CAT 2020 Question Paper | CAT LR/DI
CAT Question Paper | CAT Previous Year Paper

Out of orders booked on 11th, 12th and 13th in this order – the number of orders that took two days to deliver were 4, 6 and 8 respectively.

Out of all orders booked on a particular day, some will be delivered next day, some will be delivered next to next day and others will be considered as ‘lost’.

There are a number of conclusions we can make about the orders from the given data:

First of all, we can have a count of orders booked per day from day 14th onwards by looking at the second column ‘cumulative orders booked’.

Eg, 219 orders were booked till 13th day and 249 orders were booked till 14th day, which means 249 – 219 = 30 orders must have been booked on 14th day.

Similarly 277 – 249 = 28 orders must have been booked on 15th day. And so on.

In the same manner, we can also get a count of number of orders lost per day from day 14th onwards, if we look at the column ‘cumulative orders lost’.

We can see that number orders lost onday 14th = 92 – 91 = 1; the numbers of orders lost on day 15th = 94 – 92 = 2, and so on.

And since an order is considered lost at the end of the second day of booking, we can say that 1 lostorder on day 14th was actually from the orders booked two days before,

i.e., 12th. Similarly, 2orders ‘lost’ on 15th were actually part of orders booked on 13th. And so on.

We can make a new table based on this information as follows:

11 orders were delivered on day 13th, out of which 4 must be the ones booked on day 11th (as it is given that out of orders booked on 11th, 4 orders took two days to deliver).

That means remaining11 – 4 = 7 orders delivered on day 13th were those which were booked the previous day on 12th.

Also, out of 27 orders delivered on day 14th, 6 must be those which were booked on 12th (given inquestion), thus remaining 21 must be those which were booked on 13th.

Similarly, out of 23 orders delivered on day 15th, 8 must be the ones booked on 13th (two days before as given in question)and so remaining 15 must be those which were booked on 14th.

We can make a new table in corporating this information in new columns as follows:

So, out of the orders booked on day 12th, 7 were delivered on day 13th while 6 were delivered on day 14th.

And we can also see that 1 order booked on day 12th was lost (last column on day 14th)because it will be counted as lost at the end of two days.

Thus we can conclude that total ordersbooked on day 12th = 7 + 6 + 1 = 14.

Similarly, out of orders booked on day 13th, 21 were delivered on day 14th, 8 were delivered on day15th, and 2 got lost (end of day 15th).

Thus we can conclude that total orders booked on day 13th = 21 + 8 + 2 = 31.

We can update this information:

Now, from day 16th onwards, we can fill all the columns in yellow.

E.g., orders booked on day 14th = 30, out of which orders delivered on day 15th = 15, orders lost on day 16th = 12, thus orders booked on day 14th which were delivered on day 16th = 3.

And we can see that 11 orders were delivered on day 16th, now if we know that out of those 11, 3 delivered were those that were booked 2 days back, then 11 – 3 = 8 must be those that were booked one day back.

And in this way, we can complete the table further as follows:

Hence, option 4.

Out of orders booked on 11th, 12th and 13th in this order – the number of orders that took two days to deliver were 4, 6 and 8 respectively.

Out of all orders booked on a particular day, some will be delivered next day, some will be delivered next to next day and others will be considered as ‘lost’.

There are a number of conclusions we can make about the orders from the given data:

First of all, we can have a count of orders booked per day from day 14th onwards by looking at the second column ‘cumulative orders booked’.

Eg, 219 orders were booked till 13th day and 249 orders were booked till 14th day, which means 249 – 219 = 30 orders must have been booked on 14th day.

Similarly 277 – 249 = 28 orders must have been booked on 15th day. And so on.

In the same manner, we can also get a count of number of orders lost per day from day 14th onwards, if we look at the column ‘cumulative orders lost’.

We can see that number orders lost on day 14th = 92 – 91 = 1; the numbers of orders lost on day 15th = 94 – 92 = 2, and so on.

And since an order is considered lost at the end of the second day of booking, we can say that 1 lost order on day 14th was actually from the orders booked two days before, 

i.e., 12th. Similarly, 2 orders ‘lost’ on 15th were actually part of orders booked on 13th. And so on.

We can make a new table based on this information as follows:

11 orders were delivered on day 13th, out of which 4 must be the ones booked on day 11th (as it is given that out of orders booked on 11th, 4 orders took two days to deliver).

That means remaining11 – 4 = 7 orders delivered on day 13th were those which were booked the previous day on 12th.

Also, out of 27 orders delivered on day 14th, 6 must be those which were booked on 12th (given in question), thus remaining 21 must be those which were booked on 13th.

Similarly, out of 23 orders delivered on day 15th, 8 must be the ones booked on 13th (two days before as given in question)and so remaining 15 must be those which were booked on 14th.

We can make a new table in corporating this information in new columns as follows:

So, out of the orders booked on day 12th, 7 were delivered on day 13th while 6 were delivered on day 14th.

And we can also see that 1 order booked on day 12th was lost (last column on day 14th)because it will be counted as lost at the end of two days.

Thus we can conclude that total orders booked on day 12th = 7 + 6 + 1 = 14.

Similarly, out of orders booked on day 13th, 21 were delivered on day 14th, 8 were delivered on day15th, and 2 got lost (end of day 15th).

Thus we can conclude that total orders booked on day 13th =21 + 8 + 2 = 31.

We can update this information:

Now, from day 16th onwards, we can fill all the columns in yellow.

E.g., orders booked on day 14th= 30, out of which orders delivered on day 15th= 15, orders lost on day 16th = 12, thus orders booked on day 14th which were delivered on day 16th = 3.

And we can see that 11 orders were delivered on day 16th, now if we know that out of those 11, 3 delivered were those that were booked 2 days back, then 11 – 3 = 8 must be those that were booked one day back.

And in this way, we can complete the table further as follows:

Orders booked on day 13th were 31, out of which orders delivered next day were 21 and thosedelivered next to next day (or on second day) were 8.

Thus the delivery ratio for day 13th = 21/8.

Similarly, delivery ratio for day 14th = 15/3 or 5, for day 15th = 8/8 or 1, and for day 16th = 13/10.

Clearly, the highest ratio is that for day 14th.

Hence, option 4.

Out of orders booked on 11th, 12th and 13th in this order – the number of orders that took two days to deliver were 4, 6 and 8 respectively.

Out of all orders booked on a particular day, some will be delivered next day, some will be delivered next to next day and others will be considered as ‘lost’.

There are a number of conclusions we can make about the orders from the given data:

First of all, we can have a count of orders booked per day from day 14th onwards by looking at the second column ‘cumulative orders booked’.

Eg, 219 orders were booked till 13th day and 249 orders were booked till 14th day, which means 249 – 219 = 30 orders must have been booked on 14th day.

Similarly 277 – 249 = 28 orders must have been booked on 15th day. And so on.

In the same manner, we can also get a count of number of orders lost per day from day 14th onwards, if we look at the column ‘cumulative orders lost’.

We can see that number orders lost onday 14th = 92 – 91 = 1; the numbers of orders lost on day 15th = 94 – 92 = 2, and so on.

And since an order is considered lost at the end of the second day of booking, we can say that 1 lost order on day 14th was actually from the orders booked two days before,

i.e., 12th. Similarly, 2 orders ‘lost’ on 15th were actually part of orders booked on 13th. And so on.

