On January 1, Ashutosh opened a current account by investing $300. On

given 1st Jan 300$
8th and 15th Jan ; n$+n$; 2n$
upto 20th ; 300+2n
21st ; 300+2n-250 ; 50+2n $
so avg
(50+2n)*31/31 = 300
solve for n =125
~ for two days 125/2 ; 61 IMO C


On January 1, Ashutosh opened a current account by investing $300. On each of 8th January and 15th January, he invested $n on that account. However, on 22nd January he withdrew $250 from the account due to an emergency need. If the balance on that account at the end of 31st January was $300, and the average balance for each day was also $300, then what is the approximate value of n?

A. 43
B. 55
C. 61
D. 72
E. 97

IMO answer is B

average = 300 = (300*7+(300+n)*7+(300+n+n)*7+(300+n+n-250)*9+300)/31, solving this we get n as approx 297
so imo answer is B

Assuming each day he is getting some minor interest of $x. As we need to find approximate value , so taking interest of each day as constant
Jan1=300
Jan8=300+n+7x
Jan15=300+2n+14x
Jan22=50+2n+21x
Jan31=300=50+2n+30x

Now resolving, 300=50+2n+30x
2n+3x=250
n=125.5-\(\frac{3}{2}\)x

In order to keep x as minimum from all the provided options
n=97 gives the minimum value of x
Hence E

for 1-7 --> 7 days --> amount --> 300
8-14 --> 7 days --> amount(300+n)
15-21 --> 7 days --> amount(300+2n)
22-31 --> 10 days --> amount(300-250+2n)

The equation would be
[300(7) +7(300+n) + 7(300+2n) +10(50+2n)]/31 = 31(300)

Solving the equation gives approx n = 61.

Hence, C is the answer.

On January 1, Ashutosh opened a current account by investing $300. On each of 8th January and 15th January, he invested $n on that account. However, on 22nd January he withdrew $250 from the account due to an emergency n

[300]*7 + [300 +n]7 + [300 + 2n]7 + [50 + 2n]10 = 9300

=> n = 61 (approx)

Answer C

Given: average of 31 days is 300

Sum of daily balance in the periods
1-7th: 7(300)
8-14th: 7(300+n)
15-21th: 7(300+2n)
22-30th: 9(300-250+2n=50+2n)
31st: 300
Total: 7050+39n

Average=Sum•N…300=(7050+39n)/31…31*300=(7050+39n)…n=(9300-7050)/39=2250/39=aprox.57 or 55.

Answer (B).

We can create an equation as we read the question -
300*7 +(300+n)*7+(300+2n)*7+(300+2n-250)*10=31*300
=>41n+9300-2500=9300
=>n=2500/41
~61
IMO C is the correct choice.

On January 1, Ashutosh opened a current account by investing $300. On each of 8th January and 15th January, he invested $n on that account. However, on 22nd January he withdrew $250 from the account due to an emergency need. If the balance on that account at the end of 31st January was $300, and the average balance for each day was also $300, then what is the approximate value of n?

A. 43
B. 55
C. 61
D. 72
E. 97

Simple question.

Starting Balance = $300
8th Jan = 300+ N
15th Jan = 300+ N + N
22nd Jan = 300+N+N-250
31st Jan = 300

Lets put a simple equation: 300+ 2N - 250 = 300
N = 125 .

Seeing the answer choices 61*2 = 122 Approximately 125

Hence the answer is C.

i guessed it and marked 55.
don't know how to approach

Posted from my mobile device

Avg for 31 days is 300 so total=300*31=9100
So we can divide this in 4 dates
1st Jan to 7th Jan ->7 days
300*7=2100
8th to 14th Jan ->7 days
(300+n)*7=2100+7n
15th to 21st jan-> 7 days
(300+2n)*7=2100+14n
Remaining 10 days from 22nd to 31st=(300-250+2n)*10=500+20n
So,
2100+2100+7n+2100+14n+500+20n=9300
=>41n=9300-6800=>n=~61

Posted from my mobile device

Option C

in 1st 7 days, the balance was 300 --> sum = 2100
in 2nd 7 days, the balance was (300+n) --> sum = 2100 + 7n
in 3rd 7 days, the balance was (300+2n) --> sum = 2100 + 14n
in last 10 days, the balance was (50+2n) --> sum = 500 + 20n

the average of the 31 days = \(\frac{{6800+41n}}{31} = 300\)
so \(41n = 2500\), \(n ≃ 61\) C

C I think.

We need to get monthly average based on transaction across the month.

So, we get: 7*300+8*(300+n) + 6*(300+2n)+10*(50+2n) = 31*300
Solve for n, we get n = 62.5

the average balance for each day was also $300, Hence - 31*300(Sum of all day balances)= 7*300<first 7 days>+7*300+7n <2nd 7 days> + 7*300+7*2n <3rd 7 days> + 10*50 + 10*2n <For balance 10 days, after 250 withdrawal>

9300=2100+2100+7n+2100+14n+500+20n
3000=500+41n
41n=2500
n=2500/41~2500/40=62.something

IMO C = 61 (Less than 62 as denominator was reduced in the above calculation)

On January 1, Ashutosh opened a current account by investing $300. On each of 8th January and 15th January, he invested $n on that account. However, on 22nd January he withdrew $250 from the account due to an emergency need. If the balance on that account at the end of 31st January was $300, and the average balance for each day was also $300, then what is the approximate value of n?

Since the balance at the end of 31st Jan is the same as his initial deposit of $300
The two $n is equal to the withdrawal amount
.: 2n = 250 —> n= 125
Now we also know that averagely he deposits $300
.: approximate value of n has to be a multiple of 5
From answer choice only answer B is a multiple of 5

.: Answer B

Posted from my mobile device

On January 1, Ashutosh opened a current account by investing $300. On each of 8th January and 15th January, he invested $n on that account. However, on 22nd January he withdrew $250 from the account due to an emergency need. If the balance on that account at the end of 31st January was $300, and the average balance for each day was also $300, then what is the approximate value of n?

A. 43
B. 55
C. 61
D. 72
E. 97

Rs 300 deposited for 21 days (on 1st) = 300*21
Rs n deposited for 24 days (on 8th) = 24n
Rs n deposited for 17 days (on 15th)
Rs 50 deposited for 10 days (on 22nd) = 500

Average balance for the month = (300*21 + 24n + 17n + 500)/31 = 300
300*21 + 41n + 500 = 300 * 31
41n = 2500
n = 2500/41 = 61 approx

IMO C