
About the Course 0
No items in this section 
Alligations and Mixtures 4

Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4


Train Problems 5

Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5


Ages Problems 5

Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5


Average 5

Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5


Boats and Streams 4

Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4


Calender Problems 3

Lecture7.1

Lecture7.2

Lecture7.3


Chain Rule 3

Lecture8.1

Lecture8.2

Lecture8.3


Clock Problems 5

Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5


HCF and LCM 5

Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5


Percentage 8

Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8


Permutation and Combination 5

Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5


Pipes and Cistern 5

Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5


Probability 4

Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4


Profit and Loss 5

Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5


Ratio and Proportion 8

Lecture16.1

Lecture16.2

Lecture16.3

Lecture16.4

Lecture16.5

Lecture16.6

Lecture16.7

Lecture16.8


Simple Interest and Compound Interest 7

Lecture17.1

Lecture17.2

Lecture17.3

Lecture17.4

Lecture17.5

Lecture17.6

Lecture17.7


Time and Work 5

Lecture18.1

Lecture18.2

Lecture18.3

Lecture18.4

Lecture18.5


Verbal Reasoning 10

Lecture19.1

Lecture19.2

Lecture19.3

Lecture19.4

Lecture19.5

Lecture19.6

Lecture19.7

Lecture19.8

Lecture19.9

Lecture19.10


Verbal Ability Notes 5

Lecture20.1

Lecture20.2

Lecture20.3

Lecture20.4

Lecture20.5


[Reasoning] Blood Relations 5

Lecture21.1

Lecture21.2

Lecture21.3

Lecture21.4

Lecture21.5


[Reasoning] Number Series 6

Lecture22.1

Lecture22.2

Lecture22.3

Lecture22.4

Lecture22.5

Lecture22.6


[Reasoning] Seating Arrangement 8

Lecture23.1

Lecture23.2

Lecture23.3

Lecture23.4

Lecture23.5

Lecture23.6

Lecture23.7

Lecture23.8


[Reasoning] Directions 7

Lecture24.1

Lecture24.2

Lecture24.3

Lecture24.4

Lecture24.5

Lecture24.6

Lecture24.7


[Reasoning] Syllogism 6

Lecture25.1

Lecture25.2

Lecture25.3

Lecture25.4

Lecture25.5

Lecture25.6


[Reasoning] Venn Diagram 5

Lecture26.1

Lecture26.2

Lecture26.3

Lecture26.4

Lecture26.5


[Reasoning] Paper Folding & Cutting 1

Lecture27.1


[Reasoning] Mathematical Operations 1

Lecture28.1


[Reasoning] Statement & Conclusion 2

Lecture29.1

Lecture29.2


[Reasoning] Word Pattern 1

Lecture30.1


[Reasoning] Data Sufficiency 3

Lecture31.1

Lecture31.2

Lecture31.3


[Reasoning] Visual Reasoning 5

Lecture32.1

Lecture32.2

Lecture32.3

Lecture32.4

Lecture32.5


[Reasoning] Coding Decoding 7

Lecture33.1

Lecture33.2

Lecture33.3

Lecture33.4

Lecture33.5

Lecture33.6

Lecture33.7


[Reasoning] Cubes 4

Lecture34.1

Lecture34.2

Lecture34.3

Lecture34.4


Number System 10

Lecture35.1

Lecture35.2

Lecture35.3

Lecture35.4

Lecture35.5

Lecture35.6

Lecture35.7

Lecture35.8

Lecture35.9

Lecture35.10


Arithmetic Progression & Geometric Progression 5

Lecture36.1

Lecture36.2

Lecture36.3

Lecture36.4

Lecture36.5


True discount 2

Lecture37.1

Lecture37.2


Statement and Assumption 3

Lecture38.1

Lecture38.2

Lecture38.3


Partnership 3

Lecture39.1

Lecture39.2

Lecture39.3


Data Interpretation 9

Lecture40.1

Lecture40.2

Lecture40.3

Lecture40.4

Lecture40.5

Lecture40.6

Lecture40.7

Lecture40.8

Lecture40.9

If it was Tuesday on 3rd Jan, 2006. What was the day on Jan 3, 2010?
A) Saturday  B) Tuesday 
C) Wednesday  D) Sunday 
D) Sunday
Explanation:
It was Saturday on 31st December, 2005.
Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.
Hence on December 31, 2009, it was Thursday. Thus, on Jan 3, 2010 it was Sunday.
