NSF Math Column – Volume 6
Speed Math Techniques
Reciprocals:
This article explains a method to calculate
decimal equivalent of a fraction. This is based
on an approach described in Vedic
Mathematics. In this article we’ll look at how to
find reciprocals for numbers ending in 9:
What is 1/29?
We all know through straight division that this is
a recurring decimal 0.0344827586….. Now let’s
try the new method to calculate the decimal
equivalent. In most cases, we only need to find
decimals rounded to 2 or 3 decimals. We’ll
demonstrate up to 4 decimal places but the
same steps can be continued to derive the
remaining decimal digits.
Step 1: Place a decimal point
0.
Step 2: Add one to the digit before the last digit
in the denominator. This gives us 2 + 1 = 3. We
call this our divisor.
Step 3: Divide our numerator by the divisor. We
get quotient of 0 and remainder of 1. Add the
quotient to our decimal answer.
0.0
Step 4: Prepend the remainder in previous step
to the quotient. We get 10. Now divide this by
our by the divisor. We get 3 as the quotient and
1 as the remainder. Add quotient to our
decimal. 0.03
Step 5: Prepend the remainder in previous step
to the quotient. We get 13. Now divide this by
our by the divisor. We get 4 as the quotient and
1 as the remainder. Add quotient to our
decimal. 0.034
Volume 1-6
Step 6: Prepend the remainder in previous step
to the quotient. We get 14. Now divide this by
our by the divisor. We get 4 as the quotient and
2 as the remainder. Add quotient to our
decimal. 0.0344
We can continue this process of prepending the
reminder to the quotient and dividing by the
divisor until we start seeing repeated pattern of
decimals. This might look like a lengthy process
but with lots of practice you will be able to
quickly calculate first 2 to 3 decimals places
mentally in a few seconds!
Let’s try another example. What is 1/129?
Step 1: Place a decimal point
0.
Step 2: Add one to the number before the last
digit in the denominator. This gives us 12 + 1 =
13. We call this our divisor.
Step 3: Divide our numerator by the divisor. We
get quotient of 0 and remainder of 1. Add the
quotient to our decimal answer. 0.0
Step 4: Prepend the remainder in previous step
to the quotient. We get 10. Now divide this by
our by the divisor. We get 0 as the quotient and
10 as the remainder. Add quotient to our
decimal. 0.00
Step 5: Prepend the remainder in previous step
to the quotient. We get 100. Now divide this by
our by the divisor. We get 7 as the quotient and
9 as the remainder. Add quotient to our
decimal. 0.007
Step 6: Prepend the remainder in previous step
to the quotient. We get 97. Now divide this by
our by the divisor. We get 7 as the quotient and
6 as the remainder. Add quotient to our
decimal. 0.0077
Thus, the answer is 0.0077 (up to 4 decimals)
June 2011
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NSF Math Column – Volume 6
Practice Problems:
1. 1/39
3. 1/99
5. 1/199
2. 1/59
4. 1/169
6. 1/139
Area of one equilateral triangle in the regular
hexagon = (√3/4)(x/2)2
= (√3/4)(x2/4)
= (√3/16)x2
This is nothing but ¼ the area of the
independent equilateral triangle = 2/4 = 1/2.
Competitive Math
(
Area of the regular hexagon = 6 * (1/2) = 3.
indicates difficulty level)
An equilateral triangle and a regular hexagon
have equal perimeters. If the area of the
triangle is 2, what is the area of the regular
hexagon?
This problem highlights couple of important
facts to remember.
Area of an equilateral triangle with side x
is (√3/4)x2.
A regular hexagon is made up of 6
equilateral triangles (shown below)
1
6
5
This is an example of “Conditional Probability”.
Conditional probability is the probability of an
event A occurring given the occurrence of
another event B. We write that as:
2
P(A|B) = P(A∩B)/P(B)
3
Basically, we find out the probability of event B
occurring. Then we find out the subset in which
both A and B occur, and find the ratio to get the
probability.
4
Using the above two facts, we can now solve
our problem. In our case, we are given that the
perimeter of the equilateral triangle and the
regular hexagon are the same. Hence, length of
a side of the regular hexagon is x/2 where x is
the length of a side of the equilateral triangle.
Now it is easy to find the area of the regular
hexagon. Find the area of one equilateral
triangle part of the regular hexagon and
multiply by 6!
Volume 1-6
On an exam day, 10 boys and 6 girls write the
test during the morning session. During the
afternoon session, 8 boys and 16 girls write the
test. A session is randomly chosen and a test
score is selected for review. It happens to be a
boy’s score. What is the probability that the
selected test was from the morning session?
In our example, let’s have the following
definitions.
P(M) = probability that the session selected is
the morning session.
P(A) = probability that the session selected is
the afternoon session.
