Saturday 7 December 2013

Week 4, 07 December 2013

Before getting Answer of Week 3, Lets Practice some Analytical Portion

 Analytical

Below are some Question with Explaination, try to solve it.

Six actors ---- Bob, Carol, Dave Ed, Frank, and Grace audition for a part in an
off-Broadway play.  The auditions will take place over four consecutive days,
starting on a Thursday.  Each actor will have one audition; the days on which
the different actors will audition must conform to the following conditions.

i.   At least one audition will take place each day.
ii.   No more than two auditions will take place on any day.
iii.   No more than three auditions will take place on any two
consecutive days.
iv.   Bob’s audition must take place on Saturday.
v.   Carol’s audition must take place on the same day as another
audition.
vi.   Frank’s auditions must take place on the day before Grace’s
audition.
vii.   Dave’s audition must take place on a day after Ed’s audition.

1  If only one audition takes place on Thursday which actor could have that
audition?
(A) Bob (B) Carol (C) Dave (D)  Frank (E) Grace

2  If Bob’s and Frank’s auditions are on the same day, which of the following
must be true
(A) Dave’s audition will take place on Thursday
(B) Dave’s audition will take place on Friday
(C) Grace’s audition will take place on Thursday
(D)  Carol’s audition will  take place on Sunday
(E) Ed’s audition will take place on Sunday

3  If the director decides to hold two auditions on Thursday and two on
Sunday, how many actors would be eligible to audition on Friday?
(A)  1 (B)  2 (C)  3 (D)  4 (E) 5

4  If Ed and Grace have their auditions on the same day which of the following
must be true?
(A)  Ed’s audition will take place on Thursday.
(B)  Frank’s audition will take place on Friday.
(C)  Carol’s audition will take place on Saturday.
(D)  Grace’s audition will take place on Saturday.
(E) Carol’s audition will take place on Sunday.

5  If Ed’s audition is on Saturday, which of the following actors cannot
audition on the same day as any other actor?
(A) Bob
(B) Carol
(C)  Ed
(D)  Frank
(E)  Gr
ace
Questions 6-10:
During the first half of the year, from January through June, the chairperson of
the mathematics department will be on sabbatical.  The dean of the college has
asked each of the six professors in the department --- Arkes, Borofsky, Chang,
Denture, Hobbes, and Lee--- to serve as acting chairperson during one ofthose months.  The mathematicians can decide the order in which they will
serve, subject only to the following criteria established by the dean.
i.   Chang will serve as chairperson in February.
ii.   Arkes will serve as chairperson before Hobbes does.
iii.   Borofsky and Dexter will serve as chairpersons in consecutive
months.

6  Which of the following professors could serve as chairperson in January?

(A) Borodfsky (B) Chang (C) Dexter (D)  Hobbes (E) Lee

7  In how many ways can the schedule be made up if Lee has to serve as
chairperson in May?
(A)  1
(B)  2
(C)  3
(D)  4
(E) 6

8  If Lee serves in April, all of the following could be true EXCEPT
(A)  Arkes serves in January
(B) Hobbes serves in march
(C)  Borofsky serves in may
(D)  Borofsky serves in June
(E) Hobbes serves in June


9  If Borofsky serves in May, what is the latest month in which Arkes could
serve?
(A)  January
(B)  February
(C) March
(D)  April
(E)  June

10 Which of the following CANNOT be true?
(A)  Arkes and Lee serve in consecutive months.
(B)  Lee and Hobbes serve in consecutive months.
(C) Hobbes and Dexter serve in consecutive months.
(D)  Arkes and Chang serve in consecutive months.
(E) Borofsky and Chang serve in consecutive months.

Solutions 1-5:
First express each of the conditions symbolically:
B, C, D, E, F, and G: 1 audition each
Days: Thu, Fri, Sat, Sun
Each day: 1 or 2 auditions
2 consecutive days: 2 or 3 auditions
B=Sat   Cx  F<G  E<D
1  A violates the condition that Bob’s audition will take place on Saturday
(B=Sat).  B violates the condition that Carol’s audition cannot be the only
audition on a particular day (Cx). Choices C and E are impossible.  Since
Dave’s audition must take place on a day after Ed’s audition (E<D) and Grace’s
audition must take place on a day after Frank’s audition (F<G) neither can
take place on Thursday.  Only choice D does not violate any of the given
conditions, so this is the correct answer.

2  The condition that Bob’s and Frank’s auditions are on the same day completely
determines the schedule.  They must take place on Saturday (B=Sat).  To
avoid having more than three auditions  on two consecutive days, there can be
only one audition on Friday and one on Sunday, which means there will be two
on Thursday.  Since Frank must have to precede Grace (F<G), Grace’s audition
will take place on Sunday.  Since Ed must precede Dave, Ed’s audition will take
place on Thursday and Dave’s audition on Friday.  Finally, Carol’s audition will
be the second audition on Thursday.  The final schedule is “C and E on
Thursday, D on Friday, B and F on Saturday and G on Sunday”.  Only choice B
is consistent with this schedule, so “B” is the correct choice.

