NDA Mock Test: Mathematics - 1 - NDA MCQ

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.

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.

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.

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 ⁴P₀ = 1
at x = 4 ³P₁ = 3 
at x = 5 ²P₂ = 2

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.

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.

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

ax + by = u , cx +dy = v , 

since the solution is unique in integers.

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

⇒ (x-y)(y-z) (z-x) (x+y+z)

=1(-5) -2(10) + 3(11)
=-5-20+33 = 8

=1(-30) - 6(20) + 3(66)

= -30-120+198= 48

D = 8 ⇒ 6D = 48

Apply C₂ → C₂ + C_3,

because , the value of the determinant is zero only when , the two of its rows are identical., Which is possible only when Either x = 3 or x = 4 .

∣z−5i∣/∣z+5i| = 1
 ∣z−5i∣=∣z+5i| (Using definition ∣z−z1∣=∣z−z2∣ given Perpendicular bisector of z1 and z2.)
⇒ Perpendicular bisector of points (0,5) and (0,−5). which lies on x-axis.

 Let, z = x + iy
Putting in the given inequality,
|(x−4)−iy| < |(x−2)+iy|
⇒ [(x−4)² + y²]11/2 < [(x−2)² + y²]^1/2
squaring both sides,
⇒ (x−4)² + y² < (x−2)² + y²
⇒ −8x+16 < −4x+4
⇒ 4

The lady can select one cotton saree out of 15 cotton varieties in 15 ways since
any of 15 varieties can be selected. Corresponding to each selection of a cotton saree, she can
choose a polyester saree in 13 ways. Hence the two sarees (one cotton and one polyester), by
multiplication principle of counting, can be selected in 15 x 13= 195 ways

Out of 30 Boys,
No. of ways to select 1 boy = ³⁰C₁ = 30
Out of 18 Girls,
No. of ways to select 1 boy = ¹⁸C₁ = 18
No. of ways of selecting 1 boy and 1 girl =
30*18 = 540

n = 10

Middle term = (n/2) + 1
= (10/2) + 1
= 6th term

T(6) = T(5+1)
= ¹⁰C₅[(2x²)/3]⁵ [(3/2x²)]⁵
= ¹⁰C₅
= 252

The total number of terms in the binomial expansion of (a + b)ⁿ is n + 1, i.e. one more than the exponent n.

a = 4, l = 31, n = 8
(l-a)/(n+1) = (31-4)/(8+1)
= 27/9 
= 3

(aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) is mean between a & b
Mean of a & b =  ( a + b)/2
=> (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = ( a + b)/2
=> 2aⁿ⁺¹ + 2bⁿ⁺¹  = aⁿ⁺¹  + bⁿ⁺¹  + baⁿ + bⁿa
=> aⁿ⁺¹ + bⁿ⁺¹ = baⁿ + bⁿa
=> aⁿ(a - b) = bⁿ(a - b)
=> aⁿ = bⁿ
=> (a/b)ⁿ = 1
=> n = 0    or a  = b

Let the digits at ones, tens and hundreds place be (a−d)a and (a+d) respectively. The, the number is
(a+d)×100+a×10+(a−d) = 111a+99d
The number obtained by reversing the digits is
(a−d)×+a×10+(a+d) = 111a−99d
It is given that the sum of the digits is 21.
(a−d)+a+(a+d) = 21                        ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a−99d = 111a+99d−396          ...(ii)
⟹ 3a = 21 and 198d = 396
⟹ a = 7 and d = -2
Original number = (a−d)×+a×10+(a+d)
= 100(9) + 10(7) + 5
= 975

Let A(1,3), B(-2,9), and C(x,-1)
For to be points collinear,
x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)=0
⇒ 1(9-(-1)) + (-2)(-1-3) + x(3-9)=0
⇒ 1(10)+(-2)(-4)+x(-6)=0
⇒ 10+8-6x=0
⇒ 18 = 6x
⇒ x = 3

P(2,3) Q(3,5) R(1,2)
R is at centre between P and Q, using section formula for internal division
Therefore, (1,2) = ((3λ+2)/(λ+1), (5λ+3)/(λ+1))
1 = (3λ+2)/(λ+1)
(λ+1) = (3λ+2)
λ = -1/2
- sign indicates the external division

Given circle = (2,3)
Given line (x+y-1) = 0
Distance between point to line is:
d = |ax₁ + by₁ + c|/√(a² + b²)
where a = 1, b = 1, c = -1 and x1 = 2, y1 = 3
d = |1(2) + 1(3) - 1|/√(1+1)
d = 4/(√2)
d = 2√2

 Equation : 3x + 4y = 15
Points : (1,-3)
3(x-1) + 4(y-(-3)) = 15
3(x-1) + 4(y+3) = 15
3x - 3 + 4y + 12 = 15
3x + 4y = 6

Equation : 5x + 8y = 10
Points (2, -3)
(x-2, y+3)
⇒ 5(x-2) + 8(y+3) = 10
= 5x - 10 + 8y + 24 = 10
⇒ 5x + 8 = - 4