AP Police Constable Mock Test - 2 Free Online Test 2026

Between 5 and 78:
First number = 6
Last number = 72

Hence, Option D is correct.

The sentence is erroneous due to the use of ‘was’. The sentence is talking about two people, Will and George, so the helping verb used should be ‘were’.

Correct Sentence:

Will and George were gamblers and often tried to entice their brother into the joys and sorrows of venture.

Hence, option B is correct.

Perimeter of the rectangle
= 72/2 = 36 cm
Let the width of the rectangle = x
Length = x + 2
2 × (x + x + 2) = 36
2x + 2 = 18
x = 8 cm
Hence, Option A is correct.

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM
⇒ 864 × Second number
= 96 × 1296 ⇒ Second number

X : Y = 4% : 6% = 2 : 3
Hence, Option B is correct.

x = 12 and y = 4
∴  (x + y)^x/y = (12 + 4)^12/4 = (16)³
= 16 × 16 × 16 = 4096 

According to the question,
Every day adding 1 rs extra to previous day.
Let us assume after n day, total saving will become perfect square.
1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn.
Apply the algebra A.P formula,
sum of total rupees after n days = n(n+1)/2
Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.
If n = 2
n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square.
If n = 3
n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square.
If n = 4
n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square.
If n = 5
n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square.
If n = 6
n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square.
If n = 7
n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square.
If n = 8
n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square.

n(n+1)/2 should be a perfect square . The first value of n when this occurs would be for n = 8. thus , on the 8th of March of the required condition would come true.

I. x² – 14x + 45 = 0
x² – 9x – 5x + 45 =0
x = 9 and 5
II. y² + 2y – 35 = 0
y² + 7y – 5y – 35 = 0
y = – 7 and 5
Answer: x ≥ y
Hence, Option C is correct.

Here 11^th term = 1/13
13^th term = 1/11
Let the first term of the AP be a and common difference be d

Subtracting (1) from (2)
We get,

Now, substituting value of d in equation 1
We get,

25 : 25 = 1 : 1

Hence, Option D is correct.

Equation of line in normal form can be written as

First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like →10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number t_n = 94, Common Difference d = 7 ;
Using the Formula for Last Number t_n = a + (n - 1) x d;
⇒ a + (n - 1) x d = t_n
Put the value of a , d and t_n in above equation.
⇒ 10 + (n - 1) x 7 = 94
⇒ 10 + 7n - 7 = 94
⇒ 7n + 3 = 94
⇒ 7n = 94 - 3 = 91
⇒ n = 91/7
⇒ n = 13
Using the formula for the sum of Arithmetic Progression.
S_n = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number, we will get,
S_n = 13/2 [ 10 + 94 ]
S_n = 104 x 13/2
S_n = 52 x 13
S_n = 676 

I. x² – 4x – 96 = 0
x² – 12x + 8x – 96 = 0
x = 12 and – 8

II. 4y² – 109y + 741 = 0
4y² – 52y – 57y + 741 = 0 

Answer: x < y
Hence, Option D is correct.

40 cows eat 40 bags of husk in 40 days.
∴ 1 cows eat 40 bags of husk in 40 x 40 days.
∴ 1 cows eat 1 bags of husk in 40 x 40 / 40 days.
∴ 1 cows eat 1 bags of husk in 40 days.

The 5^th term is given as 31, and the 25^th term is 140 more than the 5^th term, which makes it 31 + 140 = 171. Using the formula for the nth term of an A.P., a + (n-1)d, we have:

a + 4d = 31
a + 24d = 171

Solving these equations:

From the first equation: a + 4d = 31 => a = 31 - 4d

Substituting in the second equation: 31 - 4d + 24d = 171

20d = 140

d = 7

Substitute d = 7 back into a = 31 - 4d:

a = 31 - 28 = 3

Thus, the A.P. is 3, 10, 17, 24, 31...., 

Reqd. time = 60 × (1 + 3) = 240 seconds
Hence, Option D is correct.

Using Rule 7,
Part of the cistern filled by pipe Q in 1 minute

Required time = 60 minutes

When B runs 25 m, A runs 45/2 m.
When B runs 1000 m,

∴ B beats A by 100 m.

Let B joined for x months.
Thus their one month equivalent capitals
A = 85000*12
B = 42500*x = 42500x
Therefore, (85000*12)⁄42500x = 3/1
=> x = 8 months (option ‘D’)

Let CP = 3x
SP = 2x

Hence, Option B is correct.

There should be 230 in place of 232.
Hence, Option D is correct. 

Let AB be the tower and C and D be the two positions of the car.
Then, from figure
AB/AC = tan 60 = √3 => AB = √3AC
AB/AD = tan 45 = 1 => AB = AD
AB = AC + CD
CD = AB - AC = √3AC - AC = AC (√3 - 1)
CD = AC (√3 - 1) => 10 min
 AC => ?
AC/(AC(√3 - 1)) x 10 =? = 10/(√3 - 1) = 13.66 = 13 min 20 sec(approx)

Total possible numbers = 4 × 3 × 2 × 1 = 24
Possibility of 5921= 1

∵ Distance covered in 20 min = 5 km
∴ Distance covered in 1 min = 5/20 km
∴ Distance covered in 100 min = (5/20) x 100
= 5 x 5 = 25 km

CD || AB (Given)
Produce RQ to meet AB in S
∠CRS = ∠PSR (at. int. ∠s)
But ∠CRS = 55°
∴ ∠PSR = 55°
Now in QSP
∠QSP + ∠QPS + ∠PQS = 180°
55° + 38° + ∠SQP = 180°
∴ ∠SQP = 180° – 93° = 87°
But angle a and ∠PQS are linear
∠a = 180° – 87°
∠a = 93°

Hence, Option B is correct.

18% of x = 13.50% of y

50% of x = 37.50% of y
Hence, Option D is correct.

Sum of 100 numbers =100×210=21000
Sum of 120 numbers =120×200=24000
Sum of 20 numbers =24000−21000=3000
∴ Average of 20 new numbers =3000/20=150

Hence, Option D is correct.

Area of an equilateral triangle


Hence, Option A is correct.