## QUANTITAITVE APTITUDE CLUBBED MCQs

Total Questions : 1394 | Page 6 of 140 pages
Question 51. G(x) is defined as the sum of all the divisors of a natural number n.
If g(x) is less than 2n then it is placed in category A.
If g(x) is greater than 2n it is placed in category B.
Of the following numbers, which pair consists of numbers one of which can be placed in A and another in B?
1.    32,64
2.    6,64
3.    7,28
4.    14,84
:
D
g(32)=1+2+4+8+16+32=63<2×32 (Category A)
g(64)=1+2+4+8+16+32+64=127<2×64 (Category A)
g(6)=1+2+3+6=12=2×6 (neither A nor B)
g(7)=1+7=8<2×7 (Category A)
g(28)=1+2+4+7+14+28=56=2×28 (Neither A nor B)
g(14)=1+2+7+14=24<2×14 (Category A)
g(84)=1+2+3+4+6+7+12+14+21+28+42+84=224>2×84 (Category B)
Hence, option (d).
Question 52. If a, b, c are three consecutive natural numbers, then the expression (a + b - c) (a + c - b) (b + c - a) is:
1.    positive
2.    negative
3.    non - positive
4.    non - negative
Answer: Option D. -> non - negative
:
D
Take any three consecutive natural numbers as: n-1, n, n + 1
Now add any two and subtract third from the sum.
n - 1 + n - (n + 1) = n - 2; n + n + 1 - (n - 1) = n + 2; n - 1 + n + 1 - n = n
So you will get either n - 2 or n + 2 or n. As you are considering natural numbers, smallest natural number is 1 n -1 = 1 or n = 2. (So that n - 1 is a natural number)
So n - 2, n, n + 2 all three will be non - negative. So the given expression is also non - negative.
Question 53. Arjun can kill a bird once in 3 shots. On the assumption that he fires 3 shots, the probability that the bird is killed is x27. Find the value of x.
___

:
P(A) = 13
P(A') = 23
P(Bird Killed) = 1 - P(None of 3 shots hit)
1 23 ×23 × 23 = 1927
Question 54. In a school of 150 students, 50 students did not play football, 90 played cricket and 20 neither played football nor cricket
How many students played only football?
1.    40
2.    50
3.    60
4.    Cannot be determined
:
A

b + c = 150 - 50 = 100.
a + d = 50 (those who do not play football)
a + b = 90, d = 20
a + 20 = 50 a = 30
b = 90 - 30 = 60, c = 100 - 60 = 40.
Question 55. The sum of the first 100 odd numbers is divisible by:
1.    2, 4 and 8
2.    2 and 4
3.    2 only
4.    None of these
Answer: Option A. -> 2, 4 and 8
:
A
The sum of n consecutive odd numbers is given as n2 . Hence 1002 is divisible by 2, 4 and 8. Answer is option (a).
(Sum of n consecutive odd numbers =n2 is derived from unitary method as follows : 1 + 3 + 5 + 7 + 9 +...... is a series (AP) with a common difference of 2. Therefore, the 100th term would be 99×2+1= 199. Therefore the sum = (1+199)1002=10000 or n2.
Question 56. A rectangle is divided into four rectangles, and their respective areas are shown in the figure. (Figure is not drawn to scale). What is the missing area 'X' of the fourth rectangle?
25205x

1.    5sq units
2.    2sq units
3.    4sq units
4.    2.5 sq units
Answer: Option C. -> 4sq units
:
C
Approach 1:

You can directly mark the answer as 25 × x = 20 × 5 x=4
Proof:
Observe that
ac = 25 sq units ....(1)
bc= 20 sq units ... (2)
ad = 5 sq units ... (3)
Dividing (1) by (3), we get: cd = 5 units ... (4)
Dividing (1) by (2) we get ab =5/4 units ... (5)
​St Multiplying (4) and (5) we get:
Sit acbd = 254 sq units
Since its already given that ac = 25 sq units, therefore the missing area 'X' =4 square units.
Approach 2:
Use the simple number approach to break up the values in each cell as shown below

thus x=4x1=4
Question 57. In how many ways can 80 be written as the difference of squares of two natural numbers?
1.    2
2.    3
3.    4
4.    5
:
B
To write a number as a difference of squares of 2 natural numbers, we need to find the 2 factor products of the number, such that both the factors are either even or odd.
This is because if we are writing a number N as a difference of squares, then -
N=a2b2N=(a+b)(ab). If (a+b, a-b) is a pair of one odd and one even number, 'a' and 'b' can't be natural numbers.
Hence, both the factors need to be either even or odd.
In the case of 80, the possible cases are 80=2×40;=4×20;=8×10.
Thus, there are only three instances where both are even. Hence, the answer is option (b).
Question 58. There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where n>6 such that, for k = 1, 2, 3, 4, ....  2n, there are exactly k boxes labelled k. The total number of boxes is N. Now, consider the following two cases for the number of footballs (fi) in the box labelled i:
Case 1: fi=i+1, if i is odd; = i + 2 if i is even.
Case 2: fi=i+2, if i is odd; = i + 1 if i is even.
The difference between the total number of footballs in the N boxes in Case 1 and Case 2 is:
1.    √N
2.    √8N+1−14
3.    N(N+1)6
4.    √8N+1−12
:
B
Method 1- Conventional Approach
Clearly, in both cases, fi=(i+1) or (i+1) + 1.
Therefore in both the cases, (i+1) is common.
As we need to find the difference in number of footballs, we can ignore (i+1).
Thus, effectively in each box, there will be 0 or 1 football.
We get:
Case 1: fi=0 if i is odd, = 1 if i is even.
Case 2: fi=1 if i is odd, = 0 if i is even.
Upon observation, we can conclude that the required difference is nothing but the difference in the number of boxes with odd numbered labels and the number of boxes with even numbered labels.
Also, number of boxes with label (x) is 1 more than number of boxes with label (x-1).
From 2n labels, we get n pairs of labels with consecutive numbers.
Therefore, required difference is n×1=n.
Or Case 1 - Case 2 = [{2+4+6+....+(2n)}-{1+3+5+....+(2n-1)}]
=2n(n+1)2n2=n
Total number of boxes = 1+2+3+....+2n= n(2n+1) = N
Therefore, 2n2+nN=0
n=1±1+8N4
Since n cannot be negative, n=(8N+1)14.
Method 2- Assumption
This is a question based on Variables to Variables.
Assume n=1, hence the number of boxes labelled 1, will be 1 and the number of boxes labelled 2 will be 2. Total number of boxes (N) = 3.
Case 1
When i=1,f1=2 and when i=2,f2=4
Case 2
When i=1,f1=3 and when i=2,f2=3
Difference in total number of footballs in both cases = (2+4+4) - (3+3+3) = 1
Now, on substitution of N=3 in options, eliminate those options where you are not getting 1.
Only option (b) satisfies the required condition.
Question 59. Find the minimum value of the function f(x) = |x-1| + |x-2| + |x-3| + ... + |x-20|
1.    190
2.    172
3.    96
4.    100
:
D
The minimum value of such an expression occurs when x takes the middle value in the given range i.e. when x is between 10 and 11. And the value of the function f(x) at that value becomes:
f(x=10) = 9+8+7+6+5+4+3+2+1+0+1+2+3+4+5+6+7+8+9+10 = 100
Also,f(x=11) = 10+9+8+7+6+5+4+3+2+1+0+1+2+3+4+5+6+7+8+9=100
Question 60. How many positive even integers less than 50 can be written as a sum of three consecutive positive integers?
1.    7
2.    4
3.    10
4.    8