Quant Boosters  Kamal Lohia  Set 3

Number of Questions  30
Topic  Quant Mixed Bag
Solved ? : Yes
Source :

Q1) The remainder when 7^2020 is divided by 2400 is
A. 0
B. 1
C. 31
D. 1201

Sol: You just need to observe that 7⁴ = 2401 = 1 mod 2400
So 7^2020 = (7⁴)^505 = 1^505 mod 2400 = 1 mod 2400

Q2) Consider a sequence of 9 consecutive integers. The average of first 7 integers is N. The average of last 7 integers is
A. N
B. N + 2
C. 3N/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.

Q3) 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)
A. 80
B. 60
C. 50
D. 30

Base7 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.

Q4) Let G be an odd positive integer greater than 99 and T = G²  G. Then T²  2T is certainly divisible by
A. 9
B. 24
C. 105
D. 1024

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!

Q5) How many positive integers less than 210 are divisible by neither 2, 3, 5 nor 7?
A. 24
B. 48
C. 120
D. 162

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.

Q6) How many three digit numbers are there which have at least one digit as 3 and are divisible by 3?
A. 100
B. 94
C. 90
D. 84

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.

Q7) The number of zeroes after decimal and before first nonzero digit in the decimal representation of 5^( 20) is
A. 7
B. 8
C. 12
D. 13

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 nonzero digit are = 20  7 = 13.

Q8) 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 straightline 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 AMGM 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.

Q9) If a, b and c are the sides of a triangle, and a² + b² + c² = bc + ca + ab, then the triangle is
a. equilateral
b. isosceles
c. rightangled
d. obtuseangled

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.

Q10) 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)?
a. 0
b. 4
c. 2n – 1
d. Cannot be determined