Select Page

This is the first post in the CAT 2016 Sprint Preparation Series – Geometry. We have posted 20 questions from previous year CAT papers, forums, mock tests, and other entrances that are on par with the level of difficulty you can expect in CAT 2016. We will be posting the solutions and traps/things to look at while solving similar questions so that you are avoid making silly mistakes during the test.

# Quant | Set 1 | Geometry

## CAT 2016 Sprint Preparation Series

Q1-3. Consider a cylinder of height h cms and radius r = 2/π cms as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B where A and B are on the same vertical line but on the bottom and top edges respectively, gives a maximum of n turns (in other words, the string’s length is the minimum length required to wind n turns.)

1. What is the vertical spacing in cms between two consecutive turns?

(1) h/n

(2) h/√n

(3) h/n2

(4) Cannot be determined

2. The same string, when wound on the exterior four walls of a cube of side n cms, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale). The length of the string, in cms, is

(1) √2n

(2) √17 n

(3) n

(4) √13 n

3. In the setup of the previous two questions, how is h related to n?

(1) h = √2 n

(2) h = √17 n

(3) h = n

(4) h = √13 n

4. In the figure, ACB is a right-angled triangle. CD is the altitude. Circles are inscribed within the ΔACD and ΔBCD. P and Q are the centres of the circles. The distance PQ is

(1) 5

(2) sqrt(50)

(3) 7

(4) 8

5. The length of the common chord of two circles of radii 15 cm and 20 cm, whose centres are 25 cm apart, is

(1) 24 cm

(2) 25 cm

(3) 15 cm

(4) 20 cm

6. Consider a circle with unit radius. There are seven adjacent sectors, S1, S2, S3, …, S7, in the circle such that their total area is 1/8th of the area of the circle. Further, the area of the jth sector is twice that of the (j – 1)th sector, for j = 2, …, 7. What is the angle, in radians, subtended by the arc of S1 at the centre of the circle?

7. There is a square field of side 500 m long each. It has a compound wall along its perimeter. At one of its corners, a triangular area of the field is to be cordoned off by erecting a straight-line fence. The compound wall and the fence will form its borders. If the length of the fence is 100 m, what is the maximum area that can be cordoned off?

(1) 2,500 sq m

(2) 10,000 sq m

(3) 5,000 sq m

(4) 20,000 sq m

8. How many triangles of integral sides and perimeter = 18 cm can be formed such that no two sides are equal?

(1) 3

(2) 4

(3) 7

(4) None of these

9. Euclid has a triangle in mind. Its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side?

(1) √260

(2) √250

(3) √240

(4) √270

10. Measure of angle EFD is 1200 and that of angle EBD is 400. What is the sum of the measures of angle BAE and angle BCD (in degrees)?

(1) 70

(2) 80

(3) 90

(4) 100

11. Find the circumradius of a triangle whose sides are 21, 72 and 75 units.

12. ABC forms an equilateral triangle in which, B is 2 km from A. A person starts walking from B in a direction parallel to AC and stops when he reaches a point D directly to the east of C. He then reverses his direction and walks till he reaches a point E directly south of C. Then D is ______ km from A. Additional information √3 = 1.732, √3=1.414. Calculate the exact answer using the additional information.

Q.13-14

13. What is the area that can be grazed by the cow if the length of the rope is 8 m?

14. What is the area that can be grazed by the cow if the length of the rope is 12 m?

15. The figure shows the rectangle ABCD with a semicircle and a circle inscribed inside in it as shown. What is the ratio of the area of the circle to that of the semicircle?

16. The length of a ladder is exactly equal to the height of the wall it is learning against. If lower end of the ladder is kept on a stool of height 3 m and the stool is kept 9 m away from the wall, the upper end of the ladder coincides with the top of the wall. Then the height of the wall is

(1) 12 m

(2) 15 m

(3) 18 m

(4) 11 m

17. Building A and building B are on horizontal ground. The angle of elevation of the top of building A from that of the lower building B is 15o. Building A is 100 m in height and B is 80 m in height. Find the distance between both the buildings. (tan 15o = 0.26)

18. A certain city has a circular wall around it, and this wall has four gates pointing north, south, east and west. A house stands outside the city, 3 km north of the north gate, and it can just be seen from a point 9 km east of the south gate. What is the diameter of the wall that surrounds the city?

