Jharkhand Police Constable (Paper-II) Mock Test - 4 - Jharkhand Police Constable MCQ

Lalloo is unfortunate that the friend is moving away from him.

(Because the friend moves in same direction as Lalloo).

relative speed= 20- 15= 5,kmph. distance= 200 m.

Thus, Lalloo will meet his friend when he gains 200 m over him.

=> time required = distance / speed = 0.2/5 = 1/25 hrs.

=>Distance travelled by the friend in 1/25 hrs. (when Lalloo catches up him)

=> Time x Speed = 1/25 x 15 = 3/5 km = 600 m.

Let
P's capital = p,
Q's capital = q and
R's capital = r.
Then

A:B:C
=(10×7):(12×5):(15×3)
=70:60:45
=14:12:9

► Cost Price(CP) of Type 1 material is Rs. 15 per kg
► Cost Price(CP) of Type 2 material is Rs. 20 per kg
Type 1 and Type 2 are mixed in the ratio of 2 : 3

Therefore, rice per kg of the mixed variety of material = Rs.18

Part filled by (A + B) in 1 minute = 
Suppose the tank is filled in x minutes.
Then, 
⇒ x = 30 min.

Part emptied by the leak in 1 hour = 1/6
Net part emptied by the leak and the inlet pipe in 1 hour = 1/24
Part filled by the inlet pipe in 1 hour 
i.e., inlet pipe fills the tank in 8 hours = (8 * 60) minutes = 480 minutes
Given that the inlet fills water at the rate of 4 liters a minute
Hence, water filled in 480 minutes = 480 * 4 = 1920 litre
i.e., The cistern can hold 1920 litre

Cost price for 100 articles = Rs. 1900
Let the market price per article be Rs. M.
So, sales revenue = 95 × M × 0.7 ... (i)
As profit = 40%, sales revenue = Rs. 1.40 × 1900 = Rs. 2660
So, 95M × 0.7 = 2660
Or M = 40
Thus, market price = Rs. 40
Hence, answer option (1) is correct.

The SP = 106.25% of the CP. Thus, CP = 29.75/1.0625
= ₹ 28.

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x)

⇒ 28x - 22x = 350800 - (13900 * 22)
⇒ 6x = 45000
⇒ x = 7500
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400

Simple Intrest = SI = PRT/100
Where P = Principal, R = Rate of intrest and T = time period
Total SI = SI from SBI and SI from ICICI
∴ 220 = (500 x R x 2)/100 + (300 x R x 4)/100
⇒ 220 = 10R + 12R
⇒ R = 10% = Rate of intrest

Semi-perimeter (2a) = 18 cm
Area (a × a) = 81 cm²
Solving both, we get a = 9 cm.
Now, circumference = 2 × 18 = 36
So, r = 
Area of circle = 

Volume of water displaced by 150 men = 150 × 8 = 1200 m3
Let the rise in water level be x.
90 × 40 × x = 1200
⇒ x = 12/36 = 1/3 or 0.33 m = 33.33 cm (Approx.)

x² - ix + 6 = 0
x² - 3ix + 2ix - 6i² = 0    { i² = -1}
x(x-3i) + 2i(x-3i) = 0
(x+2i) (x-3i) = 0
x = -2i, 3i 

{(1 + i)/(1 - i)}ⁿ = 1
multiply (1 + i) numerator as well as denominator .
{(1 + i)(1 + i)/(1 - i)(1 + i)}ⁿ = 1
{(1 + i)²/(1² - (i)²)}ⁿ = 1
{(1 + i² +2i)/2 }ⁿ = 1
{(2i)/2}ⁿ = 1
{i}ⁿ = 1
we know, i⁴ⁿ = 1 where , n is an integer.
so, n = 4n where n is an integers
e.g n = 4 { because least positive integer 1 }
hence, n = 4

Let salary of A last year be Rs. 3s.
Salary of B last year = Rs. 4s
Let present salary of A be Rs. A and that of B be Rs. B.
So, 3s/A = 4/5 or A = 15s/4
4s/B = 2/3 or B = 6s
So, adding up the present salaries, we get (15s/4) + 6s = 4160
Or, s = 426.667
Present salary of A = Rs. A = Rs. (15/4)s = Rs. (15/4) × 426.667 = Rs. 1600
Thus, answer option (2) is correct.

Suppose the shares of A, B and C are Rs. 2x, Rs. 3x and Rs. 4x, respectively.
When C gives Rs. 20 to B:
Amount with C = Rs. (4x - 20) and amount with B = (Rs. 3x + 20)
Now, B gives half of his amount to A.
Amount with A = Rs.

According to the question,

Or 4x + 3x + 20 = 8x - 40 or x = 60
Hence, total sum of money = Rs. (2x + 3x + 4x)
= Rs. 9x = Rs. (9 × 60) = Rs. 540

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 - x) respectively. Then,

Savings in Public Provident Fund = Rs. (150000 - 90000) = Rs. 60000

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

N = 7+2 =9
3n = 27
27/7 = 6

2*3*5*7*11*13*17 = 510510

Change in total weight of 10 students = difference in weight of the student who joined and the student

=> weigth of first student who left = 30 + (10×2) = 50

weight of the student who joined last = 50 – 10 = 40...
Thus change in average weight = (40 – 30)/10 = 1...
 

When we multiply each number by 9, the average would also get multiplied by 9.

Hence, the new average = 18 X 9 = 162.