Logical Reasoning & DI (LRDI) Test - 10

3A = B + C    ....(i)

10A = 4B + C    ....(ii)

(ii) - (i)

7A = 3B

A:B = 3:7 

10(i) - 3(ii)

10B + 10C - 12B - 3C = 0

7C = 2B

B:C = 7:2

A:B:C = 3:7:2

Let the value of 1 note of A be 3 Rs, 1 note of B be 7 Rs and 1 note of C be 2 Rs.

6 silver coins = B + C

6 silver coins = 9 Rs.

1 silver coin = 9/6 = 1.5 Rs

366 silver coins = 366 x 1.5 = 549 Rs.

For the maximum number of notes, we will try to maximise the number of notes of C since the value of 1 note of C is 2 Rs.

549/2 = 274.5

If we use 274 notes of C, we will have to pay the remaining 1 Rs but none of the notes is of Rs 1 denomination, so we have to use 273 notes of C and 1 note of A. Hence, total = 274

For the minimum number of notes, we will try to maximise the number of notes of B, since the value of each note of B = 7 Rs.

549/7 = 78.xx

So, if we use 78  notes of B, we are left with 549 - (78 x 7) = 3 Rs, so we can pay 1 note of A. Hence, total = 78 + 1 = 79.

Difference = 274 - 79 = 195.

3A = B + C ....(i)

10A = 4B + C ....(ii)

(ii) - (i)

7A = 3B

A:B = 3:7

10(i) - 3(ii)

10B + 10C - 12B - 3C = 0

7C = 2B

B:C = 7:2

A:B:C = 3:7:2

Let the value of 1 note of A be 3 Rs, 1 note of B be 7 Rs and 1 note of C be 2 Rs.

Let the number of notes of each type be x.

Now, there can be 3 possible cases.

Case A: We exchange a note of A. So, we should get back 3 Rs, but no other note or combination of notes can give 3 Rs. So, this case is invalidated. 

Case B: We exchange a note of B. So, we should get back Rs 7, if Kylian is returned 1 note of A and 2 notes of C, that equals Rs 7. There is no other way a sum of 7 can come up.

Case C: We exchange a note of C. So, we should get back 2 Rs, but no other note or combination of notes can give 2 Rs. So, this case is invalidated.

Hence, he exchanges one note of B for 2 notes of C and 1 note of A.

Initially, number of notes of 

A - x

B - x

C - x

After the transaction, the number of notes of 

A - x + 1

B - x - 1

C - x + 2

Arithmetic mean of x - 1 and x + 2 is x + 0.5

y = x + 0.5

Since x is a natural number, 

$$\left \lfloor y \right \rfloor$$ = x

Hence, the number of notes of A = x + 1 = $$\left \lfloor y \right \rfloor$$ + 1

3A = B + C ....(i)

10A = 4B + C ....(ii)

(ii) - (i)

7A = 3B

A:B = 3:7

10(i) - 3(ii)

10B + 10C - 12B - 3C = 0

7C = 2B

B:C = 7:2

A:B:C = 3:7:2

Let the value of 1 note of A be 3 Rs, 1 note of B be 7 Rs and 1 note of C be 2 Rs.

3B + 2A = 9 silver coins

21 + 6 = 9 coins

27 Rs = 9 coins

1 coin = 3 Rs

Alternative 1: 2 coins = 6 Rs = 3 x 2 Rs or 2 x 3 Rs. Hence, he can pay using 3 notes of C.

Alternative 2: 3 coins = 9 Rs = 7 Rs + 2 Rs or 3 x 2 Rs + 3 Rs or 3 x 3 Rs. Hence, he can pay using 1 note of B and 1 note of C.

Alternative 3: 5 coins = 15 Rs = 7 Rs + 4 x 2 Rs. Hence, he can pay using 1 note of B and 4 notes of C.

Alternative 4: 11 coins = 33 Rs = 3 x 7 Rs + 6 x 2 Rs[other alternatives are also there, but we just need to find a minimum of 1, if possible]. Hence, he can pay using 3 notes of B and 6 notes of C.

Hence, all 4 are possible alternatives. Hence Option (D).

3A = B + C ....(i)

10A = 4B + C ....(ii)

(ii) - (i)

7A = 3B

A:B = 3:7

10(i) - 3(ii)

10B + 10C - 12B - 3C = 0

7C = 2B

B:C = 7:2

A:B:C = 3:7:2

Let the value of 1 note of A be 3 Rs, 1 note of B be 7 Rs and 1 note of C be 2 Rs.

