Since an equal quantity (15 years) is to be added to both Jeremy and his aunt, we can add the 15 to the front of the bars to make the comparison easier.

Thanks.. but can tell me how u derive 3u =15?

And why one bar split into 5 equal parts?

Because the information given for their age after 15 years is 8 : 13, so the 'extra' bar is 13 - 8 = 5 units. That means each blue bar is 5 units, leaving the green to be 3 units.

Oh ya

Thanks!

A similar question was posted on Tuesday: https://www.facebook.com/photo.php?fbid=10153061406553093&set=gm.509550745868253&type=1&theater

Thanks, will let my son try this out too.

25-8 = 17

Oops... Thanks, Kelvin Koh, it should be 17 years. :)

It is difference unchanged.

yes how u get 3u is 15

ill let my daughter try. thks

Izam, your solution is easy n clear to me. Thanks!

thank you so much

Something is wrong with the numbers. At the beginning the numbers differ by 2, but at the end they differ by 1. As it stands it doesn't make sense.

Yah hor. So it's all the odd numbers from 3 to 99 inclusive plus 100?

Also your formula a bit salah. It's (L+a)*n/2, Where a = 1st term. It's not 1 cos doesn't start from 1 L = last term n = number of terms In this question, n and L are not equal

the qn starts from 3..

i got the formula from one of the posts to find the total of 1 to 99.

then someone use this formula (N+1)/ N/2

It's not divide. It's (n+1)*n/2

oh ya, i saw wrongly. thanks Ariel Lim Ah Ping

For 1 to 99, a= 1, L = n = 99 So it becomes (n+1)*n/2

Pri kids do not learn formula

You cannot assume it is the odd numbers plus 100. Unless the child understands the formula, I would definitely not allow its use. Also that formula is for the numbers 1 + 2 + 3 + 4 + ..., for numbers increasing by 1. So it will not work in this case.

this qn i know. how about the above one cos i saw this qn so i tried to use this formula.

so use the rainbow method?

Yes, use the rainbow method, if you can figure out the correct last number.

so the rainbow method be solved for my qn posted? pls enlighthen.

Rainbow method???

O. U mean 3+99, 5+97, n so on?

(a +L)*n/2 is that rainbow method loh.

If sum of odd number 1 to 99, the sum is square numbers. Try use pattern

Yes, Ariel, that is the rainbow method.

Some visuals for sunny's suggestion. http://www.9math.com/book/sum-first-n-odd-natural-numbers

Consecutive number 1 to 99, use rainbow method

Thats why must understand the formulas n don't use blindly.

Aiyoh. Any arithmetic progression (,I.e. got common difference, be it 2 or 1) can use rainbow.

For consecutive number 1 to 99

For consecutive even numbers, reduce to consecutive numbers, use rainbow method and workback by time 2

Exposure the child to all these variation using rainbow method

Mr tan, so if the numbers are odd so divide by 3 first .....then answer multiple by 3 ?

If multiples of 3... Like 3+6+9+12....

do u have any other example..

Try these variation

Important note for the child is using same technique to look for pattern...There is no formula to memorise

for odd number 1+3 +5+7...99 same method?

answer for this qn is it 4950?

soryy Mr tan, what do we do if it is not starting with 1 but 80?

to find how many terms.

Yes..Sum of odd numbers = square numbers (using pattern)

use this formula. (n+a/2) (n-a+d/d). do u know meaning.a-first no. n-last no. d-common difference

If number do not start from 1 but 80... Find sum of 1 to 250 minus 1 to 79

Mr tran, do u mind to show the steps for this qn 80 +81+82...250?

Sorry, should be first 79 term exclude 80

This shld be the correct version

Mr Tan can i Pm u my workings based on fnd the sum of 1 to 250 - 1 to 79?

28215. Poh poh

lol

Poh Poh, it takes a long while for me to understand the rainbow method.

Finally I got it.

I will ask my friend to make a slide for u ^_^

Thank you Shu Hao

Poh Poh , but must try to understand it.

Foong Foong Ling answer is 28 215?

Yes

Ermmmm

I asked so many qns over here n m so determined to find out the steps n finally understood the steps.

Can you WA me see how you do it? Clever you

Sure

Thank you 😘

Watsapp

how we do it guys??

