IPMAT Indore 2023 · Quantitative Ability · Question 27

Step 1 — cross-multiply.

\left( a+b \right)\left( d+a \right)=\left( c+d \right)\left( b+c \right)

Step 2 — expand both sides.

ad+{a}^{2}+bd+ab=bc+{c}^{2}+bd+cd

The bd terms cancel:

{a}^{2}+ad+ab={c}^{2}+bc+cd

Step 3 — bring everything to one side and group.

\left({a}^{2}-{c}^{2}\right)+\left( ad-cd \right)+\left( ab-bc \right)=0

\left( a-c \right)\left( a+c \right)+d\left( a-c \right)+b\left( a-c \right)=0

Step 4 — factor out \left( a-c \right).

\left( a-c \right)\left( a+b+c+d \right)=0

Step 5 — read off both possibilities. A product is zero when either factor is:

a=c \text{or}a+b+c+d=0

Why the other options fail.

(a) too strong — it misses the second branch. Take a=1,b=2,c=3,d=-6: the sum is 0, the equation holds (3/5=-3/-5), yet a\ne c.

(b) b=d is never forced. Take a=c=1 with b=5, d=9: both sides are 6/6 and 10/10 — the equation holds with b\ne d.

(d) too strong in the other direction — a=c alone suffices without the sum vanishing.

Answer: Option 3 — a=c, or a+b+c+d=0

Note on the supplied key. The file supplied for this paper marks Option 2 (“a=c and b=d”), and at least one online question bank repeats it. The counterexample above (a=c=1, b=5, d=9) satisfies the equation with b\ne d, so Option 2 cannot be “always true”. The verified answer is Option 3.

Shortcut. Cross-multiply and look for the common factor — do not try to “solve”. The whole question is the single factorisation

\left( a-c \right)\left( a+b+c+d \right)=0

and once you see it, the answer is forced to be the option containing the word “or”.

A faster route if you are short of time: test the options with numbers.

Put a=c=1, b=5, d=9: 6/6=10/10 ✔ — so (d) is out, and (b) is out because b\ne d here.

Put a=1,b=2,c=3,d=-6: 3/5=-3/-5 ✔ — so (a) is out.

Both branches are real, so only the “or” option survives. Two substitutions, no algebra.