If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r

Question

GMATBusters’ Quant Quiz 2.0 Question -3 
 For questions from previous quizzes click   here

If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r+s) satisfies the relation:
A. 0 <= A <=1
B. 1 <= A <=2
C. 2<= A <=3
D. 3<= A <=4
E. 4<= A <=5

willacethis · Answer

We know that 4 numbers add to 2 and all numbers are positive. Now, if (p+q) = x and (r+s) = y,

xy = A (according to question) can't have a value bigger than 1 (when both x and y are 1) because if we increase x further, y is bound to decrease (and vice versa) to have a sum of 2.

For instance, if x = 1.5, y = 0.5 and xy = A = 0.75 which falls under the equation 0 <= A <=1

Hence, answer is option A.

numb007 · Answer

ANS:A 
 value of (p+q)(r+s) will always be < 1

rvarora · Answer

Answer A:
I have attached the solution herewith

pk123 · Answer

If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r+s) satisfies the relation:
A. 0 <= A <=1
B. 1 <= A <=2
C. 2<= A <=3
D. 3<= A <=4
E. 4<= A <=5

Given p,q,r,s>0 and p+q+r+s=2
A=(p+q)(r+s)
Thinking of extreme possibilities on the left side
p+q=2
q+r=0, so A=0
if p+q=0.01 
q+r=1.99
then A=1.99*0.01=0.199

None of the option except A capture this global minimum of A expression
A is the answer

shiva1325 · Answer

let x= p+q
y=r+s
This implies x+y=2
we need to find x*y
we know that AM>=GM
from this we get (x+y)/2 >= sqrt(x*y) and the product x*y must be greater than or equal to 0
x*y <=1 (since x+y=2)
which gives 0<=A<=1

globaldesi · Answer

f p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r+s) satisfies the relation:

possible values of p,q,r,s are 
0,0,1,1
0,0,0,2
min values
(p+q)(r+s) = (0+0)(0+2) or (0+0)(0+2) = 0
max value will be 
to form factors
(p+q)(r+s) = (1+0)(0+1) =1

thus A will be of the form
0 <= A <=1

Hence A

QuantMadeEasy · Answer

If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r+s) satisfies the relation:

When (p+q) = (r+s) = 1; A =1
When (p+q) = 0.1; (r+s) = 1.9; A = 0.19
Therefore 0 <= A <=1

IMO A

fauji · Answer

Approach:

Given: p,q,r,s are positive numbers such that p+q+r+s= 2

- Sample Case 1: p,q,r,s value .5 each, we will get largest possible value: (0.5+0.5)(0.5+0.5) = 1*1 =  1 
- Sample Case 2: p = 1.9, q = 0.05, r = 0.025, s = 0.025 : we get (1.9+.05)(0.025+0.025) =  0.0975

IMO  Option A  satisfies.

SUNNYRHODE002 · Answer

Solution :

(p+q) +(r+s) =2

Product : (p+q)(r+s) will be maximum when p+q=r+s=1

Therefore, Option A

#Shortcut
A + B =10;

1+9 ; 9
4+6 ;24
5+5 ;25 (PRODUCT AB MAX WHEN TERMS ARE EQUAL)

Jawad001 · Answer

We are given that,

p+q+r+s= 2, then A = (p+q)(r+s)

0 +1/2 +1/2 + 1 =2, where p = 0
 q = 1/2
 r = 1/2
 s = 1

p+q =0 + 1/2 = 1/2
r + s = 1/2 + 1/2 =1

Now,(p+q)(r+s) = 1/2 + 1
 = 3/2(B)

vishumangal · Answer

Ans Should be A

As Max value for (P+Q)(R+S) will be 1 when each number will be 0.5

Archit3110 · Answer

p+q+r+s= 2
given all are +ve no
so we can have 0+0+0+2 , 0+0+1+1, 1/2+1/2+1/2+1/2
for A=(p+q)(r+s)
possible values = 1 and 0
IMO A; 0 <= A <=1

If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r+s) satisfies the relation:
A. 0 <= A <=1
B. 1 <= A <=2
C. 2<= A <=3
D. 3<= A <=4
E. 4<= A <=5

Aviral1995 · Answer

since p,q,r,s are positive numbers . For example consider any value for p,q,r,s. Let p=q=3/4; r=s= 1/4

=(p+q)(r+s)
=(3/4 + 3/4) (1/4 + 1/4)
=(3/2) (1/2)
=3/4=0.75 . which lies between 0 and 1

Therefore A

Raxit85 · Answer

If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r+s) satisfies the relation:
A. 0 <= A <=1
B. 1 <= A <=2
C. 2<= A <=3
D. 3<= A <=4
E. 4<= A <=5

p,q,r,s each are greater than 0.
p+q+r+s = 2
Let's assume, 0.5+0.5+0.5+0.5 = 2, then A = 1 * 1 = 1
0.1+0.2+0.3+1.4 = 2, then A = 0.3*1.7 = 0.51
0.001+0.002+ 1.000+0.997 = 2, then A = 0.003*1.997=0.005... (must be greater than 1)
Still if we take lower values of p,q,r and S, then A can be lowest near to 0 but never 0 or below 0. So, 0 <= A <=1

Hence, Ans is A.

lacktutor · Answer

p,q,r,s -- positive numbers such that p+q+r+s= 2, 
A = (p+q)(r+s)=???
--> the maximum value of (p+q)(r+s) is 1 (0.5+0.5)*(0.5+0.5)= 1*1=1
Only answer choice A satisfies the relation

The answer is A

mohagar · Answer

Option A would be the correct answer

hiranmay · Answer

If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r+s) satisfies the relation:
A. 0 <= A <=1 -->  correct 
B. 1 <= A <=2
C. 2<= A <=3
D. 3<= A <=4
E. 4<= A <=5

Solution :
p+q+r+s= 2
if p+q=1=r+s, then A = (p+q)(r+s) = 1*1=1
if p+q=3(r+s),r+s=1/2 & q+q=3/2, then A = (p+q)(r+s) = (3/4)*(1/2)=3/4<1
So maximum value of A=1
So 0<A<=1

AndreV · Answer

Since p,q,r,s are all positive number, any of them is greater than 0 and less than 2. Clearly, A>0, as (p+q) and (r+s) are greater than 0.
p+q+r+s=2. This can be rearranged as (p+q)+(r+s)=2. It follows that (p+q)=2-(r+s)
Plugging this equation into A, we have:
A=(p+q)(r+s)=[2-(r+s)](r+s)=2(r+s)-(r+s)^2=-{(r+s)^2-2(r+s)}. Adding and subtracting one to expression in brackets.
A=-{(r+s)^2-2(r+s)+1-1}=-{(r+s)^2-2(r+s)-1}+1=-(r+s-1)^2+1
Since r+s>0, it follows that r+s-1>-1. Therefore, (r+s-1)^2>=0 (The expression is equal to zero when r+s=1).
Multiplying both members of the inequality by -1 yields:
-(r+s-1)^2<=0
Consequently,
A=-(r+s-1)^2+1<=1
Answer: A [0<A<=1]

exuberantvivek · Answer

For the given sum the product will be maximum when the number are same so the max value A can have is:
 1*1=1.
Hence answer is A.

If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r

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If p,q,r,s are positive numbers such that p+q+r+s= 2, then A = (p+q)(r

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