CAT 2021 Question Paper Quant Slot 1

Speed of the faster train = $$\frac{160}{12}=\frac{40}{3}\ $$ m/s

Speed of the slower train = $$\frac{40}{3}-\left(6\times\ \frac{5}{18}\right)=\frac{35}{3}$$ m/s

Sum of speeds (when the trains travel towards each other) = $$\frac{40}{3}+\frac{35}{3}=25$$ m/s

Let the slower train be $$x$$ metres long; then: $$\frac{160+x}{25}=14$$

On solving, $$x=190\ m$$

Area of a regular hexagon = $$\frac{3\sqrt{3}}{2}x^2$$

Area of an equilateral triangle = $$\frac{\sqrt{3}}{4}\left(a\right)^2$$ ; where a = side of the triangle

Since the area of the two figures are equal, we can equate them as folllows: $$\frac{3\sqrt{3}}{2}x^2=\frac{\sqrt{3}}{4}\left(12\right)^2$$

On simplifying: $$x^2=24\ $$

$$\therefore\ x=2\sqrt{6}$$

Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.

Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC is $$60^{\circ\ }$$.

We can find out the value of AT using cosine formula:

$$AT^2=\ 4^2+1^2-2\times\ 1\times\ 4\cos60$$

$$AT^2=\ 17-4=13$$

$$AT=\sqrt{\ 13}$$

Let the total amount of work be 60 units.

Then Anu, Vinu, and Manu do 4, 5, and 3 units of work per day respectively.

On the 1st day, Anu and Vinu work. Work done on the 1st day = 9 units

On the 2nd day, Manu and Vinu work. Work done on the 2nd day = 8 units

This cycle goes on. And in 6 days, the work completed is 9+8+9+8+9+8 = 51 units.

On the 7th day, again Anu and Vinu work and complete the remaining 9 units of work. Thus, the number of days taken is 7 days.

Let the amounts Neeta, Geeta, and Sita earn in a day be n, g, and s respectively. 

Then, it has been given that:

n+g=6s -i

s+n=2g -ii

ii-i, we get: s-g = 2g-6s

7s = 3g.

Let g be 7a. Then s earns 3a.

Then n earns 6s-g = 18a-7a = 11a.

Thus, the ratio is 11a:3a = 11:3

Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.

Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.

Total amount bought = 22kg.

Total amount spent = 100+100+100+50+50 = 400.

Average expense = $$\frac{400}{22}=Rs.18.18\approx\ 18$$

The possible arrangements are of the form 

35 _ Can be chosen in 6 ways.

35 _ _ We can choose 2 out of the remaining 6 in $$^6C_2=15$$ways. We remove 1 case where 7 and 8 are together to get 14 ways.

35 _ _ _We can choose 3 out of the remaining 6 in $$^6C_3=20$$ways. We remove 4 cases where 7 and 8 are together to get 16 ways.

35 _ _ _ _We can choose 4 out of the remaining 6 in $$^6C_4=15$$ways. We remove 6 case where 7 and 8 are together to get 9 ways.

35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.

Thus, total number of cases = 6+14+16+9+2 = 47.

Alternatively,

The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.

Considering the cases, only 7 is selected.

We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.

Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have

Each of the numbers can either be selected or not selected and we have 4 numbers :

Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.

SImilarly, including only 8, we have 16 more possibilities.

Cases including neither 7 nor 8.

We must have 3 and 5 in the group but there must be no 7 and 8 in the group.

Hence we have 3 5 _ _ _ _.

For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.

= 16 possibilities

But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.

Hence 16 - 1 = 15 cases.

Total = 16+15+16 = 47 possibilities

Let the number of pens purchased be n. Then the cost price is 8n. The total expenses incurred would be 8n+W, where W refers to the wage.

Then SP in the first case = $$12\times\ 100+11\times\ \left(n-100\right)$$

Given profit is 300 in this case: 1200+11n-1100-8n-W=300 =>3n-W = 200

In second case: 1200+9n-900-8n-W=-300 (Loss). => W-n = 600.

Adding the two equations: 2n = 800

n = 400.

Thus W = 600 + 400 = 1000

Let the cost of an apple, an orange and a mango be a, o, and m respectively.

Then it is given that:

2a+4o+6m = a+4o+8m

or a = 2m.

Also, a+4o+8m = 8o + 7m

10m-7m = 4o

3m = 4o.

We can now express the cost of a basket in terms of mangoes only:

2a+4o+6m = 4m+3m+6m = 13m.

We have $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.

Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.

Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 40$$, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference. Thus we have the condition that n>40.

The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 60$$, as then it would be closer to 20 than 100. Thus we have the condition that n>60.

