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**ganesh****Administrator**- Registered: 2005-06-28
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Answer to #7 :- Take a look at the graphs on the cartesian plane. Now, do you know why they're grouped in that way?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
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#19. A person walking 5/6 of his usual rate is 40 minutes late. What is his usual time?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**JohnnyReinB****Member**- Registered: 2007-10-08
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*"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted" *

Nisi Quam Primum, Nequequam

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**ganesh****Administrator**- Registered: 2005-06-28
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Absolutely right, JohnnyReinB!!!**Congratulations!!!**

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**JaneFairfax****Member**- Registered: 2007-02-23
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**ganesh****Administrator**- Registered: 2005-06-28
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Answer to #18:-

Thats right, Janefairfax!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
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#20. A shopkeeper marks his goods at such a price that after allowing a discount of 12½% on the marked price, he still earns a profit of 10%. Find the marked price of an article which costs him $4900.

#21. Sam borrowed a sum of money and returned it in three equal instalments of $35,152 each. If the rate of interest charged was 16% per annum compounded quarterly, find the sum borrowed by him.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**JaneFairfax****Member**- Registered: 2007-02-23
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**ganesh****Administrator**- Registered: 2005-06-28
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Answer to #20:-

Perfectly right, Janefairfax!

#22. Find the area of a trapezium whose parallel sides are 77 cm and 60 cm; and the other sides are 25 cm and 26 cm.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**JaneFairfax****Member**- Registered: 2007-02-23
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**ganesh****Administrator**- Registered: 2005-06-28
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Answer to #22:-

Perfectly right, Janefairfax!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**JaneFairfax****Member**- Registered: 2007-02-23
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**ganesh****Administrator**- Registered: 2005-06-28
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Answer to #17:-

Please check your answer as I am getting a different one!

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**ganesh****Administrator**- Registered: 2005-06-28
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#22. What is the number of solutions of the following equation?

#23. What are the values of x if

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**ganesh****Administrator**- Registered: 2005-06-28
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#24. If

are in arithmetic progression, then what is the value of x?Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
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#25. What is the value of x if

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**JaneFairfax****Member**- Registered: 2007-02-23
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Argh!

Hopefully this one is correct.

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**ganesh****Administrator**- Registered: 2005-06-28
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JaneFairfax,

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**JaneFairfax****Member**- Registered: 2007-02-23
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I know. I divided by 2 twice.

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**ganesh****Administrator**- Registered: 2005-06-28
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Yet another chance for you, Jane!

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**JaneFairfax****Member**- Registered: 2007-02-23
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I beg to differ this time, Ganesh. I do believe my last answer for #17 is the correct one.

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**Identity****Member**- Registered: 2007-04-18
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Amazing ganesh, I don't think i have ever seen you use the same picture twice

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**JaneFairfax****Member**- Registered: 2007-02-23
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**ganesh****Administrator**- Registered: 2005-06-28
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Answer to #17:-

The parallel sides are 60 cm and 46 cm.

The non-parallel sides are 13 cm and 15 cm.

Let the Trapezium be named ABCD where AB=46 cm, BC=15 cm, CD=60 cm, and DA=13 cm.

AB || CD.

Draw AE parallel to BC where E is a point on CD. Therefore, ABCE is a parallelogram, where AB=CE=46cm, BC=AE=15 cm. Area of the parallelogram = 46 x 15 = 690 sq.cm.

In triangle ADE, AD=13 cm, AE=15 cm, and DE=60cm - 46 cm = 14 cm.

Area of triangle ADE, where S=(13+14+15)/2=21 cm is

Therefore, total area of Trapezium ABCD is 690 sq.cm + 84 sq.cm = 774 sq.cm.

**To JaneFairfax : Please correct me if I am wrong. I tend to make a lot of mistakes in a hurry. Its too difficult to detect these mistakes when I check the second time.**

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 25,334

Answer to #22:-

Correct, Identity!

Identity, I try not to use the same image twice. But then, there are two factors I wish to tell you. My fading memory as a result of my age fails sometimes. Second, there's a limited stock to choose from, and I keep rotating! Thanks for your observation, I shall try my best to make amends

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