# Free DI and LR - 28 Practice Test - CAT

### Question 1

What is the sum of all the numbers allocated to saplings in row Marigold?

#### SOLUTION

Solution :Sum of numbers allocated to West-end slots (all rows) = 119; Sum of numbers allocated to East-end slots (all rows) = 109; Hence sum of ‘numbers’ allocated to all slots between East and West end = 119-109 = 10.

Now, no two rows were equally long. Hence the no of slots between east and west ends for all five rows should be 0, 1, 2, 3, 4 (for a sum of 10). Hence there are 1,2,3,4 and 5 slots (inbetween) in rows 1, 2, 3, 4 and 5 respectively.

The sum of numbers allocated to Row 2 is 81. This can be sum of 1, 2, 3, 4 or 5 natural numbers.

If row 2 has only 1 or 2 slots, then it is impossible to have a sum of 109 by addition of all numbers allocated to East-ends of all rows. So the numbers allocated to Row 2 can only be 26, 27 and 28 (3 slots).

Since B has 3 slots, A and C cannot have 2 or 4 slots.

Assuming Row 1 has only 1 slot, it should be numbered as 21.

Two West-end slots, Row 2’s and Row 1’s are numbered 21 and 28. Hence two east end slots have to be numbered as 22 and 29. So the four East-end slots are numbered: 21, 22, 26, and 29. Hence the fifth West-end number is: 109-(21+22+26+29) = 11, which is not possible because in that case the sum of allocated numbers of all West-End slots would not be 119.

Hence Row 1 has 5 slots and Row 3, Row 4 and Row 5 have 1, 4 and 2 slots respectively.

Row 1 has a west-end number of 21 and Row 2 has an east-end number of 26. Thus there is a row in between Row 1 and Row2 that has 4 slots. So the other East-end numbers allocated are 17, 22 and 26. Hence the other two East-end allocated numbers are 29 and 15.

So the plot of land looks like:

RowSlot numbers:EastWestJasmine17 18 19 2021Lotus26 27 28Rose29Lavender22 23 24 25Marigold15 16

### Question 2

What is the number of saplings in row Rose?

#### SOLUTION

Solution :Sum of numbers allocated to West-end slots (all rows) = 119; Sum of numbers allocated to East-end slots (all rows) = 109; Hence sum of ‘numbers’ allocated to all slots between East and West end = 119-109 = 10.

Now, no two rows were equally long. Hence the no of slots between east and west ends for all five rows should be 0, 1, 2, 3, 4 (for a sum of 10). Hence there are 1,2,3,4 and 5 slots (inbetween) in rows 1, 2, 3, 4 and 5 respectively.

The sum of numbers allocated to Row 2 is 81. This can be sum of 1, 2, 3, 4 or 5 natural numbers.

If row 2 has only 1 or 2 slots, then it is impossible to have a sum of 109 by addition of all numbers allocated to East-ends of all rows. So the numbers allocated to Row 2 can only be 26, 27 and 28 (3 slots).

Since B has 3 slots, A and C cannot have 2 or 4 slots.

Assuming Row 1 has only 1 slot, it should be numbered as 21.

Two West-end slots, Row 2’s and Row 1’s are numbered 21 and 28. Hence two east end slots have to be numbered as 22 and 29. So the four East-end slots are numbered: 21, 22, 26, and 29. Hence the fifth West-end number is: 109-(21+22+26+29) = 11, which is not possible because in that case the sum of allocated numbers of all West-End slots would not be 119.

Hence Row 1 has 5 slots and Row 3, Row 4 and Row 5 have 1, 4 and 2 slots respectively.

Row 1 has a west-end number of 21 and Row 2 has an east-end number of 26. Thus there is a row in between Row 1 and Row2 that has 4 slots. So the other East-end numbers allocated are 17, 22 and 26. Hence the other two East-end allocated numbers are 29 and 15.

So the plot of land looks like:

RowSlot numbers:EastWestJasmine17 18 19 2021Lotus26 27 28Rose29Lavender22 23 24 25Marigold15 16

### Question 3

What is the highest number assigned to any slot by Jennifer?

