Quant with Kamal Lohia - Part 7
Faculty and Content Developer at Tathagat | Delhi College of Engineering
Q1) Consider a sequence of 9 consecutive integers. The average of first 7 integers is N. The average of last 7 integers is
B. N + 2
D. N + (7/9)
Average of first seven integers (i.e. 1st, 2nd, 3rd, 4th, 5th, 6th, 7th) was the middle one i.e. 4th integer which is given N.
And average of last seven integers (i.e. 3rd, 4th, 5th, 6th, 7th, 8th, 9th) will also be the middle one i.e. 6th integer i.e. N + 2.
Q2) On planet LOGIKA, there are only 100 inhabitants out of which 30 are adults and 40 are not adults. 20 inhabitants died because of failure/crash of an earthly spacecraft on the planet. How many inhabitants are still there on the planet LOGIKA? (Whole incidence is narrated on planet LOGIKA only)
Base-7 is being used on the planet LOGIKA so that they write 30 + 40 = 100.
So after 20 inhabitants die, remaining ones are simply 30 + 40 - 20 = 50.
Q3) Let G be an odd positive integer greater than 99 and T = G² - G. Then T² - 2T is certainly divisible by
T² - 2T = T(T - 2) = (G² - G)(G² - G - 2) = (G - 2)(G - 1)(G)(G + 1) i.e. product of four consecutive integers which is certainly divisible by 4! = 24.
G is odd integer, only tells that first of these four consecutive integers is odd which doesn't affect the final result.
Remember that: Product of any 'n' consecutive integers is divisible by n!
Q4) How many positive integers less than 210 are divisible by neither 2, 3, 5 nor 7?
Out of every 2 consecutive integers, 1 is divisible by 2 and 1 not.
Out of every 3 consecutive integers, 1 is divisible by 3 and 2 not.
Out of every 5 consecutive integers, 1 is divisible by 5 and 4 not.
Out of every 7 consecutive integers, 1 is divisible by 7 and 6 not.
And here we are looking for the numbers which are not divisible any of 2, 3, 5, 7..i.e. (1/2)(2/3)(4/5)(6/7) of 210 = 48.
Q5) How many three digit numbers are there which have at least one digit as 3 and are divisible by 3?
Total three digit numbers = 9 * 10 * 10 = 900 and divisible by 3 among them are = 900/3 = 300
Now three digit number which do not contain 3 are = 8 * 9 * 9 = 648 and divisible by 3 among them are = 648/3 = 216
So the required numbers, here, are = 300 - 216 = 84.
Q6) The number of zeroes after decimal and before first non-zero digit in the decimal representation of 5^(- 20) is
5^(- 20) = 1/5^20 = 2^20/10^20 = (1024)(1024)/10^20 = (7digit number)/10^20
i.e. number of zeroes after decimal and before first non-zero digit are = 20 - 7 = 13.
Q7) There is a square field of side 500 m long each. It has a compound wall along its perimeter. At one of its corners, a triangular area of the field is to be cordoned off by erecting a straight-line fence. The compound wall and the fence will form its borders. If the length of the fence is 100 m, what is the maximum area that can be cordoned off?
a. 2,500 sq m
b. 10,000 sq m
c. 5,000 sq m
d. 20,000 sq m
Maximum Area for a right triangular with hypotenuse 'L' is equal to = L²/4 (it can be easily derived by using AM-GM inequality or could be understood by symmetry also that two legs of right triangle should be of equal length for maximum area with fixed hypotenuse length)
So answer is 100²/4 = 2,500 sq m.
Q8) If a, b and c are the sides of a triangle, and a² + b² + c² = bc + ca + ab, then the triangle is
a² + b² + c² = bc + ca + ab
i.e. 2(a² + b² + c²) = 2(bc + ca + ab)
i.e. (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²) = 0
i.e. (a - b)² + (b - c)² + (c - a)² = 0
i.e. a = b = c
i.e. triangle is equilateral.
Alternately, you can easily verify by putting a = b = c directly that whether triangle is equilateral or not and so on.
Q9) ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. From any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and stays there. Let a(n) be the number of distinct paths of exactly n jumps ending in E. Then what is the value of a(2n – 1)?
c. 2n – 1
d. Cannot be determined
E is four steps ahead of A. So it is not possible to reach E from A in (2n - 1) i.e. odd number of steps.
Q10) A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was 5 less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them 8 m apart. What is the length of the side of his property and how many posts did he buy?
a. 100 m, 15
b. 100 m, 16
c. 120 m, 15
d. 120 m, 16
Let initially n + 1 posts were to be used i.e. the length of the fence must have been 6n as they were to be posted at a gap of 6 m from each other. So according to given scenario, we get
8(n - 5) = 6n
i.e. n = 20
i.e. fence length = 6n = 120 m
and number of posts bought = (n + 1) - 5 = 21 - 5 = 16.