@tmohan02 said:The L.C.M of 5/2, 8/9, 11/14 ???
440?
@IIM-A2013 Let us assume that the person lent out x on both occasions.
In the first case, the rate of interest is 3% per half-year. The money will become
1.03x after 6 months from t = 0
1.06x after 12 months from t = 0
1.09x after 18 months from t = 0
1.12x after 24 months from t = 0 and so on....
In the second case, the rate of interest is 6% per half-year. The money is given out at t = 18.
The money will become:
1.12x after 6 months from t = 18, which is 24 months from t = 0.
So after 2 years the returns from both the investments will be same. They will be 1.12x
This was the point at which he gets back 4704 Rs.
=> 1.12x = 4704
=> x = 4200 Rs.
@tmohan02 In the case of fractions,
LCM = LCM of Numerator / HCF of Denominator
HCF = HCF of Numerator / LCM of Denominator
So, in this case LCM (5/2, 8/9, 11/14) = LCM (5,8,11) / HCF (2,9,14) = 440 / 1 = 440
@Aman.Malhotra said:a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
I think a lot of people have answered it correctly but let me just add my two cents to it.
Roots of 3x^2 + 4x + 5 = 0 are complex i.e. they are of the form (p + iq) & (p - iq) where i is iota = sqrt(-1)
For a quadratic equation, complex roots occur in conjugate pairs if the coefficients are real.
In the equation, ax^2 + bx + c = 0, the coefficients are sides of a triangle and hence real. So, if one of roots is common with the other equation say (p+ iq) then the other root will also be compulsorily common. This implies that both equations have the same roots.
This implies that ax^2 + bx + c = 0 is nothing else but the same equation
i.e. 3x^2 + 4x + 5 = 0
Now, we know that the sides of the triangle are 3,4 & 5 where c = 5.
3,4 & 5 form a Pythagorean triplet making the triangle a right angled triangle.
The angle opposite to the biggest side is 90 degrees in a right angled triangle, which in this case is C.
Hence, C = 90 degrees.
Let N be the number of ways in which 2010 can be written in the form,
2010 = 1000a3 + 100a2 + 10a1 + a0, where ai, for 0 ≤ i ≤ 3, is an integer and 0 ≤ ai ≤99.
An example of such representation is, 2010 = 1 × 103 + 9 × 102 + 11 × 101 + 0 × 100
Find N.
OPTIONS
1) 2
2) 3
3) 52
4) 202
5) None of the above
2010 = 1000a3 + 100a2 + 10a1 + a0, where ai, for 0 ≤ i ≤ 3, is an integer and 0 ≤ ai ≤99.
An example of such representation is, 2010 = 1 × 103 + 9 × 102 + 11 × 101 + 0 × 100
Find N.
OPTIONS
1) 2
2) 3
3) 52
4) 202
5) None of the above
i dont understand this question...pls help me in solving it
@pavimai said:Let N be the number of ways in which 2010 can be written in the form,2010 = 1000a3 + 100a2 + 10a1 + a0, where ai, for 0 ≤ i ≤ 3, is an integer and 0 ≤ ai ≤99.An example of such representation is, 2010 = 1 × 103 + 9 × 102 + 11 × 101 + 0 × 100Find N.OPTIONS1) 2 2) 3 3) 52 4) 202 5) None of the above i dont understand this question...pls help me in solving it
This question has been solved already :
if a=pie/2012
If 2[cos a × sin a + cos 4a × sin 2a + cos 9a × sin 3a + … + cos n^2a × sin na] is an integer, what is the smallest possible positive value of n?OPTIONS
1) 503
2) 1006
3) 2012
4) 502
@sujamait said:if a=pie/2012If 2[cos a × sin a + cos 4a × sin 2a + cos 9a × sin 3a + … + cos n^2a × sin na] is an integer, what is the smallest possible positive value of n?OPTIONS1) 503 2) 1006 3) 2012 4) 502
T(n) = sin( n*(n+1) a) - sin( n*(n-1) a)
S = sin( n*(n+1) a) =sin ( n*(n+1)*(pi/2012) )
For n = 502 (lowest from the options) => S = sin (251pi/2) = 1
so, 502 ?
