CMAT 2018 Slot 2

Probability = Expected number of outcomes/ Total number of outcomes.

Total number of outcomes we get in a single throw of two dice = $$6\times 6$$ = 36.

Possible cases of getting 'nine' in a single throw of two dice:

                  dice 1           dice 2

                  three             six..................(1)

                  six                three................(2)

                  four              five..................(3)

                  five              four...................(4)

So, total of 4 cases.

Possible cases of getting 'ten' in a single throw of two dice:

                dice 1            dice 2

                four                six.....................(1)

                 six                 four....................(2)

                five                 five....................(3)

So, total of 3 cases.

Expected number of outcomes = Total possible cases of getting 'nine' or 'ten'  in a single throw of two dice = 4 + 3 = 7 .

So, Probability = $$\frac {7} {36}$$

Let the breadth(b) of the room be 'x' metres.

then, length(l) of the room = x+2 metres.

Area(A) = $$l\times b$$ = x(x+2) $$m^2$$

Given, length is increased by 4 meters and the breadth decreased by 2 meters

Then, new length(l') of the room = x+6 metres

new breadth(b') of the room = x-2 metres

New Area(A') of the room = $$l'\times b'$$ = (x+6)(x-2) $$m^2$$

Also given that, A = A'

$$\Rightarrow x(x+2) = (x+6)(x-2)$$

$$\Rightarrow x^2+2x = x^2+4x-12$$

$$\Rightarrow 2x = 12$$

$$\Rightarrow x = 6$$

Therefore the length of the room (l) = 8 metres

and breadth of the room (b) = 6 metres

and given height of the room (h) = 3 metres

Since the room will be in the shape of a cuboid, Surface area = 2 ($$l\times b+b\times h+l\times h$$)

But the Surface area of Walls = Total Surface area - Area of Roof and Floor = 2 ($$l\times b+b\times h+l\times h) - 2 (l\times b) = 2 (8\times 3+6\times 3) = 84 m^2$$

Hence, Surface Area of walls = 84 $$m^2$$.

Average  Speed = Total  distance  covered $$\div$$ Total  time  taken

Total distance travelled = 50 + 50 + 30 = 130 km.

Total time taken = Time taken to travel first 50 km + Time taken to travel next 50 km + Time taken to travel next 30 km = $$40 + 50\div 2 + 30\div 1 = 95$$ minutes = $$\frac{95}{60}$$ hours.

$$\Rightarrow$$ Average Speed = $$130\div \frac{95}{60} = 82.1$$ kmph

First wheel makes 60 revolutions in 1 minute

$$\Rightarrow$$ It makes 60 revolutions in 60 seconds

$$\Rightarrow$$ It makes 1 revolution in 1 second.

This implies, after every 1 second the certain point at which the wheel started its revolution reaches its initial position.

Similarly, Second wheel and Third wheel makes 36 and 24 revolutions in 1 minute respectively.

$$\Rightarrow$$ Second and Third wheel makes 1 revolution in $$\frac{5}{3} and \frac{5}{2}$$ seconds respectively.

So for all the multiples of $$\frac{5}{3} and \frac{5}{2}$$ seconds the certain point of second wheel and third wheel reaches its initial position respectively.

After LCM {1, $$\frac{5}{3}, \frac{5}{2}$$} seconds all the three wheels will come together in the same position.

LCM of fractions = LCM of numerators/ HCF of denominators 

$$\Rightarrow$$ LCM {1,$$\frac{5}{3}, \frac{5}{2}$$} = LCM {1,5,5}$$\div$$HCF {1,3,2} = 5$$\div$$1 = 5.

Hence, after 5 seconds all the wheels will come again together in the same position.

If the  principle amount 'P' when compounded annually for 'n' years at 'R%" interest rate per annum becomes P'.

Then $$P' = P {[1 +\frac{R}{100}]}^{n}$$

Given P' = 2000, n = 2 years, R = 10% 

$$\Rightarrow P = P'\div {[1 +\frac{R}{100}]}^{n}$$

$$\Rightarrow P = 2000\div {[1 +\frac{10}{100}]}^{2}$$ 

$$\Rightarrow P = 2000\div 1.21$$

$$\Rightarrow P = 1653$$

Hence the initial amount P = Rs. 1,653.

The Volume of the right circular cone of base radius 'r' and height 'h' is given by 'V' = $$\frac{1}{3} \pi r^2h$$

Given 'h' has been increased by 200%

$$\Rightarrow$$ New height h' = h[1 +$$\frac{200}{100}$$] = 3h

also,radius of the base is reduced by 50%

$$\Rightarrow$$ New base radius r' = r[1 - $$\frac{50}{100}$$] = $$\frac{r}{2}$$  

New Volume of the cone with new base radius r' and new height h' is given by V' = $$\frac{1}{3} \pi r'^2h'$$ = $$\frac{1}{3} \pi (\frac{r}{2})^2(3h) = \frac{3V}{4}$$.

Change in Volume = $$\frac{New Volume - Old Volume}{Old Volume}\times 100  = \frac{\frac{3V}{4} - V}{V}\times 100 = -25$$

Hence the new volume decreased by 25 %.

Initial price is given as 'I' = Rs. 600

After the first reduction, the initial price is reduced by 10%

$$\Rightarrow$$ the new price I' = $$600[1 - \frac{10}{100}] = 540$$

After second reduction, I' is reduced by 10%

$$\Rightarrow$$ the new price I'' = $$540[1 - \frac{10}{100}] = 486$$

After third reduction, I'' is reduced by 10%

$$\Rightarrow$$ the new price I''' = $$486[1 - \frac{10}{100}] = 437.4$$

Hence the Current price after three successive reductions is Rs. 437.4

Let say,

The number mobiles sold in last year of the brands LG, SAMSUNG, NOKIA, SONY, MICRO-MAX be A, B, C, D, and E respectively.

Given that A+B+C+D+E = 1 crore.

Out of these 1 crore mobiles, the number of mobile sets of LG sold  are 15% of A = $$\frac{15}{100}\times A$$.

But from the given data, the values of A, B, C, D, and E cannot be found out.

So the number of LG sets sold last year cannot be determined.

Note that the 15% does not represent the percentage of LG mobiles among the ones that are 1 yr old, but the percentage of 1 yr old mobiles among LG mobiles.

$$\sqrt{188+ \sqrt{51+ \sqrt{169}}}$$ = $$\sqrt{188+ \sqrt{51+13}}$$ = $$\sqrt{188+ \sqrt{64}}$$ = $$\sqrt{188+ 8}$$ = $$\sqrt{196}$$ = 14

If the principle amount 'P' when compounded half-yearly at R% interest rate per annum for 'n' years, the new amount is P'.

then $$P' = P{[1 + \frac{R}{2\times 100}]}^{n}$$

Given P' = 6,632.55, P = 6,250 and R = 4%

$$\Rightarrow 6,632.55 = 6,250{[1 + \frac{4}{2\times 100}]}^{n}$$

$$\Rightarrow 1.061 = {1.02}^{n}$$

Taking logarithm on both sides we get,

n = log(1.061)$$\div$$log(1.02) = 3

Since n refers to half a year in this case, the number of years will be $$\frac{3}{2}$$ years.