The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be $$\ge80$$, as then it would be closer to 100 than 60. Thus we have the condition that n<80.

The number which satisfies the conditions are 61, 62, 63, 64......79. Thus, a total of 19 numbers.

Alternatively

as per the given condition : $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.

Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100 )

1) For n < 20.

|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n 

considering the inequality part :$$\left|n-100\right|<\ \left|n\ -20\right|$$

100 -n < 20 -n,

No value of n satisfies this condition.

2) For 20 < n < 60.

|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.

60- n < 100 - n and 100 - n < n - 20

For 100 -n < n - 20.

120 < 2n and n > 60. But for the considered range n is less than 60.

3) For 60 < n < 100

|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n

n-60 < 100-n and 100-n < n-20.

For the first part 2n < 160 and for the second part 120 < 2n.

n takes values from 61 ................79.

A total of 19 values

4) For n > 100

|n-20| = n-20, |n-60| = n-60, |n-100| = n-100

n-60 < n - 100.

No value of n in the given range satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.

Then (100c+10b+a)-(100a+10b+c)=198

99c-99a=198

c-a = 2.

Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.

Thus, there can be 7 cases.

B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be $$7\times\ 10=70$$.

Let the principal amount be P and the interest rate be r.

Then $$P\left(1+r\right)^2-P\left(1+r\right)=806.25$$ -(1)

$$P\left(1+r\right)^3-P\left(1+r\right)^2=866.72$$ -(2)

Dividing (2) by (1), we get:

$$\frac{\left(P\left(1+r\right)^3-P\left(1+r\right)^2\right)}{P\left(1+r\right)^2-P\left(1+r\right)}=\frac{866.72}{806.25}$$

$$\frac{\left(\left(1+r\right)^2-1-r\right)}{1+r-1}=1.075$$

$$\frac{r^2+r}{r}=1.075$$

r=0.075 or 7.5%

$$\frac{\left(Interest\ accrued\ in\ 4th\ yr\right)}{Interest\ accrued\ in\ 3rd\ yr}=\frac{X}{866.72}$$

$$\frac{\left(P\left(1+r\right)^4-P\left(1+r\right)^3\right)}{P\left(1+r\right)^3-P\left(1+r\right)^2}=\frac{X}{866.72}$$

Dividing numerator and denominator by $$P\left(1+r\right)^2$$

$$\frac{r^2+2r+1-1-r}{1+r-1}=\frac{X}{866.72}$$

$$r+1=\frac{X}{866.72}$$

$$X=1.075\times\ 866.72=931.72$$

Let the number of Covid patients in Hospitals A and B be x and x+21 respectively. Then, it has been given that:

$$\frac{200}{x}-\frac{152}{x+21}=3$$

$$\frac{\left(200x+4200-152x\right)}{x\left(x+21\right)}=3$$

$$\frac{\left(48x+4200\right)}{x\left(x+21\right)}=3$$

$$16x+1400=x\left(x+21\right)$$

$$x^2+5x-1400=0$$

(x+40)(x-35)=0

Hence, x=35.

We know that Inradius=$$\frac{\left(Perpendicular+Base-Hypotenuse\right)}{2}$$

$$4=\frac{\left(p+10-h\right)}{2}$$

h-p = 2 or h= p+2.

Now, $$p^2+100\ =\ h^2$$

$$p^2+100\ =\ \left(p+2\right)^2$$

$$p^2+100\ =\ p^2+4p+4$$

4p = 96

p=24.

Hence, Area = $$\frac{1}{2}\times\ 10\times\ 24\ =\ 120$$.

$$f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18}$$<0

$$\frac{\left(x+5\right)\left(x-3\right)}{\left(x-9\right)\left(x+2\right)}<0$$

We have four inflection points -5, -2, 3, and 9.

For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.

When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.

We are left with the range (-5,-2) and (3,9) where the expression will be negative.

$$5 - \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \frac{1}{\sqrt{1 - x^2}}$$

We can re-write the equation as: $$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\left(\sqrt{1+x}\times\ \sqrt{1-x}\right)^{-1}$$

$$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\left(-1\right)\log_{10}\left(\sqrt{1+x}\right)+\left(-1\right)\log_{10}\left(\sqrt{1-x}\right)$$

$$5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}$$

$$5=-5\log_{10}\sqrt{1-x}$$

$$\sqrt{1-x}=\frac{1}{10}$$

Squaring both sides: $$\left(\sqrt{1-x}\right)^2=\frac{1}{100}$$

$$\therefore\ $$ $$x=1-\frac{1}{100}=\frac{99}{100}$$

Hence, $$100\ x\ =100\times\ \frac{99}{100}=99$$

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.