8B - 3C = 8 X 7 Rs - 3 x 2 Rs = 50

All we need to find is the number of triplets(a,b,c) that satisfy 

3a + 7b + 2c = 50

We will start with b = 1 and see how many ways are possible, we will continue with b = 2, 3 and so on.

Hence, there are 23 ways.

3A = B + C ....(i)

10A = 4B + C ....(ii)

(ii) - (i)

7A = 3B

A:B = 3:7

10(i) - 3(ii)

10B + 10C - 12B - 3C = 0

7C = 2B

B:C = 7:2

A:B:C = 3:7:2

Let the value of A, B and C be 3k, 7k and 2k.

Cost of 1 kg cashew nuts = 15A - B - C = 36k

Cost of 1 kg Brazilian nuts = 15C + 10A = 60k

Cost of 1 kg cashew nuts and 100 gms Brazilian nut = 36k + 6k = 42k

Let the number of A notes, B notes, C notes used be x, y and z respectively.

3k(x) + 7k(y) + 2k(z) = 42k    $$x,y,z\ge2$$

3x + 7y + 2z = 42    $$x,y,z\ge2$$

Case 1: y = 2

3x + 2z = 28

(x,z) = (8,2), (6,5), (4,8), (2,11) ---> 4 ways

Case 2: y = 3

3x + 2z = 21

z can take values 3 and 6

(x,z) = (5,3), (3,6) ---> 2 ways

Case 3: y = 4

3x + 2z = 14

(x,z) = (2,4) ---> 1 ways

Total number of ways = 4 + 2 + 1 = 7 ways

Each flavour was eaten by 1 to 3 children and butterscotch flavor is eaten by 3 students. So one flavour out of Vanilla or Chocolate must be eaten by just 2 children.
Using c and d, we can say that the three children who ate butterscotch ice-cream must have roll number 2,3,4 or 3,4,5 or 4,5,6.
Let v represent vanilla, c represent chocolate and b represent butterscotch.
We can make following table,

According to given condition we have, the students with roll number 1, 3 and 7 ate from shop E, D and C respectively. From this we can say that possibility 3 is not possible because then both shops D and E will be serving chocolate flavour ice creams.

Using data from option A can get following possibility -

Thus option A can be right and using the same possibility we can say that option D is right.
Using data from option B can get following possibility -

Using data from option C -

Condition that no two neighbouring shops sell same flavor of ice-cream is violated.
Hence, option C is the right choice.

Each flavour was eaten by 1 to 3 children and butterscotch flavor is eaten by 3 students. So one flavour out of Vanilla or Chocolate must be eaten by just 2 children.  Using c and d, we can say that the three children who ate butterscotch ice-cream must have roll number 2,3,4 or 3,4,5 or 4,5,6.  Let v represent vanilla, c represent chocolate and b represent butterscotch.  We can make following table,

According to given condition we have, the students with roll number 1, 3
and 7 ate from shop E, D and C respectively. From this we can say that
possibility 3 is not possible because then both shops D and E will be
serving chocolate flavour ice creams.

Thus in the other two possibilities the student with roll number 4 eats Butterscotch ice-cream.
Hence, option C is the right answer.

Each flavour was eaten by 1 to 3 children and butterscotch flavor is eaten by 3 students. So one flavour out of Vanilla or Chocolate must be eaten by just 2 children.  Using c and d, we can say that the three children who ate butterscotch ice-cream must have roll number 2,3,4 or 3,4,5 or 4,5,6.  Let v represent vanilla, c represent chocolate and b represent butterscotch.  We can make following table,

According to given condition we have, the students with roll number 1, 3
and 7 ate from shop E, D and C respectively. From this we can say that
possibility 3 is not possible because then both shops D and E will be
serving chocolate flavour ice creams.

Using data from the question, we have -

or 

This case can be eliminated as shop B and C cannot sell same flavoured ice-cream.

or

This case can be eliminated as shop B and C cannot sell same flavoured ice-cream.

Thus, we have,

We cannot assign shops to students with roll number 4, 6 and 8 without violating any condition.

None of the statements are correct. 

Hence, option D is the right choice. 

According to given condition we have, the students with roll number 1, 3
and 7 ate from shop E, D and C respectively. From this we can say that
possibility 3 is not possible because then both shops D and E will be
serving chocolate flavour ice creams.

