MAT 2025 PBT Question Paper May 31(Available): Download Solution PDF with Answer Key

Step 1: Determine speeds in still water and with the current.

Speed in still water = \( \frac{9 km}{1 hr} = 9 km/hr \)

Speed with the current = \( \frac{9 km}{45/60 hr} = \frac{9}{0.75} = 12 km/hr \)

Step 2: Find the speed of the current.

Let the speed of the current be \( c \). Then: \[ 9 + c = 12 \quad \Rightarrow \quad c = 3 km/hr \]

Step 3: Calculate upstream speed (against the current).
\[ Upstream speed = 9 - 3 = 6 km/hr \]

Step 4: Calculate time to row 36 km downstream and return 36 km upstream.

Downstream time = \( \frac{36}{12} = 3 hrs \)

Upstream time = \( \frac{36}{6} = 6 hrs \)

Step 5: Total time taken:
\[ Total time = 3 + 6 = 9 hrs \] Quick Tip: Use the relation: \[ Speed with current = b + c, \quad Speed against current = b - c \] where \( b \) is the speed in still water and \( c \) is the speed of the current.

Step 1: Assume the Marked Price (M.P.) = Rs.100.


When discount is 22%, Selling Price (S.P.) = \( 100 - 22\) = Rs.78
Let Cost Price (C.P.) be \( x \). Given profit = 17%:
\[ \frac{78 - x}{x} \times 100 = 17 \quad \Rightarrow \quad \frac{78 - x}{x} = \frac{17}{100} \]
\[ \Rightarrow 100(78 - x) = 17x \quad \Rightarrow \quad 7800 - 100x = 17x \]
\[ \Rightarrow 7800 = 117x \quad \Rightarrow \quad x = \frac{7800}{117} = Rs.66.67 \]


Step 2: When discount is 18%, new S.P. = Rs.82
\[ New profit% = \frac{82 - 66.67}{66.67} \times 100 \approx \frac{15.33}{66.67} \times 100 \approx 23% \]


Step 3: Difference in profit percentage = \( 23% - 17% = \boxed{6%} \) Quick Tip: Always take the marked price as Rs.100 to simplify problems involving discount and profit. Then calculate the selling price and compare with cost price to find profit%.

Step 1: Identify the digits allowed.

Digits given: \( \{1, 2, 3, 4, 5, 6\} \) \(\Rightarrow\) Total digits = 6

Digit 4 is not allowed, so remaining digits = \( \{1, 2, 3, 5, 6\} \) \(\Rightarrow\) 5 digits


Step 2: Count valid 4-digit numbers.

We have to form 4-digit numbers using 5 digits (excluding 4), without repetition.

So, total such numbers = Number of 4-digit permutations from 5 digits: \[ ^5P_4 = 5 \times 4 \times 3 \times 2 = 120 \] Quick Tip: For non-repeating digit problems: \[ ^nP_r = \frac{n!}{(n - r)!} \] Here, exclude restricted digits first, then apply permutation.

Step 1: Use the concept of successive replacement.

Each time, \( \frac{1}{6} \) of the juice is removed and replaced with water. So, \( \frac{5}{6} \) of the juice remains after each operation.

Step 2: Use the general formula for remaining quantity of original liquid: \[ Final amount of juice = Initial quantity \times \left( \frac{5}{6} \right)^n \]
where \( n = 3 \) (process repeated 3 times).

Step 3: Let initial quantity of juice = \( V \) ml.

We are given: \[ \left( \frac{5}{6} \right)^3 = \frac{125}{216}, \quad and final juice left = 1250 \, ml \]
So, \[ 1250 = V \cdot \frac{125}{216} \]

Step 4: Solve for \( V \). \[ V = \frac{1250 \times 216}{125} = \frac{270000}{125} = 2160 \, ml \]

Step 5: Convert to liters. \[ 2160 \, ml = 2.160 \, liters \] Quick Tip: When liquid is replaced repeatedly, use this formula:
\[ Final quantity = Initial quantity \times \left(1 - \frac{x}{V}\right)^n \] where \( x \) is the amount replaced each time, and \( n \) is the number of repetitions.

Step 1: Convert the increase in average age to years.
\[ 2\frac{1}{2} months = \frac{5}{2} months = \frac{5}{24} years \]

Step 2: Calculate total increase in age for all 36 students. \[ Total increase = 36 \times \frac{5}{24} = \frac{180}{24} = 7.5 years \]

Step 3: Since one 10-year-old child is replaced,

The new child must be older than the previous one by 7.5 years: \[ Age of new child = 10 + 7.5 = \boxed{17.5 years} \] Quick Tip: Multiply the change in average by the number of members to find the total age difference, then apply it to the replaced individual.

Let total distance = 108 km (LCM of 9 and 12 for easier calculation).
\[ Given speed = 54 km/hr \quad \Rightarrow \quad Total time = \frac{108}{54} = 2 hours \]

Step 1: In \( \frac{1}{4} \) of the time i.e. \( \frac{1}{2} \) hour, distance covered = \( \frac{2}{9} \times 108 = 24 \) km

Step 2: Remaining distance = \( 108 - 24 = 84 \) km

Remaining time = \( 2 - \frac{1}{2} = \frac{3}{2} \) hours

Step 3: Required speed = \( \frac{84}{3/2} = \frac{84 \times 2}{3} = 56 \) km/hr Quick Tip: When a question involves time and fractional distance, assume a total distance that's a multiple of denominators to simplify calculations.

