@vbhvgupta said:solve Q 31
1/a+1/b+1/c = 11/20 --- (1)
1/c = 2/9(1/a+1/b) => 1/c = 2/9(11/20-1/c) => 1/c = 1/10 -- (2)
1/a = 6/5(11/20-1/a) => 1/a = 3/10 --- (3)
putting 2 and 3 in 1
1/10+3/10+1/b=11/20
=> b=20/3
hence option 3...
@chandrakant.k said:sum till second tap = 2k kaisa?? only k is ther before that na?? and in the question it is given that sum of 1 to (n-1)
han k hoga...srry for the typo..
@dream_IIMA said:The number 123456789012...890 is four thousand digit long. first u removed all the digit in odd numbered place starting at left most place. next u removed the digits in odd numbered place in the remaining 2000 digits. you perform same operation untill no digit remain. what digit is to be last removed? expln plz
ans 4?
@chandrakant.k said:no no it is twice.. i misread again :bangheaD:
rat wld be twice bhai...sum k hi hoga...rate of second tap wld be 2k...
Q 4
@ishu1991 said:puyss help me in this
according to me the explanation can be put x=-x there are no sign changes so the number of negative roots can be 0. now put x=+x again there is no sign change so number of positive roots can be 0. Hence by descarts sign change rule the minimum no of roots can be 0
@vbhvgupta said:Q 4
bhai question samjh nahi aa raha muje.. 😞
what does "had their efficiency remained constant...
dekho..
lets say ram does 1 unit on day one....so on second day he must have done 2 units
so in 2 days he does 3 units of work...
saleem does 3 units in 40% less time ie 6/5 days...
so in 2 days saleem does 5 units of work...
hence total work done by saleem and ram in 2 days is 8 units...
abb what does it mean by their efficiency remains constant? is it 1 unit by each on each day? if yes that they do 2 units each day...therefore total of 4 days wld be needed...is the answer 4 days??
@vbhvgupta said:Q 4
If Ram does the work in R days
saleem does it in .6R
saleem does it in .6R
W/R + W/.6R + 2W/R +W/.3R =W
W/R(3+5)=W
R=8
W/R(3+5)=W
R=8
if efficiencies had remained constant
(W/8 +10W/48)n=W
(W/3)n=W
n=3
(W/8 +10W/48)n=W
(W/3)n=W
n=3
@Logrhythm said:bhai question samjh nahi aa raha muje.. what does "had their efficiency remained constant...dekho..lets say ram does 1 unit on day one....so on second day he must have done 2 unitsso in 2 days he does 3 units of work...saleem does 3 units in 40% less time ie 6/5 days...so in 2 days saleem does 5 units of work...hence total work done by saleem and ram in 2 days is 8 units...abb what does it mean by their efficiency remains constant? is it 1 unit by each on each day? if yes that they do 2 units each day...therefore total of 4 days wld be needed...is the answer 4 days??
@audiq7 said:6?
Ans 3
@Subhashdec2 said:If Ram does the work in R dayssaleem does it in .6RW/R + W/.6R + 2W/R +W/.3R =WW/R(3+5)=WR=8if efficiencies had remained constant(W/8 +10W/48)n=W(W/3)n=Wn=3
@Subhashdec2 said:If Ram does the work in R dayssaleem does it in .6RW/R + W/.6R + 2W/R +W/.3R =WW/R(3+5)=WR=8if efficiencies had remained constant(W/8 +10W/48)n=W(W/3)n=Wn=3
#18704 wale solutions dekhna...kya galti kari hai meine usme??
smallest positive integer n for which (2^2-1) (3^2-1)....(n^2-1) is perfect square
@vbhvgupta @Logrhythm hlp plz
@vijay_chandola bhai hlp karo na
@Exodia said:Function f(x) is a continuous function defined for all real values of x, such that f(x) = 0 only for two distinct real values of x. It is also known thatf(6) + f(8) = 0f(7).f(9) > 0f(6).f(10) 0 and f(1) 0III. f(7).f(8) 0(a) 1 (b) 2 (c) 3 (d) 4Plz give answer alongwith explanation.
OA :f(0) > 0 and f(1) lies between x = 0 and x = 1.
f(6) + f(8) = 0 implies that f(6) and f(8) are of opposite
sign but same absolute value. Hence another root for
f(x) = 0 must lie between x = 6 and x = 8. As f(1) f(6) must also be less than zero, otherwise we €™ll have
more than 2 roots for f(x) = 0.