We can make a new table based on this information as follows:

11 orders were delivered on day 13th, out of which 4 must be the ones booked on day 11th (as it is given that out of orders booked on 11th, 4 orders took two days to deliver).

That means remaining11 – 4 = 7 orders delivered on day 13th were those which were booked the previous day on 12th.

Also, out of 27 orders delivered on day 14th, 6 must be those which were booked on 12th (given inquestion), thus remaining 21 must be those which were booked on 13th.

Similarly, out of 23 orders delivered on day 15th, 8 must be the ones booked on 13th (two days before as given in question)and so remaining 15 must be those which were booked on 14th.

We can make a new table in corporating this information in new columns as follows:

So, out of the orders booked on day 12th, 7 were delivered on day 13th while 6 were delivered on day 14th.

And we can also see that 1 order booked on day 12th was lost (last column on day 14th)because it will be counted as lost at the end of two days.

Thus we can conclude that total orders booked on day 12th = 7 + 6 + 1 = 14.

Similarly, out of orders booked on day 13th, 21 were delivered on day 14th, 8 were delivered on day15th, and 2 got lost (end of day 15th).

Thus we can conclude that total orders booked on day 13th =21 + 8 + 2 = 31.

We can update this information:

Now, from day 16th onwards, we can fill all the columns in yellow.

E.g., orders booked on day 14th= 30, out of which orders delivered on day 15th= 15, orders lost on day 16th = 12, thus orders booked on day 14th which were delivered on day 16th = 3.

And we can see that 11 orders were delivered on day 16th, now if we know that out of those 11, 3 delivered were those that were booked 2 days back, then 11 – 3 = 8 must be those that were booked one day back.

And in this way, we can complete the table further as follows:

The number of orders booked on day 12th, 13th, 14th and 15th were 14, 31, 30 and 28 respectively.

Hence, option 1.

Out of orders booked on 11th, 12th and 13th in this order – the number of orders that took two daysto deliver were 4, 6 and 8 respectively.

Out of all orders booked on a particular day, some will be delivered next day, some will be delivered next to next day and others will be considered as ‘lost’.
There are a number of conclusions we can make about the orders from the given data:

First of all, we can have a count of orders booked per day from day 14th onwards by looking at the second column ‘cumulative orders booked’. Eg, 219 orders were booked till 13th day and 249 orders were booked till 14th day, which means 249 – 219 = 30 orders must have been booked on 14th day.Similarly 277 – 249 = 28 orders must have been booked on 15th day. And so on.

In the same manner, we can also get a count of number of orders lost per day from day 14th onwards, if we look at the column ‘cumulative orders lost’. We can see that number orders lost on day 14th = 92 – 91 = 1; the numbers of orders lost on day 15th = 94 – 92 = 2, and so on.

Andsince an order is considered lost at the end of the second day of booking, we can say that 1 lostorder on day 14th was actually from the orders booked two days before, i.e., 12th. Similarly, 2orders ‘lost’ on 15th were actually part of orders booked on 13th. And so on.
We can make a new table based on this information as follows:

11 orders were delivered on day 13th, out of which 4 must be the ones booked on day 11th (as it is given that out of orders booked on 11th, 4 orders took two days to deliver). That means remaining11 – 4 = 7 orders delivered on day 13th were those which were booked the previous day on 12th.

Also, out of 27 orders delivered on day 14th, 6 must be those which were booked on 12th (given in question), thus remaining 21 must be those which were booked on 13th. Similarly, out of 23 orders delivered on day 15th, 8 must be the ones booked on 13th (two days before as given in question)and so remaining 15 must be those which were booked on 14th.

We can make a new table in corporating this information in new columns as follows:

So, out of the orders booked on day 12th, 7 were delivered on day 13th while 6 were delivered onday 14th. And we can also see that 1 order booked on day 12th was lost (last column on day 14th)because it will be counted as lost at the end of two days. Thus we can conclude that total orders booked on day 12th = 7 + 6 + 1 = 14.

Similarly, out of orders booked on day 13th, 21 were delivered on day 14th, 8 were delivered on day15th, and 2 got lost (end of day 15th). Thus we can conclude that total orders booked on day 13th =21 + 8 + 2 = 31.

We can update this information:

Now, from day 16th onwards, we can fill all the columns in yellow. E.g., orders booked on day 14th= 30, out of which orders delivered on day 15th= 15, orders lost on day 16th = 12, thus orders booked on day 14th which were delivered on day 16th = 3.

And we can see that 11 orders weredelivered on day 16th, now if we know that out of those 11, 3 delivered were those that werebooked 2 days back, then 11 – 3 = 8 must be those that were booked one day back.

And in this way, we can complete the table further as follows:

From the table, we can see that:
31 orders were booked on 13th, out of which 2 orders got lost (end of day 15th). Fraction is 2/31.
30 orders were booked on 14th, out of which 12 orders got lost (end of day 16th). Fraction is 12/30.
28 orders were booked on 15th, out of which 12 orders got lost (end of day 17th). Fraction is 12/28.
25 orders were booked on 16th, out of which 2 orders got lost (end of day 18th). Fraction is 2/25.
Out of these four, clearly the fraction of day 15th (12/28) is largest.

Hence, option 4.

Let A, B, C, D stand for Abha, Bina, Chitra and Dipti respectively.

As each of A, B, C, D got even number of plots, using point (4) it can be concluded that the only way to divide the 12 plots among these four daughters is to give 2 plots to C and D each and 4 plots to Aand B each. So we can say that the number of plots with A, B, C, D respectively are 4, 4, 2 and 2.

Using (6) and (8), the two adjacent plots of D would be in row X-columns 3 and 4.

Now using (5) and (7), row X column 2, row Z column 4 and row Y column 3 would be plots of B.

Again looking at point 9, in column 4, above two plots are occupied by D and A, thus last plot should be occupied by B. We are now done with 2 plots of D and C each, and also 3 plots of B.

Therefore, A’s plots are row Y column 1, row Y column 2, row Y column 4 and row Z column 3.

Now we need to fill in with number of trees for each plot.

Let’s assume the number of teak trees in column 2 as x; then as per point (3), number of teak trees in column 3 and column 4 will be 2x and 4x respectively.

Thus, we can have a diagram as follows:

In this case, total number of teak trees (i.e., trees in row Y) = 21 + x +2x + 4x = 7x + 21.

Thus, asper point (9), total number of mango trees (i.e., trees in row X) = 2 × (7x + 21) = 14x + 42.

Therefore, the total number of pine trees (i.e., trees in row Z) = 205 – [7x + 21 + 14x + 42] = 142 – 21x.

We know that there are 28 + 9 = 37 trees in last two columns of row Z.

∴, the total number of trees in the first two columns of row Z = 142 – 21x – 37 = 105 – 21x

We know that x is a non-zero multiple of 3 or 4. As 105 – 21x > 0 ⇒ 5 > x ⇒ x is either 3 or 4.

CASE 1:

If we take value of x as 3, then the number of teak trees in columns 2, 3 and 4 will be 3, 6and 12 respectively.
Total number of trees with A = 21 + 3 + 12 + 9 = 45, then as per point 3,number of trees with C = 45 – 20 = 25.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 25 – 12 = 13. As 13 is neither a multiple of 3 nor of 4, this case is not valid.

