# Free DI and LR - 28 Practice Test - CAT

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

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

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

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

#### 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

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

#### 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

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.

A. 2:1
B. 4:3
C. 3:2
D. 8:5

#### 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.

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.

A. 18
B. 21
C. 24
D. 27

#### 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.

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.

A. 36
B. 40
C. 45
D. 48

#### 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

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

A. I and II
B. I only
C. I and IV
D. I and III

#### SOLUTION

Solution : D

Considering both cases, CASE 1 = 2100-1600 = 500,
CASE II- 1600-1200 = 400.

Which of the following statements is definitely true?

A. The sales value of Iron fortified drink is the highest.
B. The sales value of Vit K fortified drink is more than that of calcium fortified drink.
C. The sales value of Vit B fortified drink is the least.
D. The sales value of Calcium fortified drink is more than that of Vit E fortified drink.

#### SOLUTION

Solution : D

Only 4th statement is true.