for the 1st question--> i think continued proportion means somthing like 1:2:3:4.
then a-d will always result in 3 and b-c in 1 .Hence x will always be 3 . option(b).......
What d concept of continued proportion is given in d A.S. it is not like dat as u took.Plz,interpret from the concept i m writing If three quantities a,b and c are in continued proportion , then a:b=b:c.
So,plz check it again
and for the 2nd Q. if u take 1 year then, clearly money(principal) is not compounded so difference between S.I. and C.I. is zero.bcoz both interest is (P*r)/100 which is not in d option. If u follow A.S.'s ans(which is Rs.10000) den frm ans u can guess the t=2 yrs. And I hav already wriiten the formula of 2 yrs diff. in the b4 post.What do u think ?
I want to confirm am i goin rite in 2nd Q(dat it is wrong).
And for 1st Q chk out d concept as I hav riten the concept in bold letters.And think again
now here k has to b a positive no. as these nos. are in continous proportion , hence it will always be greater than 2.
I hope im write this time :)........do tell wat the answer is!!
Thanx, fro goin through the Q. again
see one continued proportion 2/(-4)=(-4)/8=8/(-16)=k here k= - (1/2) , a negative number. So k can be negative number or positive number.So kindly go through again.I m in middle of solution.
I wud like to giv my position in dis Q.
f(k)>=x , where f(k)=(K^2 + k +1)/(k+1)is already known
frm the above expr. we can conclude dat x is the minimum value of the function f(k), from the concept of Maxima and Minima.
therefore we have to find the min. value of f(k) either by using differentiation or any simple algebraic approach.I m still doin.And will post soon.And What about the 2nd Q which I was posted?
R anybody dere to solve dis confusion(Q. 2,which I was posted)
well i guess thts correct ... k>=2..so tht equation k^2+k+1 / (k+1) will fetch LHS>=2 ...
hence x is 2...
How u can take k >=2 . To make the L.H.S >=2. From the above statement i can conclude either u missed the expression or bit confused.If missed den,I wud like to remind u.In the above exp. L.H.S is a function of k and R.H.S function of x. I think in my previous Quote I m goin rite.Plz go through it. What do u think ?
as (a-d)/(b-c)>=x or f(k) >=x the above statement implies x is the minimum value of function f(k),we got dis concept frm maxima and minima Dis can be in 2 ways. by Differentiation or algebra.
I done through both ways and got d same ans.Solving by alebra Finding the min. value of f(k) f(k)=k + 1/k +1= ((k)^(1/2))^2 + 1/((k)^(1/2))^2 + 1
=(((k)^(1/2))+1/((k)^(1/2)))^2 - 1 (It's very easy)
for minimum (((k)^(1/2))+1/((k)^(1/2)))^2 = 0, solving we get k=1, putting dis in f(k) we get f(k)=3, Hence 3 is the ans.
It does not tak more time as it seems. It was just doe to un touch frm the concepts.It takes fraction of seconds.
Somebody plz think on the 2nd Q which I have posted earlier.
Q).1 The difference between simple and compound interest on a some of money at 5% per annum is Rs. 25. What is the sum. a) Rs. 5000 b)Rs 10,000 c) Rs 4000 d) Data Insufficient
Q).2 If Rs. 58 is divided among 150 children such that each girl and each boy gets 25 p and 50 p respectively. Then how many girls is there ? a) 52 b)54 c)68 d)62 (Solve using concepts of Ratio,Proportion and Variation.As this Q is frm this chapter.I have solved it without using R,P&V;,eqns r 25x+50y=5800,x+y=150)
----------------------------------------------------------------------- What i think on the First Q. is that it is wrong. As we know the diff is (P*r^2)/(100^2) for t=2 yrs( Which is not given). If we use it we will get option b) as ans, which is the ans in A.S.------------------------------------------------------------------------- If anybody would like to discuss on this problem, i m readily available.
as (a-d)/(b-c)>=x or f(k) >=x the above statement implies x is the minimum value of function f(k),we got dis concept frm maxima and minima Dis can be in 2 ways. by Differentiation or algebra.
I done through both ways and got d same ans.Solving by alebra Finding the min. value of f(k) f(k)=k + 1/k +1= ((k)^(1/2))^2 + 1/((k)^(1/2))^2 + 1
=(((k)^(1/2))+1/((k)^(1/2)))^2 - 1 (It's very easy)
for minimum (((k)^(1/2))+1/((k)^(1/2)))^2 = 0, solving we get k=1, putting dis in f(k) we get f(k)=3, Hence 3 is the ans.
It does not tak more time as it seems. It was just doe to un touch frm the concepts.It takes fraction of seconds.
Somebody plz think on the 2nd Q which I have posted earlier.
ANS TO QUES.-1 IS (D)- DATA INSUFFICIENT AS WE DONT KNOW THE TIME PERIOD ANS.TO QUES.-2 IS (C)-68. SOLN.: Rs.58 is divided between 150 students.so each student gets 5800/150=116/3 paise. but each girl got 25p and each boy got 50p. hence ratio of boy:girl=(50-116/3) : (116/3-25) =41:34 hence no. of girls=34*150/41+34=68. option(c)
In a triangle ABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium DE:BC = 3:5. Calculate the ratio of the area of traingle ADE and the trapezium BCED. (a) 3:4 (b) 9:16 (c) 3:5 (d) 9:25
In a triangle ABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium DE:BC = 3:5. Calculate the ratio of the area of traingle ADE and the trapezium BCED. (a) 3:4 (b) 9:16 (c) 3:5 (d) 9:25
If BCED is Trapezium, therefore DE parallel to BC, therefore triangle ADE ~ triangle ABC or (area of triangle ADE)/(area of triangle ABC)=(DE^2)/(BE^2)
or (area of triangle ADE)/(area of triangle ABC)=9/25
ans is a) sqrt2-1 Rfer the attachment let x be the radius of the small Circle say C1 Given radius of Bigger Circle=1 let O be the centre of bigger circle
now OA=1-x using pythagoras theorem (1-x)pow2=xpow2 + xpow2 ==> x=sqrt2-1
ans is a) sqrt2-1 Rfer the attachment let x be the radius of the small Circle say C1 Given radius of Bigger Circle=1 let O be the centre of bigger circle
now OA=1-x using pythagoras theorem (1-x)pow2=xpow2 + xpow2 ==> x=sqrt2-1
Plz refer to Geometyr LOD I Q. 25. (Diagram is given in the just above post)
Q. In the figure,AB is parallel to CD and RDSL||TMAN, & BR:RS:ST:TA=3:5:2:7.If it is known that CN=1.333BR. FInd the ratio of BF:FG:GH:HI:IC (Sorry In my earlier post gave the fig but forget to write d Q.,here it is)
I solved in dis way,any other easier way,or any other way in tr.NCI and tr.RBF
given CN/RB=4/3 (i.e.1.333) CI/BF=CN/RB=4/3........1
and, BF/FG=BR/RS=3/5........2 FG/GH=RS/ST=5/2.......3 GH/HI=ST/TA=2/7........4 frm 1 n 2 ,CI/FG=4/5 similarly doin CI/HI=4/7 or HI/CI=7/4......5
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