CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[Aarav]

Originally Posted by JaiRamJiKi

Jai Ram Ji Ki,


Hum bhi aa sakta hu is discussion me ??

Vaise to hum aa hi gaya hu...

Koi hame bhi bataiye k Quant ki Basics kaha se padhe jaye??

TIA

Welcome to PG -> the official language of QQAD threads is Numbers and a bit of English. Remember this.

Arithmetic -> Refer coaching institutes material
Number Systems -> PG is the best source, dig for number system and related threads
Geometry -> 9th and 10th standard books along with basics of Trigo and coordinate will help
Logic -> Book by George Summers and Ravi Narula
Algebra -> CAT related topics in TMH for JEE

[Aarav]

Originally Posted by spidy_rules

hey guys,

Please make a choice b/w the following 2 books on quant

1.Trishna's Quantitative ability
2.Arun Sharma.

Please justify!!..

Not here spidy -> please use some other quant thread for this query on choices of books.

[aditya_r]

Hey PUYS,
I joined the Thread today itself, could anyone of you please explain the answer to this questionin detail , this is the first question in thread, Q-1.
Let the sum S = 20 of four natural numbers a, b, c, d be such that a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312. Which among the a, b, c, d is/are uniquely determinable ?

(1) None if a = b (2) Atleast 2 if a ≠b (3) All if a > b (4) All of the foregoing (5) Exactly 2 of the foregoing

I am able to understand till here ..

Given a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312 => (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2 = 312 - 3S + 4 = 256. Let a-1 = A, b-1 = B, c-1 = C, d-1 = D

After that I am unable to understand how..

A^2 + B^2 + C^2 + D^2 = (A+B+C+D)^2 (the LHS <= RHS always) is done..

Please help me out PUYS ..

Originally Posted by Aarav

The problem has been solved correctly by most of you in the 1st attempt. Very well done

Here is the official solution to 001 - we might change this if by end of the day someone comes with a better approach

Solution:

Given a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312 => (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2 = 312 - 3S + 4 = 256. Let a-1 = A, b-1 = B, c-1 = C, d-1 = D => we have non-negative integers A, B, C, D such that A^2 + B^2 + C^2 + D^2 = (A+B+C+D)^2 (the LHS <= RHS always) and can only be true if all but one number is zero => three among a, b, c, d are 1 and fourth number is 17.

Thus (3) holds true - (2) can guarantee the values for c and d. For a=b we have exactly 2 determinable values - thus (1) is not true.

=> choice (5) is the right answer

[Ravi_07]

The question is simple. There are three groups of natural nos which can satisy the condition
(according to all the three rules)
14,7,11,13 - group 1
12,6,3,9 - group 2
8,4,2,1 - group 3
i.e. In all of the above case if you tkae a no in a group, the other three has to be present.
The only exceptions are 10 or 5 (agin according to the rule)
Therefor the answer will be (1).

[Ravi_07]

Quantitative Question # 008
------------------------------------------------------


Let S be the set of first 14 natural numbers. A special subset of S is a subset S' which satisfies the following three properties

a) S' has exactly 8 elements
b) If x belonging to S is even, then x is in S' if and only if x/2 is in S'
c) If y belonging to S is odd, then y is in S' if and only if (y+15)/2 is in S'

Let X denotes elements of S that cannot be the part of special subset. Then n(X) (i.e. number of elements in X) equals

(1)2 (2) 3 (3) 5 (4) 6 (5) none of these

The question is simple. There are three groups of natural nos which can satisy the condition
(according to all the three rules)
14,7,11,13 - group 1
12,6,3,9 - group 2
8,4,2,1 - group 3
i.e. In all of the above case if you tkae a no in a group, the other three has to be present.
The only exceptions are 10 or 5 (agin according to the rule)
Therefor the answer will be (1)2.

[Ravi_07]

Originally Posted by twinkle2

option 1
is d ans acc 2 me
1,2,8, 4 occur in one pair
7 11 13 14 occur in anodr pair
3,9,12,6 in anothr pair
and 5,10 in one pair

and for s` to have 8 elements 5, 10 will always b excluded so answer is 2 nos can never b included..
plzz tell if my logic is ryt. didnt c nybodys else`s reasong

u r right dear
never underestimate urself.
believe that u r d bst n u will be....

[Aarav]

34 students answered QQAD #008 correctly, has to be a record for any of the question including last 2 years

[implex]

Originally Posted by Aarav

34 students answered QQAD #008 correctly, has to be a record for any of the question including last 2 years

Records are meant to be broken!! . May be the students have improved, j/k

[anukab]

Lets make groups which we can put into S'.If we start with 1 subsequently it would take 8,4,2 and again with 3 it will take 9,12,6.So we have exactly 8 elements 1,2,3,4,6,8,9,12.Like this we can take any number,can have a try with 3 ,7 etc etc .So lets make it 4 groups 1,2,4,8 & 3,9,12,6 & 7,11,13,14 and 5,10.Now try to take 5 or 10 into S' and it will in turn leads you to have more than 8 elements or less which again violates the 1st condition of S' having exactly 8 elements.So the answer would be 2.

[sushantgupta]

I may be the last person to solve this problem....but god kasam...i got time from office work only now....
so here's my approach...
according to the given conditions ...the NO'S on THE LHS can be part of S' ..only if
EVEN ODD
2--->1 1---->8
4--->2 3---->9
6--->3 5----->10
8---->4 7----->11
10--->5 9----> 12
12---->6 11---->13
14---->7 13---->14


NOW BY OBSERVING THE CONDITIONALTIES....
we get 4 GROUP of numbers...
1....13---->14---->7---->11
2....3---->9----->12--->6
3....2---->1---->8--->4
4.....5---->10

Now we can have only 8 elements in S'....
so clearly 5 and 10 CANNOT be part of S'....
hence answer is TWO.....(2)
correct me if i am wrong