ISI CMI IMPORTANT AND COMPLICATED QUESTION WITH SOLUTION BY SOURAV SIR'S CLASSES 9836793076

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SOME IMPORTANT
AND COMPLICATED
QUESTIONS FOR
CMI ENTRANCE
EXAMINATION
BY
SOURAV SIR’S CLASSES
KOLKATA & NEW DELHI
9836793076

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Question
A system of equations has 3 equations
3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.
If the system of equations has no solution, find k.
Explanation
This is a very interesting question. If we can generate one of the equations from the other
two, we can then say that the system has infinite solutions. If we can generate one of the
equations from the other two, but with the constant part alone being different, this would
be akin to having parallel lines when we are dealing with 2 variables and that would
result in the system having no solutions. So, lets look for that.
So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) = 2x + 3y + kz.
We need to find k. In other words, we need to find a, b such that a(3x + 4y) + b(4x + 5y )
= 2x + 3y . Then said a, b would give us k.
Or, we are effectively solving for
3a + 4b = 2
4a + 5b = 3
Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = -1. a = 2.
Now, k = 5a + 3b = 10 -3 = 7.
If k = 7, this system of equations would result in no solutions.
If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and
2x + 3y + 7z = 6.
First equation * 2 - second equation would give us the equation 2x + 3y + 7z = 8.
Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has
no solution.

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Yeah, the more mechanical, rather deceptively straight-forward-looking determinant
method is also there. But where is the joy in that.
1. a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab,
bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an
integer and so is the arithmetic mean of a, b and c. How many such triplets are
possible (unordered triplets)
Exactly one of ab, bc and ca is odd => Two are odd and one is even
abc is a multiple of 4 => the even number is a multiple of 4
The arithmetic mean of a and b is an integer => a and b are odd
and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3
c can be 4 or 8.
c = 4; a, b can be 3, 5 or 5, 9
c = 8; a, b can be 3, 7 or 7, 9
Four triplets are possible
2. A seven-digit number comprises of only 2's and 3's. How many of these are
multiples of 12?
Number should be a multiple of 3 and 4. So, the sum of the digits should be a
multiple of 3. WE can either have all seven digits as 3, or have three 2's and four
3's, or six 2's and a 3. (The number of 2's should be a multiple of 3).
For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us
combine these two.
All seven 3's - No possibility

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Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some
order. No of possibilities = 5!/3!*2! = 10
Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number
2222232
3. How many factors of 2^5 * 3^6 * 5^2 are perfect squares?
Any factor of this number should be of the form 2^a * 3^b * 5^c. For the factor to
be a perfect square a,b,c have to be even. a can take values 0, 2, 4. b can take
values 0,2, 4, 6 and c can take values 0,2. Total number of perfect squares = 3 * 4 *
2 = 24
4. How many factors of 2^4 * 5^3 * 7^4 are odd numbers?
be an odd number, a should be 0. b can take values 0,1, 2,3, and c can take values
0, 1, 2,3, 4. Total number of odd factors = 4 * 5 = 20
5. A number when divided by 18 leaves a remainder 7. The same number when
divided by 12 leaves a remainder n. How many values can n take?
Number can be 7, 25, 43, 61, 79.
Remainders when divided by 12 are 7 and 1.
Question
A system of equations has 3 equations
3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.
If the system of equations has no solution, find k.
Explanation
This is a very interesting question. If we can generate one of the equations from the other
two, we can then say that the system has infinite solutions. If we can generate one of the
equations from the other two, but with the constant part alone being different, this would
be akin to having parallel lines when we are dealing with 2 variables and that would

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result in the system having no solutions. So, lets look for that.
So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) = 2x + 3y + kz.
We need to find k. In other words, we need to find a, b such that a(3x + 4y) + b(4x + 5y )
= 2x + 3y . Then said a, b would give us k.
Or, we are effectively solving for
3a + 4b = 2
4a + 5b = 3
Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = -1. a = 2.
Now, k = 5a + 3b = 10 -3 = 7.
If k = 7, this system of equations would result in no solutions.
If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and
2x + 3y + 7z = 6.
First equation * 2 - second equation would give us the equation 2x + 3y + 7z = 8.
Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has
no solution.
Yeah, the more mechanical, rather deceptively straight-forward-looking determinant
method is also there. But where is the joy in that.
Question
Product of the distinct digits of a natural number is 60. How many such numbers are
possible?
Explanation
60 = 2^2 * 3 * 5
60 cannot be written as a product of two single digit numbers. SO, the number in question
should either have 3 or more digits.

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Three-digit numbers
The digits could be 345 or 265
Digits being 345 - there are 3! such numbers
There are six numbers for each of these outlines. So, there are 3! + 3! = 12 three-digit
numbers
Four-digit numbers
The digits could be 1345, 1265 or 2235.
Digits being 2235 - this is not possible as digits have to be distinct.
So, there are totally 24 + 24 four-digit numbers possible. 48 four-digit numbers.
Total number of numbers = 12 + 48 = 60.
John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of
types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If
there are more chocolates of type C than of type B, and more of type B than of type A,
what is the minimum possible number of chocolates overall?
Explanation
Again, big thanks to Mukund Sukumar for excellent solution.
Let john have 3x chocolates of type A and 7x of type B
Let Mike have 5y chocolates of type B and 4y of type C
Let john have 3z chocolates of type C and 5z of type A
So in total A=3x+5z ; B=7x+5y ; C=4y+3z
Since C>B we get solving y<3z-7x --->(1)
Since B>A we get solving 5y>5z-4x ---> (2)
What gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only y=1
as choice, for which second condition doesnt satisfy.

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So, when x=1,z=4, we get y<5 from first condition and when y > 3.2 from second
condition. so which gives choice the only y=4.
Hence x=1,y=4 and z=4 works and is the best possible answer.
For these values, we get A=23,B=27,C=28.
Minimum possible number of chocolates overall is 76
A wonderful, but very tough question from Permutation and Combinations.
In how many ways, can we rearrange the word MONSOON such that no two adjacent
positions are taken by the same letter? (Tough one. Tougher than what we will see in
CAT)
Explanation
First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2
N’s.
Let us focus on the three O’s.
We can place the three O’s in some blanks around the other letters. Or, three O’s can be
placed in 3 slots out of the 5 in __M__N__S__N__ . This can be done in 5
C3, or 10 ways.
Or, the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1, 4, 7} or {1,
5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7} - Whew.
Now, for each of these arrangements, there are 4!/2! = 12 arrangements for the other 4
letters. But the one thing we need to keep in mind now is the fact that 2 N’s could be
adjacent in these arrangements. We will need to eliminate these.
O’s in slots {1, 3, 5} or O__O__O__ __ - Ns could be in the last two slots. There are 2! =
2 words like this. So, number of words that we need to count = 12 – 2 = 10
O’s in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are 2! = 2
words like this. So, number of words that we need to count = 12 – 2 = 10

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O’s in slots {1, 3, 7} or O__O__ __ __O - Ns could be in the slots {4, 5} or {5, 6}. There
are totally 4 words that we need to subtract. So, number of words that we need to count =
12 – 4 = 8
O’s in slots {1, 4, 6} or O__ __O__O__ - Ns could be in the slots {2, 3}. There are 2! =
2 words like this. So, number of words that we need to count = 12 – 2 = 10
O’s in slots {1, 4, 7} or O__ __O__ __O - Ns could be in slots {2, 3} or {5, 6}. There are
totally 4 words that we need to subtract. So, number of words that we need to count = 12
– 4 = 8
O’s in slots {1, 5, 7} or O__ __ __O__ O - Ns could be in the slots {2, 3} or {3, 4}.
There are totally 4 words that we need to subtract. So, number of words that we need to
count = 12 – 4 = 8
O’s in slots {2, 4, 6} or __O__O__O__ - Tehre are no possible slots for N. So, we count
all 12 words on this list.
O’s in slots {2, 4, 7} or __O__O__ __ O - Ns could be in slots {5, 6}. There are 2! = 2
words like this. So, number of words that we need to count = 12 – 2 = 10
O’s in slots {2, 5, 7} or __O__ __O__ O - Ns could be in slots {3, 4}. There are 2! = 2
words like this. So, number of words that we need to count = 12 – 2 = 10.
O’s in slots {3, 5, 7} or __ __ O__O__O - Ns could be in the first two slots. There are 2!
= 2 words like this. So, number of words that we need to count = 12 – 2 = 10
Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.
Phew!.
There is a far more elegant method for accounting for the words where the 2 N’s appear
together. This one came from Mukund Sukumar.
We need to account for the number of possibilities of N,N being together. So to subtract
that part, consider 'NN' being together as one letter and place O's. In 3 out of the 4 slots in
_M_NN_S_
The O’s can be selected in 4
C3 ways. The MNNS can be rearranged in 3! Ways if the N’s
have to appear together.

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Or, we get 4
C3 * 3! = 24 ways. So, we have 120-24 = 96 ways totally.
Questions
x^2 - 17x + |p| = 0 has integer solutions. How many values can p take?
Explanation
To begin with, if we assume roots to be a and b, sum of the roots is 17 and product of the
roots is |p|. Product of the roots is positive and so is the sum of the two roots.
So, both roots need to be positive.
So, we are effectively solving for
Number of positive integer solutions for a + b = 17.
We could have (1,16), (2, 15), (3, 14)......(8, 9) - There are 8 sets of pairs of roots. Each of
these will yield a different product.
So, |p| can take 8 different values. Or, p can take 16 different values.
is that it? Or, are we missing something? Can p be zero? What are the roots of x^2 - 17x
= 0. This equation also has integer solutions.
So, p can also be zero.
So, number of possible values of p = 16 + 1 = 17.
Wonderful question - chiefly because there are two really good wrong answers one can
get. 8 and 16. So, pay attention to detail. No point telling yourself "Just missed, I just
didnt think of zero. I deserve this mark". Being just wrong, will earn us a -1 instead of the
honourable 0.
Question
How many integer solutions exist for the equation x2
- 8|x| - 48 =0?

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Explanation
One approach is to solve when x > 0 and then solve for x < 0. However, there is a slightly
simpler method.
Note that x2
is the same as |x|2
, so we can treat the equation as a quadratic in |x|.
Or, |x|2
– 8|x| - 48 = 0
(|x| - 12|) (|x| + 4) = 0
|x| could be 12. |x| cannot be -4.
If |x| could be 12, x can be -12 and 12.
Question
There are n pipes that fill a tank. Pipe 1 can fill the tank in 2 hours, Pipe 2 in 3 hours,
Pipe 3 in 4 hours and so on. Pipe 1 is kept open for 1 hour, pipe 2 for 1 hour, then pipe 3
and so on. In how many hours will the tank get filled completely?
Explanation
Almost all questions can be approached well by asking the question "What happens in 1
hour" (or 1 day, or 1 minutes)
So, let us start with that
In 1 hour, pipe 1 fills 1/2 of the tank. So, in the first hour, the tank will not be filled
In 1 hour, pipe 2 fills 1/3 of the tank. So, in two hours we would have filled 1/2 + 1/3 of
the tank, or 5/6 of the tank. So, by the end of the second hour, the tank would still not be
filled.
Let us move to the third hour. In 1 hour, pipe 3 fills 1/4 of the tank. So, by the end of the
third hour, we should have filled 5/6 + 1/4 = 13/12 of the tank.
Oops, one cannot fill 13/12 of a tank. What this tells us is that the tank gets filled in the
3rd hour.
When exactly during the third hour?

