This set contains Geometrical Figures, Chessboard and Grid Questions with Solutions — Set 5 (Q41-Q50) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.
Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.
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Geometrical Figures, Chessboard and Grid Questions 41 to 50 with Solutions
41. In how many ways 2 + and 2 – signs are filled into 4×4 cell where each cell can contain maximum one character such that each row & column can not contain same sign.
42. There are ‘n’ points in a plane. No three of which are on same straight line. All the possible straight lines are made by joining these ‘n’ points.
(i) What is the maximum number of point of intersection of these straight lines?
(ii) Taking point of intersection of these straight lines as vertices of triangles then what is the maximum number of triangles that can be formed?
43. Consider 5 points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or co-incident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have?
44. Consider a 6×6 square which is dissected into 9 rectangles by lines parallel to its sides such that all the rectangles have integer sides. Out of 9 rectangles what is the maximum number of congruent rectangle ?
45. Consider a decomposition of an 8×8 chessboard into p non overlapping rectangles with the following condition:-
Condition (i):- Number of white and number of black squares are same
Condition (ii):- If aᵢ is the number of white squares in the iᵗʰ rectangle then
a₁ < a₂ < a₃ < ………<aₚ .
(i). Find the maximum possible value of p.
(ii). How many such different cases are possible if p is maximum.
46. In how many ways two kings one black & one white can be placed on a 8×8 chess board such that they are not on adjacent squares?
47. In how many ways two identical kings can be placed on a 8×8 chess board such that they are not on adjacent squares?
48. In how many ways two queens can be placed on a 8×8 chess board such that they are not able to attack each other (Queens can attack in the same row/column/diagonal)?
(i). one queen is black & other queen is white
(ii). both queens are identical.
49. In how many ways can two squares be chosen on a 8×8 chess board such that they have only one corner common?
50. How many regular polygons can be formed by joining the vertices of a 36 sided regular polygon?
Geometrical Figures, Chessboard and Grid Questions 41 to 50 — Step-by-Step Solutions
41. In how many ways 2 + and 2 – signs are filled into 4×4 cell where each cell can contain maximum one character such that each row & column can not contain same sign ?
Solution:-
1ˢᵗ ‘+’ sign can be arrranged in ¹⁶C₁ = 16 ways
2ⁿᵈ ‘+’ sign can be arranged in ⁹C₁ = 9 ways
Since both ‘+’ signs are indistinguishable so both ‘+’sign can be arranged in \(\frac{{16 \times 9}}{2}\) = 72 ways
similarly both ‘-‘ sign can be arranged in \(\frac{{16 \times 9}}{2}\)= 72 ways
Hence 2 ‘+’ & 2 ‘-‘ sign can be arranged in 72×72 ways.
From this valuewe have to exclude the following cases to get the final answer:-
case (i): two group (+, -) & (+, -) can be arranged in \(\frac{{16 \times 9}}{2}\) = 72 ways
case(ii):
Hence symbol of case (ii) according to condition can be arranged in 16×9×8 ways
Hence required Answer is:
= 72×72 – 72 – 16×9×8
=3960 Answer
42. There are ‘n’ points in a plane. No three of which are on same straight line. All the possible straight lines are made by joining these ‘n’ points.
(i) What is the maximum number of point of intersection of these straight lines?
(ii) Taking point of intersection of these straight lines as vertices of triangles then what is the maximum number of triangles that can be formed?
Solution:-
(i) Number of straight lines that can be drawn is ⁿC₂ = \(\frac{{n(n – 1)}}{2}\) = k (say)
Now from k straight lines for maximum number of point of intersection condition is, no two of them are parallel to each other and no three of them are concurrent. So number of point of intersection of these k straight lines is = ᵏC₂ = \(\boldsymbol{\frac{{n(n – 1)}}{2}}\)C₂ Answer
(ii). From above number of point of intersection of these straight lines = \(\boldsymbol{\frac{{n(n – 1)}}{2}}\)C₂ = p (say)
Since every straight line is intersected by every other remaining (k-1) straight lines at (k-1) points & these (k-1) points on being in same line are collinear.
Now number of triangles
43. Consider 5 points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or co-incident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have?
Solution:-
Consider 5 points A, B, C, D, E. Total ⁵C₂ = 10 lines can be formed by joining these points. Consider a line AB. In this line total 3 perpendiculars can be drawn from remaining 3 points C, D, E.
