Mathematics Test

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Mathematics Test - 1

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Question 1
2.5 / -0.83

Let X = {x | x = 2 + 4k, where k = 0, 1, 2, 3,...24}. Let S be a subset of X such that the sum of no two elements of S is 100. What is the maximum possible number of elements in S ?

A
10

B
11

C
13

D
12

Solution

Calculation:
The set X is given by
{2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98}.
We want to find the maximum size of a subset S of X such that no two
elements sum to 100.
The pairs in X that sum to 100 are
(2, 98), (6, 94), (10, 90), (14, 86), (18, 82), (22, 78), (26, 74), (30, 70), (34,
66), (38, 62), (42, 58), (46, 54).
Therefore,
S = {2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50}
∴The maximum possible number of elements in S be 13.
Question 2
2.5 / -0.83

In a party of 150 persons, 75 persons take tea, 60 persons take coffee and 50 persons take milk. 15 of them take both tea and coffee, but no one taking milk takes tea. If each person in the party takes at least one drink, then what is the number of persons taking milk only ?

A
50

B
40

C
30

D
20

Solution

Let, x number of persons taking milk only.
According to the question
60 + 15 + (x - 5) + (50 - x) + x = 150
120 + x = 150
x = 150 - 120 = 30
∴The required value is 30.
Question 3
2.5 / -0.83

If R is a relation from a non –empty set A to a non –empty set B, then

A
R ⊆A ×B.

B
R = A ∪B

C
R = A ∩B

D
R =A ×B

Solution

Let A and B be two sets. Then a relation R from set A to set B is a subset of A ×B. Thus, R is a relation from A to B ⇔R ⊆A ×B.
Question 4
2.5 / -0.83

The range of the function f(x) = ^7-x P_x-3 is

A
{1, 2, 3}

B
{1, 2, 3, 4, 5, 6}

C
{1, 2, 3, 4}

D
{1, 3}

Solution

Here, 0 ≤x- 3 ≤7 - x
⇒0 ≤x - 3 and x - 3 ≤7 - x
By solvation, we will get 3 ≤x ≤5
So x = 3,4,5 find the values of ^7-x P_{x - 3} by substituting the values of x
at x = 3 ⁴ P₀ = 1
at x = 4 ³ P₁ = 3
at x = 5 ² P₂ = 2
Question 5
2.5 / -0.83

Let R be the relation over the set of straight lines of a plane such that l₁ R l₂ ⇔l₁ ⊥l₂ . Then, R is

A
Symmetric

B
Reflexive

C
Transitive

D
An equivalence relation

Solution

To be reflexive, a line must be perpendicular to itself, but which is not true. So, R is not reflexive
For symmetric, if l₁ R l₂ ⇒l₁ ⊥l₂ .
⇒ l₂ ⊥l₁ ⇒l₁ R l₂ hence symmetric
For transitive, if l₁ R l₂ and l₂ R l₃
⇒l₁ R l₂ and l₂ R l₃ does not imply that l₁ ⊥l₃ hence not transitive.
Question 6
2.5 / -0.83

Let f be a function such that f(mn) = f(m) f(n) for every positive integers m and n. If f(1), f(2) and f(3) are positive integers, f(1)
(2019)

A
12

B
13

C
14

D
15

Solution

Given f(mn ) = f(m) f(n) and f(24) = 54
⇒f (24) = 2 ×3 ×3 ×3
⇒f(2 ×12) = f(2) f(12) = f(2) f(2 ×6)
= f(2) f(2) f(6) = f(2) f(2) f(2 ×3)
= f(2) f(2) f(2) f(3) = 2 ×3 ×3 ×3
Given that f(1). f(2) and f(3) are all positive integers hence by comparison, we get
f(2) = 3 and f(3) = 2
Hear we may safely consider f(1) = 1
Now, f(18 ) = f(2) (9) = f(2) f(3 ×3)
= f(2) f(3) f(3) = 3 ×2 ×2 = 12.
Question 7
2.5 / -0.83

Let f(x) = min{2x² , 52 −5x}, where x is any positive real number. Then the maximum possible value of f(x) is
(2018)

A
29

B
30

C
31

D
32

Solution

The maximum value of f(x) will occur when
2x² = 52 –5x i.e. when 2x² + 5x −52 = 0
⇒2x² + 13x −8x −52 = 0
⇒(2x + 13) (x –4) = 0 ⇒x = –13/2 or 4. But x is any positive real number. So, x = 4.
Hence, maximum value of f (x) = 2(4² ) = 32
Question 8
2.5 / -0.83

A

B

C

D
none of these.

Solution
Question 9
2.5 / -0.83

Let a, b, c, d, u, v be integers. If the system of equations, a x + b y = u, c x + dy = v, has a unique solution in integers, then

A
ad –bc = ±1

B
ad –bc = - 1

C
ad –bc = 1

D
ad - bc need not be equal to ±1.

Solution

ax + by = u , cx +dy = v ,

since the solution is unique in integers.
Question 10
2.5 / -0.83

A
(x –y)(y –z)(z –x)

B
2(x –y)(y –z)(z –x)

C
(x –y)(y –z)(z –x)(x+y+z)

D
none of these

Solution

⇒(x-y)(y-z)[y² + yz+z² - x² - xy - y² ]
⇒(x-y)(y-z) (z-x) (x+y+z)