Jharkhand Police Constable (Paper-II) Mock Test - 1 - Jharkhand Police Constable MCQ

SI = PRT / 100
⇒ (SI x 100) / RT = P
⇒ P = (6200*100) / 32
⇒ P = 19375

Curved surface area of a cylinder = 2πrh
Given: h = 14 cm and curved surface area = 264 cm²
∴ 2πrh = 264
r = 3 cm

Here, 'l' is slant height,

As both have the same radius, therefore: 


Squaring both sides:

h/r = 3/1

Explanation:

Let one multiple of 5 be x then the next consecutive multiple of will be (x+5) According to question,

Then the number are 30 and 35.

V α T/P
Since temperature is constant, therefore
V α T/P
⇒ V₁P₁ = V₂P₂
400P₁ = 4P₂
P₂ = 100P₁
So, percentage increase 

Cost of a giraffe = P
10 giraffes = 24 dogs
16 dogs = 4 oxen
6 oxen = 4 elephants
10 elephants = Rs. 1,70,000
P = Rs. [(24 × 4 × 4 × 1,70,000)/(10 × 16 × 6 × 10)]
P = Rs. (6,52,80,000/9600)
P = Rs. 6800

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 .... (i)

  and 2x + 4y = 140   ⇒   x + 2y = 70 .... (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

∴ The required answer = 26.

85*85 = 7225

a³ + b³– a² + b² + 2a – 3b
(-2)³ + (-5)³ – (-2)² + (-5)² + 2(-2) – 3(-5) = -8 -125-4+25-4+15 = -101

Let initial average be x.
Then the initial total is 20x and the New average will be (x – 4),

The new total will be:
20(x – 4) = 20x – 80.

The reduction of 80 is created by the replacement. Hence, the new student has 80 marks less than the student he replaces. Hence, he must have scored 10 marks.

Short Cut:
The replacement has the effect of reducing the average marks for each of the 20 students by 4. Hence, the replacement must be 20 X 4 = 80 marks below the original.

Hence, answer = 10 marks. 

The first ten composite numbers are: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18. 

Required average:

= (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18) / 10
= 112 / 10
= 11.2

Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.

Number of such number =14

x%ofy=x/100×y
=y/100×x
=y%ofx
∴A=B

Initial distance = 25 dog leaps
Per-minute dog makes 5 dog leaps and cat makes 6 cat leaps = 3 dog leaps
⇒  Relative speed = 2 dog leaps / minutes
⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes.

So Option D is correct

Let distance between the two places = d km
Let total time taken by faster horse = t hr
⇒ Total time taken by slower horse = (t + 5) hr,

Therefore,
speed of the faster horse = d/t km/hr
speed of the slower horse = d/(t + 5) km/hr 
The two horses meet each other in 3 hour 20 min i.e. in 3(1/3) hr = 10/3 hr
In this time, total distance travelled by both the horses together is d. 

∴ d/(t+5) * 10/3 + d/t * 10/3 = d
⇒ 10/(3(t+5)) + 10/3t = 1
⇒ 10t + 10(t+5) = 3t(t+5)
⇒ 20t + 50 = 3t² + 15t
⇒ 3t² − 5t − 50 = 0
⇒ 3t² + 10t − 15t − 50 = 0
⇒ t(3t + 10) − 5(3t + 10) = 0
⇒ (3t + 10)(t − 5) = 0
⇒ t = 5 (ignoring -ve value) 

Thus, Total time taken by slower horse = 5 + 5 = 10 hr

So Option B is correct

It is given that, excluding stoppages, the speed of a bus is 54 kmph.
⇒ Distance travelled by bus in 1 hour, excluding stoppages = 54 km.
Also, it is given that including stoppages the speed of the bus is 45 km/hr.
⇒ Distance travelled by the bus in 1 hour, including stoppages = 45 km.
(Distance travelled by bus in 1 hour excluding stoppages – distance travelled by bus in 1 hour including stoppages)
⇒ (54km−45km) ⇒ 9 km
Due to stoppages, it covers 9 km less in an hour.
Students can easily solve this question by using this trick.
Required time for the stoppage per hour,

Let time taken to travel the first half = x hr 
⇒ Time taken to travel the second half = (10 - x) hr 

∵ Distance = Time * Speed
Distance covered in the first half = 21x 
Distance covered in the the second half = 24(10 - x)

∵  Distance covered in the the first half = Distance covered in the the second half
⇒ 21x = 24(10 - x)
⇒ 45x = 240
⇒ x = 16/3

Total Distance = 2 * 21(16/3) = 224 Km [∵ multiplied by 2 as 21x was the distance of half way]

Now as per question, Kamal invested for 12 months and Sameer invested for 7 months.
So Kamal : Sameer = (9000 × 12) : (8000 × 7)
= 108 : 56
= 27 : 14
Sameer Ratio in profit will be 

Let the amount invested by Q = q.

8 litres are drawn from a cask full of wine and is then filled with water.

This operation is performed three more times.

The ratio of the quantity of wine now left in cask to that of the water is 16 ∶ 65
Calculations:

Let the quantity of the wine in the cask originally be x litres.

After first operation, wine left in the cask = (x - 8) litres So, the ratio of the quantity of wine now left in cask to that of filled cask = 

This operation is performed three more times. Then, Wine : Water = 16 : 65 or wine : water = 16 : (16 + 65) = 81

Correct answer is 24.

By the rule of alligation we have 
► So ratio of 1st and 2nd quantities = 7:14
= 1:2
∴ Required quantity replaced = 2/3