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
13 Kigs & 6 Libs –> 510tors in 10hrs
” –> 51 tors/hr
14 Libs –> 484 tors in 12hrs
” –> 121/3 tors/hr
1 Lib –> 121/42 tors/hr
==> 6 Libs –> 121/7 tors/hr
Now,
13 kigs and 6 Libs –> 51 tors/hr
Only 13 Kigs –> 51 – (121/7) tors/hr
= 236/91 tors/hr
Answer:
24
Step-by-step explanation:
A + B = 40.
And at Rs 7 a kg for 40 kg, you want a total of Rs.280.
So the second equation is 9A + 4B = 280
From the first equation: B = 40 – A
and sub into the second equation:
9A + 4(40-A) = 280
9A + 160 – 4A = 280
5A = 120
A = 24.
And you should check: B should equal 40-24 = 16. Check with the final equation: 9*24 + 4*16 = 216 + 64 = 280. So it works.
Your answer, of course, is A = 24
( c ) R or T
10 Camels = 68000
X = 4
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$57.30
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28
to arrange m objects in n places => nCm
i.e. for this example m=2(+,- sign) and n = 8(places between two number from 1 to 9)
SO, answer is : 8C2 = (8*7) / 2 = 28
How will you know the odd is in lighter one or heavier one from only one weighing. It will require 2 weighing to find the odd set and one weighing for odd coin in that set i.e total 3 weighings.
1. Statements :
No bat is ball. No ball is wicket.
Conclusions :
I. No bat is wicket.
II. All wickets are bats.
21
12