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At the beginning of the third hour, we still have 1/6 of the tank still to fill. Pipe 3 can fill
at the rate of 1/4 of the tank per hour. Or, pipe 3 will take 2/3 hours to fill the tank.
Or, the tank will be filled in 2 hours and 40 minutes.
This is an example of a sequence that is a harmonic progression. The formulae for HP are
needlessly confusing. So, simple step-by-step approach works best.
Question
How many functions can be defined from Set A -- {1, 2, 3, 4} to Set B = {a, b, c, d} that
are neither one-one nor onto?
Explanation
To start with, if you do not know the meaning of one-one or onto, look these up. For
good measure know the meanings of the terms surjective, injective etc also. CAT tests
these terms.
Let us start by answering a far simpler question. How many functions are possible from
Set A to Set B. This is equal to 4^4 = 256.
Note that any function from Set A to Set B that is one-one will also be onto and vice
versa. How? Why? - Think about this. Remember, tutors can only take the horse to the
pond. :-)
So, we need to subtract only those functions that are one-one AND onto. Or, effectively
we have to eliminate those functions where {1, 2, 3, 4} are mapped to {a, b, c, d} such
that each is mapped to a different element. This is effectively same as the number of
ways of rearranging 4 elements. Or, number of ways of doing this is 4! .
So, total number of functions that are neither one-one nor onto = 256 - 24 = 232.
Question
If we list all the words that can be formed by rearranging the letters of the word
SLEEPLESS in alphabetical order, what would be the rank of SLEEPLESS?

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Explanation
First let us think about the number of possible rearrangements. SLEEPLESS can be
rearranged in 9!/ (2! 3! 3!) = 5040.
Now, if rearrange these, we would have words starting with E, L, P and S.
Now, SLEEPLESS starts with S. So, let us think about how many words start with S.
Number of words starting with S = 8!/ (2! 2! 3!) = 1680.
So, there would have been 5040 - 1680 words that have gone by before the first word
starting with S. Or, 3360 words start with E, L or P.
The first word with S is SEEELLPSS. This would have a rank of 3361.
Now, let us go step by step.
Words starting with SE____ - Number of such words = 7!/(2!*2!*2!) = 630 words
Next we have words starting with SL, but our word also starts with SL, so let us go
deeper
Words starting with SLE would be the next step, but our word starts with SLE as
well.SO, let us go one further step deeper
Words starting with SLEE. The first such word would be SLEEEPLSS
So, let us think about words starting with SLEEE - there would be 4!/(2!) words like this
= 12 words like this
words starting with SLEEL - there would be 4!/(2!) words like this = 12 words like this
So, we have accounted for 3360 + 630 + 12 + 12 = 4014 words thus far.
Now, on to words starting with SLEEP - there are again 4!/(2!) words like this = 12
words like this
Our word is within these 12 words

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Words starting with SLEEPE - there are 3!/2! words like this, or 3 words like this
Our word comes in the next bunch.
Words starting with SLEEPL - SLEEPLESS is the first such word. Or, the rank fo
SLEEPLESS = 4014 + 3 + 1 = 4018.
A very good question to get lots of practice on letters rearrangement. However, it is
unlikely that you will face a question this time-consuming in CAT. Wonderful question
to practice though.
Question
A 3-digit Armstrong number is a three-digit number where the number is equal to the
sum of the cubes of the three digits. Give a few Armstrong numbers.
Explanation
To start with, be clear that CAT does not ask questions like these. You do not need to
know what Armstrong number are; nor do you need to know how to get these to crack
these exam. This is just a fun question to do some trial and error with.
The 3-digit Armstrong numbers are 153, 370, 371 and 407, obtained mostly by trial and
error (although a lil scientifically)
Question
How many trailing zeroes will be present in the base 12 representation of 55!?
Explanation
The question can be restated as follows - "What is the highest power of 12 that divides
55!"
Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.

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So, in order to find the highest power of 12 that divides 55!, we need to look at the
highest powers of 2 and 3 that divide 55!.
Successive division by 2 will give us the highest power of 2 that divides 55!:
55/2 = 27, 27/2 = 13, 13/2 = 6, 6/2 = 3, 3/2 = 1.
Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 = 50
Successive division by 3 will give us the highest power of 3 that divides 55!:
55/3 = 18, 18/3 = 6, 6/3 = 2.
Highest power of 3 that divides 55! = 18 + 6 + 2 = 26.
So, 55! is a multiple of 2^50 and 3^26. What is the highest power of 12 that will divide
55!. Or, what is the highest value n can take such that (2^2 * 3)^n is a factor of 55!
We haev 26 threes's; but we can accommodate 2^2 only 25 times. Or, 12^25 will be a
factor of 55!, while 12^26 will not, since we need 52 2's for 12^26. We have only 50 2's.
Thus 12^25 will divide 55! - there are 25 trailing zeros in base 12 representation of 55!.
Question
A bus service runs from Chennai to Blr every third day with the first bus starting on Jan
1st. Another bus service runs from Chennai to Mumbai every 5th day starting from Jan
2nd. A third bus service from Chennai to Cochin runs on every 7th day starting from Jan
4th. In that year (which is not a leap year), on how many different days will a bus run
from Chennai to all three cities?
Explanation
Let us call Jan 1st as day 1, jan 2nd as day 2 and so on.
So,
The Chennai-Blr services runs on days 1, 4, 7, 10, 13, 16, 19,......

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The Chennai-Mumbai services runs on days 2, 7, 12, 17, 22, 27,......
The Chennai-Cochin services runs on days 4, 11, 18, 25, ......
First up, let us see if we can find one day where all three buses ply.
Chennai Blr runs on days of the form 3a + 1, Chennai_Mumbai on 5b + 2, and Chennai-
Cochin on 7c + 4. If N can be written as 3a + 1, 5b + 2 and 7c + 4, then we should be able
to get n as 105d + ___ . (105 is the LCM of 3, 5 and 7)
This is a very important idea. Get lots of practice on this idea. Wrap your head around
this idea very clearly.
First let us combine 3a + 1 and 5b + 2.
A number of the form 3a + 1, on division by 15 can have one of the following remainders
{ 1, 4, 7, 10, 13}
A number of the form 5b + 2, on division by 15 can have one of the following remainders
{ 2, 7, 12}
Or, if a number is of the form 3a + 1 and 5b + 2, it has to be of the form 15k + 7.
Now, N = 15K + 7 and 7c + 4.
Number 15k + 7, on division by 105 can have one of the following remainders { 7, 22,
37, 52, 67, 82, 97}}
Number 7c + 4, on division by 105 can have one of the following remainders { 4, 11, 18,
25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95, 102, 109}
Or, the number is of the form 105d + 67.
So, all three services will run on days 67, 67 + 105, 67 + 105*2 and so on.
Or, on days 67, 172 and 277. (Any other day would go beyond 365)
So, all three services would run together 3 days of the year.
Wonderful question testing concepts of LCM and Remainders.

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Question
How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x -
12).......(x - 100) < 0?
Answer: 30
Explanation
Let us try out a few values to see if that gives us anything.
When x = 8, 10, 12, ....100 this goes to zero. So, these cannot be counted.
When x = 101, 102 or beyond, all the terms are positive, so the product will be positive.
So, straight-away we are down to numbers 1, 2, 3, ...7 and then odd numbers from there
to 99.
Let us substitute x =1,
All the individual terms are negative. There are totally 47 terms in this list (How? Figure
that out). Product of 47 negative terms will be negative. So, x = 1 works. So, will x =2, 3,
4, 5, 6, and 7.
Remember, product of an odd number of negative terms is negative; product of even
number of negative terms is positive. Now, this idea sets up the rest of the question.
When x = 9, there is one positive terms and 46 negative terms. So, the product will be
positive.
When x = 11, there are two positive terms and 45 negative terms. So, the product will be
negative.
When x = 13, there are three positive terms and 44 negative terms. So, the product will be
positive.
and so on.
Essentially, alternate odd numbers need to be counted, starting from 11.
So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7...and then 11, 15,

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19, 23, 27, 31,..... and so.
What will be the last term on this list?
99, because when x = 99, there are 46 positive terms and 1 negative term.
So, we need to figure out how many terms are there in the list 11, 15, 19,....99. These can
be written as
4 * 2 + 3,
4 * 3 + 3,
4 * 4 + 3
4 * 5 + 3
4 * 6 + 3
...
4 * 24 + 3
A set of 23 terms. So, total number of values = 23 + 7 = 30. 30 positive integer values of
x exist satisfying the condition.
Question:
A merchant can buy goods at the rate of Rs. 20 per good. The particular good is part of an
overall collection and the value is linked to the number of items that are already on the
market. So, the merchant sells the first good for Rs. 2, second one for Rs. 4, third for Rs.
6…and so on. If he wants to make an overall profit of at least 40%, what is the minimum
number of goods he should sell?
A. 24
B. 18
C. 27
D. 32
Correct Answer: (C)
Explanation:
Let us assume he buys n goods.
Total CP = 20n
Total SP = 2 + 4 + 6 + 8 ….n terms
Total SP should be at least 40% more than total CP
2 + 4 + 6 + 8 ….n terms > 1.4 * 20 n
2 (1 + 2 + 3 + ….n terms) > 28n

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n(n + 1) > 28n
n2
+ n > 28n
n2
- 27n > 0
n > 27
He should sell a minimum of 27 goods.
Answer Choice (C)
Question:
P is x% more than Q. Q is (x - 10)% less than R. If P > R, what is the range of values x
can take?
A. 10% to 28%
B. 10% to 25%
C. 10% to 37%
D. 10% to 43%
Correct Answer: (C)
Explanation:
P = Q
Q = R
R =
P > R
(100 + x) (110 – x) > 100 x 100
11,000 + 110x – 100x – x2
> 10000

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1000 + 10x – x2
> 0
x2
– 10x – 1000 < 0
x2
– 10x + 25 < 1000 + 25
(x – 5)2
< 1025
x – 5 < 32
x < 37
x could range from 10% to 37%
Answer Choice (C)
Question:
How many numbers with distinct digits are possible product of whose digits is 28?
A. 6
B. 4
C. 8
D. 12
Correct Answer: (C)
Explanation:
Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74
Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.
We cannot have three digits as (2, 2, 7) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.
Answer Choice (C)

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Question:
If log2X + log4X = log0.25and x > 0, then x is
A. 6-1/6
B. 61/6
C. 3-1/3
D. 61/3
Correct Answer: (A)
Explanation:
log2x + log4x = log0.25
log2x + = log0.25
log2x * = log0.25
log2x * 3 = 2log0.25
log2x3
= log0.256
log2x3
= -log46
log2x3
=
log2x3
=
2log2x3
= -log26
2log2x3
+ log26 = 0
log26x6
= 0
6x6
= 1
x6
=
x = Choice (A).

