Logical Reasoning & DI (LRDI) Test

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Logical Reasoning & DI (LRDI) Test - 6

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Question 1
3 / -1

Directions For Questions

Members of Greenland Association Aashi, Puja, Manu and Nitya have planted 48, 31, 29 and 28 trees in any order. They have planted six varieties of plants: Banyan, Mango, Coconut, Neem, Teak and Rose. Each of them has planted a minimum of two varieties of plants and a maximum of six varieties of plants. The minimum number of plants planted by anyone for a type of plant is one, and the maximum number of plants planted is 16.
Each of them has planted a different number of varieties. For example, if Puja has planted three varieties of plants, none of the remaining members has planted three varieties.
No two have planted the same number of plants for any two varieties. For example, if Puja has planted three mango plants, none of them has planted three plants for any variety.
1. Maximum difference between the total number of plants of any two varieties is one.
2. The number of plants planted by a person for each variety is in A.P., and the number of plants planted by another person for each variety is in G.P. Both of them have planted more than three varieties of plants.

...view full instructions

How many arrangements is/are possible considering all the attributes?

A
4

B
44

C
47

D
49

Solution

Total number of plants planted by all of them = 48 + 31 + 29 + 28 = 136
136 is sum of first 16 natural numbers, i.e.
1 + 2 + 3 + .............+ 15 + 16 = 136
The minimum number of plants planted for a variety is 1, and the maximum is 16.
Therefore, the total number of varieties of plants cannot exceed 16, but if it is less than 16, i.e. if it is 15 or 14, it doesn’t satisfy the sum value of 136, or the maximum number of plants planted for a variety is 16. This implies that all of them have planted 16 varieties of plants.
It is mentioned that each one has planted a minimum of two varieties of plants and a maximum of six varieties.
Sum = 2 + 3 + 4 + 5 + 6 = 20
Removing 4, we will get 16 total varieties for four members.
Therefore, four of them planted 2, 3, 5, and 6 varieties of plants in any order.
It is given that,
The number of plants planted by a person for each variety is in A.P. The number of plants planted by another person for each variety is in G.P. They have planted more than three varieties of plants. This means that they have planted 5 and 6 varieties in any order.

Sum of A.P series = $$\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$
The Sum of an A.P series is always divisible by $$\frac{n}{2}$$ (n - number of terms)
Out of the given values, the only possible value is 48, which is divisible by 3 (6/2)
Therefore, a person has planted 48 plants of 6 varieties.
A person has planted 48 plants of 6 varieties in A.P
Let us assume she has planted a-5d, a-3d, a-d, a+d, a+3d, and a+5d plants of each variety.
Sum = 48
6a = 48
a = 8
Number of plants planted are 8-5d, 8-3d, 8-d, 8+d, 8+3d, 8+5d
The only possible value for d is 1
Therefore, the only possibility is that she has planted 3, 5, 7, 9, 11, and 13 (6) plants of each variety.

The person who planted the plants of different varieties in G.P has planted five varieties.
Sum of G.P series = $$\ \frac{\ a\left(r^n-1\right)}{r-1}=\ \frac{\ a\left(r^5-1\right)}{r\ -1}=a\left(r^4+r^3+r^2+r+1\right)$$
Out of the given values, only possible value is 31 i.e. a = 1 and r = 2.
The number of plants planted in each variety is 1, 2, 4, 8, 16 (5) in any order.
In the table, it is given that Aashi has planted two rose plants. This implies that she has planted the plants in G.P.

Remaining numbers: 6, 10, 12, 14, 15
A person has planted two varieties of plants, and the number of plants planted is either 28 or 29. The only possibility is she has planted 14, 15 plants of each variety in any order. It is mentioned that Nitya has planted 6 Banyan trees. This implies that Nitya has planted 6, 10, and 12 plants of each variety in any order.

It is mentioned that the maximum difference between the total number of plants of any two varieties is one. Therefore, let us assume there are x varieties with y plants and (6-x) with y-1 plants, i.e.
xy + (6-x)(y-1) = 136
xy + 6y - 6 - xy + x = 136
x + 6y = 142 (x<6)
The only possibility is x = 4, y = 23
Therefore, there are 23, 23, 23, 23, 22 and 22 plants of each variety in any order.
The only possibility for the number of neem plants is 23 (9+0+14+0)
The number of coconut plants can be 22 or 23. Therefore, the remaining sum should be 11 or 12
For 11: (1, 0, 10)
For 12: (2, 0, 10)
we cannot consider 12, as Aashi has planted two mango plants. Therefore, it should be 11 (1, 0, 10).

The number of rose plants planted by Nitya cannot be 12. If it is 12, the remaining sum should be either 8 or 9, which is impossible. Therefore, the number of rose plants planted by Nitya will be 0. If it is 0, the remaining sum should be 20 or 21.
For 21: 7, 14, 0
For 20: 5, 15, 0
It cannot be 21 as 14 neem plants are planted; therefore, the remaining sum will be 20(5,15,0)

The remaining sum for the number of Banyan trees should be 17. The only possibility is (13,4)

Final arrangements:

There are two possibilities for Manu and Puja. Therefore, the total number of possible arrangements is 4.
Question 2
3 / -1

Directions For Questions

Members of Greenland Association Aashi, Puja, Manu and Nitya have planted 48, 31, 29 and 28 trees in any order. They have planted six varieties of plants: Banyan, Mango, Coconut, Neem, Teak and Rose. Each of them has planted a minimum of two varieties of plants and a maximum of six varieties of plants. The minimum number of plants planted by anyone for a type of plant is one, and the maximum number of plants planted is 16.
Each of them has planted a different number of varieties. For example, if Puja has planted three varieties of plants, none of the remaining members has planted three varieties.
No two have planted the same number of plants for any two varieties. For example, if Puja has planted three mango plants, none of them has planted three plants for any variety.
1. Maximum difference between the total number of plants of any two varieties is one.
2. The number of plants planted by a person for each variety is in A.P., and the number of plants planted by another person for each variety is in G.P. Both of them have planted more than three varieties of plants.

...view full instructions

Find the minimum possible number of Mango plants planted by Aashi and Puja?