If 13th March, 2005 was Wednesday, what was the day of the week on 13th March, 2004?
A) Rs. Monday  B) Rs. Wednesday 
C) Rs. Friday  D) Rs. Tuesday 
D) Rs. Tuesday
Explanation:
We know year 2004 is a leap year. Therefore, it has 2 odd days.
Feb 2004 was not included as we are calculating from March 2004 to March 2005.
So it has 1 odd day.
The day on March 13, 2005 will be 1 day ahead of the day on March 13, 2004.
Given that, March 13, 2005 is Wednesday.
So March 13, 2004 is Tuesday (1 day before to 13th March, 2005).
The last day of a century cannot be:
A) Wednesday  B) Friday 
C) Saturday  D) Monday 
C) Saturday
Explanation:
100 years contain 5 odd days.
Therefore, Last day of 1st century is Friday
200 years contain (5 x2) is 3 odd days (approx.)
Last day of 2nd century is Wednesday
300 years contain ( 5x3) = 15 i.e. 1 odd day( approx.)
Last day of 3rd century is Monday 400 years contain 0 odd day
Last day of 4th century is Sunday.
Thus, the cycle is repeated.
Last day of a century cannot be Tuesday or Thursday or Saturday.
What was the day on 16th July, 1776?
A) Saturday  B) Sunday 
C) Monday  D) Tuesday 
D) Tuesday
Explanation:
Step 1 : Take the last 2 digit of year : 76
Step 2 : Divide step 1 digit by 4 and take the quotient 76/4 we get 19
Step 3 : Take the day of the Date : 16
Step 4 : Code of the month : 6
Step 5 : Code of the year : 4
Step 6 : Add all of them and divide it by 7 remainder is the answer : 121/7 = 2 i.e Tuesday
[ It take only 20  30 sec to solve this sum using above trick given in the video ]
What was the day on 15th August, 2010 ?
A) Saturday  B) Sunday 
C) Monday  D) Tuesday 
B) Sunday
Explanation:
Step 1 : Take the last 2 digit of year : 10
Step 2 : Divide step1 digit by 4 and take the quotient 10/4 we get 2
Step 3 : Take the day of the Date : 15
Step 4 : Code of the month : 2
Step 5 : Code of the year : 6
Step 6 : Add all of them and divide it by 7 remainder is the answer : 35/7 = 0 i.e Sunday
If 7th Dec, 2007 was Friday. What day of the week was on 7th Dec, 2006?
A) Saturday  B) Thursday 
C) Monday  D) Sunday 
B) Thursday
Explanation:
The year 2006 was an ordinary year. So, it has 1 odd day.
The day on 7th Dec, 2007 will be 1 day beyond the day on 7th Dec, 2006.
But, 7th Dec, 2007 is Friday
So, 7th Dec, 2006 is Thursday.
January 1, 2007 was Monday. What was the day on Jan. 1, 2008?
A) Saturday  B) Friday 
C) Tuesday  D) Monday 
C) Tuesday
Explanation:
The year 2007 is an ordinary year. So, it has 1 odd day.
1st day of the year 2007 was Monday
1st day of the year 2008 will be 1 day beyond Monday
Hence, it will be Tuesday.
On what dates of July 2003 did Sunday fall?
A) 5th,10th,24th,17th  B) 6th,13th,20th,27th 
C) 5th, 12th, 19th, 26th  D) 12th,7th,19th,28th 
B) 6th,13th,20th,27th
Let us find the day on 1st July, 2003.
Explanation:
Step 1 : Take the last 2 digit of year : 03
Step 2 : Divide step1 digit by 4 and take the quotient 03/4 we get 00
Step 3 : Take the day of the Date : 01
Step 4 : Code of the month : 6
Step 5 : Code of the year : 6
Step 6 : Add all of them and divide it by 7 remainder is the answer : 16/7 = 2 i.e Tuesday
Thus, 1st Sunday in July 2003 is 6th July.
So, during July 2003, Sunday fell on 6th, 13th, 20th and 27th.
What was the day on 16th August, 1947?
A) Tuesday  B) Monday 
C) Saturday  D) Sunday 
C) Saturday
Explanation:
Step 1 : Take the last 2 digit of year : 47
Step 2 : Divide step1 digit by 4 and take the quotient 47/4 we get 11
Step 3 : Take the day of the Date : 16
Step 4 : Code of the month : 2
Step 5 : Code of the year : 0
Step 6 : Add all of them and divide it by 7 remainder is the answer : 76/7 = 6
i.e Saturday
The calendar for the year 2007 is the same for the year
A) 2018  B) 2019 
C) 2022  D) 2025 
A) 2018
Explanation:
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Sum = 14 odd days 0 odd days