3  Since only one audition can take place on Friday, it cannot be Carol’s (Cx);
and, of course, it cannot be Bob’s (B = Sat).  Any of the other four actors
could audition on Friday as indicated in the following schedules:
E/F on Thu, D on Fri, B on Sat, C/G on Sun
C/F on Thu, E on Fri, B on Sat, D/G on Sun
C/E on Thu, E on Fri, B on Sat, D/G on Sun
E/F on Thu, G on Fri, B on Sat, C/D on Sun
So the correct choice is D.

4  The only schedule that fulfils the conditions is “F on Thu, E/G on Fri, B on Sat,
and C/D on Sun”.  Only choice E is consistent with this schedule.

5  Since Ed and Bob’s auditions are both taking place on Saturday, eliminate
choices A and C.  Since Carole must audition on the same day as another
actor, eliminate B.  Finally, since Dave’s audition must take place on Sunday
(E < D), Frank’s audition must take place on Thursday and Grace’s audition on
Friday (F < G).  Eliminate choice D. The complete schedule is: “C/F on Thu, G
on Fri, B/F on Sat, and D on Sun.”
Solutions 6-10:
Let A, B, C, D, H, L represents professor names.
C=February, A<H,  B<<D and D<<B


6  Only choice E is there, which does not violate any of the conditions, so is the
correct choice.

7  With C serving in February and L in May, the only consecutive months
available for B and D are March and April.  Then since A must serve is before H
in June.  There are two possible schedules, depending on the order of B and D,
so the correct choice is B.

8  If L serves in April, the consecutive months available for B and D are May and
June; so choices C and D could be true.  Since A must serve before H, choices
A and B must be true, only choice E cannot be true.

9  Since A must serve before H does, A cannot serve in June.  Can A serve in
April? No, because then, D would serve  in June (B<<D or D<<B), and again A
would not precede H.  The latest that A could serve in March, which could
occur in the final order: L, C, A, D, B and H.

10 The only professors that can serve in  January are A and L, so, one of them
must serve in January, and neither serves in February.  So choice A cannot be
true.

Note:  if you have any question or you want furhter explaination then kindly comment below or messgae on info.learning2013@gmail.com .
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ADMIN SKOnlineLearning    

Friday 29 November 2013

Week 3, 29th Nov,2013

 Quantitative section


This Section is about how to solve problem rleated to Fractions.

                                                    Ponits to Remember:


  • 2/5 :  Then 2 is Numerator and 5 is denominator
  • 1/5 th of 2/7 th means : 1/5 * 2/7.
  • Equatio related to fraction involves brackets and +/-* signs so there are some rules which are given below:
  1. Remove Bracket first
  2. Then quantities which are connected by of.
  3. Then / and *
  4. Then + and -.

Problems:

1. If 4/13 of number is 39 then, what is 8/13 of that number?
A. 39/4
B. 78
C. 16
D. 39/8 

2. Which of following is less then 5/11?
A. 3/2
B. 2/3
C. 1/2
D. 2/5


3. 3/4 of 28 is equal to 30/7 of what number?
A. 90
B. 45
C. 30 
D. 56
E. None

4. There are 20 boys in class. Five of them are left handed. What fraction of class is left handed?
A. 1/5
B. 1/2
C. 1/4
D. 2/11

5. A chemical solution contain 8% of acid. If their is 15ml of acid, what is volume of solution.
A. 125.5ml
B. 187.5ml
C. 225.5ml
D. 171.5ml

6. What fractional part of week is 98 hour?
A. 7/98
B. 7/12
C. 1/20
D. 1/7

7. what fraction is midway between 1/3 and 1/4?
A. 7/12
B. 7/24
C. 29/11
D. 1/2

8. 3/8 of number is 10 , what is number?
A. 91
B. 81
C. 23
D. 27

Note:  if you have any question or you want furhter explaination then kindly comment below or messgae on info.learning2013@gmail.com .

