Quant with Kamal Lohia  Part 12

Q1) Which is larger 5^7 or 7^5?
You can remember a general rule for this type of problems as:
a^b > b^a where a < b.
There are exactly 2 exceptions to this rule i.e. 2^3 < 3^2 and 2^4 = 4^2. Except these two, always (smaller number)^(larger number) > (larger number)^(smaller number).Q2) A man builds 1/8th of a wall everyday. out of the length of the wall built par day , 20% falls off each day(including last day's work). In how many days he can complete the wall ?
The man builds 1/8 of the wall every day i.e. 12.5% of total. Simultaneously, 20% of the total wall built by that day day falls each day.
So on first day, amount of wall built = 12.5% and amount that fell = 20% of 12.5% = 2.5%. So remaining amount of built wall after first day = 12.5%  2.5% = 10% of total.
Now second day, amount of wall built = 12.5% of total and amount that fell = 20% of (10 + 12.5)% = 4.5%. So remaining amount of wall after second day = 10% + 12.5%  4.5% = 18% of total.
Same way, remaining amount of wall after third day = 18% + 12.5%  20% of (18% + 12.5%) = 24.4% of total.... and so on.
You can easily see that amount of wall being built per day is constant i.e. = 12.5% BUT amount of wall that fell each day is increasing as it is 20% of the wall built by that day.
So the moment when 20% of wall built by that day (say x) = 12.5%, from that day demolition will overpower construction.
And you can easily calculate the value of this 'x' above as 20% of x = 12.5% of total, so 100% of x = x = 5*12.5% = 62.5% of totalQ3) 4 vehicles are travelling on a straight line with constant speeds. truck overtook car at 2:00 pm and then met the scooter at 4:00pm and motorcycle at 6:00pm. the motorcycle met the car at 7:00 pm and then it overtook scooter at 8:00pm. At what time scooter and car meet ?
Let T, C, S and M be the speeds of truck, car, scooter and motorcycle respectively.
Scooter and Car will meet after 4 pm. Let they meet 'x' hours after 4 pm.
Now equating the distances travelled by the vehicles we get
=> (2 + x)C + xS = 2T (between 2 and 4 pm)
and 2T = 2M  4S (between 4 and 6 pm)
and 4T = 5C + M (between 2 and 6 pm)
Solving the three equations, we get x = 4/3.
Time is 4 + 4/3 pm = 16/3 pm = 5 : 20 pmQ4) How many sets of two numbers will have the LCM as P1^a * P2^b, where P1 & P2 are distinct prime numbers and a, b are natural numbers?
Now both numbers in the set will contain some powers of the two prime numbers. Only important point to note that is at least one of the two numbers in the set SHOULD contain the all powers of P1 i.e. P1^a and same way at least one number SHOULD contain all powers of P2 i.e. P2^b.....as we need these in the LCM.
Rest is child's play. We just need to distribute powers of P1 in both the numbers (that can vary from 0 to a), which can be done in (a + 1)^2 ways..(a + 1) each for both numbers.
Now subtracting the ways when none of the numbers contain all 'a' powers i.e. the two numbers contain powers from 0 to (a  1) only i.e. in 'a' ways each i.e. in total a2 ways, we are left with a total of (a + 1)^2  a^2 = (2a + 1) possible ways.
Exactly in same way there are (2b + 1) ways to distribute powers of P2 in the two numbers such that at least one of the number contains all the 'b' powers of P2.
So total cases become = (2a + 1)(2b + 1). smile
Still, this is not the end as this is ordered counting i.e. (1, N) and (N, 1) have been counted twice. Actually all cases except the (N, N) one have been counted twice.
So unordered number of sets/pairs are = {(2a + 1)(2b + 1) + 1}/2
Q 5) In the image below:  AB and AD are tangent to the circle  BC and AD are parallel. What is the length of AC?
Solution:
As BC  AD, ∠CDT = ∠DCB = θ
Also ∠DBC = ∠CDT = θ [Alternate segment theorem]
So triangle BDC is isosceles and BD = CD = 28.