A
8

B
84

C
87

D
89

Solution

Total number of plants planted by all of them = 48 + 31 + 29 + 28 = 136
136 is sum of first 16 natural numbers, i.e.
1 + 2 + 3 + .............+ 15 + 16 = 136
The minimum number of plants planted for a variety is 1, and the maximum is 16.
Therefore, the total number of varieties of plants cannot exceed 16, but if it is less than 16, i.e. if it is 15 or 14, it doesn’t satisfy the sum value of 136, or the maximum number of plants planted for a variety is 16. This implies that all of them have planted 16 varieties of plants.
It is mentioned that each one has planted a minimum of two varieties of plants and a maximum of six varieties.
Sum = 2 + 3 + 4 + 5 + 6 = 20
Removing 4, we will get 16 total varieties for four members.
Therefore, four of them planted 2, 3, 5, and 6 varieties of plants in any order.
It is given that,
The number of plants planted by a person for each variety is in A.P. The number of plants planted by another person for each variety is in G.P. They have planted more than three varieties of plants. This means that they have planted 5 and 6 varieties in any order.

Sum of A.P series = $$\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$
The Sum of an A.P series is always divisible by $$\frac{n}{2}$$ (n - number of terms)
Out of the given values, the only possible value is 48, which is divisible by 3 (6/2)
Therefore, a person has planted 48 plants of 6 varieties.
A person has planted 48 plants of 6 varieties in A.P
Let us assume she has planted a-5d, a-3d, a-d, a+d, a+3d, and a+5d plants of each variety.
Sum = 48
6a = 48
a = 8
Number of plants planted are 8-5d, 8-3d, 8-d, 8+d, 8+3d, 8+5d
The only possible value for d is 1
Therefore, the only possibility is that she has planted 3, 5, 7, 9, 11, and 13 (6) plants of each variety.

The person who planted the plants of different varieties in G.P has planted five varieties.
Sum of G.P series = $$\ \frac{\ a\left(r^n-1\right)}{r-1}=\ \frac{\ a\left(r^5-1\right)}{r\ -1}=a\left(r^4+r^3+r^2+r+1\right)$$
Out of the given values, only possible value is 31 i.e. a = 1 and r = 2.
The number of plants planted in each variety is 1, 2, 4, 8, 16 (5) in any order.
In the table, it is given that Aashi has planted two rose plants. This implies that she has planted the plants in G.P.

Remaining numbers: 6, 10, 12, 14, 15
A person has planted two varieties of plants, and the number of plants planted is either 28 or 29. The only possibility is she has planted 14, 15 plants of each variety in any order. It is mentioned that Nitya has planted 6 Banyan trees. This implies that Nitya has planted 6, 10, and 12 plants of each variety in any order.

It is mentioned that the maximum difference between the total number of plants of any two varieties is one. Therefore, let us assume there are x varieties with y plants and (6-x) with y-1 plants, i.e.
xy + (6-x)(y-1) = 136
xy + 6y - 6 - xy + x = 136
x + 6y = 142 (x<6)
The only possibility is x = 4, y = 23
Therefore, there are 23, 23, 23, 23, 22 and 22 plants of each variety in any order.
The only possibility for the number of neem plants is 23 (9+0+14+0)
The number of coconut plants can be 22 or 23. Therefore, the remaining sum should be 11 or 12
For 11: (1, 0, 10)
For 12: (2, 0, 10)
we cannot consider 12, as Aashi has planted two mango plants. Therefore, it should be 11 (1, 0, 10).

The number of rose plants planted by Nitya cannot be 12. If it is 12, the remaining sum should be either 8 or 9, which is impossible. Therefore, the number of rose plants planted by Nitya will be 0. If it is 0, the remaining sum should be 20 or 21.
For 21: 7, 14, 0
For 20: 5, 15, 0
It cannot be 21 as 14 neem plants are planted; therefore, the remaining sum will be 20(5,15,0)

The remaining sum for the number of Banyan trees should be 17. The only possibility is (13,4)

Final arrangements:

The number of mango plants planted by Aashi and Puja

Case 1a: 7 + 16 = 23
Case 1b: 16 + 0 = 16
Case 2a: 3 + 8 = 11
Case 2b: 8 + 0 = 8

Therefore, the minimum possible sum = 8
Question 3
3 / -1

Directions For Questions

Members of Greenland Association Aashi, Puja, Manu and Nitya have planted 48, 31, 29 and 28 trees in any order. They have planted six varieties of plants: Banyan, Mango, Coconut, Neem, Teak and Rose. Each of them has planted a minimum of two varieties of plants and a maximum of six varieties of plants. The minimum number of plants planted by anyone for a type of plant is one, and the maximum number of plants planted is 16.
Each of them has planted a different number of varieties. For example, if Puja has planted three varieties of plants, none of the remaining members has planted three varieties.
No two have planted the same number of plants for any two varieties. For example, if Puja has planted three mango plants, none of them has planted three plants for any variety.
1. Maximum difference between the total number of plants of any two varieties is one.
2. The number of plants planted by a person for each variety is in A.P., and the number of plants planted by another person for each variety is in G.P. Both of them have planted more than three varieties of plants.

...view full instructions

If Puja planted 7 mango plants, then what is the sum of Banyan plants planted by Aashi and Teak plants planted by Nitya?

A
4

B
8

C
12

D
16

Solution

Total number of plants planted by all of them = 48 + 31 + 29 + 28 = 136
136 is sum of first 16 natural numbers, i.e.
1 + 2 + 3 + .............+ 15 + 16 = 136
The minimum number of plants planted for a variety is 1, and the maximum is 16.
Therefore, the total number of varieties of plants cannot exceed 16, but if it is less than 16, i.e. if it is 15 or 14, it doesn’t satisfy the sum value of 136, or the maximum number of plants planted for a variety is 16. This implies that all of them have planted 16 varieties of plants.
It is mentioned that each one has planted a minimum of two varieties of plants and a maximum of six varieties.
Sum = 2 + 3 + 4 + 5 + 6 = 20
Removing 4, we will get 16 total varieties for four members.
Therefore, four of them planted 2, 3, 5, and 6 varieties of plants in any order.
It is given that,
The number of plants planted by a person for each variety is in A.P. The number of plants planted by another person for each variety is in G.P. They have planted more than three varieties of plants. This means that they have planted 5 and 6 varieties in any order.

Sum of A.P series = $$\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$
The Sum of an A.P series is always divisible by $$\frac{n}{2}$$ (n - number of terms)
Out of the given values, the only possible value is 48, which is divisible by 3 (6/2)
Therefore, a person has planted 48 plants of 6 varieties.
A person has planted 48 plants of 6 varieties in A.P
Let us assume she has planted a-5d, a-3d, a-d, a+d, a+3d, and a+5d plants of each variety.
Sum = 48
6a = 48
a = 8
Number of plants planted are 8-5d, 8-3d, 8-d, 8+d, 8+3d, 8+5d
The only possible value for d is 1
Therefore, the only possibility is that she has planted 3, 5, 7, 9, 11, and 13 (6) plants of each variety.

