CAT 2020 Question Paper | CAT QA
CAT Question Paper | CAT Previous Year Paper

A, B and C are three positive integers

Adding the two equations we get 3(A + B) + 2C = 24 ⇒ 2C = 24 – 3(A + B)

Therefor (A + B) has to be even. Thus, options [1] and [4] are eliminated.

Subtracting (i) from (ii), B – A = 4 ⇒ (A + B) cannot be 4. Thus, option [2] is eliminated.

Therefore, the sum of A and B is 6.

Hence, option 3.

Case I:

(x + y = 1 is a line with x-intercept = -1 and y-intercept = -1)

(x – y = 1 is a line with x-intercept = 1 and y-intercept = -1)

With this, we can draw a figure given below

AD = |-2-2| = 4 units and BC = |-1-1| = 2 units

A(â–¡ABCD) = ½ x (4 + 2) x 1 = 3 sq. units

Answer: 3

The gentleman exhausts his stock after he takes care of 5 children.

Assume that the gentle man has ‘x’ toffees.

Therefore, the number of toffees remaining with him

Now the toffees left with the gentleman after giving toffees to the fourth child = 2

Therefore, the toffees with the gentleman before giving toffees to the fourth child = (2 + 1) × 2 = 6

Similarly,

• the toffees with the gentleman before giving toffees to the third child = (6 + 1) × 2 = 14
• the toffees with the gentleman before giving toffees to the second child = (14 + 1) × 2 = 30
• the toffees with the gentleman before giving toffees to the first child = (30 + 1) × 2 = 62

Answer: 62.

AD is tangent to the circle. Therefore, AD ⊥ EO.

The diagonals of a rhombus bisect each other.

∴ AC = 16 cm and DB = 12 cm ⇒ AO = OC = 8 cm and DO = BO = 6 cm

In ∆AOD, AO = 8 cm and DO=6 cm ⇒AD = 10 cm

∆AOD ~ ∆AEO

The area of circle= π × (4.8)²

The area of rhombus = 1/2 × 12 × 16

Hence, option 1.

Let usual speed of the train be ‘s’ meters per minute and time taken by the train to cover the distance be ‘t’ minutes.

Given: A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time.

∴ distance = 15s

On its return journey, distance covered in first 5 minutes at its usual speed = 5s.

As the train stopped for 4 minutes, the remaining distance of (15s– 5s=) 10s is to be covered in 6 minutes.

Hence, option 2.

For a^y = 1, either [a = 1] or [y = 0] or for [y = 2x for some integer x, a = (-1)]

X² – 13x + 42 = x(x – 13) + 42. Note that x(x – 13) is divisible by 2.

Terefore x² – 13x + 42 is divisible by 2.

Case (i): (x² – 7x + 11) = 1 ⇒ x² – 7x + 10 = 0

⇒ (x – 5)(x – 2) = 0 ⇒ x = 2 or 5

Case (ii): (x² – 13x + 42) = 0

⇒ (x – 6)(x – 7) = 0 ⇒ x = 6 or 7

Case (iii): (x² – 7x + 11) = -1 ⇒ (x² – 7x + 12) = (x – 3)(x – 4) = 0

⇒ x = 3 or 4

Thus, x = 2, 3, 4, 5, 6, 7

Hence, option 2.

Let there be 100x people in the group. There are 28x young members and 72x old members. There are 65x literates and 35x illiterates.

Out of 65x literate members, (100 – 25 =) 75% are old members i.e., 48.75x are old members.

∴72x - 48.75x = 23.25x illiterates are old.

Hence, option 3.

The even natural numbers are of the form ‘aabb’, where a > 0.

As the numbers are even natural numbers, ‘b’ can take 5 values, i.e., 0, 2, 4, 6, 8. ‘a’ can take 9 values, 1 through 9.

i.e., there are 9 × 5 = 45 such numbers

Sum of the numbers = 1100 + 1122 + 1144 + 1166 + 1188 + 2200 + 2222 + 2244 + 2266 + 2288 + …

+ 9900 + 9922 + 9944 + 9966 + 9988

= 1100 × 5 + 2200 × 5 + … + 9900 × 5 + 00 × 9 + 22 × 9 + 44 × 9 + … + 88 × 9

= (1100 + 2200 + … + 9900) × 5 + (22 + … + 88) × 9

= 5500 + 44 = 5544

Hence,option 1.

Hence, option 1.

Therefore, the given equation can be written as

As (-1)² -4(1)(1) Ë‚ 0, the two roots of x² – x + 1 = 0 are complex.

Thus, x = 1 is the only real roots.

Answer: 1.