Since student with roll number 5 eats from shop B, possibility 1 and 3 can be eliminated as neighbouring shop cannot sell same flavor ice-cream.

The case given above represents one of two possibilities (G and H can be interchanged).
Therefore, in all possible combinations, A will be selling chocolate flavoured ice cream. 

As we can see, shop A sells ice cream of chocolate flavour. Therefore, option A is the right answer. 

Each flavour was eaten by 1 to 3 children, and three students ate the butterscotch flavour. So just two children must consume one flavour out of Vanilla or Chocolate.
Using c and d, we can say that the three children who ate butterscotch ice cream must have roll numbers 2,3,4 or 3,4,5 or 4,5,6.
Let v represent vanilla, c represents chocolate, and b represent butterscotch.
We can make the following table,

According to our given condition, the students with roll numbers 1, 3 and 7 ate from shops E, D and C, respectively. From this, possibility 3 is impossible because shops D and E will serve chocolate flavour ice creams.

So only cases 1 and 2 are possible. Now if we look at the options, we can easily infer that the students with roll number 3 ate butterscotch. ( Statement one is incorrect)

If we analyze Statement 2, we can see that In case 1, the Student with roll number 5 ate vanilla, so this is a possible case. ( it is not incorrect. It can be possible.)

For Statement 3, the Student with roll number 8 ate chocolates in both cases. If he ate chocolate ice cream from shop F, Then roll number 1, who ate chocolate from shop E and roll number 8, who ate ice cream from shop F, will contradict the fact that the same flavoured ice cream was not eaten in neighbouring shops. ( Statement 3 is also incorrect). 

Hence, from these 3 statements, 2 statements are incorrect.

The Correct answer is option C

The following observations we have from the data given above:

1) Mode 1: Booked and delivered in the same month.

2) Mode 2: Booked and delivered in the next month.

3) Mode 3: Booked and delivered in the next to next month.

Thus, the number of deliveries in a particular month will be the sum of bookings done in mode 1 in that month, with bookings done with mode 2 in the previous month and bookings done with mode 3 in two months back.

i.e., No. of deliveries = Mode 1(current month) + Mode 2(previous month) + Mode 3(two months before)

Also, it is given that except for January, the number of bookings made through Mode 2 is 10% of the total bookings.

Thus, the table will become as follows:

Consider the deliveries in the month of August:

The number of bookings made in either July or August is 0. Thus, the number of deliveries in this month are of the bookings made in the month of June through mode 3.

Thus, the number of bookings made in June with mode 3 = 500.

Now, consider the deliveries in the month of July:

The number of bookings made in July is also 0. Thus, the number of deliveries will be the bookings made in the month of June through mode 2 and the ones made in May through mode 3.

We know that bookings made through mode 2 in the month of June = 100.

Thus, the bookings made in May through mode 3 will be 350 - 100 = 250

Now, consider the month of June:

Total number of bookings = 1000

Mode 2 = 100 and Mode 3 = 500.

Thus, Mode 1 = 1000 - (100 + 500) = 400

Similarly, the bookings made through Mode 1 in May = 250

Again, consider the deliveries done in June:

The number of deliveries will be the bookings made in the month of June through mode 1, the ones made in May through mode 2, and the ones made in April through mode 3.

Mode 1(June) = 400, Mode 2(May) = 50.

Thus, Mode 3(April) = 770 - (400 + 50) = 320

Similarly, in this manner, we can backtrack all the values, and the final table will be:

Thus, from the table, it is evident that the highest number of bookings done through Mode 3 is in June.

 The correct option is D.

The following observations we have from the data given above:

1) Mode 1: Booked and delivered in the same month.

2) Mode 2: Booked and delivered in the next month.

3) Mode 3: Booked and delivered in the next to next month.

Thus, the number of deliveries in a particular month will be the sum of bookings done in mode 1 in that month, with bookings done with mode 2 in the previous month and bookings done with mode 3 in two months back.

i.e., No. of deliveries = Mode 1(current month) + Mode 2(previous month) + Mode 3(two months before)

Also, it is given that except for January, the number of bookings made through Mode 2 is 10% of the total bookings.

Thus, the table will become as follows:

Consider the deliveries in the month of August:

The number of bookings made in either July or August is 0. Thus, the number of deliveries in this month are of the bookings made in the month of June through mode 3.