The person who planted the plants of different varieties in G.P has planted five varieties.
Sum of G.P series = $$\ \frac{\ a\left(r^n-1\right)}{r-1}=\ \frac{\ a\left(r^5-1\right)}{r\ -1}=a\left(r^4+r^3+r^2+r+1\right)$$
Out of the given values, only possible value is 31 i.e. a = 1 and r = 2.
The number of plants planted in each variety is 1, 2, 4, 8, 16 (5) in any order.
In the table, it is given that Aashi has planted two rose plants. This implies that she has planted the plants in G.P.

Remaining numbers: 6, 10, 12, 14, 15
A person has planted two varieties of plants, and the number of plants planted is either 28 or 29. The only possibility is she has planted 14, 15 plants of each variety in any order. It is mentioned that Nitya has planted 6 Banyan trees. This implies that Nitya has planted 6, 10, and 12 plants of each variety in any order.

It is mentioned that the maximum difference between the total number of plants of any two varieties is one. Therefore, let us assume there are x varieties with y plants and (6-x) with y-1 plants, i.e.
xy + (6-x)(y-1) = 136
xy + 6y - 6 - xy + x = 136
x + 6y = 142 (x<6)
The only possibility is x = 4, y = 23
Therefore, there are 23, 23, 23, 23, 22 and 22 plants of each variety in any order.
The only possibility for the number of neem plants is 23 (9+0+14+0)
The number of coconut plants can be 22 or 23. Therefore, the remaining sum should be 11 or 12
For 11: (1, 0, 10)
For 12: (2, 0, 10)
we cannot consider 12, as Aashi has planted two mango plants. Therefore, it should be 11 (1, 0, 10).

The number of rose plants planted by Nitya cannot be 12. If it is 12, the remaining sum should be either 8 or 9, which is impossible. Therefore, the number of rose plants planted by Nitya will be 0. If it is 0, the remaining sum should be 20 or 21.
For 21: 7, 14, 0
For 20: 5, 15, 0
It cannot be 21 as 14 neem plants are planted; therefore, the remaining sum will be 20(5,15,0)

The remaining sum for the number of Banyan trees should be 17. The only possibility is (13,4)

Final arrangements:

It is mentioned that Puja planted 7 mango plants. This refers to case 1b, i.e.

Banyan plants planted by Aashi = 4

Teak plants planted by Nitya = 12

Sum = 4 + 12 = 16

Answer is option D.
Question 4
3 / -1

Directions For Questions

Members of Greenland Association Aashi, Puja, Manu and Nitya have planted 48, 31, 29 and 28 trees in any order. They have planted six varieties of plants: Banyan, Mango, Coconut, Neem, Teak and Rose. Each of them has planted a minimum of two varieties of plants and a maximum of six varieties of plants. The minimum number of plants planted by anyone for a type of plant is one, and the maximum number of plants planted is 16.
Each of them has planted a different number of varieties. For example, if Puja has planted three varieties of plants, none of the remaining members has planted three varieties.
No two have planted the same number of plants for any two varieties. For example, if Puja has planted three mango plants, none of them has planted three plants for any variety.
1. Maximum difference between the total number of plants of any two varieties is one.
2. The number of plants planted by a person for each variety is in A.P., and the number of plants planted by another person for each variety is in G.P. Both of them have planted more than three varieties of plants.

...view full instructions

How many Teak plants and Rose plants are planted?

A
45

B
46

C
47

D
Can't be determined

Solution

Total number of plants planted by all of them = 48 + 31 + 29 + 28 = 136
136 is sum of first 16 natural numbers, i.e.
1 + 2 + 3 + .............+ 15 + 16 = 136
The minimum number of plants planted for a variety is 1, and the maximum is 16.
Therefore, the total number of varieties of plants cannot exceed 16, but if it is less than 16, i.e. if it is 15 or 14, it doesn’t satisfy the sum value of 136, or the maximum number of plants planted for a variety is 16. This implies that all of them have planted 16 varieties of plants.
It is mentioned that each one has planted a minimum of two varieties of plants and a maximum of six varieties.
Sum = 2 + 3 + 4 + 5 + 6 = 20
Removing 4, we will get 16 total varieties for four members.
Therefore, four of them planted 2, 3, 5, and 6 varieties of plants in any order.
It is given that,
The number of plants planted by a person for each variety is in A.P. The number of plants planted by another person for each variety is in G.P. They have planted more than three varieties of plants. This means that they have planted 5 and 6 varieties in any order.

Sum of A.P series = $$\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$
The Sum of an A.P series is always divisible by $$\frac{n}{2}$$ (n - number of terms)
Out of the given values, the only possible value is 48, which is divisible by 3 (6/2)
Therefore, a person has planted 48 plants of 6 varieties.
A person has planted 48 plants of 6 varieties in A.P
Let us assume she has planted a-5d, a-3d, a-d, a+d, a+3d, and a+5d plants of each variety.
Sum = 48
6a = 48
a = 8
Number of plants planted are 8-5d, 8-3d, 8-d, 8+d, 8+3d, 8+5d
The only possible value for d is 1
Therefore, the only possibility is that she has planted 3, 5, 7, 9, 11, and 13 (6) plants of each variety.

The person who planted the plants of different varieties in G.P has planted five varieties.
Sum of G.P series = $$\ \frac{\ a\left(r^n-1\right)}{r-1}=\ \frac{\ a\left(r^5-1\right)}{r\ -1}=a\left(r^4+r^3+r^2+r+1\right)$$
Out of the given values, only possible value is 31 i.e. a = 1 and r = 2.
The number of plants planted in each variety is 1, 2, 4, 8, 16 (5) in any order.
In the table, it is given that Aashi has planted two rose plants. This implies that she has planted the plants in G.P.

Remaining numbers: 6, 10, 12, 14, 15
A person has planted two varieties of plants, and the number of plants planted is either 28 or 29. The only possibility is she has planted 14, 15 plants of each variety in any order. It is mentioned that Nitya has planted 6 Banyan trees. This implies that Nitya has planted 6, 10, and 12 plants of each variety in any order.

It is mentioned that the maximum difference between the total number of plants of any two varieties is one. Therefore, let us assume there are x varieties with y plants and (6-x) with y-1 plants, i.e.
xy + (6-x)(y-1) = 136
xy + 6y - 6 - xy + x = 136
x + 6y = 142 (x<6)
The only possibility is x = 4, y = 23
Therefore, there are 23, 23, 23, 23, 22 and 22 plants of each variety in any order.
The only possibility for the number of neem plants is 23 (9+0+14+0)
The number of coconut plants can be 22 or 23. Therefore, the remaining sum should be 11 or 12
For 11: (1, 0, 10)
For 12: (2, 0, 10)
we cannot consider 12, as Aashi has planted two mango plants. Therefore, it should be 11 (1, 0, 10).

