CAT 2018 Question Paper Slot 1

It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number.
There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $$9C_2$$ ways.
3 digit numbers can be formed in $$9C_3$$ ways.
..................................................
..9 digit number can be formed in 9C9 ways. 

We know that $$nC_0+nC_1+nC_2+.........+nC_n =2^n$$
=> $$9C_0 + 9C_1+9C_2+....9C_9 = 2^9$$
$$9C_0 + 9C_1 + ...9C_9 = 512$$

We have to subtract $$9C_0$$ and $$9C_1$$ from both the sides of the equations since we cannot form single digit numbers. 
=> $$9C_2 + 9C_3+...+9C_9=512-1-9$$
$$9C_2+9C_3+...+9C_9=502$$

Therefore, $$502$$ is the right answer. 

We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft 

In triangle OAB and OCD

$$\dfrac{OA}{AB} = \dfrac{OC}{CD}$$

$$\Rightarrow$$ AB = $$\dfrac{3*4}{12}$$ = 1 ft.

Therefore, the volume of remaining part = Volume of entire cone - Volume of smaller cone

$$\Rightarrow$$ $$\dfrac{1}{3}*\pi*4^2*12-\dfrac{1}{3}*\pi*1^2*3$$

$$\Rightarrow$$ $$\dfrac{1}{3}*\pi*189$$

$$\Rightarrow$$ $$\dfrac{22}{7*3}*189$$

$$\Rightarrow$$ $$198$$ cubic ft

Let the number of marbles with Raju and Lalitha initially be 4x and 9x. 
Let the number of marbles that Lalitha gave to Raju be a.

It has been given that (4x+a)/(9x-a) = 5/6
24x + 6a = 45x - 5a
11a = 21x
a/x = 21/11

Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).
a/9x = 21/99
 = 7/33.

Therefore, option D is the right answer. 

Let us draw a Venn diagram using the information present in the question. 

It is given that the number of students studying H equals that studying E. 

Let 'x' be the total number of students who studied H, and H and P but mot E.We can also say that the same will be the number of students who studied E, and E and P but not H.Therefore,

x + 20 + 10 + x = 74 

x = 22

Hence, the number of students studying H = 22 + 10+ 20 = 52

Let the price of peanuts be Rs. 100x per kg
Then, the price of walnuts = Rs. 300x per kg
Cost price of peanuts for the shopkeeper = Rs. 110x per kg
Cost price of walnuts for the shopkeeper = Rs. 360x per kg
Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x
Since, 5kg walnut and 3kg peanuts are lost in transit, the shopkeeper will be remained with (16-5)+(8-3)=16kgs of nuts
Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656
Profit = 25%
So, cost price = Rs. 2124.80
Therefore, 6640x = 2124.80
On solving, we get x = 0.32
Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.
Hence, option A is the correct answer

It has been given that among 200 students, 105 like pizza and 134 like burger.
The question asks us to find out the number of students who can be liking only burgers among the given values. 

The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too. 
In this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.

The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible. 
There are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95. 

The number of students who like only burger should lie between 29 and 95 (both the values are included). 
93 is the only value among the given options that satisfies this condition and hence, option D is the right answer. 

$$f(x + 2) = f(x) + f(x + 1)$$
As we can see, the value of a term is the sum of the 2 terms preceding it. 

It has been given that $$f(11) = 91$$ and $$f(15) = 617$$.
We have to find the value of $$f(10)$$.

Let $$f(10)$$ = b
$$f(12)$$ = b + 91
$$f(13)$$ = 91 + b + 91 = 182 + b
$$f(14)$$ = 182+b+91+b = 273+2b
$$f(15)$$ = 273+2b+182+b = 455+3b
It has been given that 455+3b = 617
3b = 162
=> b = 54

Therefore, 54 is the correct answer. 

The possible average age of people whose ages are below 51 years will be maximum if the average age of the number of people aged 51 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years. 

Let 'x' be the maximum average age of people whose ages are below 51. 

Then we can say that, 

$$\dfrac{51*30+39*x}{30+39} = 38$$

$$\Rightarrow$$ $$1530+39x = 2622$$

$$\Rightarrow$$ $$x = 1092/39 = 28$$

Hence, we can say that option D is the correct answer. 

Let the price of paint B be x. 
Price of paint A = x+8

We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture. 

The seller sells 10 litres of paint at Rs.264 earning a profit of 10%. 
=> The cost price of 10 litres of the paint mixture = Rs. 240

Therefore, the cost of 1 litre of the mixture = Rs.24

We have to find the highest possible cost of paint B. 
When we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.
The highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal. 

=> (x+x+8)/2 = 24
2x = 40
x = Rs. 20

Therefore, option C is the right answer. 

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

Given that area of parallelogram = 72 sq cm

Area of triangle ABC = $$\dfrac{1}{2}$$*area of parallelogram

(1/2)*AB*BC*sinABC = $$\dfrac{1}{2}$$*72

sinABC = $$\dfrac{1}{2}$$

$$\angle$$ ABC = 30°

Let us draw a perpendicular CQ from C to AB. 

$$\angle\ B\ =\ \angle\ D\ =\ 30^{\circ\ }$$ ( opposite angles in a parallelogram)

$$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$$  ( Sum of angles in a triangle = 180)

In triangles APD and CQB, AP= CQ , AD=CB ( opposite sides of a parallelogram) , $$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$$

Hence APD and CQB are congruent triangles using SAS property.

Therefore, we can say that area of triangle APD = area of triangle CQB

In right angle triangle CQB,

QB = CBcos30° = $$16*\dfrac{\sqrt{3}}{2}$$ = $$8\sqrt{3}$$ cm

CQ = CBsin30° = $$16*\dfrac{1}{2}$$ = $$8$$ cm

Therefore, area of triangle CQB = 1/2*CQ*QB = $$1/2*8*8\sqrt{3}$$ = $$32\sqrt{3}$$

Hence, we can say that area of triangle APD = $$32\sqrt{3}$$.