M4maths Previous Puzzles

08 Jul 2022 Average

A, B, C and D are four positive numbers such that A is 2/3 times of B, B is 5/6 times of C and C is 3/5 times of D. If the average of 3 times of A and 4 times of D is 900, then the average of all four numbers A, B, C and D is:

OPtion

1) 241.5
2) 246.5
3) 215
4) 244
5) 224
6) 208
7) 204
8) 219
9) 236
10) None of these

Solution

A=(2/3)*(5/6)*(3/5)*D
D=3A
Also, average of 3 times A and 4 times D = 900
⇒ 3A + 4D = 1800
Now substituting D=3A in above equation, we get, D + 4D = 1800
D=360
.'. A=360/3=120
B=(3/2)*A=(3/2)*120=180
C=(3/5)*D=(3/5)*360=216
Hence, average of A, B, C & D = (120+180+216+360)/4 = 876/4 = 219
Correct Option 8)

Correct Option: 8 Best Solution (1)

G.S.Kantharao   2 years AGO

A=2B/3
B=5C/6
C=3D/5 & D=D
A=2B/3=2*5C/3*6=2*5*3D/3*6*5=>A=D/3,B=D/2,
C=3D/5
(3A+4D)/2=900 (Given)
3*D/3+4D=1800
D=360,A=120,B=180,C=216
Now Average of A,B,C,D=
(120+180+216+360)/4=219
My option is 8👈

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