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For a certain savings account, the table shows the three transactions
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For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was $1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was $1,000, what was the amount of the deposit on June 21?
Okay, so june 1 - june 10, the sum of the daily balance in this time period here is 10,000 (because initial balance is 1000, and nothing changes) june 11-15 - the sum here is 3250 (1000-350)*5 june 15-20 the sum here is 750 (1000-350-500)*5
the sum of all those is 14,000.
Since we want to find a deposit, whose sum makes the total sum of all the balances 30,000 / 30 = 1,000
I did 30,000 - 14,000 which gives me 16,000.
so the sum of the balances for rest of the days (10) must equal 16,000. I do 16,000/10
and that's 1,600. E. But thats wrong. The answer is D
What am I donig wrong here? and what is the false logic?
For a certain savings account, the table shows the three transactions
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23 Nov 2012, 03:10
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For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was $1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was $1,000, what was the amount of the deposit on June 21?
A. $1,000 B. $1,150 C. $1,200 D. $1,450 E. $1,600
The daily balance from June 1 to June 10 (10 days) was $1,000 --> the sum of the balances 10*$1,000 = $10,000; The daily balance from June 11 to June 15 (5 days) was $1,000-$350=$650 --> the sum of the balances 5*$650 = $3,250 ; The daily balance from June 16 to June 20 (5 days) was $650-$500=$150 --> the sum of the balances 5*$150 = $750; The daily balance from June 21 to June 30 (10 days) was $150 --> the sum of the balances 10*($150 + x).
We need the sum of the balances for 30 days to be $30,000 (since the average for 30 days was $1,000), thus we need \(10,000+3,250+750+10*($150+x)=30,000\); \(x=$1,450\).
Answer: D.
OR: For 5 days the total balance was less by 5*$350 = $1,750; For another 5 days the total balance was less by 5*($350 + $500) = $4,250;
The final 10 days of June should compensate \($1,750 + $4,250 = $6,000\), thus the sum of the balances for the final 10 days must be \($10,000+$6,000=$16,000\): \(($100-($350+$500)+x)*10=$16,000\); \(x=$1,450\).
Re: For a certain savings account, the table shows the three transactions
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24 Dec 2012, 00:46
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liarish wrote:
Sorry for not being clear. I mean that the table says that the date of transaction is June 11. So what I am thinking is that balance on June 11 is 1000-350 = 650. Not that balance from June 11-June 16 is is (1000-350= 650 )*5. What language in the question is telling us that its for 5 continuos days (June 11-16)? ie. I am confused about the "Each" part in the following statement.
Each day from June 11 to June 15 (5 days) the daily balance was $1,000-$350=$650, thus the sum of the balances 5*$650=$3,250.
Hi,
I think I got your point Liarish, Let me try to explain.. the point to understand is that we are dealing with account balance"so if someone withdarws money on day 1, the balance will remain the same unless there is an other inflow/outflow cash transaction.. keep that in mind.. from june 1 -10 the account had Rs 1000/- on all 10 days -> so balance for the above 10 days = 1000x10 --------- (1) Now June 11 the guy withdraws 350/- bucks.. so the balance in his account is 650. and since no transaction happens in the account for the next 5 days the balance will remain the same.. hence the balance for the above 5 days = 5x650 --------- (2) Now june 16th he withdraws another 500 , so now his account bal = 150 and it will remain the same for the next 5 days.. so balance for the above 5 days = 150x5 --------- (3) and on 21st he puts "X".. so the amount in his account is 150+x and it will remain the same till 30th... so total balance for the last 10 days= (150+X)x10 --------- (4)
no the sum of all the (1,2,3,4) would be equal to 1000x30 as the average balance for the month is 1000(given) on solving for X we get the answer.. Gud Work Bunuel
Re: For a certain savings account, the table shows the three transactions
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20 Dec 2012, 22:49
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Bunuel, I am confused about this part:
The daily balance from June 11 to June 15 (5 days) was $1,000-$350=$650 --> the sum of the balances 5*$650=$3,250 ; The daily balance from June 16 to June 20 (5 days) was $650-$500=$150 --> the sum of the balances 5*$150=$750;
why are we multiplying by 5? Aren't the withdrawals/deposit made on a particular day ? how do we know that the withdrawals were made for 5 continuos days?
Re: For a certain savings account, the table shows the three transactions
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21 Dec 2012, 04:30
liarish wrote:
Bunuel, I am confused about this part:
The daily balance from June 11 to June 15 (5 days) was $1,000-$350=$650 --> the sum of the balances 5*$650=$3,250 ; The daily balance from June 16 to June 20 (5 days) was $650-$500=$150 --> the sum of the balances 5*$150=$750;
why are we multiplying by 5? Aren't the withdrawals/deposit made on a particular day ? how do we know that the withdrawals were made for 5 continuos days?