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ADMIN SKOnlineLearning    

Saturday 26 October 2013

Week 2, 26/10/2013 (updated on 26 nov,2013)

 Quantitative section

In today's section let us have practice on some Problems Related to Week 1:

1. A neon sign flashes every 3 seconds, another sign flashes every 5 seconds and a 3rd flashes every 7 seconds.If they all flash togather how many seconds will pass before they all flashes simultaneously again.
A. 15  sec
B.  35   sec
C. 105  sec
D. 21    sec
Solution:
Lets be simple : Question is telling us that at  X time they all have been flashes. Now at which time supose Y, they again flashes togather.
Supose A,B&C Flashes togather at 1 PM  then , 
A                     B                        C
1,0 sec             1,0sec                 1,0sec
1,3 sec             1,5sec                 1,7sec
1,6 sec             1,10sec               1,14sec
1,9 sec             1,5sec                 1,21sec
|                           |                             |
|                           |                             |
|                           |                             |
1,105 sec         1,105sec               1,105sec

So most simple answer will be through LCM= 3*5*7=105


2.The greatest number which exactly divides 1155 and 735 is.
A. 25
B. 5
C. 15
D. 105

Solution:
Greatest number which divide both mean it is asking about common factor which is highest in value in both, simply it is asking about HCF(highest common factor)
735 ) 1155  ( 1
         0735
        _________
           420 ) 735 ( 1
                    420
               _________
                 315 ) 420 ( 1
                          315
                    _________
                         105

3. It take Riaz 30 mint to make a paper .Razi only need 25 minute to make a paper. If they both start marking paper at 11:00 AM, What is first time they will finish marking a paper at same time?
A. 12:30
B. 12:45
C. 1:30
D. 12:25
Solution:
Question clearly asking about LCM.
so LCM of 25 and 30 is 150 minute so 
Answer is 2:30 hour and it mean 1:30 PM (C)

4. Beena buys two off-cuts of ribbon in a sale. One is 153 cm long the other is 204 cm long. She cuts them so that she ends up with a number of piece all the same length. What is the greatest length each piece can be?
A. 39
B. 6
C. 17
D. 51
Solution :
Simple answer to this Question is find HCF of 153 and 204 which is:
153: 3*3*17
204: 2*2*3*17
so 3 *17=51

5.Ahmed has a rectangular garden measuring 4.32 m by 3.36m. He wants to divide it into square plots of equal size . What is largest sized square he can use?
A. .24
B. .48
C. .16
Solution:
1st find hcf of 4.32 and 3.36 which is .96 . Now simply Divide it by 2 
.96/2= .48

6. Three bells toll after intervals of 6, 9 and 15 minute ,respectively. If they toll togather at 5 p.m, when will they toll togather next?
A. 6:30
B. 5:30
C. 6:45
D. 5:45
Solution:
It is same question as question 1. so simply LCM will give the answer
6,9,15 = 3*2*5*3=90 minute means 6.30 PM


7. The chairs in school hall can be set out in 25 equal rows or in 40 equal  rows or in 120 equal rows are:
A. 600
B. 400
C. 40
D. 80
Solution:
LCM of 25 , 40 and 120 =  5 * 5 * 8* 3= 600


Note:  if you have any question or you want furhter explaination then kindly comment below .

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ADMIN SKOnlineLearning   




Thursday 17 October 2013

Week 1, 17th OCT,2013

Quantitative section

Use of LCM 

In today's lesson you shall learn about question related to LCM.

Example: A Shining light shine at every 3 sec, another at every  5 sec and third one at 7 sec. If all shines togather, then how many seconds will pass before they all shines simultaneously together?

The answer to this question is simply by LCM of three time periods i-e 105 sec.

LCM
Resolve the given numbers into its prime factors. In each number take the prime factor with highest number and then multiply each of highest power factor.

Example: 10, 20, 30
10:               2, 5
20:               22, 5
30:               2, 3, 5
Then LCM is simply: 2x 3x 5= 60

Examples:
1. Love Kumar takes 30 minutes to check a paper, Imran takes 25 minutes to check a paper. If they both starts checking paper at 11:00 AM.  What is the first time they will finish marking a paper at same time.

1st time they complete checking together will be simply LCM of 25 and 30

25:               5, 5
30:               2, 3, 5
Then LCM = 5 x 5 x 3 x 2 = 150 minute so time will be 1:30 PM

Note: List of other questions will updated soon.

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ADMIN SKOnlineLearning


Saturday 13 July 2013

Introduction to Blog

This Blog has been created for "Cracking GAT General test"

GAT general contain 3 sections:
  1. English
  2. Quantitative
  3. Analytical
  •  What will be the format of posts
There will be one post per week . Each post will contain an trick about Quantitative ,Analytical and English section. And in the end you will be provided a practice test.

  •   How will it be help full for me
If you are busy in your jobs then at weekend you can just go through one simple trick , and related sample questions.If you have any query related to post you can leave a comment below the post and you will be answered.

Find the Attachments of  BOOK
https://docs.google.com/document/d/1ifJ63Mq2TKqJg-oJt9BXxDcaYVx8Jf3c-LG36QRg2T8/edit

or from
https://www.dropbox.com/s/qc07g1z8m917hw8/GAT%20by%20NTS.PDF

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e8239eecaef9a0b2a2318d43094571a2e3e07216a55febc3b1 Regards

Admin SKOnlineLearning