Quants 9 - Problem on Trains

Key Points to Remember:

a km/hr = a * (5/18) m/s
a m/s = a * (18/5) km/hr
Time taken by a train of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.
Time taken by a train of length l metres to pass a stationary object of length b metres is the time taken by the train to cover ( l + b ) metres.
Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed = u - v m/s.
Suppose two trains or two bodies are moving in the opposite direction at u m/s and v m/s, where u > v, then their relative speed = u + v m/s.
If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then the time taken by the train to cross each other = (a+b) / (u+v) s.
If two trains of length a metres and b metres are moving in same directions at u m/s and v m/s, then the time taken by the faster train to cross the slower train = (a+b) / (u-v) s.
If two trains start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then A's speed : B's speed = √b : √a


Practice Questions:

Question 1:

A train 100 m long is running at the speed of 30 km/hr. Find the time taken to pass a man standing near the railway line.

Speed of the train = 30 * ( 5/18 ) m/s = 25/3 ms/s
Distance moved in passing the standing man = 100 m
Time Taken = Distance / Speed =u 100 / (25/3) = 100 * ( 3/25 ) = 12 sec

Question 2:

A train is moving at a speed of 132 km/hr. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metres long?

Speed of the train = 132 km/hr = 132 * (5/18) m/s = 110/3 m/s
Distance covered in passing the platform = 110 + 165 m = 275 m
Time Taken = 275 / (110/3) = 275 * ( 3/110 ) = 15/2 = 7(1/2) sec

Question 3:
A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed.

Let the length of the train be x metres.
Then, the train covers x metres in 8 seconds and x+180 metres in 20 seconds.
=> x/8 = (x+180)/20
=> 20x = 8(x+180)
=> 5x = 2x + 360
=> 3x = 360
=> x = 120 m
Speed of the train = Distance / Speed = 120 / 8 m/s = 15 * (18/5) = 54 kmph


Question 4:
A train 150 m long is running with a speed of 68 kmph. In what time will it pass a man who is running at 8 kmph in the same direction in which the train is going?

Speed of the train relative to man = 68 - 8 = 60 kmph = 60 * (5/18) = 50/3 m/s
Time taken by the train to cross the man = Time taken by it to cover 150 m at 50/3 m/s
=> 150 * (3/50) = 9 sec


Question 5:

Two trains 137 metres and 163 metres in length are running towards each other on parallel lines, one at the rate of 42 kmph and another at 48 kmph. In what time will they be clear of each other from the moment they meet?

Relative Speed of the train = 48 + 42 kmph = 90 kmph = 90 * (5/18) = 25 m/s
Time taken to pass each other = Time taken to cover (137 + 163) metres at 25 m/s => 300 metres = 300/25 = 12 sec

Time to Think:
A train 220 m long is running with a speed of 59 kmph. In what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going?

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Quants 8 - Percentage

Key Points to Remember:

By a certain percent, we mean that many hundredths. Thus, x percent means, x hundredths, written as x%.
To express x% as a fraction, we have x% = x/100
To express a/b as a percent, we have a/b = (a/b) * 100%


Practice Questions:

Question 1:

Express as fraction.
(i) 58% (ii) 8% (iii) 0.4% (iv) 0.09%

(i) 58% = 58/100 = 29/50
(ii) 8% = 8/100 = 2/25
(iii) 0.4% = 0.4/100 = 4/1000 = 1/250
(iv) 0.09% = 0.09/100 = 9/10000

Question 2:

Express as decimal.
(i) 7% (ii) 25% (iii) 0.3% (iv) 0.02%

(i) 7% = 7/100 = 0.07
(ii) 25% = 25/100 = 1/4 = 0.4
(iii) 0.3% = 0.3/100 = 3/1000 = 0.003
(iv) 0.02% = 0.02/100 = 0.002%

Question 3:
Express as rate percent.
(i) 23/36 (ii) 6(3/4) (iii) 0.007

(i) 23/36 = 23/36 * 100 = 575/9 = 63(8/9) %
(ii) 6(3/4) = 27/4 = 27/4 * 100 = 675%
(iii) 0.007 = 7/1000 = 0.7%


Question 4:
(i) 28% of 450 + 45% of 280
(ii) 16(2/3) of 600 gram - 33(1/3) of 180 gram

(i)
= (28/100) * 450 + (45/100) * 250
= 126 +126
= 252
(ii)
= (50/3) * (1/100) * 600 - (100/3) * (1/100) * 180
= 100 - 60
= 40 gram


Question 5:

(i) 2 is what percent of 50?
(ii) What % of 7 is 84?
(iii) What % of 2 metric tonnes is 40 quintals?

(i) (2/50) * 100 = 4%
(ii) 84/7 * 100 = 1200%
(iii) 40/(2*10) * 100 = 200%

Time to Think:
(i) 1/2 is what % of 1/3?
(ii) What % of 6.5 litres is 130 ml?

Logical 7 - Statement and Assumptions

Practice Questions:

Question 1:

Statement - 
If you have any problems, bring them to me.
Assumptions - 
I. You have some problems.
II. I can solve any problem.

The word 'if' shows that 'you' do not necessarily have problems.
So, I is not implicit.
Also, the statement states that problems will be solved by 'me'.
So, II is implicit.
Hence, answer is (b) - only assumption II is implicit.

Question 2:

Statement - 
Detergents should be used to clean clothes.

Assumptions - 
I. Detergents form more lather.                    

II. Detergents help to dislodge grease and dirt.

Nothing is mentioned about the lather formation by the detergent.
So, I is not implicit.
Also, the detergents should be used as they clean clothes better and more easily.
So, II is implicit.
Hence, answer is (b) - only assumption II is implicit.

Question 3:

Statement - 
Like a mad man, I decided to follow him.

Assumptions - 
I. I am not a mad man.
II. I am a mad man.

The words 'Like a mad man' show that either a person is really mad or he is not mad but acted like mad.
So, either I or II is implicit.
Hence, answer is (c) - either assumption I or II is implicit.


Question 4:

Statement - 
The government has decided to pay compensation to the tune of Rs. 1 lakh to the family members of those who are killed in railway accidents. 

Assumptions - 
I. Prayer makes a man more humane.
II. Prayer atones all of our misdeeds.

Clearly, the amount of compensations must have been decided keeping in mind the monetary position of the Government.
So, I is implicit.
However, nothing can be said about the frequency of railway accidents in future.
So, II is not implicit.
Hence, answer is (a) - only assumption I is implicit.


Question 5:

Statement - 
Never before such a lucid book was available on the topic.