The number of rose plants planted by Nitya cannot be 12. If it is 12, the remaining sum should be either 8 or 9, which is impossible. Therefore, the number of rose plants planted by Nitya will be 0. If it is 0, the remaining sum should be 20 or 21.
For 21: 7, 14, 0
For 20: 5, 15, 0
It cannot be 21 as 14 neem plants are planted; therefore, the remaining sum will be 20(5,15,0)

The remaining sum for the number of Banyan trees should be 17. The only possibility is (13,4)

Final arrangements:

Total number of teak plants and rose plants planted = 23 + 22 = 45

Answer is option A.
Question 5
3 / -1

Directions For Questions

Members of Greenland Association Aashi, Puja, Manu and Nitya have planted 48, 31, 29 and 28 trees in any order. They have planted six varieties of plants: Banyan, Mango, Coconut, Neem, Teak and Rose. Each of them has planted a minimum of two varieties of plants and a maximum of six varieties of plants. The minimum number of plants planted by anyone for a type of plant is one, and the maximum number of plants planted is 16.
Each of them has planted a different number of varieties. For example, if Puja has planted three varieties of plants, none of the remaining members has planted three varieties.
No two have planted the same number of plants for any two varieties. For example, if Puja has planted three mango plants, none of them has planted three plants for any variety.
1. Maximum difference between the total number of plants of any two varieties is one.
2. The number of plants planted by a person for each variety is in A.P., and the number of plants planted by another person for each variety is in G.P. Both of them have planted more than three varieties of plants.

...view full instructions

If Puja has planted 7 teak plants, which of the following has maximum value?

A
Total number of mango plants

B
Teak plants planted by Puja + Neem plants planted by Manu

C
Banyan plants planted by Puja + Mango plants planted by Nitya

D
Mango plants planted by Aashi + Rose plants planted by Manu

Solution

Total number of plants planted by all of them = 48 + 31 + 29 + 28 = 136
136 is sum of first 16 natural numbers, i.e.
1 + 2 + 3 + .............+ 15 + 16 = 136
The minimum number of plants planted for a variety is 1, and the maximum is 16.
Therefore, the total number of varieties of plants cannot exceed 16, but if it is less than 16, i.e. if it is 15 or 14, it doesn’t satisfy the sum value of 136, or the maximum number of plants planted for a variety is 16. This implies that all of them have planted 16 varieties of plants.
It is mentioned that each one has planted a minimum of two varieties of plants and a maximum of six varieties.
Sum = 2 + 3 + 4 + 5 + 6 = 20
Removing 4, we will get 16 total varieties for four members.
Therefore, four of them planted 2, 3, 5, and 6 varieties of plants in any order.
It is given that,
The number of plants planted by a person for each variety is in A.P. The number of plants planted by another person for each variety is in G.P. They have planted more than three varieties of plants. This means that they have planted 5 and 6 varieties in any order.

Sum of A.P series = $$\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$
The Sum of an A.P series is always divisible by $$\frac{n}{2}$$ (n - number of terms)
Out of the given values, the only possible value is 48, which is divisible by 3 (6/2)
Therefore, a person has planted 48 plants of 6 varieties.
A person has planted 48 plants of 6 varieties in A.P
Let us assume she has planted a-5d, a-3d, a-d, a+d, a+3d, and a+5d plants of each variety.
Sum = 48
6a = 48
a = 8
Number of plants planted are 8-5d, 8-3d, 8-d, 8+d, 8+3d, 8+5d
The only possible value for d is 1
Therefore, the only possibility is that she has planted 3, 5, 7, 9, 11, and 13 (6) plants of each variety.

The person who planted the plants of different varieties in G.P has planted five varieties.
Sum of G.P series = $$\ \frac{\ a\left(r^n-1\right)}{r-1}=\ \frac{\ a\left(r^5-1\right)}{r\ -1}=a\left(r^4+r^3+r^2+r+1\right)$$
Out of the given values, only possible value is 31 i.e. a = 1 and r = 2.
The number of plants planted in each variety is 1, 2, 4, 8, 16 (5) in any order.
In the table, it is given that Aashi has planted two rose plants. This implies that she has planted the plants in G.P.

Remaining numbers: 6, 10, 12, 14, 15
A person has planted two varieties of plants, and the number of plants planted is either 28 or 29. The only possibility is she has planted 14, 15 plants of each variety in any order. It is mentioned that Nitya has planted 6 Banyan trees. This implies that Nitya has planted 6, 10, and 12 plants of each variety in any order.

It is mentioned that the maximum difference between the total number of plants of any two varieties is one. Therefore, let us assume there are x varieties with y plants and (6-x) with y-1 plants, i.e.
xy + (6-x)(y-1) = 136
xy + 6y - 6 - xy + x = 136
x + 6y = 142 (x<6)
The only possibility is x = 4, y = 23
Therefore, there are 23, 23, 23, 23, 22 and 22 plants of each variety in any order.
The only possibility for the number of neem plants is 23 (9+0+14+0)
The number of coconut plants can be 22 or 23. Therefore, the remaining sum should be 11 or 12
For 11: (1, 0, 10)
For 12: (2, 0, 10)
we cannot consider 12, as Aashi has planted two mango plants. Therefore, it should be 11 (1, 0, 10).

The number of rose plants planted by Nitya cannot be 12. If it is 12, the remaining sum should be either 8 or 9, which is impossible. Therefore, the number of rose plants planted by Nitya will be 0. If it is 0, the remaining sum should be 20 or 21.
For 21: 7, 14, 0
For 20: 5, 15, 0
It cannot be 21 as 14 neem plants are planted; therefore, the remaining sum will be 20(5,15,0)

The remaining sum for the number of Banyan trees should be 17. The only possibility is (13,4)

Final arrangements:

It is mentioned that Puja has planted 7 teak plants, this refers to case 2b, i.e.

A) Total number of mango plants = 23
B) Teak plants planted by Puja + Neem plants planted by Manu = 7 + 14 = 21
C) Banyan plants planted by Puja + Mango plants planted by Nitya = 13 + 12 = 25
D) Mango plants planted by Aashi + Rose plants planted by Manu = 8 + 15 = 23
Option C has maximum value.