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Question:
3x + 4|y| = 33. How many integer values of (x, y) are possible?
A. 6
B. 3
C. 4
D. More than 6
Correct Answer: (D)
Solution:
Let us rearrange the equation:
3x = 33 – 4|y|
Since x and y are integers, and since |y| is always positive regardless of the sign of y, this
means that when you subtract a multiple of 4 from 33, you will get a multiple of 3.
Since 33 is already a multiple of 3, in order to obtain another multiple of 3, you will have
to subtract a multiple of 3 from it. So, y has to be a positive or a negative multiple of 3.
y = 3, -3, 6, -6, 9, -9, 12, -12...etc.
For every value of y, x will have a corresponding integer value.
So there are infinite integer values possible for x and y.
Question:
3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can
to take?
A. 5
B. -5
C. 4
D. 3
Correct Answer: (A)
Explanation:
3sinx + 4cosx ≥ -r
≥ -r
= cosA => sinA =
5(sinx cosA + sinA cosA) ≥ -r
5(sin(x + A) ≥ -r

22
5sin (x + A) ≥ -r
-1 < sin (angle) < 1
5sin (x + A) ≥ -5
rmin = 5
Question:
Solve the inequality x3
– 5x2
+ 8x – 4 > 0.
A. (2, )
B. (1, 2) U (2, )
C. (-, 1) U (2, )
D. (-, 1)
Correct Answer : (B)
Explanation:
Let a, b, c be the roots of this cubic equation
a + b + c = 5
ab + bc + ca = 8
abc = 4
This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic
equations}
The other approach is to use polynomial remainder theorem
If you notice, sum of the coefficients = 0
=> P(1) = 0
=> (x - 1) is a factor of the equation. Once we find one factor, we can find the other two
by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic
equation.
(x - 1) (x - 2) (x - 2) > 0
Let us call the product (x - 1)(x - 2)(x - 2) as a black box.
If x is less than 1, the black box is a –ve number
If x is between 1 and 2, the black box is a +ve number
If x is greater than 2, the black box is a +ve number

23
Since we are searching for the regions where black box is a +ve number, the solution is as
follows:
1 < x < 2 OR x > 2
Answer Choice (B)
Question:
An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has
Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the
overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy.
What is the range of values x can take?
A. 30% < x < 40%
B. 32.5% < x < 42%
C. 33.33% < x < 40%
D. 32.5% < x < 40%
Correct Answer : (A)
Explanation:
Alloy 1 Copper Aluminium
2x 3x
Alloy 2 Copper Zinc
2y 7y
If Aluminium and Zinc have to be equal, 3x = 7y. The simplest case occurs at the LCM,
when both Aluminium and Zinc are 21kg.
Alloy 1 Copper Aluminium
14 21
Alloy 2 Copper Zinc
6 21
Even a gram of alloy 1 above this level would mean that there is more Aluminium than
Zinc in the total alloy. So, this table is a limit on the percentages of different metals.
The percentage of Copper in the total alloy would be (14 + 6)* = 32.25%

24
On the other hand, if we want more Aluminium than Zinc, we can use tons and tons of
alloy 1, and only a microgram of Alloy 2. The quantity of alloy 2 can be made so small
that its presence can be almost neglected, as it will have as good as zero impact on the
overall percentage of copper. This is another extreme case, in which Copper will have
40% weight in the alloy.
Thus, the percentage of Copper ranges from 32.25% to 40% in this alloy.
Answer Choice (A)
Probability Question
Question:
[x] = greatest integer less than equal to x. If x lies between 3 and 5. What is the
probability than [x2
] = [x]2
?
A. Roughly 0.64
B. Roughly 0.5
C. Roughly 0.14
D. Roughly 0.36
Correct Answer: (C)
Explanation:
Let us take a few examples.
[32
] = [3]2
[3.52
] =12 [3.5]2
= 9
[42
] = 16 [4]2
= 16
For x  (3, 5). [x]2
can only take value 9, 16 and 25.
Let us see when [x2
] will be 9, 16 or 25
If [x2
] = 9, x2
[9, 10)
=> x[3, )
[x2
] = 16 => x2
[16, 17)
=> x [4, )
In the given range [x2
] = 25 only when x = 5
So [x2
] = [x]2
when x [3, ] or [4, ) or 5.
Probability =
= = 0.14
Questions and solutions on Probability

25
Question:
John looks at the clock at some arbitrary time between 4pm and 6 pm and notices that the
angle between the two hands is an acute angle. He looks at the clock exactly 15 minutes
later, what is the probability that during this observation also he sees that the angle
between the hour and minute hands to be an acute-angle?
A. 1/2
B. 6/11
C. 13/24
D. 11/19
Correct Answer : (C)
Explanation:
At 4:00 pm the angle between the hands is 1200. The minutes hand is 1200 behind the
hour hand. Now, let us first see when the angle will become an acute angle. The angle
between he hands will be acute when the minutes hand gains more than 300. Or, after
30/330 * 60 minutes.
Somewhere between 4:05 and 4: 10.From now, the angle will be acute till the minute
hand reaches a point where it is 90 degree ahead of the hour hand. Or, till the time the
minute and gains 1800 (Going from 900 degree behind to 900 ahead of hour hand), the
angle will be acute. Or, for a spell of 180/330 * 60 minutes, the angle will be acute. Or
for a spell of 360/11 minutes, the angle will be acute. For a spell of 32 8/11 minutes the
angle will be acute.
So, if John had seen the clock at a time from 4 pm to 5pm, he would have seen it during
this spell. Now, if his second viewing had also happened within this spell, he would have
seen an acute angle second time.
Or, t, t + 15 should have both been within this 32 8/11 range for him to have seen two
acute angle observations. Or, t should have happened at least 15 minutes before the angle
become obtuse again. If we have the time range for it being acute as (x, x + 32 8/11), The
question can be restated as if t belongs to this range, what it is the probability that t + 15
also belongs to this range.
For t + 15 to be within the range, t + 15 < x + 32 8/11, or t < 32 8/11 – 15 = 17 8/11
Probability should be 195/11 divided by 360/11 = 195/360 = 39/72 = 13/24

26
In the 5pm to 6pm range we get an identical probability. So the overall probability is
13/24
Question:
John was born on Feb 29th
of 2012 which happened to be a Wednesday. If he lives to be
101 years old, how many birthdays would he celebrate on a Wednesday?
A. 3
B. 4
C. 5
D. 1
Correct Answer : (B)
Explanatory Answer:
Let us do this iteratively. Feb 29th
2012 = Wednesday => Feb 28th
2012 = Tuesday
Feb 28th
2013 = Thursday (because 2012 is a leap year, there will be 2 odd days)
Feb 28th
2014 = Friday, Feb 28th
2015 = Saturday, Feb 28th
2016 = Sunday, Feb 29th
2016 = Monday
Or, Feb 29th
to Feb 29th
after 4 years, we have 5 odd days
So, every subsequent birthday, would come after 5 odd days.
2016 birthday – 5 odd days
2020 birthday – 10 odd days = 3 odd days
2024 birthday – 8 odd days = 1 odd day
2028 – 6 odd days
2032 – 11 odd days = 4 odd days
2036 – 9 odd days = 2 odd days
2040 – 7 odd days = 0 odd days. So, after 28 years he would have a birthday on
Wednesday
The next birthday on Wednesday would be on 2068 (further 28 years later), the one after
that would be on 2096. His 84th
birthday would again be a leap year.
Now, there is a twist again, as 2100 is not a leap year. So, he does not have a birthday in
2100. His next birthday in 2104 would be after 9 odd days since 2096, or 2 odd days
since 2096, or on a Thursday.
From now on the same pattern continues. 2108 would be 2 + 5 odd days later = 7 odd
days later. Or, 2108 Feb 29th
would be a Wednesday.
So, there are 4 occurrences of birthday falling on Wednesday – 2040, 2068 and 2096,
2108.

27
Question
What is the remainder when (13100
+1 7100
) is divided by 25?
A. 2
B. 0
C. 15
D. 8
Correct Answer: Choice (A)
Explanation:
What is the remainder when (13100
+1 7100
) is divided by 25?
(13100
+1 7100
) = (15 – 2)100
+ (15 + 2)100
Now 52
= 25, So, any term that has 52
or any higher power of 5 will be a multiple of 25.
So, for the above question, for computing remainder, we need to think about only the
terms with 150
or 151
.
(15 – 2)100
+ (15 + 2)100
Coefficient of 150
= (-2)100
+ 2100
Coefficient of 151
= 100
C1 * 151
* (-2)99
+ 100
C1 * 151
* (-2)99
. These two terms cancel each
other.So, the sum is 0.
Remainder is nothing but (-2)100
+ 2100
= (2)100
+ 2100
2101
Remainder of dividing 21
by 25 = 2
by 25 = 4
by 25 = 8
by 25 = 16
by 25 = 32 = 7
by 25 = 72
= 49 = -1
by 25 = (-1)2
= 1
by 25 = Remainder of dividing 2100
by 25 * Remainder of
dividing 21
by 25 = 1 * 2 = 2
Question
A drain pipe can drain a tank in 12 hours, and a fill pipe can fill the same tank in 6 hours.
A total of n pipes – which include a few fill pipes and the remaining drain pipes – can fill
the entire tank in 2 hours. How many of the following values could ‘n’ take?
a) 24
b) 16

28
c) 33
d) 13
e) 9
f) 8
A. 3
B. 4
C. 2
D. 1
Correct Answer: (A)
Explanatory Answer:
Two drain pipes can drain the same volume that one fill pipe fills. This means that a D-D-
F combination has to have a net volume effect of 0.
In spite of this, the tank still gets filled. Only the fill pipes can manage to fill the tank. In
addition to all the net zero effect pipes, we need three more fill pipes in order to fill the
tank in 2 hours.
So, we can have as many D-D-Fs as we want, but we need one F-F-F at the end to ensure
that the tank gets filled in 2 hours.
So the number of pipes will be → (D – D - F).......(D – D - F) + (F – F - F)
The number of pipes has to be a multiple of 3. Only options A, C and E fit the
description.
Answer Choice (A)
Question

29
Traders A and B buy two goods for Rs. 1000 and Rs. 2000 respectively. Trader A marks
his goods up by x%, while trader B marks his goods up by 2x% and offers a discount of
x%. If both make the same profit, find x
A. 25%
B. 12.5%
C. 37.5%
D. 40%
Answer Choice: (A)
Explanatory Answer:
SP of trader A = 1000 (1 + x)
Profit of trader A = 1000 (1 + x) - 1000
MP of trader B = 2000 (1 + 2x)
SP of trader B = 2000 (1 + 2x) (1 - x)
Profit of trader B = 2000(1 + 2x) (1 - x)- 2000
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 - x)- 2000
1000x = 2000 – 4000x2
+ 4000x – 2000x - 2000
4000x2
-1000x = 0
1000x (4x - 1) = 0
x = 25%
Answer Choice (A)
Question
(|x| -3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?
A. 4

30
B. 10
C. 6
D. 8
Answer: Choice (B)
Explanatory Answer
Product of two integers is 12. Further |y| + 4 > 4 . So, we know that one of the numbers
has to be greater than 4.
So, we can have
1 x 12 => x = +4, y = +8
2 x 6 => x = +5, y = +2
3 x 4 => x = +6, y = 0
4 + 4 + 2 = 10 possible solutions
Question
How many of the following statements have to be true?
I. No year can have 5 Sundays in the month of May and 5 Thursdays in the
month of June
II. If Feb 14th
of a certain year is a Friday, May 14th
of the same year cannot be
a Thursday
III. If a year has 53 Sundays, it can have 5 Mondays in the month of May
A. 0
B. 1
C. 2
D. 3

31
Answer: Choice (B)
Explanation
I. No year can have 5 Sundays in the month of May and 5 Thursdays in the
month of June
A year has 5 Sundays in the month of May => it can have 5 each of Sundays, Mondays
and Tuesdays, or 5 each of Saturdays, Sundays and Mondays, or 5 each of Fridays,
Saturdays and Sundays. Or, the last day of the Month can be Sunday, Monday or
Tuesday.
Or, the 1st
of June could be Monday, Tuesday or Wednesday. If the first of June were a
Wednesday, June would have 5 Wednesdays and 5 Thursdays. So, statement I need not
be true.
II. If Feb 14th
of a certain year is a Friday, May 14th
of the same year cannot
be a Thursday
From Feb 14 to Mar 14, there are 28 or 29 days, 0 or 1 odd day
Mar 14 to Apr 14, there are 31 days, or 3 odd datys
Apr 14 to May 14 there are 30 days or 2 odd days
So, Feb 14 to May 14, there are either 5 or 6 odd days
So, if Feb 14 is Friday, May 14 can be either Thursday or Wednesday. So, statement 2
need not be true.
III. If a year has 53 Sundays, it can have 5 Mondays in the month of May
Year has 53 Sundays => It is either a non-leap year that starts on Sunday, or leap year
that starts on Sunday or Saturday.
Non-leap year starting on Sunday: Jan 1st
= Sunday, jan 29th
= Sunday. Feb 5th
is Sunday.
Mar 5th
is Sunday, Mar 26 is Sunday. Apr 2nd
is Sunday. Apr 30th
is Sunday, May 1st
is
Monday. May will have 5 Mondays.
So, statement C can be true.
Only one of the three statements needs to be true. Answer Choice (B)
Question
x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.
A. 2

