Serene Koon
Asked 8 years ago

SG chevron_right Primary 6 chevron_right Number and Algebra

Please help. Thanks

Replies 1

Serene Koon

(a) 20% of 600 chairs ---- 120 20÷3 = 40 chairs (b) 600 ÷40 = 15 rows

5 years ago

Serene Koon
Asked 8 years ago

SG chevron_right Primary 6 chevron_right Measurement

Please help. Thanks

Replies 2

Tjew Sheng Wei

5 years ago
Yeo See Yeong

5 years ago

Jesline Goh
Asked 8 years ago

SG chevron_right Primary 5 chevron_right Number and Algebra

Saw a similar question posted but I still don't quite get it. Please help with models please. Thanks!

Replies 10

BA Poh Ann

5 years ago
Yueh Mei Liu

Assume you first give each pupil 5 candies and have 3 candies left. Now you decide that you want to give each pupil 6 candies. You would take the extra 3 candies and give each pupil a candy (from 5 to 6), but there will be some pupils who could not have 6 candies because the question says that there will be 7 candies short. So, if we are able to get 7 candies somewhere, then we will be able to give all the pupils a candy each. 3 candies + 7 candies = 10 candies. Since I give 1 candy (6 - 5) to each pupil, I would have given to 10 pupils. Hence there are 10 pupils. If you know the number of pupils, you can find the number of candies in total.

5 years ago
Yueh Mei Liu

Above is the reasoning and explanation for understanding. The working steps are just 3 +7 = 10; 6-5=1; 10/1 = 10 pupils.

5 years ago
KH Puah

5 years ago
AisinGioro YongZhen

Suppose we have n units of 5 candies where each unit denotes a student receiving it with a surplus of 3 candies after distributing all to the student. Now when 6 candies are distributed to each students we have a shortage of 7, so we consider the first case where there is 3 surplus candies, we take 1 candy and give to the 1st student and he has 6 candies with 2 surplus, and we take the 2nd candy and give to the 2nd student and he has 6 candies with 1 surplus, and 3rd student has 3 candies with 0 surplus, and 4th student has 6 candies with -1 surplus (i.e shortfall of 1), and 5th student has 6 candies with -2 surplus, so on until the 10th student with 6 candies with -7 surplus ( i,e shortfall of 7).

5 years ago
AisinGioro YongZhen

Hope this helps, students might have issues with adding of -1 seven times to become -7.

5 years ago
Jesline Goh

Thanks all!

5 years ago
Ye Wint

7+3=10 as 6 and 5 candy differeblnce is 1 so there are 10 pupils and 10×5=50 and plus 3 that is the total number of sweet

5 years ago
Mike Tan

Easiest method i can think of

5 years ago
Tan SUnny

Model

5 years ago

Magdalene Sim
Asked 8 years ago

SG chevron_right Primary 5 chevron_right Number and Algebra

Please help thanks :) pri 5

Replies 1

KH Puah

5 years ago

Magdalene Sim
Asked 8 years ago

SG chevron_right Primary 5 chevron_right Fractions

Please help thanks :)

Replies 1

Ida Tan

5 years ago