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+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.