32
B. 3
C. 1
D. More than one possible value of z exists
Answer: Choice B
Explanatory Answer:
xy = 4
xy could be 2 x 2 or 4 x 1
221
222 These are the possible triangles
223
441
22x will be a triangle if x is 1, 2 or 3 (trial and error)
44x is a triangle only if x is 1.
221 is acute. 12
+ 22
> 22
222 is equilateral. So acute.
223 is obtuse. 22
+ 22
< 32
144 is acute. 12
+ 42
> 42
Only triangle 223 is obtuse. Hence, the third side has to be 3.
Answer Choice (B)
Question

33
Sides of a triangle are 6, 10 and x for what value of x is the area of the  the
maximum?
A. 8 cms
B. 9 cms
C. 12 cms
D. None of these
Correct Answer: D
Explanatory Answer
Side of a triangle are. 6, 10, x.
Area = 1/2 x 6 x 10 sin BAC.
Area is maximum, when BAC = 90o
x = sqrt(100 + 36) = sqrt(136)
There is a more algebraic method using hero’s formula. Try that also.
Answer Choice (D)
Question
Consider a class of 40 students whose average weight is 40 kgs. m new students join this
class whose average weight is n kgs. If it is known that m + n = 50, what is the maximum
possible average weight of the class now?
A. 40.18 kgs
B. 40.56 kgs
C. 40.67 kgs

34
D. 40.49 kgs
Correct Answer: Choice (B)
Explanation:
If the overall average weight has to increase after the new people are added, the average
weight of the new entrants has to be higher than 40.
So, n > 40
Consequently, m has to be <10 (as n + m = 50)
Working with the “differences” approach, we know that the total additional weight added
by “m” students would be (n - 40) each, above the already existing average of 40. m(n -
40) is the total extra additional weight added, which is shared amongst 40+m students.
So, m (n – 40)/ (m + 40) has to be maximum for the overall average to be maximum.
At this point, use the trial and error approach (or else, go with the answer options) to
arrive at the answer.
The maximum average occurs when m = 5,and n = 45
And the average is 40 + (45 – 5) * 5/45 = 40 + 5/9 = 40.56
Answer Choice (B)
Question
1. 5. Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90
and 91-100. The number of students who have scored in each range is given below
51 -60 - 3 students
61 -70 - 8 students
71 -80 - 7 students
81 -90 - 4 students
91 -100 - 3 students

35
Furthermore, we know that at least as many students scored 76 or more as those who
scored below 75. What is the minimum possible average overall of this class?
A. 72
B. 71.2
C. 70.6
D. 69.2
Answer: Choice (C)
Explanatory Answer:
Let's employ the idea of a total of 25 students (all of the same weight) sitting on a see-
saw, which has numbers from 51 to 100 marked on it. At least as many students are
sitting on 76 (or to its right), as there are sitting to the left of 75. Now this means that you
can have only one person sitting to the left of 75 and all the rest sitting beyond 76. But
you can't do that, as you have other constraints as well.
First of all, you have to seat 3 students from 51 to 60, and 8 students from 61 to 70.
Secondly, you also have to make sure that the average is the least. This means that the
see-saw should be tilting as much to the left as possible, which in turn means that the
number of people sitting to the left of 75 should be the highest possible.
This makes it 12 students to the left of 75, and the remaining 13 students on 76 or to its
right.
Next, how do you ensure that the average is least, i.e. how do you ensure that the balance
tilts as much as possible to the left? Make each student score as little as possible given
the constraints.

36
So, the first 3 students only score 51 each. The next 8 students score only 61 each. 11
Students are now fixed. The 12th
student has to be below 75, so seat him on 71. The
remaining 6 students (who are in the 71 to 80 range) have to score 76. The next 4 score
81 and the next 3 score 91.
This would give you the least average.
The lowest possible average would be:
= [(3 * 51) + (8 * 61) + 71 + (6 * 76) + (4* 81) + (3 * 91)]/25
= 70.6
Answer Choice (C)
Question
How many onto functions can be defined from the set A = {1, 2, 3, 4} to {a, b, c}?
A. 81
B. 79
C. 36
D. 45
Answer: Choice (C)
Explanatory Answer:
First let us think of the number of potential functions possible. Each element in A has
three options in the co-domain. So, the number of possible functions = 34
= 81.

37
Now, within these, let us think about functions that are not onto. These can be under two
scenarios
Scenario 1: Elements in A being mapped on to exactly two of the elements in B (There
will be one element in the co-domain without a pre-image).
Let us assume that elements are mapped into A and B. Number of ways in which this can
be done = 24
– 2 = 14
o 24
because the number of options for each element is 2. Each can be mapped on to either
A or B
o -2 because these 24
selections would include the possibility that all elements are mapped
on to A or all elements being mapped on to B. These two need to be deducted
The elements could be mapped on B & C only or C & A only. So, total number of
possible outcomes = 14 * 3 = 42.
Scenario 2: Elements in A being mapped to exactly one of the elements in B. (Two
elements in B without pre-image). There are three possible functions under this scenario.
All elements mapped to a, or all elements mapped to b or all elements mapped to c.
Total number of onto functions = Total number of functions – Number of functions
where one element from the co-doamin remains without a pre-image - Number of
functions where 2 elements from the co-doamin remain without a pre-image
Total number of onto functions = 81 – 42 – 3 = 81 – 45 = 36
Answer Choice (C)

38
Question
1. x + |y| = 8, |x| + y = 6.How many pairs of x, y satisfy these two equations?
A. 2
B. 4
C. 0
D. 1
Answer: Choice (D)
Explanatory Answer
We start with the knowledge that the modulus of a number can never be negative, though
the number itself may be negative.
The first equation is a pair of lines defined by the equations
y = 8 – x ------- (i) {when y is positive}
y = x – 8 ------- (ii) {when y is negative}
With the condition that x ≤ 8 (because if x becomes more than 8, |y| will be forced to be
negative, which is not allowed)
The second equation is a pair of lines defined by the equations:
y = 6 – x ------- (iii) {when x is positive}
y = 6 + x ------- (iv) {when x is negative}
with the condition that y cannot be greater than 6, because if y > 6, |x| will have to be
negative.
On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and
(iv) are parallel (same slope).

39
Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv)
holds good only when x is negative
Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1. This
satisfies both equations.
Only one solution is possible for this system of equations.
Question
Qn. f(x + y) = f(x)f(y) for all x, y, f(4) = +3 what is f(-8)?
A. 1/3
B. 1/9
C. 9
D. 6
Answer Choice: B
Explanatory Answer
f(x + 0) = f(x) f(0)
f(0) = 1
f(4 + -4) = f(0)
f(4 + -4) = f(4) f(-4)
1 = +3 x f(-4)
f(-4) = 1/3
f(-8) = f(-4 + (-4)) = f(-4) f(-4)
f(-8) = 1/3 x 1/3 = 1/9
Answer Choice (B)

40
Question
Second term of a GP is 1000 and the common ratio is where n is a natural number. Pn is
the product of n terms of this GP. P6 > P5 and P6 > P7, what is the sum of all possible
values of n?
A. 5
B. 9
C. 8
D. 11
Correct Answer
Choice B
Explanatory Answer
Common ratio is positive, and one of the terms is positive => All terms are positive
P6 = P5 * t6 => If P6 > P5, t6 > 1
P7 = P6 * t7 => If P6 > P7, t7 < 1
T6 = t2 * r4
= 1000r4
;
T7 = t2 * r5
= 1000r5
1000r4
> 1 and 1000r5
< 1
1/r^4 < 1000 and 1/r^5 > 1000
1/r = n
n4
< 1000 and n5
> 1000, where n is a natural number
n4
< 1000 => n < 6
n5
> 1000 => n > 4
n could be 4 or 5. Sum of possible values = 9
Answer Choice (B)
Question

41
If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically,
what would be the rank of “ACDEF’?
A. 4716
B. 4720
C. 4718
D. 4717
Correct Answer
Choice C
Explanatory Answer
The first word would be ABCDE. With 2 distinct vowels, 3 distinct consonants, this is
the first word we can come up with.
Starting with AB, we can have a number of words
AB __ __ __. The next three slots should have 2 consonants and one vowel. This can be
selected in 20
C2 and 4
C1 ways. Then the three distinct letters can be rearranged in 3!
Ways.
Or, number of words starting with AB = 20
C2 * 4
C1 * 3! = 190 * 4 * 6 = 4560
Next, we move on to words starting with ACB
ACB __ __. The last two slots have to be filled with one vowel and one consonant. = 19
C1
* 4
C1. This can be rearranged in 2! Ways.
Or, number of words starting with ACB = 19
C1 * 4
C1 * 2 = 19 * 4 * 2 = 152
Next we move on words starting with ACDB: There are 4 different words on this list –
ACDBE, ACDBI, ACDBO, ACDBU
So, far number of words gone = 4560 + 152 + 4 = 4716
Starting with
AB 4560
Starting with
ACB 152
Starting with 4

42
ACDB
Total words
gone 4716
After this we move to words starting with ACDE, the first possible word is ACDEB.
After this we have ACDEF.
So, rank of ACDEF = 4718
Answer Choice (C)
Question
If we listed all numbers from 100 to 10,000, how many times would the digit 3 be
printed?
A. 3980
B. 3700
C. 3840
D. 3780
Correct Answer
Choice A
Explanatory Answer
We need to consider all three digit and all 4-digit numbers.
Three-digit numbers: A B C. 3 can be printed in the 100’s place or10’s place or units
place.
Ø 100’s place: 3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed
in the 100’s place 100 times
Ø 10’s place: A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in
the 10’s place 90 times
Ø Unit’s place: A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed
in the unit’s place 90 times

43
So, 3 gets printed 280 times in 3-digit numbers
Four-digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or10’s
place or units place.
Ø 1000’s place: 3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take
values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.
Ø 100’s place: A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets
printed in the 100’s place 900 times.
Ø 10’s place: A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets
printed in the 10’s place 900 times.
Ø Unit’s place: A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets
printed in the unit’s place 900 times.
3 gets printed 3700 times in 4-digit numbers.
So, there are totally 3700 + 280 = 3980 numbers
Answer Choice (A)
Question
A is x% more than B and is x% of sum of A and B. What is the value of x? Give
approximate answer.
Correct Answer
Roughly 62%
Explanatory Answer
a = b (1 + x) => a/b = 1 + x
a = x (a + b) , dividing by a through out
1 = x (1 + b/a)
1 = x (1 + 1/(1 + x))
1 = x (x + 2) / (x + 1)
x + 1 = x2
+ 2x

44
=> x2
+ x - 1 = 0
Now, we need to solve this equation. Using the discriminant method, when we solve this,
x turns out to be {-1 + sqrt (5)}/2
x has to lie between 0 and 1 and therefor cannot be { -1 - sqrt (5) }/2.
So, the only solution is { -1 + sqrt (5) }/2. This is roughly 0.62.
Or, x has to be 62% approximately. The ration 1.618 is also called the golden ratio, and is
the conjugate and reciprocal of 0.618.
The golden ratio finds many mentions, from the Fibonacci series to Da Vinci. So, it is a
big favourite of mathematician.
Question
The sum of the factors of a number is 124. What is the number?
Question
The sum of the factors of a number is 124. What is the number?
Correct Answer
Number could be 48 or 75
Explanatory Answer
This video gives the solution for this question. Given below the video is the explanation
(in words)
Any number of the form pa
qb
rc
will have (a+1) (b+1)(c+1) factors, where p, q, r are
prime. (This is a very important idea)
For any number N of the form pa
qb
rc
, the sum of the factors will be (1 + p1
+ p2
+ p3
+ …+
pa
) (1 + q1
+ q2
+ q3
+ …+ qb
) (1 + r1
+ r2
+ r3
+ …+ rc
).
Sum of factors of number N is 124. 124 can be factorized as 22
* 31. It can be written as
4 * 31, or 2 * 62 or 1 * 124.
2 cannot be written as (1 + p1
+ p2
…pa
) for any value of p.
4 can be written as (1 + 3)
So, we need to see if 31 can be written in that form.
The interesting bit here is that 31 can be written in two different ways
31 = (1 + 21
+ 22
+ 23
+ 24
)

