The weight of X, in kg, is denoted by X. The weights of A, B, C, D, P, Q, R and S are measured. Given:

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

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): $P+Q+Q+D=15$ => $P+2Q+D=15$
From (4): $P+Q+P+B=17$ => $2P+Q+B=17$

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

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 $P+2Q+D=15$ and $2P+Q+B=17$, 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.