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income tax
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vivek.kapoor83
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sonu_thekool
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I think the answer is C - 50 + 3I/100
This is how I approached it.
Lets call the Income of the person as I.
2% of I would be 2I/100
Arithmetic Mean (Average) of 100 units of currency = Sum of 100 units / 100 = 50*100 / 100 = 50
1% of I would be I/100
Adding them all up = 50 + 3I/100
Is this the right approach ? What is the OA ?
This is how I approached it.
Lets call the Income of the person as I.
2% of I would be 2I/100
Arithmetic Mean (Average) of 100 units of currency = Sum of 100 units / 100 = 50*100 / 100 = 50
1% of I would be I/100
Adding them all up = 50 + 3I/100
Is this the right approach ? What is the OA ?
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vivek.kapoor83
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sonu_thekool wrote:I think the answer is C - 50 + 3I/100
Arithmetic Mean (Average) of 100 units of currency = Sum of 100 units / 100 = [color=red]50[/color]*100 / 100 = 50
? What is the OA ?
From where did u got this 50 ???????????????
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sonu_thekool
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Sorry, I meant B.iamcste wrote:Did you mean Bsonu_thekool wrote:I think the answer is C - 50 + 3I/100
What is the OA ?
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amitabhprasad
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I am getting "C"
50+I/40
Av of 100+1%of I = 50+I/200
Add 2% of I to it = 2*I/100+I/200+50
= I/40+50
Whats OA ?
50+I/40
Av of 100+1%of I = 50+I/200
Add 2% of I to it = 2*I/100+I/200+50
= I/40+50
Whats OA ?
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sonu_thekool
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You have the same approach as mine butamitabhprasad wrote:I am getting "C"
50+I/40
Av of 100+1%of I = 50+I/200
Add 2% of I to it = 2*I/100+I/200+50
= I/40+50
Whats OA ?
Av of 100 + 1% of I = 50 + I/100 (Not 50 + I/200)
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vivek.kapoor83
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vivek.kapoor83 wrote:sonu_thekool wrote:I think the answer is C - 50 + 3I/100
Arithmetic Mean (Average) of 100 units of currency = Sum of 100 units / 100 = [color=red]50[/color]*100 / 100 = 50
? What is the OA ?
From where did u got this 50 ???????????????
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vivek.kapoor83
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sonu_thekool
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The Average of a set = Sum of all the numbers / no. of terms in the set
A = S/n
So, Average of 100 items in the set = Sum of 100 terms / 100
Sum of consecutive numbers can be found by multiplying the number in the middle of the sequence (50) with the number of terms in the sequence
==> 50 * 100 ==> 5000
Average = 5000 / 100 = 50
This is how I got 50.
A = S/n
So, Average of 100 items in the set = Sum of 100 terms / 100
Sum of consecutive numbers can be found by multiplying the number in the middle of the sequence (50) with the number of terms in the sequence
==> 50 * 100 ==> 5000
Average = 5000 / 100 = 50
This is how I got 50.
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vivek.kapoor83
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sonu_thekool wrote:The Average of a set = Sum of all the numbers / no. of terms in the set
A = S/n
So, Average of 100 items in the set = Sum of 100 terms / 100
Sum of consecutive numbers can be found by multiplying the number in the middle of the sequence (50) with the number of terms in the sequence
==> 50 * 100 ==> 5000
Average = 5000 / 100 = 50
This is how I got 50.
Im just readin it 1st time..strange. i didnt knw that
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nervesofsteel
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vivek.kapoor83 wrote:sonu_thekool wrote:The Average of a set = Sum of all the numbers / no. of terms in the set
A = S/n
So, Average of 100 items in the set = Sum of 100 terms / 100
Sum of consecutive numbers can be found by multiplying the number in the middle of the sequence (50) with the number of terms in the sequence
==> 50 * 100 ==> 5000
Average = 5000 / 100 = 50
This is how I got 50.
Im just readin it 1st time..strange. i didnt knw that
Vivek
is 50+3I/100 the OA..??
Approach by Sonu sounds good
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vivek.kapoor83
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sonu_thekool
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Can you please post any explanation given to show how the answer is 'C' -vivek.kapoor83 wrote:OA is C
50 + I/40 ?
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vivek.kapoor83
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