So total number of perpendiculars
Hence maximum number of points of intersection of these perpendiculars = ³⁰C₂ = 435
but on close observation we found that these 435 points are not distinct means out of 30 perpendicular straight lines not all of them are non concurrent so we have to remove some values out of 435 points. Consider following cases:-
case(i): Total number of triangle that can be formed out of these 5 points = ⁵C₃ = 10
Now consider a triangle △ABC
perpendiculars are drawn from point A, B, C to respective vertices. These perpendiculars intersect at one point known as orthocentre but in calculation of 435 points we have counted number of intersection of these perpendiculars as ³C₂ = 3. But instead of 3 they intersect at one point. So for one triangle we lose 2×10 = 20 points.
Hence we have to subtract 20 points ut of 435 points.
Case(iii):- Consider point A. Remember foour points B, C, D, E ⟹ out of these ⁴C₂ = 6 lines can be drawn.
So from point A 6 perpendiculars will be drawn on these 6 lines but these perpendiculars are concurrent i.e. they interect at one point but in calculation of 435 intersection point for these 6 perpendiculars we have counted point of intersection as ⁶C₂ = 15 but actually they intersect at only 1 point so 14 intersection points are counted extrafor point A. So overall 14×5 = 70 points are counted extra which we have to remove from original calculation.
Case(ii): Consider line AB
A case may arise when perpendiculars drawn on AB from C, D & E are parallel to each other i.e. they do not intersect but we have counted their number of point of intersection as ³C₂ = 3
So we have to subtract 3×10 = 30 from original calculation
↓
10 lines from
points A, B, C, D, E
& No other case is possible
Hence maximum number of points of intersection of perpendiculars = 435 – case(i) – case(ii) – case(iii)
= 435 – 20 – 70 – 30
= 315 Answer
44. Consider a 6×6 square which is dissected into 9 rectangles by lines parallel to its sides such that all the rectangles have integer sides. Out of 9 rectangles what is the maximum number of congruent rectangle ?
Solution:-
Let us solve this problem diagramatically:-
Since we are to dissect 6×6 square with 9 rectangles so make the 9 rectangles starting with least possible areas:-
1 ⟹ 1×1
2 ⟹ 1×2
3 ⟹ 1×3
4 ⟹ 1×4 or 2×2
5 ⟹ 1×5
6 ⟹ 1×6 or 2×3
8 ⟹ 2×4
When we make diagram of these rectangle then we see that we are bond to get 2 congruent rectangles for 1×5 & 2×4.
So we will get 2 congruent Rectangles Answer
45. Consider a decomposition of an 8×8 chessboard into p non overlapping rectangles with the following condition:- Condition (i):- Number of white and number of black squares are same Condition (ii):- If aᵢ is the number of white squares in the iᵗʰ rectangle then a₁ < a₂ < a₃ < ………<aₚ .
(i). Find the maximum possible value of p.
(ii). How many such different cases are possible if p is maximum.
Solution:-
Number of white square in 2ⁿᵈ rectangle = a₂
Number of white square in 3ʳᵈ rectangle = a₃
.
.
.
.
.
Number of white square in pᵗʰ rectangle = aₚ
Since these rectangles are non-overlapping & number of white square and black square are same in each rectangle. So addition of above all number of white square & number of black squares is equal to total number of squares in chessboard because rectangles are non-overlapping.
∴ 2(a₁ + a₂ + a₃ + ……… + aₚ) = 64
⇒ a₁ + a₂ + a₃ + ……… + aₚ = 32 ………….(i)
(i). Now to get maximum value of aₚ ⇒ the values of a₁, a₂, a₃, ………. should be minimum.
now it is given that
a₁<a₂<a₃<…………<aₚ
& a₁ will have minimum value of 1.
∴ a₁≥ 1
Hence a₂ ≥ 2
a₃ ≥ 3
Put these vallues in equation (i)
1 + 2 + 3 + …………. + P ≤ 32
⟹ \(\frac{{P(P + 1)}}{2}\) ≤ 32
⟹ P(P+1) ≤ 64
Hence P = 7
∴ maximum value of p = 7 Answer
(ii). So we are to divide 8×8 chess board into 7 non-overlapping rectangles such that
⦿ number of white squares is same as number of black squares i.e. each rectangle has Even number of squares
⦿ & a₁<a₂<a₃<a₄<a₅<a₆<a₇
i.e. 2a₁<2a₂<2a₃<2a₄<2a₅<2a₆<2a₇
i.e number of squares in 1ˢᵗ rectangle < number of squares in 2ⁿᵈ rectangle < number of squares in 7ᵗʰ rectangle.
& 2a₁+2a₂+2a₃+2a₄+2a₅+2a₆+2a₇ = 64
now the different cases can be:-
No other division is possible.
