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	<title>Comments on: Quick Multiplication by 5</title>
	<atom:link href="http://www.quickermaths.com/quant-gyan-of-the-day/feed/" rel="self" type="application/rss+xml" />
	<link>http://www.quickermaths.com/quant-gyan-of-the-day/</link>
	<description>Vedic Maths Tricks &#124; Puzzles, Brainteasers &#38; Riddles</description>
	<lastBuildDate>Fri, 14 Jun 2013 12:08:15 +0000</lastBuildDate>
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		<title>By: joolie</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-17464</link>
		<dc:creator>joolie</dc:creator>
		<pubDate>Mon, 28 Jan 2013 15:35:41 +0000</pubDate>
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		<description>if sides are 4,3,5 ...den according to ur formula p= 180/12 = 15..
angles are 4*15=60
                  3*15=45
                  5*15=75
but in real 4,3,5 is a right angle tringle..(4^2 + 3^2 = 5^2)..</description>
		<content:encoded><![CDATA[<p>if sides are 4,3,5 &#8230;den according to ur formula p= 180/12 = 15..<br />
angles are 4*15=60<br />
                  3*15=45<br />
                  5*15=75<br />
but in real 4,3,5 is a right angle tringle..(4^2 + 3^2 = 5^2)..</p>
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		<title>By: priya</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-17333</link>
		<dc:creator>priya</dc:creator>
		<pubDate>Sun, 13 Jan 2013 15:45:33 +0000</pubDate>
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		<description>I think there&#039;s a alternative to the trick of multiplying by 5.

132/5
1/2 of 132 =66 and multiply by 10 which gives 660

133/5
1/2 of 133=66.5 and multiply by 10 which gives 665

Hence there is no need to remember the trick for even and odd numbers separately. Hope you agree with it . @vineet Patawari Sir</description>
		<content:encoded><![CDATA[<p>I think there&#8217;s a alternative to the trick of multiplying by 5.</p>
<p>132/5<br />
1/2 of 132 =66 and multiply by 10 which gives 660</p>
<p>133/5<br />
1/2 of 133=66.5 and multiply by 10 which gives 665</p>
<p>Hence there is no need to remember the trick for even and odd numbers separately. Hope you agree with it . @vineet Patawari Sir</p>
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		<title>By: mahesh</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-15998</link>
		<dc:creator>mahesh</dc:creator>
		<pubDate>Thu, 16 Aug 2012 10:05:59 +0000</pubDate>
		<guid isPermaLink="false">http://www.quickermaths.com/?p=392#comment-15998</guid>
		<description>can you please send me some more tricks about the multiplication of any two numbers</description>
		<content:encoded><![CDATA[<p>can you please send me some more tricks about the multiplication of any two numbers</p>
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		<title>By: santhakumar</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-14552</link>
		<dc:creator>santhakumar</dc:creator>
		<pubDate>Fri, 29 Jun 2012 06:06:21 +0000</pubDate>
		<guid isPermaLink="false">http://www.quickermaths.com/?p=392#comment-14552</guid>
		<description>I solved it by Venn diagram method.</description>
		<content:encoded><![CDATA[<p>I solved it by Venn diagram method.</p>
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		<title>By: FlameFletcher</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-14505</link>
		<dc:creator>FlameFletcher</dc:creator>
		<pubDate>Wed, 27 Jun 2012 14:18:15 +0000</pubDate>
		<guid isPermaLink="false">http://www.quickermaths.com/?p=392#comment-14505</guid>
		<description>Ok well let me submit something.
I hope you will like it vineet.
In a triangle.
we know the relation ship between sides and angles with trignometry.
but how can we find out angles if we know the sides in a simple method?
use ‘&amp;’ as the proportionality symbol.
in a triangle ABC the side opposite to angleA is BC , side opposite to angleC is AB etc.
so as the side AB increases then angleC definitely increases. right?
so
AngleA &amp; BC (&amp; —&gt; means directly proportional to)
AngleB &amp; AC
AngleC &amp; AB
Let ‘P’ be a constant.
AngleA = P(BC) —–&gt; 1
AngleB = P(AC) ——&gt;2
AngleC = P(AB) ——&gt;3
Adding 1,2,3 we get……
AngleA+AngleB+AngleC = P(BC+AC+AB)
But AngleA+AngleB+AngleC = 180
therefore, 180= P(BC+AC+AB)
=&gt; P= 180/(BC+AC+AB)
from formula if we know the 3 sides of a triangle we can find the constant ‘P’.
and to get an angle.
AngleA = P(BC)
so multiply P with a side opposite to the angle to get that angle.
here is an Example sum.
2,6,10 are the 3 sides of a triangle. Find out all their angles.
From my formula above, P = 180/(AB+BC+CA)
AB+BC+CA = 2+6+10 = 18
=&gt; P=180/18
=10
therefore P=10 degrees per unit distance.
that means if there is an increase of 1 unit distance then the angle opposite to that side increases by 10 degrees.
AngleA = 10(BC) = 10(2) = 20 degrees
AngleB = 10(CA) = 10(6) = 60 degrees
AngleC = 10(AB) = 10(10)=100 Degrees
Therefore the angles are found.
Note: This formula can be used for any type of triangle and this formula is 100% usable and proved.
Note:The value of P is different for different triangles. So we always have to find P before finding angles.
Feel Free to use it and please publish this on this site.</description>
		<content:encoded><![CDATA[<p>Ok well let me submit something.<br />
I hope you will like it vineet.<br />
In a triangle.<br />
we know the relation ship between sides and angles with trignometry.<br />
but how can we find out angles if we know the sides in a simple method?<br />
use ‘&amp;’ as the proportionality symbol.<br />
in a triangle ABC the side opposite to angleA is BC , side opposite to angleC is AB etc.<br />
so as the side AB increases then angleC definitely increases. right?<br />
so<br />
AngleA &amp; BC (&amp; —&gt; means directly proportional to)<br />
AngleB &amp; AC<br />
AngleC &amp; AB<br />
Let ‘P’ be a constant.<br />
AngleA = P(BC) —–&gt; 1<br />
AngleB = P(AC) ——&gt;2<br />
AngleC = P(AB) ——&gt;3<br />
Adding 1,2,3 we get……<br />
AngleA+AngleB+AngleC = P(BC+AC+AB)<br />
But AngleA+AngleB+AngleC = 180<br />
therefore, 180= P(BC+AC+AB)<br />
=&gt; P= 180/(BC+AC+AB)<br />
from formula if we know the 3 sides of a triangle we can find the constant ‘P’.<br />
and to get an angle.<br />
AngleA = P(BC)<br />
so multiply P with a side opposite to the angle to get that angle.<br />
here is an Example sum.<br />
2,6,10 are the 3 sides of a triangle. Find out all their angles.<br />
From my formula above, P = 180/(AB+BC+CA)<br />
AB+BC+CA = 2+6+10 = 18<br />
=&gt; P=180/18<br />
=10<br />
therefore P=10 degrees per unit distance.<br />
that means if there is an increase of 1 unit distance then the angle opposite to that side increases by 10 degrees.<br />
AngleA = 10(BC) = 10(2) = 20 degrees<br />
AngleB = 10(CA) = 10(6) = 60 degrees<br />
AngleC = 10(AB) = 10(10)=100 Degrees<br />
Therefore the angles are found.<br />
Note: This formula can be used for any type of triangle and this formula is 100% usable and proved.<br />
Note:The value of P is different for different triangles. So we always have to find P before finding angles.<br />
Feel Free to use it and please publish this on this site.</p>
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	<item>
		<title>By: manjusha</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-14483</link>
		<dc:creator>manjusha</dc:creator>
		<pubDate>Tue, 26 Jun 2012 13:37:14 +0000</pubDate>
		<guid isPermaLink="false">http://www.quickermaths.com/?p=392#comment-14483</guid>
		<description>@Santhakumar 
  from stmnt1+stmnt2 ie some boats are camels +all camels are tiles =&gt; some boats are tiles (cncln 2)

