XAT 2026 Question Paper - QADI

Three concentric circles C1, C2 and C3 with the radius OA, OB and OC are drawn. 

OA, OB and OC are in AP (given). 

OA = 70m, OB = (70 + d)m, OC = (70 + 2d)m , where, d is the common difference. 

OC = 70+2d = 140 (given) 

d = 35m 

OB = 70 + 35 = 105m  

The soldiers are standing on the circumference of these three circles at a distance of 1m. Hence, to find the total number of soldiers, we have the find the length of the circumference of all three circles. 

Total circumference = $$2\pi\ \left(r1+r2+r3\right)$$

= 2 $$\times\ $$ $$\ \frac{\ 22}{7}\times\ \left(70+105+140\right)$$  = 1980

Hence, the total number of soldiers are 1980.

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

In the above figure, let the number of students specializing in : 

Only marketing = a, Only finance = b, Only operations = c 

Both marketing and finance but not operations = d, Both marketing and operations but not finance = e, Both finance and operations but not marketing = f

All three subjects = g, None of the subjects = h

Now it is given that only students having an specialization in marketing or finance gets a consulting shortlist. So, we can say that students doing specialization only in operations and students not doing specialization in any of the 3 subjects will not get a shortlist. 

Hence, c + h = 45 (given) $$\longrightarrow\ i$$

Number of students doing specialization in marketing = a + d + e + g = 60 (given) $$\longrightarrow\ ii$$

Number of students doing specialization in finance = b + d + g + f = 50 (given) $$\longrightarrow\ iii$$

Number of students doing specialization in operations = c + e + g + f = 30 (given) $$\longrightarrow\ iv$$

Number of students who received the consulting shortlist = a + b + d + e + f + g =  75  $$\longrightarrow\ v$$

By using equation iv and v, we get, 

a + b + d + 30 - c = 75

a + b + d = 45 + c $$\longrightarrow\ vi$$

Add equation ii and iii, 

a + b + d + (d + g) + e + f + g = 110 $$\longrightarrow\ vii$$

Now, using equation iv, vi and vii, we get, 

45 + c + (d + g) + 30 - c = 110 

d + g = 35 

Hence, the number of students specializing in marketing and finance = 35

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

Let the tree be PC and the archer be RA. 

Given : Height of the tree = 56ft, height of the archer up to the eye level = 6ft, angle of elevation at point A = 45$$^{\circ\ }$$, angle of elevation at point B = 30$$^{\circ\ }$$  

To find : Length of AB

In $$\triangle\ $$PQR, $$\tan\ 45^{\circ\ }=\ \frac{\ PQ}{QR}=\ \frac{\ 50}{QR}$$

QR = 50ft (as $$\tan\ 45^{\circ\ }=\ 1$$)

Now, in $$\triangle\ PQS$$, 

$$\tan\ 30^{\circ\ }=\ \frac{\ PQ}{QS}=\ \frac{\ PQ}{QR+RS}=\ \frac{\ 50}{50+RS}$$

$$\ \frac{\ 1}{\sqrt{3\ }}=\frac{\ 50}{50+RS}$$

RS = $$50\left(\sqrt{3\ }-1\right)$$ ft

Since RS = AB, AB = $$50\left(\sqrt{3\ }-1\right)$$ ft

Hence, the length of AB is $$50\left(\sqrt{3\ }-1\right)$$ feet

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

Let the maximum allowed speed from Jamshedpur to Kolkata be s1 kmph and the maximum allowed speed from Kolkata to Jamshedpur be s2 kmph. Since the bus is moving at the maximum allowed speed then we can say that the speed of the bus moving from Jamshedpur to Kolkata (Bus 1) is s1 kmph and the speed of the bus moving from Kolkata to Jamshedpur (Bus 2) is s2 kmph.

Time taken by Bus 1 to travel from Jamshedpur to Kolkata = 4 hrs

Distance travelled by Bus 1 = 4 $$\times\ $$s1

Time taken by Bus 2 to travel from Kolkata to Jamshedpur = 3 hrs

Distance travelled by Bus 1 = 3 $$\times\ $$s2

Since both the bus is travelling on the same road, the distance travelled by both the buses will be same. 

4 $$\times\ $$s1 = 3 $$\times\ $$s2

$$\ \frac{\ s1}{s2}=\ \frac{\ 3}{4}$$

The ratio of speed of Bus 1 to the speed of Bus 2 is 3:4.

Hence, let the speed of Bus 1 be 3x kmph and the speed of Bus 2 be 4x kmph.