Thus, the number of bookings made in June with mode 3 = 500.

Now, consider the deliveries in the month of July:

The number of bookings made in July is also 0. Thus, the number of deliveries will be the bookings made in the month of June through mode 2 and the ones made in May through mode 3.

We know that bookings made through mode 2 in the month of June = 100.

Thus, the bookings made in May through mode 3 will be 350 - 100 = 250

Now, consider the month of June:

Total number of bookings = 1000

Mode 2 = 100 and Mode 3 = 500.

Thus, Mode 1 = 1000 - (100 + 500) = 400

Similarly, the bookings made through Mode 1 in May = 250

Again, consider the deliveries done in June:

The number of deliveries will be the bookings made in the month of June through mode 1, the ones made in May through mode 2, and the ones made in April through mode 3.

Mode 1(June) = 400, Mode 2(May) = 50.

Thus, Mode 3(April) = 770 - (400 + 50) = 320

Similarly, in this manner, we can backtrack all the values, and the final table will be:

 The correct option is B.

The following observations we have from the data given above:

1) Mode 1: Booked and delivered in the same month.

2) Mode 2: Booked and delivered in the next month.

3) Mode 3: Booked and delivered in the next to next month.

Thus, the number of deliveries in a particular month will be the sum of bookings done in mode 1 in that month, with bookings done with mode 2 in the previous month and bookings done with mode 3 in two months back.

i.e., No. of deliveries = Mode 1(current month) + Mode 2(previous month) + Mode 3(two months before)

Also, it is given that except for January, the number of bookings made through Mode 2 is 10% of the total bookings.

Thus, the table will become as follows:

Consider the deliveries in the month of August:

The number of bookings made in either July or August is 0. Thus, the number of deliveries in this month are of the bookings made in the month of June through mode 3.

Thus, the number of bookings made in June with mode 3 = 500.

Now, consider the deliveries in the month of July:

The number of bookings made in July is also 0. Thus, the number of deliveries will be the bookings made in the month of June through mode 2 and the ones made in May through mode 3.

We know that bookings made through mode 2 in the month of June = 100.

Thus, the bookings made in May through mode 3 will be 350 - 100 = 250

Now, consider the month of June:

Total number of bookings = 1000

Mode 2 = 100 and Mode 3 = 500.

Thus, Mode 1 = 1000 - (100 + 500) = 400

Similarly, the bookings made through Mode 1 in May = 250

Again, consider the deliveries done in June:

The number of deliveries will be the bookings made in the month of June through mode 1, the ones made in May through mode 2, and the ones made in April through mode 3.

Mode 1(June) = 400, Mode 2(May) = 50.

Thus, Mode 3(April) = 770 - (400 + 50) = 320

Similarly, in this manner, we can backtrack all the values, and the final table will be:

 The correct option is C.

The following observations we have from the data given above:

1) Mode 1: Booked and delivered in the same month.

2) Mode 2: Booked and delivered in the next month.

3) Mode 3: Booked and delivered in the next to next month.

Thus, the number of deliveries in a particular month will be the sum of bookings done in mode 1 in that month, with bookings done with mode 2 in the previous month and bookings done with mode 3 in two months back.

i.e., No. of deliveries = Mode 1(current month) + Mode 2(previous month) + Mode 3(two months before)

Also, it is given that except for January, the number of bookings made through Mode 2 is 10% of the total bookings.

Thus, the table will become as follows:

Consider the deliveries in the month of August:

The number of bookings made in either July or August is 0. Thus, the number of deliveries in this month are of the bookings made in the month of June through mode 3.

Thus, the number of bookings made in June with mode 3 = 500.

Now, consider the deliveries in the month of July:

The number of bookings made in July is also 0. Thus, the number of deliveries will be the bookings made in the month of June through mode 2 and the ones made in May through mode 3.

We know that bookings made through mode 2 in the month of June = 100.

Thus, the bookings made in May through mode 3 will be 350 - 100 = 250

Now, consider the month of June:

Total number of bookings = 1000

Mode 2 = 100 and Mode 3 = 500.

Thus, Mode 1 = 1000 - (100 + 500) = 400

Similarly, the bookings made through Mode 1 in May = 250

Again, consider the deliveries done in June:

The number of deliveries will be the bookings made in the month of June through mode 1, the ones made in May through mode 2, and the ones made in April through mode 3.

Mode 1(June) = 400, Mode 2(May) = 50.

