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JKSSB Social Forest Worker 2026 Previous Year Questions

Practice authentic exam questions with answers and explanations

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2026
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Reasoning 18 questions

2026 Prelims Reasoning
In a research facility, there are 122 scientists. Some specialize in Quantum Physics (Q), Molecular Biology (M), and Organic Chemistry (O), while a few have no specialization in any of these fields. There are 65 in Q, 45 in M, and 42 in O. The intersection of Q and M is 20, Q and O is 25, and M and O is 15. The number of scientists in all three fields is exactly half the number of scientists who have no specialization in any of these three fields. How many scientists are specialized only in Organic Chemistry? 


A 10
B 12
C 15
D 18
Correct Answer: Option B
Here is the complete solution for the 122 scientists problem explained without using any math symbols:

Step-by-Step Breakdown
1. Understand the Overlaps
Total scientists in the facility: 122

Scientists in Quantum Physics: 65

Scientists in Molecular Biology: 45

Scientists in Organic Chemistry: 42

Scientists in both Quantum Physics and Molecular Biology: 20

Scientists in both Quantum Physics and Organic Chemistry: 25

Scientists in both Molecular Biology and Organic Chemistry: 15

2. Find the Number of Scientists in All Three Fields
Adding the three individual groups together gives sixty-five plus forty-five plus forty-two, which equals 152.

Adding the three two-field overlaps together gives twenty plus twenty-five plus fifteen, which equals 60.

Subtracting the two-field overlaps from the individual total gives one hundred fifty-two minus sixty, which equals 92.

This value of ninety-two accounts for everyone in at least one field, except that the group in all three fields has been removed one extra time. Therefore, the total number of scientists in at least one field is ninety-two plus the number of scientists in all three fields.

Since the total number of scientists (122) is equal to those in at least one field plus those in no field:

One hundred twenty-two equals ninety-two plus the number in all three fields plus the number in no field.

The problem states that the number in no field is double the number in all three fields.

Combining these means ninety-two plus three times the number in all three fields equals one hundred twenty-two.

Subtracting ninety-two from one hundred twenty-two leaves 30.

Dividing thirty by three gives 10.

So, exactly 10 scientists specialize in all three fields.

3. Calculate Scientists Specialized ONLY in Organic Chemistry
To find those specialized solely in Organic Chemistry:

Start with the total in Organic Chemistry: 42.

Subtract those who also do Quantum Physics: forty-two minus twenty-five leaves 17.

Subtract those who also do Molecular Biology: seventeen minus fifteen leaves 2.

Add back the scientists who do all three fields (since they were subtracted twice): two plus ten equals 12.

The correct option is B) 12.
2026 Prelims Reasoning
A circle is divided into 5 equal slices. In each figure, x slices are removed, and the remaining slices form a pattern that corresponds to a number. Observe the following: 

 Figure 1:When 2 slices are removed, the remaining pattern corresponds to 211. Figure 2: When I slice is removed, the remaining pattern corresponds to 1023.

When 0 slices are removed, what number will the remaining figure represent? 
A 3125
B 7776
C 6561
D 2401
Correct Answer: Option A
The pattern is formed by taking the number of remaining slices raised to the fifth power (the total number of slices) and subtracting the number of removed slices raised to the fifth power.

Figure 1 (2 slices removed, 3 remaining):
Three to the power of five is two hundred forty-three. Two to the power of five is thirty-two. Two hundred forty-three minus thirty-two gives 211.

Figure 2 (1 slice removed, 4 remaining):
Four to the power of five is one thousand twenty-four. One to the power of five is one. One thousand twenty-four minus one gives 1023.

Target Figure (0 slices removed, 5 remaining):
Five to the power of five is three thousand one hundred twenty-five. Zero to the power of five is zero. Three thousand one hundred twenty-five minus zero gives 3125.
2026 Prelims Reasoning
Rehman travelled 32km north, 17km south and then 8km east. How far is he away from his starting point and in which direction?  
A 13km, North
B 17km, North-East
C 13km, North-East
D 17km, East
Correct Answer: Option B
The correct option is B) 17km, North-East.

Step-by-Step Breakdown
Calculate Net North or South Movement:

Travels 32 km North

Travels 17 km South

Subtracting 17 from 32 gives 15 km North

Calculate Net East or West Movement:

Travels 8 km East

Find the Shortest Distance:

The square of 15 is 225

The square of 8 is 64

Adding 225 and 64 gives 289

The square root of 289 gives 17 km

Determine the Direction:

Since he is 15 km North and 8 km East from where he started, he is in the North-East direction.
2026 Prelims Reasoning
A man walks 57 m, turns right and walks 76 m, then turns right again and walks 57 m. If at the time of sunset, his shadow is to his left, in which direction did he start? 
A South
B East
C West
D North
Correct Answer: Option A
The correct option is A) South.

Step-by-Step Breakdown
Determine Final Facing Direction:

At sunset, the sun is in the West, so all shadows fall towards the East.

His shadow is to his left, which means his left side is pointing East.

If facing East is to his left, he must be facing North at the end of his walk.

Trace the Movements Backwards:

Final Direction: Facing North after making two right turns.

Two right turns equal a 180-degree change in direction (facing the exact opposite way from where he started).

Since he ended up facing North, he must have started walking facing South.

Verification
Start: Walk 57 m South.

Turn Right: Walk 76 m West.

Turn Right: Walk 57 m North.

Final state: Facing North at sunset, with East (and his shadow) to his left.
2026 Prelims Reasoning
At 9:15 am, the minute hand of a clock faces west. Then, at 6 pm on the same day, in which direction will the hour hand point? 
A South
B West
C East
D North
Correct Answer: Option D
Step-by-Step Breakdown
Analyze the Clock's Orientation:

At 9:15 am, the minute hand points at the 3 o'clock position.

Normally, on a standard compass/clock, 3 o'clock points East.

The question states that 3 o'clock faces West (rotated 180 degrees from standard).

Determine the Directions for the Clock Positions:

3 o'clock = West (Opposite of East)

9 o'clock = East (Opposite of West)

12 o'clock = South (Opposite of North)

6 o'clock = North (Opposite of South)

Find the Hour Hand Direction at 6:00 pm:

At 6:00 pm, the hour hand points directly at the 6 o'clock position.

Based on the rotated directions above, the 6 o'clock position points North.
2026 Prelims Reasoning
A man is facing north. He turns 135 degrees in the anti-clockwise direction. Then he turned 180 degrees in a clockwise direction. In which direction is he facing now?  
A North-west
B South-east
C North-east
D South-west
Correct Answer: Option C
Step-by-Step Breakdown
Initial Direction: Facing North

First Turn (135 degrees anti-clockwise):
Turning 135 degrees anti-clockwise from North puts him facing South-west.

Second Turn (180 degrees clockwise):
Turning 180 degrees from South-west puts him directly opposite, facing North-east.

Alternative Method (Net Calculation)
Clockwise movement: 180 degrees

Anti-clockwise movement: 135 degrees

Net turn: 180 minus 135 = 45 degrees clockwise

45 degrees clockwise from North is North-east.
2026 Prelims Reasoning
Find the odd number in the set 515, 732, 1725, 346 
A 515
B 732
C 1725
D 346
Correct Answer: Option C
The odd number in the set is C) 1725.

Pattern Explanation
All other numbers in the set follow the rule of a whole number cubed plus three:

346 is seven cubed (343) plus three.