Assumptions - 
I. Some other books were available on this topic.
II. You can write lucid books on very few topics.

It follows from the statement that books on this topic were available before also but they were not 'lucid'.
So, I is implicit.
But a general comment as II cannot be made from the given statement.
So, II is not implicit.
Hence, answer is (a) - only assumption I is implicit.


Time to Think:

Statement - 
"I have not received telephone bills for nine months inspite of several complaints" - A telephone customer's letter to the editor of a daily.

Assumptions - 
I. Every customer has a right to get bills regularly from the telephone company.
II. The customer's complaints point to defect in the services which is expected to be corrected.

Logical 6 - Number Test

Practice Questions:

Question 1:

If it is possible to form a number with the second, the fifth and the eighth digits of the number 31549786, which is the perfect square of the two digit even number, which of the following will be the second digit of that even number?

The second, fifth and eighth digits of the number 31549786 are 1, 9 and 6 respectively.
The perfect square of the two digit even number formed using the digits 1, 9 and 6 is 196 and 196 = 14².
So, the required even number is 14 and the second digit is 4.

Question 2:

The positions of the first and the second digits in the number 94316875 are interchanged. Similarly, the positions of the third and the fourth digits are interchanged and so on. Which of the following will be the third to the left of the seventh digit from the left end after the rearrangement?

Changing the positions of the digits as mentioned, we get, 49138657
The seventh digit from the left end of this number is 5.
The third digit to the left of 5 is 3.
So, the the third to the left of the seventh digit from the left end after the rearrangement is 3.

Question 3:

How many such pairs of digits are there in the number 531268947 each of which has as many digits between them in the number as when they are arranged in descending order?

The given number is 531268947
Arranging the number in descending order, we get, 987654321
The digits satisfying the given condition are,

5312 - 5432
3126 - 6543
12 - 21
1268947 - 7654321
2268947 - 765432
89 - 98

We have six pairs such as, (5,2), (3, 6), (1, 2), (1, 7), (2, 7), (8, 9).


Question 4:

In a school, the following codes were used during physical exercise. '1' means 'start walking', '2' means 'keep standing', '3' means 'start running at the same spot', '4' means 'sit down'. How many times will a student who performs the following sequence without error from the beginning to the end have to sit down?

1 2 3 4 2 3 1 4 4 3 2 2 1 2 4 3 1 4 4 1 2

The student will have to sit down at the places,
1 2 3 4 2 3 1 4 4 3 2 2 1 2 4 3 1 4 4 1 2
The student has to sit down 4 times.


Question 5:

Thirty six vehicles are parked in a parking lot in a single row. After the first car, there is one scooter. After the second care, there are two scooters. After the third care, there are three scooters and so on. Work out the number of scooters in the second half of the row.

Let C and S denote car and scooter respectively.
The sequence of parking is,
C S C S S C S S S C S S S S C S S S S S C S S S S S S C S S S S S S S C
The parking lot has been divided into two.
C S C S S C S S S C S S S S C S S S || S S C S S S S S S C S S S S S S S C

Thus, the number of scooters in the second half of the row is 15.


Time to Think:

If it is possible to make a number which is perfect square of a two digit odd number with the second, the sixth and the ninth digits of the number 187642539, which of the following is the digit in the unit's place of that two-digit odd Number?

Answers Week 3

Logical 5 - Number Test
How many 5s are there in the following number sequence which are immediately proceeded by 7 and immediately followed by 6?

7 5 5 9 4 5 7 6 4 5 9 8 7 5 6 7 6 4 3 2 5 6 7 8 

7 5 5 9 4 5 7 6 4 5 9 8 7 5 6 7 6 4 3 2 5 6 7 8

There is only one 5 immediately preceded by 7 and immediately followed by 6.

Quants 5 - Calendar

What will be the day on October 15, 2020?

15th October 2020 = 2019 years + (Jan 1, 2020 to Oct 15, 2020)

Odd Days:

2000 = 0 Odd Days

19 Years = 4 Leap Years + 15 Ordinary Years

               = (4 * 2) + (15 * 1) = 23 days = 3 weeks + 2  day = 2 Odd Days

So, 0+2 = 2 Odd Days till 2019.

For 2020, counting days from Jan to Oct,

31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 15 = 289 days = 41 weeks + 2 days = 2 Odd Days

So, 2 + 2 = 4 Odd Days

As Odd day 4 corresponds to Thursday, 15th October, 2020 is THURSDAY.



Verbal 3 - Sentence Completion 
He is my best friend, _______ I can't avoid inviting him. (Identify the part of speech to complete the sentence)
so - conjunction

Quants 6 - Alligation

In what ratio must water be mixed with milk costing ₹12 per litre to obtain a mixture worth of ₹8 per litre?

Cost Price of water (c) = 0

Cost Price of milk (d) = 12

Mean price (m) = 8

Cheaper quantity (d-m) = 12 - 8 = 4

Dearer quantity (m-c) = 8 - 0 = 8

Required Ratio of Water to Milk = 4 : 8 = 1 : 2

Quants 7 - Clocks

At 3.40, the hour hand and the minute hand of the clock form an angle of ___.

Angle traced by the hour hand in 12 hours = 360˚.

Angle traced by it in 3.40 hours ie 3 (2/3) => 11/3 = (360/12) * ( (11/3) = 110˚

Angle traced by minute hand in 60 minutes = 360˚.

Angle traced by it in 40 minutes = (360/60) * 40 = 240˚.

Required angle = 240˚ - 110˚ = 130˚.

Quants 7 - Clocks

Key Points to Remember:

The face or dial of a watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.
A clock has two hands. The smaller one is called, the hour hand or the short hand and the larger one is called the minute hand or the large hand.

In 60 minutes, the minute hand gains 55 minutes on the hour hand.
In every hour, both the hands, coincide once.
The hands are in the same straight line, when they are coinciding or on opposite to each other.
When the two hands are at right angles, they are 15 minute spaces apart.
When the hands are in opposite directions, they are 30 minute spaces apart.
Angle traced by hour hand in 12 hours = 360˚.
Angle traced by minute hand in 60 minutes = 360˚.

Too Fast and Too Slow - 
If the clock indicates 7.15, when the correct time is 7, it is said to be 15 minutes too fast.
If the clock indicates 6.45, when the correct time is 7, it is said to be 15 minutes too slow.

Practice Questions:

Question 1:

Find the angle between the hour hand and the minute hand of the clock when the time is 3.25.