Therefore, answer is option C.
Question 6
3 / -1

Directions For Questions

The figure below shows the route map of 13 major junctions namely A, B, C, D, E, F, G, I, J, K, L, M, N in Hyderabad.

Additional information regarding the traffic on a particular day is given below:

1. The number of vehicles at a junction is equal to the sum of the number of vehicles to that junction, for example, the number of vehicles at F = number of vehicles from A + number of vehicles from D

2. All the vehicles at a junction reach the next junctions except at junctions E, I, G where the number of vehicles reaching K, L, M respectively reduces by 2/3rd.

3. A, B, C are D are the origins, and N is the endpoint of Hyderabad.

4. The number of vehicles at I and J are in the ratio of 3:2.

5. The number of vehicles at N= 35 and the number of vehicles at K, L, M are in the ratio of 2:5:2.

6. The number of vehicles travelling from A to E is the same as that from B to E.

7. The number of vehicles travelling from A to F = 2* The number of vehicles travelling from D to F.

8. The number of vehicles travelling from D to G = 2* The number of vehicles travelling from C to G.

...view full instructions

What is the sum of the number of vehicles at A, B, C, and D?

A
75

B
85

C
65

D
95

Solution

From point 2, we can see that the number of vehicles at the origins A,B,D,C will be equal to the number of vehicles going out of those junctions.
Number of vehicles at N = 35

The number of vehicles at K, L, M are in the ratio of 2:5:2.

Also, The number of vehicles at N = The number of vehicles from L to N + The number of vehicles from K to N.

The number of vehicles from L to N and the number of vehicles from K to N is equal to the number of vehicles at L, K respectively.

So, 35 = 2x + 5x

35 = 7x

x =5

The number of vehicles at K= 10

The number of vehicles at L= 25

The number of vehicles at M= 10

From 2,

The number of vehicles at M = The number of vehicles at G * 1/3

10 = The number of vehicles at G * 1/3

The number of vehicles at G = 30

The number of vehicles at K =The number of vehicles at E * 1/3

10 = The number of vehicles at E * 1/3

The number of vehicles at E = 30

From 8,

The number of vehicles at I and J be 3y, 2y respectively.

So, 25 = 3y*1/3 + 2y+10

y =5

The number of vehicles at J = 10

The number of vehicles at I = 15

The number of vehicles at I = The number of vehicles at F

The number of vehicles at F =15

The number of vehicles at E is equally divided from A and B.

So The number of vehicles from A and B towards E= 15 each

The number of vehicles at F is in ratio 2:1 from A and D respectively.

So, the number of vehicles at A and D towards F are 10 and 5 respectively.

The number of vehicles at G is in ratio 1:2 from C and D respectively.

So The number of vehicles at C and D towards G are 10 and 20 respectively.

.'. The number of vehicles at A, B, C, and D = 25, 25, 25, and 10 respectively.

so A+B+C+D=85
Question 7
3 / -1

Directions For Questions

The figure below shows the route map of 13 major junctions namely A, B, C, D, E, F, G, I, J, K, L, M, N in Hyderabad.

Additional information regarding the traffic on a particular day is given below:

1. The number of vehicles at a junction is equal to the sum of the number of vehicles to that junction, for example, the number of vehicles at F = number of vehicles from A + number of vehicles from D

2. All the vehicles at a junction reach the next junctions except at junctions E, I, G where the number of vehicles reaching K, L, M respectively reduces by 2/3rd.

3. A, B, C are D are the origins, and N is the endpoint of Hyderabad.

4. The number of vehicles at I and J are in the ratio of 3:2.

5. The number of vehicles at N= 35 and the number of vehicles at K, L, M are in the ratio of 2:5:2.

6. The number of vehicles travelling from A to E is the same as that from B to E.

7. The number of vehicles travelling from A to F = 2* The number of vehicles travelling from D to F.

8. The number of vehicles travelling from D to G = 2* The number of vehicles travelling from C to G.

...view full instructions

What are the number of vehicle at I?

A
15

B
10

C
5

D
20

Solution

From point 2, we can see that the number of vehicles at the origins A,B,D,C will be equal to the number of vehicles going out of those junctions.
Number of vehicles at N = 35

The number of vehicles at K, L, M are in the ratio of 2:5:2.

Also, The number of vehicles at N = The number of vehicles from L to N + The number of vehicles from K to N.

The number of vehicles from L to N and the number of vehicles from K to N is equal to the number of vehicles at L, K respectively.

So, 35 = 2x + 5x

35 = 7x

x =5

The number of vehicles at K= 10

The number of vehicles at L= 25

The number of vehicles at M= 10

From 2,

The number of vehicles at M = The number of vehicles at G * 1/3

10 = The number of vehicles at G * 1/3

The number of vehicles at G = 30

The number of vehicles at K =The number of vehicles at E * 1/3

10 = The number of vehicles at E * 1/3

The number of vehicles at E = 30

From 8,

The number of vehicles at I and J be 3y, 2y respectively.

So, 25 = 3y*1/3 + 2y+10

y =5

The number of vehicles at J = 10

The number of vehicles at I = 15

The number of vehicles at I = The number of vehicles at F

The number of vehicles at F =15

The number of vehicles at E is equally divided from A and B.

So The number of vehicles from A and B towards E= 15 each

The number of vehicles at F is in ratio 2:1 from A and D respectively.

So, the number of vehicles at A and D towards F are 10 and 5 respectively.

The number of vehicles at G is in ratio 1:2 from C and D respectively.

So The number of vehicles at C and D towards G are 10 and 20 respectively.

The number of vehicles at A, B, C, and D = 25, 25, 25, and 10 respectively.

Ans: 15
Question 8
3 / -1

Directions For Questions

The figure below shows the route map of 13 major junctions namely A, B, C, D, E, F, G, I, J, K, L, M, N in Hyderabad.

Additional information regarding the traffic on a particular day is given below:

1. The number of vehicles at a junction is equal to the sum of the number of vehicles to that junction, for example, the number of vehicles at F = number of vehicles from A + number of vehicles from D

2. All the vehicles at a junction reach the next junctions except at junctions E, I, G where the number of vehicles reaching K, L, M respectively reduces by 2/3rd.

3. A, B, C are D are the origins, and N is the endpoint of Hyderabad.

4. The number of vehicles at I and J are in the ratio of 3:2.

5. The number of vehicles at N= 35 and the number of vehicles at K, L, M are in the ratio of 2:5:2.

6. The number of vehicles travelling from A to E is the same as that from B to E.

7. The number of vehicles travelling from A to F = 2* The number of vehicles travelling from D to F.

8. The number of vehicles travelling from D to G = 2* The number of vehicles travelling from C to G.

...view full instructions

Which junction has the least number of vehicles?

