CMAT 2018 Slot 2

Let the maximum number of marks be 'x' and minimum passing marks be 'y'.

Given, A student who gets 20% marks fails by 20 marks.

$$\Rightarrow \frac{20}{100}\times x = y - 20$$

$$\Rightarrow 20x = 100y - 2000$$.................(1)

Also given that, another student who gets 36% marks gets 44 marks more than minimum passing marks.

$$\Rightarrow \frac{36}{100}\times x = y + 44$$

$$\Rightarrow 36x = 100y + 4400$$....................(2)

(2) - (1) $$\Rightarrow$$ 16x = 6400

$$\Rightarrow$$ x = 400

From (1) or (2), we get y = 100

Hence, maximum number of marks = x = 400

Percentage necessary for passing = $$\frac{y}{x}\times 100$$ = 25%

Let amount eaten by each horse and each bullock in one day be 'h' units and 'b' units respectively.

So Total fodder = Total effciency$$\times$$Total number of days 

$$\Rightarrow$$ Total work =  26$$\times$$h$$\times$$170 = 20$$\times$$b$$\times$$170

$$\Rightarrow$$ b = 1.3h............(1)

The amount of fodder eaten by 10 horses and 8 bullocks in one day = 10h + 8b = 10h + 8(1.3h) = 20.4h

Time taken by them to eat the same amount of fodder = total fodder/amount eaten by them in one day

= $$\frac{ 26\times h\times 170}{20.4h}$$

= 216.67 days

Let the speed of the boat in still water be V and speed of the stream be V'.

Relative speed of boat in upstream = V - V', as water stream flows against the direction of boat.

whereas Relative speed of boat in downstream = V + V', as water stream flows in the direction of boat.

Case (1)

Given Total time taken = Time taken during upstream + Time taken during downstream = 8 hours.

$$\Rightarrow \frac{24}{V - V'} + \frac{72}{V + V'} = 8$$

$$\Rightarrow 3 [\frac{1}{V - V'} + \frac{3}{V + V'}] =1$$.............(1)

$$\Rightarrow 3 [4V - 2V'] = V^2 - V'^2$$......................(2)

Case (2)

Given Total time taken = Time taken during upstream + Time taken during downstream = 14 hours

$$\Rightarrow \frac{48}{V - V'} + \frac{108}{V + V'} = 14$$

$$\Rightarrow 6 [\frac{4}{V - V'} + \frac{9}{V + V'}] = 7$$

$$\Rightarrow 6[13V - 5V'] = 7 [V^2 - V'^2]$$...................(3)

Dividing equation (3) by (2), we get

$$\frac{2 [13V - 5V']}{4V - 2V'} = 7$$

$$\Rightarrow 26V - 10V' = 28V - 14V'$$

$$\Rightarrow V = 2V'$$

Substituting this value in equation (1) we get,

$$\Rightarrow 3 [\frac{1}{V'} + \frac{1}{V'}] =1$$

$$\Rightarrow V' = 6$$

$$\Rightarrow V = 12$$

Hence, Speed of the boat in still water = 12 km/h.

and Speed of the stream = 6 km/h.

Let say cost price(C.P) of 1 kg(1000 g) of rice be Rs. 100

Given Shopkeeper is selling rice  at cost price,

$$\Rightarrow$$ Selling price(S.P) = C.P = Rs. 100

If he had used correct weight of 1000 g then C.P would have also been Rs. 100.

But given that he uses false weight. Let the weight he had used be  'x' g.

For 1000 g of rice  the C.P = Rs. 100

$$\Rightarrow$$ For '1' g of rice the C.P will be Rs. $$\frac{1}{10}$$

$$\Rightarrow$$ For 'x' g of rice the C.P will be Rs. $$\frac{x}{10}$$

Given that, by using this false weight the shop keeper gains 20%.

Gain percentage = $$\frac{S.P - C.P}{C.P}\times 100$$

$$\Rightarrow \frac{20}{100} = \frac{S.P - C.P}{C.P}$$

$$\Rightarrow S.P = 1.2\times C.P$$

$$\Rightarrow 100 = 1.2\times \frac{x}{10}$$

$$\Rightarrow x = 833.33$$ 

Hence the false weight used is 833.33 g 

Let the efficiencies of Rakesh, Prakash, and Ashok be 'r' 'p' and 'a' respectively.

GIven that Rakesh can do a job an hour quicker than Prakash.

So let time taken by Prakash be 't' hours, then time taken by Rakesh will be 't-1' hours.

Total work(W) = Efficiency$$\times$$ Time taken = p$$\times$$ t = r$$\times$$ (t-1)

$$\Rightarrow$$ t = $$\frac{r}{r-p}$$....................(1)

Given that, Working together, Rakesh, Prakash and Ashok can finish the same job in an hour.

$$\Rightarrow$$ Total work(W) = (r+p+a) (1) units.............(2)

Also given that,  if Prakash works for an hour, and then Ashok works for four hours, the job will be completed.

$$\Rightarrow$$ Total work(W) = p(1) + a(4) units...........(3)

Equating (2) and (3), we get

(r+p+a) (1) = p(1) + a(4)

$$\Rightarrow$$ r = 3a..............(4)

Substituting this value in equation (1), we get

t = $$\frac{3a}{3a-p}$$.............(5)

As the Total work is always constant, p$$\times$$ t = p(1) + a(4)

$$\Rightarrow$$ t = 1 + 4$$\frac{a}{p}$$.......(6)

Equating (5) and (6), we get

 $$\frac{3a}{3a-p} = 1 + 4\frac{a}{p}$$

Let $$\frac{a}{p}$$ = 'k'

$$\Rightarrow \frac{3k}{3k-1} = 1 + 4k$$

$$\Rightarrow 3k = 12k^2 + 3k - 4k -1$$

$$\Rightarrow 12k^2 - 4k - 1 = 0$$

Solving for k, we get k = $$\frac{1}{2} or -\frac{1}{6}$$[which is not possible]

Hence k = $$\frac{1}{2}$$

$$\Rightarrow$$ p =2a.............(7) 

Substituting (4) and (7) in equation (2) we get,

Total work(W) = 6a units.

Time taken by Ashok alone to do the job = Total work/ Efficiency of Ashok

= 6a/a

=6 hours.

Elder brother of Raj's father = Raj's uncle

Now, Sweety is grand-daughter of Raj's uncle.

=> Sweety is Raj's niece.

=> Ans - (A)

The minimum number of people taking classes after school is 3.

The minimum number of people taking classes before school is 2.

Hence the maximum number of people taking classes during lunch must be 2.

Hence the number of people taking classes during lunch and after school cannot be equal.

Option D is the correct answer.

To find : A's wife = ?

It is given that R and S are A's sisters.

Also, W is wife of E and V is wife of B. 

Thus, we have : (A,) (B,V) (C,) (D,) (EW)

Also, T is married neither to C nor to D, => T is wife of A.

=> Ans - (C)

If southeast becomes east and northwest becomes west, thus we have to tilt the direction $$45^\circ$$ clockwise. (to the right)

Thus, direction of north will be northeast.

=> Ans - (D)

Case 1 : If Kalia speaks truth

=> Raza is the thief, which means Raza's first statement is a lie.

Case 1(a) : Raza speaks lie and truth alternatively.

=> Second statement cannot be true.

Case 1(b) : Raza always lies.

=> Raza is a thief, and the other ones are not.

This can only mean Shera speaks truth and lie alternatively, which positively concludes above statements.

$$\therefore$$ Raza is the thief.

=> Ans - (C)