Ab = 432 and bc = 96

As c Ë‚ 9, c = 2, 4,6,8 accordingly, a = 9, 18, 27, 36

As bc = 96, for c = 2, 4,6,8, b = 48, 24, 16, 12

(a, b, c) = (9, 48, 2) and a + b + c = 59

(a, b, c) = (18, 24, 4) and a + b + c = 46

We do not need to check remaining values as the least value given in the options is 46.

Hence, option 2.

Let the length and breadth of the rectangular metal sheet be ‘â„“’and ‘b’ respectively.

Area of the rectangular metal sheet = â„“ × b = 135

The circle touches two opposite sides of the metal sheet. Therefore, the diameter of the circle must be same as the breadth of the rectangle.

By the given condition,

Hence, option 2.

Hence, option 4.

Answer: 36.

Hence, option 1.

Let time taken by Amal to reach office at 8 km/hr

Therefore, if he travels at 15 km/hr, the time taken by him

Now if he leaves home at 9 : 10 am, in order to reach office at 10 am

Hence, option 3.

Let the cost price of the desktop computer and the laptop computer be ‘A’ and ‘B’ respectively.

He sold the desktop at 20% profit and the laptop at 10% loss.

⇒ Selling Price of desktop = 1.2A and Selling Price of the laptop = 0.9B

By the given conditions,

A + B = 50000                                     …(i)

1.2A + 0.9B = 1.02 × 50000 = 51000 …(ii)

Solving the two equations, A = Rs. 20,000

Answer: 20000.

For x = 2, 2^x + 2^-x = 4 + ¼

Hence, option 3.

To find smallest possible value of x₀, 100 should be distributed equally as far as possible

Therefore, as x₀ = max(x₁, x₂, …, x₁₂), the smallest possible value of x₀ = 9.

Hence, option 1.

Assume that ‘t’ years after Veeru’s investment, balances of Veeru and Joy will be equal.

By the given condition,

Answer: 12

40 litres of the solution has 16 litres (i.e., 2 parts) dye and 24 litres (i.e., 3 parts) water. After adding water to 2 : 3 solution, the resultant solution changed to 2 : 5 i.e., 2 : (3 + 2). i.e., 2 parts of water i.e.,

16 litres of water was added to 40 litres.

Resultant solution: 16 litres dye and 40 litres water

After taking out 1/4th of the solution, the solution now

In order to bring the proportion back to 2 : 3, we need to add some amount of dye to this solution. It is clear that the water in the solution remains same i.e., 30 litres which will be 3 parts. So dye in the solution has to be 20 litres.

So, (20 – 12 =) 8 litres of dye has to be added to the solution.

Answer: 8.

Therefore, in 130 kg of alloy, the weight of the metal C

Hence, option 3.

Δ ABC ∼ Δ ADE ⇒ AB : AD = BC : DE ⇒ 9 : 27 = r : R ⇒ 1 : 3 =  r : R ⇒ R = 3r

Volume of the part of the bigger cone remained after cutting the smaller cone  

= 81πr² – 3πr² = 78πr²

By the given condition,

78πr² – 3πr² – 225 ⇒ 75πr² = 225

∴ 75πr² = 225 ⇒ 81πr² = 243

The volume of the original cone = 243 cc 

Hence, option 4.

Let the speed of the train be ‘s’ and length of the train be ‘d’.

∴ The time taken by the train to cross an electric post 

Hence, option 1.

As the product of the digits is more than 2 but less than 7, the digits can be (1, 1, 3), (1, 1, 4), (1, 1, 5), (1, 1, 6), (1, 2, 2), (1, 2, 3).

If digits of a number are (1, 1, 2),

we have 3!/2! = 3 possible arrangements,

i.e., numbers can be 112, 121, 211 i.e., 3 possible numbers.

Similarly, 3 numbers are possible for each of the following combination of digits. (1, 1, 4), (1, 1, 5), (1, 1, 6), (1, 2, 2).

If digits of a number are (1, 2, 3), there are 3! = 6 possible numbers.

Thus, there are 3 × 5 + 6 = 21 such numbers.

Answer: 21.

Let Car 1 and Car 2 meet at C between A and B.

The time taken by Car 1 to cover CB, t₁ = 45 minutes and the time taken by Car 2 to cover CA, t₂ = 20 minutes

As the speed of Car 1 is 60 km/hr, the speed of Car 2 = 90 km/hr

Hence, option 4.

Alternatively,

Let the two cars meet after ‘t’ minutes.

Let Car 1 and Car 2 meet at C between A and B.

The time taken by Car 1 to cover CB, t₁ = 45 minutes and the time taken by Car 2 to cover CA, t₂ = 20 minutes.