45
31 = ( 1 + 5 + 52
)
Or, the number N can be 3 * 24
or 3 * 52
. Or N can be 48 or 75.
Question
Consider functions f(x) = x2
+ 2x, g(x) = √(x +1) and h(x) = g(f(x)). What are the
domain and range of h(x)?
Correct Answers:
Domain = ( - infinity, +infinity)
Range - [0, infinity]
Explanatory Answer
h(x) = g(f(x)) = g(x2
+ 2x) = ( x2
+ 2x + 1) = (x+1)2
= | x + 1|
This bit is very important, and often overlooked
sqrt(9) = 3, not +3
If x2
= 9, then x can be +3, but 9) is only +3.
So, x2
)= |x|, not +x, not + x
Domain of | x + 1| = ( -infinity, + infinity), x can take any value.
As far as the range is concerned, | x + 1| cannot be negative. So, range = [0, infinity)
Question
log5x = a (This should be read as log X to the base 5 equals a)log20x = b
What is logx10 ?
Correct Answer: (a + b)/2ab
Explanation
log5x = a
log20x = b
logx5 = 1/a

46
logx20 = 1/b
logx100/5 = 1/b
logx100 – logn5 = 1/b
2logx10 – 1/a = 1/b
logx10 = ½(1/a + 1/b)
logx10 = (a + b)/2ab
Questions
1. Find the range of x for which (x + 2) (x + 5) > 40.
2. How many integer values of x satisfy the equation x( x + 2) (x + 4) (x + 6) < 200
3. Find the range of x where ||x - 3| - 4| > 3.
Correct Answers
1. x < -10 or x > 3
2. 9 different values
3. (-infinity, -4) or (2,4) or ( 10, infinity)
Explanatory Answers
1. Find the range of x for which (x + 2) (x + 5) > 40.
There are two ways of trying this one. We can expand and simplify this algebraically. x^2
+ 7x + 10 > 40 or
x^2 + 7x - 30 > 0
(x + 10) (x -3) > 0
The roots are -10 and +3.
=> x should lie outside the roots.
Now, what is this based on?
There is a simple thumb rule for solving quadratic inequality
For any quadratic inequality ax2
+ bx + c < 0
Factorize it as a(x - p) ( x - q) < 0
Whenever a is greater than 0, the above inequality will hold good if x lies between p and
q.

47
a(x - p) (x - q) will be greater than 0, whenever x does not lie between p and q. In other
words x should lie in the range ( -infinity, p) or (q, infinity)
Now, coming back to the question (x+10) (x -3) > 0
Or,x < -10 or x > 3
Second method: 5 * 8 = 40, -8 * -5 = 40
So, if x + 2 > 5 this will hold good => x > 3
If x + 2 is less than -8 also, this will hold good => x < -10. The first method is far more
robust.
2. How many integer values of x satisfy the equation x( x + 2) (x + 4) (x + 6) < 200
To begin with - 0, -2, -4 and -6 work. These are the values for which the left-hand side
goes to zero.
There are 4 terms in the product. If all 4 are positive or all 4 are negative the product will
be positive
The product can be negative only if exactly 1 or exactly 3 are negative. When 1 or 3
terms are negative, the product is clearly less than 200.
When x = -1, one term is negative
When x = -5, three terms are negative
So, adding these two numbers also to the set of solutions {-6, -5, -4, -2, -1, 0} satisfy the
inequality.
Beyond this it is just trial and error.
Let us try x = 1. Product is 1 * 3 * 5 * 7 = 105. This works
x = -7 gives the same product. So, that also works.
So, the solution set is now refined to {-7, -6, -5, -4, -2, -1, 0, 1}
x = 2 => Product is 2 * 4 * 6 * 8 = 8 * 48. Not possible. Any x greater than 1 does not
work
x = -8 is also not possible. Any value of x less than -7 does not work.

48
So, the solution set stays as {-7, -6, -5, -4, -2, -1, 0, 1}
The one missing value in this sequence is -3. When x = -3, product becomes -3 * -1 * 1 *
3. = 9. This also holds good.
So, values {-7, -6,-5, -4, -3, -2, -1, 0, 1} hold good. 9 different values satisfy this
inequality
3. Find the range of x where ||x - 3| - 4| > 3
If we have an inequality |y| > 3, this will be satisfied if => y > 3 or y < -3.
So, the above inequality simplifies to two inequalities
Inequality I: | x - 3| - 4 > 3
| x - 3| - 4 > 3 => | x - 3 | > 7
x - 3 > 7 or x - 3 < -7
Or, x > 10 or x < -4
Of, x lies outside of -4 and 10. Or, x can lie in the range ( - infinity, -4) or ( 10, infinity)
Inequality II: |x -3| - 4 < -3
|x - 3 | - 4 < - 3
=> | x - 3 | < 1
=> -1 < x - 3 < 1 or x lies between 2 and 4
So, final solution is given by the range
( - infinity, -4) or (2,4) or ( 10, infinity)
NOTE:
Competitive exams often ask questions with a 'wrapper' around them. Its important to get
to the right question quickly. Give the underlying questions for the following -
1. Give the smallest 4-digit number with an odd number of factors (easy one)

49
2. Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13, 420, ..........},
how many elements of Set S are integers?
3. f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such that we end up with
a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2
Discussion:
The questions sitting underneath the above statements are as follows
1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is
the smallest 4-digit perfect square?
how many elements of Set S are integers? => How many odd factors does 6300 have?
a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2? => How
many 5-digit number exist that when divided by 9 leave a remainder of 2?
Explanatory Answers
1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is
the smallest 4-digit perfect square? 1024
how many elements of Set S are integers? => How many odd factors does 6300 have?
6300 = 2^2 * 3^2 * 5^2 * 7. Number of odd factors of this number = (2+1) * (2+1) *
(1+1) = 18. For discussion on number of odd factors, look here .
a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2? => How
many 5-digit number exist that when divided by 9 leave a remainder of 2?
There are 90000 5-digit numbers. There will be 10000 numbers that leave a remainder of
2 on division by 9 within these. Answer = 10000.

50
g(n) mentioned above is identical to the remainder when a number is divided by 9. Once
you pick that, this question become a sitter. Competitive exams are good at masking
questions. As much as possible, learn from first principles. If one had thought about why
the test of divisibility for 9 works, this bit would have been clear.
NOTE 2;
Key formulae
For a1 + a2 + a3 ....ar = n. The number of solutions where a1,a2,a3....ar all take natural
numbers is (n-1) C (r-1).
For a1 + a2 + a3 ....ar = n. The number of solutions where a1,a2,a3....ar can take all
whole number values is (n+r-1) C (r-1).
Similar questions on the same theme
The proofs are just an extension of the idea discussed in the two posts. Now, this kind of
question can get asked in different frameworks. Few examples are given below.
1. 10 identical toys need to be placed in 3 distinct boxes. In how many ways can this be
done?
2. Mark needs to pick up exactly 40 fruits for his family. He has to pick at least 8 Apples,
at least 6 Oranges and at least 5 mangoes. He should not pick up any other fruit. In how
many ways can this be done?
Question
Sum of three whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

51
Correct Answer
12C2
Explanatory Answer
a + b + c = 10. a,b,c are whole numbers. Now this is similar to the previous question that
we solved by placing 10 sticks and simplifying. The discussion can be seen here .
We cannot follow a similar approach as a,b,c can be zero. Let us modify the approach a
little bit. Let us see if we can remove the constraint that a,b,c can be zero.
If we give a minimum of 1 to a,b,c then the original approach can be used. And then we
can finally remove 1 from each of a,b, c. So, let us distribute 13 sticks across a, b and c
and finally remove one from each.
a + b + c = 13. Now, let us place ten sticks in a row
| | | | | | | | | | | | |
This question now becomes the equivalent of placing two '+' symbols somewhere
between these sticks. For instance
| | | | + | | | | | + | | | |, This would be the equivalent of 4 + 5 + 4.
or, a = 4, b = 5, c = 4.
There are 12 slots between the sticks, out of which one has to select 2 for placing the '+'s.
The number of ways of doing this is 12C2.

52
Questions
1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c)
exist?
2. In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is
empty?
Correct Answers:
1. 9C2
2. 10C2
Explanatory Answers
1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c)
exist?
a + b + c = 10. Now, let us place ten sticks in a row
| | | | | | | | | |
This question now becomes the equivalent of placing two '+' symbols somewhere
between these sticks. For instance
| | | | + | | | | | + |, This would be the equivalent of 4 + 5 + 1. or, a = 4, b =
5, c = 1.
There are 9 slots between the sticks, out of which one has to select 2 for placing the '+'s.
The number of ways of doing this would be 9C2. Bear in mind that this kind of
calculation counts ordered triplets. (4,5,1) and (1, 4, 5) will both be counted as distinct
possibilities.
We can also do a + b + c = n where a, b, c have to be whole numbers (instead of natural

53
numbers as in this question) with a small change to the above approach. Give it some
thought.
2. In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is
empty?
This is nothing but number of ways of having a,b,c such that a + b + c = 11, where a, b, c
are natural numbers.
10C2.
1. (|x| - 2) ( x + 5) < 0. What is the range of values x can take?
2. a and b are roots of the equation x^2 - px + 12 = 0. If the difference between the roots
is at least 12, what is the range of values p can take?
Solutions to the above questions
1. (|x| - 2) ( x + 5) < 0 -
This can be true in two scenarios
Scenario I - (|x| - 2) < 0 and ( x + 5) > 0
Or |x| < 2 and x > -5.This gives us the range (-2,2)
Scenario II - (|x| - 2) > 0 and ( x + 5) < 0
Or |x| > 2 and x < -5. This gives us the range (-Infinity, -2)
So, the overall range is (-infinity, -2) or (-2,2)
2. a and b are roots of the equation x^2 - px + 12 = 0. If the difference between the
roots is at least 12, what is the range of values p can take?

54
The roots are a and b
a + b = p ab = 12
(a + b )^2 = p^2
(a -b)^2 = (a + b ) ^2 - 4ab
=> (a-b) ^2 = p^2 - 12*4 = p^2 - 48
If |a-b| > 12 { Difference between the roots is at least 12}
then, (a-b)^2 > 144
p^2 - 48 > 144
p^2 > 192
P > 8sqrt(3) or P < -8 sqrt(3)
Questions:
1. Sum of three distinct natural numbers is 25. What is the maximum value of their
product?
2. x ( x + 3) ( x + 5) (x + 8) < 250. How many integer values can x take?
Correct Answers:
Question 1: 560
Question 2: 11 values
Explanatory Answers
Qn 1. Sum of three distinct natural numbers is 25. What is the maximum value of their
product?
Let the three natural numbers be a, b and c. We know that a + b + c = 25.