CASE 2:

If we take x = 4, then the number of teak trees in columns 2, 3 and 4 will be 4, 8 and 16 respectively.

The number of teak trees (i.e., trees in row X) = 21 + 4 + 8 + 16 = 49, The number of mango trees(i.e., trees in row X) = 2 × 49 = 98.
The number of pine trees = 205 – 49 – 98 = 58.

Here, number of trees with A = 21 + 4 + 16 + 9 = 50, and as per point 1, the number of trees with C = 50 – 20 = 30, number of trees with D = 50 + 6 = 56.
Thus the number of trees with B = 205 –50 – 30 – 56 = 69.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 30 – 12 = 18.

Now the number of trees in row Z = 58 = the number of trees in B’s plot in row Z + 18 + 9 + 28

Therefore, the number of trees in B’s plot in row Z = 3

B has 3, 8 and 28 trees in two plots. Therefore, in the fourth plot, i.e., in row Z column 2, B has 69 –3 – 8 – 28 = 30 trees.

Now, we have to find the number of trees in D’s plots. So far the maximum number of trees are in B’s plot and that is 30. Therefore, from point (2), one of D’s plots must have 32 trees and the other plot has 56 – 32 = 24 trees in some order.

We now have all the information and we can plot that as follows:

Number of trees in column 1 = 12 + 21 + 3 = 36, in column 2 = 30 + 4 + 18 = 52, in column 3 = either 24 + 8 + 9 = 41 or 32 + 8 + 9 = 49,

and in column 4 = either 32 + 16 + 28 = 76 or 24 + 16+ 28 = 68.

In either case, column 4 had the highest number of trees.

Hence, option 1.

Let A, B, C, D stand for Abha, Bina, Chitra and Dipti respectively.

As each of A, B, C, D got even number of plots, using point (4) it can be concluded that the only way to divide the 12 plots among these four daughters is to give 2 plots to C and D each and 4 plots to Aand B each. So we can say that the number of plots with A, B, C, D respectively are 4, 4, 2 and 2.

Using (6) and (8), the two adjacent plots of D would be in row X-columns 3 and 4.

Now using (5) and (7), row X column 2, row Z column 4 and row Y column 3 would be plots of B.

Again looking at point 9, in column 4, above two plots are occupied by D and A, thus last plot should be occupied by B. We are now done with 2 plots of D and C each, and also 3 plots of B.

Therefore, A’s plots are row Y column 1, row Y column 2, row Y column 4 and row Z column 3.

Now we need to fill in with number of trees for each plot.

Let’s assume the number of teak trees in column 2 as x; then as per point (3), number of teak trees in column 3 and column 4 will be 2x and 4x respectively.

Thus, we can have a diagram as follows:

In this case, total number of teak trees (i.e., trees in row Y) = 21 + x +2x + 4x = 7x + 21.

Thus, asper point (9), total number of mango trees (i.e., trees in row X) = 2 × (7x + 21) = 14x + 42.

Therefore, the total number of pine trees (i.e., trees in row Z) = 205 – [7x + 21 + 14x + 42] = 142 – 21x.

We know that there are 28 + 9 = 37 trees in last two columns of row Z.

∴, the total number of trees in the first two columns of row Z = 142 – 21x – 37 = 105 – 21x

We know that x is a non-zero multiple of 3 or 4. As 105 – 21x > 0 ⇒ 5 > x ⇒ x is either 3 or 4.

CASE 1:

If we take value of x as 3, then the number of teak trees in columns 2, 3 and 4 will be 3, 6and 12 respectively.
Total number of trees with A = 21 + 3 + 12 + 9 = 45, then as per point 3,number of trees with C = 45 – 20 = 25.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 25 – 12 = 13. As 13 is neither a multiple of 3 nor of 4, this case is not valid.

CASE 2:

If we take x = 4, then the number of teak trees in columns 2, 3 and 4 will be 4, 8 and 16 respectively.

The number of teak trees (i.e., trees in row X) = 21 + 4 + 8 + 16 = 49, The number of mango trees(i.e., trees in row X) = 2 × 49 = 98.
The number of pine trees = 205 – 49 – 98 = 58.

Here, number of trees with A = 21 + 4 + 16 + 9 = 50, and as per point 1, the number of trees with C = 50 – 20 = 30, number of trees with D = 50 + 6 = 56.
Thus the number of trees with B = 205 –50 – 30 – 56 = 69.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 30 – 12 = 18.

Now the number of trees in row Z = 58 = the number of trees in B’s plot in row Z + 18 + 9 + 28

Therefore, the number of trees in B’s plot in row Z = 3

B has 3, 8 and 28 trees in two plots. Therefore, in the fourth plot, i.e., in row Z column 2, B has 69 –3 – 8 – 28 = 30 trees.

Now, we have to find the number of trees in D’s plots. So far the maximum number of trees are in B’s plot and that is 30. Therefore, from point (2), one of D’s plots must have 32 trees and the other plot has 56 – 32 = 24 trees in some order.

We now have all the information and we can plot that as follows:

Number of trees in column 1 = 12 + 21 + 3 = 36, in column 2 = 30 + 4 + 18 = 52, in column 3 = either 24 + 8 + 9 = 41 or 32 + 8 + 9 = 49,

and in column 4 = either 32 + 16 + 28 = 76 or 24 + 16+ 28 = 68.

In either case, column 4 had the highest number of trees.

Hence, option 1.

Let A, B, C, D stand for Abha, Bina, Chitra and Dipti respectively.

As each of A, B, C, D got even number of plots, using point (4) it can be concluded that the only way to divide the 12 plots among these four daughters is to give 2 plots to C and D each and 4 plots to Aand B each.

So we can say that the number of plots with A, B, C, D respectively are 4, 4, 2 and 2.

Using (6) and (8), the two adjacent plots of D would be in row X-columns 3 and 4.

Now using (5) and (7), row X column 2, row Z column 4 and row Y column 3 would be plots of B.

Again looking at point 9, in column 4, above two plots are occupied by D and A, thus last plot should be occupied by B. We are now done with 2 plots of D and C each, and also 3 plots of B.

Therefore, A’s plots are row Y column 1, row Y column 2, row Y column 4 and row Z column 3.

Now we need to fill in with number of trees for each plot.

Let’s assume the number of teak trees in column 2 as x; then as per point (3), number of teak trees in column 3 and column 4 will be 2x and 4x respectively.

Thus, we can have a diagram as follows:

In this case, total number of teak trees (i.e., trees in row Y) = 21 + x +2x + 4x = 7x + 21.

Thus, asper point (9), total number of mango trees (i.e., trees in row X) = 2 × (7x + 21) = 14x + 42.

Therefore, the total number of pine trees (i.e., trees in row Z) = 205 – [7x + 21 + 14x + 42] = 142– 21x.

We know that there are 28 + 9 = 37 trees in last two columns of row Z.

∴, the total number of trees in the first two columns of row Z = 142 – 21x – 37 = 105 – 21x

We know that x is a non-zero multiple of 3 or 4. As 105 – 21x > 0 ⇒ 5 > x ⇒ x is either 3 or 4.

