PLAN YOUR WEEK WELL IN ADVANCE !
THE ADURO CLASS SCHEDULE FOR THE WEEK IS HERE!
PLEASE CALL / WHATSAPP 09674924341 FOR INVITES
http://detafb.blogspot.in/2015/02/adurovarc-class-schedule-7th-12th-feb.html
Friday, 6 February 2015
ADURO/VARC CLASS SCHEDULE - 7TH - 12TH FEB
|
DATE | DAY | TIMING HOURS | SUBJECT | CLASS MODE | TOPIC | FACULTY | ||||||
| 07.02.15 | SATURDAY | 2000 - 2200 | LR | FB EVENT | INTRO | XL ALUMNUS | ||||||
| 08.02.15 | SUNDAY | 2200 - 23:59 | RC | FB EVENT | RC - TYPES OF PASSAGES | SXC ALUMNUS | ||||||
| 09.02.15 | MONDAY | 2200 - 23:59 | VA | FB EVENT | VA - VOCAB INCREASE TRICKS | SXC ALUMNUS | ||||||
| TBA | TBA | 2200 - 23:59 | QA | NUMBER THEORY 1 | NIT TRICHY ALUMNUS | |||||||
| TBA | TBA | 2200 - 23:59 | QA | FB EVENT | AVERAGE MIXTURES ALLIGATION | IIFT ALUMNUS | ||||||
| TBA | TBA | 2200 - 23:59 | DI / QA | FB EVENT | INTRO TO CAT DI | TISS | ||||||
| TBA | TBA | 2200 - 23:59 | WHY MBA | FB EVENT | GDPI | XL IIM ALUMNI | ||||||
| 11.02.15 | WEDNESDAY | 2200 - 23:59 | SC | NOUNS | MDI ALUMNUS | |||||||
| 12.02.15 | THURSDAY | 2200 - 23:59 | VA | WIZIQ | COLLOCATION | SXC ALUMNUS | ||||||
| 13.02.15 | FRIDAY | 2200 - 23:59 | QA | WIZIQ | PERCENTAGE PROFIT LOSS | IIM C ALUMNUS | ||||||
| *TBA = To Be Announced |
| ENROL FOR DEMO CLASSEShttp://bit.do/aduroll |
(16^3 + 17^3+ 18^3+ 19^3), divided by 70 leaves a remainder ?
(16^3 + 17^3+ 18^3+ 19^3), divided by 70 leaves a remainder ?
There's no remainder, and here's how you can prove it without using a
calculator *or* even bothering to work out all the numbers. It depends
on knowing this key factoring pattern, commonly taught in algebra
classes: for any pair of numbers a and b,
a³ + b³ = (a + b)(a² - ab + b²)
So we start by recognizing that 70 = 35·2. Next, we rearrange the
terms in the problem to look like this:
16³ + 19³ + 17³ + 18³
We can group the terms in pairs like this:
(16³ + 19³) + (17³ + 18³)
Now we factor each pair, using the pattern above:
(16 + 19)(16² - 16·19 + 19²) + (17 + 18)(17² - 17·18 + 18²)
now, 16 + 19 and 17 + 18 both equal 35, so:
(35)(16² - 16·19 + 19²) + (35)(17² - 17·18 + 18²)
We can now factor out the 35:
(35)(16² - 16·19 + 19² + 17² - 17·18 + 18²)
Now, let's look at that long expression in the parentheses. It's the
sum of four even numbers (16², 16·19, 17·18, and 18²) and two odd
numbers (19² and 17²). Since any two odd numbers always add up to an
even number, the terms in the parentheses all add up to some even
number. We don't know what it is, but we don't care -- all we care is
that it's even, which means it contains a factor of 2.
So our number has a factor of 2 and a factor of 35 -- and since 70 =
35·2, our number has a factor of 70. Which means that dividing it by
70 leaves no remainder.
I'm betting your teacher will give more credit for an answer like that
for one like "the calculator says so." :-)
Hope that helps!
