly help me be clear with a concept.3x + 2y + z = 15how do i find the positive integral solutions, non negative integral solutions and integral solutions of this question?
You will have to consider cases when x = 0, 1, 2, 3, 4, 5 (for non-negative integral solutions)
and x = 1, 2, 3, 4, 5 (for positive integral solutions)
In case of integral solutions, there will be infinite no of solutions
@pankaj1988 said:check again bro OA is 10:00 p.m.
oh yes its 10.00 P.M.
It gains 20 mins in 8 hours i.e. 10 A.M to 6P.M.
in 1 hour it gain 20/8= 2.5 mins
in 12 hour from 10A.M to 10P.M it gains 2.5*12=30 mins. so it will be 10 P.M
@pankaj1988 said:Please tell your approach.
8 hrs ---------------> 20 Min
12hrs----------------> 20/8*12=30 Min
@chillfactor said:ax + by = 7ax^2 + bxy = 7xaxy + by^2 = 7yAdd last two to get=> 49 + (a + b)xy = 7(x + y) ...........(1)ax^2 + by^2 = 49ax^3 + bxy^2 = 49xax^2y + by^3 = 49yAdd last two to get133 + 7xy = 49(x + y) ............(2)ax^3 + by^3 = 133ax^4 + bxy^3 = 133xax^3y + by^4 = 133yAdd last two to get406 + 49xy = 133(x + y) ...............(3)From (2) and (3) get xy and (x + y), then from (1) get (a + b) and you are done
very neat working .... one needs real superior intuitive thinking capabilities to get this kind of clean solution.
Lovely!!
ATDH.
Q. two square prisms wid length of sq faces 6 n 2 units . n height 24 units . these prisms r kept on rectangular faces such tht their rectangular faces touch each other to form a step like structure . find the max dis btw vertices of two prisms .
sol needed .
Let A be the set of 4-digit numbers a1,a2,a3,a4 where a1>a2>a3>a4,then how many values of A are possible?