Angle traced by the hour hand in 12 hours = 360˚.
Angle traced by it in 3.25 hours ie 3 (25/60) => 205/60 => 41/12 hrs = (360/12) * ( (41/12) = 205/2 = 102 (1/2)˚
Angle traced by minute hand in 60 minutes = 360˚.
Angle traced by it in 25 minutes = (360/60) * 25 = 150˚.
Required angle = 150˚ - 102 (1/2)˚ = 47 (1/2)˚.

Question 2:

A clock is started at noon. By 10 minutes past 5, the hour hand has turned through ___.

Angle traced by hour hand in 12 hours = 360˚.
Angle traced by hour hand in 5 hours 10 minutes i.e., 5(10/60) -> 31/6 hours = (360/12) * (31/6) = 155˚

Question 3:
At what time between 2 and 3 o'clock, will the hands of the clock be together?

At 2'o clock, the hour hand is at 2 and the minute hand is at 12, i.e., they are 10 minute spaces apart.
To be together, the minute hand must gain 10 minutes over the hour hand.
Now 55 minutes are gained by it in 60 minutes.
10 minutes will be gained in (60/55) * 10 = 120/11 = 10(10/11) minutes.
The hands will coincide at 10 10/11 minutes past 2.

Question 4:
An accurate clock shows 8 o'clock in the morning. Through how many degrees, will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

Angle traced by hour hand in 6 hours (8.00 am to 2.00 pm) = 60 * 6 = 360˚.


Question 5:

The angle between the minute hand and the hour hand of a clock when the time is 8.30, is ___.

Angle traced by the hour hand in 12 hours = 360˚.
Angle traced by hour hand in 8.30 hours i.e., 17/2 hours = (360/12) * (17/2) = 255˚
Angle traced by minute hand in 60 minutes = 360˚.
Angle traced by minute hand in 30 minutes = (360/60) * 30 = 180˚
Required angle = 255˚ - 180˚ = 75˚


Time to Think:

At 3.40, the hour hand and the minute hand of the clock form an angle of ___.

Quants 6 - Alligations

Key Points to Remember:

Alligation - It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce the mixture of a desired price.
Mean Price -  The cost price of a unit quantity of the mixture is called the mean price
Rule of Alligation - If two ingredients are mixed, then
Quality of Cheaper / Quality of Dearer = [ CP of dearer - Mean Price ] / [ Mean Price - CP of cheaper ]

Practice Questions:

Question 1:

In what ratio must rice at ₹9.30 per kg be mixed with rice at ₹10.80 per kg so that the mixture be worth ₹10 per kg?

Cost Price of 1st rice in paise (c) = 930
Cost Price of 2nd rice in paise (d) = 1080
Mean price in paise (m) = 1000
Cheaper quantity (d-m) = 1080 - 1000 = 80
Dearer quantity (m-c) = 1000 - 930 = 70
Required Ratio = 80 : 70 = 8 : 7

Question 2:

In what ratio must water be mixed with milk to gain 20% by selling the mixture at cost price

Let the cost price of milk be ₹1 per litre.
Then selling price of 1 litre of mixture be ₹1.
Gain obtained = 20%

Cost price of 1 litre of mixture = ₹ (100/120) * 1 = 5/6

Cost Price of water (c) = 0
Cost Price of milk (d) = 1
Mean price (m) = 5/6
Cheaper quantity (d-m) = 1 - 5/6 = 1/6
Dearer quantity (m-c) = 5/6 - 0 = 5/6
Required Ratio of Water and Milk = 1/6 : 5/6 = 1 : 5

Question 3:
In what ratio must a grocer mix two varieties of pulses costing ₹15 and ₹20 per kg respectively so as to get a mixture worth ₹16.50 per kg?

Cost Price of 1st pulse (c) = 15
Cost Price of 2nd pulse (d) = 20
Mean price (m) = 16.50
Cheaper quantity (d-m) = 20 - 16.50 = 3.50
Dearer quantity (m-c) = 16.50 - 15 = 1.50
Required Ratio = 3.50 : 7.50 = 7 : 3 


Question 4:
Find the ratio in which rice at ₹7.20 a kg be mixed with rice at ₹5.70 a kg to produce a mixture worth ₹6.30 a kg.

Cost Price of 1st rice in paise (c) = 720
Cost Price of 2nd rice in paise (d) = 570
Mean price in paise (m) = 630
Quantity 1 (d-m) = 720 - 630 = 90
Quantity 2  (m-c) = 630 - 570 = 60
Required Ratio = 60 : 90 = 2 : 3


Question 5:

In what ratio must tea at ₹62 per kg be mixed with tea at ₹72 per kg so that the mixture must be worth ₹64.50 per kg?

Cost Price of 1st tea (c) = 62
Cost Price of 2nd tea (d) = 72
Mean price (m) = 64.50
Cheaper quantity (d-m) = 72 - 64.50 = 7.50
Dearer quantity (m-c) = 64.50 - 62 = 2.50
Required Ratio = 7.50 : 2.50 = 3 : 1 


Time to Think:

In what ratio must water be mixed with milk costing ₹12 per litre to obtain a mixture worth of ₹8 per litre?

Verbal 3 - Sentence Completion

Key Points to Remember

Proactive Solving
Identifying clues
Sign posts

• Support Sign - and, similarly, in addition, since, also, thus, because etc.
• Contrast Sign - but, despite, yet, however, unless, rather, although etc.

Root Word Clues

• Prefixes

1. Positive - pro, bene, philo etc.
2. Negative - non, un, im, in, mal etc.

• Suffixes 

1. Positive - ble, ant, ent, en etc.
2. Negative - less, cide etc.

Sentence Structure Clues

• Cause and Effect - accordingly, consequently, because, thus, for, since
• Definition / Restatement - comma, semicolon or colon
• Parts of Speech

Instructions:

Fill the blanks proactively. (1 - 3)
Which part of speech would fill the blank? (4 - 6)

Practice Questions:

Question 1:

He thought his problems were too _______.
complicated

Question 2:
I'm not totally _______ to the idea; however I do have some _______.
against; opinions

Question 3:
She couldn't believe that she had scored 90% in Physics, _______.
the toughest

Question 4:
We had won the game _______.
quickly - adverb

Question 5:
On this _______ occasion, I would like to introduce my fiancee to you all.
special - adjective

Time to Think:

He is my best friend, _______ I can't avoid inviting him.

Quants 5 - Calendar

Practice Questions:

Question 1:

What was the day on September 22, 1973?