515 is eight cubed (512) plus three.

732 is nine cubed (729) plus three.

1725 breaks this pattern because twelve cubed plus three is 1731 (or twelve cubed minus three is 1725).
2026 Prelims Reasoning
Which letter cluster is the odd one out? 
A KMI
B TVR
C FHE
D MOK
Correct Answer: Option C
Here is the pattern in simple terms

A KMI
K to M moves forward two letters
M to I moves back four letters

B TVR
T to V moves forward two letters
V to R moves back four letters

C FHE
F to H moves forward two letters
H to E moves back three letters

D MOK
M to O moves forward two letters
O to K moves back four letters

All other options move forward two positions and back four positions while FHE moves forward two positions and back three positions
2026 Prelims Reasoning
Which pair is the odd one out 
A 32-18
B 63-42
C 77-42
D 95-68
Correct Answer: Option D
Reason
Calculate the difference between the two numbers in each pair:

A) 32 - 18 = 14 (Multiple of 7)

B) 63 - 42 = 21 (Multiple of 7)

C) 77 - 42 = 35 (Multiple of 7)

D) 95 - 68 = 27 (Not a multiple of 7)

In pairs A, B, and C, the difference between the numbers is a multiple of 7, whereas in pair D, the difference is 27, which is not divisible by 7.
2026 Prelims Reasoning
Find the odd one out 
A Tsunami
B Cyclone
C Nuclear explosion
D Volcano
Correct Answer: Option C
Reason
Tsunami, Cyclone, and Volcano are all natural disasters caused by natural environmental and geological processes.

Nuclear explosion is a man-made disaster resulting from human activity.
2026 Prelims Reasoning
"My father is the only son of your Father", a man said to a woman. How is the woman related to the man?
A Sister-in-law
B Paternal aunt
C Mother
D Sister
Correct Answer: Option B
Step-by-Step Breakdown
"Your father": Refers to the woman's father.

"The only son of your father": Since the woman's father has only one son, that son must be the woman's brother.

"My father is...": Substituting the above into the statement gives: "My father is your brother."

Conclusion: Since the woman is the sister of the man's father, she is his paternal aunt.
2026 Prelims Reasoning
Arpan is the son of Barkha. Barkha has a sister Charu, who has a son Dharmesh and a daughter Aditi. Dharmesh has a paternal uncle Feroz. How is Arpan related to Dharmesh? 
A Father
B Son
C Uncle
D Cousin
Correct Answer: Option D
Step-by-Step Breakdown
Mother's Side: Barkha and Charu are sisters.

Arpan's Relation: Arpan is Barkha's son.

Dharmesh's Relation: Dharmesh is Charu's son.

Conclusion: Since Arpan and Dharmesh are the children of two sisters, they are cousins.

(Note: Feroz being Dharmesh's paternal uncle is extra information about Dharmesh's father's family and does not change the relation between Arpan and Dharmesh.)
2026 Prelims Reasoning
Introducing a woman, Rohan said, "Her mother is the only daughter of my mother-in-law". How is the woman related to Rohan?
A Wife
B Niece
C Daughter
D Mother
Correct Answer: Option C
Step-by-Step Breakdown
My mother-in-law: Refers to Rohan's mother-in-law.

The only daughter of my mother-in-law: Since Rohan's mother-in-law has only one daughter, that daughter must be Rohan's wife.

Her mother is...: Substituting the result above into the statement gives: "The woman's mother is Rohan's wife."

Conclusion: Since the woman's mother is Rohan's wife, the woman is Rohan's daughter.
2026 Prelims Reasoning
Pointing to a photograph, Anisha said, "I did not have any siblings, but that woman's father is the only son of my grandfather". Whose photograph was it? 
A Her mother
B Herself
C Her cousin
D Her daughter
Correct Answer: Option B
Step-by-Step Breakdown
The only son of my grandfather: Anisha's grandfather has only one son, which means that person must be Anisha's father.

That woman's father: Substituting the phrase above into the sentence gives us: "That woman's father is Anisha's father."

No siblings: Since Anisha stated that she has no brothers or sisters, the woman whose father is Anisha's father can only be Anisha herself.
2026 Prelims Reasoning
IfRAM-16 and DAM-9, then DREAMS=? 
A 36
B 30
C 28
D 32
Correct Answer: Option B
Pattern Explanation
The code value is calculated by adding up the alphabetical position of each letter in the word, then dividing that total by two.

Example 1: RAM
R is letter 18

A is letter 1

M is letter 13

Sum of positions: 18 plus 1 plus 13 gives 32

Calculation: 32 divided by 2 gives 16

Example 2: DAM
D is letter 4

A is letter 1

M is letter 13

Sum of positions: 4 plus 1 plus 13 gives 18

Calculation: 18 divided by 2 gives 9

Word: DREAMS
D is letter 4

R is letter 18

E is letter 5

A is letter 1

M is letter 13

S is letter 19

Sum of positions: 4 plus 18 plus 5 plus 1 plus 13 plus 19 gives 60

Calculation: 60 divided by 2 gives 30
2026 Prelims Reasoning
a certain language, 'CHANGE' is written as 'BDFHIO' and 'AMEND' is written as 'BEFNO'. How will 'SHIFT" be coded in that language? 
A GIJTU
B TFHIS
C TIJGU
D FHSIT
Correct Answer: Option A
The correct answer is A) GIJTU.

Pattern Explanation
Shift every letter of the word forward by one position in the alphabet.

Arrange the resulting letters in alphabetical order.

Example 1: CHANGE

C shifts to D

H shifts to I

A shifts to B

N shifts to O

G shifts to H

E shifts to F

Arranging D, I, B, O, H, F alphabetically gives BDFHIO.

Example 2: AMEND

A shifts to B

M shifts to N

E shifts to F

N shifts to O

D shifts to E

Arranging B, N, F, O, E alphabetically gives BEFNO.

Applying to SHIFT

S shifts to T

H shifts to I

I shifts to J

F shifts to G

T shifts to U

Arranging T, I, J, G, U alphabetically gives GIJTU.
2026 Prelims Reasoning
WOMEN: EOMNW:: APPLE: ? 
A EALSP
B AELPP
C LPPEA
D EALPP
Correct Answer: Option C
To find the pattern, look at the positions of the letters in the word WOMEN:

W (Position 1)

O (Position 2)

M (Position 3)

E (Position 4)

N (Position 5)

Reordering these letters to form EOMNW uses the position sequence: 4, 2, 3, 5, 1.

Now, apply the same positional sequence (4, 2, 3, 5, 1) to the word APPLE:

A (Position 1)

P (Position 2)

P (Position 3)

L (Position 4)

E (Position 5)

Position 4 = L

Position 2 = P

Position 3 = P

Position 5 = E

Position 1 = A

Putting them together gives LPPEA.
2026 Prelims Reasoning
HUMAN: NOSE:: FROGS: ?
A NOSE
B GILLS
C SKIN
D LUNGS
Correct Answer: Option B
This analogy compares the primary organ used for breathing/gas exchange:

Human : Nose — Humans primary entry point for drawing air into the respiratory system is through the nose.

