pensiveowl:

I need help with this problem - can someone pls help solve it - thanks


http://www.positivelyquit.org/images/math.jpg\">

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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:

Please 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?

If Sam had 100u at first
Spent 1/5 of sofa => 20u
Spent on washing machine => $800
Remainder => 100u – (20u +$800) = 80u - $800
Spent 1/8 of remainder on dining table; so he’ll be left with 7/8 of remainder i.e. (70u - $700)
(70u - $700) – (100u/2) = $100
20u - $700 = $100
20u = $800
1u = $40
Amt San had at first = 100u = $4000

bobsg01:

Need 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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Ratio of X : Y : Z = 3 : 4 : 9
Ratio of Y (Shaded) : Y(Unshaded) = 1 : 3
Ratio of X (Shaded) : X (Unshaded) = 1 : 2
Total area of figure = 2u +1u + 3u + (9u – 4u) = 11u
Total area of shaded portion = 1u
Fraction of unshaded portion = (11u -1u)/11u = 10/11

zsqchx:

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?

For a fixed length of candle;
Time ratio for burning => Candle A : Candle B = 5 : 4
Ratio of burning rate => Candle A : Candle B = 4 : 5

Consider a fixed length time when the unburn section of Candle A is 2 times of that of Candle B
Ratio of burnt length : Candle A : Candle B = 4 : 5
4u + 2p = 5u + 1p [Total length of Candle A = Total length of Candle B]
1p = 1u
So, the total length of candles is 6u (4u + 2u = 6u)
For Candle A => Time taken to burn 6u is 5hrs
Time taken to burn 4u of Candle A = 4 x 5/6 = 10/3 hrs = 200 minutes
For Candle B => Time taken to burn 6u is 4hrs
Time taken to burn 5u of Candle B = 5 X 4/6 = 10/3 hrs = 200 mins

raysusan:

please 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 :?:

20 yrs ago
Diff in age = 4u – 3u = 1u

16 yrs ago
Diff in age = 9p – 7p = 2p
Difference in age between 2 persons is always the SAME.
1u = 2p
3u + 4 = 7p
6p + 4 = 7p
1p = 4
Jane’s age now = 9p + 16 = 36 + 16 = 52 yrs old

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:

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

I like this shorter approach! :imcool:

Glad you like it, ksi.

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:

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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I remembered this question. Your previous solution used the concept of similar triangles. Thank you for coming up with an alternative solution. 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.
Nonetheless, I am grateful to you and all those who have helped to solve the many difficult sums found in this thread. :thankyou:

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.
Line AC is a diagonal of ABCD which is a square. Line AF bisects AHFG which is also a square.
Area b = Area of AFE and Area c = Area of AFB
Since AE = ½ of AB; Area b = ½ x Area of c (Perpendicular height is the same; GF = HF)
Area of b = 1/3 of Area of ABE = 1/3 X ¼ X 12 X 12 = 12