22nd September 1973 = 1972 years + (Jan 1, 1973 to Sep 22, 1973)
Odd Days:
1600 = 0 Odd Days
300 Years = 1 Odd Day
72 Years = 18 Leap Years + 54 Ordinary Years
               = (18 * 2) + (54 * 1) = 90 days = 12 weeks + 6  day = 6 Odd Days
So, 0+1+6 = 7 = 0 Odd Days till 1972.
For 1973, counting days from Jan to Sep,
31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 22 = 265 days = 37 weeks + 6 days = 6 Odd Days
So, 0 + 6 = 6 Odd Days
As Odd day 6 corresponds to Saturday, 22nd September, 1973 is SATURDAY.

Question 2:

The Last day of the century cannot be?

100 years = 5 Odd Days. So, last day of the 1st century is Friday.

200 years = 10 = 3 Odd Days. So, the last day of the 2nd century is Wednesday.

300 years = 15 = 1 Odd Day. So, the last day of the 3rd century is Monday.

400 years = 20 + 1 = 21 = 0 Odd Days. So, the last day of the 4th century is Sunday. 

The same cycle of Friday, Wednesday, Monday and Sunday gets repeated.

Hence, the last day of the century cannot be a TUESDAY or a THURSDAY or a SATURDAY. 

Question 3:

On what dates of April 2001, did Sunday fall?

Let's find the day of April 1, 2001.

2000 = 0 Odd Days

For 2001, counting days from Jan to April,

31 + 28 + 31 + 1 = 91 days = 13 weeks = 0 Odd Days

So, 0 + 0 = 0 Odd Days = Sunday.

Hence, 1st Sunday in April 2001 was April1.

Other Sundays were 8, 15, 22, 29.

Thus, April 1, April 8, April 15, April 22 and April 29 of 2001 were Sundays.

Question 4:

The Calendar for the year 2005 is the same for the year?

Counting the number of days from 2005 onwards to get 0 Odd Day,
2005 + 2006 + 2007 + 2008 + 2009 + 2010 + 2011 = 1 + 1 + 1 + 2 + 1 + 1 = 7 = 0 Odd Days
Therefore, the calendar of 2011 is same as the year of 2005.

Question 5:

Today Friday, after 62 days, it will be?

Each day of the week is repeated after 7 days. So, after 63 days, it will be Friday. 

Hence, after 62 days, it will be a THURSDAY.

Time to Think:

What will be the day on October 15, 2020?

Logical 5 - Number Test

Key Points to Remember:

A group of numerals is given and the task is to trace out numerals following certain given conditions or lying at specific mentioned positions after shuffling according to certain given pattern. 

Practice Questions:

Question 1:

How much such 5s are there in the following number sequence each of which is immediately preceded by 3 or 4 but not immediately followed by 8 or 9?

3 5 9 5 4 5 5 3 5 8 4 5 6 7 3 5 7 5 5 4 5 2 3 5 1 0

A number which comes after the given number is said to be follow it while the one which comes before the given number precedes it.
Thus, the numbers satisfying the given conditions are,
3 5 9 5 4 5 5 3 5 8 4 5 6 7 3 5 7 5 5 4 5 2 3 5 1 0
Here five 5s are preceded by 3 or 4 and not following 8 or 9.
Hence, answer is 5.

Question 2:

In the series given below, how many 8s are there each of which is exactly divisible by its immediate preceding as well as succeeding numbers?

2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6 

We can mark such sets of three numbers in which the middle number is 8 and each of the two numbers on both sides of it is a factor of 8 as,
2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6 
There are two such 8s.
Hence, answer is 2.

Question 3:

Which is the third number to the left of the number which is exactly in the middle of the following sequence of numbers?

1 2 3 4 5 6 7 8 9 2 4 6 8 9 7 5 3 1 9 8 7 6 5 4 3 2 1

There are 27 numbers in the given sequence.

So, the middle number = 14th number = 9

Hence, the third number to the left of this 9 is 4.

Question 4:

How many 6s are there in the following number series, each of which is immediately preceded by 1 or 5 and immediately followed by 3 or 9? 

2 6 3 7 5 6 4 2 9 6 1 3 4  1 6 3 9 1 5 6 9 2 3 1 6 5 4 3 2 1 6 5 4 3 2 1 9 6 7 1 6 3 

2 6 3 7 5 6 4 2 9 6 1 3 4  1 6 3 9 1 5 6 9 2 3 1 6 5 4 3 2 1 6 5 4 3 2 1 9 6 7 1 6 3
Here three 6s are preceded by 1 or 5 and immediately followed by 3 or 9.
Hence, answer is 3.


Question 5:
How many 7s immediately preceded by 6 but not immediately followed by 4 are there in the following series?

7 4 2 7 6 4 3 6 7 5 3 5 7 8 4 3 7 6 7 2 4 0 6 7 4 3 

7 4 2 7 6 4 3 6 7 5 3 5 7 8 4 3 7 6 7 2 4 0 6 7 4 3 
Here two 7s are immediately preceded by 6 but not immediately followed by 4.
Hence, answer is 2.

Time to Think:

How many 5s are there in the following number sequence which are immediately proceeded by 7 and immediately followed by 6?

7 5 5 9 4 5 7 6 4 5 9 8 7 5 6 7 6 4 3 2 5 6 7 8 

Answers Week 2

Quants 3 - Calendar

On 8th March, 2005, Wednesday falls. What day of the week was it on 8th March, 2004?
The year 2004, being a Leap Year, it has 2 Odd days.
So, the day on March 8, 2005 will be two days beyond the day on March 8, 2004.
But, 8th March 2005 is Wednesday.
So, 8th March 2004 is MONDAY.


Verbal 2 - Sentence Correction
Neither the boys, nor the girl get the credit for doing this job.
Neither the boys, nor the girl gets the credit for doing this job.
(The second noun following nor here is the girl and as it is singular, the verb is gets.)

Logical 3 - Direction Sense

Rohan walks a distance of 3 km towards north, then turns to his left and walks for 2 km. He again turns left and walks for 3 km. At this point, he turns to his left and walks for 3 km. How many kilometers is he from the starting point?

Rohan moves from A to B, Bto C, C to D and C to D
AD = BC = 2 km
Distance = AE = DE - AD = 3 - 2 = 1 km
Hence, Rohan is 1 km from his starting point.

Quants 4 - Average

David obtained 76, 65, 82, 67 and 85 marks out of 100 in English, Math, Physics, Chemistry and Biology. What is his average mark?

Average = ( 76 + 65 + 82 + 67 + 85 ) / 5 = 375 / 5 = 75

David's average mark is 75.

Logical 4 - Statement and Assumptions

Statement - 
"Use of cell phones and pagers is not allowed inside the auditorium. Please switch off such devices while you are inside the auditorium." - A Notice.

