AP Police SI Mains 2019 Arithmetic and Reasoning

If a is first term and d is common difference of an AP.

Then nth term of that AP is$$=a+(n-1)d.$$

Sum of n terms of an AP is$$=\ \frac{\ n}{2}\left\{2a+\left(n-1\right)d\right\}.$$

So, from statement I wecan not say the value of a and n.

II.

11th term of that AP is 31, 

So, $$a+\left(11-1\right)3=31.$$

or,a=1.

Sum of first 21 terms is $$=\ \frac{\ 21}{2}\left\{2\times1+\left(21-1\right)\times3\right\}.$$

$$\ =\frac{\ 21}{2}\left\{2+60\right\}.$$

$$\ =\ 21\times\ 31=651.$$

So, B is correct choice.

I.

Let say value of n is 9 which is divisible by 9.

So,$$n+5=9+5=14$$,10 is not a factor of 14.

But, if value of n is 45 which is also divisible by 9, gives $$n+5=45+5=50$$.

10 is a factor of 50.

So, from statement I ,we can not say that whether 10 is a factor n+5 or not.

II.

n is even and also divisible 5 , this statement is only possible when the last digit of the number is 0.

In that case 10 is never be a factor of n+5.

So, B is correct choice.

I.

let put $$m=1 and m=2$$

when $$m=1$$ ,$$m^2+2m=1^2+2=3.$$ (which is an odd number.)

when$$m=2$$ ,$$m^2+2m=2^2+4=8.$$(which is an even number.)

So we can say that $$m^2+2m$$ is only even when m is an even number.

II.

let put $$m=1 and m=2$$

when $$m=1$$ ,$$m^2+m=1^2+1=2.$$

when $$m=2$$ ,$$m^2+m=2^2+2=6.$$

both the numbers are even numbers. So we can not say that whether m is odd or even.

So, A is correct choice.

I.

we know that $$\sec^2\theta\ -\tan^2\theta\ =1.$$

or,$$\left(\sec\theta\ -\tan\theta\ \right)\left(\sec\theta\ +\tan\theta\ \right)=1.$$

here,$$\sec\theta\ +\tan\theta\ =5.$$

So,$$\sec\theta\ -\tan\theta\ =1\div5=0.2.$$

So, $$2\sec\theta\ =5+0.2=5.2.$$

or,$$\sec\theta\ =2.6.$$

or,$$\cos\theta\ =\frac{\ 10}{26}=\frac{\ 5}{13}.$$

II.

we know that $$\sin^2\theta\ +\cos^2\theta\ =1.$$

or,$$\cos\theta\ =\pm\ \sqrt{1-\sin^2\theta\ \ }.$$

here,$$\sin\theta\ =\ \frac{\ 12}{13}.$$

So,$$\cos\theta\ =\pm\ \sqrt{1-\ \frac{\ 144}{169}\ }=\pm\ \sqrt{\ \frac{\ 25}{169}\ }=\ \pm\ \frac{\ 5}{13}.$$

So, Option A is correct choice.

Arithmetic mean of 4,20,40 and x is $$\left(4+20+40+x\right)\div4.$$

So,$$m=\left(64+x\right)\div4=\left(16+\ \frac{\ x}{4}\right).$$

I.

$$\ \frac{\ m}{x}=8+5=13.$$

or,$$\ \ m=13x.$$

So,$$\ 13x=16+\ \frac{\ x}{4}.$$

or,$$\ 51x=64.$$

or,$$\ x=\ \frac{\ 64}{51}.$$

So,$$\ m=13x=13\times\ \frac{\ 64}{51}.$$

II.  x is less than m, does not signifies any value of m.

So,A is correct choice.

Surface area of sphere is$$=4\pi\ r^2$$.

I. $$r=5$$.

Area of sphere$$=4\pi\ \times\ 5^2=100\pi\ .$$

II.

Perimeter of square $$=4a.\left(a=length\ of\ side\right).$$

So, from statement II we cant say what is the value of radius of sphere.

So, Option A is correct.

$$13\le\ x+y\le\ 19$$  and  $$-5\le\ x-y\le4$$

So, $$13+5\le\left(x+y\right)-\left(\ x-y\right)\le19-4.$$

or,$$9\ge\ y\ge\ 7.5.$$

So, both the equation needed.

C is correct choice.

We know that if Principal amount of P is compounded in n years at the rate of r percent,

Compounded Amount,A$$=P\times\ \left(1+\ \frac{\ r}{100}\right)^n$$.

But ,here P and A is given . r and n is missing.

So, it is not possible to calculate rate of interest.

D is correct choice.

We know that the nth odd number is 2n−1 and the nth even number is 2n

$$Now, let's assume the sum of first n odd numbers to be S i.e. S=1+3+5+…+(2n−1)$$

Now let us add 1 n times to the right side,

$$S+n=(1+1)+(3+1)+(5+1)+…+(2n−1+1)$$

$$or,S+n=2+4+6+…+2n$$

Now adding these 2 equations, we get,

$$2S+n=1+2+3+…+(2n−1)+2n$$

The RHS is the sum of first 2n natural numbers which is as below,

$$2S+n=2n(2n+1)/2$$

$$or,2S+n=2n^2+n$$

$$or,2S=2n^2$$

$$or,S=n^2$$

$$So the sum of first n odd integers is n^2.$$

So, Condition R is true.

But Condition A is not true.

So, Option D is correct choice.

Geometric mean of numbers a and b is 

$$=√ab.$$ and Arithmetic mean

$$=(a+b)/2$$.

we know that ,Geometric mean of two positive numbers is less than or equal to their Arithmetic mean.

So, Condition R is correct.

Condition A:

Let say ,length and breadth of the rectangle are l & b respectively.

So, $$2(l+b)=12.$$

Or,$$(l+b)=6.$$

And Area $$=lb$$.

multiplication of two numbers achieve when they are both equal in equation 1.

So,$$l=b=3.$$

Then only it can attain maximum value of 

$$3×3=9$$m^2.

So, condition 'A' is true and correct explanation made by 'R'.

So,Option A is correct choice.