CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

[Aarav]

Originally Posted by thebornattitude

Solved it somewhere before too ....

The speed with which it was solved last night forced me to extend the question Earlier, it was evolving around 3rd option only.

[Aarav]

Originally Posted by gripened

Answer :

(2)
Let their ages be : A, B and C.
Given : A, B, C <= 99
And some of any two is reverse of the 3rd ==>
sum of two is 9's complement of the third ==> A+B = 99 - C
==> A+B+C = 99.
Hence i) is true.

Hence the first and second digits of the sum of two have to be 9's complement of each other.
Thus, we have only 10 possibilities :
(1st Digit of A+B, 2nd Digit of A+B) = (SUM) ==> (C) - (Yuvraj,B)
(0,9-0) = (09) ==> (90) --> Not a solution, since we can't split 9 further (only poss. is 9 and 0, but that is trivial, hence not valid)
(1,9-1) = (18 ) ==> (81) - (9,9) Hence Yuvraj cannot be 9
(2,9-2) = (27) ==> (72) - (18,9)
(3,9-3) = (36) ==> (63) - (27,9) {(18,18 )isn't valid for Yuvraj}
(4,9-4) = (45) ==> (54) - (27,18 ),(36,9)
(5,9-5) = (54) ==> (45) - (36,18 ){(27,27) again isn't valid)
(6,9-6) = (63) ==> (36) - In this case, Yuvraj is 36, *****
(7,9-2) = (72) ==> (27) - Again Yuvraj is 27
(8,9-1) = (81) ==> (18 )- (27,54), And Yuvraj can be 18
(9,9-9) = (90) ==> (09) --> Again not a valid solution...

Hence we have total 8 possibilities of Cs age.
therefore, at max, Yuvraj (strictly aged in between)
can have 8 different ages , sounds weird right ...
But from the above deductions, for him to be strictly aged in between,
he cannot take 8 different ages,
Implies (2) is false,

Now for (3), ***** clearly implies that if , A, B and C have distinct ages,
then (3) is true...But it is also given that Rohit is older than either of the two, hence his age isn't the same as Yuvraj's (the middle aged guy )...Hence, (3) is true...

Note, if I have interpreted (3) correctly, then the answer is (2),
Hence answer : (2)

This solution deserves 5. The complement part was stunning to start with.

[arbit_rageur]

Let their ages be 10a+b,10c+d and 10e+f.

Then 10(a+c)+(b+d)=10f+e....(1)
10(c+e)+(f+d)=10b+a
and 10(e+a)+(f+b)=10d+c.

19(a+e+c)=8(f+b+d).

Implies a+e+c duvisible by 8 implies a+e+c=8 or 16.

and f+b+d=19 or 38.

We find that only 8,19 combo is possible implies sum is 99.

a+e+c=8.
f+b+d=19.

Reaaranginbg eqn 1 we get 10(8-e)+19-f=10f+e.

Implies 99-10e-f=10f+e.

Implies e+f=9and similarly a+b=9, c+d=9.

Implies statement 3 is true on combing this eqn with the eqn a+e+c=8.

As for statement 2, Yuvraj's age can take either the values of 27,18,36...so only 3 values are possible.

So answer is (2)

Originally Posted by sabsebadapaagal

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Quantitative Question # 074
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Yuvraj, Rohit and Mahender each had age (always considered an integer) less than 100, such that sum of the ages of any two of them is same as reverse of third's age. Which among the following can not be true?



(1) The sum of the ages of the three is always 99.

(2) Yuvraj's age, strictly middle in age among the three, can assume 8 values

(3) If Rohit was older than either of the others, the youngest he could be is 45

(4) At least two of the above

(5) none of the above

[ravishah17]

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Quantitative Question # 074
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Yuvraj, Rohit and Mahender each had age (always considered an integer) less than 100, such that sum of the ages of any two of them is same as reverse of third's age. Which among the following can not be true?



(1) The sum of the ages of the three is always 99.

(2) Yuvraj's age, strictly middle in age among the three, can assume 8 values

(3) If Rohit was older than either of the others, the youngest he could be is 45

(4) At least two of the above

(5) none of the above

Tried by trial and error and found that the ages should be multiple of 9
(1) is true (3) is true as Rohit is older than other two
Yuvraj's age is in middle then, he cannot take 8 values.