@sujamait said:if a=pie/2012If 2[cos a × sin a + cos 4a × sin 2a + cos 9a × sin 3a + … + cos n^2a × sin na] is an integer, what is the smallest possible positive value of n?OPTIONS1) 503 2) 1006 3) 2012 4) 502
503?
@YouMadFellow said:How ? Share the approach naa
yar approach yehi thi ki..saste wale petrol pump par petrol fill karao..jaha ho sakta hai.. :)
cost is (13.75 × 45 + 6.25 × 40 + 25 ×
35) = 1743.75
35) = 1743.75
@YouMadFellow said:T(n) = sin( n*(n+1) a) - sin( n*(n-1) a)S = sin( n*(n+1) a) =sin ( n*(n+1)*(pi/2012) )For n = 502 (lowest from the options) => S = sin (251pi/2) = 1 so, 502 ?
@maddy2807 said:503?
no OA..wid me. @maddy2807 bhai approach ?
@sujamait said:yar approach yehi thi ki..saste wale petrol pump par petrol fill karao..jaha ho sakta hai.. cost is (13.75 × 45 + 6.25 × 40 + 25 ×35) = 1743.75
Ohh no, that question was not my doubt 😃 . It was asked by someone else, but krum said that he couldn't see the question, so I shared the image 😃 . I don't know the OA
@YouMadFellow said:T(n) = sin( n*(n+1) a) - sin( n*(n-1) a)S = sin( n*(n+1) a) =sin ( n*(n+1)*(pi/2012) )For n = 502 (lowest from the options) => S = sin (251pi/2) = 1 so, 502 ?
YMF yar yeh kaun sa formula lagaya hua hai ?...trig ke formulas yad nhn hein..
ohh i'd scrolled back and found that Q...din follow d history of same anyways...
anyways...@sujamait said:no OA..wid me. @maddy2807 bhai approach ?
yaar i calculated it with values of sina* cosa
if these values are 1/2
then we could hv an integer value.
503 would give sin 45 * cos 45 as 1/2
But it is an absurd approach.
@ravihanda sir help out with it
@sujamait said:YMF yar yeh kaun sa formula lagaya hua hai ?...trig ke formulas yad nhn hein..
Check below
@gautam22 said:S = sin( n*(n+1) a) sir ye kaise aaya?
sin(A +B) - sin(A-B) = 2*cosA*sin(B)
so, T(n) = 2*cos(n^2 * a) * sin(n*a) ... Break this according to the formula above
=> T(n) = sin( n*(n+1)*a) - sin(n*(n-1)*a)
Now, T(1) = sin(2a)
T(2) = sin(6a) - sin(2a)
T(3) = sin(12a) - sin(6a) ...
and so on
So, S = sin (n*(n+1)*a)
n = 502 (from options)
sin( 502*(503)*pi/2012) = sin(251*pi/2) = sin ( 125 pi + pi/2) = -1 .. integer
@YouMadFellow said:Check below sin(A +B) - sin(A-B) = 2*cosA*sin(B)so, T(n) = 2*cos(n^2 * a) * sin(n*a) ... Break this according to the formula above=> T(n) = sin( n*(n+1)*a) - sin(n*(n-1)*a)Now, T(1) = sin(2a)T(2) = sin(6a) - sin(2a)T(3) = sin(12a) - sin(6a) ...and so onSo, S = sin (n*(n+1)*a)
A potter asked his two sons to sell some pots in
the market. The amount received for each pot
was same as the number of pots sold. The two
brothers spent the entire amount on some
packets of potato chips and one packet of banana
chips. One brother had the packet of banana chips
along with some packets of potato chips, while
the other brother just had potato chips. Each
packet of potato chips costs 10/- and the packet
of banana chips costs less than 10/-. The
packets of chips were divided between the two
brothers so each brother received equal number
the market. The amount received for each pot
was same as the number of pots sold. The two
brothers spent the entire amount on some
packets of potato chips and one packet of banana
chips. One brother had the packet of banana chips
along with some packets of potato chips, while
the other brother just had potato chips. Each
packet of potato chips costs 10/- and the packet
of banana chips costs less than 10/-. The
packets of chips were divided between the two
brothers so each brother received equal number
of packets. How much money should one brother
give to the other to make the division financially
equitable?
give to the other to make the division financially
equitable?