
Reya
Member •
Jul 19, 2011
No takers??
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I found this part a bit confusing:
sells the mixture at Rs.8 per kg with tea costing Rs.7 per kg. and sells the mixture at Rs.8 per kg
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Is the answer 19 : 24 ??
If yes I will give you the detailed explanation.
If my answer is incorrect, I dont want to confuse others with a invalid solution.
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Reya
Member •
Jul 20, 2011
@sandy: Your answer is wrong!
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if the profit is 15/2 (7.5%) then i also get the same answer as ES 19:24
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Reya
Member •
Jul 20, 2011
No the answer is wrong! Anyone there to give the correct answer?
If not,I ll post the answer here.
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Is the answer 2:15? 2 parts of Rs. 8 tea and 15 parts of Rs. 7 tea?
I will post the solution if you say it is correct.
Besides I found this part confusing too.
sells the mixture at Rs.8 per kg with tea costing Rs.7 per kg. and sells the mixture at Rs.8 per kg
Please clarify. 😀
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Reya
Member •
Jul 21, 2011
@Akd,Ishu: I just repeated the line again!
The question should be like
A dealer mixes tea costing Rs.8 per kg with tea costing Rs.7 per kg and sells the mixture at Rs.8 per kg and earns a profit of Rs 15/2% on his sale price.In what proportion does he mix then?
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Is the answer 611/16?
P.S: This is my last try. 😉
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Reya
Member •
Jul 21, 2011
Wrong!!
@CT & CP: where are you?
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is it 2:3 ratio?
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Reya
Member •
Jul 21, 2011
@Madhu : You are rocking!
Can you please post your method of approach?
2.Find the no of triangles in an octagon having no side common with the octagon?
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Method:
A dealer mixes tea costing Rs.8 per kg with tea costing Rs.7 per kg and sells the mixture at Rs.8 per kg and earns a profit of Rs 15/2% on his sale price.In what proportion does he mix then?
i think ES and Yadav made mistake in that bold part only...
15/2 = (SPCP)/SP*100
so putting SP=8 => CP=7.4
then it is the usual allegation procedure to get the ratio 😀
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is it 10? i'm not sure though...
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Is it 6? I am not sure.
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Reya
Member •
Jul 21, 2011
Both of them are wrong!
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The answer is 16.
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It too think its 16.
8C3  328 = 5640= 16.
@CP : Yes you are correct, the mistake was done in the bolded part 😀
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Reya
Member •
Jul 21, 2011
@ishu,sandy:Right!
3.
This expression is valid for a particular value of X? value of 5X?
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Reya
Member •
Jul 21, 2011
4. The ratio upto sum of n terms of the 2 A.Ps is 3n+2/4n13.Find the ratio of 17th term of 2 A.Ps?
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The value of 5X is > 5*46656^(a square).
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Reya
Member •
Jul 21, 2011
ishutopre
The value of 5X is > 5*46656^(a square).
Wrong ishu!
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@Praveena: My last chance, either it is 2.78 or 2.18.
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Reya
Member •
Jul 21, 2011
@Ishu :wrong! Can you give me the value of x you arrived at?
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2.78 is the value of X, which is also the value of natural base 'e'. So 5X is 13.591. 😔
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Reya
Member •
Jul 21, 2011
Ha Ha 😀
@ishu : Wrong!
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101:119 ??
Logarithms are my least priority!!
I have to practise a lot to get to my old momentum! I dont want to see my score card with less marks than previous years in DI and Quant.
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Reya
Member •
Jul 21, 2011
Your answer is right sandy!
Give me your method of approach!
Try to solve 3rd ques too.
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Approach:
we need , a1+16d1/a2+16d2 (1)
Given that ratio of sum of n numbers in both the series,
so, [2a1+(n1)d1]/[2a2+(n1)d2] = 3n+2/4n13
To get (1) we have to substitute n=33 , which gives 101:119
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Reya
Member •
Jul 22, 2011
3.
This expression is valid for a particular value of X? value of 5X?
No takers for 3rd quest??