Calendar for the year 2018 will be the same as for the year 2007
What was the day of the week on 16th June, 1999?
A) Saturday  B) Monday 
C) Wednesday  D) Thursday 
C) Wednesday
Explanation:
Step 1 : Take the last 2 digit of year : 99
Step 2 : Divide step1 digit by 4 and take the quotient 99/4 we get 24
Step 3 : Take the day of the Date : 16
Step 4 : Code of the month : 4
Step 5 : Code of the year : 0
Step 6 : Add all of them and divide it by 7 remainder is the answer : 143/7 = 3 i.e wednesday
If it was Thursday on Aug 15, 2012, then what was the day on June 11, 2013?
A) Wednesday  B) Monday 
C) Saturday  D) Tuesday 
A) Wednesday
Explanation:
First, we count the number of odd days for the left over days in the given period.
Here, given period is from 15.8.2012 to 11.6.2013
Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun
16 30 31 30 31 31 28 31 30 31 11(left days)
Therefore, 2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd days
So, the given day Thursday + 6 = Wednesday.
It was Tuesday on Feb 8, 2005. What was the day of the week on Feb 8, 2004?
A) Monday  B) Thursday 
C) Friday  D) Sunday 
D) Sunday
Explanation:
The year 2004 was a leap year. So, it had 2 odd days.
The day on Feb 8, 2004 must be 2 days before the day on Feb 8, 2005.
Hence, this day was Sunday
What was the day on June 17, 1998?
A) Monday  B) Sunday 
C) Wednesday  D) Friday 
C) Wednesday
Explanation:
Step 1 : Take the last 2 digit of year : 98
Step 2 : Divide step1 digit by 4 and take the quotient 98/4 we get 24
Step 3 : Take the day of the Date : 17
Step 4 : Code of the month : 4
Step 5 : Code of the year : 0
Step 6 : Add all of them and divide it by 7 remainder is the answer : 143/7 = 3 i.e wednesday
What was the day on 28th May, 2006?
A) Wednesday  B) Sunday 
C) Saturday  D) Thursday 
B) Sunday
Explanation:
Step 1 : Take the last 2 digit of year : 06
Step 2 : Divide step1 digit by 4 and take the quotient 06/4 we get 01
Step 3 : Take the day of the Date : 28
Step 4 : Code of the month : 1
Step 5 : Code of the year : 6
Step 6 : Add all of them and divide it by 7 remainder is the answer : 42/7 = 0 i.e Sunday
What was the day on February 9, 1979?
A) Tuesday  B) Saturday 
C) Friday  D) Thursday 
C) Friday
Explanation:
Step 1 : Take the last 2 digit of year : 79
Step 2 : Divide step1 digit by 4 and take the quotient 79/4 we get 19
Step 3 : Take the day of the Date : 09
Step 4 : Code of the month : 3
Step 5 : Code of the year : 0
Step 6 : Add all of them and divide it by 7 remainder is the answer : 110/7 = 5 i.e Friday
If 10th May, 1997 was a Monday, what was the day on Oct 10, 2001?
A) Saturday  B) Sunday 
C) Thursday  D) Wednesday 
D) Wednesday
rather than finding odd days between both the dates i will use the trick and directly find the day on Oct 10, 2001 because it will save time.
Explanation:
Step 1 : Take the last 2 digit of year : 01
Step 2 : Divide step1 digit by 4 and take the quotient 01/4 we get 00
Step 3 : Take the day of the Date : 10
Step 4 : Code of the month : 0
Step 5 : Code of the year : 6
Step 6 : Add all of them and divide it by 7 remainder is the answer : 17/7 = 3 i.e Wednesday
If April 11, 1911 was a Tuesday, what was the day on September 17, 1915?
A) Friday  B) Thursday 
C) Sunday  D) Tuesday 
A) Friday
rather than finding odd days between both the dates i will use the trick and directly find the day on September 17, 1915 because it will save time.
Explanation:
Step 1 : Take the last 2 digit of year : 15
Step 2 : Divide step1 digit by 4 and take the quotient 15/4 we get 03
Step 3 : Take the day of the Date : 17
Step 4 : Code of the month : 5
Step 5 : Code of the year : 0
Step 6 : Add all of them and divide it by 7 remainder is the answer : 40/7 = 5 i.e Friday
Tuesday fell on which of the following dates of April, 2002?
A) 3rd, 10th, 17th, 24th  B) 1st, 8th, 15th, 22nd, 29th 
