Quant Boosters  Soumil Jain, CAT Quant 100 percentiler, IIM Calcutta  Set 1

Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?
Theoretically, the least value of c is when c = b.
Therefore, a + (a + 11) + (a + 11) = 60 (b and c are equal and b, the median, is a + 11)
Or 3a = 38 or a = 12.66
So, b = c = 12.66 + 11 = 23.66However, we know that these numbers are all integers.
Therefore, a, b, and c cannot take these values.
So, the least value for c with this constraint is NOT likely to be when c = bLet us increment c by 1. Let c = (b + 1)
In this scenario, a + (a + 11) + (a + 12) = 60
Or 3a = 37. The value of the numbers is not an integer in this scenario as well.Let us increment c again by 1. i.e., c = b + 2
Now, a + (a + 11) + (a + 13) = 60
Or 3a = 36 or a = 12.
If a = 12, b = 23 and c = 25.
The least value for c that satisfies all these conditions is 25A sequence of numbers is defined as 2 = a_n – a_(n1) . S_n is sum upto n terms in this sequence and a_3 = 5. How many values m, n exist such than S_m – S_n = 65?
a3 = 5, a4 = 7, a5 = 9, a2 = 3, a1 = 1
So, the sequence is nothing but 1, 3, 5, 7, 9....
S1 = 1
S2 = 4
S3 = 9
S4 = 16
Sn = n2
Sm – Sn
=> m2 – n2 = 65
=> (m + n) ( m – n) = 65, m, n are natural numbers.
=> This could be 65 × 1 or 13 × 5. Two possibilities. Note that this cannot be written as 1 x 65 or 5 x 13 as m + n > m  n {as m, n are natural numbers}4x^3 + ax^2 – bx + 3 divided by x – 2 leaves remainder 2, divided by x + 3 leaves remainder 3. Find remainder when it is divided by x + 2
Let f(x) = 4x3 + ax2 – bx + 3.
Remainder on division of f(x) by x  2 = f(2)
Or, f(2) = 32 + 4a – 2b + 3 = 2
4a – 2b = 33 Eqn (1)
Remainder on division of f(x) by x + 3 = f(3)
Or, f(3) = –108 + 9a + 3b + 3 = 3
9a + 3b = 108
Dividing by 3, the above equation becomes
3a + b = 36. Multiplying this equation by 2, we get
6a + 2b = 72 Eqn (2)
Adding equations (1) and (2), we have 10a = 39
a = 3.9
4(3.9) – 2b = –33
b = 24.3
f(x) = 4x3 + 3.9x2 – 24.3x + 3
Remainder on division of f(x) by x + 2 = f(2)
f(–2) = 4(–8) + 3.9(4) – 24.3(–2) + 3 = 35.2If x is a positive integer such that (x1)(x3)(x5)....(x93) < 0, how many values can x take?
The expression takes negative values when x = 4, 8, 12, 16 ..... 92 (multiples of 4).
92, the last value that x can take is the 23rd multiple of 4. Hence, number of such values of x = 23A 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.
n can take exactly 2 valuesIf a three digit number ‘abc’ has 3 factors, how many factors does the 6digit number ‘abcabc’ have?
abc’ has exactly 3 factors, so ‘abc’ should be square of a prime number. (This is an important inference, please remember this).
Any number of the form p^a q^b r^c will have (a + 1) (b + 1) (c + 1) factors, where p, q, r are prime. So, if a number has 3 factors, its prime factorization has to be p2.
‘abcabc’ = ‘abc’ * 1001 or abc * 7 * 11 * 13 (again, this is a critical idea to remember)
Now, ‘abc’ has to be square of a prime number. It can be either 121 or 169 (square of either 11 or 13) or it can be the square of some other prime number.
When abc = 121 or 169, then ‘abcabc’ is of the form p^3q^1r^1 1, which should have 4 * 2 * 2 = 16 factors.
When ‘abc’ = square of any other prime number (say 172 which is 289) , then ‘abcabc’ is of the form p^1q^1r^1s^2 , which should have 2 * 2 * 2 * 3 = 24 factors
So, ‘abcabc’ will have either 16 factors or 24 factors
Working alone, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?
A starts and works for 2 days. So, A will work on day 1 and day 2.
Then B will work for the next 2 days. B will work on day 3 and day 4.
A will continue for the next 2 days. i.e., on day 5 and day 6.
B will work on day 7 and day 8.
A will for the last 2 days i.e., day 9 and day 10.Therefore, A will work on day 1, day 2, day 5, day 6, day 9, and day 10. i.e., for 6 days.
And B will work on day 3, day 4, day 7, and day 8. i.e., for 4 days.In 6 days A will complete 6/a of the task.
In 4 days B will complete 4/b of the task.With A working 6 days and B working 4 days, the task is completed.
i.e., 6/a + 4/b = 1 .... eqn (1)Therefore, B will work on day 1, day 2, day 5, day 6, day 9, and day 10. i.e., for 6 days.
And A will work on day 3, day 4, day 7, day 8 and half a day on day 11. i.e., for 4.5 days.In 6 days B will complete 6/b of the task.
In 4.5 days A will complete 4.5/b of the task.With B working 6 days and A working 4.5 days, the task is completed.
i.e., 4.5/a + 6/b = 1 .... eqn (2)Solving the two equations we get a = 9 days and b = 12 days.
The question: How long does it take to complete the task if they both work together?
Working together A and B will complete 7/36th of the task in a day.
Hence, they will complete the task in 36/7 days
What is the area enclosed in the region defined by y = x – 1 + 2, line x = 1, X–axis and Y–axis?
The point of intersection of y = x – 1 + 2 where the y–axis can be found by substituting x = 0 in the equation. Thus we get y = 3.
Required area = Area of the trapezium formed by the points (0, 0), (1, 0), (1, 2) and (0, 3).
Area of a trapezium = 12 × height × sum of the parallel sides = 12 × 1 × (3 + 2) = 52 sq units.
If x and y are integers and x  y = 12, what is the minimum possible value of xy?
Square both sides and solve
Squaring both sides, we get (x  y)^2 = 144
x^2 + y^2  2xy = 144Add, 4xy to both sides of the equation.
x^2 + y^2  2xy + 4xy = 144 + 4xy
x^2 + y^2 + 2xy = 144 + 4xy
Or (x + y)^2 = 144 + 4xy(x + y)^2 will NOT be negative for real values of x and y.
i.e., (x + y)^2 ≥ 0∴ 144 + 4xy ≥ 0
Or 4xy ≥ 144
So, xy ≥ 36The least value that xy can take is 36
If  r6 = 11 and  2q12 = 8, then what is the minimum possible value of q/r ?
There are clearly only 4 combinations, r = 17, q = 10 # r = 5 , q= 10 # r = 17, q= 2 and r = 5, q = 2. Now the minimum value therefore would be 10/(5) = 2