Thus, Mode 3(April) = 770 - (400 + 50) = 320

Similarly, in this manner, we can backtrack all the values, and the final table will be:

 The correct option is A.

The following observations we have from the data given above:

1) Mode 1: Booked and delivered in the same month.

2) Mode 2: Booked and delivered in the next month.

3) Mode 3: Booked and delivered in the next to next month.

Thus, the number of deliveries in a particular month will be the sum of bookings done in mode 1 in that month, with bookings done with mode 2 in the previous month and bookings done with mode 3 in two months back.

i.e., No. of deliveries = Mode 1(current month) + Mode 2(previous month) + Mode 3(two months before)

Also, it is given that except for January, the number of bookings made through Mode 2 is 10% of the total bookings.

Thus, the table will become as follows:

Consider the deliveries in the month of August:

The number of bookings made in either July or August is 0. Thus, the number of deliveries in this month are of the bookings made in the month of June through mode 3.

Thus, the number of bookings made in June with mode 3 = 500.

Now, consider the deliveries in the month of July:

The number of bookings made in July is also 0. Thus, the number of deliveries will be the bookings made in the month of June through mode 2 and the ones made in May through mode 3.

We know that bookings made through mode 2 in the month of June = 100.

Thus, the bookings made in May through mode 3 will be 350 - 100 = 250

Now, consider the month of June:

Total number of bookings = 1000

Mode 2 = 100 and Mode 3 = 500.

Thus, Mode 1 = 1000 - (100 + 500) = 400

Similarly, the bookings made through Mode 1 in May = 250

Again, consider the deliveries done in June:

The number of deliveries will be the bookings made in the month of June through mode 1, the ones made in May through mode 2, and the ones made in April through mode 3.

Mode 1(June) = 400, Mode 2(May) = 50.

Thus, Mode 3(April) = 770 - (400 + 50) = 320

Similarly, in this manner, we can backtrack all the values, and the final table will be:

The highest difference is for January. ( 175 - 25 = 150).

The answer is Option (D).

Let R be the total amount spent by all the friends on Rent, G be the total amount spent by all the friends on Groceries and T be the total amount spent by all the friends on Travelling.
The given information in table form is as shown below:-

The total amount of money spent by exactly 3 friends is equal.
There are 10 possible combinations for this.
But as can be observed the amount spent by Anay, Bhola and Chintu on each of Rent, Groceries and Travel is greater than the amount spent by Eshwar on each of Rent, Groceries and Travel.
Thus, Amount of money spent by Eshwar cannot be equal to the amount of money spent by 2 other friends.
Thus, only 4 possible combinations now remain.
Out of these 4, the amount of money spent by Anay and Bhola on each of Rent, Groceries and Travel is greater than the amount spent by Dinu on each of Rent, Groceries and Travel
Thus, Amount of money spent by Dinu cannot be equal to the amount of money spent by 2 other friends.
Thus, the amount spent by Anay, Bhola and Chintu is equal.
Thus, 0.26R+0.24G+0.22T=0.22R+0.24G+0.24T
=> 0.04R = 0.02T
Thus, 2R=T
Also, 0.22R+0.24G+0.24T=0.18R+0.28G+0.24T
=> R=G
Thus, the table becomes:-

Thus, if R = 10000 then Dinu spent 0.36*10000 = 3600 Rupees on Travelling.
Hence, option B is the correct answer.

Let R be the total amount spent by all the friends on Rent, G be the total amount spent by all the friends on Groceries and T be the total amount spent by all the friends on Travelling.
The given information in table form is as shown below:-

The total amount of money spent by exactly 3 friends is equal.
There are 10 possible combinations for this.
But as can be observed the amount spent by Anay, Bhola and Chintu on each of Rent, Groceries and Travel is greater than the amount spent by Eshwar on each of Rent, Groceries and Travel.
Thus, Amount of money spent by Eshwar cannot be equal to the amount of money spent by 2 other friends.
Thus, only 4 possible combinations now remain.
Out of these 4, the amount of money spent by Anay and Bhola on each of Rent, Groceries and Travel is greater than the amount spent by Dinu on each of Rent, Groceries and Travel
Thus, Amount of money spent by Dinu cannot be equal to the amount of money spent by 2 other friends.
Thus, the amount spent by Anay, Bhola and Chintu is equal.
Thus, 0.26R+0.24G+0.22T=0.22R+0.24G+0.24T
=> 0.04R = 0.02T
Thus, 2R=T
Also, 0.22R+0.24G+0.24T=0.18R+0.28G+0.24T
=> R=G
Thus, the table becomes:-

Thus, the required ratio is = 0.18R:0.24R = 3:4
Hence, option B is the correct answer.