A
C,J,K,M

B
C,K,M

C
C,J

D
J,K,M

Solution

From point 2, we can see that the number of vehicles at the origins A,B,D,C will be equal to the number of vehicles going out of those junctions.

Number of vehicles at N = 35

The number of vehicles at K, L, M are in the ratio of 2:5:2.

Also, The number of vehicles at N = The number of vehicles from L to N + The number of vehicles from K to N.

The number of vehicles from L to N and the number of vehicles from K to N is equal to the number of vehicles at L, K respectively.

So, 35 = 2x + 5x

35 = 7x

x =5

The number of vehicles at K= 10

The number of vehicles at L= 25

The number of vehicles at M= 10

From 2,

The number of vehicles at M = The number of vehicles at G * 1/3

10 = The number of vehicles at G * 1/3

The number of vehicles at G = 30

The number of vehicles at K =The number of vehicles at E * 1/3

10 = The number of vehicles at E * 1/3

The number of vehicles at E = 30

From 8,

The number of vehicles at I and J be 3y, 2y respectively.

So, 25 = 3y*1/3 + 2y+10

y =5

The number of vehicles at J = 10

The number of vehicles at I = 15

The number of vehicles at I = The number of vehicles at F

The number of vehicles at F =15

The number of vehicles at E is equally divided from A and B.

So The number of vehicles from A and B towards E= 15 each

The number of vehicles at F is in ratio 2:1 from A and D respectively.

So, the number of vehicles at A and D towards F are 10 and 5 respectively.

The number of vehicles at G is in ratio 1:2 from C and D respectively.

So The number of vehicles at C and D towards G are 10 and 20 respectively.

The number of vehicles at A, B, C, and D = 25, 25, 25, and 10 respectively.

Junctions C, J, K, M have 10 vehicles each.
Question 9
3 / -1

Directions For Questions

The figure below shows the route map of 13 major junctions namely A, B, C, D, E, F, G, I, J, K, L, M, N in Hyderabad.

Additional information regarding the traffic on a particular day is given below:

1. The number of vehicles at a junction is equal to the sum of the number of vehicles to that junction, for example, the number of vehicles at F = number of vehicles from A + number of vehicles from D

2. All the vehicles at a junction reach the next junctions except at junctions E, I, G where the number of vehicles reaching K, L, M respectively reduces by 2/3rd.

3. A, B, C are D are the origins, and N is the endpoint of Hyderabad.

4. The number of vehicles at I and J are in the ratio of 3:2.

5. The number of vehicles at N= 35 and the number of vehicles at K, L, M are in the ratio of 2:5:2.

6. The number of vehicles travelling from A to E is the same as that from B to E.

7. The number of vehicles travelling from A to F = 2* The number of vehicles travelling from D to F.

8. The number of vehicles travelling from D to G = 2* The number of vehicles travelling from C to G.

...view full instructions

At how many junctions the number of vehicles must be 25?

A
3

B
2

C
4

D
Cannot be determined

Solution

From point 2, we can see that the number of vehicles at the origins A,B,D,C will be equal to the number of vehicles going out of those junctions.
Number of vehicles at N = 35

The number of vehicles at K, L, M are in the ratio of 2:5:2.

Also, The number of vehicles at N = The number of vehicles from L to N + The number of vehicles from K to N.

The number of vehicles from L to N and the number of vehicles from K to N is equal to the number of vehicles at L, K respectively.

So, 35 = 2x + 5x

35 = 7x

x =5

The number of vehicles at K= 10

The number of vehicles at L= 25

The number of vehicles at M= 10

From 2,

The number of vehicles at M = The number of vehicles at G * 1/3

10 = The number of vehicles at G * 1/3

The number of vehicles at G = 30

The number of vehicles at K =The number of vehicles at E * 1/3

10 = The number of vehicles at E * 1/3

The number of vehicles at E = 30

From 8,

The number of vehicles at I and J be 3y, 2y respectively.

So, 25 = 3y*1/3 + 2y+10

y =5

The number of vehicles at J = 10

The number of vehicles at I = 15

The number of vehicles at I = The number of vehicles at F

The number of vehicles at F =15

The number of vehicles at E is equally divided from A and B.

So The number of vehicles from A and B towards E= 15 each

The number of vehicles at F is in ratio 2:1 from A and D respectively.

So, the number of vehicles at A and D towards F are 10 and 5 respectively.

The number of vehicles at G is in ratio 1:2 from C and D respectively.

So The number of vehicles at C and D towards G are 10 and 20 respectively.

The number of vehicles at A, B, D and L is 25.
Question 10
3 / -1

Directions For Questions

The figure below shows the route map of 13 major junctions namely A, B, C, D, E, F, G, I, J, K, L, M, N in Hyderabad.

Additional information regarding the traffic on a particular day is given below:

1. The number of vehicles at a junction is equal to the sum of the number of vehicles to that junction, for example, the number of vehicles at F = number of vehicles from A + number of vehicles from D

2. All the vehicles at a junction reach the next junctions except at junctions E, I, G where the number of vehicles reaching K, L, M respectively reduces by 2/3rd.

3. A, B, C are D are the origins, and N is the endpoint of Hyderabad.

4. The number of vehicles at I and J are in the ratio of 3:2.

5. The number of vehicles at N= 35 and the number of vehicles at K, L, M are in the ratio of 2:5:2.

6. The number of vehicles travelling from A to E is the same as that from B to E.

7. The number of vehicles travelling from A to F = 2* The number of vehicles travelling from D to F.

8. The number of vehicles travelling from D to G = 2* The number of vehicles travelling from C to G.

...view full instructions

The government plans to set up a few new junctions between the origin and end point of Hyderabad so that all the vehicles from origin can reach the end point . If each junction has a capacity of 10 cars, what will be the minimum number of junctions that the government would have to set up?

A
7

B
3

C
5

D
9

Solution

Total number of vehicles at origin = 85
Total number of vehicles at end point = 35

Hence , 50 vehicles cannot reach the end point from the origin , so we require 5 junctions.
Question 11
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Directions For Questions

Read the following information carefully and answer the given questions.