Total distance travelled by the buses = $$4\times\ 3x\ =3\times\ 4x=12x$$

Now, both the buses started at the same time and travelled at the maximum allowed speed for 1 hr. 

Distance travelled by Bus 1= JA = $$3x\times\ 1=3x$$

Distance travelled by Bus 2= KB = $$4x\times\ 1=4x$$

Distance left to be travelled= AB = $$12x-\left(3x+4x\right)=5x$$

Now, the speed of Bus 1 is reduced to half due to congestion. 

New speed of Bus 1 = $$\ \frac{\ 3x}{2}$$

After the reduction of speed, let the two buses meet at point C and the time taken by both the buses to reach C is the same. 

Let the distance travelled by Bus 1 from A to C be y km. 

BC = 5x-y km

Hence, $$\ \frac{\ y}{\ \frac{\ 3x}{2}}=\ \frac{\ 5x-y}{4x}$$

y = $$\ \frac{\ 15x}{11}$$

Time required = $$\ \frac{\ \ \frac{\ 15x}{11}}{\ \frac{\ 3x}{2}}=\ \frac{\ 10}{11}$$ hrs

Total time required for both the buses to meet = $$\frac{\ 10}{11}+1=\ \frac{\ 21}{11}$$ hrs

Hence, both the buses will meet after $$\frac{\ 21}{11}$$ hrs from starting. 

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

For Chhuttan,

Perimeter of his land = 2 (l + b) = 250m 

l + b = 125m $$\longrightarrow\ i$$ 

$$\ \frac{\ l}{b}=\ \frac{\ 4}{1}$$ (Given) $$\longrightarrow\ ii$$

From i and ii, we get, 

4b + b = 125

b = 25m  

l = 100m 

Area of Chhuttan's land = 100 $$\times\ $$25 = 2500 $$m^2$$

Loan given to Chhuttan by Brijbhushan = 2500 $$\times10$$ = Rs 25000

Interest rate of the loan = 10%

Interest owed by Chhuttan at the end of one year = 10% $$\times\ $$Rs 25000 = Rs 2500

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

Let the name of farmers be A,B,C,D,G for simplicity. 

Let the area of G's plot be x $$m^2$$.

Area of A's plot = 3x $$m^2$$

Let area of D's plot be y $$m^2$$. 

The area of A's plot is the average of the area of D's and G's plot. 

3x = $$\ \frac{\ x+y}{2}$$

y = 5x $$m^2$$ 

The width of the plot of A,B and D is the same and the lengths are different. Let the length of the plot of A,B and D be l1,l2 and l3 respectively and the width be y. 

We got the following table. 

From A's plot, $$l1\times\ y=3x$$

$$l1=\ \frac{\ 3x}{y}$$

From D's plot, $$l3\times\ y=5x$$

$$l3=\ \frac{\ 5x}{y}$$

The length of B's plot is the sum of the length of A's plot and D's plot.

l2 = l1 + l3 = $$\frac{\ 8x}{y}$$

Hence, the area of B's plot = $$l2\times\ y=8x$$

The required final table is : 

 Total area of all plots = 2500 + 17x

Total loan given by Brijbhushan = $$Rs\ \left(2500+17x\right)\times\ 10$$

Since, we have to maximize the loan amount, we have to maximize the value of x. 

The largest area of a land = $$10000\ m^2$$ (Given)

Among the farmers, B will have the land with the largest area. 

8x = 10000 $$m^2$$

x = 1250 $$m^2$$

Hence, maximum loan given by Brijbhushan = $$Rs\ \left(\left(1250\times\ 17\right)+2500\right)\times\ 10$$ = Rs 237500

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

Let the name of farmers be A,B,C,D,G for simplicity.

Let the area of G's plot be x $$m^2$$.

Area of A's plot = 3x $$m^2$$

Let area of D's plot be y $$m^2$$.

The area of A's plot is the average of the area of D's and G's plot.

3x = $$\ \frac{\ x+y}{2}$$

y = 5x $$m^2$$

The width of the plot of A,B and D is the same and the lengths are different. Let the length of the plot of A,B and D be l1,l2 and l3 respectively and the width be y.

We got the following table.

From A's plot, $$l1\times\ y=3x$$

$$l1=\ \frac{\ 3x}{y}$$

From D's plot, $$l3\times\ y=5x$$

$$l3=\ \frac{\ 5x}{y}$$

The length of B's plot is the sum of the length of A's plot and D's plot.

l2 = l1 + l3 = $$\frac{\ 8x}{y}$$

Hence, the area of B's plot = $$l2\times\ y=8x$$

The required final table is :

The width of A's plot = 25m 

y = 25m 

For B's plot, $$l2\times\ 25=8x$$

To minimize the value of l2, we have to minimize the value of x.  