 Now stmnt 1 can be converted into &#039;some camels are boats &#039;+ no boats is a soil(stmnt2) = some camels are not soils(concln 4) {I+E=O}

 SO ANSWER IS 2 &amp; 4.</description>
		<content:encoded><![CDATA[<p>@Santhakumar<br />
  from stmnt1+stmnt2 ie some boats are camels +all camels are tiles =&gt; some boats are tiles (cncln 2)</p>
<p> Now stmnt 1 can be converted into &#8216;some camels are boats &#8216;+ no boats is a soil(stmnt2) = some camels are not soils(concln 4) {I+E=O}</p>
<p> SO ANSWER IS 2 &amp; 4.</p>
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	</item>
	<item>
		<title>By: Santhakumar</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-14482</link>
		<dc:creator>Santhakumar</dc:creator>
		<pubDate>Tue, 26 Jun 2012 10:55:56 +0000</pubDate>
		<guid isPermaLink="false">http://www.quickermaths.com/?p=392#comment-14482</guid>
		<description>How</description>
		<content:encoded><![CDATA[<p>How</p>
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	<item>
		<title>By: manjusha</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-14350</link>
		<dc:creator>manjusha</dc:creator>
		<pubDate>Sat, 16 Jun 2012 05:36:18 +0000</pubDate>
		<guid isPermaLink="false">http://www.quickermaths.com/?p=392#comment-14350</guid>
		<description>i think u got 1 &amp; 2   by ven method. can u depict it.</description>
		<content:encoded><![CDATA[<p>i think u got 1 &amp; 2   by ven method. can u depict it.</p>
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	<item>
		<title>By: manjusha</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-14349</link>
		<dc:creator>manjusha</dc:creator>
		<pubDate>Sat, 16 Jun 2012 05:30:22 +0000</pubDate>
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		<description>thank u sir</description>
		<content:encoded><![CDATA[<p>thank u sir</p>
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	<item>
		<title>By: Santhakumar</title>
		<link>http://www.quickermaths.com/quant-gyan-of-the-day/comment-page-1/#comment-14340</link>
		<dc:creator>Santhakumar</dc:creator>
		<pubDate>Fri, 15 Jun 2012 10:17:04 +0000</pubDate>
		<guid isPermaLink="false">http://www.quickermaths.com/?p=392#comment-14340</guid>
		<description>1&amp;2</description>
		<content:encoded><![CDATA[<p>1&amp;2</p>
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