Assumptions - 
I. All those who have such devices will switch them off before they take their seat in the auditorium.
II. Generally, people do not bring such devices when they come to attend functions in the auditorium.

The notice has clearly put up to make it clear that use of cell phones and pagers would create disturbance in the auditorium.

So, I is implicit.

Further, such notice has been issued keeping in mind that lot of people come with such devices to the auditorium.

So, II is not implicit.

Hence, answer is (a) - only assumption I is implicit.

Logical 4 - Statement and Assumptions

Key Points to Remember:

An ASSUMPTION is something supposed or taken for granted, i.e., a fact that can be accepted as true on considering the contents of the given statement.

In this type, a statement is given, followed by two assumptions. The task is to assess the given statement and then decide which of the given assumptions is implicit in the statement.

Give answers as
(a) - if only assumption I is implicit
(b) - if only assumption II is implicit
(c) - if either I or II is implicit
(d) - if neither I nor II is implicit
(e) - if both I and II are implicit

Practice Questions:

Question 1:

Statement - 
It is faster to travel by air to Delhi from Bangalore.
Assumptions - 
I. Bangalore and Delhi are connected by air.
II. There are no other means of transport available to Delhi from Bangalore.

The statement advises to travel by air between the two cities.
So, I is implicit.
Besides, it talks of air transport being a faster means of travel. This means that other means of transport, slower than air transport, are available.
So, II is not implicit.
Hence, answer is (a) - only assumption I is implicit.

Question 2:

Statement - 
Bank A has announced reduction of half percentage on the interest rate on retail lending with immediate effect

Assumptions - 
I. Other banks may also reduce the retail lending rates to be in competition.                       

II. Bank A may be able to attract more customers for availing retail loans.

Reducing interest rate on loans is surely a step a draw in more customers.
But, the implications of such a policy on other banks cannot be ascertained, as interest rate is not the only criterion to lure customers.
So, II is not implicit.
Hence, answer is (b) - only assumption II is implicit.

Question 3:

Statement - 
Even with the increase in the number of sugar factories in India, we will still continue to import Sugar.

Assumptions - 
I. The consumption of sugar per capita has increased in India.
II. Many of the factories are not in a position to produce sugar to their fullest capacity.

Clearly, the need to import sugar could be either due to increase in consumption or the inefficiency of the factories to produce sugar to their fullest capacity.
So, either I or II is implicit.
Hence, answer is (c) - either assumption I or II is implicit.


Question 4:

Statement - 
Who rises from the prayer a better man, his prayer is answered.

Assumptions - 
I. Prayer makes a man more humane.
II. Prayer atones all of our misdeeds.

The fact that only persons who become better by saying prayer are responded to, shows that prayer does not necessarily make man humane.
So, I is not implicit.
Nothing is mentioned as regards the fruitfulness of prayer.
So, II is not implicit.
Hence, answer is (d) - neither assumption I nor II is implicit.


Question 5:

Statement - 
We must settle all the payment due to our suppliers within three working days.

Assumptions - 
I. We will always have necessary funds in our account to settle the bills.
II. We are capable of verifying and clearing the bills in less than three working days.

Since, the statement talks of making all payments within three days, it is evident that the company has the necessary funds and the bills can be verified and cleared within the stipulated time.
So, both I and II are implicit.
Hence, answer is (e) - both assumptions I and II are implicit.


Time to Think:

Statement - 
"Use of cell phones and pagers is not allowed inside the auditorium. Please switch off such devices while you are inside the auditorium." - A Notice.

Assumptions - 
I. All those who have such devices will switch them off before they take their seat in the auditorium.
II. Generally, people do not bring such devices when they come to attend functions in the auditorium.

Quants 4 - Average

Key Points to Remember:

Average = Sum of observations / Number of observations
Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then the average speed during the whole journey is 2xy / (x + y) kmph

Practice Questions:

Question 1:

Find the average of all prime numbers between 30 and 50.

There are 5 prime numbers between 30 and 50.
They are 31, 37, 41, 43 and 47.
Average = ( 31 + 37 + 41 + 43 + 47 ) / 5 = 199 / 5 = 39.8
Thus, average of all prime numbers between 30 and 50 is 39.8.

Question 2:

Find the average of the first 40 natural numbers.

Sum of n natural numbers = [ n(n+1) ] / 2
Sum of first 40 natural numbers = ( 40 * 41 ) / 2 = 820
Average = 820 / 40 = 20.5
The average of the first 40 natural numbers is 20.5.


Question 3:

Find the average of first 20 multiples of 3.
Required average = [ 3 * ( 1 + 2 + 3 + ... + 18 + 19 + 20 ) ] / 20 = ( 3 * 20 * 21 ) / ( 20 * 2 )
= 1260 / 40 = 31.5
The average of first 20 multiples of 3 is 31.5.


Question 4:
The average of four consecutive even numbers is 27. Find the largest of these numbers.
Let the numbers be x, x+2, x+4, x+6.
[ x + (x+2) + (x+4) + (x+6) ] / 4 = 27
x + (x+2) + (x+4) + (x+6) = 108
4x + 12 = 108
4x = 96
x = 96 / 4 = 24
Hence, the largest number is x + 6 = 24 + 6 = 30.
The largest of the 4 consecutive numbers whose average is 27 is 30.


Question 5:

There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of section A is 40 kg and that of section B is 35 kg. Find the average weight of the whole class.

Total weight of 36 + 44 students = ( 36 * 40 ) + ( 44 * 35 ) = 1440 + 1540 = 2980 kg
Average of whole class = 2980 / (36 + 44) = 2980 / 80 = 37.25 kg
The average weight of the whole class is 37.25 kg.


Time to Think:

David obtained 76, 65, 82, 67 and 85 marks out of 100 in English, Math, Physics, Chemistry and Biology. What is his average mark?

Logical 3 - Direction Sense

Instructions:

The questions consist of a sort of direction puzzle. A successive follow up of directions is formulated and the task is to ascertain the final direction or the distance between the two points. This test judges the ability to trace and follow correctly and sense the direction correctly. 

The figure show the four main directions - North, South, East and West. The four cardinal directions are North East (NE), North West (NW), South East (SE), South West (SW).

Practice Questions:

Question 1:

One day, Ravi left home and cycled 10 km southwards, turned right and cycled 5 km and turned right and cycled 10 km and turned left and cycled 10 km. How many kilometers will he have to cycle to reach his home straight?

Ravi starts from home and moves 10 km southwards upto B
He then turns right and moves 5 k upto C
Then turns right again and moves 10 km upto D
Finally turns left and moves 10 km upto E
Total distance from initial position
A = AE = AD + DE = BC + DE = 5 + 10 = 15 km.
Hence, he needs to cycle 15 km to reach his home straight.

