Jharkhand Police Constable (Paper-II) Mock Test - 2 - Jharkhand Police Constable MCQ

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.

(a*25)/100 = b(2*65)/100
a*25 = b*130
a/b = 130/25 = 26/5
b:a = 5:26

Square number can never end in 2,7

Normal process:
Runs in 26th inning = Runs total after 26 innings – Runs total after 25 innings
= 26 X 58 – 25 X 56

For Easy calculation use:

= (56 + 2) X 26 – 56 X 25 )
= 2 X 26 + (56 X 26 – 56 X 25)
= 52 + 56 = 108

Since the average increases by 2 runs per innings, it is equivalent to 2 runs being added to each score in the first 25 innings. Now, since these runs can only be added by the runs scored in the 26th inning, the score in the 26th inning must be 25 X 2 = 50 runs higher than the average after 26 innings (i.e. new average = 58).

Hence, runs scored in 26th inning:
= New Average + Old innings X Change in average

= 58 + 25 X 2
= 108 

In order to find the average age of the family before Ram was born, we need to know the age of the youngest member of the family. 
Since, we do not know the age of the youngest member, we can not calculate the total age of the family before Ram was born.
Hence, we can not calculate the answer with the given conditions.

Thus, D is the right choice.

X+ y=Rs. 550
x = 120% of y
120% * y+ y =550
120/100*y+y = 550
6/5y+y =550
6y+5y/5 = 550
11/5y = 550
11y = 550*5
y = 250

Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water

Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

Each transfer can be captured through the following table.

Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4

So Option A is correct

A and B complete a work in = 15 days
⇒ One day's work of (A + B) = 1/ 15

B complete the work in = 20 days
⇒ One day's work of B = 1/20

⇒ A's one day's work = 1/15 − 1/20 = (4−3)/6 = 1/60

Thus, A can complete the work in = 60 days.

So Option A is correct

Let Anil takes x hrs.
⇒ Arun takes x+2

If Arun doubles the speed, he will take x-1 hrs
⇒ He needs 3 hours less.
∵ Double speed means half time.
∴ Half of the time required by Arun to cover 30 km = 3 hours

⇒ Time required by Arun to cover 30 km = 6 hour
Thus, Arun's speed = 30/6 = 5 km/h

Time taken = Distance/Speed
Total time taken = (160/64) + (160/80) = 9/2 hrs
Average Speed = Distance/Time
∴ Average speed = 320 / (9/2) = 320 x (2/9) = 71.11 km/hr.

P : Q : R
= (25 × 1 + 35 × 2) : (35 × 2 : 25 × 1) : (30 × 3)
= 95 : 95 : 90
= 19 : 19: 18

Let C = x.
Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000
So, x + x + 5000 + x + 9000 = 50000
⇒ 3x = 36000
⇒ x = 12000
A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12
So A's Share

Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid

Hence milk now contained by the container = 40

► Since first and second varieties are mixed in equal proportions.So, their average price = Rs.
= Rs. 130.50

► So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.

By the rule of alligation, we have:

Let capacity of bucket P = x
Then capacity of bucket Q =  x/3
Given that it takes 80 turns for bucket P  to fill the empty drum
⇒ capacity of the drum = 80x
Number of turns required if both P and Q are used = 

(A + B)'s 1 hour's work =  = 9/60 = 3/20.
(A+C)'s hour's work =   = 8/60 = 2/15.
Part filled in 2 hrs = 

Part filled in 6 hrs = 
Remaining part = 
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
∴ Total time taken to fill the tank = (6+1) hrs = 7 hrs

CP = 69/1.15 = 60. Thus, the required SP for 20%
profit = 1.2 x 60 = 72.

Sales tax = 150/5 = 30. Thus, the SP contains ₹ 30
component of sales tax. Of the remainder (150 – 30
= 120) 1/3rd is the profit. Thus, the profit = 120/3 =
40. Cost price = 120 – 40 = 80.

Given, P = Rs.900

SI = Rs.81

T = ?

R = 4.5%

Simple Interest (SI) = P N R / 100

P is the Principal loan amount = Rs.1400

N is the number of years of deposit

R is the rate of interest

It is given that the loan period is as many years as the rate of interest.

So, N = R

Interest at the end of the loan period (SI ) = Rs.686

So,

686 = 1400 * R * R /100

R² = 686*100 /1400

R² = 49

R = 7%

Let each side of the cube be x cm.
Therefore, surface area = 6x²
Surface area of the cube after making the side of cube double = 6(2x)² = 24x²
So, clearly, it becomes 4 times of the previous one.
Percentage increase 

Height of the cylinder = 15 m
Height of the cone, 'h' = 19 - 15 = 4 m
Radius =  6/2 = 3 m
Slant height of the cone = 
Canvas required = C.S.A. of (cone + cylinder)
= πrl + 2πrH
= πr(5 + 2 × 15)

= 330 m²

Here, the degree is 2, therefore it is a quadratic equation.