Frogs : Skin — While adult frogs have basic lungs, they rely heavily on cutaneous respiration, absorbing oxygen and releasing carbon dioxide directly through their moist skin

Maths 32 questions

2026 Prelims Maths
A right circular cylinder and a sphere have the same radius r. If their total surface areas are also exactly equal, what is the ratio of the volume of the cylinder to the volume of the sphere?
A 3:4
B 2:3
C 9:8
D 3:2
Correct Answer: Option A
Explanation
To find the ratio of their volumes, we first determine the height of the cylinder relative to its radius using their surface areas:

Equating Total Surface Areas:

The total surface area of a sphere is four multiplied by pi multiplied by the radius squared.

The total surface area of a cylinder is two multiplied by pi multiplied by the radius multiplied by the sum of its height and radius.

Setting these equal to each other and simplifying shows that the height of the cylinder must be equal to its radius.

Comparing Volumes:

The volume of the cylinder (with height equal to radius) is pi multiplied by the radius cubed.

The volume of the sphere is four-thirds multiplied by pi multiplied by the radius cubed.

Finding the Ratio:

Dividing the volume of the cylinder by the volume of the sphere gives one divided by four-thirds.

Simplifying this fraction yields three-fourths, which gives a volume ratio of 3:4.
2026 Prelims Maths
A wooden toy is shaped like a rocket with a cone mounted on a cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 8 cm. The base of the conical portion has a diameter of 12 cm, while the base diameter of the cylindrical portion is 6 cm. The conical portion is to be painted orange and the cylindrical portion yellow. Calculate the total cost of painting the orange portion in terms of a, if the cost of painting is 0.1 Rs per mm².
A 870π
B 8700π
C 2025 π
D 202.5π
Correct Answer: Option A
The correct option is A) 870π.

Explanation
To find the total cost of painting the orange portion, we first find the total surface area of the cone exposed to be painted, convert it to square millimeters, and multiply by the unit cost:

Dimensions of the Cone:

Radius of the cone base is six centimeters (half of twelve centimeters).

Height of the cone is eight centimeters.

Using the Pythagorean relationship, the slant height is ten centimeters.

Surface Area of the Orange Region:

Curved surface area of the cone equals six multiplied by ten multiplied by pi, which is sixty pi square centimeters.

Base area of the cone equals six squared multiplied by pi, which is thirty-six pi square centimeters.

Base area of the attached cylinder equals three squared multiplied by pi, which is nine pi square centimeters.

The exposed base area around the cylinder is thirty-six pi minus nine pi, which equals twenty-seven pi square centimeters.

Total exposed orange surface area is sixty pi plus twenty-seven pi, which equals eighty-seven pi square centimeters.

Converting Units and Calculating Cost:

Since one square centimeter equals one hundred square millimeters, eighty-seven pi square centimeters equals eight thousand seven hundred pi square millimeters.

Multiplying eight thousand seven hundred pi square millimeters by zero point one Rupees per square millimeter gives a total cost of eight hundred seventy pi Rupees.
2026 Prelims Maths
A wooden toy is shaped like a rocket with a cone mounted on a cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 8 cm. The base of the conical portion has a diameter of 12 cm, while the base diameter of the cylindrical portion is 6 cm. The conical portion is to be painted orange and the cylindrical portion yellow. Which of the following statements is correct regarding the orange-painted area? 
A The orange area is simply the Curved Surface Area (CSA) of the cone
B The orange area is the sum of the CSA of the cone and the area of the base of the cone
C The orange area is calculated as (CSA of cone) + (Base area of cone - Base area of cylinder)
D The orange area is calculated as (CSA of cone) + (CSA of cylinder)
Correct Answer: Option C
The correct statement is C) The orange area is calculated as (CSA of cone) + (Base area of cone - Base area of cylinder).

Explanation
To understand why option C is correct, consider which parts of the conical top are exposed to be painted orange:

Curved Surface Area (CSA) of the Cone: The entire slanted outer surface of the cone is exposed and must be painted orange.

Base of the Cone: Since the base of the cone (12 cm diameter) is wider than the top of the cylinder (6 cm diameter), the cylinder does not cover the entire bottom of the cone.

Overlapping Base Region: The base of the cylinder attaches to the center of the cone's base. The region where they overlap is hidden and attached to the cylinder, leaving a ring-shaped surface around the cylinder that is exposed on the underside of the cone.

Therefore, the total area to be painted orange equals the curved surface area of the cone plus the exposed ring at the bottom, which is calculated by subtracting the base area of the cylinder from the base area of the cone.
2026 Prelims Maths
A solid metallic sphere of radius R is melted and recast into n identical small solid spheres of radius r. Let S be the surface area of the original large sphere and s be the sum of the surface areas of all n small spheres. Identify the single correct relationship between s and S: 
A s = S
B s = n ^ (1/3) * S
C s = n ^ (2/3) * S
D s = nS
Correct Answer: Option B
To find the relationship between the surface area of the original sphere and the combined surface area of the smaller spheres, we look at volume conservation:

Volume Conservation:
The total volume remains unchanged during melting. The volume of the large sphere equals n times the volume of one small sphere. Since volume depends on the cube of the radius, the cube of the large radius R equals n multiplied by the cube of the small radius r. Taking the cube root, the large radius R equals the small radius r multiplied by n raised to the power of one-third. Therefore, the ratio of the large radius R to the small radius r is n raised to the power of one-third.

Surface Area Comparison:
Surface area depends on the square of the radius. The surface area S of the large sphere is proportional to the square of R. The surface area of a single small sphere is proportional to the square of r, so the sum of the surface areas s of all n small spheres is n multiplied by the square of r.

Ratio of Surface Areas:
Dividing the total small surface area s by the large surface area S gives n multiplied by the square of the ratio of r to R. Since R divided by r equals n raised to the power of one-third, r divided by R equals one divided by n raised to the power of one-third. Squaring this fraction gives one divided by n raised to the power of two-thirds.

Final Relation:
Multiplying n by one divided by n raised to the power of two-thirds simplifies to n raised to the power of one-third (since one minus two-thirds equals one-third).

Thus, the sum of the surface areas s equals n raised to the power of one-third multiplied by S.
2026 Prelims Maths
The LCM of two numbers is 180, and their HCF is 6. One of the numbers is 18 and the other number can be written in the form 6k What is the value of k? 
A k = 6
B k = 10
C k = 8
D k = 12
Correct Answer: Option C
To find the missing value, we use the property that the product of two numbers equals the product of their Highest Common Factor and Lowest Common Multiple.

Multiply the Highest Common Factor and the Lowest Common Multiple: six multiplied by one hundred eighty equals one thousand eighty.

The product of the two numbers must also equal one thousand eighty.

Divide one thousand eighty by the given number, eighteen, which gives sixty for the second number.

Since the second number is expressed as six times k, divide sixty by six to get ten.

Therefore, the value of k is ten.
2026 Prelims Maths
The LCM of two expressions is 3x ^ 3 + 18x ^ 2 + 30x + 12 and their HCF is x + 2 One of the expressions is x + 2 What is the sum of the coefficients (a,b), if the other expression can be represented as k(x ^ 2 + ax + b) ?  
A 3
B 4
C 5
D 6
Correct Answer: Option D
To find the other expression, divide the Lowest Common Multiple by the Highest Common Factor.

Dividing three x cubed plus eighteen x squared plus thirty x plus twelve by x plus two gives three x squared plus twelve x plus six.

Factoring out the constant three from this expression gives three times the quantity x squared plus four x plus two.

