## Question of the Week – 08 : Probability

Ishan being dyslexic, gets confused between the following pairs/sets of letters and writes them interchangeably: (B & D); (U, V & W); (E & F). He gets punished by his teacher whenever he makes three or more mistakes in a single word. Find the probability that he’ll get punished if he’s asked to write the word ‘OLIVEBOARD’. (Assume he makes no other mistake)

Options:

a) 3/8

b) 5/24

c) 1/12

d) None of these

Solution:

As per the given conditions, he can possibly make mistake at four places

A – Letter V

B – Letter E

C – Letter B

D – Letter D

Let P(A), P(B), P(C) and P(D) denote the probabilities that he makes the above mistakes respectively.

P(A) = 2/3, P(B) = 1/2, P(C) = 1/2 and P(D) = 1/2.

Probability that he gets punished = P(A B C D) + P(A B C D) + P(A B C D) + P(A B C D) + P(A B C D)

= (2/3*1/2*1/2*1/2) + (2/3*1/2*1/2*1/2) + (2/3*1/2*1/2*1/2) + (1/3*1/2*1/2*1/2) + (2/3*1/2*1/2*1/2)

= 3/8

## Question of the Week – 07 : Numbers

What is the largest three-digit factor of N = 9603 – 6203 – 3403?

Options:

a) 996

b) 992

c) 960

d) None of these

Solution:

We can factorize the number as:

N = (960 – 620) (9602 + 960×620 + 6202) – 3403

= 340 x (9602 + 960×620 + 6202 – 3402)

= 340 x (1300×620 + 960×620 + 6202)

= 340 x 620 x (1300 + 960 + 620)

= 340 x 620 x 2880

= (22 x 5 x 17) x (22 x 5 x 31) x (26 x 32 x 5)

= 210 x 32 x 53 x 17 x 31

The largest three-digit number that can be formed using the combination of these factors is 992(25 x 31)

Alternate solution:

If we observe the given numbers, we can see that 960 = 620 + 340

The expression can be written as N = A3 – B3 – C3, where A = B + C

N = (B+C)3 – B3 – C3

= B3 + 3B2C + 3BC2 + C3 – B3 – C3

= 3BC (B+C)

= 3ABC

Here, A = 960, B = 620, C = 340

So, N = 3 x 960 x 620 x 340

Then we can proceed as above.

## Question of the Week – 06 : Triangles

In  triangle ABC, D, E and F are points on BC, AC and AB respectively such that BD: DC = 1:1, CE: EA = 1:2 and AF: FB = 1:3. If the area of  triangle ABC = 48 cm2, find the area of triangle DEF.

Options:

a) 14 cm2

b) 16 cm2

c) 18 cm2

d) None of these

Solution:

Area of triangle = ½ bc sin A

Using this theorem, we’ll calculate the area of the smaller triangles.

## Question of the Week – 05 : Mixtures and Alligations

As per new regulations, wine that contains more than 10% of alcohol cannot be sold. A winemaker has three wines X, Y and Z that contain 30%, 20% and 5% of alcohol (by volume) respectively. As a measuring equipment, he has only a beaker which has a capacity of 50 ml. What is the least volume of wine which he can prepare that contains exactly 10% of alcohol using all the three wines?

Options:

a) 350 ml

b) 400 ml

c) 500 ml

d) 450 ml

Solution:

Wines X and Y when mixed in any proportion would yield a mixture that contains alcohol between 20% and 30%.

So the resultant wine must be prepared using X and Z separately and Y and Z separately and then adding both of those mixtures.

From X and Z,

So, X and Z when mixed in 5:20 or 1:4, yield a mixture (say A) that has 10% of wine.

From Y and Z,

So, Y and Z when mixed in 5:10 or 1:2, yield a mixture (say B) that has 10% of wine.

Now mixtures A and B contain 10% of alcohol each. So they need to be mixed in equal volumes to produce the final mixture.

Therefore the resultant mixture will have 1 part of X, 1 part of Y and 4+2=6 parts of Z.

Hence, X:Y:Z = 1:1:6

Now he has a beaker that can measure only 50 ml. So the minimum volume that he can prepare is 50 x 8 = 400 ml.

## Question of the Week – 04 : Equations

How many real roots does the below equation have?

x4 + 10|x3| + 35x2 + 50|x| + 24 = 0

Options:

a) 2

b) 4

c) 8

d) None of these

Solution:

Let f(x) = x4 + 10|x3| + 35x2 + 50|x| + 24

For all real values of x, x4 and x2 both are non-negative.