55
Let us substitute some values and see where this is headed
a = 1, b = 2 c = 22, Product = 44
a = 2, b = 3 c = 20, product = 120
a = 5, b = 6 c = 14, product = 420
The numbers should be as close to each other as possible. (Just trial and error and one can
figure this out). This property is an extension of the AM-GM Inequality.
25/3 ~ 8. 25 divided by 3 is approximately equal to 8. So, we should choose a, b, and c
close enough to 8
8 + 8 + 9 = 25. But the numbers need to be distinct. 8 * 7 * 10 come closest. The
maximum product would be 560.
Qn 2. x ( x + 3) ( x + 5) (x + 8) < 250. How many integer values can x take?
Straight away we can see that x = 0 works. The product goes to zero in this case. When x
takes values -3, -5 or -8, the product will go to zero and the inequality holds good.
Now, let us substitute some other values
When x = 1 => Product = 1 * 4 * 6 * 9 = 216 < 250 So, x =1 holds good
When x = 2, the product is clearly greater than 250.
So, thus far, we have seen that for x = 0 , -3 -5 -8 or 1; the above inequality holds good.
There are 4 terms in this product. If all 4 are positive or all 4 are negative, the product
will be positive. If exactly one term is positive or exactly one term is negative, the
product will be negative.
Whenever the product is negative, the inequality will hold good. So, let us find the values
of x for which the product will be negative
x = -1 or -2, the product is negative, so the inequality will hold good

56
Let us think of other values of x for which the product is negative. For the product to be
negative, either 1 or 3 of the four terms should be negative.
When x is -6 or -7, three of the terms are negative and the product is negative.
So, for x = 0, 1, -3, -5, -8, -1, -2, -6 or -7 this holds good. We have seen 9 values thus far.
In ascending order, the values are -8, -7, -6, -5, -3, -2, -1, 0, 1
The value in between that we have thus far not verified is -4. Let us try -4 as well. In this
case the product is -4 * -1 * 1 * 7 < 250
For x = -9, product is -9* -6 * -4 * 1 = 216 < 250
Clearly for x = -10 or lesser the inequality does not hold good.
So, the inequality holds good for 11 terms. x can take all integers from -9 to +1 (both
inclusive).
1. Give the domain and range of the following functions
i. f(x) = x2
+ 1
Domain = All real numbers (x can take any value)
Range [1, infinity). Minimum value of x2
is 0.
ii. g(x) = log (x + 1)
Domain = Log of a negative number is not defined so (x + 1) > 0 or x > -1
Domain ( -1, infinity)
Range = (-infinity, +infinity)
Note: Log is one of those beautiful functions that is defined from a restricted domain to
all real numbers. Log 0 is also not defined. Log is defined only for positive numbers
iii. h(x) = 2x
Domain - All real numbers.

57
Range = (0, infinity)
The exponent function is the mirror image of the log function.
iv. f(x) = 1/ (x+1)
Domain = All real numbers except -1
Range = All real numbers except 0
v. p(x) = | x + 1|
Domain = All real numbers
Range = [0, infinity) Modulus cannot be negative
vi. q(x) = [2x], where [x] gives the greatest integer less than or equal to x
Domain = All real numbers
Range = All integers
The range is NOT the set of even numbers. [2x] can be odd. [2*0.6] = 1. It is very
important to think fractions when you are substituting values.
1. Give the domain and range of the following functions
i. f(x) = x2
+ 1
ii. g(x) = log (x + 1)
iii. h(x) = 2x
iv. f(x) = 1/ (x+1)
v. p(x) = | x + 1|
vi. q(x) = [2x], where [x] gives the greatest integer less than or equal to x
1. A point P is identified as P(m,n). What is the ratio AP:BP given that the points A
and B are identified as A(5,-4) and B(1,6)?
Stmt 1. m = 3. Not sufficient.
Stmt 2. n = 2.5 Not sufficient.
Either statement alone is not sufficient. Both put together, we can answer the
questions. Ans B

58
2. A point P is identified as P(m,n). What is the ratio AP:BP given that the points
A and B are identified as A(5,-4) and B(5,6)?
Stmt 1. m = 5. Not sufficient. It tells us that all points lie on the line x = 5, but this
is not enough
Stmt 2. n = 3 Not sufficient.
Both put together, this is enough.
In the above question, would n = 1 have been sufficient? Think about that.
3. A survey of 100 people tried to find the number of people who can write with
both their left and right hands. What is the maximum number of people who could
write left-handed and right-handed?
Stmt 1. 50 people can write only with their left hand. 40 people can write only with
their right hand. Sufficient: Maximum of 10 people could write left-handed and
right-handed.
Stmt 2. 50 people can write with their left hand. 40 people can write with their
right hand. Sufficient: Maximum of 40 people could write left-handed and right-
handed.
Answer Choice C
4. What is the slope of a line?
Stmt 1: The line makes 135 degrees with the negative direction of x - axis.
Sufficient: The line makes 45 degrees with positive x axis. This should be enough.
Stmt 2: The line makes an isosceles right triangle with the coodinate axes and the
product of the intercepts is negative. Sufficient: Either both intercepts are positive
and equal or negative and equal. Slope = -1
1. Perimeter of a triangle with integer sides is equal to 15. How many such triangles
are possible?

59
This is just a counting question, with the caveat that sum of two sides should be
greater than the third. Let us assume a < b < c
a = 1, Possible triangle 1, 7, 7
a = 2, possible triangle 2, 6, 7
a = 3, possible triangles 3, 6, 6 and 3, 5, 7
a = 4, possible triangles 4, 4, 7 and 4, 5, 6
Again, from comments,
a = 5, possible triangle is 5,5,5,
There are totally 7 triangles possible
2. Triangle ABC has integer sides x, y, z such that xz = 12. How many such
triangles are possible?
xz = 12
x,z can be 1, 12 or 2, 6 or 3, 4
Possible triangles
1-12-12
2-6-5; 2-6-6; 2-6-7
3-4-2; 3-4-3; 3-4-5; 3-4-6.
As pointed out in the comments section, I have missed the triangle 3-4-4.
There are totally 9 triangles.
3. Triangle has sides a^2, b^2 and c^2. Then the triangle with sides a, b, c has to be
- a) Right angled b) Acute-angled c) Obtuse angled d) can be any of these three
Assuming a < b < c, we have a^2 + b^2 > c^2. This implies the triangle with sides

60
a, b, c has to be acute-angled.
P.S: Big thanks to 'maniac' for pointing out the errors
Exactly one of ab, bc and ca is odd => Two are odd and one is even
abc is a multiple of 4 => the even number is a multiple of 4
The arithmetic mean of a and b is an integer => a and b are odd
and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3
c can be 4 or 8.
c = 4; a, b can be 3, 5 or 5, 9
c = 8; a, b can be 3, 7 or 7, 9
Four triplets are possible
multiples of 12?
Number should be a multiple of 3 and 4. So, the sum of the digits should be a
multiple of 3. WE can either have all seven digits as 3, or have three 2's and four
3's, or six 2's and a 3. (The number of 2's should be a multiple of 3).
For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us
combine these two.
All seven 3's - No possibility
Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some

61
order. No of possibilities = 5!/3!*2! = 10
Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number
2222232
be a perfect square a,b,c have to be even. a can take values 0, 2, 4. b can take
values 0,2, 4, 6 and c can take values 0,2. Total number of perfect squares = 3 * 4 *
2 = 24
be an odd number, a should be 0. b can take values 0,1, 2,3, and c can take values
0, 1, 2,3, 4. Total number of odd factors = 4 * 5 = 20
Number can be 7, 25, 43, 61, 79.
Remainders when divided by 12 are 7 and 1.
NOTE 3
multiples of 12?

62
NOTE 4
1. Any two triangles that are congruent to each other will also be similar to each other
2. If in a triangle with sides a, b and c, if a^2 + b^2 > c^2 the triangle has to be
acute-angled
3. Any parallelogram inscribed inside a circle has to be a rectangle
4. If in a triangle the orthocenter, incenter and circumcenter are collinear the
triangle has to be isosceles.
5. There will be a unique circle passing through any three points
6. If two circles with centers A and B and radii r and R intersect, then AB > R - r
7. Circumradius of a triangle cannot be greater than the three sides of the triangle
8. For any obtuse-angled triangle, the orthocenter and circumcenter will lie outside
the triangle
9. In a scalene triangle, the sum of the three medians will be greater than the sum
of the three altitudes.
10. In circumcircle and incircle are concentric, the triangle has to be equilateral
1. From the digits 2,3,4,5,6 and 7, how many 5-digit numbers can be formed that
have distinct digits and are multiples of 12?
Any multiple of 12 should be a multiple of 4 and 3. First, let us look at the
constraint for a number being a multiple of 3. Sum of the digits should be a
multiple of 3. Sum of all numbers from 2 to 7 is 27. So, if we have to drop a digit
and still retain a multiple of 3, we should drop either 3 or 6.

63
So, the possible 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.
When the digits are 2, 4, 5, 6, 7. the last two digits possible for the number to be a
multiple of 4 are 24, 64, 52, 72, 56, 76. For each of these combinations, there are 6
different numbers possible. So, with this set of 5 digits we can have 36 different
numbers
When the digits are 2, 3, 4, 5, 7. the last two digits possible for the number to be a
multiple of 4 are 32, 52, 72, 24. For each of these combinations, there are 6
different numbers possible. So, with this set of 5 digits we can have 24 different
numbers
Overall, there are 60 different 5-digit numbers possible
2. All numbers from 1 to 200 (in decimal system) are written in base 6 and base 7
systems. How many of the numbers will have a non-zero units digit in both base 6
and base 7 notations?
If a number written in base 6 ends with a zero, it should be a multiple of 6. In other
words, the question wants us to find all numbers from 1 to 200 that are not
multiples of 6 or 7. There are 33 multiples of 6 less than 201. There are 28
multiples of 7 less than 201. There are 4 multiples of 6 & 7 (or multiple of 42)
from 1 to 200.
So, total multiples of 6 or 7 less than 201 = 33 + 28 - 4 = 57. Number of numbers
with non-zero units digit = 200-57 = 143.
3. All numbers from 1 to 150 (in decimal system) are written in base 6 notation.
How many of these will not contain any zero?
Any multiple of 6 will end in a zero. There are 25 such numbers. Beyond this, we
can have zero as the middle digit of a 3-digit number. This will be the case for
numbers from 37-41, 73-77, 109-113 and 145-149. There are 20 such numbers.