Not sure that I understand your question. Anyway:
Each day from June 11 to June 15 (5 days) the daily balance was $1,000-$350=$650, thus the sum of the balances 5*$650=$3,250. Each day from June 16 to June 20 (5 days) the daily balance was $650-$500=$150, thus the sum of the balances 5*$150=$750.
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Re: For a certain savings account, the table shows the three transactions
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21 Dec 2012, 05:50
Sorry for not being clear. I mean that the table says that the date of transaction is June 11. So what I am thinking is that balance on June 11 is 1000-350 = 650. Not that balance from June 11-June 16 is is (1000-350= 650 )*5. What language in the question is telling us that its for 5 continuos days (June 11-16)? ie. I am confused about the "Each" part in the following statement.
Each day from June 11 to June 15 (5 days) the daily balance was $1,000-$350=$650, thus the sum of the balances 5*$650=$3,250.
Re: For a certain savings account, the table shows the three transactions
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23 Dec 2012, 07:48
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liarish wrote:
Sorry for not being clear. I mean that the table says that the date of transaction is June 11. So what I am thinking is that balance on June 11 is 1000-350 = 650. Not that balance from June 11-June 16 is is (1000-350= 650 )*5. What language in the question is telling us that its for 5 continuos days (June 11-16)? ie. I am confused about the "Each" part in the following statement.
Each day from June 11 to June 15 (5 days) the daily balance was $1,000-$350=$650, thus the sum of the balances 5*$650=$3,250.
Since no transactions were made from June 11 till June 16, then the daily balance from June 11 to June 15 was $650 (on each day).
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Re: For a certain savings account, the table shows the three transactions
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24 Dec 2012, 07:12
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Great !! Thank you Bunuel and SpotlessMind.
This is exactly what I was confused about : "so if someone withdarws money on day 1, the balance will remain the same unless there is an other inflow/outflow cash transaction.. keep that in mind.. " that the previous balance remains same i.e continues until the next transaction.
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25 Dec 2012, 07:32
SpotlessMind wrote:
liarish wrote:
Sorry for not being clear. I mean that the table says that the date of transaction is June 11. So what I am thinking is that balance on June 11 is 1000-350 = 650. Not that balance from June 11-June 16 is is (1000-350= 650 )*5. What language in the question is telling us that its for 5 continuos days (June 11-16)? ie. I am confused about the "Each" part in the following statement.
Each day from June 11 to June 15 (5 days) the daily balance was $1,000-$350=$650, thus the sum of the balances 5*$650=$3,250.
Hi,
I think I got your point Liarish, Let me try to explain.. the point to understand is that we are dealing with account balance"so if someone withdarws money on day 1, the balance will remain the same unless there is an other inflow/outflow cash transaction.. keep that in mind.. from june 1 -10 the account had Rs 1000/- on all 10 days -> so balance for the above 10 days = 1000x10 --------- (1) Now June 11 the guy withdraws 350/- bucks.. so the balance in his account is 650. and since no transaction happens in the account for the next 5 days the balance will remain the same.. hence the balance for the above 5 days = 5x650 --------- (2) Now june 16th he withdraws another 500 , so now his account bal = 150 and it will remain the same for the next 5 days.. so balance for the above 5 days = 150x5 --------- (3) and on 21st he puts "X".. so the amount in his account is 150+x and it will remain the same till 30th... so total balance for the last 10 days= (150+X)x10 --------- (4)
no the sum of all the (1,2,3,4) would be equal to 1000x30 as the average balance for the month is 1000(given) on solving for X we get the answer.. Gud Work Bunuel
Hey Guys, I'm sorry but I'm not getting it.. In bank account, if I deposit $1000.00 on say Jan-01 and if for next 30 days there is no transaction then at the end 30 days my bank account will have $ 30000.00 as per your explanations.Does that make sense guys ?Keeping money idle in the bank and its getting multiplied with no. of days kept in the account...I mean certainly there will be addition of Interest but that will not fetch you 30 times your deposit in 30 days.
Then, why we're considering this theory..Bunuel I would request you help.
Please come up with a deep analysis on this.
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Re: For a certain savings account, the table shows the three transactions
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26 Dec 2012, 04:25
debayan222 wrote:
SpotlessMind wrote:
liarish wrote:
Sorry for not being clear. I mean that the table says that the date of transaction is June 11. So what I am thinking is that balance on June 11 is 1000-350 = 650. Not that balance from June 11-June 16 is is (1000-350= 650 )*5. What language in the question is telling us that its for 5 continuos days (June 11-16)? ie. I am confused about the "Each" part in the following statement.