Comparing this to the form k times the quantity x squared plus a x plus b, we get:

The constant k equals three

The coefficient a equals four

The constant b equals two

The sum of the coefficients a and b is four plus two, which equals six.
2026 Prelims Maths
Identify the single correct statement regarding the properties of HCF and LCM:  
A If the product of two numbers is 2028 and their HCF is 13, there are exactly 2 such pairs of numbers possible
B The HCF of any two numbers always divides their LCM
C If the HCF of two numbers is 8, then 60 can never be their LCM
D All of the above statements are correct
Correct Answer: Option D
Statement A is correct:
If the Highest Common Factor is 13, the two numbers can be written as 13 multiplied by a and 13 multiplied by b, where a and b share no common factors. Multiplying these together gives 169 multiplied by a and b, which equals 2028. Dividing 2028 by 169 gives a multiplied by b equal to 12. The only pair of factors of 12 with no common factors are (1, 12) and (3, 4). This gives exactly two valid pairs of numbers: (13, 156) and (39, 52).

Statement B is correct:
It is a fundamental mathematical rule that the Highest Common Factor of any set of numbers always divides their Lowest Common Multiple completely without leaving a remainder.

Statement C is correct:
Since the Highest Common Factor must divide the Lowest Common Multiple, and 8 does not divide 60 evenly (60 divided by 8 leaves a remainder of 4), 60 can never be the Lowest Common Multiple for two numbers whose Highest Common Factor is 8.

Since statements A, B, and C are all true, Option D is the single correct answer.
2026 Prelims Maths
A florist is preparing for a large event and has 200 roses and 180 orchids. She wants to create the maximum possible number of identical bouquets such that every bouquet has the same number of roses and orchids, and no flowers are left over. How many roses will be present in each bouquet? 
A 10
B 20
C 9
D 18
Correct Answer: Option A
Step 1: Find the maximum number of bouquets
To make the maximum possible number of identical bouquets with no leftover flowers, we need to find the largest number that divides evenly into both two hundred and one hundred eighty.

The numbers that divide evenly into two hundred are one, two, four, five, eight, ten, twenty, twenty-five, forty, fifty, one hundred, and two hundred.

The numbers that divide evenly into one hundred eighty are one, two, three, four, five, six, nine, ten, twelve, fifteen, eighteen, twenty, thirty, forty-five, sixty, ninety, and one hundred eighty.

The largest number present in both lists is twenty.

Therefore, the florist can make a maximum of 20 identical bouquets.

Step 2: Calculate the number of roses per bouquet
The florist has two hundred roses in total.

Dividing two hundred roses equally among twenty bouquets gives 10 roses in each bouquet.

Correct Answer:
A) 10
2026 Prelims Maths
Two numbers are chosen at random from the set {1, 2, 3, 4, 5, 6) without replacement. What is the probability that their sum is prime? 
A 1/3
B 2/5
C 7/15
D 8/15
Correct Answer: Option C
Here is the step-by-step breakdown using plain text without any mathematical symbols:

Step 1: Find the total number of unique pairs
The set contains six numbers: one, two, three, four, five, and six.

Choosing any two numbers from these six gives a total of fifteen unique combinations:

Pairs starting with 1: (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) — 5 pairs

Pairs starting with 2: (2, 3), (2, 4), (2, 5), (2, 6) — 4 pairs

Pairs starting with 3: (3, 4), (3, 5), (3, 6) — 3 pairs

Pairs starting with 4: (4, 5), (4, 6) — 2 pairs

Pairs starting with 5: (5, 6) — 1 pair

Step 2: Identify pairs with a prime sum
A prime number is a number greater than one that can only be divided by one and itself. The possible prime sums we can reach are three, five, seven, and eleven.

Pairs that sum to three: (1, 2) — 1 pair

Pairs that sum to five: (1, 4) and (2, 3) — 2 pairs

Pairs that sum to seven: (1, 6), (2, 5), and (3, 4) — 3 pairs

Pairs that sum to eleven: (5, 6) — 1 pair

Adding these together gives one plus two plus three plus one, which equals seven favorable pairs.

Step 3: Calculate the probability
Favorable outcomes: 7

Total possible pairs: 15

The probability of choosing two numbers whose sum is prime is 7 out of 15.

Correct Answer:
C) 7/15
2026 Prelims Maths
A bag contains 4 red, 3 blue, and 3 green balls. Three balls are drawn without replacement. What is the probability that exactly two balls are of the same colour and one is different? 
A 18/20
B 13/45
C 30/45
D 13/20
Correct Answer: Option D
Here is the step-by-step breakdown using plain text without any mathematical symbols:

Step 1: Find the total number of ways to pick three balls
The bag contains four red, three blue, and three green balls, making a total of ten balls.

The number of ways to select any three balls out of ten without regard to order is one hundred twenty ways.

Step 2: Calculate the favorable ways for each color pair
We need three balls where exactly two share one color and the third ball is a different color.

Case 1: Two Red balls and one non-Red ball

Choosing two red balls out of four gives six ways.

Choosing one non-red ball from the remaining six balls (three blue and three green) gives six choices.

Combining these yields six times six, which equals thirty-six ways.

Case 2: Two Blue balls and one non-Blue ball

Choosing two blue balls out of three gives three ways.

Choosing one non-blue ball from the remaining seven balls (four red and three green) gives seven choices.

Combining these yields three times seven, which equals twenty-one ways.

Case 3: Two Green balls and one non-Green ball

Choosing two green balls out of three gives three ways.

Choosing one non-green ball from the remaining seven balls (four red and three blue) gives seven choices.

Combining these yields three times seven, which equals twenty-one ways.

Step 3: Calculate the total favorable ways and probability
Adding all favorable outcomes together: thirty-six plus twenty-one plus twenty-one gives seventy-eight favorable ways.

The probability is therefore seventy-eight out of one hundred twenty.

Dividing both seventy-eight and one hundred twenty by six reduces the fraction to thirteen out of twenty.

Correct Answer:
D) 13/20
2026 Prelims Maths
A die is thrown 3 times. What is the probability that the sum is even, given that at least one 6 appears and all three outcomes are not the same?
A 1/2
B 3/5
C 4/7
D 5/9
Correct Answer: Option A
Here is the step-by-step breakdown using plain text without any mathematical symbols:

Step 1: Count the total valid outcomes (the given condition)
We are looking for outcomes where at least one 6 appears, but all three dice are not identical (which excludes the combination of three 6s).

Case 1: Exactly one 6 and two non-6s

The 6 can land on the first, second, or third die (3 possible positions).

The other two dice can be any number from 1 through 5 (5 choices each).

3 positions times 5 choices times 5 choices gives 75 outcomes.

Case 2: Exactly two 6s and one non-6

The non-6 can land on the first, second, or third die (3 possible positions).

The non-6 die can be any number from 1 through 5 (5 choices).

3 positions times 5 choices gives 15 outcomes.

Adding both cases together gives 90 total possible outcomes under the given condition.

Step 2: Count the outcomes where the sum is even
Since 6 is an even number, adding a 6 to a total does not change whether the total sum is even or odd.

For Case 1 (One 6 and two non-6s):

The sum will be even if the remaining two non-6 dice sum to an even number.

Two numbers sum to an even number if both are even or both are odd.

Both even (chosen from 2 or 4): 2 choices times 2 choices gives 4 ways.

Both odd (chosen from 1, 3, or 5): 3 choices times 3 choices gives 9 ways.

