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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
24
2, 5, 10, 50, 500, 5000
5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000
x=12;y=7
MONKEY=YEKNOM then each letter is moved backward =XDJMNL so for TIGER=REGIT the each of the reversed letters are moved backward=QDFHS
What was his salary to begin with?
Assume his salary was Rs. X
He earns 5% raise. So his salary is (105*X)/100
A year later he receives 2.5% cut.
So his salary is ((105*X)/100)*(97.5/100)
which is Rs. 22702.68
So Ans is Rs.22176
2000
39/50
Correct option is C)
Let be students are consider as child
Let the age of child added later be x years.
average age of 12 children =20 years
∴ Total age of 12 children =20×12=240 years
After one more child is added-
Average of 13 children =20−1=19 years
Average age of 13 children =
13
sum of 12 children+x
19=
13
240+x
⇒240+x=247
⇒x=247−240=7 years
Hence the age of child added later is 7 years.
125% of s = 60
=> 1.25s = 60
=> s = 60*4/5
=> s = 48
60 – s = 60 – 48 = 12
13 and 15
Grandma