CAT 2017 Question Paper | CAT QA
CAT Question Paper | CAT Previous Year Paper

Given volume of the cylinder is 9π cm³

Now, if ‘r’ and ‘h’ are the radius and height of the cylinder then

Now using the information given in the question let us construct a 2-D view of the ball on the top of the cylinder.

Here AB represents the diameter of the ball and FC and ED the diameter of cylinder. O is the centre of the sphere. Now let us drop a perpendicular from O FC and name the point at which the perpendicular meets FC as P. Now if we join OP, PF and OF we get a triangle OFP.

 
Also OF being the radius of the ball is 2 cm and FP being the radius of the cylinder is

As ∆ OFP is a right angled triangle

OP2 = OF² – FP²

OP = 4 – 3 = 1 ⇒ OP = 1 unit

Now the vertical distance from the topmost point of the ball is the radius of the ball + length (OP) Height of the cylinder ⇒ 2 units + 1 unit + 3 units = 6 units

Answer: 6 

We know that the large cube is melted into 5 smaller cubes i.e. cube 1, cube 2, cube 3, cube 4 and cube 5.

5 solid cubes i.e. cube 1, cube 2, cube 3, cube 4 and cube 5 are 1 cubic unit, 1 cubic unit, 8 cubic units, 27 cubic units and 27 cubic units respectively.

1 side of cube 4 and cube 5



Now surface area of cube 1 and cube 2 is 6 × 12 = 6 sq.units.

Also, the volume of the larger cube = sum of the volumes of all 5 smaller cubes

⇒1 + 1 + 8 + 27 + 27 = 64 cubic units



Surface area of the larger cube = 6 × 4 × 4 = 96 sq.units

Total surface area of 5 smaller cubes - surface area of larger cube = 144 – 96 = 48 sq.units.

So percentage by which surface area of the larger cube exceeds the total surface area of the 5 smaller cubes



Hence, option 2.

Let us depict the given information using diagram shown below:



We need to find the area depicted by the shaded region.

As per given information, ABC is an isosceles right angled triangle with BC being the hypotenuse. This implies AB = AC. So if AB = 6cm, then AC will also be 6cm and BC will be

Now Arc BPC is a quarter circle with radius 6cm as the central angle of APC is 90^°

So area of Arc BPC

Now Area of triangle ABC

Area of arc of triangle BPC = Area of ∆ABC + Area if Arc BPCB

Area of BPCB = (9π – 18) sq.cm

Now area of semi circle BQC

So area of region enclosed by BQC and BPC

= 9π – (9π – 18) = 18 sq.cm.

Hence, option 2.

Semi-perimeter (s)



Now Area of ∆ABC. This means that the remaining area of the triangle (apart from ∆GBC) will be two third of the area of the ∆ABC. So the remaining area of ∆ABC (apart from ∆GBC)



Hence, option 2.

The closed region formed by the equation

| x | + | y | = 2 will be a rhombus as shown below:

In the above figure AC and BD are the diagonals of the rhombus that intersect at right angles

Length of AC = Length of BD = 4 units

Area of Rhombus ABCD

Hence, option 3.

Let ‘x’ and ‘x + 5’ be the average of the scores of the boys and girls respectively in the mid-semester exam. Now as the ratio of boys to girls is 2 : 3, then average of the scores of all students in the mid-semester exam then

As per the given information, the average score of the girls dropped by 3 in the final semester exam. So the average score of the girls now decreases to ‘x + 2’ in the final semester. Also as the overall average of the entire class in the final semester, increases by 2, the new overall average becomes ‘x + 5’. If ‘a’ is the overall average of the boys in the final semester exam, then

This means that the average of the boys in the final semester exam increases by 9.5.

Hence, option 1.

Now as ‘a’ and ‘b’ are integers of different signs, they will have a minimum difference of 2 (with the higher number being 1 and lower number being-1). Also, as the modulus of the ratio(a - 1): (b - 1) is 2 : 1, it implies that neither of (a - 1) or (b - 1) is 0. Further, for the sign of the greater of the 2 numbers to change it has to reduce by the least 2. This implies that (a - 1) and (b - 1) will also have opposite signs.

So the ratio (a - 1) : (b – 1) will be negative

Now the modulus of the ratio (a + 3) : (b) will be 3 : 1

However, since a is an integer, this case is not possible

On solving (I) & (III) we get a = 15 and b = -6

So a : b = 15: -6 or 5: -2

Hence, option 4.

Let the printed or marked price of each toy be Rs. ‘x’. At 40% discount S.P of each toy will be 0.6x. Total S.P of all 60 toys = 0.6x × 60

= 36x

So at 20 percent profit, C P of all 60 toys will be calculated as shown below:

⇒3600x –100 C.P = 20 C.P

⇒120 C.P = 3600x or C.P = 30x

Now, if the same amount of profit i.e.

36x – 30x = 6x is to be made, then the S.P of 36x is to be earned out of 60 – 10 or 50 toys.

∴ The discount on printed price of

50 items will be

Hence, option 4.

Let the C.P of good quality mangoes be 

Rs.‘x’per kg.

So the total C.P of good quality mangoes is Rs. 80x

Now as the C.P of good quality mangoes is Rs ‘x’ per kg, the C. P of medium quality mangoes will be Rs. ‘0.5x’ per kg.