Question 2:

Kailash faces towards north. Turning to his right, he walks 25 meters. He then turns to his left and walks 30 meters. Next, he moves 25 meters to his right. He then turns to his right again and walks 55 meters. Finally, he turns to the right and moves 40 meters. In which direction, is he now from his starting point?

Kailash turns towards right from North direction.
He walks 25 m towards east upto B
Turns left and moves 30 m upto C
Turns right and goes 25 m upto D
At D, he turns to right towards the South and walks 55 m upto E
He again turns to right and walks 40 m upto F, the final position
F is to the South - east of A
Hence, he is in the direction of South east from his starting point.


Question 3:
Deepa moved a distance of 75 meters towards the north. She then turned to the left and walking for about 25 meters, turned left again and walked 80 meters. Finally, she turned to the right at an angle of 45˚. In which direction was she moving finally?

Deepa started from A, moved 75 m upto B
Turned left and walked 25 m upto C
Turned left again and moved 80 m upto D
Turning to the right at an angle of 45˚, she was finally moving in the direction of DE ie South - west
Hence, the direction she was moving finally was South west.


Question 4:
Kunal walks 10 km towards north. From there, he walks 6 km towards south. Then, he walks 3 km towards east. How far and in which direction is he with reference to his starting point?

Kunal moves from A 10 km northwards upto B
Then he moves 6 km southwards upto C
Turns towards East and walks 3 km upto D
AC = AB - BC = 10 - 6 = 4 km; CD = 3 km
AD = √ [(AC)² + (CD)²] = √ (4² + 3²) = 5 km
D is in the direction of North - east of A
Hence, he is 5 km North east from his starting point.


Question 5:
Johnson left for his office in his car. He drove 15 km towards north and then 10 km towards west. He then turned to the south and covered 5 km. Further, he turned to the east and moved 8 km. Finally, he turned right and drove 10 km. How far and in which direction is he from his starting point?

Johnson drove 15 km from A to B northwards
Then 10 km from B to C towards west
Then moves 5 km southwards from C to D
Then 8 km eastwards upto E
Finally, he turned right and moved 10 km upto F
A and F lie in the same straight line and F lies to the west of A
Distance = AF = BC - DE = 10 - 8 = 2 km
Hence, he is 2 km far in the West from his starting point.


Time to Think:

Rohan walks a distance of 3 km towards north, then turns to his left and walks for 2 km. He again turns left and walks for 3 km. At this point, he turns to his left and walks for 3 km. How many kilometers is he from the starting point?

Verbal 2 - Sentence Correction

Instructions:
Find the mistakes in the sentences and correct them.

Practice Questions:

Question 1:

Ram and I is running.

Ram and I are running

Question 2:
The Chennai Super Kings Team have won today.
The Chennai Super Kings Team has won today.

Question 3:
The parents, along with Surya, are coming to school today.

The parents, along with Surya, is coming to school today.

Question 4:
Either the boys or the girl was managing the show.
No error.
(The verb takes the form of the second noun. Here it's the girl, so was is right.)  #Remember

Question 5:
Everyone have to be present tomorrow for the class.

Everyone has to be present tomorrow for the class.

(Everyone always takes the singular form.)

Time to Think:

Neither the boys, nor the girl get the credit for doing this job.

Quants 3 - Calendar

Practice Questions:

Question 1:

On what dates of July 2004, did Monday fall?

Let's find the day of July 1, 2004.
2000 = 0 Odd Days
3 Years = 3 Ordinary Years
             = (3 * 1) = 3 Odd Days
For 2004, counting days from Jan to Jul,
31 + 29 + 31 + 30 + 31 + 30 + 1 = 26 weeks + 1 day = 1 Odd Day
So, 3 + 1 = 4 Odd Days = Thursday.
Hence, 1st Monday in July 2004 was July 5.
Other Mondays were 12, 19, 26.
Thus, July 5, July 12, July 19 and July 26 of 2004 were Mondays.

Question 2:

Prove that the Calendar for the year 2003 will serve for the year 2014.

In order for both the calendars of 2003 and 2014 to be same, both the Jan 1st must be on same date and they both must be either Leap years or Ordinary Years.
Here, both are Ordinary Years.
For the first days to be the same, the odd days between Dec 31, 2002 and Dec 31, 2013 must be zero.
During this period, there were, 3 Leap Years and 8 Ordinary Years.
So, the Odd Days are (3 * 2) + ( 8 * 1) = 14 Days = 2 Weeks = 0 Odd Days.
Thus, it is proven that the Calendar for 2004 will serve for 2014.

Question 3:
Prove that any date in March of a year is the same day of the week as the corresponding date in November that year.

The number of Odd days between last day of February and last day of October should be zero for both the months to be same.
Counting days from March to October,
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 245 = 35 weeks = 0 Odd Days
As the period between February and November has 0 Odd Days,
It is proven that any date in March of a year is the same day of the week as the corresponding date in November that year.

Question 4:
January 1, 2004 was a Thursday. What day of the week lies on January 1, 2005?

The year 2004 being a Leap year, it has 2 Odd days. So, the first day of 2005 will be 2 days beyond Thursday, and so it is a SATURDAY.


Question 5:

The first Republic day of India was celebrated on January 26, 1950. It was?

26th January 1950 = 1949 years + (Jan 1, 1950 to Jan 26, 1950)

Odd Days:

1600 = 0 Odd Days

300 Years = 1 Odd Day

49 Years = 12 Leap Years + 37 Ordinary Years

               = (12 * 2) + (37 * 1) = 61 days = 8 weeks + 5 day = 5 Odd Days

So, 0+1+5 = 6 Odd Days till 1949.

For 1950, counting days from Jan 1 to Jan 26 = 26 days = 3 weeks + 5 days = 5 Odd Days

So, 6 + 5 = 11 = 4 Odd Days

As Odd day 4 corresponds to Tuesday, 26th January, 1950 is THURSDAY.

Time to Think:

On 8th March, 2005, Wednesday falls. What day of the week was it on 8th March, 2004?

Answers Week 1

Quants 1 - Time and Work

A man can do a job in 15 days. His father takes 20 days to complete and his son takes 25 days to do the same. How long it takes, if they all work together?
Man's 1 day work = 1/15
Father's 1 day work = 1/20
Son's 1 day work = 1/25
Man + Father + Son 's 1 hour work = (1/15) + (1/20) + (1/25) = (47/300)
Hence it takes 300/47 = 6.38 = 6.4 days

(Task for the readers, find the days for Gracia to complete the work)

Ziva is twice as good as Gracia and together they finish the piece of work in 20 days. In how many days will Ziva alone do the work?

