GATE 2018 IN GA Q.7: solution and answer

GATE 2018 · INGA Q.7 · 2 marks · MCQ

A set of 4 parallel lines intersect with another set of 5 parallel lines. How many parallelograms are formed?

(A)
20
(B)
48
(C)
60
(D)
72

See the question in the paper

Answer and solution

ANSWER

(C) 60. Choosing any 2 of the 4 lines and any 2 of the 5 lines fixes exactly one parallelogram, so there are $(2 4) \times (2 5) = 6 \times 10 = 60$ .

Official answer: CSource: Official answer key, GATE 2018 IN (IIT Guwahati)

A parallelogram needs two pairs of parallel sides: one pair of lines from the first set and one pair from the second set. Any two lines of one set together with any two lines of the other set enclose exactly one parallelogram, and different choices give different parallelograms.
Ways to pick 2 of the 4 parallel lines: $(2 4) = \frac{4 \times 3}{2} = 6$ .
Ways to pick 2 of the 5 parallel lines: $(2 5) = \frac{5 \times 4}{2} = 10$ .
The two choices are independent, so by the multiplication principle: $Parallelograms = (2 4) \times (2 5) = 6 \times 10 = 60$
This matches option (C).

KEY IDEA

A parallelogram is fixed by choosing its two sides from each set of parallel lines, so the count is $(2 m) \times (2 n)$ for $m$ lines crossing $n$ lines. This counts every parallelogram of every size, not only the smallest cells of the grid.

WATCH OUT FOR

Counting only the smallest cells of the grid, $(4 - 1) \times (5 - 1) = 12$ , which leaves out the larger parallelograms made up of several cells. Another mistake is multiplying $4 \times 5 = 20$ (option A), which is the number of intersection points, not the number of parallelograms.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2018 IN GA Q.7: solution and answer

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