CAT 2021 Question Paper Slot 1

Let the total work be 48 units. Let Amar do 'm' work, Akbar do 'k' work, and Anthony do 'n' units of work in a month.

Amar and Akbar complete the project in 12 months. Hence, in a month they do $$\frac{48}{12}$$=4 units of work.

m+k = 4.

Similarly, k+n = 3, and m+n = 2.

Solving the three equations, we get $$m=\frac{3}{2},\ k=\frac{5}{2},\ n=\frac{1}{2}$$.

Here, Amar works neither the fastest not the slowest, and he does 1.5 units of work in a month. Hence, to complete the work, he would take $$\frac{48}{1.5}=32$$months.

$$x_0=1$$

$$x_1=2$$

$$x_2=\frac{\left(1+x_1\right)}{x_0}=\frac{\left(1+2\right)}{1}=3$$

$$x_3=\frac{\left(1+x_2\right)}{x_1}=\frac{\left(1+3\right)}{2}=2$$

$$x_4=\frac{\left(1+x_3\right)}{x_2}=\frac{\left(1+2\right)}{3}=1$$

$$x_5=\frac{\left(1+x_4\right)}{x_3}=\frac{\left(1+1\right)}{2}=1$$

$$x_6=\frac{\left(1+x_5\right)}{x_4}=\frac{\left(1+1\right)}{1}=2$$

Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are 1, 5n+1 are 2... and so on, where n=0,1,2,3,....

2021 is of the form 5n+1. Hence, its value will be 2.

Let Bottle A have an indigo solution of strength 33% while Bottle B have an indigo solution of strength 17%.

The ratio in which we mix these two solutions to obtain a resultant solution of strength 21% : $$\frac{A}{B}=\frac{21-17}{33-21}=\frac{4}{12}or\ \frac{1}{3}$$

Hence, three parts of the solution from Bottle B is mixed with one part of the solution from Bottle A. For this process to happen, we need to displace 600 cc of solution from Bottle A and replace it with 600 cc of solution from Bottle B {since both bottles have 800 cc, three parts of this volume = 600cc}.As a result, 200 cc of the solution remains in Bottle B.

Hence, the correct answer is 200 cc.  

Let the selling price of the Large and Small boxes of chocolates be Rs.200 and Rs.100 respectively. Let us consider that the Large box has $$L$$ grams of chocolate while the Small box has $$S$$ grams of chocolate. 

The relation between the selling price per gram of chocolate can be represented as: $$\frac{200}{L}=0.88\times\ \frac{100}{S}$$

On solving we obtain the ratio of the amount of chocolate in each box as: $$\frac{L}{S}=\frac{25}{11}$$ 

The percentage by which the weight of chocolate in the large box exceeds that in the small box = $$\left(\frac{25}{11}-1\right)\times\ 100\approx\ 127\%$$

The first number in each group: 1, 2, 5, 10, 17.....

Their common difference is in Arithmetic Progression. Hence, the general term of the series can be expressed as a quadratic equation.

Let the quadratic equation of the general term be $$ax^2+bx+c$$

1st term = a+b+c=1

2nd term = 4a+2b+c=2

3rd term = 9a+3b+c=5

Solving the equations, we get a=1, b=-2, c=2.

Hence, the first term in the 15th group will be $$\left(15\right)^2-2\left(15\right)+2=197$$

We can see that the number of terms in each group is 1, 3, 5, 7.... and so on. These are of the form 2n-1. Hence, the number of terms in 15th group will be 29. Hence, the last term in the 15th group will be 197+29-1 = 225.

Sum of terms in group 15= $$\frac{29}{2}\left(197+225\right)\ =\ 6119$$

Alternatively,

The final term in each group is the square of the group number.

In the first group 1, second group 4, ............

The final element of the 14th group is $$\left(14\right)^2=\ 196$$, similarly for the 15th group this is : $$\left(15\right)^2=\ 225$$

Each group contains all the consecutive elements in this range.

Hence the 15th group the elements are:

(197, 198, ................................225).

This is an Arithmetic Progression with a common difference of 1 and the number of element 29.

Hence the sum is given by :  $$\frac{n}{2}\cdot\left(first\ term\ +last\ term\right)$$

$$\frac{29}{2}\cdot\left(197+225\right)$$

6119.

The quadratic equation of the form $$\mid x^2 - 4x - 13 \mid = r$$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

Hence at x = 2 the quadratic equation has its minimum.

Considering the quadratic part : $$\left|x^2-4\cdot x-13\right|$$. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the function $$\left|x^2-4\cdot x-13\right|$$. Because of the modulus function, the representation of the quadratic equation becomes :

ABC'DE. 

There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.

The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :

The point C which is the value of the function at x = 2, = $$2^2-8-13$$

= -17, the reflection about the x-axis is 17.

Alternatively,

$$\mid x^2 - 4x - 13 \mid = r$$ .

This can represented in two parts :

$$x^2-4x-13\ =\ r\ if\ r\ is\ positive.$$

$$x^2-4x-13\ =\ -r\ if\ r\ is\ negative.$$

Considering the first case : $$x^2-4x-13\ =r$$

The quadraticequation becomes : $$x^2-4x-13-r\ =\ 0$$

The discriminant for this function is : $$b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$$

SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

For the second case :

$$x^2-4x-13+r\ =\ 0$$ the function inside the modulus is negaitve

The discriminant is $$16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$$

In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.

Hence $$\ 68-4r\ =\ 0$$

r = 17, for r = 17 we can have exactly 3 roots.