(1) 6 km

(2) 9 km

(3) 12 km

(4) None of these

19.

20. What is the number of distinct triangles with integral valued sides and perimeter 14?

(1) 6

(2) 5

(3) 4

(4) 3

Solutions follow on the next page.

<!––nextpage––>

1. The best thing to do in a CAT question that you cannot lay your hands on is to simplify things and then see if you are going anywhere. Letting go of a question simply because it ‘looks’ difficult is not really something that a good candidate does and so, you need to be aware of these traps. The second thing to remember is that, any question that has variables in the question and in the options (or in other words, is not text-input based) can be solved by simple substitution.

As we can see here, there are n turns. Now, if there were say 2 turns instead of a generic number like n, how would the figure look?

Doesn’t it look friendlier now?

If you just look at the figure, you will get it that if there are n turns distributed over height h, the distance between the two successive turns will be h/n. Simple, isn’t it?

2. Whenever you have questions that involve wrapping or winding a wire around another object, opening the object up is a great idea. This is particularly applicable in questions like these or in questions involving movement along the sides of a cuboid or a cylinder or a cone. So, in this case, the opened up figure would look like this:

As each side of the square face is n, we get that the base will be 4n and the height will be n. So, the total distance to be travelled will be √{(4n)2 +n2} or n√17.

3. The first thing that needs to be understood is that the question is indeed weird. There cannot be a specific relation between the height and the number of turns that are present. So, it is obvious that the person who set this question was not able to word this properly. The only clue here is that, there HAS been a relation given in each of the options and so, we need to figure out the answer.

Now that there has been an option given that we can consider the two cases above, the second question is the indicator. As obvious as it may sound, the height of the cube is n and there are n turns and so, the answer will be h = n.

4.

The first trap that you need to notice is that the two circles do not intersect at the point R. If that were the case, you would have gone on the wrong path and wasted precious time.

The first thing to see here is that ACB is a right angled triangle. So, if AC and BC are 15 and 20 units, AB will have to be 25 units and CD will have to be 12 units (by comparing the area of the triangle ABC through two.

What next?

Now, an average candidate will see two circles but a good candidate will see two incircles. That’s where the difference lies.

Now, we know that CD is equal to 12 and so, we can figure out that AD will be equal to 9 units and BD will be equal to 16 units.

The killer blow will be to relate the area of a triangle to the inradius. So, as the area of the triangle is given by the product of the inradius and the semiperimeter.

So, for the smaller (9-12-15) triangle, we get the inradius to be 3 units and for the larger (12-16-20) triangle, we get the inradius to be 4 units.

Is the answer 7 units? Nope. Again, if we look at the starting comment, the line joining the centers of the two circles will pass through two different points. So what next?

Let’s look at the relevant part of the diagram:

Now does it look friendly?

Of course the answer is √50.

So, unless you are able to see these things right at the start, it doesn’t make sense even attempting this one. This can be classified as a question that will create the difference between a 99.5%ile in QA and a 99.8%ile in QA.

5. Now, it is important to remember that, if in a triangle, a^2 + b^2 = c^2 then that triangle has to be a right angled triangle. The second property to remember is that, the line joining the centers of two circles will be the perpendicular bisector of the common chord.

Simple enough now?

You have to compare the two ways of finding area of the right angled triangle with sides 15-20-25.

½ * 15 * 20 = ½ * x * 25 (where x is half the length of the common chord)

x = 12 cm. So, length of the common chord will be 24 cm.

6. Looks difficult at first glance. But it is certainly not so.

Area of a sector is dependent on the angle subtended at the center. So, area of S2 = 2*area of S1, area of S3 = 2*area of S2 and so on. So, we get:

A(S1) + 2A(S1) + 4A(S1) … 64A(S1) = 1/8 A(Circle)

So, 127*(angle subtended by S1) = 1/8 (2π) … as the entire circle will span 2π radians

Angle subtended by S1 = π/508

7. Nothing more to think than the fact that product of two elements is the maximum when they are equal to each other. So, both the sides making the right angle will be equal and so, the area cordoned off will be ½ * 50 √2 * 50 √2 = 2500 sq. m

8. Again the options tell us that the cases are limited so we write them down:

(7, 6, 5)

(8, 7, 3)

(8, 6, 4)

9.

