XAT 2026 Question Paper

Let the number of correct and wrong questions be x and y respectively. 

Number of unattempted questions = 20 - (x+y) $$\longrightarrow\ i$$

4 marks are awarded for every correct questions and 2 and 1 marks are deducted for every wrong and unattempted questions respectively. 

Total original marks scored = 4x-2y-((20-(x+y)) = 5x-y-20 $$\longrightarrow\ ii$$

But the teacher deducted 1 and 2 marks for every wrong and unattempted questions respectively by mistake. 

Total marks = 4x-y-2(20-x-y) = 6x+y-40

Total marks = 46 (given)

6x+y-40 = 46

6x+y = 86 

The possible pairs of (x,y) are (14,2), (13,8), (12,14), ........ , (1,80), (0,86). 

x+y = 16, 21, 26, ........ , 81,86

Only one case will be possible as (x+y) $$\le\ $$ 20 (as equation i $$\ge\ 0$$)

Hence, Number of correct questions = x = 14

Number of wrong questions = y = 2 

Number of unattempted questions = 20 - (14+2) = 4

Hence, final marks = $$\left(5\times\ 14\right)-2-20$$ = 48 (from equation ii) 

$$\therefore\ $$ The required answer is C.

Given : f(0,y) = y+1 and f(x+1,y) = f(x,f(x,y)) + x

To find : f(2,2)

f(x+1,y) = f(x,f(x,y)) + x

Let x=0

f(1,y) = f(0,f(0,y))

f(1,y) = f(0,y+1)  (as f(0,y) = y+1 (given))

f(1,y) = (y+1)+1 = y+2

Hence, f(1,y) = y+2 $$\longrightarrow\ i$$

Now, 

f(x+1,y) = f(x,f(x,y)) + x

Let x=1

f(2,y) = f(1,f(1,y)) + 1

f(2,y) = f(1,y+2) + 1 (from equation i)

f(2,y) = (y+2) + 2 + 1 = y+5

Put y=2

f(2,2) = 2+5 =7

$$\therefore\ $$ The required answer is B.

List of all prime numbers less than 25 = 2,3,5,7,11,13,17,19,23 = 9 numbers

$$a_1$$ = 2, $$a_2$$ = 3, $$a_3$$ = 5, $$a_4$$ = 7, $$a_5$$ = 11, $$a_6$$ = 13, $$a_7$$ = 17, $$a_8$$ = 19, $$a_9$$ = 23

Sum of all the above prime numbers ($$a_1$$ + $$a_2$$ + ...... + $$a_9$$) = 100

$$X_i=\ \frac{\ b_i}{a_i}$$ ,where, $$b_i$$ = Sum of all prime numbers $$a_1\ to\ a_{n\ }$$ except $$a_i$$

Example : $$b_3$$ is the sum of all the prime numbers $$a_1$$ to $$a_9$$ except $$a_3$$. 

$$X_1$$ = $$\ \frac{\ b_1}{a_1}$$ = $$\ \frac{\ 100-a_1}{a_1}$$ = $$\ \frac{\ 100-2}{2}$$ = 49

$$X_2$$ = $$\ \frac{\ b_2}{a_2}$$ = $$\ \frac{\ 100-a_2}{a_2}$$ = $$\ \frac{\ 100-3}{3}$$ = $$\ \frac{\ 97}{3}$$

$$X_3$$ = $$\ \frac{\ b_3}{a_3}$$ = $$\ \frac{\ 100-a_3}{a_3}$$ = $$\ \frac{\ 100-5}{5}$$ = 19

$$X_4$$ = $$\ \frac{\ b_4}{a_4}$$ = $$\ \frac{\ 100-a_4}{a_4}$$ = $$\ \frac{\ 100-7}{7}$$ = $$\ \frac{\ 93}{7}$$

$$X_5$$ = $$\ \frac{\ b_5}{a_5}$$ = $$\ \frac{\ 100-a_5}{a_5}$$ = $$\ \frac{\ 100-11}{11}$$ = $$\ \frac{\ 89}{11}$$

$$X_6$$ = $$\ \frac{\ b_6}{a_6}$$ = $$\ \frac{\ 100-a_6}{a_6}$$ = $$\ \frac{\ 100-13}{13}$$ = $$\ \frac{\ 87}{13}$$

$$X_7$$ = $$\ \frac{\ b_7}{a_7}$$ = $$\ \frac{\ 100-a_7}{a_7}$$ = $$\ \frac{\ 100-17}{17}$$ = $$\ \frac{\ 83}{17}$$

$$X_8$$ = $$\ \frac{\ b_8}{a_8}$$ = $$\ \frac{\ 100-a_8}{a_8}$$ = $$\ \frac{\ 100-19}{19}$$ = $$\ \frac{\ 81}{19}$$

$$X_9$$ = $$\ \frac{\ b_9}{a_9}$$ = $$\ \frac{\ 100-a_9}{a_9}$$ = $$\ \frac{\ 100-23}{23}$$ = $$\ \frac{\ 77}{23}$$

B is the set of all integer-valued $$X_i$$ = {$$X_1$$, $$X_3$$} = {49, 19}

The smallest element of B = 19

$$\therefore\ $$ The required answer is B.

