pensiveowl:I need help with this problem - can someone pls help solve it - thankshttp://www.positivelyquit.org/images/math.jpg\">
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pensiveowl:I need help with this problem - can someone pls help solve it - thankshttp://www.positivelyquit.org/images/math.jpg\">
Fig 1 => 1
Fig 2 => 5 = 4 + 1
Fig 3 => 13 = 9 + 4
Fig 4 => 25 = 16 + 9
Fig 5 => 41 = 25 + 16
Fig n => n^2 + (n - 1)^2
Fig 87 => 87^2 + 86^2 = 14965 ovals
If number of ovals is 5305
Take the average of n^2 and (n-1)^2 = 5305/2 = 2652.5
Square root of 2652.5 = 51.5
n is the next whole number > 51.5 (n = 52)
52^2 + 51^2 = 5305 (Fig. 52 has 5305 ovals)
zsqchx:If Sam had 100u at firstPlease help to solve the following question. Thanks.
Sam went shopping at a department store. He spent 1/5 of his money on a soft set and $800 on a washing machine. Later, he spent 1/8 of his remaining amount on dining table. If the amount of money left was $100 more than half of the amount he had at first, how much money did Sam have at first?
bobsg01:Ratio of X : Y : Z = 3 : 4 : 9Need some engligtenment on this answer.
In Total number of parts the figure has is 2 + 9 = 11 units
Why does it used 9 instead of 8 with minus 1 for Z? Confusing because they used 2 for X (which is minus 1 instead of 3). :?
Thanks for your kind assistant.
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zsqchx:For a fixed length of candle;Can someone help for this question. Thanks.
Candle A and Candle B are of the same length. Candle A, Which is broader, can burn for 5h while Candle B, the thinner candle, can burn for 4h. If both candles are lighted at the same time, how long does it take for Candle A to be twice as long left as Candle B?
raysusan:20 yrs agoplease help me on this Qns
jean's age was 3/4 lily's age 20 years ago. jean's age was 7/9 lily's age 16 years ago. how old is lily now??
TIA :?:
Q2)
Number of households : number of computers = 3:5
5 = 2 + 2 + 1 = ( 1 x 2) + ( 1 x 2) + ( 1 x 1)
For every 3 households, 2 households have 2 computers each and the remaining household has 1 computer.
Fraction of household that have 1 computer
= 1/3
ksi:Glad you like it, ksi.I like this shorter approach! :imcool:Dharma:When distance is fixed; speed and time are inversely proportional.
Speed ratio => 80 : 60 = 4 : 3
Time ratio => 3: 4
4u – 3u = 1u = (3/4 – 1/3) = 5/12 hrs
Total distance = 80 x 3 x 5/12 = 100km
Time taken when speed is 90km/hr = 100/90 = 10/9 hrs
When distance is fixed; speed and time are inversely proportional.
Speed ratio => 80 : 60 = 4 : 3
Time ratio => 3: 4
4u – 3u = 1u = (3/4 – 1/3) = 5/12 hrs
Total distance = 80 x 3 x 5/12 = 100km
Time taken when speed is 90km/hr = 100/90 = 10/9 hrs
PiggyLalala:If we project a line from F vertically upwards until it intersects line AB and call it H and project a line from F horizontally towards the left until it intersects line AD and call it G.I remembered this question. Your previous solution used the concept of similar triangles. Thank you for coming up with an alternative solution.Dharma:
Glad that you liked it.
There is a another area qn in the other thread that can be solved as follows.
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However, this alternative solution makes use of the fact that the height of triangle b and c are the same. I feel that this assumption should not be made as the concept of congruent triangles is not taught in the primary school. Or maybe you can provide a better explanation as to why the heights have to be the same without using congruent triangles.
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Nonetheless, I am grateful to you and all those who have helped to solve the many difficult sums found in this thread. :thankyou: