AP Police SI Mains 2019 Arithmetic and Reasoning

Volume of well$$=π2^2×35=440m^3.$$

Volume of truck$$=5×2×0.5=5m^3.$$

but the truck is only able to fill its 80% of capacity to fill the space.

So, effective amount was$$=0.8×5=4m^3.$$

So,20% of wasted amount is

$$=0.2×440=88m^3.$$

So, number of trips required are

$$=(440/4)+(88/4)=110+22=132.$$

D is correct choice.

Let say, Volume of sphere$$=9πk.$$ and Volume of cube$$=2k.$$

So,$$(4/3)πr^3=9πk$$

or,$$r=3k/2^(2/3).$$

and edge of cube$$=k2^(1/3).$$

So, required ratio would be$$=3:2.$$

B is correct choice.

According to the question,

length of the path would be 40m and width of the path would be 2m by the length side(area of that path is$$=40×2=80m^2$$).

And ,length of the path would be 20m and width of the path would be 2m by the breadth side(area$$=20×2=40m^2$$).

If it cost 500 Rs. to laying the path, total cost of laying the path would be

$$=80×500+40×500=60000.$$

But,in corner a reapetated cost is taken during the path ,whose length and breadth would be of 2m respectively.

So,area of that portion is$$=2×2=4m^2.$$

So,cost of that path$$=4×500=2000Rs.$$

So,Total cost$$=60000-2000=58000.$$

A is correct choice.

He can carry maximum number of boxes ,only if he put one box over another .

So, maximum number boxes he can carry is

$$=15×15×15=3375.$$

C is correct choice.

Length of the arc is$$=8π.$$

So, perimeter of the base of cone is$$2πr$$.

So,$$2πr=8π.$$

or,$$r=4.$$

radius of the circle is 12 cm.

So,slant height of the cone is 12 cm.

So, height of the cone$$=√(12^2-4^2)$$

$$=√128=8√2.$$

So,V$$=(1/3)π4^2×8√2.$$

and S$$=π×4×√(h^2+r^2)=4π×12.$$

So,V:S$$=8√2:9.$$

C is correct choice.

Volume of the sphere is

$$=(4/3)πr^3.$$

height of cone r/2.

let say radius of cone is a.

So, volume of cone is 

$$=(1/6)πa^2r.$$

side of cube is $$=r/2.$$

volume of cube$$=r^3/8.$$

So,$$(4/3)πr^3=(1/3)πa^2r+r^3/8.$$

So,$$a=(√11/2)r$$.

So, height: radius$$=√11:1$$.

B is correct choice.

Radius of circle is$$=14/2=7cm.$$

From the picture we can say that area of the shaded region is$$=(1/3)×3×area$$.

So, required area$$=π×7^2=154$$.

D is correct choice.

Let say ,a be the length of the side conjugate to the right angle .

so,length of the diameter of the circle is

$$=√2a$$.

so,radius$$=(√2/2)a$$.

so,$$∆=πa^2/2$$.

Now,∆1=∆2=

$$π(a/2)^2=πa^2/4.$$

So,∆1+∆2=$$πa^2/2.$$

So,∆=∆1+∆2.

C is correct choice.

If 4 squares of 4×4 is removed from each corner then length of the rectangle become

$$36-4-4=28cm$$ , breadth of the rectangle become$$=24-4-4=16cm$$ and height of the box is 4 cm.

So,volume of box is$$=28×16×4=1792.$$

A is correct choice.

So,length of the diagonal of the square=$$\sqrt{2}r+\sqrt{2}r+\left(4\times r\right)=\left(4r+2\sqrt{2}r\right)$$cm.

But ,diagonal of Square=$$\sqrt{a^2+a^2\ }=\sqrt{2}a$$=$$8\sqrt{2}.$$(square of side 8 cm)

So,

$$8\sqrt{2}=4r+2\sqrt{2}r.$$

or,$$r=\frac{8\sqrt{2}}{4+2\sqrt{2}}.$$

or,$$r=\frac{8\sqrt{2}\left(4-2\sqrt{2}\right)}{16-8}=\sqrt{2}\left(2\left(2-\sqrt{2}\right)\right)=2\left(2\sqrt{2}-2\right)=4\left(\sqrt{2}-1\right).$$

C is correct choice.