Jharkhand Police Constable (Paper-II) Mock Test - 9 - Jharkhand Police Constable MCQ

Let salary = 20,000 + k (experience)
If experience = 2 years
Salary = 50,000
50,000 = 20,000 + k(2)
⇒ k = 15,000
So, for 5 years of experience, salary will be:
20,000 + 15,000(5)
= 95,000

3a - 4b + 2c = 0 ... (1)
4a - 2b - c = 0 ... (2)
From equation (2),
c = 4a - 2b
Substituting the value of c = 4a - 2b in (1), we get
3a - 4b + 8a - 4b = 0
Or 11a = 8b
Further, c = 4a - 2b
Multiplying by 4 on both sides, we get
4c = 16a - 8b
Substituting 8b = 11a, we get
4c = 16a - 11a = 5a
Or a : b : c = a : 11a/8 : 5a/4 = 8 : 11 : 10 or 16 : 22 : 20
So, option 1 is the answer.

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

∴ 160 x 2.40 + x x 3.20 = 432

⇒ 3.20x = 432 - 384 = 48

⇒ x = 15.

Hence, total hours of work = (160 + 15) = 175.

Let total number of children be x.

 Then, x x(1/8)x = (x/2)x 16   ⇔  x = 64

 Number of notebooks =(1/8)x² =(1/8x 64 x 64)= 512

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

X+x+2+x+4 = 40+(x+x+2+x+4)/3
3x+6 = 40+ (3x+6)/3
(9x+18 – 3x-6)/3= 40
6x + 12 = 120
X = 108/6 = 18
18+2 = 20

X+x+1+x+2 = 1251
3x+3 = 1251
X = 1248/3 = 416
X+2 = 416+2 = 418

David's Weight = (4 x 70) – (3 x 74) = 280 – 222 = 58
Eric’s weight = 58 + 3 = 61

Now, we know that:
Ross + Joey + Chandler + David = 4 x 70 = 280
Joey + Chandler + David + Eric = 75 x 4 = 300.

Hence, Ross’s weight is 20 kg less than Eric’s weight. Ross = 61 - 20 = 41 kg. 

Mon + Tue + Wed = 35*3 = 105  ---------(1)
Tue + Wed + Thu = 30*3 = 90  -------------(2)
Thu = (1/2) Mon  ------------(3)

Eqn (1)-(2):
Mon-Thu = 15 ------------(4)

⇒ Mon - (1/2) Mon = 15
⇒ (1/2) Mon = 15
⇒ Mon =30
⇒ Thu = 30/2=15

Initially number of matches = 40 
Now matches won = 12 
Now let remaining matches be x 
Now number of matches won = 0.6x
Now as per the condition :

​
 
24 +1.2x=40+x
0.2x=16
x=80
Now when they won 90% of remaining = 80(0.9) =72
So total won = 84

Time take for reaching B for ram is T = 5/5 = 1hr
Time taken for reaching B for Shyam is T = 5/10 = 1/2 hr

Its 10 am when ram reaches B and 10: 15 when Shyam reaches B , so they must meet each after 10 and before 10:15 for sure as ram starts back after reaching B

if we see for options 10: 10 am is the only answer

Since the second rabbit covers 7/120 of the distance in 2 hours 30 minutes
⇒ it covers 8.4 / 120 = 7% of the distance in 3 hours.

Thus, in 3 hours both rabbits together cover 15% of the distance which means 5% per hour so they will meet in 20 hours.

The ratio of speeds = 8 : 7.
⇒ the second rabbit would cover 700 ft to the meeting point in 20 hours and its speed would be 35 feet/hr.

So Option D is correct

To find the speed of the man in km per hour, we have to convert the given distance to km and also convert the given time to hour.
First, let us see the conversion of the given distance into km. We know that 1km = 1000m.
So, to convert 600m to km, we can write, 600m = x km.
x = 600/1000 km  = 0.6km 

Now, let us convert the given time into hours. We know that 1 hour = 60 minutes.
So, to convert 5 minutes to hours, we can write, 5 minutes = y hours.
y = 5/60 hrs = 0.083 hrs

Now we know that speed can be found out using the formula, 
Speed = Distance/Time
           = 0.6/ 0.083
           = 6×1000 / 83 × 10
           = 6000/ 830
           = 7.2 km/h

Therefore, we get the speed of the man crossing a 600m long street in 5 minutes as 7.2km/h.
Hence, option (D) is the correct answer.

Let the distance be x km. Then,
(Time taken to walls x km) + (Time taken to ride x km) = 23/4 hrs
⇒  (Time taken to walls 2x km) + Time taken to ride 2x km  =  23/2 hrs
but time taken to ride 2x km = 15/4 hrs 
Therefore, Time taken to walk 2x km = (23/2 - 15/4) hrs = 31/4 hrs =7 hr 45 min.

Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.
Then, 14x : 8y : 7z = 5 : 7 : 8.
Now,

Suppose C's capital = x then
B's capital = 4x (Since B's Capital is four times C's capital)
A's capital = 6x ( Since twice A's capital is equal to thrice B's capital)
A : B : C =6x : 4x : x
= 6 : 4 : 1
B's share

► S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.

By the rule of allilation, we have:

 Ratio of quantities of 1^st and 2^nd kind = 14 : 6 = 7 : 3.

► Let x kg of sugar of 1^st be mixed with 27 kg of 2^nd kind.

Then, 7 : 3 = x : 27

Alcohol content in the alcohol to be added = 100%
Therefore, the allegation diagram for this problem can be drawn as:

Therefore, ratio of the mixture to alcohol = 70/5 = 14/1
So, alcohol to be added = 100/14

Hence, the answer is "7.14 litres".

Choice B is the correct answer.

Suppose the slower pipe can fill the tank in x minutes.
Then the faster pipe can fill in x/4 minutes.
Part filled by the slower pipe in 1 min =1/x
Part filled by the faster pipe in 1 min =4/x
Part filled by the both the pipes in 1 min =1/x +4/x
 
Therefore, both the pipes can fill together in 36 minutes.
Part filled by both in 1 minute = 1/36.
1/x + 4/x = 1/36
5/x = 1/36
 x = 5 * 36 = 180

Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour = 
Remaining part = 

 hours i.e, 45 mins.
So, total time taken = 3 hrs. 45 mins.

CP = $20,000
SP for Harish = CP for Monika = 1.2 × 20,000 = $24,000
SP for Monika = CP for Sonu = 0.95 × 24,000 = $22,800

We need to know the S.I, principal and time to find the rate. Since the principal is not given, so data is inadequate.

The Theory of Comparative Advantage, proposed by David Ricardo, suggests that a country should specialize in producing goods and services in which it has a lower opportunity cost and trade those goods with other countries. By doing so, all countries can benefit from trade and overall efficiency increases.

The Industrial Policy of 1991 aimed to encourage foreign competition and attract foreign investment to boost industrial growth in India.

Economies of scale refer to the cost advantages that a firm experiences when it increases its scale of production. Large-scale industries benefit from economies of scale due to higher production levels, leading to lower average costs.