CASE 1:

If we take value of x as 3, then the number of teak trees in columns 2, 3 and 4 will be 3, 6 and 12 respectively.

Total number of trees with A = 21 + 3 + 12 + 9 = 45, then as per point 3,number of trees with C = 45 – 20 = 25.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 25 – 12 = 13. As13 is neither a multiple of 3 nor of 4, this case is not valid.

CASE 2:

If we take x = 4, then the number of teak trees in columns 2, 3 and 4 will be 4, 8 and 16 respectively.

The number of teak trees (i.e., trees in row X) = 21 + 4 + 8 + 16 = 49, The number of mango trees(i.e., trees in row X) = 2 × 49 = 98.

The number of pine trees = 205 – 49 – 98 = 58.

Here, number of trees with A = 21 + 4 + 16 + 9 = 50, and as per point 1, the number of trees with C = 50 – 20 = 30, number of trees with D = 50 + 6 = 56.

Thus the number of trees with B = 205 –50 – 30 – 56 = 69.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 30 – 12 = 18.

Now the number of trees in row Z = 58 = the number of trees in B’s plot in row Z + 18 + 9 + 28

Therefore, the number of trees in B’s plot in row Z = 3

B has 3, 8 and 28 trees in two plots. Therefore, in the fourth plot, i.e., in row Z column 2, B has 69 –3 – 8 – 28 = 30 trees.

Now, we have to find the number of trees in D’s plots. So far the maximum number of trees are in B’s plot and that is 30.

Therefore, from point (2), one of D’s plots must have 32 trees and the other plot has 56 – 32 = 24 trees in some order.

We now have all the information and we can plot that as follows:

Number of pine trees with Bina is (3 + 28) = 31, not 32. So the second statement is not true, while all the other statements are true.

Hence, option 2.

Let A, B, C, D stand for Abha, Bina, Chitra and Dipti respectively.

As each of A, B, C, D got even number of plots, using point (4) it can be concluded that the only way to divide the 12 plots among these four daughters is to give 2 plots to C and D each and 4 plots to Aand B each.

So we can say that the number of plots with A, B, C, D respectively are 4, 4, 2 and 2.

Using (6) and (8), the two adjacent plots of D would be in row X-columns 3 and 4.

Now using (5) and (7), row X column 2, row Z column 4 and row Y column 3 would be plots of B.

Again looking at point 9, in column 4, above two plots are occupied by D and A, thus last plot should be occupied by B.

We are now done with 2 plots of D and C each, and also 3 plots of B.

Therefore, A’s plots are row Y column 1, row Y column 2, row Y column 4 and row Z column 3.

Now we need to fill in with number of trees for each plot.

Let’s assume the number of teak trees in column 2 as x; then as per point (3), number of teak treesin column 3 and column 4 will be 2x and 4x respectively.

Thus, we can have a diagram as follows:

In this case, total number of teak trees (i.e., trees in row Y) = 21 + x +2x + 4x = 7x + 21.

Thus, asper point (9), total number of mango trees (i.e., trees in row X) = 2 × (7x + 21) = 14x + 42.

Therefore, the total number of pine trees (i.e., trees in row Z) = 205 – [7x + 21 + 14x + 42] = 142– 21x.

We know that there are 28 + 9 = 37 trees in last two columns of row Z.

∴, the total number of trees in the first two columns of row Z = 142 – 21x – 37 = 105 – 21x

We know that x is a non-zero multiple of 3 or 4. As 105 – 21x > 0 ⇒ 5 > x ⇒ x is either 3 or 4.

CASE 1:

If we take value of x as 3, then the number of teak trees in columns 2, 3 and 4 will be 3, 6 and 12 respectively.

Total number of trees with A = 21 + 3 + 12 + 9 = 45, then as per point 3, number of trees with C = 45 – 20 = 25.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 25 – 12 = 13. As13 is neither a multiple of 3 nor of 4, this case is not valid.

CASE 2:

If we take x = 4, then the number of teak trees in columns 2, 3 and 4 will be 4, 8 and 16 respectively.

The number of teak trees (i.e., trees in row X) = 21 + 4 + 8 + 16 = 49, The number of mango trees(i.e., trees in row X) = 2 × 49 = 98.

The number of pine trees = 205 – 49 – 98 = 58. 

Here, number of trees with A = 21 + 4 + 16 + 9 = 50, and as per point 1, the number of trees with C = 50 – 20 = 30, number of trees with D = 50 + 6 = 56.

Thus the number of trees with B = 205 –50 – 30 – 56 = 69.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 30 – 12 = 18.

Now the number of trees in row Z = 58 = the number of trees in B’s plot in row Z + 18 + 9 + 28

Therefore, the number of trees in B’s plot in row Z = 3

B has 3, 8 and 28 trees in two plots. Therefore, in the fourth plot, i.e., in row Z column 2, B has 69 –3 – 8 – 28 = 30 trees.

Now, we have to find the number of trees in D’s plots. So far the maximum number of trees are in B’s plot and that is 30.

Therefore, from point (2), one of D’s plots must have 32 trees and the other plot has 56 – 32 = 24 trees in some order.

We now have all the information and we can plot that as follows:

Bina has 3 pine trees in column 1, which is the least in any of these plots.

Hence, option 2.

Let A, B, C, D stand for Abha, Bina, Chitra and Dipti respectively.

As each of A, B, C, D got even number of plots, using point (4) it can be concluded that the only way to divide the 12 plots among these four daughters is to give 2 plots to C and D each and 4 plots to Aand B each.

So we can say that the number of plots with A, B, C, D respectively are 4, 4, 2 and 2.

Using (6) and (8), the two adjacent plots of D would be in row X-columns 3 and 4.

Now using (5) and (7), row X column 2, row Z column 4 and row Y column 3 would be plots of B.

Again looking at point 9, in column 4, above two plots are occupied by D and A, thus last plot should be occupied by B.

We are now done with 2 plots of D and C each, and also 3 plots of B.

Therefore, A’s plots are row Y column 1, row Y column 2, row Y column 4 and row Z column 3.

Now we need to fill in with number of trees for each plot.

Let’s assume the number of teak trees in column 2 as x; then as per point (3), number of teak treesin column 3 and column 4 will be 2x and 4x respectively.

Thus, we can have a diagram as follows:

In this case, total number of teak trees (i.e., trees in row Y) = 21 + x +2x + 4x = 7x + 21.

Thus, asper point (9), total number of mango trees (i.e., trees in row X) = 2 × (7x + 21) = 14x + 42.

Therefore, the total number of pine trees (i.e., trees in row Z) = 205 – [7x + 21 + 14x + 42] = 142– 21x.

We know that there are 28 + 9 = 37 trees in last two columns of row Z.

∴, the total number of trees in the first two columns of row Z = 142 – 21x – 37 = 105 – 21x

We know that x is a non-zero multiple of 3 or 4. As 105 – 21x > 0 ⇒ 5 > x ⇒ x is either 3 or 4.

CASE 1:

If we take value of x as 3, then the number of teak trees in columns 2, 3 and 4 will be 3, 6and 12 respectively.

Total number of trees with A = 21 + 3 + 12 + 9 = 45, then as per point 3,number of trees with C = 45 – 20 = 25.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 25 – 12 = 13. As13 is neither a multiple of 3 nor of 4, this case is not valid.