So answer is option (2)

[deepakraam]

My take,
Let Yuvaraj = 10a+b

Rohit = 10c+d

Mahendar = 10e+f ,where a,b,c,de,f <= 9

Solving with equations, we get a+b = c+d = e+f ----(1)

Since a-f < 9 we can get a solution as 18,36,45

From the options we can easily rule out 1,3,4 and 5.So option 2 can't be true.

My answer is 2.

-Deepak.

[WannaBThere]

Let age of 3 players be 10a+b, 10c+d and 10x+y
According to the given condition,
(10a+b) + (10c+d) = (10y+x)
(10c+d) + (10x+y) = (10b+a)
(10x+y) + (10a+b) = (10d+c)
Solving above equations;
a=-b, c=-d, x=-y
i.e. numbers would be of form (10a-a), (10c-c), (10x-x) i.e. ages would be multiples of 9
Also age of any player can not be 99 or 90, since in that case given condition is not satisfied.
Thus, minimum possible age=9 and maximum possible age=81
Thus, possible ages of three players will look like this: (9,9,81),...
In any of these case sum of three ages is 99..(1) true
If Yuvraj's age is the middle one, then age of youngest one will be 9 (to get the maximum possible ages of Yuvraj). In that case following combinations are possible:
(Youngest Player, Yuvraj, Eldest player)
(9, 18, 72)
(9, 27, 63)
(9, 36, 54)
=>(2) false
(3) is also true with following combination
(Youngest Player, Other Player, Rohit)
(9, 45, 45)
Thus, answer is (2)

[v-factor]

Let the ages be :
10*a+b
10*c+d
10*e+f

10*(a+c)+(b+d) = 10*f+e
10*(c+e)+(d+f) = 10*b+a
10*(e+a)+(f+b) = 10*d+c

Adding we get 19*(a+c+e) = 8*(b+d+f)
The ages must be multiples of 9.
The various combination of ages are
(09,45,45), (18,18,63), (18,36,45), (18,27,54), (27,27,45), (27,36,36), (81,09,09), (09,18,72)

Now the sum has to be 99
If Yuvraj's age is middle then there will be 2 combinations each for 18,27 and 36. So only 8 distinct values are not possible.
Raina if older than other 2 will have youngest age as 45.

So, only option 2 is false
Ans is (2)

[Aarav]

Congratulations to all who got today's QQAD right, though some of the answers/approaches were not all that convincing. We will use gripend's solution (with some correction needed) as the official solution for this problem.

[saurabhsmit2004]

Hey Aarav and all guys out there!!!
Is it possible for ny 1 of u to post the link to the pdf of mock CAT of any one of
the standard test taking body, like TIME , CL or IMS......
I guess this can help guys who don't hav access to MOck CATs........

[javed_nitb]

Yuvraj, Rohit and Mahender each had age (always considered an integer) less than 100, such that sum of the ages of any two of them is same as reverse of third's age. Which among the following can not be true?



(1) The sum of the ages of the three is always 99.

(2) Yuvraj's age, strictly middle in age among the three, can assume 8 values

(3) If Rohit was older than either of the others, the youngest he could be is 45

(4) At least two of the above

(5) none of the above

The correct answer is option (2)
Let yuvraj's age Y = 10a1 + b1
Rohit R = 10a2 + b2
Mahendra M = 10a3 + b3

since sum of the ages of any two of them is same as reverse of third's age= gives

10(a1 + a2) + (b1 + b2) = 10b3 + a3....
...adding 3 such equations..
19(a1 + a2 +a3) = 8(b1 + b2 +b3)

Sum of ages = 99/8(a1 + a2 + a3)

value of (a1 + a2 +a3) can only be 8 so 1 is true...

also b>=5 only
and for rohit(45) to be the eldest 18 + 36 = 54 is the least possible
...so 3 is also true...
for option 2... (10a1 + 2b1 = 10(a1 + a2) + (b1 + b2))/2
or 19a1 = 8b1....not at all possible...for any possible value of a1 and ba1...hence 2 is not true...

so option 2 is the correct option...