#### SOLUTION

Solution :Sum of numbers allocated to West-end slots (all rows) = 119; Sum of numbers allocated to East-end slots (all rows) = 109; Hence sum of ‘numbers’ allocated to all slots between East and West end = 119-109 = 10.

Now, no two rows were equally long. Hence the no of slots between east and west ends for all five rows should be 0, 1, 2, 3, 4 (for a sum of 10). Hence there are 1,2,3,4 and 5 slots (inbetween) in rows 1, 2, 3, 4 and 5 respectively.

The sum of numbers allocated to Row 2 is 81. This can be sum of 1, 2, 3, 4 or 5 natural numbers.

If row 2 has only 1 or 2 slots, then it is impossible to have a sum of 109 by addition of all numbers allocated to East-ends of all rows. So the numbers allocated to Row 2 can only be 26, 27 and 28 (3 slots).

Since B has 3 slots, A and C cannot have 2 or 4 slots.

Assuming Row 1 has only 1 slot, it should be numbered as 21.

Two West-end slots, Row 2’s and Row 1’s are numbered 21 and 28. Hence two east end slots have to be numbered as 22 and 29. So the four East-end slots are numbered: 21, 22, 26, and 29. Hence the fifth West-end number is: 109-(21+22+26+29) = 11, which is not possible because in that case the sum of allocated numbers of all West-End slots would not be 119.

Hence Row 1 has 5 slots and Row 3, Row 4 and Row 5 have 1, 4 and 2 slots respectively.

Row 1 has a west-end number of 21 and Row 2 has an east-end number of 26. Thus there is a row in between Row 1 and Row2 that has 4 slots. So the other East-end numbers allocated are 17, 22 and 26. Hence the other two East-end allocated numbers are 29 and 15.

So the plot of land looks like:

RowSlot numbers:EastWestJasmine17 18 19 2021Lotus26 27 28Rose29Lavender22 23 24 25Marigold15 16

### Question 4

What is the average number assigned to a slot in row Lotus?

#### SOLUTION

Solution :

Now, no two rows were equally long. Hence the no of slots between east and west ends for all five rows should be 0, 1, 2, 3, 4 (for a sum of 10). Hence there are 1,2,3,4 and 5 slots (inbetween) in rows 1, 2, 3, 4 and 5 respectively.

The sum of numbers allocated to Row 2 is 81. This can be sum of 1, 2, 3, 4 or 5 natural numbers.

If row 2 has only 1 or 2 slots, then it is impossible to have a sum of 109 by addition of all numbers allocated to East-ends of all rows. So the numbers allocated to Row 2 can only be 26, 27 and 28 (3 slots).

Since B has 3 slots, A and C cannot have 2 or 4 slots.

Assuming Row 1 has only 1 slot, it should be numbered as 21.

Two West-end slots, Row 2’s and Row 1’s are numbered 21 and 28. Hence two east end slots have to be numbered as 22 and 29. So the four East-end slots are numbered: 21, 22, 26, and 29. Hence the fifth West-end number is: 109-(21+22+26+29) = 11, which is not possible because in that case the sum of allocated numbers of all West-End slots would not be 119.

Hence Row 1 has 5 slots and Row 3, Row 4 and Row 5 have 1, 4 and 2 slots respectively.

Row 1 has a west-end number of 21 and Row 2 has an east-end number of 26. Thus there is a row in between Row 1 and Row2 that has 4 slots. So the other East-end numbers allocated are 17, 22 and 26. Hence the other two East-end allocated numbers are 29 and 15.

So the plot of land looks like:

RowSlot numbers:EastWestJasmine17 18 19 2021Lotus26 27 28Rose29Lavender22 23 24 25Marigold15 16

### Question 5

What is the average of all the numbers allocated to the slots?

#### SOLUTION

Solution :

Now, no two rows were equally long. Hence the no of slots between east and west ends for all five rows should be 0, 1, 2, 3, 4 (for a sum of 10). Hence there are 1,2,3,4 and 5 slots (inbetween) in rows 1, 2, 3, 4 and 5 respectively.

