Bingo!
You are right.
Please post your explanation.
Thanks
use the property tangents from an external point are equal
sides are x+6 x+8 and 14
area = rs s - semiperimeter r = inradius
use the property tangents from an external point are equal
sides are x+6 x+8 and 14
area = rs s - semiperimeter r = inradius
Hi all,
Pls help with this question.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
6346
Regards
See the above file once X=7
sides are 7+6=13
7+8=15
See the above file once X=7
sides are 7+6=13
7+8=15
Sharad, good explaination that the lengths of tangents from external point will be same.
So, the sides have to be x+6, x+8 ;
but then the later explaination in the solution diagram assumes that the extension of the
angular bisectors are going to become parpenticulars onto the apposite side, which may not hold good ( just think about it , perpenticulars are different from bisectors ).
This qn could be solved by Area= rs
Also from the answer choices we can eliminate the combinations that don't differ
by 2 , and the order given ( which one is shorter/longer ).
Cheers,
ChivaS
but then the later explaination in the solution diagram assumes that the extension of the
angular bisectors are going to become parpenticulars onto the apposite side, which may not hold good ( just think about it , perpenticulars are different from bisectors ).
This qn could be solved by Area= rs
Also from the answer choices we can eliminate the combinations that don't differ
by 2 , and the order given ( which one is shorter/longer ).
Cheers,
ChivaS
Nooooo Noooooo Noooooo
I think diagram was a bit unclear and u did not get me. The two lines are different.
one is the perp from center to the side. which is a tangent and hence angle is 90.
second joins center to the vertex and hence is the angle bisector.
Tan are taken for smaller triangles
though may seem same from the diagram but are different lines.
hope that clarifies . Please state if there is stilll a confusion.
Hey Puys, one small DS question,
Give it a try , I will post the answer with explaination by evening today!
If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?
(1) The largest number of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3
A, B , C , D , E are,
as per DS standard answer choices. (1 alone, 2 along etc....)
Give it a try , I will post the answer with explaination by evening today!
If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?
(1) The largest number of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3
A, B , C , D , E are,
as per DS standard answer choices. (1 alone, 2 along etc....)
Hey Puys, one small DS question,
Give it a try , I will post the answer with explaination by evening today!
If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?
(1) The largest number of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3
A, B , C , D , E are,
as per DS standard answer choices. (1 alone, 2 along etc....)
It is clearly mentioned that the integers are positive. So since the square root is equal to the largest of the numbers, the product of the other two has to be equal to the largest number.
So (1) is enough to answer this question.
Now, the only combination x,y,z such that x+y+z=20 and xy=z is x=2,y=6,z=12. I am not able to think of any other combination.
Hence (2) alone is also enough.
Hence the answer is "Can be answered using either one of the two." (D)
Hey Puys, one small DS question,
Give it a try , I will post the answer with explaination by evening today!
If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?
(1) The largest number of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3
A, B , C , D , E are,
as per DS standard answer choices. (1 alone, 2 along etc....)
As told ealier , posting the explaination for this question :
If you call the three numbers x, y and z, and let x be the biggest, then the question tells you that x = square root of the product of x, y and z. If you square both sides of this equation you get:
x = xyz (which gets rid of the square root). Dividing both sides by x gives you:
x = yz.
So, if you can find a value for x, you can answer the question, which ONLY asks for 'the product of the smaller numbers'.
Statement (1) tells you that x is 12, plug that into x = yz and you get: yz = 12 (because the question asks "what is the product of the two smaller numbers?", this is what you are looking for and it is sufficient.)
Statement (2) tells you that (x + y + z)/3 = 20/3. Now, if x = yz and x + y + z = 20, then the only possibility is (2,6) and (12) as the (y, z) and (x). Every other combination of three integers summing up to 20 fails. 1,2,17; 1,3,16; 1,4,15; 1,5,14; 1,6,13; 1,7,12; 1,8,11; 1,9,10; 2,3,15; 2,4,14; 2,5,13; 2,6,12; 2,7,11; 2,8,10; 3,4,13; 3,5,12; 3,6,11; 3,7,10; 3,8,9; 4,5,11; 4,6,10; 4,7,9; 5,6,9; 5,7,8; and that's the end of the possible combinations of three integers whose sum equals 20. The only combination that works is 2,6,12. If it were 1,9,10, then yz would not equal x. If it were any other combination, such as 5,7,8, then yz would not equal x. The only solution is 2,6,12. This means that yes, Statement 2 is sufficient because the product of yz is 12. Note: this is an extremely hard question that few students get right.
Since both statements are sufficient, the answer is D.
Anyother time saving tips/tricks are welcome.
Puys ! one more DS qn!
x > y > 0. If x and y are integers, is y even?
(1) x - y = 1
(2) 2x + y = 32
A,B,C,D,E , just the usual DS answer choices ( 1 alone, 2 alone etc ... )
Cheers,
ChivaS
x > y > 0. If x and y are integers, is y even?
(1) x - y = 1
(2) 2x + y = 32
A,B,C,D,E , just the usual DS answer choices ( 1 alone, 2 alone etc ... )
Cheers,
ChivaS
Puys ! one more DS qn!
x > y > 0. If x and y are integers, is y even?
