CAT 2018 Question Paper Slot 1

It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x : 1. Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.

Let us assume that 'x+1' is the length of side of square ABCD. 

Area of square ABCD = $$(x+1)^2$$ sq. units.

Therefore, area of square EFGH = $$\dfrac{62.5}{100}*(x+1)^2$$ = $$\dfrac{5(x+1)^2}{8}$$ ... (1)

In right angle triangle EBF,

$$EF^2 = EB^2 + BF^2$$

$$\Rightarrow$$ $$EF = \sqrt{1^2+x^2}$$

Therefore, the area of square EFGH = $$EF^2$$ = $$x^2+1$$ ... (2)

By equating (1) and (2),

$$x^2+1 = \dfrac{5(x+1)^2}{8}$$

$$\Rightarrow$$ $$8x^2+8 = 5x^2+10x+5$$

$$\Rightarrow$$ $$3x^2-10x+3 = 0$$

$$\Rightarrow$$ $$(x - 3)(3x - 1) = 0$$

$$\Rightarrow$$ $$x = 3$$ or $$1/3$$

The ratio of length of EB to that of CG = 1 : x 

EB : CG = 1 : 3 or 3 : 1. Hence, option D is the correct answer. 

Let the distance between A and B be 4x. 
Length of BP is thrice the length of AP.
=> AP = x and BP = 3x

Let the speed of car 1 be s and the speed of car 2 be 0.5s. 
Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.

=> x/s + 60 = 3x/0.5s
x/s + 60 = 6x/s
5x/s = 60
x/s = 12

Time taken by car 1 in reaching P from A = x/s = 12 minutes.
Therefore, 12 is the correct answer.  

f(x) = min ($${2x^{2},52-5x}$$)
The maximum possible value of this function will be attained at the point in which $$2x^2$$ is equal to $$52-5x$$.

$$2x^2 = 52-5x$$
$$2x^2+5x-52=0$$
$$(2x+13)(x-4)=0$$
=> $$x=\frac{-13}{2}$$ or $$x = 4$$
It has been given that $$x$$ is a positive real number. Therefore, we can eliminate the case $$x=\frac{-13}{2}$$.
$$x=4$$ is the point at which the function attains the maximum value. $$4$$ is not the maximum value of the function. 
Substituting $$x=4$$ in the original function, we get, $$2x^2 = 2*4^2= 32$$.
f(x) = $$32$$.
Therefore, 32 is the right answer.

We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20

Product of 2 real numbers is 20. 
We have to find the minimum possible value of the sum of the squares of the 2 numbers. 
Let x=a*b
It has been given that a*b=20

The least possible sum for a given product is obtained when the numbers are as close to each other as possible. 
Therefore, when a=b, the value of a and b will be $$\sqrt{20}$$.

Sum of the squares of the 2 numbers = 20 + 20 = 40.

Therefore, 40 is the correct answer. 

Train T starts at 3 PM and train S starts at 4 PM. 
Let the speed of train T be t.
=> Speed of train S = 0.75t.

When the trains meet, train t would have traveled for one more hour than train S. 
Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours. 

Distance traveled by train T = xt
Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t

We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities. 

=> (xt)/(0.75xt-0.75t) = 3/2
2xt = 2.25xt-2.25t
0.25x = 2.25
x = 9 hours. 

Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3) = 15 hours. 
Therefore, 15 is the correct answer.

The selling price of the mixture is Rs.40/kg. 
Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B. 
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x. 
It has been given that 1.1x = 40
x = 40/1.1
Price per kg of the mixture in ratio 3:2 = $$\frac{3a+2b}{5} $$
$$\frac{3a+2b}{5} = \frac{40}{1.1}$$
$$3.3a+2.2b=200$$ --------(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
Price per kg of the mixture in ratio 2:3 = $$\frac{2a+3b}{5}$$
$$\frac{2a+3b}{5} = \frac{40}{1.05}$$
$$2.1a+3.15b=200$$ ------(2)

Equating (1) and (2), we get, 
$$3.3a+2.2b = 2.1a+3.15b$$
$$1.2a=0.95b$$
$$\frac{a}{b} = \frac{0.95}{1.2}$$
$$\frac{a}{b} = \frac{19}{24}$$

Therefore, option C is the right answer. 

We can see that T$$_{2}$$ is formed by using the mid points of T$$_{1}$$. Hence, we can say that area of triangle of T$$_{2}$$ will be (1/4)th of the area of triangle T$$_{1}$$.

Area of triangle T$$_{1}$$ = $$\dfrac{\sqrt{3}}{4}*(24)^2$$ = $$144\sqrt{3}$$ sq. cm

Area of triangle T$$_{2}$$ = $$\dfrac{144\sqrt{3}}{4}$$ = $$36\sqrt{3}$$ sq. cm 

Sum of the area of all triangles =  T$$_{1}$$ + T$$_{2}$$ + T$$_{3}$$ + ... 

$$\Rightarrow$$ T$$_{1}$$ + T$$_{1}$$/$$4$$ + T$$_{1}$$/$$4^2$$ + ... 

$$\Rightarrow$$ $$\dfrac{T_{1}}{1 - 0.25}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*T_{1}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*144\sqrt{3}$$

$$\Rightarrow$$ $$192\sqrt{3}$$

Hence, option D is the correct answer. 

Given that: $$\log_{12}{81}=p$$

$$\Rightarrow$$ $$\log_{81}{12}=\dfrac{1}{p}$$

$$\Rightarrow$$ $$\log_{3}{3*4}=\dfrac{4}{p}$$

$$\Rightarrow$$ $$1+\log_{3}{4}=\dfrac{4}{p}$$

Using Componendo and Dividendo, 

$$\Rightarrow$$ $$\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{36}{64}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{6^2}{8^2}=\log_{6}{8}$$. Hence, option D is the correct answer.

We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years. 

John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years. 
Let the amount paid as installment every year be Rs.x.

John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid. 

=> 1.1x + x = 2,54,100
2.1x = 2,54,100
=> x = Rs. 1,21,000.

Therefore, 121000 is the correct answer. 

Let us assume that A can complete 'a' units of work in a day and B can complete 'b' units of work in a day. 
A works alone till half the work is completed. 
A and B work together for 4 days and B works alone to complete the last 5% of the work. 
=> A and B in 4 days can complete 45% of the work.

Let us assume the total amount of work to be done to be 100 units.
4a + 4b = 45 ---------(1)
B needs 25% more time than A to finish a job. 
=> 1.25*b = a ----------(2)

Substituting (2) in (1), we get, 
5b+4b = 45
9b = 45
b = 5 units/day

B alone can finish the job in 100/5 = 20 days. 
Therefore, option A is the right answer.