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22 Jul 2011, 11:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Hi club,
Well this is my 1st post and i need your help to solve these seemingly simple question which i am unable to
.
All these are fractions related question.
I need your help to understand how the answer has been arrived at.
1. A man pays off 3/20 of his debt every month.At the end of 6 months, his remaining debt is $A.How much amount has he cleared off every month?
(a) 3A/20
(b) 9A/10
(c) A/10
(d) 3A/10
(e) 3A/2
Answer:(e)
2. 3/5 part of a petrol tin is filled.If 6 bottles are taken out of it and 3 bottles are filled again, then half the tin is full. What is the capacity of the tin(in bottles)?
(a) 20
(b) 30
(c) 45
(d) 50
(e) 40
Answer:(b)
3. It takes 40 days for a pond to get filled with rain water.If the level of water doubles each day, then how long would it take to fill 1/4 of the pond
(a) 10 days
(b) 20 days
(c) 30 days
(d) 35 days
(e) 38 days
Answer:(e)
Well this is my 1st post and i need your help to solve these seemingly simple question which i am unable to
.All these are fractions related question.
I need your help to understand how the answer has been arrived at.
1. A man pays off 3/20 of his debt every month.At the end of 6 months, his remaining debt is $A.How much amount has he cleared off every month?
(a) 3A/20
(b) 9A/10
(c) A/10
(d) 3A/10
(e) 3A/2
Answer:(e)
2. 3/5 part of a petrol tin is filled.If 6 bottles are taken out of it and 3 bottles are filled again, then half the tin is full. What is the capacity of the tin(in bottles)?
(a) 20
(b) 30
(c) 45
(d) 50
(e) 40
Answer:(b)
3. It takes 40 days for a pond to get filled with rain water.If the level of water doubles each day, then how long would it take to fill 1/4 of the pond
(a) 10 days
(b) 20 days
(c) 30 days
(d) 35 days
(e) 38 days
Answer:(e)
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22 Jul 2011, 21:55
Kudos
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1. x = total debt
x - { (3/20)*(x)*(6) } = A
x = 10A
debt paid off each month;
= 3/20*x
substitute x with 10A;
= (3/20)*(10A)
= 3A / 2
2. x = Total capacity
(3/5)x - 6 + 3 = (1/2)x
x = 30
3. not sure if we need to know volume of the first day
x - { (3/20)*(x)*(6) } = A
x = 10A
debt paid off each month;
= 3/20*x
substitute x with 10A;
= (3/20)*(10A)
= 3A / 2
2. x = Total capacity
(3/5)x - 6 + 3 = (1/2)x
x = 30
3. not sure if we need to know volume of the first day
23 Jul 2011, 01:03
Kudos
Bookmarks
Welcome to club!
Let his total debt be T
Every month he pays off 3/20 of his debt = \(3/20 * T\)
After 6 months he would have paid = \(3/20 * 6 * T\)
His remaining debt after 6 months is A.
Hence Total debt = Remaining debt + Paid debt
\(T=A+3/20 * 6 * T\)
\(T-18/20 * T = A\)
\(T * 2/20 = A\)
\(T= 10A\)
Every month he pays off 3/20 of his debt = \(3/20 * T = 3/20 * 10 A = 3A/2\)
OA E.
Let the capacity of tin in bottles be x
\(3/5 x - 6 + 3 =1/2 x\)
\(3/5 x - 1/2 x = 3\)
\(x/10 = 3\)
\(x= 30\)
OA B.
Let the rate at which pond fills each DAY be x.
Water level doubles each day, hence
DAY LEVEL
1 x
2 2x
3 4x
4 8x
5 16x
Water level rises in a series which is geometric series with first term a=x and common ration r = 2
Also, it takes 40 days to fill the pond to its capacity.
According to series, the LEVEL of water in 40 days will be 40th term of the series.
\(a40 = x*2^(^40-1^)\)
\(a40= x * 2^3^9\)
Hence capacity of pond is = \(x*2^3^9\)
1/4 of that capacity = \(1/4 * x*2^3^9\)
=\(x*2^3^7\)
Now mere observation will tell you that LEVEL of pond on a given day is \(x*2^(^D^A^Y-^1^)\)
Like on day 4, level is 8x => \(2^(^4^-^1^)x\)
on day 5, level is 16x => \(2^(^5-^1^)x\)
On day 38, LEVEL will be =\(x * 2^(3^8-^1) = x*2^3^7.\)
Hence on 38th day pond will be 1/4 full.
OA E.
anuragbytes
Let his total debt be T
Every month he pays off 3/20 of his debt = \(3/20 * T\)
After 6 months he would have paid = \(3/20 * 6 * T\)
His remaining debt after 6 months is A.
Hence Total debt = Remaining debt + Paid debt
\(T=A+3/20 * 6 * T\)
\(T-18/20 * T = A\)
\(T * 2/20 = A\)
\(T= 10A\)
Every month he pays off 3/20 of his debt = \(3/20 * T = 3/20 * 10 A = 3A/2\)
OA E.
anuragbytes
Let the capacity of tin in bottles be x
\(3/5 x - 6 + 3 =1/2 x\)
\(3/5 x - 1/2 x = 3\)
\(x/10 = 3\)
\(x= 30\)
OA B.
anuragbytes
Let the rate at which pond fills each DAY be x.
Water level doubles each day, hence
DAY LEVEL
1 x
2 2x
3 4x
4 8x
5 16x
Water level rises in a series which is geometric series with first term a=x and common ration r = 2
Also, it takes 40 days to fill the pond to its capacity.
According to series, the LEVEL of water in 40 days will be 40th term of the series.
\(a40 = x*2^(^40-1^)\)
\(a40= x * 2^3^9\)
Hence capacity of pond is = \(x*2^3^9\)
1/4 of that capacity = \(1/4 * x*2^3^9\)
=\(x*2^3^7\)
Now mere observation will tell you that LEVEL of pond on a given day is \(x*2^(^D^A^Y-^1^)\)
Like on day 4, level is 8x => \(2^(^4^-^1^)x\)
on day 5, level is 16x => \(2^(^5-^1^)x\)
On day 38, LEVEL will be =\(x * 2^(3^8-^1) = x*2^3^7.\)
Hence on 38th day pond will be 1/4 full.
OA E.
. change the above question to :