Each day from June 11 to June 15 (5 days) the daily balance was $1,000-$350=$650, thus the sum of the balances 5*$650=$3,250.
Hi,
I think I got your point Liarish, Let me try to explain.. the point to understand is that we are dealing with account balance"so if someone withdarws money on day 1, the balance will remain the same unless there is an other inflow/outflow cash transaction.. keep that in mind.. from june 1 -10 the account had Rs 1000/- on all 10 days -> so balance for the above 10 days = 1000x10 --------- (1) Now June 11 the guy withdraws 350/- bucks.. so the balance in his account is 650. and since no transaction happens in the account for the next 5 days the balance will remain the same.. hence the balance for the above 5 days = 5x650 --------- (2) Now june 16th he withdraws another 500 , so now his account bal = 150 and it will remain the same for the next 5 days.. so balance for the above 5 days = 150x5 --------- (3) and on 21st he puts "X".. so the amount in his account is 150+x and it will remain the same till 30th... so total balance for the last 10 days= (150+X)x10 --------- (4)
no the sum of all the (1,2,3,4) would be equal to 1000x30 as the average balance for the month is 1000(given) on solving for X we get the answer.. Gud Work Bunuel
Hey Guys, I'm sorry but I'm not getting it.. In bank account, if I deposit $1000.00 on say Jan-01 and if for next 30 days there is no transaction then at the end 30 days my bank account will have $ 30000.00 as per your explanations.Does that make sense guys ?Keeping money idle in the bank and its getting multiplied with no. of days kept in the account...I mean certainly there will be addition of Interest but that will not fetch you 30 times your deposit in 30 days.
Then, why we're considering this theory..Bunuel I would request you help.
Please come up with a deep analysis on this.
No one is saying that.
If you deposit 1,000 on January 1 and there is no transaction after that, then you'll daily balance would be 1,000 each day but the sum of the balances would be 31*1,000, which does not mean that you'll have 31*1,000 at the end.
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Re: For a certain savings account, the table shows the three transactions
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26 Dec 2012, 05:44
[ Hey Guys, I'm sorry but I'm not getting it.. In bank account, if I deposit $1000.00 on say Jan-01 and if for next 30 days there is no transaction then at the end 30 days my bank account will have $ 30000.00 as per your explanations.Does that make sense guys ?Keeping money idle in the bank and its getting multiplied with no. of days kept in the account...I mean certainly there will be addition of Interest but that will not fetch you 30 times your deposit in 30 days.
Then, why we're considering this theory..Bunuel I would request you help.
Please come up with a deep analysis on this.[/quote]
No one is saying that.
If you deposit 1,000 on January 1 and there is no transaction after that, then you'll daily balance would be 1,000 each day but the sum of the balances would be 31*1,000, which does not mean that you'll have 31*1,000 at the end.[/quote]
Hey Guys, I apologize if I sound harsh..seriously didn't mean that way..!
But I'm still unable to get it..Are you considering that every day after Day 1, for the rest of the month you''ll be depositing $1000 each day..otherwise how can you sum up and get it $31000 at the end of the month where it's mentioned that there is no transaction hence no inflows to the account after Day1 till the end of the month.
So,$1000 deposit on Day1, no further inflow/outflow till Day31...sum is 31000.! By what logic we're concluding this? I'm completely lost.. Would request your help Bunuel...!
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Re: For a certain savings account, the table shows the three transactions
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26 Dec 2012, 06:25
debayan222 wrote:
So,$1000 deposit on Day1, no further inflow/outflow till Day31...sum is 31000.! By what logic we're concluding this? I'm completely lost.. Would request your help Bunuel...!
Again, you won't have 31,000 at the end. Each day your daily balance will be 1,000. The sum of the balances will be 31,000. Cannot explain this any better.
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Re: For a certain savings account, the table shows the three transactions
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29 Mar 2013, 01:23
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1
June has 30 days
1)for the first 10 days the amount in the deposit is 1000 2)for the next 5 days is 650 (1000-350) 3)for the next 5 days is 150 (650-500) 4)for the final 10 days is x
The median is \(\frac{1000*10+650*5+150*5+10*x}{30}=1000\) \(10x=16000, x= 1600\) So the amount for the last 10 days was 1600 $, but the question asks for the DEPOSIT, and knowing that in the account there were 150 $, the deposit is 1600-150=1450 $
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Re: For a certain savings account, the table shows the three transactions
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29 Mar 2013, 02:35
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mun23 wrote:
Need explanation of the attached file...........
Given:
1. The variable is the daily balance 2. The variable changes thrice i.e., it takes 4 values. 3. The average daily balance for the 30 days is given as 1000
Relevant topic: Weighted average. The clues for identifying this topic are the mention of arithmetic mean and the unequal days (i.e., the weights) for each value.