CASE 2:

If we take x = 4, then the number of teak trees in columns 2, 3 and 4 will be 4, 8 and 16 respectively.

The number of teak trees (i.e., trees in row X) = 21 + 4 + 8 + 16 = 49, The number of mango trees(i.e., trees in row X) = 2 × 49 = 98.

The number of pine trees = 205 – 49 – 98 = 58.

Here, number of trees with A = 21 + 4 + 16 + 9 = 50, and as per point 1, the number of trees with C = 50 – 20 = 30, number of trees with D = 50 + 6 = 56.

Thus the number of trees with B = 205 –50 – 30 – 56 = 69.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 30 – 12 = 18.

Now the number of trees in row Z = 58 = the number of trees in B’s plot in row Z + 18 + 9 + 28

Therefore, the number of trees in B’s plot in row Z = 3

B has 3, 8 and 28 trees in two plots. Therefore, in the fourth plot, i.e., in row Z column 2, B has 69 –3 – 8 – 28 = 30 trees.

Now, we have to find the number of trees in D’s plots. So far the maximum number of trees are in B’s plot and that is 30.

Therefore, from point (2), one of D’s plots must have 32 trees and the other plot has 56 – 32 = 24 trees in some order.

We now have all the information and we can plot that as follows:

The number of pine trees in Chitra’s plot = 18

Hence, option 2.

Let A, B, C, D stand for Abha, Bina, Chitra and Dipti respectively.

As each of A, B, C, D got even number of plots, using point (4) it can be concluded that the only way to divide the 12 plots among these four daughters is to give 2 plots to C and D each and 4 plots to Aand B each.

So we can say that the number of plots with A, B, C, D respectively are 4, 4, 2 and 2.

Using (6) and (8), the two adjacent plots of D would be in row X-columns 3 and 4.

Now using (5) and (7), row X column 2, row Z column 4 and row Y column 3 would be plots of B.

Again looking at point 9, in column 4, above two plots are occupied by D and A, thus last plot should be occupied by B.

We are now done with 2 plots of D and C each, and also 3 plots of B.

Therefore, A’s plots are row Y column 1, row Y column 2, row Y column 4 and row Z column 3.

Now we need to fill in with number of trees for each plot.

Let’s assume the number of teak trees in column 2 as x; then as per point (3), number of teak treesin column 3 and column 4 will be 2x and 4x respectively.

Thus, we can have a diagram as follows:

In this case, total number of teak trees (i.e., trees in row Y) = 21 + x +2x + 4x = 7x + 21.

Thus, asper point (9), total number of mango trees (i.e., trees in row X) = 2 × (7x + 21) = 14x + 42.

Therefore, the total number of pine trees (i.e., trees in row Z) = 205 – [7x + 21 + 14x + 42] = 142– 21x.

We know that there are 28 + 9 = 37 trees in last two columns of row Z.

∴, the total number of trees in the first two columns of row Z = 142 – 21x – 37 = 105 – 21x

We know that x is a non-zero multiple of 3 or 4. As 105 – 21x > 0 ⇒ 5 > x ⇒ x is either 3 or 4.

CASE 1:

If we take value of x as 3, then the number of teak trees in columns 2, 3 and 4 will be 3, 6and 12 respectively.

Total number of trees with A = 21 + 3 + 12 + 9 = 45, then as per point 3,number of trees with C = 45 – 20 = 25.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 25 – 12 = 13. As13 is neither a multiple of 3 nor of 4, this case is not valid.

CASE 2:

If we take x = 4, then the number of teak trees in columns 2, 3 and 4 will be 4, 8 and 16 respectively.

The number of teak trees (i.e., trees in row X) = 21 + 4 + 8 + 16 = 49, The number of mango trees(i.e., trees in row X) = 2 × 49 = 98.

The number of pine trees = 205 – 49 – 98 = 58.

Here, number of trees with A = 21 + 4 + 16 + 9 = 50, and as per point 1, the number of trees withC = 50 – 20 = 30, number of trees with D = 50 + 6 = 56.

Thus the number of trees with B = 205 –50 – 30 – 56 = 69.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 30 – 12 = 18.

Now the number of trees in row Z = 58 = the number of trees in B’s plot in row Z + 18 + 9 + 28

Therefore, the number of trees in B’s plot in row Z = 3

B has 3, 8 and 28 trees in two plots. Therefore, in the fourth plot, i.e., in row Z column 2, B has 69 –3 – 8 – 28 = 30 trees.

Now, we have to find the number of trees in D’s plots. So far the maximum number of trees are in B’s plot and that is 30.

Therefore, from point (2), one of D’s plots must have 32 trees and the other plot has 56 – 32 = 24 trees in some order.

We now have all the information and we can plot that as follows:

The total number of trees in plots of Abha, Bina, Chitra and Dipti are 50, 69, 30 and 56 respectively.

Hence, option 3.

Let A, B, C, D stand for Abha, Bina, Chitra and Dipti respectively.

As each of A, B, C, D got even number of plots, using point (4) it can be concluded that the only way to divide the 12 plots among these four daughters is to give 2 plots to C and D each and 4 plots to Aand B each.

So we can say that the number of plots with A, B, C, D respectively are 4, 4, 2 and 2.

Using (6) and (8), the two adjacent plots of D would be in row X-columns 3 and 4.

Now using (5) and (7), row X column 2, row Z column 4 and row Y column 3 would be plots of B.

Again looking at point 9, in column 4, above two plots are occupied by D and A, thus last plot should be occupied by B.

We are now done with 2 plots of D and C each, and also 3 plots of B.

Therefore, A’s plots are row Y column 1, row Y column 2, row Y column 4 and row Z column 3.

Now we need to fill in with number of trees for each plot.

Let’s assume the number of teak trees in column 2 as x; then as per point (3), number of teak treesin column 3 and column 4 will be 2x and 4x respectively.

Thus, we can have a diagram as follows:

In this case, total number of teak trees (i.e., trees in row Y) = 21 + x +2x + 4x = 7x + 21.

Thus, asper point (9), total number of mango trees (i.e., trees in row X) = 2 × (7x + 21) = 14x + 42.

Therefore, the total number of pine trees (i.e., trees in row Z) = 205 – [7x + 21 + 14x + 42] = 142– 21x.

We know that there are 28 + 9 = 37 trees in last two columns of row Z.

∴, the total number of trees in the first two columns of row Z = 142 – 21x – 37 = 105 – 21x

We know that x is a non-zero multiple of 3 or 4. As 105 – 21x > 0 ⇒ 5 > x ⇒ x is either 3 or 4.

CASE 1:

If we take value of x as 3, then the number of teak trees in columns 2, 3 and 4 will be 3, 6and 12 respectively.

Total number of trees with A = 21 + 3 + 12 + 9 = 45, then as per point 3,number of trees with C = 45 – 20 = 25.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 25 – 12 = 13. As13 is neither a multiple of 3 nor of 4, this case is not valid.

CASE 2:

If we take x = 4, then the number of teak trees in columns 2, 3 and 4 will be 4, 8 and 16 respectively.

The number of teak trees (i.e., trees in row X) = 21 + 4 + 8 + 16 = 49, The number of mango trees(i.e., trees in row X) = 2 × 49 = 98.