In this question :
"If x is a positive integer such that (x1)(x3)(x5)....(x93) < 0, how many values can x take? "
How did you get that for all multiples of 4 the expression will take negative value ?

@PiyusJain
(x1)(x3)(x5)....(x93) > we have 47 terms
When x is a multiple of 4, we will have an odd number of negative values which yields a negative result.
For example, if x = 4, except (x1) and (x3) all the other 45 values starting from (x5) would be negative and the overall product is a negative value < 0
Also, If x = 8, except (x1), (x3), (x5) and (x7) all the other 43 values starting from (x9) would be negative resulting in a negative value.. again < 0.
Similarly we can extend the logic to all the multiple of 4 till 92 for the given expression.
Hope it is clear.

Thanks for the explanation, what I wanted to know was how did we decide on the number 4, thought process. TIA :)

@PiyusJain
(x1)(x3)(x5)....(x93) has to be negative.First hint  x has to be less than 93 otherwise the value will not be negative
Second hint  x cannot be an odd number less than or equal to 93  as the value will be 0 in this case.
Conclusion  1  x has to be a subset of even numbers less than 93
Third hint  There are 47 terms in the expression and we need odd numbers of negative values to get a final negative value as two negatives cancels out to yield a positive value
If x = 2, 46 terms are negative (overall value  positive)
if x = 4, 45 terms are negative (overall value  negative)
if x = 6, 44 terms are negative (overall value  positive)
if x = 8, 43 terms are negative (overall value  negative)
and so on.
Conclusion 2  x can take values  4, 8, 12 ... 88, 92  23 values
Clear ? :thumbsup_tone2:

Awesome, Thanks :)