Let Ajay's present age be \( x \) years.

Then Vijay's present age = \( x + 22.5 \) years.

6.5 years ago:

Ajay's age = \( x - 6.5 \)

Vijay's age = \( x + 22.5 - 6.5 = x + 16 \)

Given: \( x + 16 = 3.5(x - 6.5) \)

Step 1: Expand the equation \[ x + 16 = 3.5x - 22.75 \]

Step 2: Simplify \[ x - 3.5x = -22.75 - 16 \Rightarrow -2.5x = -38.75 \Rightarrow x = \frac{38.75}{2.5} = 15.5 \]

Step 3: Vijay's present age \[ x + 22.5 = 15.5 + 22.5 = 38 \]

Final Answer:

Ajay = \(15\frac{1}{2}\) years, Vijay = 38 years. Quick Tip: When past age ratios and present differences are given, always use variable expressions and set up an equation based on the older statement to solve efficiently.

Let the quantity of the first variety of Moong dal be \( x \) kg.

Let the quantity of the second variety of Moong dal be \( y \) kg.


Step 1: Formulate equations based on given information.

The total weight of the mixture is 90 kg: \[ x + y = 90 \quad \cdots (1) \]
The total cost of the mixture is Rs 7443. The price of the first variety is Rs 57.50 per kg, and the second variety is Rs 111.50 per kg: \[ 57.50x + 111.50y = 7443 \quad \cdots (2) \]

Step 2: Solve the system of equations.

From equation (1), we can express \( x \) in terms of \( y \): \[ x = 90 - y \]
Substitute this expression for \( x \) into equation (2): \[ 57.50(90 - y) + 111.50y = 7443 \]
Distribute 57.50: \[ 5175 - 57.50y + 111.50y = 7443 \]
Combine the terms with \( y \): \[ 5175 + (111.50 - 57.50)y = 7443 \] \[ 5175 + 54y = 7443 \]
Subtract 5175 from both sides: \[ 54y = 7443 - 5175 \] \[ 54y = 2268 \]
Divide by 54 to find \( y \): \[ y = \frac{2268}{54} \] \[ y = 42 \]

Step 3: Determine the quantity of the second variety.

The quantity of the second variety of moong dal is \( y \).

From the calculation in Step 2, \( y = 42 \) kg.


Therefore, Ram Kumar mixes 42 kg of the second variety of moong dal. Quick Tip: For mixture problems involving different quantities and costs, set up a system of two linear equations: one for the total quantity and another for the total cost. Solve these equations simultaneously to find the unknown quantities.

Let the original cost price (CP) of the fruit items be Rs 100.

Step 1: Calculate the marked price (face value).

Bittu raised the price by 35%.

Marked Price (MP) = Cost Price + 35% of Cost Price

MP = \(100 + 35%\) of \(100 = 100 + 35 = 135\)

So, the Marked Price is Rs 135.


Step 2: Calculate the selling price (SP) after the discount.

Bittu gave a 35% discount on the face value (Marked Price).

Discount Amount = 35% of Marked Price

Discount Amount = \(35%\) of \(135 = \frac{35}{100} \times 135\)

Discount Amount = \(0.35 \times 135 = 47.25\)

Selling Price (SP) = Marked Price - Discount Amount

SP = \(135 - 47.25 = 87.75\)

So, the Selling Price is Rs 87.75.


Step 3: Calculate the percentage gain or loss.

Since the Selling Price (Rs 87.75) is less than the Cost Price (Rs 100),

there is a loss.

Loss Amount = Cost Price - Selling Price

Loss Amount = \(100 - 87.75 = 12.25\)


Percentage Loss = \(\frac{Loss Amount}{Cost Price} \times 100%\)

Percentage Loss = \(\frac{12.25}{100} \times 100%\)

Percentage Loss = \(12.25%\)

Therefore, there was a loss of 12.25%. Quick Tip: When an item's price is increased by \(x%\) and then decreased by \(x%\), there is always a net loss. The percentage loss is given by the formula \( \left(\frac{x}{10}\right)^2 % \).

Step 1: Understand the given information.

Total number of apples in the lot = 480.

Probability of getting a rotten apple = 0.125.


Step 2: Calculate the number of rotten apples.

The number of rotten apples = Probability of rotten apple \( \times \) Total number of apples

Number of rotten apples = \( 0.125 \times 480 \)

To make calculation easier, convert 0.125 to a fraction: \( 0.125 = \frac{125}{1000} = \frac{1}{8} \).

Number of rotten apples = \( \frac{1}{8} \times 480 \)

Number of rotten apples = \( \frac{480}{8} = 60 \)


Step 3: Calculate the number of good apples.