P(B) = probability that the test selected is that
of a boy.
June 2011
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NSF Math Column – Volume 6
P(Bm) = probability that the test selected is that
of a boy from morning session
In January, Winnie spent 50% of her allowance
= (50/100) * 180 = $90
P(Ba) = probability that the test selected is that
of a boy from afternoon session.
Remaining money, she deposited into the bank.
Hence current savings = 180 – 90 = $90.
Since selecting a morning or an evening session
is equally likely, selecting a session has a
probability of ½.
In the following month, her allowance increased
by 10%. New allowance = 180 + 180*(10/100) =
$198. Since her expense also increased by 15%,
her new expenses = 90 + 90*(15/100) = $103.50
P(A) = P(M) = ½
Now, probability of selecting a boy’s test score
from either the morning or afternoon session is:
P(B) = (1/2) P(Bm) + (1/2) P(Ba)
= (1/2)(10/16) + (1/2)(8/24)
= (5/16) + (1/6)
= 23/48
Probability that it’s a boy’s test score and it is
from the morning session is:
P(M ∩ B) = (1/2) (10/16) = 5/16
New savings = 198 – 103.50 = $94.50
Total savings = 90 + 94.50 = $184.50
An adult ticket cost $10 and a child ticket cost
$7. Mr. John bought 13 tickets for $106. How
many child tickets did he buy?
Let x be the number of child tickets. Number of
adult tickets is (13 – x).
Hence, probability that the selected test score is
from the morning session given it is a boy’s test
is:
(13 – x)*10 + x*7 = 106
130 – 10x +7x = 106
-3x = -24
x=8
P(M|B) = P(M ∩ B)/P(B) = (5/16)/(23/48)
Hence, number of child tickets is 8.
= 15/23
Winnie’s allowance in January was $180. She
deposited 50% of her allowance in the bank
and spent the rest of it. Her allowance was
increased by 10% the following month but her
expenses went up by 15%. What is her total
savings?
This is a problem of percentages. In order to
solve this, we need to remember the formula
that relates the “whole” and “part” as follows.
A passenger train travelling at 40 mph passes a
freight train travelling in the opposite direction
at 20 mph. Leo, riding in the passenger train,
notes that the freight train passes him in 15
seconds. What is the length of the freight
train?
This problem deals with an important concept
of relative speed. When two objects A and B
move in opposite directions at speeds s1 and s2,
it is same as one object (say A) being stationary
and the other (say B) moving at a speed equal
Part = (Percentage/100) * Whole
Volume 1-6
June 2011
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NSF Math Column – Volume 6
to s1 + s2. In other words, we say B is moving at
a speed s1 + s2 relative to A.
Now in our case, relative speed of the freight
train with reference to the passenger train is
(40 + 20) = 60 mph.
Problem of the month
How many times in a week (Monday through
Sunday) do the hour and minute hands of a
clock form a right angle?
It is given that the freight train passed Leo in 15
seconds. In order to find the length of the train,
we need the following formula:
Distance = speed x time
We can substitute the speed and time to find
the length of the train.
= (60 mph) x (15 seconds)
= (60 mph) x (15/3600 h)
= 0.25
Would you like submit your answer? Please
click on the following link:
https://spreadsheets.google.com/viewform?formkey=dHR6ek5BazVnRVM3d01nbG1fNVdybXc6MA
In other words, the freight train is 0.25 miles
long.
Names of everybody who submitted correct
answers will be published in the next edition!
Interested to know the solution for last
column’s problems? Refer to the end of this
document!
Special thanks to the following Math Column
contributors:
•
Srinivasa Rao Karanam (Sugarland, TX)
For any questions or comments, please contact the
team at [email protected]
Volume 1-6
June 2011
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NSF Math Column – Volume 6
Answer to “Can you prove why this works?
Also, can you think how to do this for
multiple numbers each having its 10’s
complement from different powers of 10?”
(Vol 1-5)
Let’s say we want to find the value of d – n,
where d and n are integers. If 10k is the nearest
power of 10 greater than n, 10’s complement of
n = 10k – n.
According to our method, we add the number
and the 10’s complement.
d + 10k – n
Approach #2: Count in fives from the beginning.
Once we reach the end of the line, go back to
the beginning of the line but continue the
counting.
Based on approach #1, you would get 98 and
approach #2 would result in 89 as the answer.
We can manually, mark 100 numbers in a sheet
of paper and start striking off every 5th number.
Another approach is to find a pattern. For
example, in approach #1 we see the following
pattern:
Round
Then we subtract the power of 10 with respect
to which we found the 10’s complement. IN
other words, we end up with:
(d + 10k – n) - 10k = d – n
When we deal with multiple numbers to
subtract and each of the 10’s complement is
from different powers of 10, we basically have
to subtract each powers of 10 after adding the
10’s complements.