Looks easy now?

½ * 20 * h = 80

So, h = 8

So, 20 will be split as 14 + 6.

So, third side will be √260.

10.

This is one of the easy ones that you should not miss.

EBD = 40

EFD = 120

DFA = 60

EFC = 60

BAE + BEA = 140

BEA = EFC + BCD

140 – BAE = 60 + BCD

BAE + BCD = 80

11. The difficult way is to apply the formula abc/4R and then calculate. The easy way is to understand that 21, 72 and 75 are in the ratio 7 : 24 : 25 which means that it will be a right angled triangle. The direct property tells us that in a right angled triangle, the circumradius is half the length of the hypotenuse and so, 37.5 units.

12.

As it is an equilateral triangle, the height will be √3 km. Also, the symmetry dictates that the horizontal distance between B and D is 1 km. So, using Pythagoras’ theorem, we get:

3 + 9 = AD^2

AD = √12 = 3.464 km.

Be careful while entering the answer if it is a text input based question.

13. Another typical trap when it comes to ropes and boundaries. If the rope is shorter than the side, the movement will be in the form of a sector. But, if the rope is longer than the side, it will cover an additional sector around the edge.

In this question, we need not worry about the movement.

The angle subtended is 330 degrees and the radius is 8 m. So, the area grazed will be:

11/12 * π * 64

14.

So, it will cover two sectors of 105 degrees (on either side) of radius 2 m more than what it would have in the previous case.

15. A simple construction would do:

Consider triangle ECB

EB = √2 * R where R is the radius of the semicircle.

Also EB = R + r + √2r = R + r(√2 + 1)

So, we get the ratio of the radii to be

r/R = (√2 – 1)/(√2 + 1)

So, area of circle to area of semicircle will be

2(r/R)^2 = 2(√2 – 1)^4

16. (x – 3)^2 + 81 = x^2

Substitute and check the options. x = 15 m.

17.

The simple approach is to use the tan given.

The distance between the two buildings be d.

tan 15 = 20/d

Or, d = 20/0.26 = 76.92 m

18.

ODA is similar to CBA

r/9 = x/(2r + 3)

x^2 = 3 * (3 + 2r)

x^2 = 3 * 9x/r

rx = 27

So,

r/9 = 27/r(2r + 3)

Substituting options, we get r = 4.5 or diameter = 9.

19.

Whenever you have multiple angles, always mark them using numbers instead of symbols. In this question, you would find it extremely easy to get mixed up with symbols and so, try representing it like this:

Now I suppose it is crystal clear.

(x) = 2 * (1)

(2) = 2 * (1)

(2) + (z) = (x) + (1)

(3) = (x) + (y) = (1) + (2), (y) = (1) + (2) – (x) = (1)

So, we get

(1) + (x) + (y) + (2) + (z) = 180

(1) + 2 * (1) + (1) + 2 * (1) + (1) = 180

7 * (1) = 180 or (1) = 25.7 degrees.

20. The normal equation based method would have taken a lot of time to finish the solution. What can be done here is a mixture of some smart work and reverse substitution.

As soon as we see the options as 3, 4, 5, 6 the first instinct should be to write down the cases instead of going for a theoretical based approach. Of course the p^2/48 formula would hold true here but assuming that we do not know this, we can still go for writing down cases.

The important thing to remember here is that, in any triangle, the longest side is less than half the perimeter of the triangle but is greater than or equal to one third of the perimeter of the triangle.

So, the longest side should be less than 7 but greater than or equal to 4.67. So, it could be either 5 or 6.

If longest side is 5, we get (5, 5, 4)

If longest side is 6, we get (6, 6, 2) (6, 5, 3) (6, 4, 4)

Total of 4 cases.

PS: [196/48] will be 4 too.

Hope you enjoyed the set. Coming with a new set tomorrow. Stay tuned!

You can go through the entire series by clicking on this link: CAT 2016 Sprint Preparation Series by Learningroots.

## You have successfully subscribed! :)

error: Content is protected !!