The number of good apples = Total number of apples - Number of rotten apples

Number of good apples = \( 480 - 60 \)

Number of good apples = \( 420 \)


Therefore, the number of good apples is 420. Quick Tip: Probability of an event occurring is \( P(E) = \frac{Number of favorable outcomes}{Total number of outcomes} \). If you know the probability and total outcomes, you can find the number of favorable outcomes. Also, \( P(Event) + P(Not Event) = 1 \).

Let the three numbers be \( N_1 = 299 \), \( N_2 = 253 \), and \( N_3 = A \).

Given:

HCF(\( N_1, N_2, A \)) = 23

LCM(\( N_1, N_2, A \)) = 16445


Step 1: Express the given numbers as multiples of their HCF.

Since the HCF of the three numbers is 23, each number must be a multiple of 23.

Let \( N_1 = 23 \times p \), \( N_2 = 23 \times q \), and \( A = 23 \times r \), where \( p, q, r \) are integers.


Calculate \( p \) and \( q \):

For \( N_1 = 299 \):
\[ 299 = 23 \times p \quad \Rightarrow \quad p = \frac{299}{23} = 13 \]
For \( N_2 = 253 \): \[ 253 = 23 \times q \quad \Rightarrow \quad q = \frac{253}{23} = 11 \]
So, the numbers are \( 23 \times 13 \), \( 23 \times 11 \), and \( 23 \times r \).

Step 2: Use the LCM property to find \( r \).

The LCM of three numbers \( k \times a, k \times b, k \times c \) is \( k \times LCM(a, b, c) \).

Here, \( k = 23 \), \( a = 13 \), \( b = 11 \), and \( c = r \).
Given LCM = 16445. \[ LCM(23 \times 13, 23 \times 11, 23 \times r) = 23 \times LCM(13, 11, r) \] \[ 16445 = 23 \times LCM(13, 11, r) \]
Divide 16445 by 23 to find \( LCM(13, 11, r) \): \[ LCM(13, 11, r) = \frac{16445}{23} = 715 \]
Now, we need to find \( r \) such that \( LCM(13, 11, r) = 715 \).
First, find the prime factorization of 715: \[ 715 = 5 \times 143 = 5 \times 11 \times 13 \]
We know that 11 and 13 are prime numbers. Since \( LCM(13, 11, r) = 5 \times 11 \times 13 \), and 11 and 13 are already factors of the first two numbers, \( r \) must contribute the factor 5.

For the HCF to be 23, \( r \) must not share any common factors with 11 or 13 (other than 1), which is true if \( r=5 \).
So, \( r = 5 \).

Step 3: Calculate the number 'A'.

We defined \( A = 23 \times r \).

Substitute the value of \( r \): \[ A = 23 \times 5 \] \[ A = 115 \]
Thus, the third number 'A' is 115. Quick Tip: For two numbers \( a \) and \( b \), \( a \times b = HCF(a,b) \times LCM(a,b) \). For three or more numbers, a direct formula doesn't exist, but if the HCF is \( k \), then each number can be written as \( k \times p, k \times q, k \times r, \dots \), and the LCM will be \( k \times LCM(p,q,r, \dots) \).

When a cone shaped toy is changed into a spherical toy using the same modelling clay, the volume of the clay remains constant. Thus, the volume of the cone is equal to the volume of the sphere.

Given:

Height of the cone, \( h = 20 \) cm

Radius of the base of the cone, \( r_{cone} = 5 \) cm


Step 1: Calculate the volume of the cone.

The formula for the volume of a cone is \( V_{cone} = \frac{1}{3}\pi r_{cone}^2 h \).

Substituting the given values: \[ V_{cone} = \frac{1}{3} \times \pi \times (5)^2 \times 20 \] \[ V_{cone} = \frac{1}{3} \times \pi \times 25 \times 20 \] \[ V_{cone} = \frac{500\pi}{3} cm^3 \]

Step 2: Equate the volume of the cone to the volume of the sphere and find the radius of the sphere.

The formula for the volume of a sphere is \( V_{sphere} = \frac{4}{3}\pi r_{sphere}^3 \), where \( r_{sphere} \) is the radius of the sphere.

Since \( V_{cone} = V_{sphere} \): \[ \frac{500\pi}{3} = \frac{4}{3}\pi r_{sphere}^3 \]
Cancel out \( \frac{\pi}{3} \) from both sides: \[ 500 = 4 r_{sphere}^3 \]
Divide by 4: \[ r_{sphere}^3 = \frac{500}{4} \] \[ r_{sphere}^3 = 125 \]
Take the cube root of both sides to find \( r_{sphere} \): \[ r_{sphere} = \sqrt[3]{125} \] \[ r_{sphere} = 5 cm \]
So, the radius of the sphere is 5 cm.

Step 3: Calculate the surface area of the sphere.

The formula for the surface area of a sphere is \( SA_{sphere} = 4\pi r_{sphere}^2 \).

