CAT 2020 Question Paper Slot 1

Let the length of the train be $$ l kms$$ and speed be $$ s kmph$$. Base on the two scenarios presented, we obtain:

$$\frac{l}{s-2}=\frac{90}{3600}$$....(i) and $$\frac{l}{s-4}=\frac{100}{3600}$$...(ii)

On dividing (ii) by (i) and simplifying we acquire the value of $$s$$ as $$22 kmph$$. Substituting this value in (i), we have $$l=\frac{90}{3600}\times\ 20\ kms$$ {keeping it in km and hours for convenience}

Since we need to find $$\frac{l}{s}$$, let this be equal to $$x$$. Then, $$x\ =\ 90\times\frac{20}{22}\ =81.81\ \approx\ 82\ \sec onds\ $$

Hence, Option B is the correct choice.

Since $$c<9$$, we can have the following viable combinations for $$b\times\ c\ =96$$ (given our objective is to minimize the sum):

$$48\times\ 2$$ ; $$32\times3$$ ; $$24\times\ 4$$ ; $$16\times6$$ ; $$12\times8$$ 

Similarly, we can factorize $$a\times\ b\ = 432$$ into its factors. On close observation, we notice that $$18\times24\ and\ 24\ \times\ 4\ $$ corresponding to $$a\times b\ and\ b\times\ c\ $$ respectively together render us with the least value of the sum of $$a+b\ +\ c\ \ =\ 18+24+4\ =46$$

Hence, Option D is the correct answer.

Let 'x' be the strength of group G. Based on the information, $$0.65x$$ constitutes of literate people {the rest $$0.35x$$ = illiterate}
Of this $$0.65x$$, 75% are old people =(0.75)0.65x old literates. The total number of old people in group G is $$0.72x$$ {72% of the total}. Thus, the total number of old people who are illiterate = $$0.72x-0.4875x\ =\ 0.2325x$$. This is $$\frac{0.2325x}{0.35x}\times\ 100\ \approx\ \ 66\%$$ of the total number of illiterates. Hence, Option D is the correct answer.

Let their individual Amounts be equal after 't' years. Let their initial investments amount to $$A_V$$ and $$A_J$$ ; 

$$A_V\ =10,000\left(1+\frac{5t}{100}\right)$$ and $$A_J\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$$

Equating both: $$10,000\left(1+\frac{5t}{100}\right)\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$$

On simplifying both sides, we get: $$15t\ =\ 180\ ;\ t\ =\ 12$$

Let 'r' be the root of the function. It follows that f(r) = 0. We can represent this as $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}$$

Based on the relation: $$f\left(5-x\right)=f\left(5+x\right)$$; $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}=f\left\{5+\left(5-r\right)\right\}$$

$$\therefore\ f\left(r\right)=f\left(10-r\right)$$

Thus, every root 'r' is associated with another root '(10-r)' [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $$r_1,\ \left(10-r_1\right),\ r_2,\ \left(10-r_2\right)$$

The sum of these roots = $$r_1\ +\left(10-r_1\right)+\ r_2+\ \left(10-r_2\right)\ =\ 20$$

Hence, Option D is the correct answer.

Given

A+(B+C)/2=5 => 2A+B+C=10....(i)

(A+C)/2 +B=7 => A+2B+C=14.....(ii)

(i)-(ii)=> B-A=4 => B=4+A.

Given, A, B, C are positive integers

If A=1=>B=5 => C=3

If A=2=>B=6 => C=0 but this is invalid as C is positive.

Similarly if A>2 C will be negative and cases are not valid.

Hence, A+B=6.