The deposit therefore is 1600-150=1450 ( because the balance was already 150 when the deposit was made and so has to be deducted to find the deposit)
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Re: For a certain savings account, the table shows the three transactions
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29 Mar 2013, 09:44
debayan222 wrote:
[ Hey Guys, I'm sorry but I'm not getting it.. In bank account, if I deposit $1000.00 on say Jan-01 and if for next 30 days there is no transaction then at the end 30 days my bank account will have $ 30000.00 as per your explanations.Does that make sense guys ?Keeping money idle in the bank and its getting multiplied with no. of days kept in the account...I mean certainly there will be addition of Interest but that will not fetch you 30 times your deposit in 30 days.
Then, why we're considering this theory..Bunuel I would request you help.
Please come up with a deep analysis on this.
Its NOT about cash inflow. ITS BECAUSE THE PROMPT MENTIONS THE AVERAGE OF BALANCES For example If I have a balance of $1000 for 3 days in my a/c the average is $(1000X3)/3 = 1000
If I had withdrawn $500 on the second day BY THE LOGIC OF THE PROMPT THE AVERAGE WILL BE $(1000+500+500)/3 = $667
Re: For a certain savings account, the table shows the three transactions
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29 Mar 2013, 16:59
Since there is only three things happening on the account...
$1000 to start off with, -$350 ($750 left), -$500 ($250 left) how much do we need to deposit to take the avg back to $1000?
1000+750+250 = 2000, divide this by 3 (2000/3) and we get roughly 666. Then $1000 - $666 = $444, so the account must go up to $1000 + 444 =1444 (since 666 is roughly) its 1450 that was deposited on that day.
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27 Sep 2016, 00:14
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anon1 wrote:
Attachment:
Untitled.png
For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was $1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was $1,000, what was the amount of the deposit on June 21?
Okay, so june 1 - june 10, the sum of the daily balance in this time period here is 10,000 (because initial balance is 1000, and nothing changes) june 11-15 - the sum here is 3250 (1000-350)*5 june 15-20 the sum here is 750 (1000-350-500)*5
the sum of all those is 14,000.
Since we want to find a deposit, whose sum makes the total sum of all the balances 30,000 / 30 = 1,000
I did 30,000 - 14,000 which gives me 16,000.
so the sum of the balances for rest of the days (10) must equal 16,000. I do 16,000/10
and that's 1,600. E. But thats wrong. The answer is D
What am I donig wrong here? and what is the false logic?
Responding to a pm: For the ease of calculations, you can use deviation from the mean concept for weighted averages.
Re: For a certain savings account, the table shows the three transactions
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12 Feb 2018, 13:19
Hi All,
For this question, we have to deal with the average the daily balance at the END of each day for the entire 30 days.
For the first 10 days, the average is $1,000 For the next 5 days, the average is $650 For the next 5 days after, the average is $150 For the final 10 days, the average is $X
We're told that the average for the ENTIRE month is $1,000, so we need to use the Average Formula:
[10(1,000) + 5(650) + 5(150) + 10(X)]/30 = 1,000
This math can be done in a couple of different ways, but you'll eventually get to….
[14,000 + 10X]/30 = 1,000
14,000 + 10X = 30,000 10X = 16,000 X = 1600
Since X is the BALANCE for the last 10 days and there was already $150 in the account BEFORE the deposit was made, the deposit must be $1,450
Re: For a certain savings account, the table shows the three transactions
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31 May 2018, 21:54
VeritasPrepKarishma wrote:
anon1 wrote:
Attachment:
Untitled.png
For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was $1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was $1,000, what was the amount of the deposit on June 21?
Okay, so june 1 - june 10, the sum of the daily balance in this time period here is 10,000 (because initial balance is 1000, and nothing changes) june 11-15 - the sum here is 3250 (1000-350)*5 june 15-20 the sum here is 750 (1000-350-500)*5
the sum of all those is 14,000.
Since we want to find a deposit, whose sum makes the total sum of all the balances 30,000 / 30 = 1,000
I did 30,000 - 14,000 which gives me 16,000.
so the sum of the balances for rest of the days (10) must equal 16,000. I do 16,000/10
and that's 1,600. E. But thats wrong. The answer is D
What am I donig wrong here? and what is the false logic?
Responding to a pm: For the ease of calculations, you can use deviation from the mean concept for weighted averages.
I read through your article and went through the solution you posted here. The only part I didn't get was : why did you multiply 350 and 500 with 20 and 15 respectively?
I multiplied 350 with 5 (as we are short of 350 for 5 days) and multiplied 850 with 5 (as we are short of 850 for 5 days)
Shortfall = Excess
350*5 + 850*5 = 10*x
x = 600
Can you tell me where I am going wrong?
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