So, the total C.P of medium quality mangoes will be 0.5x ×40 = 20x

Total C.P of all mangoes = 80x + 20x = 100x

Now the S.P per kg of all mangoes

Total S.P of all 120kg of mangoes

= 0.9x × 120 = 108x

Hence, profit percent



Hence, option 2.

Let the number of large, super and jumbo size popcorn packets be 7x, 17x and 16x respectively and the number of large, super and jumbo size chips packets be 6y, 15y and 14y respectively. Now as the total number of popcorn and chips packets is the same.

7x + 17x + 16x = 6y + 15y + 14y

⇒ 40x = 35y or 8x = 7y

Now, the number of jumbo popcorn packets is 16x and number of jumbo chip packets is 14y. But since 8x = 7y, this will also imply that 16x = 14y.

So, the required ratio of jumbo popcorn packets to jumbo chips packets will be 1 : 1.

Hence, option 1

Let the number of boys appearing for the test be x and the number of girls appearing for the test be 2x.

If 30% of the girls get admission, it implies 70% do not get admission. So 70% of 2x or 1.4x do not get admission.

Similarly. as 45% of the boys get admission, it implies 55% of the boys do not get admission. So 55% of x or 0.55x boys do not get admission.

So total number of students not getting admission = 1.4 + 0.55x = 1.95x

Hence, option 4.

Now the ratio of profits earned by C1, C2 and C3 is 9 : 10 : 8. Also ratio of the profits earned by C2, C4 and C5 is the ratio 

18 : 19 : 20. This can be represented as below:

Since C2 is common, we can find the combined ratio by finding the LCM of 10 and 18 which is 90. So all the numbers in the first row are multiplied by 9 and those in the 2nd row by 5.

So the ratio of profits made by C1, C2, C3, C4 and C5 is 81 : 90 : 72 : 95 : 100 Let us suppose profits made by C1, C2, C3, C4 and C5 is 81x, 90x, 72x, 95x and 100x.

∵ C5 has made 19 crore profit more than C1

100x -81x = 19 ⇒ x = 1

Now, sum of profits of all 5 companies (in terms of x)

= 81x + 90x + 72x + 95x + 100x = 438x

So sum of profits of all 5 companies

= 438 × 1 = 438 crores.

Hence, option 1

Let ‘d’ be the total distance for a back and forth journey.

Let ‘x’ be the original speed of the boat

So if he speed of the boat is doubled it becomes 2x.

Let ‘y’ be the speed of the river

Now ‘x + y is the relative speed of the boat travelling down stream and ‘x – y’ is the relative speed of the boat travelling upstream.

If the speed of the boat is doubled, then it’s relative speed downstream is ‘2x + y and its relative speed upstream is 2x – y’

Total time taken when speed of the boat is ‘x’

Total time taken when speed of the boat is ‘2x’

As per given information, total time taken when the speed of the motor boat is 2x reduces by 75% or becomes one-fourth of the original time taken

⇒ 4x² – y² = 8x² – 8y²

⇒ 4x² – 7y²

Hence, option 2.

Let M.P = Rs. 100

Since discount is 15%, S.P = 100 – 15 = 85

Let the C.P be Rs. ‘x’.

As the profit percentage is 2



2x = 8500 – 100x


Now let us suppose that S. P is ‘y’ To ensure profit is 20%.







⇒100y = 30000/3

100y = 10000

y = 100

So, the seller should sell at the marked price or retail to ensure a profit of 20%.

Hence, option 4.

Let us assume that total monthly savings of Ravi are 100x. If 50x is put in fixed deposits then 30% of (100x – 50x) or 15x will be put in stocks.

So the amount put in Ravi’s savings account = 100x – 50x – 15 = 35x

Now the amount in Ravi’s savings account and fixed deposits = 50x + 35 = 85

As per given information 85x = 59500

⇒ x = 700

So Ravi’s Total monthly savings is 100x or 100 × 700 = Rs. 70000

Answer: 70000 

Let ‘t’ be the time the man should take to reach the station such that he reaches just at the train is about to depart.

As per given data, if the train man walks at 12 kmph, then he will take

Also, if he walks at 15 kmps, then he will

Since the distance travelled is the same in both cases

Therefore, the required answer is 20

Answer: 20

Now if we divide 630 kg (which is the weight limit of the elevator) by 53 kg (which is the highest weight)

So the elevator will accommodated less than 12 or a maximum of 11 people.

One possible way less than 12 or a maximum of 11 people. weighing 57 kg are accommodated in the elevator. So then total weight of 11 people in the elevator

= 53 × 10 + 57 × 1 = 587 kg. So, a maximum of 11 people can be accommodated in the elevator

Therefore, the required answer is 11.

Answer: 11

Let us suppose it takes ‘n’ days to complete the required job. On day 1 we have  1 person working alone, on day 2 we have 2 people working together, on day 3 we have 3 people working together and so on till the nth day. where we have n people working together

On day 2,

On day ‘n’, n

As per the given information

Since n cannot be negative. n = 15.

So, the number of days taken to complete the job is 15.

Answer : 15

Let the present ages ages of Arun and Barun be 4x and 10x respectively. Let us suppose in ‘y’ years, Arun’s age will be half of Barun’s age

So in ‘y’ years, Barun’s age increases from 10x to 12x

Percentage increase in Barun’s age

Answer: 20