Ziva's work : Gracia's work = 2:1

Ziva + Gracia's 1 day work = 1/20

Dividing 1/20 in the ratio of 2:1

Gracia's work = (1/20) * 1/3 = 1/60

Hence Gracia takes 60 days to complete the work.

Logical 1 - Blood Relations

Pointing out to a lady, Raja said, "She is the daughter of the woman who is the mother of the husband of my mother." Who is the lady to Raja?
Mother's husband - Father; Father's mother - Grandmother; Grandmother's daughter - Aunt
Hence, the lady is Raja's Aunt.


Verbal 1 - Error Correction

The man told to her (a) / that he had not brought his dog (b) / out for a walk as he was afraid that it would rain. (c)

(a) is the error part.

Correction - The man told her. (to doesn't follow told, to is followed after said in indirect assertive sentences)


Quants 2 - Calendar

What was the day of the week on 6th February, 2015?
6th February 2015 = 2014 years + (Jan 1, 2015 to Feb 6, 2015)
Odd Days:
2000 = 0 Odd Days
14 Years = 3 Leap Years + 11 Ordinary Years
               = [ ( 3 * 2) + ( 11 * 1 ) ] = 17 days = 2 weeks + 3 days = 3 Odd Days
So, 0 + 3 = 3 Odd Days till 2014.
For 2015, counting days from Jan to Feb,
31 + 6 = 37 days = 5 weeks + 2 days = 2 Odd Days
So, 3 + 3 = 5 = 5 Odd Days
As Odd day 5 corresponds to Friday, 6th February, 2015 is FRIDAY.


Logical 2 - Coding and Decoding

If FISH is written as EHRG in a certain code, how will JUNGLE be written in that code?
Answer is ITMFKD
Each letter in the word is moved one step backward to obtain the corresponding letter of the code.

Logical 2 - Coding and Decoding

Key Points to Remember:

A CODE is a system of signals.
Coding is a method of transmitting a message between the sender and the receiver without the third person knowing it. 
Coding and Decoding test is ability to decipher the rule that codes a particular word or message and break the code to decipher the message. 

The various types of Coding and Decoding are,
Type 1 : Letter Coding
Type 2: Direct Letter Coding
Type 3 : Number / Symbol Coding
Type 4 : Matrix Coding
Type 5 : Substitution
Type 6 : Deciphering Message Word Codes
Type 7 : Deciphering Number and Symbol Codes for messages

Let's see exercises for Type 1 Coding and Decoding. 

Practice Questions:

Question 1:

If in a certain code, LUTE is written as MUTE and FATE is written as GATE, then how will BLUE be written in that code?

Answer is CLUE

The first letter of the word is moved one step forward to obtain the first letter of the code, while the other letters remain unaltered.

Question 2:

In a certain code, MADRAS is coded as NBESBT, how is RANCHI coded in that language?

Answer is SBODIJ

Each letter in the word s moved one step forward to obtain the corresponding letter of the code.

Question 3:
In a certain code language, OPERATION is written as NODQBUJPO. How is INVISIBLE written in that code?

Answer is HMUHTJCMF
Each of the four letters in the word is moved one step backward, while each of the last five letters is moved one step forward to obtain the corresponding letter of the code.

Question 4:
In a certain code, TWINKLE is written as SVHOJKD, then how would FILTERS be written in the same code?

Answer is EHKUDQR  (Have a closer look readers)
Each letter in the word, except the middle letter, is moved one step backward while the middle letter is moved one step forward to obtain the corresponding  letter of the code.

Question 5:
In a certain code, ROAD is written as URDG. How is SWAN written in that code?

Answer is VZDQ
Each letter in the word is moved three steps forward to obtain the corresponding letter of the code.

Time to Think:

If FISH is written as EHRG in a certain code, how will JUNGLE be written in that code?

Quants 2 - Calendar

Key Points to Remember:

The main objective of this exercise is to find the day of the week on a particular given date. The process involves in getting the number of odd days.

Odd Days - Number of days more than one complete number of weeks in a given period.
Leap Year - Every year which is divisible by 4. Eg - 1992, 1996, 2004, 2006, 2008 etc.
Every 4th century such as 400, 800, 1200, 1600, 2000 etc are leap years.
But other centuries such as 100, 300, 500, 900, 1000, 1900 etc are not leap years.
An year which is not a Leap year is an Ordinary Year.

Number of Days in Leap year - 366; Number of days in Ordinary year - 365

Counting of Odd Days 
1 Ordinary Year - 52 weeks + 1 day. So, an Ordinary year has 1 Odd day.
1 Leap Year - 52 weeks + 2 days. So, a Leap Year has 2 Odd days.
100 Years = 76 Ordinary years + 24 Leap Years 
                 = [ (76 * 52) weeks + 76 days ] + [ (24 * 52) weeks + 48 days ]
                 = 5200 weeks + 124 days = 5217 weeks + 5 days
100 years has 5 Odd Days
200 years has 10 Odd Days = 3 Odd Days (10 = 7 + 3; 7 days = 1 week)
300 years has 15 Odd Days = 1 Odd Day
400 years has 20 + 1 Odd Days = 21 = 0 Odd Days.
Similarly, 800, 1200, 1600, 2000 etc have 0 Odd Day.

Odd Days         Day
     0                   Sunday
     1                   Monday
     2                   Tuesday
     3                   Wednesday
     4                   Thursday
     5                   Friday
     6                   Saturday

Practice Questions:

Question 1:

What was the day of the week on 16th July, 1776?

16th July 1776 = 1775 + (Jan 1, 1776 to Jul 16, 1776)
Odd Days :
1600 = 0 Odd Days
100 Years = 5 Odd Days
75 Years = 18 Leap Years + 57 Ordinary Years
               = [ (18 * 2) + (57 * 1) ] = 93 Days = 13 weeks + 2 days = 2 Odd Days
So, 0 + 5 + 2 = 7 = 0 Odd Day till 1775.
For 1776, counting days from Jan to Jul,
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 Days = 28 weeks + 2 days = 2 Odd Days
As Odd day 2 corresponds to Tuesday, 16th July, 1776 is TUESDAY.

Question 2:

What was the day of the week on 15th August, 1947?