So there are 4 different cases possible if p is maximum Answer
46. In how many ways two kings one black & one white can be placed on a 8×8 chess board such that they are not on adjacent squares?
Solution:-
Here we have following cases for placing the 1ˢᵗ king.
Case(i): Lets 1ˢᵗ king be placed at one of the 4 corners of chess board ⟹ ⁴C₁ = 4 ways
So number of ways in this case is 4×60 = 240
case(ii): If 1ˢᵗ king is placed at a square on edge other than corner’s square then this can be done in 24 ways
So number of ways in this case is = 24×58 = 1392
case (iii): If 1ˢᵗ king is placed at an interior square then this can be done in 36 ways
So number of ways in this case is = 36×55 = 1980
So total number of ways = 240 + 1392 + 1980
= 3612 Answer
47. In how many ways two identical kings can be placed on a 8×8 chess board such that they are not on adjacent squares?
Solution:-
From the solution of previous question number of ways = 3612 (When kings are distinguishable)
now in this question kings are indistinguishable i.e. we can not differentiate between kings because both kings are identical.
So this will half the count of the previous question to get the result.
Hence final answer is \(\frac{{3612}}{2}\) = 1806 Answer
48. In how many ways two queens can be placed on a 8×8 chess board such that they are not able to attack each other (Queens can attack in the same row/column/diagonal)?
(i). one queen is black & other queen is white
(ii). both queens are identical.
Solution:- we will solve this question with the help of diagram and taking different cases:-
case(i): This 8×8 chess board has 4 center squares.
when one of the queen is placed on one of these squares then number of ways to place 1ˢᵗ queen = 4
Here 4 check (✓) marks & 24 cross (✖) marks show the moves of 1ˢᵗ queen.
Hence 2ⁿᵈ queen can not be placed at these places.
∴ Number of ways to place 2ⁿᵈ queeen = 64 – (4 + 28)
= 36
So total number of ways in this case = 4×36 = 144
case (ii): When 1ˢᵗ queen is placed at one of the squares as shown in the above figure by check marks (✓), that can be done in 12 ways.
Here dots (•) & (✖) cross marks shows the moves of the 1ˢᵗ queen.
So 2ⁿᵈ queen can be placed at 64 – 26 = 38 places
So total number of ways in this case = 12×38 = 456
case (iii): When 1ˢᵗ queen is placed at one of the squares as shown in the given figure by check marks (✓), that can be done in 20 ways.
Here dot (•) & cross marks (✖) show the moves of the first queen.|
So 2ⁿᵈ queen can not be placed at 12 (•) + 12 (✖) = 24 places
∴ 2ⁿᵈ queen can be placed at 64 – 24 = 40 places
So total number of ways in this case = 20×40 = 800
case (iv): When 1ˢᵗ queen is placed at one of the given squares as shown in the given figure by check sign (✓) that can be done in 28 ways.
Here dots (•) & cross marks (✖) show the moves of the 1ˢᵗ queen.
So 2ⁿᵈ queen can not be placed at 16 (•) + 6 (✖) = 22 places
∴ 2ⁿᵈ queen can be placed at 64 – 22 = 42 places
So total number of ways in this case = 28×42
= 1176
Hence (i). when one queen is black & other is white then desired result can be achieved in 144 + 456 + 800 + 1176
= 2576 ways Answer
(ii). When both queens are identica
Then we can not distinguish between queens. Hence in this case the count will become half of the previous count when both queens are distinguishable
Hence desired result = \(\frac{{2576}}{2}\) = 1288 Answer
49. In how many ways can two squares be choosen on a 8×8 chess board such that they have only one corner common?
Sol:-
Two squares may have one corner in common only when these squares are from adjacent columns.
Now there are total 7 adjacent columns in 8×8 chess board.
Now consider 1ˢᵗ column, marked with ✓ sign will have 1 diagonal square in common marked with ✓✓ sign & remaining other 6 squares marked with ✖ sign will have 2 diagonal squares in common marked with ✖✖ sign.
Thus from the first two columns we will get
2×1 + 6×2 = 14 pair of squares
therefore the total number of ways to chosing the squares = 14×7
↓
adjacent column
= 98 Answer
50. How many regular polygons can be formed by joining the vertices of a 36 sided regular polygon?
Sol:- Since polygon is regular, so all its 36 vertices lie on a circle & are equidistant to its adjacent vertices.
now consider 36 sided regular polygon.
a new regular polygon will be formed from its vertices only when its sides are a factor of 36.
now 36 = 2×2×3×3
∴
Hence desired number of polygon = 12 + 9 + 6 + +4 + 3 + 2 + 1
= 37 Answer
✅ Well done on completing set 5!
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