two questions ,
1 .one is based on Wted avg, but im not able to get what and how shud be the avg value and weights and ratio . please help.
2. other one im not getting the exact value as 41(3/17) %. im getting ==> (1-x%)*68% of M = 40 % of M ==>
x% = 7/17 im getting please help me.
Kartik, the first one is a simple weighted average 12%….15%….20% so the ratio of original amounts should be 5:3 i.e. A has 5/8th of 1200 = 50. After spending 12% of this i.e. 90 he should be left with 660. Another way to look at it might be to realise that if he spent 12% he is left with 88% and chances are pretty good the amount left with him should be a multiple of 11….:)
In the second, you have to note that the base changes for each percentage – so the last one must be measured on 68. Suppose the total marks are 100, then passing marks must be 40. So A gets 10% of 40 i,e, 4 less i.e 36 and B gets 11.11% of 36 i.e 4 less i.e 32. Together they score 68. So C must get not more than 28 less than 68 i.e (28/68) * 100 % i.e. 41 (3/17)% less to pass.
J Sir, Thanks a lot
2 . I did in exactly the same way : (1-x%)*68% of M = 40 % of M as i mentioned earlier. now I get x% = 28/68 = 7/17 . but we need to multiply by hundred and then we get the right answer( this is what confuses me ,all the time).
This is very silly question but please help me when shud we multiply by 100 and when not ?. Instead of x% should i have used x ( as a ratio ) , and then multiply by 100.
but what if i directly wanted to find the percentage value without multiplication.
Then it shud be : (100 – x%)*68 of M = 40 % of M still not getting the right answer.
I don’t have a rule as to when to multiply by 100…I tend to think of the underlying logic, that a percentage is just a fraction * 100. Here I want the percentage change from 68 to 40 and hence the change is 28/68 as a fraction and 28/68 * 100 as a percentage.
In your equation it would be (100-x)/100 * (68/100) * M = 40/100 * M (you forgot the % on the 68!) but it would still be painful to do. Equations should be your last resort, not the first….logic and understanding beats mechanical application of equations any day.
Why is the inradius (base+base-hypotenuse)/2 ? Is this a formula for the right angled triangles? Is there a simple proof for this?
# (base + perpendicular – hypotenuse)/2
Tanvi, we had discussed this a fortnight or so ago. Do see this post
and let me know if further clarification is needed.
Got it. Thanks.
Happy Teachers Day!! Thanks and keep sharing your knowledge with us!!
Thanks, Ravi Have a great day yourself….
how is dq in the above figure 8cm?
Sibi, BDC is a right angled triangle. So DQ is equal to the inradius, which we calculated as 8.
Sir, i will like to add one thing here. let I be the inrcentre of our primary tiangle, R be inradius, R1 inradius of right triangle and R2 of left one, then
BI= EF= R rt2 (directly 10rt2 in this case)
R1^2+R2^2 = R^2
aR2+cR1 = bR…. where a, b, c are three sides….
and happy teachers day to you!!!!
Fill in your details below or click an icon to log in:
You are commenting using your WordPress.com account. ( Log Out / Change )
You are commenting using your Twitter account. ( Log Out / Change )
You are commenting using your Facebook account. ( Log Out / Change )
You are commenting using your Google+ account. ( Log Out / Change )
Connecting to %s
Notify me of follow-up comments via email.
Get every new post delivered to your Inbox.
Join 720 other followers