# Quant with Kamal Lohia - Part 2

• Q1) A motorboat moves from point A to point B and back again. Both points being located on the river bank. If the speed of the boat in still water is doubled, then the trip from A to B and back again would take 20% of the original time. The ratio of the actual sped of motorboat to the speed of river is?
a (3/2)^1/2
b (3^1/2)/3
c 2/3
d 3/2

Let a and be the speed of motorboat in still water and that of river.
As distance travelled is same in both cases, so time taken is inversely proportional to speeds in two scenarios.
Also speed in the two scenarios is Harmonic mean of upward and downward motion in both the scenarios.
Just putting all the expressions in place, we get
2(2a-b)(2a+b)/4a = 10(a-b)(a+b)/2a
i.e. 4a² - b² = 10a² - 10b²
i.e. 6a² = 9b²
i.e. a/b = √(3/2) i.e. option (a)

Q2) If the nth day of August lies on the same day as the 2nth day of October, then how many values of n are possible?

Let's say nth Aug = 0 mod7
then nth Sep = 3 mod7
and nth Oct = 5 mod7
Given 2nth Oct = 0 mod7
=> n = 2 mod7 and also n => n = 2, 9 i.e. two values.:)

Q3) A function f(x) is defined as f(x + y) = f(x) + f(y), for all real values of x and y. What is the value of f(2/3)
a) 0
b) 2/3
c) 2/3f(1)
d) 3/2f(2)
e) None of these

As f(0) = 0
f(2) = 2f(1)
f(3) = 3f(1)
f(2/3) = 2/3f(1)
See f(x) = x is one such function which satisfy the given equation.
for f(x) = x, f(2/3) = 2/3 only.
So option 2) and 3) both are correct

Q4) How many 5 digit numbers are there such that, digit at hundredth place ,unit's place & ten thousandth place are the first three terms of Geometric progression in any order?

If I am not understanding it wrongly, then it asks about the number of 5 digit numbers say abcde such that c, e, a are three consecutive terms of a GP. If that's so, then the three terms can be (1, 1, 1),(2, 2, 2)...(9, 9, 9), (1, 2, 4), (1, 3, 9), (2, 4, 8 ), (4, 6, 9) and there reverse also. Remaining two numbers can be arranged in 10*9 = 90 ways. And these 3 GP terms can be arranged in 9 + 2 + 2 + 2 + 2 = 17 ways.
So total such five digit numbers = 1530

Q5) A two-digit number having distinct digits when divided by the sum of the digits gives the same remainder as when a two-digit number that is formed by reversing the digits of original number is divided by the sum of the digits. Out of all such possible two-digit numbers, a number is randomly picked. What is the probability that this number is divisible by 4?
a) 3/8
b) 5/12
c) 2/7
d) 7/12

Let's say the number is 10a + b and after reversing it becomes 10b + a.
Both these numbers give same remainder with a + b. That means there difference must be divisible by a + b or a + b must be a divisor of the difference of two numbers which is 9(a - b).
So a + b can be 3, 6(if a - b = 2k), 12(if a - b = 4k), 9.
Just check the cases then.
There are 16 such numbers (12, 21, 15, 51, 24, 42, 18, 81, 27, 72, 36, 63, 45, 54, 48, 84) out of which 6 numbers (12, 24, 36, 72) are divisible by 4. So the required probability is 6/16 = 3/8.

Q6) How many sets of three distinct factors of the number N = 2^6 3^4 5^2 can be made such that the factors in each set has a highest common factor of 1 with respect to every other factor in that set?
a)236
b)360
c)104
d)380

Let p1, p2, p3 represents the three prime numbers and their powers.
Let's make the cases as:
(1, p1, p2) = 24 + 12 + 8 = 44 cases
(1, p1, p2p3) = 48 + 48 + 48 = 144 cases
(p1, p2, p3) = 48 cases
Total = 44 + 144 + 48 = 236 cases.

Q7) 2 friends decide to meet between 1 & 3 p.m. Anybody who reaches first will wait for 20 min & then leave. What is the probability that they meet.

the time interval given is from 1 pm to 3 pm i.e. 2hrs or 120 minutes and 20 minutes waiting time is 1/6 part of that
Answer will be ratio of grey area/total area = (6^2 - 5^2)/6^2 = 11/36. :)

Another down to earth and in-the-box approach.
Divide the question in two parts - first when first friend is coming in first 100 minutes for sure (with a probability = 5/6) and second one is coming in next adjacent 20 minutes (with a probability = 1/6). So total becomes = 2*(5/6)*(1/6) = 10/36. (I hope you won't ask the reason for that 2 )

Second when both friends are coming in last 20 minutes with a probability (1/6)(1/6) = 1/36.

So the required answer becomes = 10/36 + 1/36 = 11/36 :)

Q8) which is Greater 200^300 or 300^200 or 400^150?

200^300 = (2^300)(100^300) = (8^100)(100^300) = 8000000^100
300^200 = (3^200)(100^200) = (9^100)(100^200) = 90000^100
That means 200^300 > 300^200

Also,
200^300 = (200^2)^150 = 40000^150 > 400^150

Q9) One day Vikram was out bicycling. After entering a one-way tunnel and after having ridden one-fourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr?
a) 32 (b) 36 (c) 40 (d) 48 (e) none of the foregoing

Let the bus is at a distance of "a" miles from the tunnel whose length is "x" miles. And also top speed of cyclist be V miles per hour.
=> a/(x/4) = 80/V (1)
Also, (a+x)/(3x/4) = 80/V (2)
Equating LHS of (1) and (2)
=> 2a = 2
Hence, V = 40 miles per hour. Option (c) is correct.

Q 10) Compute the sum, S=1 X 9 + 2 X 99 + 3 X 999 + 4 X 9999 + … + n (999…..999)
S = 1(10 – 1) + 2(100 – 1) + … + n(100…n zeroes – 1)
S = 1(10^1) + 2(10^2) + … + n(10^n) – (1 + 2 + … + n)
Now let S = P + Q
where P = 1(10^1) + 2(10^2) + … + n(10^n) and Q = (1 + 2 + … + n) = n(n + 1)/2
Now 10P = 1(10^2) + 2(10^3) + … + n(10^n+1)
=> 9P = -(10^1 + 10^2 + … + 10^n) + n(10^n+1)
=> P = n(10^n+1)/9 – 10(10^n – 1)/81

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