digital root...

Seed Number

Test Assignment

Assignment 1

Factor questions wd answers..

Factor Practice

Factor test

I am posting some questions on factors...solve them and get back to me..

Factor Practice

Algebra...A.P. >=G.P.

apgp

suggestions invited.......

Hey guys..nw i want ur participation in this blog...suggest me which topics you are facing problems in...I'll try to write those topics..thanx

Regards,

Monika Bansal

Type of numbers

Type of Numbers

Remainders....

Remainders theory

Factor theory...

Factors Theory

Logic Funda

• The 20 digit number 1920212223….2728, was obtained by listing the integers from 19 to 28. 10 digits of this number are now removed, so that the resulting number is as large as possible. What are the first 3 digits of the new 10- digit number?

A.942
B.967
C.956
D.989
E.999

AVERAGE

Average has a great significance in BANK PO. Not only in Math's but in DI section also, Average based questions are coming. I am writing this topic in layman's language. So that back-benchers also can understand. Everybody knows the meaning of average which says:

AVERAGE= Sum of observation / no. of observation

Now come to the practical meaning. Let me start with examples;

• 5 students are sitting in a room. They have average money of 45 Rs. 'A' goes out. Now the average money becomes 44 Rs. How many Rs. does 'A' have?
  

Ans. Firstly, if average money with 5 students is 45 Rs. It does not mean that every student has 45 Rs. It means, if they were having same money, then amount would have been 45 Rs. with each student. In total, they have 45x5 Rs.

Now answer to the question…one way to ans. Is:

45x5-44x4=49  

I am sure most of the students have done in this way only.

But why to calculate, if the sum can be damn easy by logic.

Smart Application:
Suppose every student has 45 Rs. in his pocket. If A were taking 45 Rs, there would have been no change in the amount of rest of the students. But since average falls by 1 Re. It means remaining 4 persons (except A) have 1 Re. less. In total 4 Rs. are missing. Where are these 4 Rs.? Obviously with A, because he is taking his own 45 Rs. + 4 Rs. = 49 Rs.  

Now try to solve following questions by using your brain. You are not allowed to write….

• Average weight of 10 men is 50 kg. Sanjay joins the group and average becomes 51 kg. What is weight of Sanjay?
  

Think this way:
Average money with 10 men is 50 Rs. Sanjay comes in and average becomes 51 Rs. What amount does Sanjay has?

If Sanjay were coming with 50 Rs., there would have been no change. Since, there is a rise of 1 Re. per head and there are 11 persons in the new group. It means Sanjay has come with 11 Rs. more. So Ans. = 50+11=61 Rs.  

• The average weight of class of 40 students is 50 kg, while the average weight of class along with Fatty and Motu is 51 kg. What is the average weight of Fatty and Motu?
  

Again be money minded….

40 students have 50 Rs. each. 2 persons are coming. Suppose they had entered with 50 Rs. each (100 Rs. In total), there would have been no change. But there is rise of 1 Rs. per head. (Now 42 students in total)

=1x42 in total

=Amount with Fatty and Motu = 100+42=142 Rs.

Average amount = 142/2= 71 Rs.  

• In the 15th inning, Sachin scored 58 runs. Thereby decreasing his average by 3 runs. What is his new average?
  

  Or
  

• A group of 14 persons has certain average money. A new student joined the group with 58 Rs. Thereby decreasing average by 3 Rs. What is the new average of the group?
  

By Equation:-

Let the old average be x

(14x+58)/15=(x-3)

=14x +58= 15x-45

= x=103

So, new average= 103-3=100  

Now we will do it as done earlier. Since the fall in average (of 15 students) is 3. i.e. every student is short of 3 Rs. It means in total, group is losing 3x15=45 Rs. Where are these Rs? Of course with new student. He came with 58 Rs. Of his own+ he stole 45 Rs from the group i.e. = with 103 Rs. = old average of the group. But average fall by 3. So new average is 100.  

• Group A of 45 students has average weight 25 kg.
  

  Group B of 30 students has average weight 26 kg.
  

