ISS 2016 Statistics Paper-1 Solution: Question 47 (Valid Pairs of Regression Equations)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, today we take up Question 47 from the 2016 paper. This one sits in the correlation and regression portion of the syllabus and trips up a lot of aspirants not because the algebra is hard, but because the logic of the test is easy to apply backwards.

Quick Summary

• Topic: Regression equations — testing validity of a pair of lines

• Question reference: ISS 2016, Statistics Paper-1, Question 47

• Correct answer: (c) P and Q only

The Question (as it appeared in the paper)

Consider the following pairs of equations:

P : X + Y = 2, 2X + 3Y = 4 Q : X − 2Y = 3, 2X − 3Y = 5 R : X + 2Y = 5, 2X − 3Y = 3

Which of the above is/are valid pair(s) of regression equations?

• (a) P only

• (b) Q only

• (c) P and Q only

• (d) P, Q and R

The Rule That Decides Everything Here

Whenever a question gives you two lines and asks "could these be a genuine pair of regression lines," there is exactly one test you need. For any pair of regression lines, one line is the regression of Y on X (slope bYX) and the other is the regression of X on Y (slope bXY). The product of these two slopes is always equal to the square of the correlation coefficient:

bYX × bXY = r²

Since r² always lies between 0 and 1 (inclusive), a genuine pair of regression equations must satisfy 0 ≤ bYX × bXY ≤ 1, and because r and the two slopes always carry the same sign, bYX and bXY must also agree in sign. If no way of labelling the two given lines satisfies both conditions, the pair cannot be a valid regression pair.

That last part is the catch: the question never tells you which equation is "Y on X" and which is "X on Y." You have to try both labellings for each pair and see if either one works.

Step 1: Testing Pair P (X + Y = 2 and 2X + 3Y = 4)

First, write each line in both possible forms.

From X + Y = 2: as Y on X, Y = 2 − X, so the slope is −1. As X on Y, X = 2 − Y, so the slope is −1.

From 2X + 3Y = 4: as Y on X, Y = 4/3 − (2/3)X, so the slope is −2/3. As X on Y, X = 2 − 1.5Y, so the slope is −1.5.

Labelling attempt 1:let "X + Y = 2" be Y on X (bYX = −1) and "2X + 3Y = 4" be X on Y (bXY = −1.5). Product = (−1) × (−1.5) = 1.5. This is greater than 1, so this labelling fails.

Labelling attempt 2:let "2X + 3Y = 4" be Y on X (bYX = −2/3) and "X + Y = 2" be X on Y (bXY = −1). Product = (−2/3) × (−1) = 2/3. This lies between 0 and 1, and both slopes are negative (same sign). This labelling works.

Since at least one valid labelling exists,P is a valid pair.

Step 2: Testing Pair Q (X − 2Y = 3 and 2X − 3Y = 5)

From X − 2Y = 3: as Y on X, Y = 0.5X − 1.5, slope = 0.5. As X on Y, X = 3 + 2Y, slope = 2.

From 2X − 3Y = 5: as Y on X, Y = (2/3)X − 5/3, slope = 2/3. As X on Y, X = 2.5 + 1.5Y, slope = 1.5.

Labelling attempt 1:let "X − 2Y = 3" be Y on X (bYX = 0.5) and "2X − 3Y = 5" be X on Y (bXY = 1.5). Product = 0.5 × 1.5 = 0.75. This lies between 0 and 1, and both slopes are positive (same sign). This labelling works immediately.

Since a valid labelling exists,Q is a valid pair.(For completeness: the other labelling gives 2/3 × 2 = 4/3, which fails — but we only need one labelling to succeed.)

Step 3: Testing Pair R (X + 2Y = 5 and 2X − 3Y = 3)

From X + 2Y = 5: as Y on X, Y = 2.5 − 0.5X, slope = −0.5. As X on Y, X = 5 − 2Y, slope = −2.

From 2X − 3Y = 3: as Y on X, Y = (2/3)X − 1, slope = 2/3. As X on Y, X = 1.5 + 1.5Y, slope = 1.5.

Labelling attempt 1:bYX = −0.5 (from the first equation), bXY = 1.5 (from the second). Product = −0.5 × 1.5 = −0.75. A negative product is impossible since r² can never be negative. This labelling fails.

Labelling attempt 2:bYX = 2/3 (from the second equation), bXY = −2 (from the first). Product = (2/3) × (−2) = −4/3. Again negative. This labelling also fails.

Neither labelling works, and notice why: in R, one equation always forces a negative slope and the other a positive slope, no matter how you assign them. Two regression lines with opposite-signed slopes can never come from the same data, because both must share the sign of the same correlation coefficient. So R is not a valid pair.

Pro Tip: Don't try to "guess" which equation is Y-on-X by looking at which variable is isolated — the paper deliberately writes both equations in general form so that guess fails. The reliable method, the way it's usually taught, is to solve each equation for Y first and for X first, write down both possible slopes, and then test both pairings against the two conditions (product between 0 and 1, same sign) rather than just one. Students who force themselves to check both labellings every single time, instead of stopping at the first one they try, stop losing marks on exactly this kind of question.

Final Answer

P is valid, Q is valid, R is not valid. So the correct choice is (c) P and Q only, which matches the official answer key for this question.

Why This Question Matters

Regression-pair validity questions show up again and again across ISS, IIT JAM, and GATE Statistics papers because they test three things at once: your algebraic fluency in rearranging linear equations, your conceptual grip on what bYX and bXY actually mean, and your understanding of why r² is bounded between 0 and 1. A student who has memorised the "product between 0 and 1" rule but forgets the "same sign" condition will get pairs like R wrong, since its product is negative rather than simply out of range — a distinction examiners deliberately exploit.

Frequently Asked Questions

Why must the product of the two regression coefficients lie between 0 and 1?

Because bYX × bXY is algebraically equal to r², the square of the correlation coefficient. Since r itself can only range from −1 to 1, its square can only range from 0 to 1. Any pair of lines whose slopes multiply to something outside this range cannot represent real regression equations.

Why do both regression coefficients need the same sign?

Both bYX and bXY are defined using the same correlation coefficient r in the numerator (bYX = r × σY/σX and bXY = r × σX/σY), and standard deviations are always positive. So the sign of both coefficients is entirely determined by the sign of r, meaning they must match.

How do I know which given equation is the regression of Y on X?

You usually don't know in advance — that's the point of this question type. You must test both possible labellings (each equation taken in turn as Y on X) and accept the pair as valid if at least one labelling satisfies both the range and sign conditions.

What if neither labelling works for a pair?

Then that pair cannot represent a genuine pair of regression lines for any data set, and you mark it invalid — exactly what happens with pair R in this question.

Is this topic important for the ISS Paper-1 syllabus specifically?

Yes. Correlation and regression is a core topic under Descriptive Statistics in Paper-1, and validity-of-regression-lines questions appear almost every year in some form, making this a high-yield area to master rather than skip.

Can both equations in a pair have the same slope sign but still be invalid?

Yes — same sign is necessary but not sufficient. The product of the slopes must also fall within the 0-to-1 range. A pair can have matching signs and still fail if the product exceeds 1, as seen in the rejected labelling of pair P.

If any step above felt unclear, drop a comment with the exact line you got stuck on — working through it together usually clears it up faster than re-reading alone. And if you know a fellow ISS aspirant grinding through the same previous-year papers, send this one their way.