Total ways for the non-6 dice to sum to even: 4 plus 9 gives 13 ways.

Multiply by the 3 positions for the 6: 3 times 13 gives 39 favorable outcomes.

For Case 2 (Two 6s and one non-6):

Two 6s sum to twelve (an even number), so the remaining single non-6 die must be even.

The remaining die must be chosen from 2 or 4 (2 choices).

Multiply by the 3 positions for the non-6 die: 3 times 2 gives 6 favorable outcomes.

Adding the favorable outcomes from both cases gives 39 plus 6, which equals 45 favorable outcomes.

Step 3: Calculate the probability
Favorable outcomes: 45

Total given outcomes: 90

Probability: 45 out of 90, which simplifies to 1 out of 2 (one half).

Correct Answer:
A) 1/2
2026 Prelims Maths
A coin is tossed 5 times. Each head contributes +1 and each tail contributes-1. What is the probability that the total sum is exactly +1? 
A 5/16
B 5/32
C 10/16
D 2/3
Correct Answer: Option A
Here is the step-by-step breakdown without using any math symbols:

Step 1: Find the total number of outcomes
Each time a coin is tossed, there are two possible outcomes: heads or tails.

For five tosses, the total number of possible sequences is calculated by doubling the possibilities for each additional toss: two, four, eight, sixteen, thirty-two.

Thus, there are 32 total possible outcomes.

Step 2: Determine the required heads and tails
Each head adds one point, and each tail subtracts one point.

To reach a final score of positive one across five tosses, you must get three heads and two tails.

Three heads contribute three points, and two tails take away two points, leaving a final score of positive one.

Step 3: Count the successful combinations
There are 10 different ways to land three heads and two tails in five tosses:

Head, Head, Head, Tail, Tail

Head, Head, Tail, Head, Tail

Head, Head, Tail, Tail, Head

Head, Tail, Head, Head, Tail

Head, Tail, Head, Tail, Head

Head, Tail, Tail, Head, Head

Tail, Head, Head, Head, Tail

Tail, Head, Head, Tail, Head

Tail, Head, Tail, Head, Head

Tail, Tail, Head, Head, Head

Step 4: Calculate the probability
There are 10 favorable outcomes out of 32 total possibilities, making the probability 10 out of 32.

Reducing this fraction by halving both numbers gives 5 out of 16.

Correct Answer:
A) 5/16
2026 Prelims Maths
The present age of a man is equal to the square of his son's age. One year ago, the man's age was exactly eight times the age of his son. Two years hence, the sum of their ages will be 6 more than six times the son's age at that time. If the son's present age is 'x', which of the following represents the man's present age? 
A 25 years
B 36 years
C 49 years
D 64 years
Correct Answer: Option C
Step 1: Set up their present ages
The son's present age is given as x.

The man's present age is the square of the son's age, which means x times x.

Step 2: Use the condition from one year ago
One year ago, the son was x minus 1 years old.

One year ago, the man was x times x, minus 1 years old.

The problem states that one year ago, the man was eight times as old as his son:

Eight times the son's past age is eight times x, minus 8.

Setting this equal to the man's past age gives: x times x, minus 1 equals eight times x, minus 8.

Rearranging this gives: x times x, minus eight times x, plus 7 equals zero.

Finding the number that satisfies this relationship:

If x is 7, then 7 times 7 is 49.

49 minus 56 (which is eight times 7) plus 7 equals zero.

Therefore, the son's present age x is 7 years old.

Step 3: Calculate the man's present age
Since the son is 7 years old, the man's present age is 7 times 7, which equals 49 years old.

Step 4: Verify with the future condition
Two years from now, the son will be 9 years old (7 plus 2).

Two years from now, the man will be 51 years old (49 plus 2).

The sum of their future ages will be 60 (51 plus 9).

Six times the son's future age is 54 (six times 9).

Six more than 54 is indeed 60, which confirms the answer is correct.

Correct Answer:
C) 49 years
2026 Prelims Maths
An HR department is auditing the ages of its 5-member executive board A, B, C, D, and E, to comply with a "balanced experience" policy. The average age of the board members is 25 years. Member D (age 30) quits and is replaced by Member F. The new average age becomes 24 years. Additionally, Member B is 5 years younger than Member C. Member A is 2 years older than Member F. Member E and Member B have same age. What is the current age of Member E? 
A 20 years
B 21 years
C 25 years
D 27 years
Correct Answer: Option B
Step 1: Find Member F's age
For five members to average 25 years, their total age combined must be 125 years.

Since Member D is 30 years old, the remaining four members (A, B, C, and E) have a combined total age of 95 years (125 minus 30).

When Member F replaces Member D, the new average age for the five members becomes 24 years, making the new total age 120 years.

Since A, B, C, and E account for 95 years of that total, Member F must be 25 years old (120 minus 95).

Step 2: Find Member A's age
Member A is two years older than Member F.

Since Member F is 25 years old, Member A is 27 years old.

Step 3: Find Member E's age
Now we look at the new board total of 120 years for members A, B, C, E, and F:

Members A and F together contribute 52 years (27 plus 25).

Subtracting 52 from the overall total of 120 leaves 68 years split among members B, C, and E.

Member C is five years older than Member B. Taking away those extra 5 years leaves 63 years.

Member E and Member B are the same age, meaning B, E, and the base portion of C are three equal shares.

Dividing 63 years into three equal parts gives 21 years for each share.

Therefore, Member E is 21 years old.

Correct Answer:
B) 21 years
2026 Prelims Maths
Six years ago, the ratio of the ages of Kunal and Sagar was 6:5. Four years hence, the ratio of their ages will be 11:10. Based on this information, identify the single correct statement from the list below: 
A Sagar is currently 18 years old
B Sum of their current ages is 32 years
C Six years ago, Kunal was 12 years old
D 2 years ago, the ratio of their ages was 5:4
Correct Answer: Option C
Step 1: Set up their ages
Six years ago:

Kunal's age was 6 parts.

Sagar's age was 5 parts.

Present day (6 years later):

Kunal is 6 parts plus 6 years.

Sagar is 5 parts plus 6 years.

Four years from now:

Kunal will be 6 parts plus 10 years.

Sagar will be 5 parts plus 10 years.

Step 2: Find the value of 1 part
Four years from now, their ages will be in a 11 to 10 ratio.

10 times Kunal's future age equals 11 times Sagar's future age.

10 times (6 parts plus 10 years) gives 60 parts plus 100 years.

11 times (5 parts plus 10 years) gives 55 parts plus 110 years.

Comparing both sides:

60 parts plus 100 equals 55 parts plus 110.

The difference of 5 parts equals 10 years.

Divide 10 by 5 to find that 1 part equals 2 years.

Step 3: Calculate their current ages
Kunal's present age: 6 parts plus 6 years = 12 plus 6 = 18 years old.

Sagar's present age: 5 parts plus 6 years = 10 plus 6 = 16 years old.

Step 4: Check the options
A) Sagar is currently 18 years old — Incorrect (he is 16).

B) Sum of their current ages is 32 years — Incorrect (18 plus 16 is 34).

C) Six years ago, Kunal was 12 years old — Correct (18 minus 6 is 12).

D) Two years ago, the ratio of their ages was 5 to 4 — Incorrect (16 to 14 reduces to 8 to 7).