Answers to Practice Problems (Vol 1-5)
1. 20
3. 17488
5. 258424
2. 269
4. 20498
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
# of
people at
start
100
80
64
52
42
34
28
23
19
16
13
11
9
8
7
6
5
# of
people
removed
20
16
12
10
8
6
5
4
3
3
2
2
1
1
1
1
1
# of people
at the end
80
64
52
42
34
28
23
19
16
13
11
9
8
7
6
5
4
Answer to Problem of the month (Vol 1-5)
89 or 98
There are two approaches to this problem.
Approach #1: Count in fives from the beginning.
Once we reach the end of the line, go back to
the beginning of the line and restart the
counting from 1.
Volume 1-6
Since, in every round people who are at a
position in line that is a multiple of 5 get
eliminated you can notice that only in round 1
and 2 the last person would get eliminated (not
considering the last round). Therefore, 98th
person from the original line will remain until
Round 17.
June 2011
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NSF Math Column – Volume 6
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Pranav Rekapalli (Atlanta, GA)
Shivani Guha (Missouri City, TX)
Siddarth Guha (Missouri City, TX)
Akaash Sanyal (Cary, NC)
Krishna Bharathala (Fremont, CA)
Ashish Kumbhardare (West Chester, PA)
Anna Nixon (Portland, OR)
Akash Karanam (Sugar Land, TX)
Desigamoorthy Nainar (Champaign, IL)
Nishanth Anand (Herndon, VA,US)
Roshan George (charlottesville)
Maya Shankar (Bridgewater, NJ)
Shreya Shankar (Bridgewater, NJ)
Sanjna Khanna (Piscataway, NJ)
Ankit Patel (Princeton, NJ)
Navya Prabhushankar (Olathe, KS)
Akshay Prabhushankar (Olathe, KS)
Sreekar chitti (bangalore, India)
Kamala Pullakhandam (Westborough, MA)
Aarush Prasad (Tampa, FL)
Hemanth Chitti (Bangalore, India)
Nikhil Parchuri (NJ)
Aayush Singh (Tampa, FL)
Shritha Gunturu (Aurora, CO)
Indumathi Prakash (Sharon, MA)
Rohan Mylavarapu (Superior, CO)
Sayuj shajith (suwanee, ga)
Akhila Mamandur (Houston,TX)
Adhith Palla (Hoffman Estates, IL)
Aditya Vargheese (Overland Park, KS)
Priya parchuri (Princeton, NJ)
Shrutika Kumareshan (Sharon, MA)
Sudeep Kumareshan (Sharon, MA)
Nymisha Mattapalli (Herndon, VA)
Shreyaa Raghavan (Sharon, MA)
Anirudh Rangaswamy (Dayton, OH)
Ananya Yammanuru (St Charles, IL)
Harshika Avula (San Antonio, TX)
Aditya Sridhar (Iselin, NJ)
Shreya shetty (Richmond, VA)
Hrishikesh Kommu (Houston, TX)
Volume 1-6
June 2011
Himanvi Kopuri (Denver, CO)
Arvind Chava (Herndon, VA)
Vamsi Subraveti (Nashville, TN)
Ashwath Raj (San Diego, CA)
Barath Raj (San Diego, CA)
Saketh Chillara (New Lenox, IL)
Pravin Mahadevan (Gainesville, VA)
Mrugank Gandhi (Aurora, Il)
Sanjana M (Herndon, VA)
Shruthi Santhanam (Suwanee, GA)
Rubesh Sivakumar (Irving, TX)
preetha sivakumar (Irving , TX)
Beena Vora (Owensboro KY)
Samhitha Somavarapu (Manassas, VA)
Sadhvika Challa (Frisco, TX)
Anjali Gupta (Edison, NJ)
Aishwarya Ilangovan (Westerville, OH)
Thushar Mahesh (Tampa, FL)
Shashank Mahesh (Tampa, FL)
Sonali Razdan (Shrewsbury, MA)
Varun Singh (Tampa, FL)
Vikas Ravi (San Ramon, CA)
Samruddhi Hande (San Diego, CA)
Praneeth Prathi (Shrewsbury, MA,U.S.A)
Pragathi Vellanki (San Ramon, CA)
Jarnail Singh (Cleveland)
Ranjan Veludandi (spring , TX)
Tarini Singh (cleveland, OH)
Tanishq Kancharla (Middlebury, CT)
Anirudh Kuchibhatla (Chicago, IL)
Anusha Vajrala (Aurora ,CO)
Keerti Vajrala (Aurora, CO)
