10%=394;50%=1970;250%=9850;30%=1182;
250%+30%=9850+1182=11032
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.
297
let x=speed
t=time taken when speed is x so…
xt=4/5x(t+40)
t=160 minutes
2 hr 40 minutes
twenty members
63:45:55
Answer:
Since the car has met the person 20 minutes before hand, it has actually saved 10 mins of journey (to and fro)
since the man has started 1.30 hrs before and car has met him 10 mins before actual time he takes to reach daily is 1hr and 20 mins
Pfull = 4 Hrs; Qfull = 5 Hrs
Assume velocity Vp = x then Vq = 4/5 x –> avg velocity = (5/5 + 4/5) / 2 = 0.9
If fastest time = 4hrs then the time it would take to fill up tank by alternating is 4/0.9 = 4.44 Hrs
only if the results are greater than my input
6, 12, 48, 100, 384, 768, 3072
B. 100
512