5.a0, a1, a2, a3..an are (n + 1) vertices of a regular polygon. And functions f(x) and g(x) are defined in such a way that. f(n) = Total number of diagonals in a regular polygon of (n + 1) sides and g(n) = Total number of triangles that can be made using the vertices of the (n + 1) sided regular polygon, as its vertices (None of the sides of the triangle should be common with that of polygon.) If f(n) = 5, then what is the value of g(n)?
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qn. 3: is it 30?
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Reya
Member •
Jul 22, 2011
@CP: Right!
Post your method of approach!
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the given expression can be written as a^(log (base 6) x ^ (log (base 4) a) ) = log (base 4) 256 = 4
(log (base 4) a)= (log (base 6) x ^ (log (base 4) a) = (log (base 4) a) * (log (base 6) x) => (log (base 6) x)=1 => x=6 ==> 5x =30 😀
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EnglishScared
It too think its 16.
8C3  328 = 5640= 16.
i don't understand the part in bold. can you please explain a little more in detail ? 😀
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If you consider a octagon, it has 8 points.
We can draw straight lines from one end to other end 5 lines, which gives you 4 traingles.
so , total 8 points which gives you 32 triangles with one side of the octagon.
Next from center you can draw 8 lines where you get 8 more triangles.
total of 40.
So, in an regular octagon you can have 8C3 triangles.
so its 16.
Hope I am clear, I am not sure of the process but this is what I got when I was trying 😀
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Reya
Member •
Jul 22, 2011
6.A car travels from B at a speed of 20 km/hr. The bus travel starts from A at a time of 6 A.M. There
is a bus for every half an hour interval. The car starts at 12 noon. Each bus travels at a speed of 25 km/hr. Distance between A and B is 100 km. During its journey, The number of buses that the car encounter is ?
PS: I'm not sure of the answer. I want to check my answer😔
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By counting I got the answer as 19 buses.
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yeah i also got 19...
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Reya
Member •
Jul 22, 2011
I too got the same answer.
7.X is a number of more than 4 digits.The digits of X increase from left to right.What can be sum of digits of number 9X?
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5 = 0.
The number of diagonals for a n side regular polygon = n(n3)/2
which gives n = 5.
So in order to draw a triangle without touching any of the side, we need more than 5 sides.
So there will be Zero triangles in this case.
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praveena211
7.X is a number of more than 4 digits.The digits of X increase from left to right.What can be sum of digits of number 9X?
is it 9? 😐
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Reya
Member •
Jul 22, 2011
yes you are right!
post your method of approach😀
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7. Answer is 9. For digits not more than 9.
Its a very big tedious process, I dont remember correctly. Will try posting it tomorrow.
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IS the answer for Question 5 correct?
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Reya
Member •
Jul 22, 2011
8.Sandeep is planning to attend wedding.He starts at a specific time and specific speed so as to be there at the exact time.he was driving for exactly 1 hour when his car had a flat tyre.He changed the tyre in 10 minutes.He started driving the remaining distance @ 40 km/h.he was late for the wedding by 30 minutes.If the puncture had happened 25 km later,he would be late for wedding by 20 min.what is the total dist. travelled by him?
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Reya
Member •
Jul 22, 2011
ASsuming the number is N.
N = a(n)a(n  1).....a(0), where 0 < a(n) < a(n  1) < .... < a(0) < 10
>0 < a(k)  a(k + 1) < 10 (1)
N = a(n)10ⁿ + .... + a(2)10² + a(1)10 + a(0)
9N = 9*10ⁿa(n) + .... + 900a(2) + 90a(1) + 9a(0)
Now, 9*10ⁿa(n) can be written as 10^(n + 1)a(n)  10ⁿa(n)
9*10^(n  1)a(n  1) can be written as 10ⁿa(n  1)  10^(n  1)a(n  1)
.
.
900a(2) can be written as 1000a(2)  100a(2)
90a(1) can be written as 100a(1)  10a(1)
9a(0) can be written as 10a(0)  a(0)
Then subtract 10 and add 10 and keep all numbers of form 10^k together
Write coefficient of 10 as a(0)  a(1)  1 and that of 1 as 10  a(0)
So, it will become
> 9N = 10^(n + 1)a(n) + 10ⁿ{a(n  1)  a(n)} + ..... + 1000{a(2)  a(3)} + 100{a(1)  a(2)} + 10{a(0)  a(1)  1} + (10  a(0)}
Using (1)equ, we will get
Sum of digits of 9N = a(n) + {a(n  1)  a(n)} + .. + {a(2)  a(3)} + {a(1)  a(2)} + {a(0)  a(1)  1} + (10  a(0))
= 10  1 = 9
@ES,CT : Is this the way you arrived at the result?