Let R be the total amount spent by all the friends on Rent, G be the total amount spent by all the friends on Groceries and T be the total amount spent by all the friends on Travelling.
The given information in table form is as shown below:-

The total amount of money spent by exactly 3 friends is equal.
There are 10 possible combinations for this.
But as can be observed the amount spent by Anay, Bhola and Chintu on each of Rent, Groceries and Travel is greater than the amount spent by Eshwar on each of Rent, Groceries and Travel.
Thus, Amount of money spent by Eshwar cannot be equal to the amount of money spent by 2 other friends.
Thus, only 4 possible combinations now remain.
Out of these 4, the amount of money spent by Anay and Bhola on each of Rent, Groceries and Travel is greater than the amount spent by Dinu on each of Rent, Groceries and Travel
Thus, Amount of money spent by Dinu cannot be equal to the amount of money spent by 2 other friends.
Thus, the amount spent by Anay, Bhola and Chintu is equal.
Thus, 0.26R+0.24G+0.22T=0.22R+0.24G+0.24T
=> 0.04R = 0.02T
Thus, 2R=T
Also, 0.22R+0.24G+0.24T=0.18R+0.28G+0.24T
=> R=G
Thus, the table becomes:-

Thus, The amount spent on traveling is 100 percent more than the amount spent on rent by all the friends combined.

Let R be the total amount spent by all the friends on Rent, G be the total amount spent by all the friends on Groceries and T be the total amount spent by all the friends on Travelling.
The given information in table form is as shown below:-

The total amount of money spent by exactly 3 friends is equal.
There are 10 possible combinations for this.
But as can be observed the amount spent by Anay, Bhola and Chintu on each of Rent, Groceries and Travel is greater than the amount spent by Eshwar on each of Rent, Groceries and Travel.
Thus, Amount of money spent by Eshwar cannot be equal to the amount of money spent by 2 other friends.
Thus, only 4 possible combinations now remain.
Out of these 4, the amount of money spent by Anay and Bhola on each of Rent, Groceries and Travel is greater than the amount spent by Dinu on each of Rent, Groceries and Travel
Thus, Amount of money spent by Dinu cannot be equal to the amount of money spent by 2 other friends.
Thus, the amount spent by Anay, Bhola and Chintu is equal.
Thus, 0.26R+0.24G+0.22T=0.22R+0.24G+0.24T
=> 0.04R = 0.02T
Thus, 2R=T
Also, 0.22R+0.24G+0.24T=0.18R+0.28G+0.24T
=> R=G
Thus, the table becomes:-

Thus, the amount spent by Anay on Travelling = 0.44*90000 = 39600 rupees

Let R be the total amount spent by all the friends on Rent, G be the total amount spent by all the friends on Groceries and T be the total amount spent by all the friends on Travelling.
The given information in table form is as shown below:-

The total amount of money spent by exactly 3 friends is equal.
There are 10 possible combinations for this.
But as can be observed the amount spent by Anay, Bhola and Chintu on each of Rent, Groceries and Travel is greater than the amount spent by Eshwar on each of Rent, Groceries and Travel.
Thus, Amount of money spent by Eshwar cannot be equal to the amount of money spent by 2 other friends.
Thus, only 4 possible combinations now remain.
Out of these 4, the amount of money spent by Anay and Bhola on each of Rent, Groceries and Travel is greater than the amount spent by Dinu on each of Rent, Groceries and Travel
Thus, Amount of money spent by Dinu cannot be equal to the amount of money spent by 2 other friends.
Thus, the amount spent by Anay, Bhola and Chintu is equal.
Thus, 0.26R+0.24G+0.22T=0.22R+0.24G+0.24T
=> 0.04R = 0.02T
Thus, 2R=T
Also, 0.22R+0.24G+0.24T=0.18R+0.28G+0.24T
=> R=G
Thus, the table becomes:-

The difference in rent paid by Dinu & Chintu = 0.22R - 0.18R = 0.04R.

From the given options, only the difference between the travel expenses of Bhola & Anay will be equal to 0.04R. (0.48R - 0.44R = 0.04R).

The answer is Option (A).