CMAT 2018 Slot 1 Question Paper

Volume(V)  of the sphere with radius R=8 cm is given by V= $$\frac{4}{3} \pi  R^{3}$$.

                                                                                                       = $$\frac{4}{3} \pi(8^{3})$$= $$\frac{4}{3} \pi (512) cm^{3}$$.                                                                                                

and Volume (v) of each iron ball with radius r= 1cm is given by v= $$\frac{4}{3} \pi r^{3}$$.

                                                                                                              = $$\frac{4}{3} \pi (1^{3})$$= $$\frac{4}{3} \pi cm^{3}$$.

Let say 'n' iron balls each of volume 'v' are required to form a sphere of volume 'V'.

=> Total volume of 'n' iron balls = Volume of the sphere

=> $$n\times v$$ = V

=> n= $$V\div v$$ = $$\frac{4}{3}  \pi (512)\div \frac{4}{3}  \pi$$ = 512. 

Hence 512 iron balls are required in total to form the sphere.

Given      0<a<1..............................................(1)

Let us go with the verification of options

Option (i):

squaring on both sides of (1), we get

0<$$a^2$$<1

Subtracting '1' in the above equation, we get

-1<$$a^2$$-1<0....................(2)

$$\Rightarrow a^2$$-1< 0, Hence option (i) is false.

Option (ii):

Squaring on both sides of (1), we get

0<$$a^2$$<1

Adding '1' on both sides of above equation, we get

1<$$a^2$$+1<2....................(3)

$$\Rightarrow a^2$$+1>0, Hence Option (ii) is true.

Option (iii):

As 0<a<1, $$a^2<a$$.............(4)

Subtracting 'a' on both sides of (4), we get

$$a^2-a<0$$, Hence Option (iii) is also false.

Let, us assume the total revenue for 2010 as 100.

Since, all the companies except T have generated the same revenue in 2011, the percentage increase in revenue for all companies will be only due to the increase in revenue of T.

Increase in revenue of T(2011) = 20% * 14 = 2.8

Total revenue(2011) = 102.8

$$\therefore $$ % increase in revenue = $$\frac{102.8-100}{100}$$ * 100% = 2.8 %

Hence, option A.

Let say after a Time period (T) 'x'  years the amount Rs. 2 lakh gets doubled at a rate of interest (R) = 11.5% per annum

Initial principle amount (P) = 2,00,000 Rs.

Simple Interest (I)  for 'x' years is given by I = $$PTR\div 100$$ = $$200000\times x\times 11.5\div 100$$

                       => I = $$23000\times x$$.

Final Principle amount (P') = P + I = $$2\times P$$ (as given) 

                                               => I = $$2\times P - P$$ = P

                                              => $$23000\times x$$ = 200000

                                              => x = $$200000\div 23000$$

                                              => x = 8.695 = between 8 and 9.                                                              

Given AB + C = D................................(1

If B = 6 and D = 30,  then from (1),

6A + C = 30........................(2)

and If B = 8 and D = 36,

8A + C = 36........................(3)

(3)-(2) => 2A = 6

=> A= 3

Substituting the value of A in equation (2) we get,

18 + C = 30

=> C = 12.

$$y^2 + 3y - 18 \geq 0$$

 $$\Rightarrow y^2 + 6y - 3y - 18 \geq 0$$

 $$\Rightarrow y (y+6) - 3 (y+6) \geq 0$$

 $$\Rightarrow (y-3) (y+6) \geq 0$$

 $$\Rightarrow y\geq3  and  y\leq-6$$

Since balls of same color should come together, let us consider 3 red balls as one unit and 2 blue balls as one unit and 4 yellow balls as another unit.

So, we have a total of 3 different units which can be arranged in '3!' ways.

these 3 red balls can internally be arranged in '3!' ways.

Similarly the blue balls and yellow balls can be arranged internally in '2!' and '4!' ways respectively.

So, total number ways = $$3!\times 3!\times 2!\times 4!= 6\times 6\times 2\times 24$$= 1728 ways

By carefully observing the given data and by observing the given options,

The charges will be least for the parcel from Kolkata to Mumbai which is Rs. 7,

Hence option (D) is the answer.

Given X, Y and Z are in the ratio 12: 15: 25

Let say X = 12k, Y = 15k and Z = 25k, where k is any constant.

Difference between Y and X = 15k - 12k = 3k..............................(1)

and Difference between Z and Y = 25k - 15k = 10k.....................(2)

Ratio of (1) and (2) = $$3k\div 10k = 3:10$$.

Given speed of Ankush = 3 rounds per hour = 1  round per 20 min

Whereas speed of Babulal = 5 rounds per hour = 1 round per 12 min.

Let us consider Ankush as stationery at the starting point.

The relative speed of Babulal with respect to Ankush = 3 rounds per hour + 5 rounds per hour = 8 rounds per hour

This implies, In one hour Babulal crosses Ankush 8 times.

So, starting from 9.00 am to 10.30 am i.e., 1.5 hours or 90 min, Babulal crosses Ankush 12 times.