Let the price of blueberries, kiwis and peaches be Rs b, k and p respectively. 

Money spent on blueberries = Rs 9b

Money spent on kiwis = Rs 22k

Money spent on peaches = Rs 50p

Since, money spent everyday is the same, 

9b = 22k = 50p = M , where, M is the total money spent everyday. 

b = $$\ \frac{\ M}{9}$$

k = $$\ \frac{\ M}{22}$$

p = $$\ \frac{\ M}{50}$$

Since, b, k and p are integers, M should be the LCM of (9,22,50).

M = LCM (9,22,50) = LCM (22,450) = Rs 4950

Total money spent on the 4th day = Rs 4950 - Rs 15 = Rs 4935

Weight of mangoes bought on the 4th day = $$\ \frac{\ Rs\ 4935}{Rs\ \frac{35}{kg}}$$ = 141 kg

Cost Price of mangoes = Rs 35/kg

Sell Price of mangoes = Rs 50/kg

Profit on 4th day = $$\ \frac{\ Rs\ \left(50-35\right)}{kg}$$ $$\times\ $$141 kg = Rs 2115

Hence, the minimum possible profit on the 4th day is Rs 2115. 

$$\therefore\ $$ The required answer is B.

$$x+y^2+z^3=50$$

Let z = 1, $$x+y^2=49$$ 

Pairs of (x,y) = (48,1),(45,2),(40,3),(33,4),(24,5),(13,6) = 6 solutions

Let z = 2, $$x+y^2=42$$

Pairs of (x,y) = (41,1),(38,2),(33,3),(26,4),(17,5),(6,6) = 6 solutions 

Let z = 3,$$x+y^2=23$$

Pairs of (x,y) = (22,1),(19,2),(14,3),(7,4) = 4 solutions 

Total possible solutions = 16 

$$\therefore\ $$ The required answer is C.

The figure of one of the structure is given above. 

Since, ABCD is a square, AB = CD = 0.5 m

Now, CE=EF=DF= (Given) = CD/3 = $$\ \ \ \frac{\ 0.5}{3}=\ \frac{\ 1}{6}m$$

Let PE = x m and EB = y m respectively. 

In $$\triangle\ $$PFE and $$\triangle\ $$PAB, 

$$\angle\ P=\angle\ P$$ (common angle)

$$\angle\ PFE=\angle\ PAB$$ (since FE is parallel to AB)

Hence, $$\triangle\ $$PFE and $$\triangle\ $$PAB are similar. 

$$\ \frac{\ PE}{PB}=\ \frac{\ FE}{AB}$$

$$\ \frac{\ PE}{PE+EB}=\ \frac{\ FE}{AB}$$

$$\ \ \ \frac{\ x}{x+y}=\ \frac{\ \ \frac{\ 1}{6}}{0.5}$$

y = 2x i.e. EB = 2 $$\times\ $$PE

Now, in $$\triangle\ $$ECB, 

$$EB^2=EC^2+CB^2$$

$$\left(2x\right)^2=\ \left(\frac{1\ }{6}\right)^2+\ \left(\frac{1\ }{2}\right)^2$$

$$x=\ \frac{\ \sqrt{10\ }}{12}m$$

PB = PE + EB = x + 2x = 3x = $$\ \frac{\ \sqrt{10}}{4}\ $$m

Length of the structure = PA + PB + AB + BC + CD + DA

= $$\ \frac{\ \sqrt{10}}{4}\ $$ + $$\ \frac{\ \sqrt{10}}{4}\ $$ + $$\ \frac{\ 1}{2}$$ + $$\ \frac{\ 1}{2}$$ + $$\ \frac{\ 1}{2}$$ + $$\ \frac{\ 1}{2}$$

= $$\left(2+\ \frac{\ \sqrt{10\ }}{2}\right)$$m

Length of 20 such structures =$$\left(2+\ \frac{\ \sqrt{10\ }}{2}\right)\times\ 20$$

= $$10\left(4+\sqrt{10\ }\right)m$$

$$\therefore\ $$ The required answer is C. 

The figure of the path is given below. Gate 1 and Gate 2 is situated at A and B respectively. 

We have drawn ED which is perpendicular to AB in order to get a right angled triangle AED where AE = 60m and ED = 80m. 

From pythagoras theorem, $$AD^2=AE^2+ED^2$$

$$AD^2=60^2+80^2$$

AD = 100m

Length of the semi circular path AD = $$\pi\ r=\pi\ \left(\frac{AD\ }{2}\right)\ $$ = $$50\pi\ \ $$ = 157 m

Length of path ABCD = 80+80+20 = 180 m

A person entered through gate 1 and travelled along path 1 and path 2 and exited from gate 1. 

Total distance covered = 180+157 = 337m

Speed of the person = 5 kmph = $$\ \frac{\ 5\times\ 1000}{60}=\ \frac{\ 250}{3}\ \frac{\ m}{\min}$$

Total time taken = $$\ \frac{\ 337}{\ \frac{\ 250}{3}}\min$$ = 4 min (approx)

So the total time taken for the person to travel across the park is 4 min. 

$$\therefore\ $$ The required answer is A.