Substituting the value of \( r_{sphere} = 5 \) cm: \[ SA_{sphere} = 4 \times \pi \times (5)^2 \] \[ SA_{sphere} = 4 \times \pi \times 25 \] \[ SA_{sphere} = 100\pi cm^2 \]
Using the approximate value \( \pi \approx 3.14 \) (as indicated by the options): \[ SA_{sphere} = 100 \times 3.14 \] \[ SA_{sphere} = 314 cm^2 \]
Therefore, the surface area of the sphere is 314 sqcm. Quick Tip: When a solid is melted and recast into another solid, the volume remains constant. Remember the formulas for volumes and surface areas of common 3D shapes.

Step 1: Set up the problem with a diagram and variables.

Let P be the point of observation, and its height above the lake surface is \( PE = 56 \) m.

Let the cloud be at point C, and its height above the lake surface be \( CE' = H \) m.

Let C' be the reflection of the cloud in the lake. The depth of the reflection below the lake surface will also be \( CE' = H \) m.

Draw a horizontal line from P parallel to the lake surface, intersecting the vertical line from C (and C') at point D.

So, \( PD = EE' \). Also, \( DE' = PE = 56 \) m.


Step 2: Use the angle of elevation to form an equation.

The angle of elevation from P to the cloud C is \( \angle CPD = 30^\circ \).
In the right-angled triangle \( \triangle PDC \): \[ \tan(30^\circ) = \frac{CD}{PD} \]
We know that \( CD = CE' - DE' = H - 56 \).
So, \[ \tan(30^\circ) = \frac{H - 56}{PD} \] \[ \frac{1}{\sqrt{3}} = \frac{H - 56}{PD} \quad \Rightarrow \quad PD = \sqrt{3}(H - 56) \quad \cdots (1) \]

Step 3: Use the angle of depression to form an equation.

The angle of depression from P to the reflection C' is \( \angle DPC' = 60^\circ \).

In the right-angled triangle \( \triangle PDC' \): \[ \tan(60^\circ) = \frac{DC'}{PD} \]
We know that \( DC' = DE' + E'C' = 56 + H \).
So, \[ \tan(60^\circ) = \frac{56 + H}{PD} \] \[ \sqrt{3} = \frac{56 + H}{PD} \quad \Rightarrow \quad PD = \frac{56 + H}{\sqrt{3}} \quad \cdots (2) \]

Step 4: Solve the system of equations to find the height H.

Equate the expressions for PD from (1) and (2):
\[ \sqrt{3}(H - 56) = \frac{56 + H}{\sqrt{3}} \]
Multiply both sides by \( \sqrt{3} \): \[ 3(H - 56) = 56 + H \]
Distribute 3 on the left side: \[ 3H - 168 = 56 + H \]
Bring all terms with H to one side and constants to the other: \[ 3H - H = 56 + 168 \] \[ 2H = 224 \]
Divide by 2: \[ H = \frac{224}{2} \] \[ H = 112 m \]
Thus, the height of the cloud from the surface of the lake is 112 meters. Quick Tip: For problems involving angles of elevation and depression with reflections in water, remember that the object and its reflection are equidistant from the surface of the water. Draw a horizontal line from the observation point to create right-angled triangles and use tangent ratios.

Step 1: Calculate the total number of beads in the packet.

Number of blue beads = 15

Number of yellow beads = 16

Number of orange beads = 19

Total number of beads = Number of blue beads + Number of yellow beads + Number of orange beads \[ Total number of beads = 15 + 16 + 19 = 50 \]

Step 2: Find the probability of drawing an orange bead.

Number of favorable outcomes (drawing an orange bead) = 19

Total number of possible outcomes (total beads) = 50
\[ P(orange bead) = \frac{Number of orange beads}{Total number of beads} = \frac{19}{50} \]

Step 3: Find the probability of drawing a blue or a yellow bead.

Number of favorable outcomes (drawing a blue or a yellow bead) = Number of blue beads + Number of yellow beads \[ Number of blue or yellow beads = 15 + 16 = 31 \]
Total number of possible outcomes (total beads) = 50 \[ P(blue or yellow bead) = \frac{Number of blue or yellow beads}{Total number of beads} = \frac{31}{50} \]

Thus, the probabilities are:

1. Probability of drawing an orange bead = \( \frac{19}{50} \)

2. Probability of drawing a blue or a yellow bead = \( \frac{31}{50} \)


Comparing these results with the given options, option (A) matches. Quick Tip: The probability of an event is calculated as \( \frac{Number of favorable outcomes}{Total number of possible outcomes} \). For the probability of event A OR event B (mutually exclusive events), it's \( P(A or B) = P(A) + P(B) \).

Step 1: Calculate the initial quantity of glycerin in the body lotion.
Given:
Initial volume of body lotion = 320 ml
Initial glycerin content = 60%

Quantity of glycerin = Initial glycerin content \( \times \) Initial volume \[ Quantity of glycerin = 60% \times 320 ml \] \[ Quantity of glycerin = \frac{60}{100} \times 320 \] \[ Quantity of glycerin = 0.6 \times 320 \] \[ Quantity of glycerin = 192 ml \]
So, there are 192 ml of glycerin in the original 320 ml body lotion.

Step 2: Set up an equation for the final state after adding water.
Let \( x \) ml be the quantity of water added to the body lotion.
The total volume of the new solution will be \( (320 + x) \) ml.
The glycerin content in the new solution is desired to be 50%.
Since only water is added, the actual quantity of glycerin (192 ml) remains unchanged.