Hence f(8) > 0 and f(6) Further f(7).f(9) > 0 implies that both f(7) and f(9) are
greater than zero.
So the second root for f(x) = 0 must lie between x = 6
and x = 7.
So f(x) would look like :
1 2 3 4 5 6
7
8 9 10
x
As f(1), f(2) and f(3) are less than zero,
f(1).f(2).f(3) As f(3), f(5) 0,
f(3).f(5).f(7).f(9) > 0 is true.
As f(7), f(8) > 0,
f(7).f(8) f(0), f(9), f(10) > 0 and f(1) know the magnitude of any of these four we cannot
judge if f(0) + f(1) + f(9) + f(10) is greater than zero or not.
f(6) + f(8) = 0 implies that f(6) and f(8) are of opposite
sign but same absolute value. Hence another root for
f(x) = 0 must lie between x = 6 and x = 8. As f(1) f(6) must also be less than zero, otherwise we €™ll have
more than 2 roots for f(x) = 0.
Hence f(8) > 0 and f(6) Further f(7).f(9) > 0 implies that both f(7) and f(9) are
greater than zero.
So the second root for f(x) = 0 must lie between x = 6
and x = 7.
So f(x) would look like :
1 2 3 4 5 6
7
8 9 10
x
As f(1), f(2) and f(3) are less than zero,
f(1).f(2).f(3) As f(3), f(5) 0,
f(3).f(5).f(7).f(9) > 0 is true.
As f(7), f(8) > 0,
f(7).f(8) f(0), f(9), f(10) > 0 and f(1) know the magnitude of any of these four we cannot
judge if f(0) + f(1) + f(9) + f(10) is greater than zero or not.
@Logrhythm said:bhai question samjh nahi aa raha muje..what does "had their efficiency remained constant...dekho..lets say ram does 1 unit on day one....so on second day he must have done 2 unitsso in 2 days he does 3 units of work...saleem does 3 units in 40% less time ie 6/5 days...so in 2 days saleem does 5 units of work...hence total work done by saleem and ram in 2 days is 8 units...abb what does it mean by their efficiency remains constant? is it 1 unit by each on each day? if yes that they do 2 units each day...therefore total of 4 days wld be needed...is the answer 4 days??
u took saleem's original efficiency to be 1 unit of work 1 day
but he does same work in 40 % less time
so he does 1 unit in .6 days
in 1 day he does 5/3 units of work
@Exodia said:ABCD is a rectangle with BC = a units and DC = 3 a units. The perpendicular dropped from point A meets BD at point F. The diagonals AC and BD intersect at point G. What is the area (in squareunits) of ΔAFG?(a) [root(3) x a2]/12(b) [root(3) x a2]/6(c) [root(3) x a2]/10(d) [root(3) x a2]/8Plz give answer along with explanation.
OA: (d) [root(3) x a2]/8
@Subhashdec2 said:u took saleem's original efficiency to be 1 unit of work 1 daybut he does same work in 40 % less timeso he does 1 unit in .6 daysin 1 day he does 5/3 units of work
nah...it is given that they complete the work in 2 days...
i took ram does 1 unit on first day...that means second day he must do 2 units...
so in 2 days he does 3 units...
now since saleem takes 40% less time to complete the work done by ram in 2 days (units)
=> 2*60/100 = in 6/5 days he does 3 units...
hence in 2 days he does -> 3*5*2/6 = 5 units
this implies that the total work would have been 3+5 = 8 units (as ram and saleem together took 2 days to complete the job)
this is what i had done.......now let me knw if something is wrong...
@vbhvgupta said:solve Q 31
time taken by A,B and C be A,B and C hours
1/C=2/9(1/A+1/B)
1/C-2/9A -2/9B=0
1/A=6/5(1/B+1/C)
1/A-6/5B-6/5C=0
1/A+1/B+1/C=11/20
2/9A +2/9B +2/9C=11/90
11/9C=11/90
C=10
6/5A +6/5B+6/5C=66/100
11/5A=66/100
1/A=6/20
A=20/6
6/20+1/B+1/10=11/20
1/B=1/4-1/10
B=20/3
B=6 (2/3)