Gopi has to arrange 10 different balls on a stand in 10 places numbered 1 to 10 from left to right. The colours of the balls are Violet, Indigo, Blue, Green, Yellow, Orange, Red, Maroon, White and Pink. Additionally, the following conditions have to be followed:
1) Yellow and Orange coloured balls have to be kept as far as possible
2) There are four balls in between Pink and Yellow coloured balls
3) Violet and Indigo coloured balls have to be kept together, either at the beginning or at the end of the stand
4) Blue and Green coloured balls have to be kept in positions 4 and 5 in any order
5) The Red coloured ball is kept to the left of the Yellow coloured ball but to the right of the White coloured ball
6) The Maroon coloured ball is kept to the right of the Green coloured ball but to the left of the Red coloured ball

...view full instructions

In which position is the yellow coloured ball?

A
3

B
6

C
7

D
8

Solution

From the information given in the question, we know that Blue and Green balls are in positions 4 and 5 in any order.
We also know that Violet and Indigo balls are in places 1 and 2 or in places 9 and 10 in any order.
So there will be two possible cases as shown below.
Case-1:

Case-2:

But in the arrangement of the first case, we cannot place exactly four balls in between Pink and Yellow balls and simultaneously satisfy the condition for the White coloured ball.
So we can rule out Case-1 and only consider Case-2.
According to the given information, the balls from left to right should be Green, Maroon, Red and Yellow. Thus yellow ball can only be in place 8. This means that Pink ball has to be in place 3.
As yellow and orange balls have to be kept as far as possible, orange ball can be in place 1.
This means that white and pink balls are in places 2 and 3 respectively.
Hence the final arrangement of all the balls is as shown below.
Question 12
3 / -1

Directions For Questions

Read the following information carefully and answer the given questions.

Gopi has to arrange 10 different balls on a stand in 10 places numbered 1 to 10 from left to right. The colours of the balls are Violet, Indigo, Blue, Green, Yellow, Orange, Red, Maroon, White and Pink. Additionally, the following conditions have to be followed:
1) Yellow and Orange coloured balls have to be kept as far as possible
2) There are four balls in between Pink and Yellow coloured balls
3) Violet and Indigo coloured balls have to be kept together, either at the beginning or at the end of the stand
4) Blue and Green coloured balls have to be kept in positions 4 and 5 in any order
5) The Red coloured ball is kept to the left of the Yellow coloured ball but to the right of the White coloured ball
6) The Maroon coloured ball is kept to the right of the Green coloured ball but to the left of the Red coloured ball

...view full instructions

Which of the following is definitely the right combination of the ball and its position from left?

A
Blue-4

B
Pink-3

C
White-1

D
Violet-9

Solution

From the information given in the question, we know that Blue and Green balls are in positions 4 and 5 in any order.
We also know that Violet and Indigo balls are in places 1 and 2 or in places 9 and 10 in any order.
So there will be two possible cases as shown below.
Case-1:

Case-2:

But in the arrangement of the first case, we cannot place exactly four balls in between Pink and Yellow balls and simultaneously satisfy the condition for the White coloured ball.
So we can rule out Case-1 and only consider Case-2.
According to the given information, the balls from left to right should be Green, Maroon, Red and Yellow. Thus yellow ball can only be in place 8. This means that Pink ball has to be in place 3.
As yellow and orange balls have to be kept as far as possible, orange ball can be in place 1.
This means that white and pink balls are in places 2 and 3 respectively.
Hence the final arrangement of all the balls is as shown below.
Question 13
3 / -1

Directions For Questions

Read the following information carefully and answer the given questions.

Gopi has to arrange 10 different balls on a stand in 10 places numbered 1 to 10 from left to right. The colours of the balls are Violet, Indigo, Blue, Green, Yellow, Orange, Red, Maroon, White and Pink. Additionally, the following conditions have to be followed:
1) Yellow and Orange coloured balls have to be kept as far as possible
2) There are four balls in between Pink and Yellow coloured balls
3) Violet and Indigo coloured balls have to be kept together, either at the beginning or at the end of the stand
4) Blue and Green coloured balls have to be kept in positions 4 and 5 in any order
5) The Red coloured ball is kept to the left of the Yellow coloured ball but to the right of the White coloured ball
6) The Maroon coloured ball is kept to the right of the Green coloured ball but to the left of the Red coloured ball

...view full instructions

If Blue ball is to the immediate right of the Pink ball, then which ball is two places to the right of red ball?

A
Violet

B
Green

C
Indigo

D
Cannot be determined

Solution

From the information given in the question, we know that Blue and Green balls are in positions 4 and 5 in any order.
We also know that Violet and Indigo balls are in places 1 and 2 or in places 9 and 10 in any order.
So there will be two possible cases as shown below.
Case-1:

Case-2:

But in the arrangement of the first case, we cannot place exactly four balls in between Pink and Yellow balls and simultaneously satisfy the condition for the White coloured ball.
So we can rule out Case-1 and only consider Case-2.
According to the given information, the balls from left to right should be Green, Maroon, Red and Yellow. Thus yellow ball can only be in place 8. This means that Pink ball has to be in place 3.
As yellow and orange balls have to be kept as far as possible, orange ball can be in place 1.
This means that white and pink balls are in places 2 and 3 respectively.
Hence the final arrangement of all the balls is as shown below.
Question 14
3 / -1

Directions For Questions

Read the following information carefully and answer the given questions.

Gopi has to arrange 10 different balls on a stand in 10 places numbered 1 to 10 from left to right. The colours of the balls are Violet, Indigo, Blue, Green, Yellow, Orange, Red, Maroon, White and Pink. Additionally, the following conditions have to be followed:
1) Yellow and Orange coloured balls have to be kept as far as possible
2) There are four balls in between Pink and Yellow coloured balls
3) Violet and Indigo coloured balls have to be kept together, either at the beginning or at the end of the stand
4) Blue and Green coloured balls have to be kept in positions 4 and 5 in any order
5) The Red coloured ball is kept to the left of the Yellow coloured ball but to the right of the White coloured ball
6) The Maroon coloured ball is kept to the right of the Green coloured ball but to the left of the Red coloured ball

...view full instructions

The balls of which two colours cannot be placed next to each other?

A
Blue-Maroon

B
Yellow-Violet

C
Blue-Red

D
Maroon-Green

Solution

From the information given in the question, we know that Blue and Green balls are in positions 4 and 5 in any order.
We also know that Violet and Indigo balls are in places 1 and 2 or in places 9 and 10 in any order.
So there will be two possible cases as shown below.
Case-1:

Case-2:

But in the arrangement of the first case, we cannot place exactly four balls in between Pink and Yellow balls and simultaneously satisfy the condition for the White coloured ball.
So we can rule out Case-1 and only consider Case-2.
According to the given information, the balls from left to right should be Green, Maroon, Red and Yellow. Thus yellow ball can only be in place 8. This means that Pink ball has to be in place 3.
As yellow and orange balls have to be kept as far as possible, orange ball can be in place 1.
This means that white and pink balls are in places 2 and 3 respectively.
Hence the final arrangement of all the balls is as shown below.
Question 15
3 / -1

Directions For Questions

Read the following information carefully and answer the given questions.

