AP Police SI Mains 2019 Arithmetic and Reasoning

Let us say,$$k=2^{350}× 5^{409}×3^{149}$$

So,$$A=4×25×27k=2700k.$$

$$B=4×5×81k=1620k.$$

$$C=4×125×1k=500k.$$

$$D=8×1×3k=24k.$$

So,$$A>B>C>D$$

A is correct choice.

We know that -

Dividend=divisor*quotient+remainder

For our question dividend must be of 5 digit.Let dividend be x.

Divisor=LCM(7,11,21)=231.

Let quotient be y.

Remainder is 3.

Substitute above values in equation-

x=231*y+3

Now we need to substitute different value of y(y=1,2,3…) in order to find smallest 5 digit x.

When y=1 then x=234

.

.

.

When y=43 then x=9936

When y=44 then x=10167

When y=45 then x=10398

So least 5 digit number(10167) is obtained at y=44.

The 5 digit number when divided by 7,11,21 which gives remainder 3 is 10167.

So,Option B is correct choice.

As per divisibility rule of 11,

$$9+7+8+7=3+x+y$$

or,$$31=3+x+y$$

or,$$x+y=28$$

but ,x and y are single digit number they can take a maximum value of 9 .

if both the digits are 9 then it can't be divisible.

So,either x or y is equal to 9 or 8.

So, the maximum value of x+y is 17.

D is correct choice.

If we substract 50 from 2035 it will become

1085.

Now when 1085 is divided by 9,10 and 15 they will give a remainder of 5 each.

A is correct choice.

Let say,

$$x=\overline{.27}\ and\ y=\overline{.94\ }\ .$$

So,we can rewrite it as,

$$x=.27\overline{27\ }\ and\ \ y=.94\overline{94\ }\ .$$

or,$$100x=27.\overline{27\ }\ and\ \ 100y=94.\overline{94\ }\ .$$

or,$$100x=27+.\overline{27\ }\ and\ \ 100y=94+.\overline{94\ }\ .$$

or,$$100x=27+x\ and\ \ 100y=94+y\ .$$  (as $$x=\overline{.27\ }\ and\ \ y=\overline{.94\ }\ .$$)

or,$$99x=27\ and\ 99y=94.$$

or,$$x=\frac{\ 27}{99}\ and\ \ y=\frac{94\ }{99}.$$

So,$$1.\overline{27\ }+.\overline{94\ }=1+x+y.$$

or,$$1.\overline{27\ }+.\overline{94\ }=1+\frac{27\ }{99}+\frac{94\ }{99}.$$

or,$$1.\overline{27\ }+.\overline{94\ }=\frac{220\ }{99}=2.22222222.....=2.\overline{2\ }.$$

B is correct choice.

It is given that ,$$c^2=a^2+b^2$$.

Now,if you compare set(a,b,c) with set(3,5,X) then we can say that a=3,b=5 & c=x.

So,$$x^2=3^2+5^2$$

or,x=√(9+25)=√34.

Similarly,

$$y=√(49-9)=√40$$ and 

$$z=√(25-1)=√24$$                           

so,$$LCM(x^2,y^2,z^2)=LCM(34,40,24)$$

$$=2^3×17×5×3.$$

so,$$p1+P2+P3+p4=2+17+5+3=27.$$

and $$a1+a2+a3+a4=3+1+1+1=6$$

so,ratio would be$$=27/6=9/2.$$

A is correct choice.

Between 100 and 1000 there are 81 numbers present which are divisible by 11.

81 can be divisible by 1,3,9,27 & 81.

So, value of x is 5 .

B is correct choice.

a) a,b are prime numbers then it will go with option(v).

b) if a and b are composite number the their multiplication shall be greater than their LCM.so,it will go with (i).

c) c will definitely go with (iv).

and d) with (iii) .

C is correct choice.

$$34864764=2^2\times3\times11\times264127.$$

So,$$p_1,p_2,p_3\ could\ have\ values\ 2,3,11.$$ and $$\alpha\ ,\beta\ ,\gamma\ \ could\ have\ values\ of\ 2,1,1.$$

So,$$\left(p_1+p_2+p_3\right)\left(\alpha\ +\beta\ +\gamma\ \right)=\left(2+3+11\right)\left(2+1+1\right)=16\times4=64.$$

A is correct choice.

From the given question we can say that,

$$x=\frac{1\ }{2+x}\left(as\ the\ series\ reapeted\ from\ the\ same\ position\right).$$

or,$$x^2+2x=1.$$

or,$$x^2+2x-1=0.$$

or,$$x^2+2x+1-2=0.$$

or,$$\left(x+1\right)^2=2.$$

or,$$x+1=\sqrt{2\ }.$$

or,$$x=\left(\sqrt{2\ }-1\right).$$

So, C is correct choice.