  What is the combined/pooled average of both these group?
  

We will think this way… (Again Monetary approach)

If every student of group B was carrying 25 Rs, then the pooled average would have been 25 Rs. Only. But now group B's students are coming with extra 1 Rs. Per head i.e. 30 Rs. In total. These 30 Rs we will divide equally among 45+30 students. So, each student's additional share will become 30/75 Rs. So, new average 25+30/75=25.4 Rs. Ans.  

• What is average of 1, 2, 3… 49?
  

Average of n Consecutive natural no.= (n+1)/2

So, there average is = (49+1)/2=25                          

Or average is nothing but middle no. (if terms are in A.P )

• On a blackboard, 76 consecutive numbers (starting with 1) were written. Yogesh come and deleted 27. What is the average of remaining numbers?
  

Average of 1,2,3,….,76=(76+1)/2 = 38.5  

(Now again this sum boils out in Money sum.)

76 persons have average money 38.5 Rs. 1 person with 27 Rs. goes away. What is average of remaining group? Yogesh is taking 27 Rs. out of 38.5 Rs. It means he is leaving 38.5-27=11.5 Rs. for the group. Rest of the 75 people equally divides this sum and new average will become 38.5+ (11.5/75).

Suppose in the above example, we have deleted 70. Now what will be the change in average?

Now instead of 38.5 Rs., the money stealer is going with 70 i.e. 31.5 Rs. extra. So, these 31.5 Rs. should be taken from remaining 75 persons. So,

New Average = 38.5-

         =38.5-0.42

         =38.08  

• Always remember:
  

1. If x is added in all items, then average increases by x.
   
2. If x is subtracted in all items, then average decrease by x.
   
3. If every item is multiplied by x, then average also gets multiplied by x.
   
4. If every item is divided by x, then average also gets divided by x.
   

 

I guess now you can relate all the above fact with many (monetary) things. Now it's your turn to turn & churn your mind wheel. I am giving you a practice exercise that, I am a damn sure, if you will do with sincerity, you will never stuck in average funda.

 

who'll solve it??????

Que: What is the sum of perimeters of all the triangles with integer length whose area is 7200 square units???

some more questions on LCM funda....

Some Questions for Practice:

1.) N = 123456……………..142143, What is the remainder when N is divided by
(a) 36
(b) 72
(c) 18
(d) 6

2.) N = 42+442+4442+……….(88 terms) is divided by 36. What is the remainder?

3.) X = 22233344455566677788, Remainder when X is divided by
(a) 888
(b) 111
(c) 333

LCM funda in remainders..

I have seen 3 types of questions on LCM..

Type 1

Que 1: Which series will give all the +ve nos. Which when divided by 15, 18, 24, remainder is 1?

Soln. Of course this is an easy nut 2 crack.....You will immediately say…
The series will be represented by L.C.M.(15, 18, 24)K + 1
Where k = 0, 1, 2, 3..........
Now the 2nd type....

Type 2

Que 2: Which series will give all the +ve nos. which when divided by 15, 18, 24, remainders are 13, 16, 22 respectively?

Soln. this is the same question as above....watch it carefully..

Divisor        remainder
15            13
18          16
24         22

Now in each case,remainder can be written as -2 (-ve remainder) also.This is what the question requires...but now since the remainders r same..We can opt 1st type..
So, it will be L.C.M. (15, 18, 24) K - 2
Where k = 1, 2, 3,4.........(neglecting zero because we want +ve nos)
Now comes d mind boggling 1....I mean 3rd type

Type 3

First of I’ll take a smaller example...

Que 3: Which series will give all the nos. which when divided by 9 and 8, remainders are 5 and 1 respectively???

Soln: I’ll obviously use Chinese Reminder theorem…but not algebraically..i have picked up d logic only…see

I’ll pick d bigger divisor which is 9 here (why???cz otherwise it’ll be more calculative).
My first condition is nos should leave remainder 5 when divided by 9.It’ll be of 9k + 5 type
K          9k + 5
0          5
1          14
2         23
3         32

and so on……..