C) 4th, 11th, 18th, 25th  D) 2nd, 9th, 16th, 23rd, 30th 
D) 2nd, 9th, 16th, 23rd, 30th
Let first find the day on 1st April, 2002.
Explanation:
Step 1 : Take the last 2 digit of year : 02
Step 2 : Divide step1 digit by 4 and take the quotient 02/4 we get 00
Step 3 : Take the day of the Date : 01
Step 4 : Code of the month : 6
Step 5 : Code of the year : 6
Step 6 : Add all of them and divide it by 7 remainder is the answer : 15/7 = 01 i.e Monday
On April 1, 2002 it was Monday.
So in April 2002, Tuesday falls on 2nd, 9th, 16th and 23rd & 30th.
What was the day on May 2, 1995?
A) Monday  B) Thursday 
C) Sunday  D) Tuesday 
D) Tuesday
Explanation:
Step 1 : Take the last 2 digit of year : 95
Step 2 : Divide step1 digit by 4 and take the quotient 95/4 we get 23
Step 3 : Take the day of the Date : 02
Step 4 : Code of the month : 01
Step 5 : Code of the year : 00
Step 6 : Add all of them and divide it by 7 remainder is the answer : 121/7 = 02
i.e Tuesday
Today is Wednesday, after 68 days, it will be
A) Friday  B) Sunday 
C) Monday  D) Thursday 
C) Monday
Explanation:
Each day of a week is repeated after 7 days, so after 70 days, it will be Wednesday.
Therefore, after 68 days, it will be Monday.
On February 5, 1998, it was Thursday. The day of the week on February 5, 1997, was
A) Wednesday  B) Monday 
C) Friday  D) Sunday 
A) Wednesday
Explanation:
1997 was an ordinary year, it had 1 odd day. So, the day on February 5, 1998, would be one day beyond the day on February 5, 1997
∴ Thursday on February 5, 1998, would be one day beyond the day on February 5, 1997, so the day on February 5, 1997, was Wednesday.
What was the day of the week on June 17, 1991 ?
A) Tuesday  B) Wednesday 
C)Friday  D) Monday 
D) Monday
Explanation:
Step 1 : Take the last 2 digit of year : 91
Step 2 : Divide step1 digit by 4 and take the quotient 91/4 we get 22
Step 3 : Take the day of the Date : 17
Step 4 : Code of the month : 04
Step 5 : Code of the year : 00
Step 6 : Add all of them and divide it by 7 remainder is the answer : 134/7 = 01.
i.e Monday
If January 1, 1996, was Monday, what day of the week was January 1, 1997?
A) Thursday  B) Wednesday 
C) Friday  D) Sunday 
B) Wednesday
Explanation:
The year 1996 is divisible by 4, so it is a leap year with 2 odd days.
As per the question, the first day of the year 1996 was Monday, so the first day of the year 1997 must be two days after Monday. So, it was Wednesday.
The first republic day of India was celebrated on January 26, 1950. What day of the week was it?
A) Wednesday  B) Friday 
C) Thursday  D) Tuesday 
C) Thursday
Explanation:
Step 1 : Take the last 2 digit of year : 50
Step 2 : Divide step1 digit by 4 and take the quotient 50/4 we get 12
Step 3 : Take the day of the Date : 26
Step 4 : Code of the month : 00
Step 5 : Code of the year : 00
Step 6 : Add all of them and divide it by 7 remainder is the answer : 88/7 = 04 i.e Thursday
How many days are there in y weeks y days?
A) 8y  B) 8y^{2} 
C) 16y  D) 21y 
A) 8y
Explanation:
There are 7 days in a week, so y weeks will contain 7y days.
∴The required number of days = 7y + y days = 8y days
The day on 5th April of a year will be the same day on 5th of which month of the same year?
A) 5^{th} July  B) 5^{th} August 
C) 5^{th} June  D) 5^{th} October 
A) 5^{th} July
Explanation:
April & July for all years have the same calendar. So, a day on any date of April will be the same day on the corresponding date in July.
∴The same day will fall on 5th July of the same year.
It was Saturday on January 1, 2005. Find the day of the week on January 1, 2010.
A) Tuesday  B) Friday 
C) Wednesday  D) Sunday 
B) Friday
Explanation:
On December 31, 2004, it was Friday.
Number of odd days from the year 2005 to the year 2009 = 1+1+1+2(leap year) +1 = 6 odd days
∴On December 31, 2009, it was Thursday, so on January 1, 2010, it was Friday.
