Jharkhand Police Constable (Paper-II) Mock Test - 3 - Jharkhand Police Constable MCQ

A tap can fill a tank in 4 hours.
Therefore the tap can fill half the tank in 2 hours.
Remaining part = 1/2
After half the tank is filled, two more similar taps are opened.
Hence, total number of taps becomes 3.
Part filled by one tap in 1 hour = 1/4
Part filled by three taps in 1 hour 
= 3 × 1/4 = 3/4
i.e., 4 taps can fill remaining half in 40 minutes.
Total time taken = 2 hour + 40 minute 
= 2 hour 40 minutes

part filled in 2 hours = 
Remaining  part =  
∴ (A + B)'s 7 hour's work = 2/3
(A + B)'s 1 hour's work = 2/21
∴ C's 1 hour's work = {(A + B + C)'s 1 hour's  work} - {(A + B)'s 1 hour's work}

∴ C alone can fill the tank in 14 hours.

Gain = 10%
Let the C.P. be x.

Let the original price be Rs. 100.
Marked down price = Rs. 80
Now, we want to raise 80 to 100.

Simple Interest = PRT / 100
Here P = Principal, R = Rate of interest and T = time period
Hence, Simple Interest ∝ T
Required Ratio = (Simple Interest for 5 years) / (Simple Interest for 15 years)
⇒ T₁/ T₂ = 5 / 15
⇒ 1 / 3 = 1 : 3

We can get SI of 3 years = 12005 - 9800 = 2205

SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]

Principal = 9800 - 3675 = 6125 

So Rate = (100*3675)/(6125*5) = 12%

According to the question, rotating square along a diagonal makes two right-circular cones.
As the diagonal makes a right-angled triangle by the sides of square, so
D = 9√2 m (Using Pythagoras theorem)

So, curved surface area of the cone = 2 πrl

According to the question:
Volume of the drum = πr²h

Volume of small can

Divide both the above volumes to get the maximum number of cans as 52.73, i.e. maximum 52 cans can be completely filled. 

Explanation:

Let the number of wickets taken by Srinath be x then the number of wickets taken by Kumble will be 2x−3
According to question, x(2x−3)=20

Ratio of amounts with Sam : Michael : Puma = 100 : 60 : 40 = 5 : 3 : 2
Let us assume that amounts spent are 5x, 3x and 2x, respectively.
A = 5x - 3x = 2x
B = 3x - 2x = x
Hence, A : B = 2 : 1 (Total expense is the redundant data in the question)

Income of Aman = Rs. 4x
Income of Raman = Rs. 3x
Expenditure of Aman = Rs. 3y
Expenditure of Raman = Rs. y
According to question:
4x - 3y = 16,000 … (i)
3x - y = 25,000 … (ii)
Multiplying (ii) by 3 and then subtracting (ii) from (i), we get

x = 11,800
Aman's income = Rs. 4x
= Rs. 4 × 11,800
= Rs. 47,200

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 - 63x    ⇔    120x = 40

x =1/3

Number of floors covered by David in (1/3) min. =((1/3)x 57)= 19

So, their paths cross at (11+19) i.e., 30^th floor.

Let C's share = Rs. x

Explanation:Let the one number be xx .As the sum  of numbers is 27 , then the other number will be (27−x)(27−x)                                                                                                                                    According to question

X+(1/X) = 2 ;

(X^2+ 1)/ X = 2 ;

X^2+1 = 2X ;

X^2 - 2X + 1= 0 ;

By factorisation method ,

X = 1 ;

Now substitute X= 1 in X +(-1/X) ;

we get , 1 - 1/1 = 0 ;

Therefore , X = 0 ;

Avg of 7 no= (7+1)/2 = 8/2 = 4
Avg of 9 no = (9+1)/2 = 5
5- 4 = 1
Hence increased by 1

Runs scored by Kohli in first 4 innings = 48*4 = 192
Average of 5 innings is 60, so total runs scored after 5 innings = 60*5 = 300
Hence runs scored by Kohli in fifth inning = 300 – 192 = 108
It is given that in 6th innings he scores 12 runs more than this, so he must score 120 in the sixth inning. Hence total runs scored in 6 innings = 300+120 = 420
Now average of last five innings is 78, so runs scored in last innings = 390
Hence runs scored in first inning = 420 – 390 = 30.

Let the score of Om =x
Total marks of all the students = 21*62 +x
As per the question, The average score of all the 22 students is one more than the average score of the 21 students other than Dev,
Or (21*62 +x)/22 -1 = (21*62 +x – 82.5)/21
(21*62 + x -22)/22 = (21*62 +x – 82.5)/21. Therefore, x =51

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

Total marks = N
Pass marks = 45% of N = 0.45N
Marks obtained = 36
It is given that, obtained marks is 68% less than that pass marks
=>the obtained marks is 32% of the pass marks.
So, 0.32 * 0.45N = 36
On solving, we get N = 250
Hence, option D is the correct answer.

Correct option is C
Time taken by first sprinter 
= 100/10 = 10sec
Time taken by second sprinter 
= 100/80 = 12.5sec
Difference = 12.5 - 10 = 2.5 sec

• X’s five day work = 5/20 = 1/4. Remaining work = 1 – 1/4 = 3/4.
• This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 × 4/3 = 20 days.  
• X & Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.

So Option C is correct

Time taken = 1 hr 40 min 48 sec = 1 hr (40 + 4/5) min = 1 + 51/75 hrs = 126/75 hrs
Let the actual speed be x km/hr
Then, (5/7) * x * (126/75) = 42
⇒ x = 42 * 7 * (75 / 5) * 126
⇒ x = 35 km/hr