Out of these nos.. I want 2 choose that no which leave remainder 1 when divided by 8…we’ll start checking…

K          9k + 5         remainder wn divided by 8
0         5                 5
1         14               6....(increase of 1.same pattern)
2        23               7
3         32               8
We’ll check for only first 2 nos for the middle column....as we can see there is increase of 1 in the last column, so we’ll keep on increasing the remainders by 1 until we get 1. (the required remainder).....and that corresponds to k = 4…
So, answer is 9×4 +5 = 41.
Now come 2 the original question..
Answer will be L.C.M. (9, 8) k+41, where k = 0, 1, 2, 3………this series will give all d nos. of above said type.

Qus 4: What is the smallest +ve no. which when divided by 72 and 10; remainders are 41 and 9 respectively?

Soln. I’ll pick the bigger divisor which is 72 here.
My first condition is that the number should leave remainder 41 when divided by 72.It’ll be of 72k + 41 type.

Out of these nos.. I want 2 choose that no. which leave remainder 9 when divided by 10…we’ll start checking…

K          72k+41         remainder wn % by 10
0 +4        41                 1
1         113               3....(increase of 2..so, +2 in each term)
2        __               5
3         __               7
4         __               9 Answer
We’ll check for only first 2 nos for d middle column....as we can see there is increase of 2 in the last column, So we’ll keep on increasing the nos. by 2 until we get 9 (the required remainder).....and that corresponds to k = 4…
So, answer is 72*4+41 = 329.

Qus 5:
What is the smallest +ve no. which when divided by 8, 9 and 10; remainders are 1, 5 and 9 respectively?
Ans. This question consists of above two questions (3 and 4). First of all we will find for 8 and 9. By above question the required number will be 41.Now question becomes

What is the smallest +ve no. which when divided by 72 and 10; remainders are 41 and 9 respectively?
And the ans. to this question is 329.

Application of type 3 in Remainder theory

Qus 6: What is the remainder when N= 777……777 (41 times) is divided by 36?
Ans: 36 = 9*4 (We will Split the divisor in to 2 or 3 factors which are Co-Prime to each other)
Now, Remainder when N%9 = 7+7+7….. (41 Times)
= 7*41
= 7*5
= 35 =8
Remainder when N%4 = 1 (We will Check for last 2 places)
Now, we are looking for the smallest positive integer which leaves remainder 8 and 1 respectively, when divided by 9 and 4.
K          9k+8         remainder wn % by 4
0         8                 0
1         17               1

Luckily, we got our ans. at the initial stage so our ans. Correspond to K=1 i.e. 17.

original CAT Geometry

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CAT Geometry

by Monika Bansal

Yet to come...

ystrdy my 1 stdnt came 2 me.He ws vry frustrated cz of CAT prep.He askd;"mam, wt is d usage f geometry in CAT...mera toh dabba gol hai ismein..". Thn he showed mw d following thng..


wn I saw at d figure...i laughed lyk nythng..i said;"if u want to go 2 rectangular premises of an elite B- school...u shold not be a dabba gol circle in geometry"no doubt GEOMETRY is beloved of CAT people.so,try 2 solve as many qustions as possible...nd to make it possible i m posting all d original CAT questions f d topic...Go throuh it nd enlighten ur spirits...

question on base system

(7⁷⁷)₁₀=(x)_7⁷, with how many zeroes will x end??

greatest integer fn...

if 76 < x< 92 and 142 < x< 2003; then how many solutions are possible for the following equation:
[x/143] + [2x/91] + [3x/77] = 68x/1001
where [x] represents greatest integer less than equal to x.

base sys....

A +ve integer x; (100 < x < 150); is represented in base 2, base 3, base 4 and base 5 notation. In all the 4 cases the last digits are 1, 2, 3 and 4 respectively. What will be the last digit when x is represented in base 6??

remainders..

Let p = x²⁵-x¹³+x⁶-1 ; where x>2007.
How many remainders are possible if p is divided by 7???