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Mathematics: Post your doubts here!

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In quadrilateral ABCD, AB produced is perpendicular to DC produced. If Angle A is 44 and angle C is 148, calculate angle D and B
 
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then please solve the question which i have asked.
sorry fr d late reply ..........welll here u go


Draw the quadrilateral ABCD and let AB produced meet DC produced at E

Given,

angle A = 44 degrees

angle C = 148 degrees



The triangle AED is a right triangle with right angle at E (since AB is perpendicular to DC) thus

angle D = 180 - (90 + angle A) = 180 - (90 + 44) = 46 degrees

angle D = 46 degrees



The sum of the angles in the quadrilateral is 360 thus

angle B = 360 - (angle A + angle C + angle D) = 360 - (44 + 148 + 46)

angle B = 122 degrees


I hope u have undestood =)
 
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X axis per 2.5 say line oper extend kro .. Us curve ko line touch kray gi .. Wahan say line full lay jana ..
Then 2 point lenay hain for gradient ..
 
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Can someone explain me Q.7 last part (b)
and last past of Q.9 (e)
 

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Ans 7 b) Mean mass of 20 oranges is 70g so total mass of 20 oranges will be 70 x 20 = 1400 g

Mean mass of 19 oranges is 70.5g so total mass of 19 oranges will be 70.5 x 20 = 1339.5 g

The mass of the orange which is eaten = 1400 - 1339.5 = 60.5 g


Ans 9 e) This question is bit hard !! But here we go

The probability of weather is fine is 3/4

The probability of weather is not fine will be 1 - 3/4

The probability of weather is not fine for 1 day in 5 days will 1 - (3/4)^5 = 781 / 1024

Note ( ^ means raise to power)
 
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Ans 7 b) Mean mass of 20 oranges is 70g so total mass of 20 oranges will be 70 x 20 = 1400 g

Mean mass of 19 oranges is 70.5g so total mass of 19 oranges will be 70.5 x 20 = 1339.5 g

The mass of the orange which is eaten = 1400 - 1339.5 = 60.5 g


Ans 9 e) This question is bit hard !! But here we go

The probability of weather is fine is 3/4

The probability of weather is not fine will be 1 - 3/4

The probability of weather is not fine for 1 day in 5 days will 1 - (3/4)^5 = 781 / 1024

Note ( ^ means raise to power)
Wonderful. ... Thanks bro
 
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Oh this question! Just cracked it yesterday, but its incredibly long!

Alright first we gather the possibilities. So to get to the 100th place, these are the possible ways:

1. Roll a 3.
2. Roll a 2 on first throw and then a 4 on second throw.
3. Doesn't roll a 2 or 3 on first throw and then rolls a 3 on second throw.

Now we gather the probabilities for each of these cases.
To roll a 3, he can either roll a 1 and 2 on die, or a 2 and 1 on die. Which is (1/6 * 1/6) + (1/6 * 1/6).
Therefore to roll a 3, the probability is 2/36.

Next, for roll a 2 on first throw and 4 on second.
To roll a 2, he needs to roll 1 and 1 on die. Therefore it is (1/6 * 1/6). which is 1/36.
Now to roll a 4 there are numerous possibilities. He can roll a 1 and 3, or a 3 and 1, or a 2 and 2. Calculate the probabilities for this as follows:
(1/6*1/6) + (1/6*1/6) + (1/6*1/6).
Therefore probability for rolling a 4 is 3/36.

Next, we find probability for not rolling a 2 or 3. Which is the same as 1 - (probability of rolling 2 or of rolling 3).
1 - (1/36 + 2/36). This is (1 - 3/36) or 33/36. same thing. And now probabibility of rolling a 3 on second throw will be 2/36 as calculated before.


FINALLY. We calculate all these.
So probabibility of getting a 100 is:
Probability of rolling a 3, OR, not rolling 2 or 3 on first throw, AND rolling 3 on second, OR probability of rolling a 2 on first throw AND 4 on second. Remember OR is adding. AND is multiplying.
[2/36] + [1-(3/36) * 2/36] + [1/36 * 3/36]

And that gives the final answer of 47/432. If you check the calculation above is similiar to one in marking scheme!

If you need any clarification on this, let me know.
Cheers
 
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I have one question on the marking of these papers, do they give all marks if answer is correct or the correct method should also be there?
What if the method is different from the one in marking scheme but answer is correct, is full credit awarded then?
 
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