(1) x - y = 1
(2) 2x + y = 32
A,B,C,D,E , just the usual DS answer choices ( 1 alone, 2 alone etc ... )
Cheers,
ChivaS
ans -> B
2x has to be even and for sum to be even y has to be even. A just gives the difference. doesnt tell things either way
Puys ! one more DS qn!
x > y > 0. If x and y are integers, is y even?
(1) x - y = 1
(2) 2x + y = 32
A,B,C,D,E , just the usual DS answer choices ( 1 alone, 2 alone etc ... )
Cheers,
ChivaS
(1) doesnt tell us anything.
(2) is enough to answer the question as (32 - 2x) = 2(16-x) has to be even.
Hence (B).
Puys ! one more DS qn!
x > y > 0. If x and y are integers, is y even?
(1) x - y = 1
(2) 2x + y = 32
A,B,C,D,E , just the usual DS answer choices ( 1 alone, 2 alone etc ... )
Cheers,
ChivaS
Answer choice as pointed by dumbJoe, man-h , is B ( 2 alone sufficient ).
The reason I shared this question is, to let you all know the following points :
1) One has to keep in mind that there is secondary information avaiable within the given
information. example (2x+ y =32) is the given info. and from here 2x is an even number, even + Y = even so Y has to be even.
2) If somebody asks an apparantly simple qn, one may tend to think a bit more, approching the question in a skeptical manner to double check if there is any hidden thing.
So, believe me, this type of questions can really trick you in a real exam, when you are hurrying to catch up.
3) General DS confusion: sometime ( this question for example) , when you read both the inputs, you kind of think that together they can get you the answer.
In this question for exaple: 1, 2 are to two equations of two variables,
solving which you can get values.
But its not enough,
particularly when it looks so very simple; one should look at each option seperately and see either of them alone ( more in particular, input 2 ) is sufficient.
I missed out 2 of this kind of questions during two of my practice exams: ( ironically my scores were 36/37, 36/37 , so understand how much I was tricked).
OK then, bahut hogaya hain !
May be everyone knows about this type of questions 😃
Haffun ,
ChivaS
Here is a question,
For every positive even integer n, the function h(n) is defined to be the product of all even intergers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is
1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40
answer is E, all prime numbers below 53 are factors of h(n) hence they cant be factor of h(n) + 1 because the smallest prime number is 2
Question
. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160
Question
. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160
from the given data 5t+s = 65
answer is 10t+2s+20 = 150
I want to practise a lot of maths first and then go indepth into the English section. My english is strong which means that I can really trump in the section with proper effort despite the section being tough.
However, as maths is comparatively easier, I do not want to play easy with maths. I want to give it proper effort but there seems to be a lack of proper maths material floating around. What stuff should I go through to make sure than my maths/ds is steel solid? Please explain in detail including some elementary books worth going through.
Thanks..
dumbJoe Saysanswer is E, all prime numbers below 53 are factors of h(n) hence they cant be factor of h(n) + 1 because the smallest prime number is 2
Hi Dumbjoe!
Can you please amplify your answer incl the methology (or the shortcuts) used.
Thanks
I want to practise a lot of maths first and then go indepth into the English section. My english is strong which means that I can really trump in the section with proper effort despite the section being tough.
However, as maths is comparatively easier, I do not want to play easy with maths. I want to give it proper effort but there seems to be a lack of proper maths material floating around. What stuff should I go through to make sure than my maths/ds is steel solid? Please explain in detail including some elementary books worth going through.
Thanks..
Hi!
I would say start with CBSE class ixth and x th books (if you have time).
Questions from CAT thread QQAD. (Very tough but excellent for getting your funda's right)
Quant Lesson thread on Totalgadga.com
Maths Book from 4gmat.com
ATB
Here is a question,
For every positive even integer n, the function h(n) is defined to be the product of all even intergers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is
1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40
it is indeed a very good questions, certainly for those targeting 51 in quant. methodology is quite simple though,
h(n) = 2 * 2x2 * 2x3 * 2x4 * 2x5 * 2x6 * 2x7 * ....................................
so if n is an even number, all numbers from 1 to n/2 are factors of h(n). This implies that for n=100. 1 to 50 are factors, the last prime number is 47 here. hence if we add 1 to h(n) we cannot get a factor among 1 to 50 for sure, because since h(n) was a multiple of each one of them individually, you need to add atleast the factor itself to get yet another multiple. Since the smallest prime number is 2, this requisite minimum accretion comes to 2.
I want to practise a lot of maths first and then go indepth into the English section. My english is strong which means that I can really trump in the section with proper effort despite the section being tough.
However, as maths is comparatively easier, I do not want to play easy with maths. I want to give it proper effort but there seems to be a lack of proper maths material floating around. What stuff should I go through to make sure than my maths/ds is steel solid? Please explain in detail including some elementary books worth going through.
Thanks..
strengths and weaknesses are relative, please give a few sectional tests of quant and assess where you need to work upon and then start with the cbse books as sailor said, in fact RD SHARMA' solved examples would be perfect for your purpose. The only catch is that you need to have time for that. If you just need questions, i can suggest a few sources
1. OG 11
2. OG MAths workbook
3. 1000PS and DS questions (though, i feel they are very lame)
4. MAnhattan GMAT archive ( an excellent source of exceptional questions)
These would provide you with a very good start, then there are always questions in practise tests sets.