The number of pine trees = 205 – 49 – 98 = 58.

Here, number of trees with A = 21 + 4 + 16 + 9 = 50, and as per point 1, the number of trees with C = 50 – 20 = 30, number of trees with D = 50 + 6 = 56.

Thus the number of trees with B = 205 –50 – 30 – 56 = 69.

As the number of mango trees in C’s plot is 12, number of pine trees in C’s plot = 30 – 12 = 18.

Now the number of trees in row Z = 58 = the number of trees in B’s plot in row Z + 18 + 9 + 28

Therefore, the number of trees in B’s plot in row Z = 3

B has 3, 8 and 28 trees in two plots. Therefore, in the fourth plot, i.e., in row Z column 2, B has 69 –3 – 8 – 28 = 30 trees.

Now, we have to find the number of trees in D’s plots. So far the maximum number of trees are in B’s plot and that is 30.

Therefore, from point (2), one of D’s plots must have 32 trees and the other plot has 56 – 32 = 24 trees in some order.

We now have all the information and we can plot that as follows:

There were 98 mango trees.

Hence, option 2.

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets + 1, the Lo gets – 3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets + 2, each Lo gets – 2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets + 3, each Lo gets – 1. In this case, total points in the round = 0.

If all four bid Lo, each gets + 1. In this case, total points in the round = + 4.
 
From fact (1), the total points of the first 3 rounds together add up to + 4 (6 – 2 – 2 + 2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds isto get  + 4 in one round and 0 in other two rounds.  And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3). i.e.,
in the first three rounds, the totals were + 4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points,each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1 + 1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = – 2, Charu = – 2 + 0 = – 2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

From facts (1) and (2), Arun, Bankim, Chanru and Dipak scored (7 – 1 =) + 1, (– 1 –(– 2) =) + 1, (– 5–(– 2) =) – 3 and (– 1 – 2 =) – 3 points respectively.

So, the total points of the last 3 rounds together addup to – 4 (1 + 1 – 3 – 3).

Now, – 4 = – 4 – 4 + 4 or – 4 + 0 + 0

If the scores of the three players together in the last three rounds are – 4, – 4, and + 4, then each player would have scored –1, –1, +1.

And total points scored by each player would be – 1. But this is not possible. So – 4 = – 4 + 0 + 0

The total number of points in a round by all the three players would be – 4 only when all the four bid Hi.

Arun had scored 3, 2 and 1 points in first three rounds. Therefore, using fact (4), in the last tworounds also, he must have scored 3 points in exactly one round.

Also, in this round remaining threeplayers have scored – 1 point each. Now we know the scores of the four players in two rounds out of the last three rounds.

Arun: 3 – 1 = 2, Bankim = Charu = Dipak = – 1 – 1 = – 2,

So, in the remaining round, points scored by

Arun: + 1 – (2) = – 1

Bankim: + 1– ( – 2) = 3

Charu: – 3– (– 2) = –1

Dipak: – 3 – (–2) = –1

Therefore, we can make a final table as follows:

Arun was the only player to bid Hi in round 2. And same thing happened in another round out of the last three but we can’t say which round number was that.

So the only round number where we canbe sure about this is the second round.

Hence, option 2.

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets + 1, the Lo gets – 3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets + 2, each Lo gets – 2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets + 3, each Lo gets – 1. In this case, total points in the round = 0.

If all four bid Lo, each gets + 1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6 – 2 – 2 + 2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds isto get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three roundsis to get + 4 in two rounds and – 4 in one round. But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were + 4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points,each of the remaining players scored – 1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1 + 1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = – 2, Charu = – 2 + 0 = – 2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

\

From facts (1) and (2), Arun, Bankim, Chanru and Dipak scored (7 – 1 =) +1, (– 1– (– 2) =) +1, (– 5– (– 2  =) –3 and (– 1– 2 =) – 3 points respectively.

So, the total points of the last 3 rounds together add up to – 4 (1 + 1– 3 – 3).

Now, – 4 = – 4 – 4 + 4 or – 4 + 0 + 0

If the scores of the three players together in the last three rounds are – 4, – 4, and + 4, then each player would have scored –1, –1, +1.

And total points scored by each player would be –1. But this is not possible. So – 4 = – 4 + 0 + 0

The total number of points in a round by all the three players would be – 4 only when all the four bid Hi.

Arun had scored 3, 2 and 1 points in first three rounds. Therefore, using fact (4), in the last two rounds also, he must have scored 3 points in exactly one round.

Also, in this round remaining three players have scored –1 point each. Now we know the scores of the four players in two rounds out ofthe last three rounds.

Arun: 3 – 1 = 2, Bankim = Charu = Dipak = – 1–1 = – 2,

So, in the remaining round, points scored by

Arun: + 1– (2) = – 1

Bankim: + 1– (–2) = 3

Charu: – 3– (–2) = – 1

Dipak: – 3 – (– 2) = – 1

Therefore, we can make a final table as follows:

We can see that Dipak got +1 only in round 3; in all other rounds he got either + 2 or – 1.

Answer: 1.

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets + 1, the Lo gets – 3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets + , each Lo gets – 1. In this case, total points in the round = 0.

If all four bid Lo, each gets + 1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6 – 2 – 2 + 2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round. But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were + 4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points,each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1  = 4, Bankim = – +1 = 0, Charu = –1+1 = 0, Dipak = –1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 +0 = – 2, Charu = – 2 + 0 = – 2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

From facts (1) and (2), Arun, Bankim, Chanru and Dipak scored (7 – 1 =) +1, (– 1– (– 2) =) +1, (– 5– (– 2) =) – 3 and (– 1– 2 =) – 3 points respectively.

So, the total points of the last 3 rounds together add up to – 4 (1 + 1– 3 – 3).

Now, – 4 = – 4 – 4 + 4 or – 4 + 0 + 0

If the scores of the three players together in the last three rounds are – 4, – 4, and + 4, then each player would have scored –1, – 1, +1.

And total points scored by each player would be –1. But this isnot possible. So – 4 = – 4 + 0 + 0

The total number of points in a round by all the three players would be – 4 only when all the four bid Hi.

Arun had scored 3, 2 and 1 points in first three rounds. Therefore, using fact (4), in the last two rounds also, he must have scored 3 points in exactly one round.

Also, in this round remaining three players have scored –1 point each. Now we know the scores of the four players in two rounds out of the last three rounds.

Arun: 3 – 1 = 2, Bankim = Charu = Dipak = – 1 – 1 = – 2,

So, in the remaining round, points scored by

Arun: + 1– (2) = – 1

Bankim: +1– (–2) = 3

Charu: – 3 – (–2) = –1

Dipak: – 3– (– 2) = –1

Therefore, we can make a final table as follows:

One of such rounds is round 3 and another such round is one of the last three rounds where everyone bid Hi. Thus, a total of 2 such rounds.

Answer: 2.

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets – 1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6–2–2+2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points, each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1+1 = 0

So in the round when Arun scored 2, Bankim = – 2 + 0 = –2, Charu = – 2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

From facts (1) and (2), Arun, Bankim, Chanru and Dipak scored (7 – 1 =) +1, (–1–(–2) =) +1, (–5–(–2) =) –3 and (–1–2 =) –3 points respectively.