In the new solution:
Quantity of glycerin = Final glycerin content \( \times \) New total volume \[ 192 = 50% \times (320 + x) \] \[ 192 = \frac{50}{100} \times (320 + x) \] \[ 192 = 0.5 \times (320 + x) \]

Step 3: Solve the equation to find the quantity of water added (\( x \)). \[ 192 = 0.5 \times (320 + x) \]
Divide both sides by 0.5: \[ \frac{192}{0.5} = 320 + x \] \[ 384 = 320 + x \]
Subtract 320 from both sides: \[ x = 384 - 320 \] \[ x = 64 ml \]
Therefore, 64 ml of water should be added to reduce the glycerin content from 60% to 50%. Quick Tip: In dilution problems where a solvent (like water) is added, the amount of the solute (the substance being diluted) remains constant. Use the formula: Initial Quantity of Solute = Final Quantity of Solute, which can be expressed as \( C_1V_1 = C_2V_2 \), where \( C \) is concentration and \( V \) is volume.

Step 1: Set up positions in a circular arrangement.

Label the five positions around the table as:

Position 1, Position 2, Position 3, Position 4, Position 5 (in clockwise order).

Step 2: Use condition 1: Chandan is third to the right of Farooq. \[ If Farooq is at Position 1 \Rightarrow Chandan is at Position 4. \]

Step 3: Use condition 2: Bidhan is second to the right of Chandan. \[ Chandan at Position 4 \Rightarrow Bidhan is at Position 1. \]

Step 4: Assign Farooq, Chandan, and Bidhan:

Position 1: Bidhan

Position 4: Chandan

So, Farooq must be at Position 5 (since Chandan is third to the right of Farooq).


Step 5: Use condition 3: Dinesh is not sitting next to Chandan.

Chandan is at Position 4 \(\Rightarrow\) Neighbors are Positions 3 and 5 \[ \Rightarrow Dinesh cannot be at Position 3 or 5 \]

Step 6: Use condition 4: Akash is not sitting next to Bidhan.

Bidhan is at Position 1 \(\Rightarrow\) Neighbors are Positions 5 and 2 \[ \Rightarrow Akash cannot be at Position 5 or 2 \]

Step 7: Fill remaining positions:

Remaining positions: 2, 3

Remaining people: Akash, Dinesh, Esha


From above:

Dinesh \(\neq\) 3 or 5 \(\Rightarrow\) Dinesh = 2

Akash \(\neq\) 1 or 2 \(\Rightarrow\) Akash = 3

Esha = 5





Step 8: Check adjacency:

Farooq (5) and Chandan (4): Adjacent

Chandan (4) and Akash (3): Adjacent

Dinesh (2) and Esha (5): Not adjacent

Bidhan (1) and Akash (3): Not adjacent


Conclusion: Only pair that is adjacent is:
\(\)
\boxed{\text{Chandan - Akash \(\) Quick Tip: For circular arrangement problems, fix one person's position first and build the rest of the arrangement step-by-step using the constraints. Always verify all conditions once an arrangement is made.

Step 1: Fix positions in a circular arrangement.

Label the six positions around the circular table clockwise as:

Position 1, Position 2, Position 3, Position 4, Position 5, Position 6.

Step 2: Use condition 1: Chandan is third to the right of Farooq.

Assume Farooq is at Position 1. Then Chandan is at Position 4 (third to the right).

Step 3: Use condition 2: Bindan is second to the right of Chandan.

Chandan is at Position 4 → Bindan is at Position 6 (second to the right).

So far:

Farooq = Position 1

Chandan = Position 4

Bindan = Position 6


Step 4: Apply condition 3: Dinesh is not sitting next to Chandan.

Chandan is at Position 4 → neighbors are Positions 3 and 5 \(\)
 ⇒Dinesh cannot be at Position 3 or 5

Step 5: Apply condition 4: Akash is not sitting next to Bindan.

Bindan is at Position 6 → neighbors are Positions 5 and 1 \(\)
⇒ Akash cannot be at Position 5 or 1 

Step 6: Assign remaining people to positions:

Remaining positions: 2, 3, 5

Remaining people: Akash, Dinesh, Esha



Dinesh \(\neq 3\) or \(5 \implies\) Dinesh \(= 2\)
Akash \(\neq 1\) or \(5 \implies\) Akash \(= 3\)
Esha \(= 5\)





Step 7: Determine Bindan's position with respect to Farooq.

Farooq is at Position 1

Bindan is at Position 6

In clockwise direction: 1 → 2 → 3 → 4 → 5 → 6

So from Farooq (1), moving counterclockwise (left), we reach Bindan (6) in one step

⇒ Bindan is 1st to the left of Farooq 

Conclusion: 1st to the left of Farooq

Quick Tip: In circular seating arrangements, fix one person first and build the rest using the constraints. Always verify all conditions once an arrangement is made.

Step 1: Fix positions in a circular arrangement.

