Tutor MathsGuru: Ask me for your burning Maths questions!
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Hi, pls help me with these questions:
1. Luke and Isaac had $95 at first. After they each spent an equal amount of money, Luke had four times as much money as Isaac. How much money did each of them spend?
2. The ratio of the number of red marbles to the number of blue marbles on a box was 7:4. When 10 red marbles were replaced by 10 blue marbles, the ratio of the number of red marbles to the number of blue marbles became 1:2. How many red marbles were there at first?
3. The ratio of John’s age to Ryan’s age is 3:1 now. 25 years later, the ratio will become 4:3. How old is John now?
4. The ratio of the number of 20-cent coins to the number of 50-cent coins to the number of $1 coins in a bag is 2:6:7. If the total amount if money in the bag is $156, how many coins are there altogether?
Thanks -
Hi, pls help me with these questions:
Hi, pls help me with these questions:
1. Luke and Isaac had $95 at first. After they each spent an equal amount of money, Luke had four times as much money as Isaac. How much money did each of them spend?
2. The ratio of the number of red marbles to the number of blue marbles on a box was 7:4. When 10 red marbles were replaced by 10 blue marbles, the ratio of the number of red marbles to the number of blue marbles became 1:2. How many red marbles were there at first?
3. The ratio of John's age to Ryan's age is 3:1 now. 25 years later, the ratio will become 4:3. How old is John now?
4. The ratio of the number of 20-cent coins to the number of 50-cent coins to the number of $1 coins in a bag is 2:6:7. If the total amount if money in the bag is $156, how many coins are there altogether?
Thanks
:welcome:
Dear Parents,
Are you frustrated/stuck when helping your child solve his/her Maths questions? Are you inclined to use Algebra most of the time? Do you have difficulty trying to use diagrams or other heuristic methods (that Primary School students learn) to solve?
:idea: Post your questions here and see how MathsGuru solve them to the best of her ability. Detailed solutions will be posted back in this thread.
So start asking and watch this space!!
Cheers :celebrate: ,
MathsGuru
P/S (Disclaimer, in case you're wondering...):
Although MathsGuru is a full-time Maths tutor, this thread is meant to be an absolutely free resource for parents (or even children) with no strings attached. Just someone who's passionate about Maths and wanna spread the fun in learning Maths with others. :D[/quote] -
Hi MathsGuru,
Need help for these Questions:
A) There were 1072 pupils in a school hall.
5/8 of them were girls. When some of the girls left the hall, the number of girls who stayed behind made 7/13 of the total number of pupils who remained in the hall.
How many girls left the hall?
B)There were some 20cents coins in Drawer A and some 50cents coins in Drawer B. The amount of money in Drawer B was $28.90 more than the amount in Drawer A. There were 46 more coins in Drawer A than in Drawer B. Find the amount of money in Drawer A.
Thanks alot.
:welcome:
Dear Parents,
Are you frustrated/stuck when helping your child solve his/her Maths questions? Are you inclined to use Algebra most of the time? Do you have difficulty trying to use diagrams or other heuristic methods (that Primary School students learn) to solve?
:idea: Post your questions here and see how MathsGuru solve them to the best of her ability. Detailed solutions will be posted back in this thread.
So start asking and watch this space!!
Cheers :celebrate: ,
MathsGuru
P/S (Disclaimer, in case you're wondering...):
Although MathsGuru is a full-time Maths tutor, this thread is meant to be an absolutely free resource for parents (or even children) with no strings attached. Just someone who's passionate about Maths and wanna spread the fun in learning Maths with others. :D[/quote] -
Bunny27:
Hi there! Hope these answers are helpful. Didn't have the time to do up a pretty model for better visualisation. By the way, your Q1 could be wrongly worded (or have missing parts) as there are literally infinite number of solutions based on your question. (you will understand the rationale when you get to Upper Sec )Hi, pls help me with these questions:
Hi, pls help me with these questions:
1. Luke and Isaac had $95 at first. After they each spent an equal amount of money, Luke had four times as much money as Isaac. How much money did each of them spend?