So, the total points of the last 3 rounds together addup to – 4 (1+1–3–3).

Now, – 4 = – 4 – 4 + 4 or – 4 + 0 + 0

If the scores of the three players together in the last three rounds are – 4, – 4, and + 4, then each player would have scored –1, –1, +1.

And total points scored by each player would be – 1. But this isnot possible. So – 4 = – 4 + 0 + 0

The total number of points in a round by all the three players would be – 4 only when all the four bid Hi.

Arun had scored 3, 2 and 1 points in first three rounds. Therefore, using fact (4), in the last tworounds also, he must have scored 3 points in exactly one round.

Also, in this round remaining threeplayers have scored – 1 point each. Now we know the scores of the four players in two rounds out ofthe last three rounds.

Arun: 3 – 1 = 2, Bankim = Charu = Dipak = – 1 – 1 = – 2,

So, in the remaining round, points scored by

Arun: + 1 – (2) = –1

Bankim: + 1 – (–2) = 3

Charu: – 3 – (–2) = –1

Dipak: – 3 – (–2) = –1

Therefore, we can make a final table as follows:

Bankim bid Lo in 4 rounds.

Answer: 4.

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo.

If all four bid Hi, each gets –1. In this case, total points in the round = – 4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = + 4.

From fact (1), the total points of the first 3 rounds together add up to + 4 (6 – 2 – 2 + 2). And we knowthat each round’s total can be either – 4 or 0 or +4.

So, one way of achieving + 4 in three rounds is to get + 4 in one round and 0 in other two rounds.

And another way of achieving + 4 in three rounds is to get + 4 in two rounds and – 4 in one round.

But in the latter case, every player will get either + 1or – 1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds,which is not possible by fact (3).

i.e., in the first three rounds, the totals were + 4, 0, 0 in some order.

When the total was + 4, every player scored 1 point. As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points,each of the remaining players scored – 1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = – 1 + 1 = 0, Charu = – 1 + 1 = 0, Dipak = – 1 + 1 = 0

So in the round when Arun scored 2, Bankim = –2 + 0 = –2, Charu = –2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

From facts (1) and (2), Arun, Bankim, Chanru and Dipak scored (7–1 =) +1, (–1–(–2) =) +1, (–5–(–2) =) –3 and (–1–2 =) –3 points respectively.

So, the total points of the last 3 rounds together add up to – 4 (1+1–3–3).

Now, – 4 = – 4 – 4 + 4 or – 4 + 0 + 0

If the scores of the three players together in the last three rounds are – 4, – 4, and + 4, then each player would have scored –1, –1, +1.

And total points scored by each player would be –1. But this isnot possible. So – 4 = – 4 + 0 + 0

The total number of points in a round by all the three players would be – 4 only when all the four bid Hi.

Arun had scored 3, 2 and 1 points in first three rounds. Therefore, using fact (4), in the last two rounds also, he must have scored 3 points in exactly one round.

Also, in this round remaining three players have scored –1 point each. Now we know the scores of the four players in two rounds out of the last three rounds.

Arun: 3 – 1 = 2, Bankim = Charu = Dipak = – 1 –1 = –2,

So, in the remaining round, points scored by

Arun: +1–(2) = – 1

Bankim: +1–(–2) = 3

Charu: –3–(–2) = –1

Dipak: –3–(–2) = –1

Therefore, we can make a final table as follows:

Arun bid Hi in 4 rounds.

Answer: 4.

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses tobid Hi or Lo.

If all four bid Hi, each gets –1. In this case, total points in the round = –4.

If three Hi and one Lo, each Hi gets +1, the Lo gets –3. In this case, total points in the round = 0.

If two Hi and two Lo, each Hi gets +2, each Lo gets –2. In this case, total points in the round = 0.

If one Hi and three Lo, the Hi gets +3, each Lo gets –1. In this case, total points in the round = 0.

If all four bid Lo, each gets +1. In this case, total points in the round = +4.

From fact (1), the total points of the first 3 rounds together add up to +4 (6 – 2 – 2 + 2). And we know that each round’s total can be either – 4 or 0 or + 4.

So, one way of achieving +4 in three rounds is to get +4 in one round and 0 in other two rounds.

And another way of achieving +4 in three rounds is to get + 4 in two rounds and –4 in one round. But in the latter case, every player willget either +1 or –1 in each of the first three rounds.

i.e., Dipak’s score would be equal in at least two rounds, which is not possible by fact (3).

i.e., in the first three rounds, the totals were +4, 0, 0 in some order. When the total was +4, every player scored 1 point.

As Arun had scored 6 points at the end of the third round, he must have scored 3 and 2 in two rounds in some order.

When Arun scored 3 points,each of the remaining players scored –1 point. Now we know score of each player in two rounds (out of the first three rounds).

Arun = 3 + 1 = 4, Bankim = –1 + 1 = 0, Charu = –1 + 1 = 0, Dipak = –1+ 1 = 0

So in the round when Arun scored 2, Bankim = –2 + 0 = –2, Charu = –2 + 0 = –2, Dipak = 2 + 0 = 2

Thus, Dipaks score in the first three rounds would be 2, 1, –1.

Therefore, using fact (3) we can conclude

From facts (1) and (2), Arun, Bankim, Chanru and Dipak scored (7–1 =) +1, (–1–(–2) =) +1, (–5–(–2) =) –3 and (–1–2 =) –3 points respectively.

So, the total points of the last 3 rounds together add up to –4 (1+1–3–3).

Now, – 4 = – 4 – 4 + 4 or – 4 + 0 + 0

If the scores of the three players together in the last three rounds are – 4, – 4, and + 4, then each player would have scored –1, –1, +1.

And total points scored by each player would be –1. But this is not possible. So – 4 = – 4 + 0 + 0

The total number of points in a round by all the three players would be – 4 only when all the four bid Hi.

Arun had scored 3, 2 and 1 points in first three rounds. Therefore, using fact (4), in the last two rounds also, he must have scored 3 points in exactly one round.

Also, in this round remaining threeplayers have scored –1 point each. Now we know the scores of the four players in two rounds out of the last three rounds.

Arun: 3 –1 = 2, Bankim = Charu = Dipak = –1–1 = –2,

So, in the remaining round, points scored by

Arun: +1–(2) = –1

Bankim: +1–(–2) = 3

Charu: –3–(–2) = –1

Dipak: –3–(–2) = –1

Therefore, we can make a final table as follows:

The bids by Arun, Bankim, Charu and Dipak, respectively in the first round were Hi, Lo, Lo and Hi respectively.

Hence, option 3.

There are two possibilities here –

(1) the two patients whose blood samples got mixed did not have the disease
(2) one patient had the disease.

Possibility 1 – if both were already negative, then the vials in which their blood was kept will not test positive.

In this case, vial containing blood of some other patient (who has the disease) will test positive. And since each patients’ blood is distributed in 4 vials, those 4 vials would test positive.

Possibility 2 – if blood of the patient with disease (let’s call this P1) got mixed with another patient(let’s call this P2), then the four vials containing blood of P1 as well as four vials containing blood of P2 will test positive.

Now, in total, how many vials will finally test positive would depend on the number of vials common between P1 and P2.