Also ∠ADB = ∠DCB = θ [Alternate segment theorem]
and ∠ADB = ∠ABD = θ [AD and AB both are tangents to circle]
So triangles ABD and DBC are similar and by comparing the ratio of corresponding sides, we get that
28/49 = BC/28
i.e. BC = 16Now in triangle ABC, cos(2θ) = (49² + 16²  AC²)/(2 * 49 * 16)
and in triangle ABD, cos(180  2θ) = (49² + 49²  28²)/(2 * 49 * 49)
As cos(2θ) + cos(180  2θ) = 0, we get
49(49² + 16²  AC²) + 16(49² + 49²  28²) = 0
(49² + 16²  AC²) + 16(49 + 49  4²) = 0
49² + 32(49) = AC²
AC² = 49(49 + 32) = 49(81) = 7²(9²) = 63²
AC = 63.Q6) In Sivkasi,each boy's quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25 , he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?
a) 200
b) 150
c) 125
d) 175Let 'b' be the number of boxes which a boy fills and 'n' be the number of match sticks he put in each box such that nb < 200.
Now given that if he reduces the number of sticks per box by 25, he can fill 3 more boxes with total number of sticks assigned to him.
So, we can write that: nb = (n  25)(b + 3)
i.e. 3n  25b = 75 where n and b both are positive integers.
So possible solutions for (n, b) are: (50, 3), (75, 6), (100, 9), .. and so on.
But we have the condition that nb < 200, so only first case satisfy i.e. nb = (50)(3) = 150.Q7) Four cups of milk are to be poured into a 2cup bottle and a 4cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4cup bottle?
(a) 2/3
(b) 7/3
(c) 5/2
(d) 8/3Total milk to be filled is 4 cups and total capacity available is 2 + 4 = 6 cups i.e. in all we can fill 4/6 = 2/3 of both vessels. Also as each vessel should be filled in equal proportion, that means both the vessel contains 2/3 of their capacity. So 4 cup vessel contains (2/3) * 4 = 8/3 cups of milk.
Q8) Find the sum of first n terms of the series 1.2.3 + 2.3.4 + 3.4.5 + ..... ?
See the pattern:
1×2×3 = (1/4) × [1×2×3×4  0×1×2×3]
2×3×4 = (1/4) × [2×3×4×5 1×2×3×4]
...
n(n + 1)(n + 2) = (1/4) × [n(n + 1)(n + 2)(n + 3)  (n  1)n(n + 1)(n + 2)]=> The required sum is = (1/4) × [n(n + 1)(n + 2)(n + 3)] = [(n² + 3n + 1)²  1]/4.
Q 9) What is the remainder when 128 ^ 1000 is divided by 153.
This is simply application of Chinese Remainder Theorem.
For two divisors D1 and D2 such that HCF(D1 , D2) = 1, first find integers x, y so that xD1 + yD2 = 1
Next let N ≡ r1 mod D1
and also N ≡ r2 mod D2
Then according to Chinese Remainder Theorem (CRT):
N ≡ (xD1r2 + yD2r1) mod D1D2.153 = 3^2 * 17 = 9 * 17
and 128^1000 = 2^1000mod9 = 2^(3 * 333 + 1)mod9 = 1 * 2mod9 = 7mod9.
also 128^1000 = (8)^1000mod17 = 2^3000mod17 = 2^(4 * 750)mod17 = 1mod17.
Also 9 * 2  17 * 1 = 1
Using Chinese Remainder Theorem
128^1000 = (9 * 2 * 1  17 * 1 * 7)mod153 = 101mod153 = 52mod153Q 10) Find the number of integer solution to the equation 1/x + 1/y = 1/48
Let x = 48 + a
and y = 48 + b
so 1/x + 1/y = 1/48
=> 1/(48 + a) + 1/(48 + b) = 1/48
=> 48(48 + a) + 48(48 + b) = (48 + a)(48 + b) = 48² + 48a + 48b + ab
=> ab = 48² = 2^8 3^2
Now we just need to find the number of ways in which RHS can be written as product of two integers.
See, number of (positive) factors of RHS = (8 + 1)(2 + 1) = 27.....so number of (ordered) pairs of positive factors possible = 27.
And including the negative values for a, b we get total 2 × 27 = 54 pairs.