CAT 2021 Slot 1 Question Paper

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