Eg., if P1 and P2 are patients 1 and 2 respectively, then after their blood samples get mixed, vials B, D, F, G, H will test positive (5 vials).

If P1 and P2 are patients 1 and 6 from the table respectively, then vials B, C, D, F, G, H will test positive (6 vials). Suppose P1 and P2 are patients 1 and 8, then 7 vials (B, C, D, E, F, G, H) will test positive.

Maximum no of vials which can test positive will be all 8 (A, B, C, D, E, F, G, H) in case P1 and P2 are let’s say patients 1 and 16.

Thus, from the above, we can say that the number of vials which can test positive after accidental mix up of two patients’ blood samples could be either 4 or 5 or 6 or 7 or 8.

Hence, option 1.

Option [1] – B tests positive, so the patient with disease can be from 1 to 8. But since C, F and H are negative; patients 1, 2, 3, 5, 6, 7, 8, CANNOT have the disease.

This leaves patient 4 who then has the disease. So this combination of the test results is possible.

Option [2] – As C tests negative, the patients 5, 6, 7, 8, 13, 14, 15 and 16 CANNOT have the disease.

As D tests negative, the patients 1, 2, 3, 4, 9, 10, 11 and 12 CANNOT have the disease. i.e., none of the patients has the disease which is not possible.

Thus, this combination of the test results is NOT possible.

Option [3] – As D tests negative, the patients 1, 2, 3, 4, 9, 10, 11 and 12 CANNOT have the disease.

As E tests negative, the patients 3, 4, 7, 8, 11, 12, 15 and 16 CANNOT have the disease. i.e., 1, 2, 3, 4, 7, 8, 9 10, 11, 12, 15 and 16 CANNOT have the disease.

Therefore, the patient with the disease is one among 5, 6, 13, 14. As A and G are positive, patient 14 has the disease. This combination of the test results is possible.

Option [4] – B and D test positive, one of the patients 1, 2, 3 or 4 has the disease. F and H test negative, so out of these, patients 1, 2 and 3 CANNOT have the disease.

This leaves patient 4 who has the disease in this case. Possible.

Hence, option 2.

If vial A tests positive, then one of the patients from 9 to 16 has the disease.

But since vials D and G test negative, patients 9, 10, 11, 12, 14 and 16 CANNOT have the disease. That leaves patients 13 and 15.

Now vials A, C and H contain bloods of both these patients, so testing any of these three vials will not help in determining which one of these patients has the disease.

The vial that is to be tested in that case should be one out of vials F (patient 13) or E (patient 15).

Hence, option 2.

If vial C tests positive, it means one out of patients 5, 6, 7, 8, 13, 14, 15 or 16 has the disease (because each of their bloods is in vial C).

But since vials A, E and H test negative, patients 5, 7, 8,13, 14, 15 and 16 CANNOT have the disease (since these patients have their blood in one or more of these vials).

Thus, patient 6 has the disease.

Hence, option 1.

This question talks about 4 different types of facilities. Therefore, we can make a Venn diagram using observations (2) to (7) as shown below:

The above diagram indicates all the areas which are possible. Eg., F1F2 indicates those schools which have facilities F1 & F2 only, F2F3F4 indicates those schools which have exactly 3 facilities F2, F3, F4 (but not F1), F2 indicates schools which have only facility F2, F1F2F3F4 indicates schools which have all 4 facilities F1, F2, F3, F4 – and so on.

Using observation (8), 30 + 26 + 45 + a = 313 – 162 ⇒ a = 50

40 + 50 + 50 + F1F2 = 162 ⇒ F1F2 = 22

Now, two values are remaining – ‘b’ and ‘c’. All other values are there, and the sum of all values should be 600. So: b + c + 25 + 50 + 313 (entire set F2 including second and third columns) + 26 + 24 + 20 + 80 = 600 or b + c + 538 = 600 or b + c = 600 – 538 = 62.

Also, as per observation (9), sum of all heads under F1 should be equal to sum of all heads under F4, i.e., sum of all elements in first and second columns should be equal to sum of all elements in third and fourth rows. Thus, we have:

b + c + 237 = c + 279 or b = 279 – 237 = 42.

Substituting this value of b in the earlier equation, we get 42 + c = 62 or c = 62 – 42 = 20.

Now all values have been found, the final diagram will be as follows:

The number of schools having only facilities F1 and F4 = 20

Answer: 20.

This question talks about 4 different types of facilities. Therefore, we can make a Venn diagra musing observations (2) to (7) as shown below:

The above diagram indicates all the areas which are possible. Eg., F1F2 indicates those schools which have facilities F1 & F2 only, F2F3F4 indicates those schools which have exactly 3 facilities F2, F3, F4 (but not F1), F2 indicates schools which have only facility F2, F1F2F3F4 indicates schools which have all 4 facilities F1, F2, F3, F4 – and so on.
Using observation (8), 30 + 26 + 45 + a = 313 – 162 ⇒ a = 50

40 + 50 + 50 + F1F2 = 162 ⇒ F1F2 = 22

Now, two values are remaining – ‘b’ and ‘c’. All other values are there, and the sum of all values should be 600.
So: b + c + 25 + 50 + 313 (entire set F2 including second and third columns) + 26 + 24 + 20 + 80 = 600 or b + c + 538 = 600 or b + c = 600 – 538 = 62.


Also, as per observation (9), sum of all heads under F1 should be equal to sum of all heads under F4, i.e., sum of all elements in first and second columns should be equal to sum of all elements in third and fourth rows. Thus, we have:

b + c + 237 = c + 279 or b = 279 – 237 = 42.

Substituting this value of b in the earlier equation, we get 42 + c = 62 or c = 62 – 42 = 20.

Now all values have been found, the final diagram will be as follows:

The number of schools having only facilities F1 and F3 = 42

Answer: 42.

This question talks about 4 different types of facilities. Therefore, we can make a Venn diagramusing observations (2) to (7) as shown below:

The above diagram indicates all the areas which are possible. Eg., F1F2 indicates those schools which have facilities F1 & F2 only, F2F3F4 indicates those schools which have exactly 3 facilities F2, F3, F4 (but not F1), F2 indicates schools which have only facility F2, F1F2F3F4 indicates schools which have all 4 facilities F1, F2, F3, F4 – and so on.
Using observation (8), 30 + 26 + 45 + a = 313 – 162 ⇒ a = 50
40 + 50 + 50 + F1F2 = 162 ⇒ F1F2 = 22

Now, two values are remaining – ‘b’ and ‘c’. All other values are there, and the sum of all values should be 600. So: b + c + 25 + 50 + 313 (entire set F2 including second and third columns) + 26 + 24 + 20 + 80 = 600 or b + c + 538 = 600 or b + c = 600 – 538 = 62.

Also, as per observation (9), sum of all heads under F1 should be equal to sum of all heads under F4, i.e., sum of all elements in first and second columns should be equal to sum of all elements in third and fourth rows. Thus, we have :
b + c + 237 = c + 279 or b = 279 – 237 = 42.
Substituting this value of b in the earlier equation, we get 42 + c = 62 or c = 62 – 42 = 20.
Now all values have been found, the final diagram will be as follows:

The total number of schools having exactly three of four facilities = 50 + 50 + 50 + 50 = 200.

Hence, option 3.