Correct Answer:
C) Six years ago, Kunal was 12 years old
2026 Prelims Maths
Match the scenarios in List-I with the correct "Present Age of the Older Person" in List-II.

List-I 
 P. The sum of the ages of a father and son is 45 years. Five years ago, the product of their ages was four times the father's age at that time. 
Q. A man is 24 years older than his son. In two years, his age will be twice the age of his son.
R. The ratio of the ages of A and B is 3:4. The sum of their ages is 28. 
S. 10 years ago, a father was 4 times as old as his son. 2 years hence, he will be twice as old as his son.  

List-II 
1) 16years
2) 36years 
3) 46years 
4) 34 years

Choose the correct match: 
A P-1, Q-2, R-3, S-4
B P-2, Q-1, R-3, S-4
C P-2, Q-3, R-1, S-4
D P-3, Q-2, R-4, S-1
Correct Answer: Option C
Here is the step-by-step breakdown without using math symbols:

1. Scenario P
The total sum of the father and son's present ages is 45 years.

Five years ago, the product of their ages was four times the father's age at that time.

This means the son's age five years ago was 4 years old.

Adding 5 years gives the son's present age as 9 years old.

Subtracting 9 from 45 gives the father's present age as 36 years (matches 2).

2. Scenario Q
A man is 24 years older than his son.

In two years, the difference in their ages remains 24 years, but the father will be twice as old as the son.

For the father to be twice as old with a difference of 24 years, the son must be 24 years old in two years, and the father must be 48 years old.

Subtracting 2 years gives the father's present age as 46 years (matches 3).

3. Scenario R
The ratio of the ages of A and B is 3 to 4, which totals 7 equal parts.

Dividing the total sum of 28 by 7 gives 4 years per part.

The older person has 4 parts, so multiplying 4 by 4 gives 16 years (matches 1).

4. Scenario S
Let the son's age ten years ago be 1 part, making the father's age 4 parts.

Twelve years later (two years from now), the father is twice as old as the son.

Working through the age gap reveals that ten years ago the son was 6 and the father was 24.

Adding 10 years gives the father's present age as 34 years (matches 4).

Conclusion
The correct matching sequence is P-2, Q-3, R-1, S-4.
2026 Prelims Maths
If x/ay/bz/ck, then find the value of: (bx + cy + az) / k
B abc
C a + b + c
D ab + bc + ac
Correct Answer: Option D
Here is the step-by-step breakdown without using math symbols:

1. Express x, y, and z in Terms of k
From the given equal ratios, x over a equals k, which means x equals a times k.

Similarly, y over b equals k, which means y equals b times k.

Likewise, z over c equals k, which means z equals c times k.

2. Substitute the Values into the Expression
The term b times x becomes b times a times k, or a times b times k.

The term c times y becomes c times b times k, or b times c times k.

The term a times z becomes a times c times k, or a times c times k.

Adding these three terms together gives k times the quantity a times b plus b times c plus a times c.

3. Divide by k
Dividing the combined total by k cancels out k.

This leaves a times b plus b times c plus a times c.

Conclusion
Correct Answer: D) ab + bc + ac
2026 Prelims Maths
A high-end jeweller manufactures gold bars where the price of a bar is directly proportional to the square of its weight. During transit, a bar of weight 5 units and worth Rs. 25,000 accidentally breaks into two pieces with weights in the ratio 2:3. What is the total reduction in the value of the gold due to this breakage? 
A Rs. 5,000
B Rs. 12,000
C Rs. 13,000
D Rs. 14,000
Correct Answer: Option B
Here is the step-by-step solution without using math symbols:

1. Determine the Value Per Weight Unit
The price of a gold bar is proportional to the square of its weight.

For the original bar of weight 5 units, the square of its weight is 25.

Given that 25 squared units of weight equal 25,000 rupees, each squared unit of weight is worth 1,000 rupees (25,000 divided by 25).

2. Calculate the Value of the Broken Pieces
The bar breaks into two pieces with weights of 2 units and 3 units.

For the first piece of weight 2 units:

Squaring the weight gives 4.

Multiplying 4 by 1,000 rupees per unit gives a value of 4,000 rupees.

For the second piece of weight 3 units:

Squaring the weight gives 9.

Multiplying 9 by 1,000 rupees per unit gives a value of 9,000 rupees.

The total value of both pieces combined is 13,000 rupees (4,000 plus 9,000).

3. Calculate the Reduction in Value
Subtracting the combined broken value of 13,000 rupees from the original value of 25,000 rupees gives a total loss of 12,000 rupees.

Conclusion
Correct Answer: B) Rs. 12,000
2026 Prelims Maths
If a:bc:d, which of the following is correct using Componendo and Dividendo? 
A (a + b) / (a - b) = (c + d) / (c - d)
B (a - b) / (a + b) = (c + d) / (c - d)
C (a + c) / (b + d) = (a - c) / (b - d)
D (a + b) / c = (b + d) / a
Correct Answer: Option A
1. Understand Componendo and Dividendo
If four quantities form a proportion such that a is to b as c is to d, then the ratio of the sum of the first two terms to their difference equals the ratio of the sum of the last two terms to their difference.

2. Apply the Rule
Take the left side: a plus b divided by a minus b.

Take the right side: c plus d divided by c minus d.

Equating both sides gives a plus b over a minus b equals c plus d over c minus d.

Conclusion
Correct Answer: A) (a + b) / (a - b) = (c + d) / (c - d)
2026 Prelims Maths
Match the mathematical descriptions (List-A) with their corresponding ratios (List-B):

List-A  
i. Sub-duplicate ratio of 25:36 
ii. Duplicate ratio of 3:4 
iii. Reciprocal ratio of 5:7 
iv. Compound ratio of 2:3 and 9:4  

List -B 
a. 9:16 
b. 3:2 
c. 5:6 
d. 7:5

 Choose the correct match:  
A i-d, ii-a, iii-b, iv-c
B i-b, ii-c, iii-a, iv-d
C i-c, ii-d, iii-a, iv-b
D i-c, ii-a, iii-d, iv-b
Correct Answer: Option D
1. Sub-duplicate ratio of 25 to 36
The sub-duplicate ratio means taking the square root of each number.

The square root of 25 is 5, and the square root of 36 is 6.

This gives the ratio 5 to 6 (matches c).

2. Duplicate ratio of 3 to 4
The duplicate ratio means squaring each number.

3 squared is 9, and 4 squared is 16.

This gives the ratio 9 to 16 (matches a).

3. Reciprocal ratio of 5 to 7
The reciprocal ratio means reversing the position of the terms.

Inverting 5 to 7 gives 7 to 5 (matches d).

4. Compound ratio of 2 to 3 and 9 to 4
The compound ratio is found by multiplying the first terms together and the second terms together.

Multiplying 2 by 9 gives 18, and multiplying 3 by 4 gives 12.

Reducing 18 to 12 gives 3 to 2 (matches b).

Conclusion
The correct matching sequence is i-c, ii-a, iii-d, iv-b.

Correct Answer: D) i-c, ii-a, iii-d, iv-b
2026 Prelims Maths
A train covers a distance S at a speed U. If its speed is increased by 10km / h the travel time reduces by 2 hours, and if increased by 20km / h the time reduces by 3 hours. Find the distance S. 
A 60 km
B 90 km
C 120 km
D 150 km
Correct Answer: Option C
1. Set up the Relationships
Let the original speed be U kilometers per hour and the original time be T hours.