The sum of numbers allocated to Row 2 is 81. This can be sum of 1, 2, 3, 4 or 5 natural numbers.

If row 2 has only 1 or 2 slots, then it is impossible to have a sum of 109 by addition of all numbers allocated to East-ends of all rows. So the numbers allocated to Row 2 can only be 26, 27 and 28 (3 slots).

Since B has 3 slots, A and C cannot have 2 or 4 slots.

Assuming Row 1 has only 1 slot, it should be numbered as 21.

Two West-end slots, Row 2’s and Row 1’s are numbered 21 and 28. Hence two east end slots have to be numbered as 22 and 29. So the four East-end slots are numbered: 21, 22, 26, and 29. Hence the fifth West-end number is: 109-(21+22+26+29) = 11, which is not possible because in that case the sum of allocated numbers of all West-End slots would not be 119.

Hence Row 1 has 5 slots and Row 3, Row 4 and Row 5 have 1, 4 and 2 slots respectively.

Row 1 has a west-end number of 21 and Row 2 has an east-end number of 26. Thus there is a row in between Row 1 and Row2 that has 4 slots. So the other East-end numbers allocated are 17, 22 and 26. Hence the other two East-end allocated numbers are 29 and 15.

So the plot of land looks like:

RowSlot numbers:EastWestJasmine17 18 19 2021Lotus26 27 28Rose29Lavender22 23 24 25Marigold15 16

### Question 6

**If the selling price of the Vit B fortified drink is 25% less than that of Vit E fortified drink. What is the ratio of the selling price of Iron fortified drink to that of V it A fortified drink. **

#### SOLUTION

Solution :B

Let 100x and 100y be the total sales value and total sales volume. The values from the graph can be deduced as:

LC TO PG RA IM TPSales Value 30% 14% 9% 28% 10% 9%Sales Vol 20% 21% 16% 21% 10% 12%Selling PriceSP 72 32 27 64 48 36Case IVit B IronVit ACalciumVit KVit ECase IICalcium Vit KVit A Vit EVit B Iron

Case II is possible. The required ratio is 36:27 = 4:3.

### Question 7

** If the selling price of Vit K fortified drink is Rs. 12 more than that of Vit E fortified drink, what is the selling price (in Rs) of Vit A fortified drink. **

#### SOLUTION

Solution :D

Let 100x and 100y be the total sales value and total sales volume. The values from the graph can be deduced as:

LC TO PG RA IM TPSales Value 30% 14% 9% 28% 10% 9%Sales Vol 20% 21% 16% 21% 10% 12%Selling PriceSP 72 32 27 64 48 36Case IVit B IronVit ACalciumVit KVit ECase IICalcium Vit KVit A Vit EVit B Iron

Case I is possible. 48-36=12, Hence A=27.

### Question 8

**If the sales value is the highest for Calcium fortified drink and is Rs. 180, what is the difference between the selling prices of Vit E fortified drink and Vit B fortified drink. **

#### SOLUTION

Solution :B

Let 100x and 100y be the total sales value and total sales volume. The values from the graph can be deduced as:

LC TO PG RA IM TPSales Value 30% 14% 9% 28% 10% 9%Sales Vol 20% 21% 16% 21% 10% 12%Selling PriceSP 72 32 27 64 48 36Case IVit B IronVit ACalciumVit KVit ECase IICalcium Vit KVit A Vit EVit B Iron

Case II is possible. Given that 72 ⇒180 = 72×2.5, Hence answer = (64-48)×2.5 = 40.

### Question 9

**The total sales volume is 10000. What is a possible difference in the sales volume of the Vitamin A fortified drink and the Iron fortified drink?
I) 500 II) 200 III) 400 IV) 800**

#### SOLUTION

Solution :D

Considering both cases, CASE 1 = 2100-1600 = 500,

CASE II- 1600-1200 = 400.

### Question 10

**Which of the following statements is definitely true? **

#### SOLUTION

Solution :D

Only 4th statement is true.