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Overall, there are 45 numbers that have a zero in them.
4. How many factors of 1080 are perfect squares?
1080 = 2^3 * 3^3 * 5. For any perfect square, all the powers of the primes have to
be even numbers. So, if the factor is of the form 2^a * 3^b * 5^c. The values 'a' can
take are 0 and 2, b can take are 0 and 2, and c can take the value 0. Totally there
are 4 possibilities. 1, 4, 9, and 36.
1. Australia, in one of their matches, scored a total that is the lowest multiple of 11
with exactly 10 factors. Against who was this?
2^4 * 11 = 176. Australia scored this against Pakistan
2. In one of the matches, the top-scorer (who scored a century) for the team batting
first scored 3/8th the team's runs, but the team still did not cross 300. The HCF of
the team's score and the top-scorers score is a prime number. Which match was
this?
Top-scorer's score is a multiple of 3 greater than 100. So, it has to be a score like
34 * 3, 35 * 3, 36 * 3,..... The overall score is less than 300. So, the overall score
has to be lower than 38 * 8. So, the overall score has to be of the form 37 * 8, 36 *
8, 35 * 8,...
The HCF is a prime number, this implies that the top scorer scored 111, the team
scored 296. India vs. South Africa. Sachin Tendulkar century
3. A 'minnow' scored a total that can be made a perfect square if it is multiplied by
23. If they had scored 1 more, then their score multiplied by 13 would have yielded
a perfect square. What was their score? Bonus point, what match was this?
The score should have been 23 * 4 or 23 * 9, 23 * 9 = 207. 207 + 1 = 208, which is

65
13 * 16.
India vs. Ireland. Ireland scored 207
4. In a particular match, team A batting first scored twice a perfect square. Team B
batting second, also scored twice a perfect square. The total score of both the teams
put together was also a perfect square. What match are we talking about here?
India 338, England 338 (169 * 2). Total 676 (26* 26)
5. Team scored a three digit score 'abc' in a match. a and b are prime numbers. a +
2b and a + 2c are also prime. b + 2a and b + 2c are also primes. a + c and b + c are
also primes. Two digit numbers 'ab' and 'ba' are both primes. Which match was
this?
a + c and b + c are also primes => a, b should be odd, c should be even
a,b cannot be 5 as ab is also prime. So, abh has to be 37.
c can only be zero.
abc = 370 (India vs. Bangladesh)
1. A number n! is written in base 6 and base 8 notation. Its base 6 representation ends
with 10 zeroes. Its base 8 representation ends with 7 zeroes. Find the smallest n
that satisfies these conditions. Also find the number of values of n that will satisfy
these conditions.
Base 6 representation ends with 10 zeroes, or the number is a multiple of 6^10. If
n! has to be a multiple of 6^10, it has to be a multiple of 3^10. The smallest
factorial that is a multiple of 3^10 is 24!. So, when n = 24, 25 or 26, n! will be a
multiple of 6^10 (but not 6^11).
Similarly, for the second part, we need to find n! such that it is a multiple of 2 ^ 21,
but not 2 ^ 24. When n = 24, n! is a multiple of 2^22. S0, when n = 24, 25, 26, 27,
n! will be a multiple of 2 ^ 21 but not 2 ^ 24.

66
The smallest n that satisfies the above conditions is 24. n = 24, 25 or 26 will satisfy
the above conditions.
2. [x] is the greatest integer less than or equal to x. Find the number of positive
integers n such that [n/11] = [n/13].
This is a classic case of brute-force counting.
When n = 1, 2, 3....10, both values will be equal to 0. 10 possibilities
When n = 13,14, ....21, both values will be equal to 1. 9 possibilities
When n = 52, 53, 54, both values will be equal to 4. 3 possibilities
when n = 65, both will be equal to 5. 1 possibility
So, totally there are 1 + 3 + 5 + 7 + 9 +10 = 35 possibilities.
3. Positive numbers 1 to 55, inclusive are placed in 5 groups of 11 numbers each.
What is the maximum possible average of the medians of the 5 groups?
We need to maximise each median in order to have the overall maximum median
possible.
The highest possible median is 50 as they should be 5 numbers higher than the
median in the group of 11. So, if we have a set that has a, b, c, d, e, 50, 51, 52, 53,
54, 55, the median will be 50. In this set, it is best not to waste any high values on
a, b, c, d or e as these do not affect the median. So, a set that reads as 1, 2, 3, 4, 5,
50, 51, 52, 53, 54, 55 will also have median 50.
The next set can be 6, 7, 8, 9, 10, 44, 45, 46, 47, 48, 49. The median will be 44.
2. 1. Average of 6 distinct positive integers is 33. The median of the three largest
numbers is 43. What is the difference between the highest and lowest possible
median of the 6 numbers?
3. Let a, b, c, d, e, f be the six numbers in ascending order
4. e = 43; a + b + c + d + e + f = 198

67
5. Median of 6 integers = (c +d)/2
6. (c +d)/2 is maximum when c and d are maximum since e = 43, dmax = 42 and cmax =
41
7. (c +d)/2 is minimum when c and d are minimum amin = 1, bmin = 2, cmin = 3 dmin = 4
8. Therefore, = (c +d)/2 min = 3.5
9. Max – min = 41.5 – 3.5 = 38
10.
11.2. In class A, the ratio of boys to girls is 2:3. In class B the ratio of boys to girls
is 4 : 5. If the ratio of boys to girls in both classes put together is 3 : 4, what is
the ratio of number of girls in class A to number of girls in class B?
12.Let us assume class A has 2x boys and 3x girls, class B has 4y boys and 5y girls.
13.(2x + 4y)/(3x + 5y) = ¾
14.8x + 16y = 9x + 15y
15.x = y
16.Reqd ratio = 3x/5y (since x = y) = 3/5
17.3. N is an 80-digit positive integer (in the decimal scale). All digits except the
44th digit (from the left) are 2. If N is divisible by 13, find the 26th digit.
18.To begin with, the question should read "find the 44th digit".
19.Any number of the form abcabc is a multiple of 1001. 1001 is 7 * 11 * 13. So, any
number of the form abcabc is a multiple of 13.
20.So, a number comprising 42 2's would be a multiple of 13, so would a number
comprising 36 2's. So, in effect, we are left with a two digit number 2a, where a is
the 44th digit. 26 is a multiple of 13, so the 44th digit should be 6.
21.
22.4. A page is torn from a novel. The sum of the remaining digits is 10000. What
is the sum of the two page-numbers on the torn page of this novel?
23.n(n+1)/2 should be nearby 10,000, so n(n+1) is somewhere near 20,000. So n
should be around sqrt(20,000) about 141. Try 142
24.142*143 = 10153
25.141 *142 = 10011
26.So, the missing pages are either 76 and 77 or 5 & 6
1) Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open
throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes
after pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe
fills an equal share of the tank. Furthermore, it is known that if pipe A and B are kept

68
open continuously, the tank would be filled completely in t minutes. Find t?
A is kept open for all t minutes and fills one-third the tank. Or, A should be able to fill the
entire tank in '3t' minutes.
A and B together can fill the tank completely in t minutes. A alone can fill it in 3t
minutes.
A and B together can fill 1/t of the tank in a minute. A alone can fill 1/3t of the tank in a
minute. So, in a minute, B can fill 1/t - 1/3t = 2/3t. Or, B takes 3t/2 minutes to fill an
entire tank
To fill one-third the tank, B will take t/2 minutes. B is kept open for t - 10 minutes.
t/2 = t - 10, t = 20 minutes.
A takes 60 mins to fill the entire tank, B takes 30 minutes to fill the entire tank. A is kept
open for all 20 minutes. B is kept open for 10 minutes.
C, which is kept open for 8 minutes also fills one-third the tank. Or, c alone can fill the
tank in 24 minutes.
2) Pipe A fills a tank at the rate of 100lit/min, Pipe B fills at the rate of 25 lit/min, pipe C
drains at the rate of 50 lit/min. The three pipes are kept open for one minute each, one
after the other. If the capacity of the tank is 7000 liters, how long will it take to fill the
tank if
i) A is kept open first, followed by B and then C
Each cycle of 3 minutes, 75 liters get filled. 100 + 25 - 50. So, after 3 minutes the tank
would have 75 liters
6 mins - 150 liters
9 mins - 225 liters
30 mins - 750 liters
In the 280th minute, pipe A would be open and it would fill the remaining 25 liters in 15
seconds. So, it would take 279 mins and 15 seconds to fill the tank.

69
The most important thing in these type of questions is to think in terms of cycles till we
reach close to the required target and then think in simple steps.
ii) B first, followed by A, and then C
Each cycle of 3 minutes, 75 liters get filled. 25 + 100 - 50. So, after 3 minutes the tank
would have 75 liters
6 mins - 150 liters
9 mins - 225 liters
In the 280th minute, pipe B would be open and it would fill the remaining 25 liters in one
minutes. So, it would take 280 mins.
iii) B first, followed by C, and then A
In this case also, Each cycle of 3 minutes, 75 liters get filled. 25 -50 + 100. But there is a
small catch here. In the first set of 3 minutes, we would fill up to about 100 liters. After 1
minute, we would be at 25 liters, after 2 minutes, we would be at 0 liters and in the third
minute, the tank would be 100 liters full.
6 mins - 170 liters
9 mins - 250 liters
279 minutes - 7000 liters
So, it would take 279 mins and 15 seconds to fill the tank.
The most important thing in these type of questions is to think in terms of cycles till we
reach close to the required target and then think in simple steps.
3) 4 men and 6 women complete a task in 24 days. If the women are at least half as
efficient as the men, but not more efficient than the men, what is the range of the number

70
of days for 6 women and 2 men to complete a task twice as difficult as this one?
4m and 6w finish in 24 days.
In one day, 4m + 6w = 1/24 of task
In these questions, just substitute extreme values to get the whole range
If a woman is half as efficient as man
4m + 3m = 1/24, 7m = 1/24, m = 1/168
6w + 2m = 3m + 2m = 5m, 5m will take 168/5 days = 33.6 days
If a woman is as efficient as a man
4m + 6w finish in 24 days
10m finish 1/24 of task in a day
6w + 2m = 8m, 8m will take 240/8 = 30 days to finish the task.
So, the range = 30 to 33.6 days. The new team will take 30 to 33.6 days to finish the task.
4) A fill pipe can fill a tank in 20 hours, a drain pipe can drain a tank in 30 hours. If a
system of n pipes (fill pipes and drain pipes put together) can fill the tank in exactly 5
hours, which of the following are possible values of n (More than one option could be
correct)?
1) 32 2) 54 3) 29 4) 40
3 fill pipes cancel out 2 drain pipes. Plus, you need an additional 4 fill pipes fill the tank
in 5 hours. so the answer has to be 5k + 4. Both 54 and 29 are possible.
1. a is x % of b, b is x% more than a. Find x?
a = bx,
b = a (1+x), substituting this in the previous equation, we get
x(x+1) = 1
x^2 + x -1 = 0
or x is approximately 0.62 or 62%

71
2. A two digit number ab is 60% of x. The two-digit number formed by reversing the
digits of ab is 60% more than x. Find x?
10a + b = 0.6x
10b + a = 1.6x
Subtracting one from the other, we have
9b - 9a = x or x = 9( b-a)
x should be a multiple of 9.
10a + b = 3x/5, this implies that x should also be a multiple of 5.Or, x should be a
multiple of 45
x should be equal to 45, ab = 27, ba = 72
3. A is x % more than B. B is y% less than C. If A,B and C are positive and A is greater
than C, find the relation between x and y
a = (1 +x)b
b = c(1-y)
c = b /(1-y)
a is greater than c
Or, b (1+x) > b/(1-y)
1 + x > 1/(1-y)
(1+x)(1-y) > 1
1 + x -y - xy > 1
x > y (1+x)
y < x/(1+x)

72
1. The median of n distinct numbers is greater than the average, does this mean that there
are more terms above the average than below it?
No.
If there are an odd number of terms, say, 2n+1, then the median is the middle term. And
if average is lesser than the middle term, there will at least be n +1 terms greater than the
average. So, there will be more terms above the average than below it.
However, this need not be the case when there are an even number of terms. When there
are 2n distinct terms, n of them will be greater than the median and n will be lesser than
the median. The average of these two terms can be such that there are n terms above the
average and n below it.
For instance, if the numbers are 0, 1, 7, 7.5. The median is 4, average is 3.875. Average is
less than the median. And there are more 2 numbers above the average and 2 below the
average.
2. In a sequence of 25 terms, can 20 terms be below the average? Can 20 terms be
between median and average?
Yes. We can have 24 zeroes and 1500 as the 25 numbers. In this case, there are 24
numbers below the average.
No. In a sequence of 25 numbers, 12 will be greater than or equal to the median and 12
will be lesser than or equal to the median. We cannot have 20 terms in between the
average and median
3. From 10 numbers, a,b,c,...j, all sets of 4 numbers are chosen and their averages
computed. Will the average of these averages be equal to the average of the average of
the 10 numbers?
Yes. All the terms appear the same number of times when we select them 4 at a time. So,
the average of averages will be equal to the overall average.