2. The ratio of the number of red marbles to the number of blue marbles on a box was 7:4. When 10 red marbles were replaced by 10 blue marbles, the ratio of the number of red marbles to the number of blue marbles became 1:2. How many red marbles were there at first?
3. The ratio of John's age to Ryan's age is 3:1 now. 25 years later, the ratio will become 4:3. How old is John now?
4. The ratio of the number of 20-cent coins to the number of 50-cent coins to the number of $1 coins in a bag is 2:6:7. If the total amount if money in the bag is $156, how many coins are there altogether?
Q2.
before
RED : BLUE = 7 : 4
after
RED : BLUE = 1 : 2
Key Concept: total number of (ratio) units must remain the same since there is NO net change in total number of Blue & Red marbles (just a mere exchange)
How to make the (ratio) units before and after the same?
Total number of (ratio) units before = 7 + 4 = 11
Total number of (ratio) units after = 3
The (lowest) common multiple = 3 x 11 = 33
before
RED : BLUE = 7 : 4 = 21 : 12 (x 3)
after
RED : BLUE = 1 : 2 = 11 : 22 (x 11)
Ans = 21 red marbles
------------------------------------------------------------------------------------------------------------
Q3.
now
John : Ryan = 3 : 1
after 25 yrs
John : Ryan = 4 : 3
Key Concept: the difference between the (ratio) units of each person (object) remains unchanged after an equal amount is added to both persons (objects) (this is a typical \"aging\" question whereby everyone of us will age by an equal amount of time (days, months, years) unless one of us is from another planet)
Difference in (ratio) units of John's & Ryan's age = 3-1 = 2
Difference in (ratio) units of John's & Ryan's age in 25 yrs' time must be 2 as well, but 4-3 =1.
We need to multiply the ratio by 2 to give 8 : 6
increase in number of units for both persons = 5U (8 - 3 = 5; 6 - 1 = 5)
so 5U = 25 yrs --> 1U = 5yrs
Ans = 3U x 5 = 15 yrs' old
------------------------------------------------------------------------------------------------------------
Q4.
ratio of the number of 20cent, 50cent, & $1 coins = 2 : 6 : 7
ratio of the worth of 20cent, 50cent, & $1 coins = 2x20 : 6x50 : 7x100 = 40 : 300 : 700 = 2 : 15 : 35
this means that for every 2 units worth of 20c coins, there are 15 units worth of 50c coins, & 35 units worth of $1 coins
Key Concept: need to distinguish between number of coins (or dollar notes) and their worth in dollars & cents, & convert them into their respective ratios
total number of units (in terms of worth) = 2 + 15 + 35 = 52U = $156
so 1U = 156 / 52 = $3
amount of 20c = $3 x 2 = $6 --> number of 20c coins = 600c / 20c = 30
amount of 50c = $3 x 15 = $45 --> number of 50c coins = 4500c / 50c = 90
amount of $1 = $3 x 35 = $105 --> number of 20c coins = $105 / $1 = 105 [check 30 : 90 : 105 = 6 : 18 : 21 = 2 : 6 : 7]
Ans = 30 + 90 + 105 = 225 coins -
Hi,
Can anyone help in this question. (Nah Hua CA1 2011 , Q12)
Jan and Kay had equal number of sweets and equal number of chocolates.
Jan ate 12 sweets and Kay ate 18 chocolates and then the ratio of Jan's sweets to chocolates became 1:7 and the ratio of Kay's sweets to chocolates became 1:4.
How many sweets did Jan have at first?
TIA
-
Bunny27:
Hi again!Hi MathsGuru,
Need help for these Questions:
A) There were 1072 pupils in a school hall.
5/8 of them were girls. When some of the girls left the hall, the number of girls who stayed behind made 7/13 of the total number of pupils who remained in the hall.
How many girls left the hall?
B)There were some 20cents coins in Drawer A and some 50cents coins in Drawer B. The amount of money in Drawer B was $28.90 more than the amount in Drawer A. There were 46 more coins in Drawer A than in Drawer B. Find the amount of money in Drawer A.
pls see image file.