Label the six positions around the circular table clockwise as:

Position 1, Position 2, Position 3, Position 4, Position 5, Position 6.

Step 2: Use condition 1: Chandan is third to the right of Farooq.

Assume Farooq is at Position 1. Then Chandan is at Position 4 (third to the right).

Step 3: Use condition 2: Bindan is second to the right of Chandan.

Chandan is at Position 4 → Bindan is at Position 6 (second to the right).

So far:

Farooq = Position 1

Chandan = Position 4

Bindan = Position 6


Step 4: Apply condition 3: Dinesh is not sitting next to Chandan.

Chandan is at Position 4 → neighbors are Positions 3 and 5 \(\)
 ⇒ Dinesh cannot be at Position 3 or 5 

Step 5: Apply condition 4: Akash is not sitting next to Bindan.

Bindan is at Position 6 → neighbors are Positions 5 and 1 \(\)
 ⇒ Akash cannot be at Position 5 or 1 

Step 6: Assign remaining people to positions:

Remaining positions: 2, 3, 5

Remaining people: Akash, Dinesh, Esha

From above:

Dinesh \(\neq 3\) or \(5\) \(\Rightarrow\) Dinesh \(= 2\)
Akash \(\neq 1\) or \(5\) \(\Rightarrow\) Akash \(= 3\)


Esha = 5

Final arrangement:




Step 7: Determine who is sitting between Bindan and Chandan.

Bindan is at Position 6

Chandan is at Position 4

In clockwise direction, the person between them is at Position 5

From the final arrangement, Position 5 is occupied by Esha

Conclusion: EshaQuick Tip: In circular seating arrangements, fix one person first and build the rest using the constraints. Always verify all conditions once an arrangement is made.

Step 1: Fix positions in a circular arrangement.

Label the six positions around the circular table clockwise as:

Position 1, Position 2, Position 3, Position 4, Position 5, Position 6.

Step 2: Use condition 1: Chandan is third to the right of Farooq.

Assume Farooq is at Position 1. Then Chandan is at Position 4 (third to the right).

Step 3: Use condition 2: Bidhan is second to the right of Chandan.

Chandan is at Position 4 → Bidhan is at Position 6 (second to the right).

So far:
Farooq = Position 1

Chandan = Position 4

Bidhan = Position 6


Step 4: Apply condition 3: Dinesh is not sitting next to Chandan.

Chandan is at Position 4 → neighbors are Positions 3 and 5 \(\)
 ⇒ Dinesh cannot be at Position 3 or 5 

Step 5: Apply condition 4: Akash is not sitting next to Bidhan.

Bidhan is at Position 6 → neighbors are Positions 5 and 1 \(\)
 ⇒ Akash cannot be at Position 5 or 1 

Step 6: Assign remaining people to positions:

Remaining positions: 2, 3, 5

Remaining people: Akash, Dinesh, Esha


From above:

Dinesh \( \neq 3 \) or \( 5 \) \( \Rightarrow \) Dinesh = 2

Akash \( \neq 1 \) or \( 5 \) \( \Rightarrow \) Akash = 3
Esha = 5


Final arrangement:





Step 7: Determine how many persons are seated to the left of Akash and to the right of Esha.

Akash is at Position 3

Esha is at Position 5


To find the people between Esha (Position 5) and Akash (Position 3), we count the positions moving clockwise from Esha to Akash:

From Esha (Position 5) to Akash (Position 3):

Position 6: Bidhan

Position 1: Farooq

Position 2: Dinesh


Thus, there are 3 people between Esha and Akash.

Conclusion: 3 Quick Tip: In circular seating arrangements, fix one person first and build the rest using the constraints. Always verify all conditions once an arrangement is made.

Step 1: Calculate the total marks obtained by Tahir.

From the table:

Tahir's marks in Thermodynamics = 50

Tahir's marks in Hydraulics = 60

Total marks obtained by Tahir = \(50 + 60 = 110\).


Step 2: Compare Tahir's total marks with the marks mentioned in each option.

(A) Quesh in Hydraulics:

Marks obtained by Quesh in Hydraulics = 110.

Is Tahir's total (110) more than Quesh in Hydraulics (110)? No, they are equal.

(B) Rao in Thermodynamics:

Marks obtained by Rao in Thermodynamics = 80.

Is Tahir's total (110) more than Rao in Thermodynamics (80)? Yes, \(110 > 80\). This option is correct.

(C) Sharma in Hydraulics:

Marks obtained by Sharma in Hydraulics = 120.

Is Tahir's total (110) more than Sharma in Hydraulics (120)? No, \(110 < 120\).

(D) Parikh in Hydraulics:

Marks obtained by Parikh in Hydraulics = 80.

Is Tahir's total (110) more than Parikh in Hydraulics (80)? Yes, \(110 > 80\). This option is also numerically correct based on the given information.

Given that this is a multiple-choice question format expecting a single correct answer, and both (B) and (D) satisfy the condition, there might be an ambiguity in the question or options. However, following the typical pattern of choosing one correct answer, and assuming one of the correct numerical comparisons is the intended answer, we select (B). Quick Tip: When comparing values from a table, ensure you are referencing the correct student and subject for each comparison. Pay close attention to keywords like "more than", "less than", "equal to", or "at least/most" to determine the correct inequality.

