"Twice more than (two times more)" vs "Twice as much"

A common mistake by my student is "twice more than", they often mistake it as "twice as much as". Actually they are different:

•   Twice as much = 2 X the base
•      Twice more than (Two times more) = 2 X the base + the base

Here is an example:

Tom's present age is twice more than Sam's age years ago. Today, Sam's age is Tom's age the same number of years ago. How old is Tom now if the sum of their present age is 30?

Solution:

In this question, "Tom's present age is twice more than Sam's age years ago", means Tom's present age is 3u, Sam's age years ago is 1u.

Same Denominator in Fraction Questions

I have received some feedback from one visitor regarding my previous post. She asked me to post an example of same denominator question. Here it is.

Question:

Adam and Ben had same amount of money. Adam spent 1/4 of his money and Ben spent 2/5 of his money. Adam spent $300 less than Ben. How much did the boys have altogether at first?

Solution:

Adam spent 1/4

Ben spent 2/5

Adam and Ben had same amount of money --> Adam's 4u has to be same as Ben's 5u

Adam 1/4 (multiply both numerator and denominator by 5) --> 5/20

Ben 2/5 (multiply both numerator and denominator by 4) --> 8/20

Adam spent 5u and Ben spent 8u

8u - 5u = 3u --> $300

1u --> $100

The boys had 16u altogether at first --> 16 x $100 = $1600

Same Numerator in Fraction Questions

Adam and Ben had $3400 altogether. After Adam spent 3/4 of his money and Ben spent 7/9 of his money, they had the same amount of money left. How much money did each boy have at first?

Solution:

Normally students tend to use model to solve this question. However, it is actually unnecessary to draw model. It can be solved just using the fraction itself.

Fraction of what Adam left --> 1/4

Fraction of what Ben left --> 2/9

They had the same amount of money left, so Adam's 1 unit needs to be changed to 2 units so as to be the same as Ben's units.

1/4 --> 2/8

At first Adam had 8 units, Ben had 9 units.
17u --> $3400
1u --> $200

Adam 8u --> $1600
Ben 9u --> $1800

Note:
The key for this kind of questions is to change either numerator or denominator to be the same depends on the questions. 

Another Ratio Question (total unchanged)

Here is another ratio question from Kiasuparents

The cost of a birthday present was shared between Benjamin and Christopher. At first, Benjamin paid 2/3 of what Christopher paid. when Benjamin paid $50 more, he ended up paying 4/5 of what Christopher paid. how much was the cost of the present?

Soultion:

Hi use total unchanged (cost of the present is constant)
First B:C = 2:3 (total 5u)
After B:C = 4:5 (total 9u)

Change total units to 45u
First B:C = 2:3 (x9) --> 18:27
After B:C = 4:5 (x5) --> 20:25

2u --> 50
1u --> 25
Cost of present 45u --> $1125 

'Set' Question

Question: 



Solution:

It can be solved using 'set'

1 set --> 8 bags of sweets and 1 bag of chocolate
Money collected for 1 set --> 8 x 5.5 + 2 = 46
No. of sets = 3128 / 46 = 68

No. of sweets sold --> 68 x 8 x 4 = 2176
No. of chocolates sold --> 68 x 8 = 544

Another Model Question

Question: 

Alex, Ben and Caleb had some stamps. Alex had 90 stamps more than Ben, and Ben had 10 stamps more than Caleb. After Alex had given Ben 95 of his stamps and Caleb bought some stamps, Alex and Caleb had the same number of stamps while Ben had thrice as much as either of them.

a) How many stamps did Caleb have at first?
b) How many stamps did they have altogether at first?

Solution:

Another common question on Ratio and Age

Question:

The ratio of the age of Mdm Huda to her son's age is 5:1 now. In 7 years time, she would be thrice as old as her son. What is the total age of Mdm Huda and her son now?

Solution:

Before ratio --> H : S = 5 : 1 (difference 4u)
After ratio --> H : S = 3 : 1 (difference 2u)
Age difference between Mdm Huda and her son is unchanged after 7 years

So make the difference same
After ratio (x2) --> H : S = 6 : 2 (difference 4u now, same as before)
1u (6u-5u) --> 7 years
Age of Mdm Huda now is 5u --> 5x7=35
Age of her son now is 1u --> 7

Two Math Questions

Two questions from Belnanna:

Question 1:

In a school science fair, there were exhibits from class A , class B and class C. Altogether, 25 exhibits came from class B and class C. In a total of 16 exhibits were not from class C and a total of 15 exhibits were not from class B.
How many exhibits were there altogether?

Solution:

16 exhibits not from C --> A + B = 16
15 exhibits not from B --> A + C = 15
2A + B + C = 31
B + C = 25
2A --> 6
A --> 3
B --> 13
C --> 12
Total exhibits --> 3 + 13 + 12 = 28


Question 2:

The length of a rectangular pond is 14 m and its breadth is 7 m. Fatimah wants to build a running path around the pond. The width of the path is 1/2 m. what is the area of the running path?

Solution:

Units and parts

Question:

Class 5A and 5B have the same number of pupils. The ratio of the number of boys in 5A to the number of boys in 5B is 2 : 3. The number of girls in 5A to the number of girls in 5B is 7 : 4. 
(A) Find the ratio of the number of boys to the number of girls in 5A.
(B) If there are 15 fewer girls than boys in 5B, how many pupils are there in 5B?

