Tutor MathsGuru: Ask me for your burning Maths questions!
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super star:
Let me try:=pls help me with this problem
fill in the blank
0.7:7/8=__ :___
2 : 3 is also equal to 2 divide by 3
so 0.7 : 7/8
= 7/10 divide by 7/8
= 7/10 x 8/7
= 4/5
= 4 : 5
another way is to make to common denomintaors:
7/10 : 7/8
= 28/40 : 35/40
= 28 : 35
= 4 : 5
AM I correct?
THanks -
super star:
Hi Superstar,hi maths guru
please help me with this problem.
Eliot can paint a room in 3 hrs.fred can paint the same room in 5 hrs.how long will they take to paint the room if they do it together?
Let me try:
Eliot can paint 1/3 of a room in 1 hr
Fred can paint 1/5 th of a room in 1 hr.
If they paint together:
in 1 hr they can paint 8/15 of a room
So,they will take 1 7/8 hrs to paint a room together.
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HI Dharma Sir,
Tom's mother told him to wash her car. She would pay him 10 cents on the first day and twice the amount of the previous day for the next 8 days.(edited for the typo error ) How much did Tom's mother pay him altogether?
Should we take the total number of days to be 8 or 9 days?
My method:
9 days - ans : $51.10
Book method:
8 days - ans: 25.50
If book answer is right,can u Please explain why it has to be taken as 8 days.
Thankyou -
starlight1968sg:
I don't understand that given solution too.Tang:
Hi,
A - largest, E - smallest
A + B = 139
A + C = 137
Based on answer above C = 62, so A = 137 - 62 = 75
and B = 139 - 75 = 64
E + D = 105
E + C = 113
Similarly, E = 113 - 62 = 51 and D = 105 - 51 = 54
A, B, C, D, E --> 75, 64, 62, 54, 51
However the mass of 2 are:
139, 137, 129, 126, 126, 118, 116, 115, 113 and 105
These do not tally with those numbers given in the question.
Any comments?
Hi,
Given solution:
Mass of all 5 boys
= (105 + 113 + 116 + 117 + 118 + 124 + 125 + 130 + 137 + 139)/4
= 306
Sum of all readings then divide by 4 because each person partnered with 4 others.
Sum of the mass of the 2 lightest boys and 2 heaviest boys
= 105 + 139
= 244
The 2 lightest boys will give the smallest reading. The 2 heaviest boys will give the largest reading.
The mass of the 3rd heaviest boy = 306 – 244 = 62
See comments underlined. -
Tang:
Hi Yu Xuan,Dharma:
[quote=\"Yu Xuan\"]
cracking my head on how I should proceed after getting 306.
I surelyyou.
Thanks but I am actually learning from all of you everyday and I hope that our sharing will be truly beneficial to parents and their P6 kids in this coming PSLE Maths exam.
Hi,
A - largest, E - smallest
A + B = 139
A + C = 137
Based on answer above C = 62, so A = 137 - 62 = 75
and B = 139 - 75 = 64
E + D = 105
E + C = 113
Similarly, E = 113 - 62 = 51 and D = 105 - 51 = 54
A, B, C, D, E --> 75, 64, 62, 54, 51
However the mass of 2 are:
139, 137, 129, 126, 126, 118, 116, 115, 113 and 105
These do not tally with those numbers given in the question.
Any comments?[/quote]
Hi Tang,
There is some difference between the question originally posted and your version. Instead of 124 and 125 kg, you have entered two 126 kgs. -
starlight1968sg:
Hi,
I don't understand that given solution too.Tang:
Hi,
A - largest, E - smallest
A + B = 139
A + C = 137
Based on answer above C = 62, so A = 137 - 62 = 75
and B = 139 - 75 = 64
E + D = 105
E + C = 113
Similarly, E = 113 - 62 = 51 and D = 105 - 51 = 54
A, B, C, D, E --> 75, 64, 62, 54, 51
However the mass of 2 are:
139, 137, 129, 126, 126, 118, 116, 115, 113 and 105
These do not tally with those numbers given in the question.
Any comments?
My DS worked it out like this:
Let the boys in the order of lightest to heaviest be v,w,x,y,z
The weight of the two lightest boys, v and w is 105
So,
v+w=105
v+x=113
v+y=115
v+z=116
w+x=118
w+y=124
w+z=125
x+y=130
x+z=137
y+z=139
So in total there would be four v, four w, four x, four y and four z.
Total weight would be 1224kg
So, 4(v+w+x+y+z) --> 1224kg
v+w+x+y+z--> 1224kg/4= 306 kg
Since the two lightest weigh 105 kg and the two heaviest weigh 139 kg, the third heaviest or the last boy weighs, 306 kg- 139 kg- 105 kg= 62 kg[/i] -
At 10 a.m., melvin started to jog from town P to town Q at an average speed of 8 km/h.30 minutes later,natalie started cycling from town P to town Q at an average speed of 16 km/h.Both melvin and natalie reached town Q at the same time. what time did melvin reach town Q?
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Hi Mathsguru, pls advise on how u interpret this question:
Peter & Ali met at 9 am. Peter’s watch is 15 mins faster than the actual time. Ali was earlier than him by 10mins.
If Peter reached on time, according to the time shown on his watch, what is the actual time Ali arrive? -
super star:
At 10 a.m., melvin started to jog from town P to town Q at an average speed of 8 km/h.30 minutes later,natalie started cycling from town P to town Q at an average speed of 16 km/h.Both melvin and natalie reached town Q at the same time. what time did melvin reach town Q?
For fixed distance ( from Town P to Town Q)
Speed ratio => Melvin : Natalie = 1 : 2
Time ratio => Melvin : Natalie = 2 : 1
2u - 1u = 1u = 1/2 hr
Time taken by Melvin = 2u = 1hr
Melvin reached Town Q at 11.00 am ( 1 hr after 10am) -
Almighty:
Hi Almighty,HI Dharma Sir,
Tom's mother told him to wash her car. She would pay him 10 cents on the first day and twice the amount of the previous day for the next 18 days. How much did Tom's mother pay him altogether?
Should we take the total number of days to be 8 or 9 days?
My method:
9 days - ans : $51.10
Book method:
8 days - ans: 25.50
If book answer is right,can u Please explain why it has to be taken as 8 days.
Thankyou
Tom’s mother will pay him 10 cents on the 1st day.
Tom’s mother will pay him 2 times the amount of the previous day.
Tom has to wash his mum’s car for 8 days starting from Day 1. (My interpretation) - maybe if \"over the next 8 days \"would have been clearer.
Day 1 => 10 = 10(2^0)
Day 2 => 20 = 10(2^1)
Day 3 => 40 = 10(2^2)
.
.
.
Day 8 => 10(2^7) = 1280
To find the sum of payment received for 8 days
S\t = 10 + 20 + 40 + ….. + 640 + 1280
2S \t= 20 + 40 + 80 + ……. + 1280 + 2560
2S – S = 2560 – 10 = 2550
Total payment received by Tom in 8 days = $25.50
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