73
1. Consider three classes A, B and C with 20, 30 and 50 students respectively. The
averages score in maths of students in class B is 16 more than that of those in class C and
the average score of those in class A is 2 more than the overall average of scores in A, B
and C. What is the difference between average of class C and class A
Let average in class C = x. Average of class B = x +16. Average of class A = y
y = 2 + (20y + 30(x+16) + 50x) /100
(y - 2) * 10 = 2y + 3x + 48 + 5x
10y -20 = 2y + 8x + 48
8y - 8x = 68
y - x = 8.5. This is the difference between average of class A and class C
2. Natural numbers 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The
medians of these 5 groups are A, B, C, D and E. If the average of these medians is m,
what are the smallest and the largest values m can take?
Let us try to construct a scenario for getting the smallest value of the average. We need to
have the minimum value for each of the 5 medians.
Now, what is the lowest value the median of any of the 5 groups can take?
The median is the middle term among the 5 terms. So, it cannot be 1 or 2, but it can be 3.
If the set were 1, 2, 3, 4, 5, the median would be 3. If we choose a sub-group such as this,
the smallest median fro the next group will be 8. The next group could be 6, 7, 8, 9, 10.
The idea is to create sub-groups in such a way so that the second group can have a
smaller median than 8.
Now, even a sub-group that has the numbers 1, 2, 3, 24, 25 would have the median as 3.
So, if we want to keep the medians as small as possible, we should choose sub-groups
that have small numbers till the median, and then very large numbers. Because, these
large numbers do not affect the median, we will still have small numbers to choose from
for the next group. So, choose sub-groups such that the first 3 numbers are small, the
final two are as large as possible.
1, 2, 3, 24, 25
4, 5, 6, 22, 23
7, 8, 9, 20, 21

74
10, 11, 12, 18, 19
13, 14, 15, 16, 17
would be an ideal set of groups. The medians would be 3, 6, 9, 12, 15, and the average of
the medians would be 9.
Similar set of groups can be found to find the highest value of the average. The medians
would be 23, 20, 17, 14, 11 and the average would be 17.
Q. Consider 4 numbers a, b, c and d. Ram figures that the smallest average of some three
of these four numbers is 30 and the largest average of some three of these 4 is 40. What is
the range of values the average of all 4 numbers can take?
We can assume a, b,c d are in ascending order (with the caveat that numbers can be equal
to each other)
a + b + c = 90
b + c + d = 120
We need to find the maximum and minimum value of a + b + c + d.
a + b + c + d = 120 + a. So, this will be minimum when a is minimum. Given a + b + c =
90. a is minimum when b + c is maximum. If b + c is maximum, d should be minimum.
Given that b + c + d = 120, the minimum value d can take is 40 as d cannot be less than b
or c. The highest value b +c can take is 80, when b = c= d = 40. When b = c = d = 40, a =
10. a + b + c + d = 130. Average = 32.5
Similarly, a + b + c + d = 90 + d. So, this will be maximum when d is maximum. Given b
+ c + d = 120. d is maximum when b + c is minimum. If b + c is minimum, a should be
maximum. Given that a + b + c = 90, the maximum value a can take is 30 as a cannot be
greater than b or c. The lowest value b +c can take is 60, when a = b= c = 30. When a = b
= c = 30, d = 60. a + b + c + d = 150. Average = 37.5

75
So, the average has to range from 32.5 to 37.5
Q. The average of 5 distinct positive integers if 33. What are the maximum and minimum
possible values of the median of the 5 numbers if the average of the three largest numbers
within this set is 39?
Let the numbers be a, b, c, d, e in ascending order. a + b + c + d + e = 165. Average of
the three largest numbers is 39, so c + d + e = 117, or a + b = 48.
We need to find the maximum and minimum possible values of c.
For minimum value, a and b have to be minimum. a +b = 48. Let us assume a =23, b =25,
ca can be as low as 26.
23, 25, 26, 45, 46 is a possible sequence that satisfies the conditions.
For maximum value, we need d + e to be minimum as c + d + e = 117. we can have c =
38, d =39 and e =40. Or, the maximum value c can take = 38
23, 25, 38, 39, 40 is a possible sequence that satisfies the conditions specified
Q. Consider 5 distinct positive numbers a, b, c, d, and e. The average of these numbers is
k. If we remove b from this set, the average drops to m (m is less than k). Average of c, b,
d and e is K. We also know that c is less than d and e is less than k. The difference
between c and b is equal to the difference between e and d. Average of a, b, c and e is
greater than m. Write down a, d, c, d and e in ascending order.
Average of a, b, c, d and e is k, Average of b, c, d and e is also k, this implies that a = k
If we remove b, the average drops, this implies that b is higher than the average
e is less than k, or e is less than a. c is less than d.
e < a < b, c < d
Average of a, b, c and e is greater than average a, c, d and e. This tells us that d < b

76
From this, we get that b is the largest number
b - c = d - e
b + e = c + d
a + b + c +d + e = 5a
Or, b + c + d + e = 4a
b + e = 2a, c + d = 2a. Or, e, a, b is an AP, c, a, d is an AP.
e is the smallest number (as b is the largest number)
Or, the ascending order should be e c a d b
1. Two friends A and B simultaneously start running around a circular track . They run in
the same direction. A travels at 6m/s and B runs at b m/s. If they cross each other at
exactly two points on the circular track and b is a natural number less than 30, how many
values can b take?
Let track length be equal to T. Time taken to meet for the first time = T / relative speed
= T/(6-b) or T/(b-6)
Time taken for a lap for A = T/6
Time taken for a lap for A = T/b
So, time taken to meet for the first time at the starting point = LCM (T/6, T/b) = T / HCF
(6,b)
Number of meeting points on the track = Time taken to meet at starting point/Time taken
for first meeting = Relative speed / HCF (6,b). For a more detailed discussion on this
have a look at the last few slides in this presentation.
So, in essence we have to find values for b such that 6-b/ HCF(6,b) = 2 or b-6/ HCF(6,b)
= 2

77
b = 2, 10, 18 satisfy this equation. So, there are three different values that b can take.
2. Three friends A, B and C decide to run around a circular track. They start at the same
time and run in the same direction. A is the quickest and when A finishes a lap, it is seen
that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact
same position on the circular track as B was when A finished 1 lap. Find the ratio of the
speeds of A,B and C?
Let track length be equal to T. When a completes a lap, let us assume B has run a
distance of (t-d). At this time, C should have run a distance of (t-2d)
After three laps C would have traveled a distance of 3 * (t-2d) = 3t - 6d.
After 3 laps C is in the same position as B was at the end one lap. So, the position after
3t-6d should be the same as t-d. Or, C should be at a distance of d from the end of the lap.
C will have completed less than 3 laps (as he is slower than A), so he could have traveled
a distance of either t-d or 2t-d.
=> 3t-6d = t-d => 2t = 5d => d = 0.4t => The distances covered by A,B and C when A
completes a lap will be t, 0.6t and 0.2t respectively. Or, the ratio of their speeds is 5:3:1
In the second scenario, 3t-6d = 2t-d => t = 5d=> d = 0.2t => The distances covered by
A,B and C when A completes a lap will be t, 0.8t and 0.6t respectively. Or, the ratio of
their speeds is 5:4:3
The ratio of the speeds of A, B and C is either 5:3:1 or 5:4:3
Station X of length 900 meters has two station masters A and B. But as the station is not a
busy one, they are mostly jobless and decide to conduct an experiment. They stand at
either end of the station and decide to note the exact time when trains cross the
stationmasters. They synchronize their watches and proceed to either end of the station.
Two trains P and Q go past the station (neither train stops here), and after having taken
down their readings, the station masters sit down to have a chat
1. A: Train P entered the station at exactly 8:00:00
2. B: Train Q entered the station at exactly 8:00:10 (10 seconds past 8)
3. A: The last carriage of train P crossed me by at 8:00:20, and precisely two seconds

78
after this, the engines of the two trains went past each other. (Engines are at the front of
the train)
4. B: The last carriage of train Q crossed me 22 seconds after the engine of P went past
me.
5. A: After the last carriage of train P crossed by me, it took 35 seconds for the engine of
train Q to cross me.
6. B: I got bored and I came here.
Now, let us try to jot down the points in a slightly different format and see if we can
make any inferences. Let us assume speed of train P = p, speed of train Q = q, length of
train P = L, length of train Q = M
Take statements 1 and 3
Train P crosses the stationmaster A entirely in 20 seconds (enters station at 8:00:00 and
last carriage passes at 8:00:20) => Length of train = 20p => L = 20p
Statement 5 tells us that engine of train Q crosses station master A at 8:00:55. Train Q
enters the station at 8:00:10, so the train takes 45 seconds to cross the entire station. Or, it
takes 45 seconds to travel 900 meters => Speed to train Q, q = 20m/s.
Statement 3 tells us that the two trains cross each other at 8:00:22. This implies train P
has traveled for 22 seconds since entering the station and train Q has traveled for 12
seconds since entering the station before they cross each other. The cumulative distance
traveled by the two trains should be equal to the length of the station = 900m.
=> 900 = 22p + 12q
q = 20m/s => p = 30m/s.
So, train P_ takes 900/p = 30 seconds to cross the station. So, engine of train P will cross
stationmaster B at 8:00:30.
Statement 4 states that the last carriage of Q went past 22 seconds after the engine of P
went by. Or, the last carriage of Q went by at 8:00:52.
Engine of train Q went by at 8:00:10, last carriage went by at 8:00:52, or train Q took 42
seconds to cross station master B. Train Q travels at 20m/s. => Length of train Q = 840m
Now, to the questions
1. What is the length of train Q?

79
Train Q is 840m long
2. At what time do the rear ends of the two trains cross each other?
The engines cross each other at 8:00:22. The relative speed of the two trains = 20+30 =
50m/s. The relative distance traveled by two trains from the time the engines cross each
other to the times the rear ends cross each other = Sum of the two lengths = 600 + 840
=1440. Time taken = 1440/50 = 28.8 seconds past 8:00:22, or at 8:00:50.8 seconds.
3. How far from station master A do the rear ends of the two trains cross each other?
At 8:00:50.8, P would have traveled 50.8 * 30m/s post entering the station. Or, train P
would have traveled 1524m. The rear end of train P would be at a point 1524-600m =
924m from stationmaster A (or 24 meters from stationmaster B and outside the station)
4. You are told that the two trains enter the station at the same times mentioned and the
length of the two trains are unchanged. Furthermore, train P continues to travel at the
same speed (as computed above). At what minimum speed should train Q travel such that
the rear ends of the two trains cross each other at a point within the length of the
platform?
Rear end of train P crosses the station completely at 8:00:50. (Train P takes 30 seconds to
travel the station and 20 seconds to travel a distance equal to its length). Train Q should
have traveled 840m by this time. => Train Q should travel 840 within 40 seconds.
Or, minimum speed of Q = 840/40 = 21m/s.
1. Two friends A and B leave City P and City Q simultaneously and travel towards Q
and P at constant speeds. They meet at a point in between the two cities and then
proceed to their respective destinations in 54 minutes and 24 minutes respectively.
How long did B take to cover the entire journey between City Q and City P?
Let us assume Car A travels at a speed of a and Car B travels at a speed of b.
Further, let us assume that they meet after t minutes.
Distance traveled by car A before meeting car B = a*t. Likewise distance traveled
by car B before meeting car A = b * t.
Distance traveled by car A after meeting car B = a *54. Distance traveled by car B
after meeting car A = 24* b

ISI CMI IMPORTANT AND COMPLICATED QUESTION WITH SOLUTION BY SOURAV SIR'S CLASSES 9836793076

ISI CMI IMPORTANT AND COMPLICATED QUESTION WITH SOLUTION BY SOURAV SIR'S CLASSES 9836793076

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ISI CMI IMPORTANT AND COMPLICATED QUESTION WITH SOLUTION BY SOURAV SIR'S CLASSES 9836793076