Minimum possible area of land among the farmers = 1000 $$m^2$$

Smallest area of land = G's land = x $$m^2$$

x = 1000

Hence, minimum value of l2 = $$\ \frac{\ 8\times\ 1000}{25}=320\ m$$

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

A seven-digit number 735x6y4 is divisible by 44. It means that the number is also divisible by 4 and 11. 

Divisibility rule of 4 = Last two digits of the number is divisible by 4. 

Divisibility rule of 11 = Sum of odd digits starting from left - Sum of even digits starting from left

For the given number to be divisible by 4, y4 should be divisible by 4. 

So, the possible values of y = 0,2,4,6,8

For the number to be divisible by 11, (7+5+6+4)-(3+x+y) = 19-(x+y) should be divisible by 11. 

Case 1 : Both x and y are even numbers. 

If y = 0, x = 8 

If y = 2, x = 6

If y = 4, x = 4

If y = 6, x = 2

If y = 8, x = 0

In this case, we got 5 different combinations of values. 

Case 2 : Both x and y are equal. 

If y = 4, x = 4

In this case, we got a unique solution. 

Hence, we only need condition 2 to uniquely determine the values of x and y. 

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

Let us denote Ayub and Rana as A and R respectively for simplicity. 

Let the speed of A and B in still water be Sa and Sr respectively and the speed of river be r. 

r = 3 kmph = 50 $$\ \frac{\ m}{\min}$$ (clockwise)

Sr = 3 kmph = 50 $$\ \frac{\ m}{\min}$$ (clockwise)

Since A is swimming in counter clockwise direction and R is swimming in clockwise direction, let C be the point where they meet for the first time. 

Distance travelled by A = AC = 60m 

Speed of A = (Sa-50) $$\ \frac{\ m}{\min}$$ (as he is travelling in counter clockwise direction which is upstream)

Time taken by A to reach C = $$\ \frac{\ 60}{Sa-50}\min$$

Speed of R = (Sr+50) $$\ \frac{\ m}{\min}$$ (as he is travelling in clockwise direction which is downstream)

= 100 $$\ \frac{\ m}{\min}$$

Since, the time taken by R to reach C is the same as the time taken by A to reach C (as they started swimming at the same time), distance travelled by R = RC = $$\ \frac{\ 100\times\ 60}{Sa-50}m$$

Total distance travelled by both A and R = AC + RC = $$\ \frac{\ 6000}{Sa-50}+60\ m$$

As A and R travelled in opposite directions from exactly diametrical ends, the distance A to R will be half of the circumference of the circle. 

$$\pi\ r$$ = $$\ \frac{\ 6000}{Sa-50}+60\ m$$ $$\longrightarrow\ i$$

Now, let's assume that they meet again at D after R travelled for 180m, 

Time taken for them to meet again = $$\ \frac{\ 180}{100}\min$$

Distance travelled by A = CD (in counter clockwise direction) = $$\ \frac{\ \left(Sa-50\right)\times\ 180}{100}m$$

Distance travelled by R = CD (in clockwise direction) = 180m 

Total distance travelled = $$\ \frac{\ \left(Sa-50\right)\times\ 180}{100}m$$ + 180 m

This distance will be equal to the circumference of the circle. 

Hence, $$2\pi\ r$$ = $$\ \frac{\ \left(Sa-50\right)\times\ 180}{100}m$$ + 180 m $$\longrightarrow\ ii$$

Take the value of $$\pi\ r$$ from equation i and substitute it in equation ii. 

$$\ \frac{\ 12000}{Sa-50}+120=\ \frac{\ 180\left(Sa-50\right)}{100}+180$$

Let the value of Sa-50 be t.

 $$\ \frac{\ 12000}{t}+120=\ \frac{\ 180\left(t\right)}{100}+180$$

$$\ 3t^2+100t-20000=0$$

t = $$\ \frac{\ 200}{3}$$, -100 (Not possible as the speed can't be negative)

Sa-50 = $$\ \frac{\ 200}{3}$$

Sa = $$\ \frac{\ 350}{3}\ \frac{\ m}{\min}$$

Use this value in equation ii to get, $$2\pi\ r$$ = 300m

Speed of A when he is swimming in clockwise direction = $$\ \frac{\ 350}{3}+50$$

= $$\ \frac{\ 500}{3}\ \frac{\ m}{\min}$$

Time taken by A to complete one round of the circular strip = $$\ \frac{300}{\ \frac{\ 500}{3}\ }\min$$

= $$\ \frac{\ 9}{5}\min$$ = 1 min 48 sec

Hence, the time taken by A to complete one round of the circular strip is 1 min 48 sec. 

$$\therefore\ $$ The answer is D.