Also |x3|≥0 and |x|≥0

Thus, f(x) ≥ 24

Hence the value of the function can never be zero or in other words, the given equation has no real roots.

## Question of the Week – 03 : Work and Time

On a scorching hot day, Arun saw a bird die of thirst. Affected by this pitiful sight, he decided that he’ll make some arrangement so that no more birds die in this manner. There’s a pit in his locality which has a cross section 2.9 m x 1.6 m and is 40 cm deep. The next morning he poured a bucket of water into the empty pit for the birds to drink. Next day, his immediate neighbors Manish and Guru were influenced by his noble deed and they also poured a bucket each along with him. On the third day, their immediate neighbors also contributed to the cause. In this way, every day two more people join and all of them pour a bucket each, into the pit. If every bucket used has a capacity of 20 liters and 16 liters of water is consumed by the birds every day, on which day will the pit be full?

Options:

a) 8th day

b) 10th day

c) 11th day

d) 12th day

Solution:

Volume of the pit = 2.9 x 1.6 x 0.4 = 1.856 m3= 1856 liters

Let the number of days in which the pit gets filled be n.

Water filled on 1st day = 20 liters

Water filled on 2nd day = 20×3 liters = 60 liters

Water filled on 3rd day = 20×5 liters = 100 liters

Water filled on nth day = 20x(2n-1) liters

This is an A.P. with common difference of 40.

Total water filled = 20 + 60 + 100 + ……….. + 20(2n-1)

= n/2 (20 + 40n – 20)                                         [∵ Sum of A.P. = n/2(a+l)]

= 20n2

But 16 liter of water is consumed by the birds every day. So on the nth day, when the people have poured water into the pit, the volume of water remaining in the pit will be 20n2 – 16(n-1)

Equating this with the volume of the pit, we get

20n2 – 16(n-1) = 1856

20n2 – 16n – 1840 = 0

5n2 – 4n – 460 = 0

Solving we get n = 10

Hence, the pit will be full on the 10th day.

## The right time to start for CAT

Many of the CAT aspirants are freshers; either they are in the final year of their under graduation or are fresh college pass outs. They are perplexed about the right time to start their preparation for CAT. Though it is always good to start as early as possible, but many aspirants tend to lose their grip due to lack of persistence in their preparation. The start should neither be too late because in that case one is unable to cover many topics for the exam. So the question arises what is the ideal time to start preparing for CAT?

As per the previous years’ results, it has been observed that most of the successful CAT aspirants start their preparation during summers (May-June). This period is neither too early nor too late. By this time all the college students across the country are done with their final exams, so they don’t have any extra burden as well. The CAT exam is held somewhere in November, so they have about 5-6 months in hand which is enough for preparation.

The next thing they are uncertain about is how to start their preparation? It is recommended to take a mock test first or go through previous years’ exam paper in order to find their own level. Once they have assessed themselves, they should prepare a fixture so as to cover various topics in a certain duration of time. Once done with the topic wise preparation, take mocks on a regular interval, preferably once or twice a week and analyze the results to improve further.

## Questions of the Week – 02 : Numbers & Parajumbles

1) At one stage of IPL, DeVilliers, Milller, Maxwell and Gayle were among the top four batsmen in the tournament run chart. The aggregates of their runs were the perfect squares of four consecutive numbers. In the encounter between Bangalore and Punjab, Gayle and DeVilliers were the star performers for Bangalore. Gayle’s score was more than 100 which is a perfect square of a number and he jumped from the 4th to the 2nd position. DeVilliers got 80 and retained his top spot. Maxwell and Miller, who play for Punjab, both scored less than 50. Coincidentally, at the end of the match, their aggregates were still perfect squares of numbers.

What was the total number of runs scored by these four batsmen in the match?

Options:

a) 252

b) 280

c) 296

d) 269

Solution:

It is clear from the given statements that prior to the match, DeVilliers was leading the chart and Gayle was at the 4th position.

It is also given that DeVilliers scored 80 and his new aggregate is also a perfect square.

So we need to express 80 as the difference of two squares.

a2 – b2 = 80

or (a+b)(a-b)=80

For (a+b) and (a-b) both to be whole numbers, either both are even or both are odd.