Shreya Bellur (Dunlap, IL)
Aarush Garg (Farmington Hills, MI)
Ilakiya Udhayakumar (Metuchen, NJ)
Aditi Mahajan (Troy, MI)
Nisha Goel (Woburn, MA)
Asha Chandaka (Hillsborough, NJ)
Lalith Goli (Stamford, CT)
Nivedhitha Sivakumar (Belle Mead, NJ)
Sruthi Parthasarathi (Mason, OH)
Arnav Singh (Tampa, FL)
Anmol Nigam (Chicago IL)
Soumika Guduru (San Diego, CA)
Shaheel Mitra (Cincinnati, OH)
Shalin Desai (Olmsted Falls, OH)
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NSF Math Column – Volume 6
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Anup Hiremath (Old Bridge, NJ)
Dhivya Senthil Murugan (Denver, CO)
Sraavya Pinjala (Salt Lake City, UT)
Meghana Kandarpa (Southington, CT)
Simoni Maniar (Grapevine, TX)
Manisha Solipuram (Novi, MI)
sanmeshkumar Udhayakumar (Metuchen,
NJ)
Tanushree Pal (Ventura, CA)
Maansa K (Atlanta, GA)
Leela Pakanati (Dunlalp, IL)
Akshay Venkat (Leawood, KS)
Harshnandan Dasika (Buffalo Grove, IL)
Rinni Bhansali (Deer Park, NY)
Deepankar Gupta (Naperville, IL)
Bhavana Muppavarapu (Buffalo Grove, IL)
Vishal Gullapalli (Wayne, NJ)
Rama Balasubramaniam (Dublin, OH)
Anurag Krishna Dasika (Tucson, AZ)
Rachana Madhukara (San Diego CA)
Rudrakshi Dasika (Tucson, AZ)
Subahshi Rajiv (San Jose, CA)
Vishnu Gopikanth (Long Grove, IL)
Pallavi Thawani (San Jose, CA)
Satvik Reddy (Jacksonville, FL)
Sahana Aiyer (Fairfax, VA)
Sneha Reddy (Jacksonville, FL)
Renae George (Charlottesville, VA)
Hemanth Bhagawatula (Austin, TX)
Sharmila Perumal (San Diego, CA)
Sneha Revanur (San Jose, CA)
Sameer Saptarshi (Thornton, CO)
Sanjana Rao (Suwanee, GA)
Kritika Rao (Suwanee, GA)
Aarthi Sridhar (Sacramento, CA)
Praneet Dara (Palatine, IL)
Sayak Chatterjee (Winchester, MA)
Shreya Rekapalli (Atlanta, GA)
Ameya Rekapalli (Atlanta, GA)
Shivansh Gupta (Naperville, IL)
Nikhil Pandeti (Weymouth, MA)
Sindhuja Karanam (Sugar Land, TX)
Akhil Chava (Herndon, VA)
Gargi Sadalgek (NJ)
Aayush Gupta (Cupertino, CA)
Shreya Ramineni (Buford, GA)
Volume 1-6
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Reshmi Ranjit (Stamford, CT)
Gayathri Srirajan (Waukegan, IL, US)
Sanjana Vadlamudi (Cary, NC)
Ankita Ramachandran (Cupertino, CA)
Anish Bose (Cedar Park, TX)
Ramanan Srirajan (Waukegan, IL)
Rohan Rege (Buford, GA)
Akshaj Kadaveru (Fairfax, VA)
Mana Singri (Southlake, TX)
Akhil Aggarwal (Buffalo Grove, IL)
Sameer Lal (Macungie, PA)
Sreeniketh Vogoti (Mooresville, NC)
Satvik Kolluri (Austin, TX)
Pavan Kumaraguru (Chappaqua NY)
Pranav Rakkappan (Canton MI)
Baskaran Rakkappan (Canton MI)
Arya Koneru (San Antonio, TX)
Adithya Mummidi (San Antonio, TX)
Saahith Mummadi (Novi, MI)
Aneesh Virjala (Santa Clara, CA)
Tharini Ramakrishnan (Portland, OR)
Aasish Virjala (Santa Clara)
Anita Virjala (Santa Clara)
Ishan Rereddy (Garland)
Vishik Bhalla (Nashua NH)
Pranav Krishna (Edison, NJ)
Nishant Chittari (Columbus, OH)
A Thakkar (IL)
Shreya Kolluri (Austin, TX)
Ananya Rege (Buford, GA)
Abhiram Vallabhaneni (Schaumburg, IL)
Manvitha Kapireddy (Charlotte, NC)
Rohin Tangirala (Campbell CA)
Pujita Tangirala (Campbell CA)
Ashwini Rege (Buford, GA)
Thanks to all who attempted to solve the
problem of the month. This month had a
record number of participants! The Math
Column team is looking forward to your
continued interest and increased
participation.
June 2011
Page 7