The total distance S is calculated as speed multiplied by time.

For the first scenario, increasing the speed by 10 kilometers per hour decreases the travel time by 2 hours. Multiplying this new speed by the new time gives the distance S, which simplifies to: 10 times T minus 2 times U equals 20.

For the second scenario, increasing the speed by 20 kilometers per hour decreases the travel time by 3 hours. Multiplying this new speed by the new time gives the distance S, which simplifies to: 20 times T minus 3 times U equals 60.

2. Solve for Time and Speed
From the first scenario, expressing U in terms of T gives: U equals 5 times T minus 10.

Substituting this expression for U into the second scenario equation gives: 20 times T minus 3 times the quantity 5 times T minus 10 equals 60.

Expanding and combining like terms yields: 5 times T plus 30 equals 60, which means 5 times T equals 30.

Dividing 30 by 5 reveals the original time T is 6 hours.

Plugging 6 back into the speed formula gives U equals 5 times 6 minus 10, which means the original speed U is 20 kilometers per hour.

3. Calculate Distance S
The total distance S is the original speed multiplied by the original time.

Multiplying 20 kilometers per hour by 6 hours yields a total distance of 120 kilometers.

Conclusion
Correct Answer: C) 120 km
2026 Prelims Maths
The speed of a car travelling from city A to city B was recorded as 60km / h and the journey time was noted as 5 hours. However, the actual speed was 50 km/h. Calculate the real time taken for the trip.  
A 4.5 hours
B 6 hours
C 5.5 hours
D 5 hours
Correct Answer: Option B
1. Calculate the Total Distance
The recorded speed was 60 kilometers per hour.

The recorded journey time was 5 hours.

Multiplying the speed of 60 by the time of 5 gives a total distance of 300 kilometers.

2. Calculate the Real Time Taken
The actual speed was 50 kilometers per hour.

The distance to cover remains 300 kilometers.

Dividing the total distance of 300 by the real speed of 50 gives 6 hours.

Conclusion
The real time taken for the trip is 6 hours.
2026 Prelims Maths
A contractor assigns a task to two workers, A and B. Worker A can complete the work in x days, while worker B takes 5 days more than A. Working together, they can finish the task in 6 days. Find the value of x.
A 12
B 8
C 10
D 6
Correct Answer: Option C
1. Define Daily Work Rates
Worker A completes the task in x days, which means worker A completes 1 over x of the task per day.

Worker B takes 5 days longer, which is x plus 5 days, meaning worker B completes 1 over x plus 5 of the task per day.

Working together, they finish the task in 6 days, making their combined daily work rate 1 over 6.

2. Combine Rates and Simplify
Adding their individual daily rates together gives 1 over x plus 1 over x plus 5, which equals 1 over 6.

Combining the left side over a common denominator gives two x plus 5 over x squared plus 5 x, equal to 1 over 6.

Cross-multiplying to eliminate the fractions gives 6 times two x plus 5, which equals x squared plus 5 x.

Expanding the left side gives 12 x plus 30 equals x squared plus 5 x.

Moving all terms to one side gives the quadratic equation: x squared minus 7 x minus 30 equals 0.

3. Solve for x
Factoring this equation yields x minus 10 times x plus 3 equals 0.

This provides two possible values for x: 10 or negative 3.

Since time in days cannot be negative, x must be 10.

Conclusion
Correct Answer: C) 10
2026 Prelims Maths
A, B, and C can complete a complex engineering project in 25 days working together. The ratio of their individual efficiencies is 4:3: 5 respectively. The project begins with all three working together. However, the following events occur: After 5 days, A leaves the project. To compensate for A's departure, B increases his efficiency 33.3%, while C continues to work at his original efficiency. C leaves d days before the completion of project. The entire project is completed in exactly 35 days. What is the value of d? 
A 4
B 6
C 5
D 8
Correct Answer: Option C
1. Calculate the Total Work Required
The daily efficiencies of A, B, and C are 4, 3, and 5 units per day.

Working together, their combined daily output is 12 units (4 plus 3 plus 5).

Since they can complete the project in 25 days, the total project size is 300 units (12 units per day multiplied by 25 days).

2. Work Completed in the First 5 Days
For the first 5 days, all three work together at their combined rate of 12 units per day.

Work completed in these 5 days is 60 units (5 multiplied by 12).

Remaining work to finish is 240 units (300 minus 60).

Remaining time to finish the project is 30 days (35 total days minus the 5 days already spent).

3. Determine Output for the Remaining 30 Days
A leaves the project.

B increases efficiency by one-third (33.3 percent), raising B's daily output from 3 to 4 units.

C remains at an efficiency of 5 units per day.

B stays for the entire remaining 30 days, completing 120 units of work (30 days multiplied by 4 units per day).

4. Calculate Days Worked by C and the Value of d
The work left for C to complete during these 30 days is 120 units (240 remaining units minus B's 120 units).

Since C contributes 5 units per day, C needs 24 days to complete 120 units (120 divided by 5).

Since the second phase lasts 30 days and C only works for 24 days, C leaves 6 days before the project ends (30 minus 24).

Conclusion
The value of d is 6.
2026 Prelims Maths
The average score of a group of n students in a mathematics competition is 60. A series of adjustments is made to the data. The top 25% of students (by score) are removed from the data set. Their average score was 80. A new group of 10 students joins the remaining group. This new group has an average score of 40. After these changes, it was discovered that a score belonging to one of the original remaining students was misrecorded as 100 instead of 80. After all departures, additions, and corrections, the final average score of the group is exactly 50. What was the original number of students (n)? 
A 36
B 40
C 44
D 48
Correct Answer: Option D
Step-by-Step Breakdown
1. Initial State
Let the original number of students be n.

The original average score is 60.

The original combined score is 60 times n.

2. Removing the Top Quarter
The top 25 percent means one-fourth of the total students, or 0.25 times n.

Their average score was 80, making their combined score 80 times 0.25 times n, which equals 20 times n.

Removing these students leaves three-fourths of the students, or 0.75 times n.

The combined score of the remaining students is 60 times n minus 20 times n, which equals 40 times n.

3. Adding New Students and Correcting the Error
A new group of 10 students with an average score of 40 joins, adding 400 total marks (10 times 40).

The total number of students becomes 0.75 times n plus 10.

A misrecorded score of 100 instead of 80 is corrected, which reduces the total score by 20 marks.

The final total score is 40 times n plus 380 (400 minus 20).

4. Solving for the Original Number of Students
The final average score of 50 means the total score equals 50 times the new group size (0.75 times n plus 10).

Multiplying 50 by 0.75 times n plus 10 gives 37.5 times n plus 500.

Equating the two total scores: 40 times n plus 380 equals 37.5 times n plus 500.

Subtracting 37.5 times n from 40 times n leaves 2.5 times n.

Subtracting 380 from 500 leaves 120.

Therefore, 2.5 times n equals 120.

Dividing 120 by 2.5 yields 48 original students.
2026 Prelims Maths
The average score of 20 students in a class is 75. It is later found that 5 students had been absent. In a re-examination, 4 of them scored 85 each. After including the marks of all 5 students, the overall average becomes 77. Determine the marks scored by the fifth student. 
A 80
B 87
C 85
D 89
Correct Answer: Option B
Step-by-Step Breakdown
1. Calculate Initial Total Marks
The average score for the initial 20 students is 75.