Step 1: Calculate total marks obtained by Parikh.

Parikh's marks:

Thermodynamics = 130

Hydraulics = 80
\(\)
Total for Parikh = 130 + 80 = 210 \(\)

Step 2: Calculate total marks obtained by Tahir.

Tahir's marks:

Thermodynamics = 50

Hydraulics = 60
\(\)
Total for Tahir = 50 + 60 = 110 \(\)

Step 3: Find the ratio of their total marks.
\(\)
Ratio = 210/110 = 21/11 \(\)

Now compare with given options:

(A) 3:2 → 3/2 = 1.5 

(B) 4:3 → 4/3  1.33

(C) 5:3 → 5/3  1.67

21/11 1.909  — does not match any option

Correct answer is (D) None of these Quick Tip: When calculating ratios from tabular data, always compute totals first, then simplify or approximate to match the closest given option. If none match exactly, choose "None of these."

Step 1: Calculate total Hydraulics marks for Quesh and Sharma.

Quesh = 110

Sharma = 120
\(\)
Total = 110 + 120 = 230 \(\)

Step 2: Calculate total Thermodynamics marks for Parikh and Rao.

Parikh = 130

Rao = 80
\(\)
Total = 130 + 80 = 210 \(\)

Step 3: Find the ratio.
\(\)
Ratio = 230/210 = 23/21 = 23:21 \(\)

Conclusion:(B) 23:21Quick Tip: When calculating ratios from tabular data, always compute totals first, then simplify to match the given options.

Step 1: Identify Tahir's original marks in Thermodynamics.

From the table:

Tahir's marks in Thermodynamics = 50



Step 2: Calculate 24% of Tahir's original marks.
\(24%\) of \(50 = \frac{24}{100} \times 50 = 12\)



Step 3: Add the increase to the original marks.

New marks \(= 50 + 12 = 62\)



Step 4: Calculate the new percentage with maximum marks of 140.

Percentage \(= \left(\frac{62}{140}\right) \times 100 = \frac{6200}{140} \approx 44.29%\)



Now compare with given options:
[6pt]
(A) 49% → too high

(B) 53% → too high

(C) 57% → too high

(D) 41% → closest lower value


(D) 41%Quick Tip: When calculating percentage after a mark increase, always clarify whether the increase is in actual marks or in percentage. In exams, choose the closest matching option if exact match isn't available.

Step 1: Use condition 8: F is the oldest member.
From condition 2: \(\)
 ⇒ F goes to Canada \(\)

Step 2: From condition 7: D is unmarried and niece of C. \(\)
 ⇒ C is male and D's uncle. \(\)

Step 3: From condition 5: E’s mother-in-law is the mother of C, who is the father of B and brother of H.

So:

C is male

C is father of B

C is brother of H

E is married to H (from condition 3)
\(\)
⇒ H is E’s wife
⇒ E’s mother-in-law = mother of H and C \(\)

Let’s denote this unknown person as X.

Step 4: From condition 6: A is the grandfather of one of the male members who goes to USA.

From condition 4:

B and D go to USA

B is male → A is B’s grandfather


Since C is B’s father → A is also C’s father \(\)
⇒ A is also grandfather of D (if D is C's niece) \(\)

Step 5: Determine identity of X (mother of C and H, i.e., E’s mother-in-law)

We know:

A is father of C and H

So X is their mother

Remaining females: F (oldest), H (wife of E), D (youngest)

From condition 8:

F is oldest

F goes to Canada

No direct info on gender of G or E

But from condition 5:

E’s mother-in-law = mother of C

And we’re told F is the oldest member — which fits better with being the grandmother (mother of C and H)
\(\)
⇒ F is likely the mother of C and H
⇒ F is E’s mother-in-law \(\)

Conclusion: (B) FQuick Tip: In logical reasoning questions involving family relationships and grouping logic, start by identifying all generations, genders, and pairings. Assign known values first, then deduce missing ones based on constraints.

Step 1: Use condition 8: F is the oldest member.
From condition 2: \(\)
⇒ F goes to Canada \(\)

Step 2: From condition 6: A is the grandfather of one of the male members who goes to USA.
From condition 4:
- B and D go to USA
- B is male → A is B’s grandfather

Since C is B’s father → A is also C’s father \(\)
 ⇒ A is part of the older generation
 ⇒ A can be paired with F \(\)

Step 3: Determine who goes to Canada.
From earlier:
- F is the oldest member → goes to Canada
- A is the only other person logically from the same generation who can pair with F

Thus: (A) A and FQuick Tip: In logical reasoning questions involving family relationships and grouping logic, start by identifying all generations, genders, and pairings. Assign known values first, then deduce missing ones based on constraints.

Step 1: Identify known genders from the conditions.

From condition 3: H is female → E is her husband → E is male

From condition 5: C is father of B and brother of H → C is male

From condition 6: A is grandfather of B, who goes to USA with D → B is male

From condition 7: D is niece of C → D is female

So far:

Male: E, C, B

Female: H, D

Step 2: Determine gender of remaining members: A, F, and G.