Gopi has to arrange 10 different balls on a stand in 10 places numbered 1 to 10 from left to right. The colours of the balls are Violet, Indigo, Blue, Green, Yellow, Orange, Red, Maroon, White and Pink. Additionally, the following conditions have to be followed:
1) Yellow and Orange coloured balls have to be kept as far as possible
2) There are four balls in between Pink and Yellow coloured balls
3) Violet and Indigo coloured balls have to be kept together, either at the beginning or at the end of the stand
4) Blue and Green coloured balls have to be kept in positions 4 and 5 in any order
5) The Red coloured ball is kept to the left of the Yellow coloured ball but to the right of the White coloured ball
6) The Maroon coloured ball is kept to the right of the Green coloured ball but to the left of the Red coloured ball

...view full instructions

If position of white ball is interchanged with position of red ball then after interchanging their position which color ball is placed to the 4th left of white ball?

A
White

B
Pink

C
Blue

D
Green

Solution

From the information given in the question, we know that Blue and Green balls are in positions 4 and 5 in any order.
We also know that Violet and Indigo balls are in places 1 and 2 or in places 9 and 10 in any order.
So there will be two possible cases as shown below.
Case-1:

Case-2:

But in the arrangement of the first case, we cannot place exactly four balls in between Pink and Yellow balls and simultaneously satisfy the condition for the White coloured ball.
So we can rule out Case-1 and only consider Case-2.
According to the given information, the balls from left to right should be Green, Maroon, Red and Yellow. Thus yellow ball can only be in place 8. This means that Pink ball has to be in place 3.
As yellow and orange balls have to be kept as far as possible, orange ball can be in place 1.
This means that white and pink balls are in places 2 and 3 respectively.
Hence the final arrangement of all the balls is as shown below.

If position of white ball is interchanged with red ball then at position 2 red ball will come & at position 7 white ball will come.

Therfore, ball that is placed to the 4th left of white ball after interchanging position is Pink ball at position 3.

Pink ball (Ans- B)
Question 16
3 / -1

Directions For Questions

Monda Car Company has five production plants A, B, C, D and E. They store the cars produced in five warehouses, P, Q, R, S and T. They have a showroom in Mumbai, Hyderabad, Bangalore, Delhi and Chandigarh. The figures in the table represent the distance in km between the production plants and warehouses and between warehouses and showrooms.

...view full instructions

What is the minimum distance that would need to be covered to deliver a car to the Mumbai showroom?

A
478

B
534

C
592

D
728

Solution

The nearest combined distance is to T warehouse + A production plant = 215+319=534
Question 17
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Directions For Questions

Monda Car Company has five production plants A, B, C, D and E. They store the cars produced in five warehouses, P, Q, R, S and T. They have a showroom in Mumbai, Hyderabad, Bangalore, Delhi and Chandigarh. The figures in the table represent the distance in km between the production plants and warehouses and between warehouses and showrooms.

...view full instructions

What is the maximum distance between any production plant to any showroom?

A
1877

B
18774

C
18777

D
18779

Solution

The longest route is A to R and R to Mumbai = 882+995=1877
Question 18
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Directions For Questions

Monda Car Company has five production plants A, B, C, D and E. They store the cars produced in five warehouses, P, Q, R, S and T. They have a showroom in Mumbai, Hyderabad, Bangalore, Delhi and Chandigarh. The figures in the table represent the distance in km between the production plants and warehouses and between warehouses and showrooms.

...view full instructions

In how many different ways can a car be sent to the Bangalore showroom so that the total distance travelled is less than 1000 km?

A
10

B
104

C
107

D
109

Solution

There are 10 routes to Bangalore such that combined distance <1000 km. There are 3 through P, 1 through Q, 3 through R, 1 through S and 2 through T.
Question 19
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Directions For Questions

Monda Car Company has five production plants A, B, C, D and E. They store the cars produced in five warehouses, P, Q, R, S and T. They have a showroom in Mumbai, Hyderabad, Bangalore, Delhi and Chandigarh. The figures in the table represent the distance in km between the production plants and warehouses and between warehouses and showrooms.

...view full instructions

Monda cars' base price is the same at each showroom. However, each showroom adds an additional charge depending on the transportation cost of getting a vehicle to a showroom. If the additional cost depends on the minimum distance that needs to be covered to get a vehicle to a showroom, in which showroom would the total cost be the minimum?

A
Chandigarh

B
Hyderabad

C
Bangalore

D
None of the above

Solution

From the table, we can see that the Delhi showroom has the least distance.
Among the warehouses that transport to Delhi, the distance from A to S is 201. So, the distance from A to S to Delhi is 201 + 290 = 491. This is the minimum distance to get a car to a showroom.
So, the minimum cost would be in the Delhi showroom.
Question 20
3 / -1

Directions For Questions

Monda Car Company has five production plants A, B, C, D and E. They store the cars produced in five warehouses, P, Q, R, S and T. They have a showroom in Mumbai, Hyderabad, Bangalore, Delhi and Chandigarh. The figures in the table represent the distance in km between the production plants and warehouses and between warehouses and showrooms.

...view full instructions

The connectivity of a showroom is defined as the average distance that must be travelled to get a car to the showroom - the lower the distance the better connected the showroom is. Which showroom is the best connected showroom?

A
Chandigarh

B
Delhi

C
Bangalore

D
None of the above

Solution

For each showroom, to calculate the total distance travelled, we need to add the distance travelled for each trio of factory - warehouse-showroom.
Hence, for Mumbai showroom, total distance travelled = Distance A - P + Distance P - Mumbai + Distance B - P + Distance P - Mumbai . . . Distance A - Q + Distance Q - Mumbai + Distance B - Q + Distance Q - Mumbai + ... + Distance E - T + Distance T-Mumbai
= (Distance A-P + Distance B-P + . . . Distance E-T) + 5 ( Distance P - Mumbai + . . .+ Distance T-Mumbai)
The first bracket contains all the distances given in the first table and this sum would be common for all showrooms. Hence, as it is common to all showrooms, this distance can be ignored while calculating the relative order of showrooms. Thus we need to only find out the total distance given in the second bracket.
The sum in the second bracket is lowest for Bangalore at = 379 + 349 + 350 + 752 + 410 = 2240.
Hence, Bangalore is the best connected showroom.