Multiplying 20 students by 75 gives a total of 1,500 marks.

2. Calculate New Total Marks for All Students
Including the 5 absent students brings the total number of students to 25 (20 plus 5).

The new overall average for all 25 students is 77.

Multiplying 25 students by 77 gives a new total of 1,925 marks.

3. Find the Total Marks Scored by the 5 Students
Subtracting the initial 1,500 marks from the new total of 1,925 marks leaves 425 marks scored by the 5 additional students.

4. Calculate the Fifth Student's Score
4 of the students scored 85 marks each, which equals 340 marks in total (4 multiplied by 85).

Subtracting 340 marks from the 425 total marks scored by the group leaves 85 marks for the fifth student.
2026 Prelims Maths
The average of 12 students in a class is 68. Later, it was found that one student's score was incorrectly recorded as 72 instead of 96. What is the correct average of the class? 
A 70
B 71.33
C 72.5
D 73.4
Correct Answer: Option A
Step-by-Step Breakdown
1. Calculate the Difference in Score
The score was recorded as 72 instead of 96.

The actual score is higher by 24 marks (96 minus 72).

2. Distribute the Extra Marks
These extra 24 marks need to be shared equally among all 12 students.

Dividing 24 marks by 12 students increases each student's share by 2 marks.

3. Calculate the Correct Average
Adding the 2 additional marks to the original average of 68 gives the correct average of 70.
2026 Prelims Maths
The average score of 10 students in a class is 72. One student scored 90 marks, and another scored 30 marks less than the first. If these two students are removed, what is the new average of the remaining students? 
A 71.2
B 72.5
C 73.4
D 74.6
Correct Answer: Option A
Step-by-Step Breakdown
1. Calculate the Total Score of All Students
The average score of 10 students is 72.

Multiplying 10 students by 72 gives a total combined score of 720 marks.

2. Find the Scores of the Two Removed Students
The first student scored 90 marks.

The second student scored 30 marks less than the first student, which gives 60 marks (90 minus 30).

Combined, these two students scored 150 marks (90 plus 60).

3. Calculate the Total Score of Remaining Students
Removing 150 marks from the total of 720 marks leaves 570 marks (720 minus 150).

Removing two students leaves 8 students remaining (10 minus 2).

4. Calculate the New Average
Dividing the remaining total of 570 marks by 8 students gives a new average of 71.25 marks (approximately 71.2).
2026 Prelims Maths
A storekeeper gives 20% discount on an item and then adds 12% sales tax on the discounted price. If the original price of the item is Rs 500, which of the following statements is INCORRECT? 
A The discounted price is Rs. 400.
B The sales tax on the discounted price is Rs. 48.
C The final price the customer pays is Rs. 450.
D The final price is less than the original price.
Correct Answer: Option C
Step-by-Step Breakdown
1. Calculating the Discounted Price
The original price is Rupees 500.

A 20 percent discount on Rupees 500 equals Rupees 100.

Subtracting Rupees 100 from Rupees 500 leaves a discounted price of Rupees 400.

Statement A is correct.

2. Calculating the Sales Tax
A 12 percent sales tax is calculated on the discounted price of Rupees 400.

12 percent of Rupees 400 equals Rupees 48.

Statement B is correct.

3. Calculating the Final Price
Adding the tax of Rupees 48 to the discounted price of Rupees 400 gives a final price of Rupees 448.

Statement C claims the final price is Rupees 450, which is wrong.

Statement C is INCORRECT.

4. Comparing Final and Original Price
The final price of Rupees 448 is less than the original price of Rupees 500.

Statement D is correct.
2026 Prelims Maths
A gadget was sold for Rs. 522 after a 10% discount and then a 10% sales tax was applied on the discounted price. What was the original price of the gadget? (Round to nearest ten)  
A Rs. 495
B Rs. 539
C Rs. 527
D Rs. 548
Correct Answer: Option B
Assume the original price of the gadget is 100 units.

1. Applying the Discount
A 10 percent discount reduces the price by 10 units.

The discounted price becomes 90 units (100 units minus 10 units).

2. Applying the Sales Tax
A 10 percent sales tax is calculated on the discounted price of 90 units.

10 percent of 90 units is 9 units.

Adding the tax gives a final selling price of 99 units (90 units plus 9 units).

3. Finding the Original Price
The final selling price is given as Rupees 522, which corresponds to 99 units.

Dividing Rupees 522 by 99 units gives approximately Rupees 5.2727 per unit.

Multiplying 100 units by Rupees 5.2727 gives an original price of approximately Rupees 527.27.

Rounding to the nearest whole rupee gives Rupees 527.
2026 Prelims Maths
A shopkeeper marks the price of an item 40% above the cost price. He then gives two successive discounts of 10% and 5% on the marked price. If the shopkeeper makes a profit of Rs.492.5, find the cost price of the item.
A Rs.2,000
B Rs.2,100
C Rs.2,500
D Rs.2,600
Correct Answer: Option C
Assume the cost price of the item is 100 units.

1. Marked Price
The marked price is 40 percent above the cost price.

40 percent of 100 units is 40 units.

The marked price is 140 units.

2. Discounts
First discount: 10 percent on 140 units is 14 units.

Price after the first discount is 140 units minus 14 units, which equals 126 units.

Second discount: 5 percent on 126 units is 6.3 units.

The final selling price is 126 units minus 6.3 units, which equals 119.7 units.

3. Profit in Units
Profit equals selling price minus cost price.

119.7 units minus 100 units gives a profit of 19.7 units.

4. Calculate the Cost Price
The actual profit is given as Rupees 492.5.

Dividing Rupees 492.5 by 19.7 units gives Rupees 25 per unit.

Multiplying 100 units by Rupees 25 gives the cost price of Rupees 2,500.
2026 Prelims Maths
In a national competitive examination, 35% of the total candidates were girls. The results revealed the following: 75% of the boys passed the exam. 80% of the girls passed the exam. Among all the candidates who passed, 20% were awarded a merit scholarship. Among all the candidates who failed, 10% were eligible to apply for secondary "recovery" test. If the number of candidates who were awarded a scholarship exceeds the number of candidates eligible for the recovery test by 2,605, what was the total number of candidates who appeared for the exam? 
A 15,000
B 18,500
C 20,000
D 22,500
Correct Answer: Option C
The correct answer is C) 20,000.

Step-by-Step Breakdown
Assume the total number of candidates is 100 units.

1. Candidate Demographics
Girls: 35 units out of 100

Boys: 65 units out of 100

2. Passed and Failed Candidates
Passed Boys: 75 percent of 65 units is 48.75 units

Passed Girls: 80 percent of 35 units is 28 units

Total Passed: 48.75 units plus 28 units is 76.75 units

Total Failed: 100 units minus 76.75 units leaves 23.25 units

3. Scholarship and Recovery Test Candidates
Merit Scholarship: 20 percent of the 76.75 passed units is 15.35 units

Recovery Test Eligible: 10 percent of the 23.25 failed units is 2.325 units

4. Finding the Value of One Unit
Subtracting the recovery test units from the scholarship units gives 13.025 units (15.35 minus 2.325).

This difference represents 2,605 candidates.

Dividing 2,605 by 13.025 gives 200 candidates per unit.

5. Calculate Total Candidates
Multiplying 100 units by 200 candidates per unit gives 20,000 total candidates.

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