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

Varun Khullar

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Quantitative Question # 059
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If p, q, r be positive numbers satisfying p + 1/q = 4, q + 1/r = 1, r + 1/p = 7/3, then pqr =

(1) 2/3 (2) 1 (3) 4/3 (4) 2 (5) 7/3

was given values so i solved the equations

q= r-1/r put that in p+1/q =4 and using thrd equation we got
4p^2-12p+9 =0 hence
p=3/2
q=2/5
r=5/3
pqr =1 ........

mba.yodha

From eqn 1 n 2 we get q<4...

Also for the 3 eqns to satisfy p,q,r cannot be intergers... for e.g if r=2, q=1/2 (frm eqn2),

But that then there is discrepancy in values for p for eqn 1 n 3.... Similary frm other integer values for p,q,r...

So p,q,r got to be fractions... Only values satisfying this is p=3/2,q=2/5 n r=5/3...

so pqr = 1 hence option(2)

Though multiplying and forming quadratic eqn in pqr is easy...

romy4allcse

If p, q, r be positive numbers satisfying p + 1/q = 4, q + 1/r = 1, r + 1/p = 7/3, then pqr =
(1) 2/3 (2) 1 (3) 4/3 (4) 2 (5) 7/3

p = 4-1/q---(1)
r = 1/(1-q)---(2)
putting values of p and r in eq---(3)
we get 25q^2 - 20q + 4 = 0
=> q = 2/5
=> p = 3/2
=> r = 5/3
hence pqr = (2/5)*(3/2)*(5/3)
= 1
hence option (2)

jangogautam

If the base 8 representation of a perfect square is ab3c, where a is non-zero, then c equals

(1) 0 (2) 1 (3) 3 (4) 4 (5) can not be uniquely determinable

CAN u tel the answer for this ..?
c+8*3+8^2*b+8^3*a
this means c can be 4 and 8..
so (5) is ans

madhulikasambit

p+1/q=4......eq 1
q+1/r=1.....eq 2
r+1/p=7/3......eq3
now multipying eq 1,2,3 we get
pqr+1/pqr+(p+q+r)+(1/p+1/q+1/r)=28/3......eq 4

again adding eq 1,2,3 we get

(p+q+r)+(1/p+1/q+1/r)=22/3.... eq 5
substitute eq 5 in 4 and put pqr as x then

x+ 1/x=28/3-22/3=6/3=2
=>x^2+1=2x
=>(x-1)^2=0
=>x=1
thus pqr=1
therfore choice 2

ravishah17 · 11:19 AM

I suppose easiest problem of QQAD

Adding all the three equations
p+q+r+1/p+1/q+1/r=22/3

Multiplying 1st and 3rd equation
pr+1+r/q+1/pq=28/3

Multiplying the above equation with 2nd equation(thus multiplying all the three equations)
pqr+p+q+r+1/p+1/q+1/r+1/pqr=28/3

pqr+22/3+1/pqr=28/3

Thus, pqr+1/pqr=2

Therefore pqr=1(from options)

So answer is (2) 1

thevoid

Hi PGuys....
Followed the same method....as everyone else....
i.e. substituting the values of 2 variables in one equation and finding out the third variable.....
in the beginning seemed tricky....
after checking answers i was apprehensive that answer can't be 1...
but the number 1 ruled..after few calculations....
hence...found out that
p=3/2
q=2/5
r=5/3..

therefore =: 1

thebornattitude

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Quantitative Question # 059
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If p, q, r be positive numbers satisfying p + 1/q = 4, q + 1/r = 1, r + 1/p = 7/3, then pqr =

(1) 2/3 (2) 1 (3) 4/3 (4) 2 (5) 7/3

SOLUTION

Multiply all the equations to get
pqr+1/pqr + p+q+r + 1/p+ 1/q+1/r=28/3

now add all to get
p+q+r + 1/p+ 1/q+1/r = 22/3
put in above equation we get

pqr+1/pqr= 2
solving we get pqr=1 ,,,,hence option 2 is the answer

vivekma

Answer to Question 59

q=1-(1/r) => qr+1=r => qr=r-1
p=4-(1/q) => pqr = pr-p

Rewrite each equation in terms or p ,
but pr+1 = (7/3)p => pr = (7/3)p-1
pqr = pr-p = (7/3)p-1-p= (4/3)p-1 -- (1)

Rewrite each equation in terms or r ,
pqr = (1-1/q)*(4-1/q)*(r)
Substitute ( q by r )
pqr = 4(r-1)*r = 3r-4 -- (2)

from 1 and 2 ,

p = 3/2 , q= 2/5 , r = 5/3

pqr = 3/2 * 2/5 * 5/3 = 1

Hence option (2)

Quite lengthy i know. Came up with a bunch of solutions but this wasn't posted yet.

killerkart19

If p, q, r be positive numbers satisfying p + 1/q = 4, q + 1/r = 1, r + 1/p = 7/3, then pqr =
(1) 2/3 (2) 1 (3) 4/3 (4) 2 (5) 7/3
ANSWER:
p = 4-1/q (1)
r = 1/(1-q) (2)
Puttin the value of p and q we got frm eq 1 and 2 in eq 3(given)
we get 25q^2 - 20q + 4 = 0

q = 2/5 hence, p = 3/2 r = 5/3

pqr = (2/5)*(3/2)*(5/3) =1

hence option 2 is correct

TORRES -El orgullo espanol