From condition 6: A is grandfather → must be male

From condition 8: F is oldest member → likely grandmother or grandfather

Since A is already grandfather, F is likely the grandmother

G's gender is not directly stated

But we have only 3 couples → 6 people in total are part of couples
So G must be single

Only one person can be left unpaired → G is female

Final list:

Male: A (grandfather), C (father), B (son), E (husband) → 4 males
Female: F (oldest), H (wife), D (niece), G (unmarried)

(C) 4Quick Tip: In logical reasoning questions involving family relationships, start by identifying all genders and relationships clearly. Use marital status and generational clues to deduce missing information.

Step 1: Identify key relationships from the conditions.

From condition 3: H is married to E → H is E’s wife

From condition 5: C is the father of B and brother of H → So H and C are siblings

From condition 7: D is niece of C
\(\)
⇒ D is child of C's sibling → D is cousin or daughter of H (since H is C's sister)
\(\)

From condition 4: B and D go to USA

From condition 6: A is grandfather of B

From condition 8: F is the oldest member

From condition 2: Oldest member goes to Canada → F goes to Canada


Step 2: Determine D’s relationship to E

We know:

E is married to H

H is sister of C

D is niece of C

→ Since H is C’s sister and D is C’s niece

→ D could be H’s daughter

→ If D is H’s daughter, then D is also E’s daughter (since E is H’s husband)


Also:

B and D go to USA

B is son of C

D is niece of C

→ Suggests both D and B are children of same generation
→ Supports that D is H’s daughter

(A) Daughter Quick Tip: In logical reasoning questions involving family relationships, always map out known relationships clearly. Use generational clues and marital links to deduce indirect relationships like parent-child or in-laws.

Step 1: Total number of items distributed = 3300

Now calculate each category:

Blankets:
\[ 24% of 3300 = 0.24 \times 3300 = 792 \]

Jackets:
\[ \frac{1}{6} \times 3300 = 550 \]

Shoes:
\[ 14% of 3300 = 0.14 \times 3300 = 462 \]

Socks and Room Heaters:

Total remaining: \[ 3300 - (792 + 550 + 462) = 3300 - 1804 = 1496 \]

Let number of socks be \(S\).

Then number of room heaters = \(S + 100\).


So: \[ S + (S + 100) = 1496 \implies 2S + 100 = 1496 \implies 2S = 1396 \implies S = 698 \]

Thus: \[ Socks = 698, \quad Blankets = 792, \quad Shoes = 462 \]

Step 2: Calculate required difference.
\[ (Socks + Blankets) - Shoes = (698 + 792) - 462 = 1490 - 462 = 1028 \]
\[ \boxed{(D) 1028} \] Quick Tip: When solving distribution problems involving percentages and unknowns, always calculate known values first, then solve for unknowns using given relationships. Match final result carefully with options provided.

From the previous problem, we have the following values:

Socks: \(S = 698\)

Shoes: \(Shoes = 462\)

Room Heaters: \(Room Heaters = S + 100 = 698 + 100 = 798\)

Now, calculate the total number of shoes and room heaters: \[ Total of Shoes and Room Heaters = Shoes + Room Heaters = 462 + 798 = 1260 \]

Next, determine what percent the number of socks is of this total: \[ Percentage = \left(\frac{Number of Socks}{Total of Shoes and Room Heaters}\right) \times 100 \]

Substitute the known values: \[ Percentage = \left(\frac{698}{1260}\right) \times 100 \]

Calculate the fraction: \[ \frac{698}{1260} \approx 0.554 \]

Convert to percentage: \[ 0.554 \times 100 \approx 55.4% \]

Rounding to the nearest whole number: \[ \Rightarrow \boxed{(C) 55} \] Quick Tip: When calculating percentages from a distribution, always ensure you are comparing the correct quantities. Use approximations carefully to match the closest given option.

From previous data:

Total items distributed = 3300

Jackets: \( \frac{1}{6} \times 3300 = 550 \)

Shoes: \( 14% \times 3300 = 0.14 \times 3300 = 462 \)

Socks and Room Heaters:

Socks = 698

Room Heaters = 698 + 100 = 798


Now calculate the total number of jackets, shoes, and room heaters: \(\)
Total = Jackets + Shoes + Room Heaters \(\) \(\)
Total = 550 + 462 + 798 = 1810 \(\)
\(\)
(A) 1810Quick Tip: In distribution problems, always cross-check totals to ensure they add up correctly across all categories. Use previously calculated values where applicable to save time.

From previous data:

Number of room heaters = \( 698 + 100 = 798 \)

Number of shoes = \( 14% of 3300 = 0.14 \times 3300 = 462 \)


Now calculate the ratio: \(\)
\text{Ratio of room heaters to shoes = \frac{798{462 \(\)

Simplify the fraction: \(\)
798/462 = 19/11 \quad \text{(after dividing both numerator and denominator by 42) \(\)

So, the required ratio is: \(\)
(A) 19:11Quick Tip: When calculating ratios from given data, always simplify the fraction to its lowest terms and match it with the closest option.