If it was Wednesday on March 1, 2006, which day was it on March 1, 2002?
A) Tuesday  B) Friday 
C) Monday  D) Sunday 
B) Friday
Explanation:
Total number of odd days between the years 2002 and 2006 = (2006  2002) + 1 = 5 odd days. The year 2004 is a leap year, it has two odd days. So, one extra odd day is added.
So, if it was Wednesday on March 1, 2006, it would be (Wednesday  5) Friday on March 1, 2002.
Which of the following years is not a leap year?
A) 800  B) 700 
C) 1600  D) 2000 
B) 700
Explanation:
A century divisible by 400 is a leap year. The years 800, 1600 and 2000 are divisible by 400, so they are leap years.
The year 700 is not divisible by 400, so it is not a leap year.
45 Comments
bro me purchase kruga aptitute course with notes ko to mujhe ye kese milega via cd ya online hi milega? or is Course ki Validity kya rhegi?
online hai, 6 months validity
Bhai notes already purchase Kiya hua he muje Khali videos chahiye
Very helpful and knowledgeable course for placement preparation .
Notes buy kiye hain sirf videos chahiye
Will you please send me the notes on this
[email protected]
if you have took the course you can download the notes from last section
add more videos on problem of ages.
we will surely try
[email protected]
Please send me notes on this email
i just want videos ?
sirf video bhi purchase karne padenge?
i paid 299, so i want to know.
I will gets the notes by delivery or else i need to download and take the print?
You have to download the printout
Saare topic ka index pehle hi diya hai description me
Sumer bhai bhot sahi kam kre ho 👍… thank you
Hamne puri engineering apke videos dekh ki nikali hai ..
Thanku sir ,i got it.
unable to buy the course,when i click buy this course button its show 404 page not found
please fix this bhai its urgent…
You can check it now the problem is solved
the problem is solved
Hi,
I want to purchase this course (Aptitude Series with Notes).
But before purchasing I wanted to confirm whether this course covers ALL aspects of Aptitude – which are Quantitative Aptitude, Logical Reasoning Ability, Verbal Ability??
Please reply, so that I can decide soon.
Thank you.
quantitative aptitude and verbal is cover and this course is really good for cracking campus placement
Thank you very much for the reply. I will buy this course for sure!!
But can you plese refer me any resources for “Logical Reasoning Ability” too?
Thanks in advance.
iss course mai video aur notes dono included hai na??? aur notes kis form mai access hoga hardcopy ya softcopy??
yes it include videos+ Notes and notes will be in form of PDF
Bhaiii Notes Online kii Soft Copy hiii milti hai kya
pdf milenga bro which you can download
OK Bro
Thank you
Any discount available for this course
yes filal sale chalu hai get the course now
Bro kab tak sale hai….
22 November tak
I am really grateful for you SUMER that you have shared your knowledge here with less amount of time.I know you from from the 1st youtube video till here.I am your biggest fan really 🙂 .Those who are thinking to buy this series ,please do. Because it almost covered all the basics that are needed in aptitude .Wish me luck for my placement 🙂
sir me 2nd year student hu ye course gate aur placment ke liye enough he ya extra resources lagana padenge except reasoning
Placement ke liye course enough hai
Excellent way of teaching 😊😍 money can’t waste here
thanks alot
ho gya bhai sort out…thanks
I have bought this course but how to download notes in notes there is not an option to download them
Hey hi Ankur ,
you can access the notes on the website but they are not downloadable
is aptitude course aur tcs nqt course ki validity kya hai bhaiya
Aptitude is for 1 year and TCS NQT is for 6 month
can we extend the validity of the course with 1 month or 2 month,please sir