AP Police SI Mains 2019 Arithmetic and Reasoning

Side of the square is 4 cm.

So,radii of the two circles are also 4 cms.

So, area of the one of the arc is 

$$(π×4^2/4)=4π.$$

Area of the square is$$=4×4=16cm^2.$$

clearly the common sector have the half of square common to it.

So,area of one arc is

$$4π-16/2=4π-8 cm^2$$

So,Total area of the arc is$$8π-16$$

$$=8(π-2).$$

A is correct choice.

As 2018,1998 and x are the side of triangle.

Then sum of two sides should be greater than the third side.

So,$$2018+1998>x$$

or,$$4016>x.$$

revesely,$$1998+x>2018.$$

So, minimum value of X should be 21 .

And maximum value of X should be 4016.

So,x can take$$(4016-21)=3995$$ values.

D is correct choice.

Minimum no. of identical small equilateral triangles that can be cut is 4. (1 at the top, 3 at the bottom). These 4 can be cut into 4 small triangles each.
Hence, the next value is 4*4 = 16 Hence, the number of identical small triangles that can be cut is of the form $$4^n. 256=4^4$$

C is correct choice.

Largest distanc in a cube is the diagonal of the cube.

if we consider a square of unit side then the diagonal of that square is

$$=√(1^2+1^2)=√2.$$

so,value of largest diagonal is

$$=√(1^2+(√2)^2)=√3.$$

D is correct choice.

The train which started from station A traveled 144 km in first 2 hours.

The train from station A and station B had speed of 72 km/h and 90 km/h respectively.

And they expected to be meet in 1:30 PM.

But from 11:00 AM they started to travel towards eachother.It means that they expected to meet eachother by 2.5 hours from 11:00AM.

Their relative velocity is=(72+90)=162 km/h.

So, in 2.5 hours they should travel=2.5×162=405 km. So they were 405 km apart from eachother at 11:00 AM.

Let say, due to signal,they reduced their speed by x km/h.

So,train from station A and station B will have (72-x) km/h and (90-x) km/h as their respective speed.

But,they shifted their speed to this reduced speed at 12 noon, which means that they travel for more 1 hour from 11:00AM at their previous speeds.

So,in that 1 hour they travel for more 72 km and 90 km further towards eachother.

Which in turn reduced their total journey of 405 km by (72+90)=162 km by this 1 hour.

So,they traveled only (405-162)=243 km in (12 noon-4:30 PM)= 4.5 hours.

But their new relative velocity is (162-2x) km/h.

So, they will take (243/(162-2x)) hours to cross eachother.

So,

$$(243/(162-2x))=4.5$$

or,$$54=162-2x$$

or,$$2x=108$$

or,$$x=54$$.

So, the train ,which left from station A,have reduced its speed to (72-54)=18 km/h.

C is correct choice.

First they go in opposite direction .

So, relative velocity of them is (40+60) km/h or 100 km/h.

So, time taken by them to meet for first time is =(120/100) h= 1.2 hours.

So,in their first meeting P travels =(1.2×40)=48 km and Q travels=(1.2×60)=72 km.

But,after first meeting P and Q reversed their speeds into 60 km/h and 40 km/h respectively.

Now,P will take (48/60)=0.8 hours to reach point A and Q will take (72/40)=1.8 hours to reach point B.

But in the mean time, P will move 60 km further from point A towards point B in next 1 hour when Q was traveling towards B.

So far ,after first meeting they traveled for 1.8 hours to get a head on move towards eachother.

So,for their second meeting they had to travel for only (120-60)=60 km.(as P already moved 60 km from point A)

So,100 km/h would be their relative velocity.

So,they will take (60/100)=0.6 hours to meet for the second time.

So,Total time taken by them to meet for the second time is=(1.2+1.8+0.6)=3.6 hours.

B is correct choice.

Let say, speed of the boat is b km/h and speed of the current is c km/h.

So, in downstream speed is (b+c) km/h and in upstream speed is (b-c) km/h.

Time taken in downstream=(12/(b+c)) h.

and time taken in upstream=(12/(b-c)) h.

So, according to question:

$$(12/(b-c))-(12/(b+c))=6.$$

or,$$(b+c-b+c)/(b^2-c^2)=(1/2).$$

$$b^2-c^2=4c.$$

$$b^2=c^2+4c.$$..............(1)

if speed of speed is doubled ,then speed in downstream =(2b+c) km/h .

and speed in upstream=(2b-c) km/h.

So, time taken in downstream 12/(2b+c) h.

And time taken in upstream 12/(2b-c) h.

So,

$$12/(2b-c)-12/(2b+c)=1$$

or,$$(2b+c-2b+c)/(4b^2-c^2)=(1/12)$$

or,$$24c=4b^2-c^2$$

or,$$b^2=(c^2+24c)/4.$$........(2)

So,from (1) & (2) :

$$(c^2+24c)/4=c^2+4c.$$

or,$$c^2+24c=4c^2+16c$$

or,$$3c^2=8c$$

or,$$c=8/3$$

So,speed of current is 8/3 km/h.

A is correct choice.

Let say, 

p,q and r are the three parts.

So,Then the simple interest would be

$$(p×0.04×1)=(q×0.04×2)=(r×0.04×3)$$

$$=k(let say).$$

$$p=25k,q=12.5k and r=(25/3)k.$$

So,

$$p:q:r=25k:12.5k:(25/3)k$$

$$=k:0.5k:(k/3)$$

$$=6:3:2.$$

A is correct choice.

At the end of first year

$$=5000×1.1=5500$$Rs would be paid.

After 1500 payment,he would be left with

$$5500-1500=4000.$$

At the end of second year

$$4000×1.1=4400.$$

After 1500 payment,he would be left with 

$$4400-1500=2900.$$

So,at the end of third year the amount would become$$=2900×1.1=3190.$$

B is correct choice.