Nishant, Rakesh, Brijesh, and Rohit bought a house for Rs.40 lakhs.

N:R:B:H
N=3/5(R+B+H)
R=2/3(B+H)
B=H

R: 2/3(B+H)=2/3(B+B)=2/3(2B)=4B/3
N: 3/5(R+B+H)=3/5(R+2B)=3/5(4B/3+2B)=3/5(10B/3)=2B

N:R:B:H=[2B:4B/3:B:B]*3=6B:4B:3B:3B=6:4:3:3

Ans (C)

Nishant, Rakesh, Brijesh, and Rohit bought a house for Rs.40 lakhs. The contribution of Nishant is 3/5thof the contribution of Rakesh, Brijesh, and Rohit taken together. The contribution of Rakesh is 2/3rd of the contribution of Brijesh and Rohit taken together. Brijesh and Rohit contributed equal amount. What is the ratio of their respective contributions?

A. 6:4:3:1
B. 8:4:2:2
C. 6:4:3:3
D. 5:4:3:4
E. 7:4:3:4

Solution:
Nishant = N, Rakesh = Ra, Brijesh = B, and Rohit = Ro
Suppose, Contribution of Brijesh = Rohit = 3x
Contribution of Rakesh = 2/3*(Brijesh + Rohit) = 2/3*(6x) = 4x
Contribution of Nishat = Contribution of (Ra + B +Ro)*3/5 = 10x*(3/5) = 6x

Ratio of contribution: N : Ra : B : Ro = 6x : 4x : 3x : 3x
= 6 : 4 : 3 : 3(C)

Nishant, Rakesh, Brijesh, and Rohit bought a house for Rs.40 lakhs. The contribution of Nishant is 3/5thof the contribution of Rakesh, Brijesh, and Rohit taken together. The contribution of Rakesh is 2/3rd of the contribution of Brijesh and Rohit taken together. Brijesh and Rohit contributed equal amount. What is the ratio of their respective contributions?

A. 6:4:3:1
B. 8:4:2:2
C. 6:4:3:3
D. 5:4:3:4
E. 7:4:3:4

Only options B and C satisfy the conditions since contribution by Rohit and Brijesh are equal.
Let Ni, Ra, Br and Ro are contributions made by Nishant, Rakesh, Brijesh, and Rohit respectively.

Now, \(Ni = \frac{3}{5}(Ra + Br + Ro)\) and \(Ra = \frac{2}{3}(Br + Ro)\) where Br = Ro
\(5Ni = 3(Ra + 2Ro)\) and 3Ra = 4Ro

So, 5Ni = 3Ra + 6Ro
5Ni = 10Ro
Ni = 2Ro and \(Ra = \frac{4}{3}Ro \)

Taking everything in terms of Ro
\(Ni:Ra:Br:Ro = 2Ro:\frac{4}{3}Ro:Ro:Ro = 6:4:3:3\)

Answer C.

Start from the end:

Brijesh = Rohit = x

Rakesh = (2/3) * (Brijesh + Rohit) = (2/3) * (x + x) = (2/3) * (2x) = (4/3)x

Nishant = (3/5) * (Rakesh + Brijesh + Rohit) = (3/5) * ((4/3)x + x + x) = (30/15)x = 2x

LCM of: 2x : (4/3)x : x : x = 6:4:3:3

Let the contribution of B = Ro = a
—> Ra = 2/3*(2a) = 4/3a
& N = 3/5(4/3a + a + a) = 2a

N:Ra:B:Ro = 2a : 4/3a : a : a = 6a : 4a : 3a : 3a
= 6 : 4 : 3 : 3

Option C

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Brijesh and Rohit contributed equal amounts so their ratios must be the same. This rules out options (A), (D) and (E) leaving only (B) and (C). According to (B), the ratio of Rakesh (4) is equal to the sum of the ratios of Brijesh and Rohit (2+2) but, as per the question, Brijesh's ratio must be 2/3 of the combined ratios of Brijesh and Rohit. So (B) is not the right option which leaves only (C) which satisfies all three stipulated conditions:
(a) Niten's contribution (6) is equal to 3/5 of the sum of the contributions of Rakesh, Brijesh and Rohit (3/5 of 10=6)
(b) Rakesh's contribution (4) is equal to 2/3 of the sum of the contributions of Brijesh and Rohit (2/3 of 6=4)
(c) Brijesh's (3) contribution is equal to that of Rohit (3).
ANS: C