Rye Tan here~~

Used the question in the main post as an example....

what is the answer everyone

Hello Foong Foong Ling. Your determination to get to the bottom of things is commendable. Let me teach you how you can modify the formula to add up the odd numbers between 1 and 100, which is 3+5+7+......+97+99. Again I will list two of the series in columns, one ascending and one descending, with a third column to show the sum of elements in the same row. 1st , 2nd , total in row 3 , 99 , 102 5 , 97 , 102 7 , 95 , 102 : : 97 , 5 , 102 99 , 3 , 102 There are altogether 49 rows or in other words there are 49 odd numbers in the series. Do you know why? So add the first and smallest number (3 ) in the series to the largest and last number (99 ) to yield 102. There will be 49 102's if you add up the two series. The sum of one single series is thus (3 + 99) × 49/2 = 102 × 49/2 = 51 × 49 = 2499 One of the contributors mentioned that the sum of odd numbers starting from 1 to N is N^2. So 3+5+7+....+97+99 = 50^2 - 1 = 2500 - 1 = 2499 Using the formula is nothing wrong, as long as you know what you are doing. The formula can help speed up problem solving, so it cannot possibly be a bad thing unless it is used blindly without comprehension. I applaud your effort to learn it. Please let me know if you have problems using it.

As for adding the sum of all multiples of 3 between 1 and 100, first we need to know there 33 multiples of 3, the smallest being 3 and the largest being 99. (3 + 99) × 33/2 = 102 × 33/2 = 51 × 33 = 1683

Oops sorry I told Rye the wrong Question...... But idea is there lah

Thanks Xavier for your compliment. Can I pm this qn n see if I m correct?

By all means please

As for the sum 80 + 81+83+....+249+250, (80 + 250) × 171/2 = 165 × 171 = 28215

Which paper will such qn be set?

p1 i think

Paper 2!

is it

It really doesn't matter which paper this type of questions appear in. If you can master it you will wish it appears always as a 5-mark question.

yes agree

In a PSLE MCQ, the question asked for the digit in the ones place for the sum of 1+2+3+....+96+97

This one I know.

Solution : The sum is (1+ 97) × 97/2 = 49 × 97 There is no need to find the sum since the student is only required to find out the ones place of 49 × 97, which is 3

I have seen this qn before n tried using rainbow method

The formula shortens the amount of time to arrive at the answer so it pays if your child can master it. I would like to stress that the formula should never be memorized and applied without understanding.

hehe still trying

No hurry. It is worth taking one's time to master the formula. This formula will be useful up to the A levels and beyond.

To find: 3 + 5 + 7 + ... + 99 + 100 Break it down as such: Find the sum of odd numbers up to 99 (1 + 3 + 5 +... + 99) Subtract 1 (get 3 + 5 + ... + 97 + 99) Then add 100 (get 3 + 5 + ... + 97 + 99 + 100) Sum of odd numbers: Number of terms = (1 + 99) / 2 = 50 Sum = 50 x 50 = 2500 2500 - 1 + 100 = 2599 Key idea here is to start off with the closest pattern relation and play around from there.

Some advice not to use formula

That is only if the student does not understand how the formula can be applied. I strongly recommend students who can understand the formula to use it. Discouraging one from using the formula when the formula can be mastered is like having a hammer but one still uses the forehead to hammer the nail into the wall.

(N+1)*N/2?

Yes. Of course the variable in the formula needs to be tweaked accordingly when applied to different questions.

I remember when my girl was in p3 she has this qn. Find out the sum of 1+2+3......+98+99+100

(1+100) × 100/2 = 101 × 50 = 5050 At P3, I have to suppose the school would want your daughter to use number bonds or the rainbow bonds to determine the sum. 1 + 100 2 + 99 3 + 98 : : : : 50 + 51 Since there are 50 pairs of number from 1 to 100 50 × 101 = 5050

Xavier Sng yes used Rainbow bonds

If a student prefers rainbow bonds, that isn't a problem. But I hope you can see how versatile and robust the formula can lead to the answer if it is mastered.

I see stars already. I need to take a break from this but will come back to it. :(

Well Jasmine, the formula applied to the summation from 1 to 151 inclusive would yield (1 + 151) × 151/2 = 76 × 151 = 11 476 So essentially the rainbow bonds is no different from the formula except that when one uses the formula there is no need to create a rainbow, which takes up time.

5 marks?!

You make it looks very easy..thanks Adrian

Thank you very much...

Thank u very much to Izam Marwasi

Thank u very much to Adrian Ng

why divide by 2 Adrian Ng

Wahidah Khan...its calculation for the area of ∆-- half x base x height

∆ADE-> base is 1u ->height is 3u Area-> 1/2 base x height -> 3u÷2= 1.5u

Wahidah Khan, This part of the figure (a rectangle) is 3u and ∆ADE is half of 3u. Agree?