125 = $$5^3$$

10 = $$2\times\ 5$$

15 = $$3\times\ 5$$

In each step, we are multiplying 125 by either 10 or 15. So, the power of 5 and the power of either 2 or 3 will only increase by one in every step. So at any point of time, the difference between the power of 5 and the power of 2 and 3 combined will remain constant (which is 3).

Option A : Power of 5 = 691

Power of 2 = 235

Power of 3 = 453

Difference of powers = 691 - (235+453) = 3 

Option B : Power of 5 = 1080

Power of 2 = 253

Power of 3 = 824

Difference of powers = 1080 - (235+824) = 3

Option C : Power of 5 = 1604

Power of 2 = 689

Power of 3 = 912

Difference of powers = 1604 - (689+912) = 3

Option D : Power of 5 = 1034

Power of 2 = 476

Power of 3 = 455

Difference of powers = 1034 - (476+455) = 103

Option E : Power of 5 = 1145

Power of 2 = 689

Power of 3 = 453

Difference of powers = 1145 - (689+453) = 3

Only in option D the difference of power is different than 3. 

$$\therefore\ $$ The answer is D. 

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.

The 5 digits which are used in the combination lock are 9,8,7,5 and 4.

The sum of first 3 digits and the last 3 digits are divisible by 3. Hence, we can say that the third digit is common in both the combinations.

The possible groups of 3 digits whose sum is divisible by 3 are (9,8,7),(9,8,4),(9,7,5),(9,5,4). 

Out of these 4 possible groups, we have to select two groups which can be used as the first and last 3 digits. 

The selection should be done in such a way that both the groups should only have one digit in common. 

Example : If we select (9,8,7) and (9,8,4) and let the middle digit be 9, then we will have 8 in both the first 3 digits and the last 3 digits which is only possible for the 3rd digit which is occupied by 9 in this case. Hence, this case is not possible. 

So the only possible combinations for the first and last 3 digits are :

Case 1: (9,8,7) and (9,5,4) where the 3rd digit will be 9. 

Case 2: (9,8,4) and (9,7,5) where the 3rd digit will be 9.

Now, it is given that the sum of the last 4 digits is divisible by 4. 

The possible cases of the above scenario are :

Case 3: (9,8,7,4) where the first digit will be 5. 

Case 4: (8,7,5,4)where the first digit will be 9 which is not possible as 9 is fixed as the 3rd digit from Case 1 and 2.

Since, the first digit is 5 and the third digit is 9, the second digit will either be 4 or 7 and the last two digits will be either (8,7) or (8,4)

So, the possible combinations of the lock are 54987, 54978, 57984 and 57948.  

Hence, the option which is definitely false is E as no possible combinations has 8 as its 2nd digit. 

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

The length, width and height of Tank 1 is m meters each. 

Volume of Tank 1 = $$m^{3\ }$$ cubic meters

The length, width and height of Tank 2 is n meters each. 

Volume of Tank 2 = $$n^{3\ }$$ cubic meters

The length, width and height of Tank 3 is m, n and 1 m respectively. 

Volume of Tank 3 = mn cubic meters 

Since, Tank 1 is filled first and after Tank 1 is full, water is flowing into Tank 2 and Tank 3, 

Volume of water left in Tank 1 after Tank 2 and Tank 3 are filled = $$\left(m^{3\ }-n^3-mn\right)$$ cubic meters

Volume of water left in Tank 1 = 85000 L (given) = 85$$m^3$$

Hence, $$\left(m^{3\ }-n^3-mn\right)$$ = 85 , where, m and n are integers. 

$$n^3+mn=m^3-85$$

From option B, we get $$n^3+7n=258$$ $$\longrightarrow\ $$ n=6

From option C, we get $$n^3+5n=40$$ $$\longrightarrow\ $$ No integer solution

From option D, we get $$n^3+6n=131$$ $$\longrightarrow\ $$ No integer solution 

From option E, we get $$n^3+10n=915$$ $$\longrightarrow\ $$ No integer solution 

Hence, we got m = 7 and n = 6. 

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