C is the answer

Replacing 40 lakh by 100

3/5 of 100 is 60

So we know answer will have 6 first

Rohit and brijesh have equal control so 3:3 at end tells us it is the correct answer

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Nishant ( N ) Rakesh (R) Brijesh ( B) Rohit ( Ro)

B = Ro
R = (2/3)(B+Ro) = (2/3) ( B + B ) = (2/3) (2B) = 4B/3
N = (3/5) ( R+B+Ro) = (3/5) (4B/3 + B + B) = (3/5)(10B/3) = 2B

N:R:B:Ro = 2B: (4B/3):B: B = 6B:4B:3B:3B = 6:4:3:3

Ans: C.

given
n+r+b+ro=4*10^6
n=3/5 * ( r+b+ro)
r=2/3*(b+ro)
b=ro
we get
8/5 * ( r+b+ro)
18/5 * ( 5/3*(b+ro)
8/3* 2B = 4*10^6
B = Ro = 75*10^4
we get R= 10^6
and N = 15*10^5
Ratio
15*10^5: 10^6: 75*10^4:75*10^4
150:100:75:75
30:20:15:15
6:4:3:3
IMO C


Nishant, Rakesh, Brijesh, and Rohit bought a house for Rs.40 lakhs. The contribution of Nishant is 3/5thof the contribution of Rakesh, Brijesh, and Rohit taken together. The contribution of Rakesh is 2/3rd of the contribution of Brijesh and Rohit taken together. Brijesh and Rohit contributed equal amount. What is the ratio of their respective contributions?

A. 6:4:3:1
B. 8:4:2:2
C. 6:4:3:3
D. 5:4:3:4
E. 7:4:3:4

Ni+Ra+Br+Ro=Rs.40 lakhs

But we are informed that:
Ni=3/5 *(Ra+Br+Ro) -------(1)
Ra=2/3 *(Br+Ro) -------(2)
Br=Ro -------(3)
So Ra = 2/3 *(2Ro) = 4Ro/3
and Ni = 3/5 *(4Ro/3 + 6Ro/2) = 3/5 * 10Ro/3 = 2Ro
Hence Ni:Ra:Br:Ro = 2:4/3:1:1 = 6:4:3:3

The answer is C.

LET B=Ro= x THEN
Ra Becomes 4/3x, N becomes 2x

N:Ra:B:Ro = 6:4:3:3

OA:C

Approach:

- Assume contribution by Brijesh = Rohit = x(both contributed equal amount), Rakesh = y, and Nishant = z

Given: y = 2/3(x+x) and z = 3/5(y+x+x) or 3/5(2/3(2x)+2x)

- Total(z+y+x+x) = 40 => 2/3(2x) + 4/5x + 6/5x + x + x

- On Solving we get x = 7.5, y = 10, z =15

- So, ratio for values 15:10:7.5:7.5 will be 6:4:3:3

Option C!

N + Ra+ B + Ro = 40
\(N= \frac{3}{5}(Ra+ B+ Ro) —> Ra+ B+ Ro=( \frac{5}{3})N\)

—> \(N+( \frac{5}{3})N = 40\)

\((\frac{8}{3})N= 40\) —> \(N = 15\)

Ra + B+Ro= 25

\(Ra=(\frac{ 2}{3})(B+ Ro)\)

—>\( B+ Ro=( \frac{3}{2})Ra\)

\(Ra+ (\frac{3}{2})Ra = 25\)
Ra= 10

—> \(B= \frac{15}{2}\), \(Ro= \frac{15}{2}\)

\(15: 10: \frac{15}{2}: \frac{15}{2} \) (dividing by 3 )

—> \(3: 2: \frac{3}{2}: \frac{3}{2}\) (multiplying by 2)

—> 6:4:3:3

The answer is C

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