Since 80 is an even number, both its factors cannot be odd. So, (a+b) and (a-b) are both even. 80 can be factorized as a product of even numbers as:

80 = 10*8 = 20*4 = 40*2

So, (a,b) = (9,1), (12,8) or (21,19)

Now we’ll examine each case.

CASE I: (a,b) = (9,1)

It means DeVilliers had 1 run at the beginning of the match and reached 81 after it.

This case is not possible as he was ranked first prior to the game.

CASE II: (a,b) = (12,8)

It means DeVilliers had 64 runs at the beginning of the match and reached 144 after it.

So the other three players must have had 49, 36 and 25 runs, Gayle being at 25.

If we add a perfect square more than hundred to 25, we observe that 25+144=169, which is also a perfect square.

But this case is also negated as DeVilliers still led the chart.

CASE III: (a,b) = (21,19)

It means DeVilliers had 361 runs at the beginning of the match and reached 441 after it.

So the other three players must have had 324, 289 and 256 runs, Gayle being at 256.

If we add a perfect square more than hundred to 256, we observe that 256+144=400, which is also a perfect square.

This case satisfies the given conditions, that DeVilliers still leads the chart and Gayle jumps from the 4th spot to the 2nd spot.

Since Maxwell/Miller were at 289 and 324, neither of them scored more than 50 and their new aggregates are still perfect squares, their new aggregates must be 324 and 361. So their scores must have been 35 and 37.

So, the total runs scored by these four players in the match = 80+37+35+144 = 296

 Player Before the match Score in the match After the match DeVilliers 361 80 441 Maxwell/Miller 324 37 361 Miller/Maxwell 289 35 324 Gayle 256 144 400

Total = 80 + 37 + 35 + 144 = 296

Verbal – Parajumbles (Sentence Exclusion)

2) The following paragraph consists of four sentences, three of which form a coherent paragraph when arranged in the correct order. Identify the one that is not part of the paragraph.

1. Living things start with this wholeness from the beginning of their career.
2. Growth is the movement of a whole towards a yet fuller wholeness.
3. The one question before all others that has to be answered by our civilizations is not what they have and in what quantity, but what they express and how.
4. A child has its own perfection as a child; it would be ugly if it appeared as an unfinished man.

Solution:
In this case, the set of connected statements is: 2-1-4 (in that order).The connected set of statements (2-1-4) explains growth and how it develops. Statement 4 develops this forward, and provides an example of a child to consolidate the concept of growth.Statement 3 is the odd one out here as it talks about a more generic concept and about civilization in general, a topic that is absent from the other three statements.

## Question of the Week – 01 : Number Systems

If A = 200

B = A – 99

C = A + B – 98

D = A + B + C – 97 and so on upto Z.

Find the value of 2-25(Z+1)

Options:

a) 50

b) 51

c) 26

d) None of these

Solution:

A = 200

B = 101

C = 203

D = 407

E = 815

This way we see that for the nth English alphabet (where n>1), the value is (2n-2 x 100) + (2n-1 – 1)

So, Z = (224x100) + (225-1)

So, Z + 1 = (224x100) + (225-1) + 1

= 225(50 + 1)

So, 2-25(Z+1) = 2-25 x 225(50 + 1)

= 51

## CAT 2015: Basics and Solved Examples for Number Systems

Today, we look at a few questions from Number Systems and provide Video solutions for the same. Register for Oliveboard’s  CAT 2015 Course here. Question 1: If x = -0.5, which of the following has the smallest value? 21/x , 1/x, 1/x2, 2x, 1/√-x Solution: In solving this problem, we will first need to understand a couple of important concepts on Powers of numbers for CAT 2015. The video below explains all the concepts and basics related to it.

Now that we have the concept clear, we can look at the solution to this particular problem.

Question 2: Which among the following is the largest? 21/2, 31/3, 41/4, 61/6, 121/12 Solution: The solution relies on another basic concept of Number systems to compare different numbers with different exponents . The concept and the solution are given below.

Question 3: How may 2 digit numbers increase by 18 when the digits are reversed? Solution: Note that 2 digit numbers can be expressed as ab, where a,b are digits. So the original number is 10a + b, reversed number 10b + a. The complete solution along with a few more basics for CAT 2015 can be viewed below.

We will be returning with our next post in a few days time! Happy Learning! To attempt more practice questions from Number Systems, Register for Oliveboard’s  CAT 2015 Course here  and take the free Number Systems test!