15th August 1947 = 1946 years + (Jan 1, 1947 to Aug 15, 1947)
Odd Days:
1600 = 0 Odd Days
300 Years = 1 Odd Day
46 Years = 11 Leap Years + 35 Ordinary Years
               = (11 * 2) + (35 * 1) = 57 days = 8 weeks + 1 day = 1 Odd Day
So, 0+1+1 = 2 Odd Days till 1946.
For 1947, counting days from Jan to Aug,
31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days = 32 weeks + 3 days = 3 Odd Days
So, 2 + 3 = 5 Odd Days
As Odd day 5 corresponds to Friday, 15th August, 1947 is FRIDAY.


Question 3:
What was the day of the week on 16th April, 2000?

16th April 2000 = 1999 years + (Jan 1, 2000 to Apr 16, 2000)
Odd Days:
1600 = 0 Odd Days
300 Years = 1 Odd Day
99 Years = 24 Leap Years + 75 Ordinary Years
               = [ ( 24 * 2 ) + ( 75 * 1 ) ] = 123 days = 17 weeks + 4 days = 4 Odd Days
So, 0 + 1 + 4 = 5 Odd Days till 1999.
For 2000, counting days from Jan to Apr,
31 + 29 + 31 + 16 = 107 days = 15 weeks + 2 days = 2 Odd Days
So, 5 + 2 = 7 = 0 Odd Days
As Odd day 0 corresponds to Sunday, 16th April, 2000 is SUNDAY.


Question 4:
What was the day of the week on 7th July, 1981?

7th July 1981 = 1980 years + (Jan 1, 1981 to Jul 7, 1981)
Odd Days:
1600 = 0 Odd Days
300 Years = 1 Odd Day
80 Years = 20 Leap Years + 60 Ordinary Years
               = [ ( 20 * 2) + ( 60 * 1 ) ] = 100 days = 14 weeks + 2 days = 2 Odd Days
So, 0 + 1 + 2 = 3 Odd Days till 1980.
For 1981, counting days from Jan to Jul,
31 + 28 + 31 + 30 + 31 + 30 + 7 = 188 days = 26 weeks + 6 days = 6 Odd Days
So, 3 + 6 = 9 = 2 Odd Days (9 - 7 = 2 )
As Odd day 2 corresponds to Tuesday, 7th July, 1981 is TUESDAY.


Question 5:
What was the day of the week on 9th August, 1997?

9th August 1997 = 1996 years + (Jan 1, 1997 to Aug 9, 1997)
Odd Days:
1600 = 0 Odd Days
300 Years = 1 Odd Day
96 Years = 24 Leap Years + 72 Ordinary Years
               = [ ( 24 * 2) + ( 72 * 1 ) ] = 120 days = 17 weeks + 1 days = 1 Odd Day
So, 0 + 1 + 1 = 2 Odd Days till 1996.
For 1997, counting days from Jan to Aug,
31 + 28 + 31 + 30 + 31 + 30 + 31 + 9 = 221 days = 31 weeks + 4 days = 4 Odd Days
So, 2 + 4 = 6 = 6 Odd Days
As Odd day 6 corresponds to Saturday, 9th August, 1997 is SATURDAY.


Time to Think:

What was the day of the week on 6th February, 2015?

Verbal 1 - Error Correction

Instructions:
Find the part of the sentence with mistake. If no error, answer (d) - No error.

Practice Questions:

Question 1:

We discussed about the problem so thoroughly (a) / on the evening of the examination (b) / that I found it easy to work out. (c)

(a) is the error part.
Correction - We discussed the problem so thoroughly.

Question 2:
You will get (a) / all the information (b) / if you read this booklet carefully. (c)

(a) is the error part.
Correction - You can get.

Question 3:
The scientist must follow (a) / his hunches and his data (b) / wherever it may lead. (c)
(c) is the error part.
Correction - wherever they may lead. (plural - hunches and data)

Question 4:
He asked me (a) / why I called (b) / him a rogue. (c)
(d) - No error

Question 5:
The customer handed over (a) / a hundred rupees note (b) / to the shopkeeper. (c)

(b) is the error part.
Correction - a hundred rupee note. (as it is one single note)

Time to Think:

The man told to her (a) / that he had not brought his dog (b) / out for a walk as he was afraid that it would rain. (c)

Logical 1 - Blood Relations

Key Points to Remember:

Some of the relations
Mother's or Father's Son - Brother
Mother's or Father's Daughter - Sister
Mother's or Father's Brother - Uncle
Mother's or Father's Sister - Aunt
Mother's or Father's Father - Grandfather
Mother's or Father's Mother - Grandmother
Son's Wife - Daughter in law
Daughter's Husband - Son in law
Husband's or Wife's Sister - Sister in law
Husband's or Wife's Brother - Brother in law
Brother's or Sister's Son - Nephew
Brother's or Sister's Daughter - Niece
Uncle's or Aunt's Son or Daughter - Cousin
Sister's Husband - Brother in law
Brother's Wife - Sister in Law
Grandson's or Granddaughter's Daughter or Son - Great Grand Children

A relation on the mother's side is called Maternal while that on the father's side is called Paternal.
The mother's brother is called Maternal Uncle and father's brother is called Paternal Uncle.

Instructions:

Here a roundabout description is given in the form of certain small relationships and the task is to analyze the whole chain of relations and decipher the direct relationship between the persons concerned.

Practice Questions:

Question 1:

Pointing towards a lady in the photograph, Shaloo said, "Her son's father is the son in law of my mother" How is Shaloo related to the lady?

Lady's son's father is lady's husband. So, the lady's husband is son-in-law of Shaloo's mother ie the lady is the daughter of Shaloo's mother. Thus, Shaloo is the lady's Sister.

Question 2:

Anil introduces Rohit as son of the only brother of his father's wife. How is Rohit related to Anil?

Father's wife - mother; Mother's brother - Uncle; Uncle's Son - Cousin.
Hence, Rohit is Anil's Cousin.

Question 3:
Pointing towards a person in the photograph, Anjali said, "He is the only son of the father of my sister's brother" How is the person related to Anjali?

Sister's Brother - Brother; Brother's Father - Father; Father's Son - Brother.
Hence, the person in the photograph is Anjali's Brother.


Question 4:
Introducing Reena, Monika said, "She is the only daughter of my father's only daughter." How is Monika related to Reena?

Father's Daughter - she herself, Monika
Her only daughter - Reena
So, Monika is Reena's Mother.


Question 5:
Pointing to a man, a woman said, "His mother the only daughter of my mother". How is woman related to the mother?

Mother's only daughter - she herself
So, he is the mother of the man.

Time to Think:

Pointing out to a lady, Raja said, "She is the daughter of the woman who is the mother of the husband of my mother." Who is the lady to Raja?