AP ICET 2013

Solution

A, B, C can individually complete a work in 20 days, 15 days, 12 days respectively

Efficiency of (A + B + C) = $$\left(\frac{1}{20}+\frac{1}{15}\ +\frac{1}{12}\right)$$ = $$\left(\frac{3+4+5}{60}\right)\ =\ \frac{12}{60\ }$$

Efficiency of only (B + C) =$$\left(\frac{1}{15}+\frac{1}{12}\right)$$ = $$\left(\frac{4+5}{60}\right)\ = \ \frac{9}{60\ }$$

So efficiency of A = $$\frac{12}{60}-\frac{9}{60}\ =\ \frac{3}{60}$$

While working, all 3 work for 3 days.

So work completed in 3 days = $$3\ \times\ \frac{12}{60}\ =\ \frac{36}{60}$$

Work Remaining = $$1\ -\frac{36}{60}\ =\ \frac{24}{60}$$ 

If A works alone @ $$\frac{3}{60}$$ work per day. = $$\frac{\left(\frac{24}{60}\right)}{\left(\frac{3}{60}\right)}$$ = 8 days.

Solution

Let Perpendicular be p, base be b and Hypotenuse be h

As it is isosceles right angle triangle, 

so P=B

According to Pythogaras Theorem 

$$h\ =\ \sqrt{\ p^{2\ }+b^2}$$

$$12=\ \sqrt{\ b^{2\ }+b^2}$$

$$12= \ \sqrt{\ 2b^{2\ }}$$

$$12=\ \sqrt{\ 2}\ b$$

$$b\ =\ 6\sqrt{\ 2}$$

Area = $$\ \frac{\ 1}{2}\ \times\ product\ of\ sides\ \times\ \sin\ of\ included\ angle\ $$

       = $$\ \frac{\ 1}{2}\ \times\ b\times\ b\ \times\ \sin\ 90^{\circ\ }$$

       =  $$\ \frac{\ 1}{2}\ \times\ \left(6\sqrt{\ 2}\ \right)^2\times\ 1$$

       = $$\ \frac{\ 1}{2}\ \times\ 72\times\ 1$$

      =36 Answer

Solution

Area of Diagonal = $$\frac{1}{2\ }\left(product\ of\ diagonals\right)$$ = $$\frac{1}{2\ }\left(18\ \times\ 24\right)$$ = 216 $$cm^2$$

Solution:

Volume of Cylindrical bar = Volume of cast sphere

$$\pi\ r^2\ h\ =\ \frac{4}{3\ }\pi\ R^3$$

$$\pi\ \left(0.75\right)^2\ 1\ =\ \frac{4}{3\ }\pi\ R^3$$

$$\left(\frac{3}{4}\right)^2\ =\ \frac{4}{3}R^3$$

$$\left(\frac{3}{4}\right)^2\ \ .\ \frac{3}{4}=\ R^3$$

$$\left(\frac{3}{4}\right)^3=\ R^3$$

R= 3/4 m 

D = 2R = 3/2 = 1.5m

Solution:

Bases (b) are in the ratio 2:1

Height (h) are in the ratio 1:2

As, volume's formula is defined as $$b^2h$$

So Volume ratio => $$\left(2\right)^21\ :\ \left(1\right)^22\ =\ 4:2\ =\ 2:1$$

Answer = 2:1

Solution:

The radius of the cone is 3 times the radius of the cylinder and their heights are same.

Volume of cone = $$\frac{1}{3}\pi\ R^2h$$

Volume of cylinder = $$\pi r^2h$$

Given, R= 3r

So,Volume of cone = $$\frac{1}{3}\pi\ (3r)^2h$$

Ratio of volumes =  $$\frac{1}{3}\pi\ (3r)^2h$$ : $$\pi r^2h$$ = 3:1 Answer

Solution:

Area of the window = Area of rectangular part + area of semicircular part

                              = $$l.b\ +\ \ \frac{\ \pi\ r^2}{2}$$          ......(Radius of semicircle = 0.5b)

                              = $$6\ \times\ \left(2\cdot2\right)\ +\ \ \frac{\pi\ \left(2\right)^{2\ }\ }{2}\ =\ 24\ +2\pi\ $$

                              = $$24\ +2\pi\ $$                                

Solution 

Volume of given steel = 11$$m^3$$

Volume of the bar = $$\pi\ \left(\frac{0.1}{2}\right)^2\ 1.4\ m^3$$ = $$\frac{22}{7}\frac{0.01}{4}\ 1.4\ m^3\ =\ 11\cdot\frac{0.01}{2}\cdot0.2$$ = $$11\cdot0.01\cdot0.1\ =\ 0.011$$

No. of bars, n =  $$\left(\frac{\left(Volume\ of\ steel\right)}{Volume\ of\ a\ bar}\right)\ =\ $$ 

$$\left(\frac{11}{11\ \times\ 10^{-3}}\right)\ =\ 10^{3\ }\ =\ 1000\ Answer$$

Solution

y will be in the form of 5x+2

For,

$$3^{48} = x(mod 10)$$

We need to find remainder when $$3^{48\ }$$ is divided by 10.

3 has a cyclicity if 4. So as 48 is a multiple of 4, so last digit will be 1.

Hence remainder on division  by 10 =1

1 Answer