For A), I have included two variations (model & ratio). For the ratio method, since there is no change in the number of boys, we must change the ratio units of boys \"before\" to 6 in order to maintain the same number of units of boys \"before\" & \"after\". [note!! without this step, we cannot equate each unit \"before\" & \"after\" to be of the same value]
cheers!
http://postimage.org/image/w1qwaokk/
http://postimage.org/image/w1sju12c/ -
fxchow:
Hi there! The steps may be simple but this question is slightly \"trickier\" in the sense that students must be able to manage and equate 2 different types of units. I hope I have presented the solution in the most easy-to-understand manner by using K (Kay) & J (Jay) instead of the usual U (Units). We can also use the ratio method but I feel it may be rather complex, hence I have not included this variation of the solution. Hope this is useful to you.Hi,
Can anyone help in this question. (Nah Hua CA1 2011 , Q12)
Jan and Kay had equal number of sweets and equal number of chocolates.
Jan ate 12 sweets and Kay ate 18 chocolates and then the ratio of Jan's sweets to chocolates became 1:7 and the ratio of Kay's sweets to chocolates became 1:4.
How many sweets did Jan have at first?
TIA
cheers!
http://postimage.org/image/ycxyk6w4/ -
Hi I’ve a question from my P6 girl;
3 men, A, B and C, worked together to paint a wall. If the painting was done by one man, the time taken to complete the wall for A, B and C would have been 6 hours, 8 hours and 12 hours respectively. A and B had painted for 3 hours after which A rested. B and C then continued with the painting. What would be the total number of hours taken to complete the wall? (Give your answer as a mixed number).
The given answer by the teacher is 3 and 3/5 hours but my answer is 3 and 3/7 hours. Pls help best with workings. Thanks. -
ADoc:
Hi ADoc,
Hi there! The steps may be simple but this question is slightly \"trickier\" in the sense that students must be able to manage and equate 2 different types of units. I hope I have presented the solution in the most easy-to-understand manner by using K (Kay) & J (Jay) instead of the usual U (Units). We can also use the ratio method but I feel it may be rather complex, hence I have not included this variation of the solution. Hope this is useful to you.fxchow:
Hi,
Can anyone help in this question. (Nah Hua CA1 2011 , Q12)
Jan and Kay had equal number of sweets and equal number of chocolates.
Jan ate 12 sweets and Kay ate 18 chocolates and then the ratio of Jan's sweets to chocolates became 1:7 and the ratio of Kay's sweets to chocolates became 1:4.
How many sweets did Jan have at first?
TIA
cheers!
http://postimage.org/image/ycxyk6w4/
Thank you very much!
:lovesite: -
chrisu:
Hi there! Here's my solution. You can advise your kid to think of this sort of problem as the distance - speed - time problem.Hi I've a question from my P6 girl;
3 men, A, B and C, worked together to paint a wall. If the painting was done by one man, the time taken to complete the wall for A, B and C would have been 6 hours, 8 hours and 12 hours respectively. A and B had painted for 3 hours after which A rested. B and C then continued with the painting. What would be the total number of hours taken to complete the wall? (Give your answer as a mixed number).
The given answer by the teacher is 3 and 3/5 hours but my answer is 3 and 3/7 hours. Pls help best with workings. Thanks.
- amount of wall to be completed ~ distance
- completion per hour ~ speed (just like km per hour)
hope it's useful!
cheers!
fraction of wall completed for every hour of work
A -> 1/6, B -> 1/8, C -> 1/12 (eg. A takes 6 hours to complete one wall on his own. In other words, A will complete 1/6 of one wall per hour... ~ \"speed\" or \"rate of completion\")
A & B worked for 3hours
A would have completed 3 x (1/6) = 1/2 of the wall
within the same 3hours, B would have completed 3 x (1/8) = 3/8 of the wall
Total fraction of the wall completed in 3 hours by A & B = (1/2) + (3/8) = 7/8
Amount left uncompleted = 1 - (7/8) = 1/8 of the wall
with B & C
every hour, they would complete (1/8) + (1/12) = (3/24) + (2/24) = 5/24 of the wall
therefore time taken to complete the 1/8 of the wall = (1/8) / (5/24) = (1/8) x (24/5) = 3/5hr [D-S-T: to get T -> D/S]
ans: total time required = 3hr + 3/5hr = 3 & 3/5 hr
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