The correct option is B - P and Q together weigh more than the total weight of A and C.

[as per provisional answerkey]

Solution

We are given the following equations based on the weights:
1) $A + B + C + D = 17$
2) $A + C = 6$
3) $P + Q + S + D = 15$
4) $P + Q + R + B = 17$
5) $P = R$ and $Q = S$

Step 1: Find the value of (B + D)
Substitute equation (2) into equation (1):
$(A + C) + B + D = 17$
$6 + B + D = 17$
$B + D = 11$

Step 2: Simplify equations (3) and (4) using the given equalities ( $P = R$ , $Q = S$ )
From (3): $P + Q + Q + D = 15$ => $P + 2 Q + D = 15$
From (4): $P + Q + P + B = 17$ => $2 P + Q + B = 17$

Step 3: Add the simplified equations from Step 2
$(P + 2 Q + D) + (2 P + Q + B) = 15 + 17$
$3 P + 3 Q + B + D = 32$
$3 (P + Q) + (B + D) = 32$

Step 4: Substitute the value of (B + D) from Step 1
$3 (P + Q) + 11 = 32$
$3 (P + Q) = 21$
$P + Q = 7$

Step 5: Compare the results
We know $A + C = 6$ and $P + Q = 7$ .
Therefore, $(P + Q) > (A + C)$ , which means P and Q together weigh more than A and C.

Why the other options are incorrect

Option (a) - B and D together weigh less than P and Q: Our calculation shows $B + D = 11$ and $P + Q = 7$ . Since 11 is greater than 7, B and D weigh more than P and Q, making this statement false.
Option (c) - P weighs more than Q: From the equations $P + 2 Q + D = 15$ and $2 P + Q + B = 17$ , we cannot determine the individual values of P and Q without knowing the individual values of B and D. There is no information to prove $P > Q$ .
Option (d) - Q weighs more than P: Similar to option (c), the individual weights of P and Q are indeterminate based on the provided system of equations. We only know their sum is 7.

Key Concept

This problem tests the ability to solve a system of linear equations using substitution and elimination to compare sums of variables when individual values are indeterminate.