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Bunuel
Chu-Li has $100 in her piggy bank. How much will she have in her bank 40 weeks from now if she puts $1 in the bank next week, $2 two weeks from now, $3 three weeks from now, and continues to increase the amount that she puts in by $1 each week?

A. $140
B. $800
C. $920
D. $1640
E. $1740

Cumulative amount of money in 40 weeks: 1+2+3+...............+40=\(\frac{40*41}{2}\)=820

Total savings=Base deposit+Additional deposit in 40 weeks=100+820=$920

Ans. (C)
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Bunuel
Chu-Li has $100 in her piggy bank. How much will she have in her bank 40 weeks from now if she puts $1 in the bank next week, $2 two weeks from now, $3 three weeks from now, and continues to increase the amount that she puts in by $1 each week?

A. $140
B. $800
C. $920
D. $1640
E. $1740

Cumulative amount of money in 40 weeks: 1+2+3+...............+40=\(\frac{40*41}{2}\)=820

Total savings=Base deposit+Additional deposit in 40 weeks=100+820=$920

Ans. (C)

Hi, what formula is this through which we get 40*41/2?

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Bunuel
Chu-Li has $100 in her piggy bank. How much will she have in her bank 40 weeks from now if she puts $1 in the bank next week, $2 two weeks from now, $3 three weeks from now, and continues to increase the amount that she puts in by $1 each week?

A. $140
B. $800
C. $920
D. $1640
E. $1740

Cumulative amount of money in 40 weeks: 1+2+3+...............+40=\(\frac{40*41}{2}\)=820

Total savings=Base deposit+Additional deposit in 40 weeks=100+820=$920

Ans. (C)

Hi, what formula is this through which we get 40*41/2?

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Hi vedanths,

The sum of first 'n' natural numbers:\(\frac{n(n+1)}{2}\)

Here we have to determine the sum of first 40 natural numbers,(1+2+...+40)
Now put n=40 in the above sum formula.We have ,sum=40*(40+1)/2.

Hope it helps.
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PKN

Hi vedanths,

The sum of first 'n' natural numbers:\(\frac{n(n+1)}{2}\)

Here we have to determine the sum of first 40 natural numbers,(1+2+...+40)
Now put n=40 in the above sum formula.We have ,sum=40*(40+1)/2.

Hope it helps.

Yes got it, thank you!
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Initial deposit = $100
Cumulative deposit over 40 weeks = (1+2+3+4+....40)$
= 40 * 41 /2 = 820
[ As we know,1+2+3+4+.....n =n(n+1) / 2 ]
Thus total deposits of Chu-Li = (100+820)$
=920$ (C)
Hope this helps :)
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It took me longer than it should have to figure this out, but this was my approach.

Every number 1 through 9 appears 4 times so 4 x each digit 1 through 9 is:
4 + 8 + 12 + 16 + 20 + 24 + 28 + 32 + 36 = 180
10, 20, 30 each appear 10 times
(10 x 10) + (10 x 20) + (10 x 30) = 600
40 appears once:
40

600 + 180 + 40 + 100 (original amount) = 920.
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Answer: C

There are 2 ways this can be solved

With the formula Sum=\(\frac{n(n+1)}{2}\)

Sum=\(\frac{40(40+1)}{2}\)=820

The other way is slightly longer but involves identifying a pattern
Let's take the first 5 weeks
1+2+3+4+5=15
The next 5 weeks
6+7+8+9+10=40
The next 5 weeks
11+12+13+14+15=65
The next 5 weeks
16+17+18+19+20=90

If you notice the increments are of 25, we can just add up the numbers now and come to the answer.
This method is best used when then numbers are small, otherwise it might take a very long time to reach the answer.
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What we have done is (1+2+3+4...+40)+ 100 as it is 40 weeks but qs stem says person will deposit $1, 1 week from so this very week there's no deposit leaves us w only 39 weeks to deposit money so it should be (1+2+3...39) isn't it ? Any experts,kindly help

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