Which of the following is the correct sequence of goals scored in matches 1, 3, 5 and 7?

	A	B	H	S	Total
M1
M2
M3
M4
M5
M6
M7
M8
Total					$12$

In this table M1, M2, M3, M4, M5, M6, M7, M8 - are all the 8 matches

And, A, B, H, S are all 4 players Amla, Bimla, Harita and Sarita respectively.

Also, we are given the total number of goals in all the 8 matches by all the 4 players is 12.

	A	B	H	S	Total (2)
M1
M2
M3
M4
M5
M6
M7
M8
Total (1)					$12$

We are given,

There was a total of 12 goals in these matches

and each of them had scored at least one goal. No two players had scored the same total number of goals.

Therefore the possibilities of the total number of goals by players in these matches is

i.e., the possible values in total (1) -

1. $1, 2, 3, 6$

2. $1, 2, 4, 5$

	A	B	H	S	Total (2)
M1		1
M2
M3
M4
M5
M6
M7
M8
Total (1)	1/2	4	5	2/1	$12$

Using statement 4,

Bimla scored a goal in Match 1 and one each in three other consecutive matches.

=> Bimla has a minimum score 4 goals.

But, clue 2 says,

Harita scored more goals than Bimla.

Now if we consider case 1 - $1, 2, 3, 6$

then, since Bimla has a score > or equal to 4

=> Bimla score will be 6

but then Harita's score can't be more then Bimla

therefore, this case is rejected

And therefore, we can say that,

total scores will be $1, 2, 4, 5$

From which Bimla's total score will be - 4 (and she will score 1 goal in match 1)

and, Harita's score will be greater then Bimla = 5

and, Amla and Sarita score will be 1 or 2.

	A	B	H	S	Total (2)
M1		1
M2					1
M3
M4					1
M5
M6					1
M7
M8					1
Total (1)	1/2	4	5	2/1	$12$

Using statement 1,

Only one goal was scored in every even numbered match

Therefore, Total (2) score in M2, M4, M6, M8 = 1

	A	B	H	S	Total (2)
M1		1
M2					$1$
M3
M4	-	-	1	-	$1$
M5
M6					$1$
M7
M8	-	-	1	-	$1$
Total (1)	$1/2$	$4$	$5$	$2/1$	$12$

Clue 3 says,

The highest goal scorer scored goals in exactly 3 matches including Match 4 and Match 8.

We know, highest goal scorer is Harita

=> Harita scored goal in M4, M8, and one other

But, we know the total score in M4 and M8 is 1 each.

=> Harita will score 1 goal each in M4 and M8 and 3 goals in one of the matches.

	A	B	H	S	Total (2)
M1		1
M2		-			$1$
M3		-
M4	-	-	1	-	$1$
M5		1
M6	-	1	-	-	$1$
M7		1
M8	-	-	1	-	$1$
Total (1)	$1/2$	$4$	$5$	$2/1$	$12$

Now, we know Bimla score 1 goal each in 3 consecutive matches

The only 3 consecutive places left is M5, M6, and M7

Therefore, Bimla scored 1 score each in M5, M6, M7.

	A	B	H	S	Total (2)
M1		1
M2		-			$1$
M3		-			$x$
M4	-	-	1	-	$1$
M5		1
M6	-	1	-	-	$1$
M7		1			$x$
M8	-	-	1	-	$1$
Total (1)	$1/2$	$4$	$5$	$2/1$	$12$

Clue 5 says,

An equal number of goals were scored in Match 3 and Match 7

Let that be $'x'$

Also, $x$ is different from the number of goals scored in either Match 1 or Match 5.

But Clue 6 says,

The match in which the highest number of goals was scored was unique and it was not Match 5.

Now, only Unique scoring matches are 1 and 5

from which 5 is not the highest scoring match

=> Match 1 had highest number of goals.

	A	B	H	S	Total (2)
M1		1			$4$
M2		-			$1$
M3		-			$1$
M4	-	-	1	-	$1$
M5		1			$2$
M6	-	1	-	-	$1$
M7	-	1	-	-	$1$
M8	-	-	1	-	$1$
Total (1)	$1/2$	$4$	$5$	$2/1$	$12$

Let the total number of goals in M1 = $y$

and total number of goals in M8 = $z$

=> $x + x + y + z = 12 -1 -1- 1- 1$

=> $2x +y+ z = 8$

If $x=1$,

then $y +z =6$

therefore possible solutions for $y$ and $z$ will be 2 and 4 only

If $x=2$,

then $y+z=4$

therefore possible solutions for $y$ and $z$ will be 1 and 3 only

but since highest goals scored is in Match 1

so then no. of goals scored in match 1 must be 3

Harita must have scored 3 goals in match 1 as Harita scored 5 goals in exactly 3 matches.

Therefore, we can see this is not possible because then the no. of goals scored in Match 1 becomes 4.

Therefore the only possible solution is $x=1, y=4$ and $z=2$

	A	B	H	S	Total (2)
M1	-	1	3	-	$4$
M2		-	-		$1$
M3		-	-		$1$
M4	-	-	1	-	$1$
M5		1	-		$2$
M6	-	1	-	-	$1$
M7	-	1	-	-	$1$
M8	-	-	1	-	$1$
Total (1)	$1/2$	$4$	$5$	$2/1$	$12$

We know, Harita has scored total of 5 goals,

From which 1 goal each was scored in M4 and M8

=> only match where she can score 3 goals is M1

	A	B	H	S	Total (2)
M1	-	1	3	-	$4$
M2		-	-		$1$
M3		-	-		$1$
M4	-	-	1	-	$1$
M5		1	-		$2$
M6	-	1	-	-	$1$
M7	-	1	-	-	$1$
M8	-	-	1	-	$1$
Total (1)	$1/2$	$4$	$5$	$2/1$	$12$

This is the final table we can make,

In LRDI sets it is many times possible that you may not be able to fill whole of the table,

in that case you should not give up thinking that you are unable to solve the set, instead you should move to the questions and try them and find out how much of them you can solve using the table you have formed.

	A	B	H	S	Total (2)
M1	-	1	3	-	$4$
M2		-	-		$1$
M3		-	-		$1$
M4	-	-	1	-	$1$
M5		1	-		$2$
M6	-	1	-	-	$1$
M7	-	1	-	-	$1$
M8	-	-	1	-	$1$
Total (1)	$1/2$	$4$	$5$	$2/1$	$12$

Correct sequence of goals is 4,1,2,1 : Option 4