AP Police Constable Mock Test - 5 - AP Police Constable MCQ

Weight of the car = (180 × 7 – 72 × 6) = 828 kg
Hence, Option B is correct.

6l + 6b = 8l
6b = 2l
l : b = 3 : 1
Hence, Option B is correct.

HCF of two-prime numbers = 1
∴ Product of numbers = their LCM = 117
117 = 13 × 9 where 13 & 9 are co-prime. L.C.M (13,9) = 117.

50 × R + 60 × 80 = 70 × (50 + 60)
50 × R + 4800 = 7700 
50R = 2,900
R = 58
Hence, Option B is correct.

On cubing both sides,


⇒  a⁶ + 1 = 0
∴  a¹⁸ + a¹² + a⁶ + 1
= a¹² (a⁶ + 1) + 1 (a⁶ + 1)
= (a⁶ + 1)(a¹² + 1) = 0 

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
n - 24 = n x 4/7
⇒ n - 4n/7 = 24
⇒ (7n - 4n)/7 = 24
⇒ 3n/7 = 24
⇒ 3n = 24 x 7
⇒ n = 24 x 7/3
⇒ n = 8 x 7
⇒ n = 56
Sum of the digits of the number = 5 + 6 = 11

I. 3x + 3y = √1764
3x + 3y = 42
x + y = 14

II. 7x – 5y = √1444
7x – 5y = 38
5 × I + II
12x = 70 + 38
x = 9
y = 14 – 9 = 5
x > y
Hence, Option A is correct.

Here 7^th term = 20
11^thterm = 40
Let the first term of the AP be a and common difference be d
T₇ = a + (n – 1)d = 20
T₇ = a + (7 – 1)d = 20
T₇ = a + 6d = 20     (1)
T₁₁ = a + (n – 1)d = 40
T₁₁ = a + (11 – 1)d = 40
T₁₁ = a + 10d = 40     (2)
Subtracting (1) from (2)
We get,
a + 10d – (a + 6d) = 40 – 20
4d = 20
d = 5 

Hence, Option D is correct.

a + b + c = 6 and ab + bc + ca = 11
∴  bc (b + c) + ca (c + a) + ab (a + b) + 3abc
= bc (b + c) + abc + ca (c + a) + abc + ab (a + b) + abc
= bc (a + b + c) + ca (a + b + c) + ab (a + b + c)
= (a + b + c) (bc + ca + ab)
= 6 × 11 = 66

Cost of 5 pens + 8 pencils = ₹31
On multiplying by 3
15 pens + 24 pencils = 3 x 31 =
₹15 pens + 24 pencils = 93

I. 144x² = 16 + 20

II. 20y = √169 – √9
20y = 13 – 3
20y = 10

Answer: x ≤ y
Hence, Option B is correct.

= log₁₀10 + log₁₀100 + log₁₀1000 + log₁₀10000 + log₁₀100000
= 1 + 2 + 3 + 4 + 5
= 15 

AP: 10, 6, 2, …
a = 10, d = - 4
Sum of first n terms = S(n) = (n/2) x [2a + (n – 1) x d]
S5 = (5/2) x [2 x (10) + (5 – 1) x (-4)]
= (5/2) x [20 + 4 x (-4)]
= (5/2) x (20 – 16)
= (5/2) x (4)
= 5 x 2
= 10

Required time 

Hence, Option A is correct.

Net part filled in 1 hr = (1/8 ? 1/16) = 1/16
Total time taken to fill the tank = 16 hrs.

Time taken by A to run 1 km = 290 sec.
Time taken by B to run 1 km = 300 sec.
A can give B a start of (300-290) = 10sec.
In 300 sec, B runs (1000/300*10) m = 100/3 m
∴ A can give B start of 100/3 m.

Simran invested Rs 50000 for 36 months and Nanda invest Rs 80000 for 30 months
Jitesh’s one month equivalent investment = 50000*36
Rahul’s one month equivalent investment = 80000*30
Thus, ratio of their investments =  = 50000*36 : 80000*30 = 3 : 4
Therefore Jitesh’s share in the profit = 24500 x 3/7 = Rs 10,500 (option ‘D’)

Let CP = 100
MP = 160
SP = 125

Hence, Option D is correct.

There should be 56 in place of 57.
Hence, Option C is correct. 

One of AB, AD and CD must have given.

So, the data is inadequate.

Total milk bought =8.5l+7.25l+9.4l=25.15l

∴ Milk left in booth = 30l−25.15l = 4.850 liters

11/17 (if the first one is black, there will be 11 black balls left out of 17)

∵ 8 m shadow means original height = 12 m
∴ 1 m shadow means original height = 12/8 m
∴ 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m

On the basis of question we draw a figure of a circle with centre O ,

Let OE = y cm
then OF = (y +1) cm
OA = OC = r cm
AE = 4 cm; CF = 3 cm
From ∆ OAE,
OA² = AE² + OE²
⇒ r² = 16 + y²
⇒ y² = r² – 16 ......(i)
From ∆OCF,
(y + 1)² = r² – 9 ..... (ii)
By equation (ii) – (i),
(y + 1)² – y² = r² – 9 – r² + 16
⇒ 2y + 1 = 7
⇒ y = 3 cm
∴ From equation (i),
9 = r² – 16
⇒ r² = 25
⇒ r = 5 cm

Let total work be LCM of (90, 60) = 180 unit
Ratio of number of days by Sarthak and Deepak = 90 : 60 = 3 : 2
Ratio of efficiency of Sarthak and Deepak = 2 : 3
Total unit in one day = 2 + 3 = 5 unit
Number of days taken = 180/5 = 36 days 
Hence, Option B is correct.

So the answer = (100 – 50)% of 700 + (100 – 60)% of 500 = 550
Hence, Option D is correct.

All even numbers between 1 and 50 = 2 to 48

Hence, Option D is correct.

Volume of the cylinder = πr²h

Hence, Option B is correct.

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
We have,
First number × second number = LCM × HCF