Solution:

5A boys : 5B boys = 2u : 3u
5A girls : 5B girls = 7p : 4p

5A and 5B same no. of pupils --> 2u + 7p = 3u + 4p --> 1u = 3p
5A boys : 5B boys = 2u : 3u = 6p : 9p

(A) 5A boys : 5A girls = 6 : 7

(B) 5B boys : 5B girls = 9 : 4 
5parts --> 15 
1p --> 3
No. of pupils in 5B is 13p --> 13 x 3 = 39

Cross Method for Fraction / Ratio Questions

Here is a question from visitor Belnanna:

There were 3/5 as many children as adults on a bus. At the next stop, 3 children boarded and 2 adults alighted from the bus.Then, there were 5/6 as many children as adults on the bus. How many children were there on the bus at first.

Solution:

Ratio, Ratio and Ratio!!

A very common Ration question!

Solution:

A P5 Question

The importance of 'same' in Fraction questions

When it comes to fraction questions, pay attention to the 'same' word if it appears in the question. It will be the key to solve the question. Here is an example.

Question:

Adam and Bella had $5100 altogether. After Adam spent 4/5 of his money and Bella spent 5/7 of her money, they had the same amount of money left. How much money did each of them have at first?
 
Solution:

Key info: 'same amount of money left' 

Adam left 1 unit and Bella left 2 units. It means Adam's 1 unit is equal to Bella's 2 units. 

Change Adam's units to Bella's Units as shown in the table below:

17 units -> $5100
1 unit -> $300
Adam 10 units -> $3000
Bella 7 units -> $2100 

Another Question from Kiasuparents

In class 5A, there were 1/2 as many girls as boys. In class 5B , there were 1/3 as many boys as girls. The number of boys in class 5B was 2/3 as many as the number of girls in class 5A. There were 32 pupils in class 5B.

A) express the number of pupils in 5B as a fraction of the number of pupils in 5A

B) after some boys were transferred from 5A to 5B , there were 1/2 as many of 5B boys as the number of 5B girls. How many pupils were there in class 5B after the transfer ?

Solution:

5B: Boys : Girls = 1 :3 so 4units -> 32, 1unit -> 8. Boys is 8 and girls is 24
5B Boys : 5A girls = 2:3, so 5A girls is 12
5A boys: 12x2=24 
5A total pupils is 36

A) 32/36=8/9

B) After transfer
No. of 5B girls remains the same -> 24
No. of 5B boys --> half of 5B girls -> 12
Total no. of pupils in 5B -> 24 + 12 = 36

LCM Method

A question from Kiasuparents:

A group of people met at a party. Each person shook hands with everyone else. Mr Ong shook hands with 4 times as many men as women and Mrs Ong shook hands with 5 times as many men as women. How many women were there at the party ?

Solution:

Number of Coins

Another question today.

Adam has 46 10-cent, 20-cent and 50-cent coins which add up to $14.60. There were 9 10-cent coins and the rest were 20-cent coins and 50-cent coins. How many 20-cent and 50-cent coins were there?

Solution:

10-cent coins --> $0.90
so 37 coins (20-cent coins + 50-cent coins) --> $14.60 - $0.90 = $13.70

Here we use assumption method to find the answer:

Assume all 37 coins are 50-cent coin --> 37 x $0.50 = $18.50
Difference --> $18.50 - $13.70 = $4.80
Exchange --> $0.50 - $0.20 = $0.30 
Number of 20-cent coin --> $4.80 ÷ $0.30 = 16
Number of 50-cent coin --> 37 - 16 = 21

Cost, Price, Discount and Profit

Here is a question from an anonymous visitor.

At Mrs Yu's shop, there were two vases for sale at $630 each. She sold one of them at this price and earned 40% of what she paid for it. She sold the other vase later at 20% discount. If the two vases had the same cost price, how much did Mrs Yu earn altogether?

Solution:

Vase 1: 140% --> $630, so 100% --> $450 (cost price), profit --> $630-$450= $180

Vase 2: 100% --> $630, so selling price is 80% --> $504,  profit --> $504 - $450= $54

Thus, total profit is $180+$54= $234

Note:

The common mistake in this question is the misunderstanding of " earned 40% of what she paid for it ". Students often use $630 x 40% to get the earning of first vase. However, this sentence actually means "what she paid for it " (cost price) is 100%. Therefore, the selling price $630 is 140%. 

Importance of 100%

Here is another question that my students always have problem with.

Allan had 40% more money than Ben. Charles had 20% more money than Allan. When Charles spend $1680, his amount of money decreased by 50%. Allan and Ben each spent 20% of their money. How much money did three of them left altogether?

The key for this question is to let Ben be 100% and be careful with Charles' percentage.

        84% --> $1680
        1% --> $20
        Total 276% --> $5520

The common mistake is the highlighted part - finding the percentage for Charles. 

Students normally just add 20% to Allan's 140% to find Charles. But this is wrong because Allan is not 100%.

**Note: We can only add the percentage directly when the base is 100% 

Importance of reading model

My student asked me this question:

If Alex uses his money to buy 8 similar notebooks, he will still have $6 more than Bella. If Bella uses some of her money to buy the same 8 notebooks, she will have $70 less than Alex. How much is the cost of 1 notebook?

This question can be solved easily using model.

16 notebooks + $6 --> $70

16 notebooks --> $64

1 notebook --> $4

Problem sums, even complex sums, can be solved easily using model drawings.  This is so as model drawing simplifies the problem and the pictorial representation helps pupils interpret the sum in a more comprehensible manner.   

However, the use of model drawing is just one of the many other heuristic strategies used in problem solving.  We will be sharing more effective methods, taught in Singapore Schools and we improvise them, to help you better guide your child in their Math homework.