From Yahoo Posts
There's no remainder, and here's how you can prove it without using a
calculator *or* even bothering to work out all the numbers. It depends
on knowing this key factoring pattern, commonly taught in algebra
classes: for any pair of numbers a and b,
a³ + b³ = (a + b)(a² - ab + b²)
So we start by recognizing that 70 = 35·2. Next, we rearrange the
terms in the problem to look like this:
16³ + 19³ + 17³ + 18³
We can group the terms in pairs like this:
(16³ + 19³) + (17³ + 18³)
Now we factor each pair, using the pattern above:
(16 + 19)(16² - 16·19 + 19²) + (17 + 18)(17² - 17·18 + 18²)
now, 16 + 19 and 17 + 18 both equal 35, so:
(35)(16² - 16·19 + 19²) + (35)(17² - 17·18 + 18²)
We can now factor out the 35:
(35)(16² - 16·19 + 19² + 17² - 17·18 + 18²)
Now, let's look at that long expression in the parentheses. It's the
sum of four even numbers (16², 16·19, 17·18, and 18²) and two odd
numbers (19² and 17²). Since any two odd numbers always add up to an
even number, the terms in the parentheses all add up to some even
number. We don't know what it is, but we don't care -- all we care is
that it's even, which means it contains a factor of 2.
So our number has a factor of 2 and a factor of 35 -- and since 70 =
35·2, our number has a factor of 70. Which means that dividing it by
70 leaves no remainder.
I'm betting your teacher will give more credit for an answer like that
for one like "the calculator says so." :-)
Hope that helps!
From Yahoo Posts
Thursday, 5 February 2015
TIME AND WORK
Event - https://www.facebook.com/events/1554467484827240/?ref_notif_type=plan_user_joined&source=1
- Work from Days:
If A can do a piece of work in n days, then A's 1 day's work = 1 . n - Days from Work:
If A's 1 day's work = 1 , then A can finish the work in n days. n - Ratio:If A is thrice as good a workman as B, then:Ratio of work done by A and B = 3 : 1.Ratio of times taken by A and B to finish a work = 1 : 3.****
#2
(M1D1HI)/ W1 = (M2D2H2)/W2
Where M1 & M2 represents no of labourers; D1 & D2 represents no of days; H1 & H2 represents no of hours; W1&W2 represents work done.
If there are 2 persons A & B such that A can do work in ‘a’ days and B can do work in ‘b’days.
Such that ‘a’ is a multiple of ‘b’, then, time taken by them to complete the work together = Bigger no/Sum of ratios
Eg: A can do work in 9 days, B can do work in 18 days. In how many days they will complete
the work together.
Bigger no=18, Ratio=9:18=1:2
No of days = 18/(1 + 2)
= 6 days
If ‘a’ is not a multiple of ‘b’, then time taken by A&B to complete the work together
= (LCM)/(Sum of ratios)
Eg: A can do a piece of work 30 days. B can do work in 45 days. In how many days they will
complete the work together?
LCM = 90, Ratio= 30:45=2:3
No of days= 90/(2 + 3) = 90/5
************
#3If there are 3 persons A, B & C whose time taken a,b,c days respectively, to complete a certain
work. Time taken by them to complete the work = LCM of (a, b, c)/(LCM/a + LCM/b + LCM/c)*********************Note: For 3 persons the common format is
Step1: Find the LCMStep2: Find the individual share of work i.e. LCM/a, LCM/b, LCM/c.Step3: Rest methods depend on the question i.e. follow the question patterns.Eg: A, B and C can do a work in 15,20,45 days respectively. In how many days they can
complete the work together.LCM=180No of days= [180/ (180/15 + 180/20 + 180/45)
= [180/ (12+9+4)]
= [180/25]
= 7.2 days*********#4Fastest Finger FirstA and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work in 23 days. The number of days after which A left the work wasA.12B.11C.10D.9*******************#5
Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank in gallons isA.100B.110C.120
***
#6
***********
Pipes & Cisterns
If there are 2 pipes A & B such that A (inlet pipe) & B (outlet pipe). Such that A can fill tank in
‘a’ minutes and B can empty the tank in ‘b’ minutes , then time taken to fill the tank if both are
operated together = Bigger no/Difference of ratios
Eg: A can fill tank in 9 minutes, B can empty the tank in 18 minutes.. In what time the tank be
filled, if both pipes work simultaneously?
Bigger no=18, Ratio=9:18=1:2
Time taken to fill the tank = 18/(2 1)
= 18 minutes
If ‘a’ is not a multiple of ‘b’, then time taken by A&B to fill the tank.
= (LCM)/(Difference of ratios)
Eg.: An inlet pipe can fill the tank in 30 minutes. B an outlet pipe can empty the tank in 45
minutes. In what time the tank be filled if both pipes work simultaneously?
Time taken to fill the tank= LCM = 90
= Ratio= 30:45=2:3
= 90/(3 2) = 90/1 = 90 minutes
If there are 3 pipes A, B & C, in which A, B are inlet pipes which takes a,b,minutes respectively
to fill the tank and C an outlet pipe which takes c minutes to empty the tankD.140****
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