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thats a good way to solve it... but no, i did not 😛 i tried it using examples like 12345, 123456, 1234567 and found all of them gave the same answer "9" 😛 trial and error method 😁
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praveena211
8.Sandeep is planning to attend wedding.He starts at a specific time and specific speed so as to be there at the exact time.he was driving for exactly 1 hour when his car had a flat tyre.He changed the tyre in 10 minutes.He started driving the remaining distance @ 40 km/h.he was late for the wedding by 30 minutes.If the puncture had happened 25 km later,he would be late for wedding by 20 min.what is the total dist. travelled by him?
is the answer 150 kms?
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Reya
Member •
Jul 23, 2011
@CP:wrong!
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i think this is wrong.. but i'm getting 106.5
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Reya
Member •
Jul 23, 2011
9.A,B and C start from a point X, speeds of A,B and C are a, a+s and a+2s respectively. if A starts first and after x hours B starts, then after how many hours from starting time of B should C start so that all of them meet at the same time( in otherwords both B and C overtake A at the same time).
(a and s are positive and nonzero values).
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After, ax/a+2s hours after B starts C should start such that they reach at the same time.
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Solution:
Few assumptions to be made,
t = time taken for A to reach the meeting point 😀
x = time at which B starts after A starts (Given)
y = time at which C starts after A starts (Given)
So, Distance = Speed * time
And, the distance travelled by all three is same.
So, at = (a+s)(tx) = (a+2s)(ty)
On solving first , we get => t = x(a+s)/s
On solving second , we get => y = 2st/(a+2s)
Putting the value of t above will give,
y = 2sx(a+s)/s(a+2s)
The question asked is yx:
So, the solution is ax/a+2s.
I think its correct 😀
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but i got the answer as x... 😐
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Explain your procedure too, lets see where did I or you went wrong 😀
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I'm not able to find a mistake in my approach: identify if any,
as B starts x minutes after A, A would have travelled some distance = ax
so now B must cover that distance + some distance that A will cover from the time B also starts = ax+D (say)
from the time B starts, both take the same time,
and time=dist/speed
so, D/a = (D+ax)/(a+s) =>x*(a)^2=Ds (1)
now let C start some time y after A starts,
so as usual, by the time C starts A would have covered ay distance,
so, D/a = (D+ya)/(a+2s) => y*(a)^2=2Ds(2)
using (1),(2) => y=2x
but that is the time at which C should start after A starts but the question is after B starts => 2xx =x
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My answer to Q8 in post number 46 by Praveena.
My answer is the distance between Sandeep's home and the wedding place is 73.333Km. I will post solution upon confirmation by Praveena that my answer is correct. 😀
P.S: It is better to reach late than meet an accident in a hurry. Drive safe Sandeep. 😛
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Reya
Member •
Jul 23, 2011
@CP: You are very close to the answer!
Post your procedure if you wanna know where you went wrong!
@Ishu: Your answer is wrong!
@Sandy : You gave the right answer 😀
@CT : Finding out your mistake!
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is the answer for the Q8 129.54?
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for qn 8)
I got (25/v)(25/40)=(10/60)
=> (25/v)=19/24 => v=600/19.
then the distance traveled in first one hour is 600/19.
if after 25 km the time delay reduced by 10 mins. after 50 km there will be no delay.
so the total distance is 600/19 + 50. so i got 106.9.
just tell me is my velocity correct...
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Reya
Member •
Jul 23, 2011
Distance travelled in first one hour is wrong!
Check it! It should be 600/11.
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@Praveena: I would request you to post few more questions till this weekend.
I would be updating another thread with 5 questions per day starting August 1st .
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Reya
Member •
Jul 25, 2011
How many 10 digit positive integers with distinct digits are multiples of 11111 ???
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none? 😕 😉
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Reya
Member •
Jul 26, 2011
@Cooltwins : wrong!
Use divisibility rule 😀
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