Quantitative Aptitude PYQs

Calculation:

Odd digits - 1, 3, 5, 7, 9

A four-digit number with all digits odd has 5 choices in the first position, and since digits can be repeated, there are 5 choices for the second, third, and fourth positions also.

So, the number of four-digit natural numbers such that all of the digits are odd = 5 × 5 × 5 × 5 = 625

Concept:

Binomial expansion of (x + y)ⁿ is given by

(x + y)n = ⁿC₀xⁿ + nC1xn-1y + nC2xn-2y²+.....+ nCn-1xy^n-1 + nCnyn  

Calculation:

Given,  \(\displaystyle\sum_{r=0}^n\)2r C(n, r)

Expanding the expression,

⇒ nC020 + nC1 2¹ + nC222+.....+ nCn-12n-1 + nCn2n  

⇒ nC01n 20 + nC1 1n-1 2¹ + nC21n-2 22+.....+ nCn-1 2n-1 + nCn2n  

Comparing with binomial expansion x = 1 and y = 2

⇒  \(\displaystyle\sum_{r=0}^n\)2r C(n, r) = (1 + 2)ⁿ

⇒  \(\displaystyle\sum_{r=0}^n\)2r C(n, r) = 3n

∴ The correct option is (2).

Concept:

The number of ways of arranging n objects out of which p are of one type, q are of the second type is \(n! \over p!q!\)

Calculation:

Letters in the word 'MATHEMATICS' in alphabetical order are

AACEHIMMTTS

So in the list, the words that start with A only lie before the words start with C.

Now the number of words start with A are: 

{fix A at first position}

= \(10! \over 2! 2!\) = 907200

∴ The correct answer is option (3).

Concept:

A relation f from set A to set B is said to be a function if every element of set A has a unique image in set B.

Calculation:

1. If f is the subset of Z × Z defined by f = {(xy, x − y); x, y ∈ Z},

let x = 2, y = 3 → (2 × 3, 2 - 3) = (6, -1) ∈ f

and for x = -2, y = -3 → (-2 × (-3), -2 + 3) = (6, 1) ∈ f

So, the element 6 ∈ Z has two different images 1 and -1 in Z 

⇒ f is not a function.

2. If f is the subset of N × N defined by f = {(xy, x + y); x, y ∈ N},

 let x = 3, y = 4→ (3 × 4, 3 + 4) = (12, 7) ∈ f

and for x = 2, y = 6 → (2 × (6), 2 + 6) = (12, 8) ∈ f

So, the element 12 ∈ N has two different images 7 and 8 in N

⇒ f is not a function.

∴ The correct option is (4).

Explanation:

Given:

Δ = \(\left|\begin{array}{lll}\text{a}_{11} & \text{a}_{12} & \text{a}_{13} \ \text{a}_{21} & \text{a}_{22} & \text{a}_{23} \ \text{a}_{31} & \text{a}_{32} & \text{a}_{33}\end{array}\right|\)

 a13 = yz, a23 = zx, a33 = xy and M₁₃ = (z − y), M₂₃ = (z − x), M₃₃ = (y − x)

Expanding the determinant along Column 3,

⇒ Δ = a13 (a₂₁a₃₂ - a₂₂a₃₁) - a23 (a₁₁a₃₂ - a₁₂a₃₁) + a33 (a11a2 - a12a1)

⇒ Δ = a₁₃M₁₃ - a₂₃M₂₃ + a₃₃M₃₃

where M_ij denotes the minor of a_ij element.

⇒ Δ = yz(z - y) - zx(z - x) + xy(y - x)

⇒ Δ = yz(z - y) - z²x + zx² + xy² - x²y

⇒ Δ = yz(z - y) + (xy2 - z2x) + (zx2 - x2y)

⇒ Δ = yz(z - y) + x(y2 - z2) + x²(z - y)

⇒ Δ = (z - y)[yz - x(y + z) + x2]

⇒ Δ = (z - y)[yz - xy - xz + x2]

⇒ Δ = (z - y)[y(z - x) - x(z - x)]

⇒ Δ = (z - y)[(z - x)(y - x)]

∴ The correct answer is option (1).

Concept:

If a matrix A = \(\left(\begin{array}{lll}\text{a}_{11} & \text{a}_{12} & \text{a}_{13} \ \text{a}_{21} & \text{a}_{22} & \text{a}_{23} \ \text{a}_{31} & \text{a}_{32} & \text{a}_{33}\end{array}\right)\) then AT = \(\left(\begin{array}{lll}\text{a}_{11} & \text{a}_{21} & \text{a}_{31} \ \text{a}_{12} & \text{a}_{22} & \text{a}_{32} \ \text{a}_{13} & \text{a}_{32} & \text{a}_{33}\end{array}\right)\)

Explanation:

If A = \(\left(\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos θ & \sin θ \ 0 & \sin θ & −\cos θ\end{array}\right)\), 

the cofactor matrix of A is given by,

cofactor(A) = \(\left(\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos θ & -\sin θ \ 0 & -\sin θ & \cos θ\end{array}\right)\)

So, adj A = \(\left(\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos θ & -\sin θ \ 0 & -\sin θ & \cos θ\end{array}\right)\) (Transpose of cofactor matrix)

⇒ adj A = - A __(i)

⇒ adj A + A = 0 is a null matrix

So, statement 1 is correct.

Now, expanding A along Row 1,

|A| = 1(-cos²θ - sin²θ) = -1

⇒ A^-1 = \(1 \over |A|\)adj A = -adj A  

⇒ A-1 + adj A = 0 __(ii)

So, statement 2 is correct.

Putting the value of (i) in (ii),

⇒ A-1 - A = 0  

or A - A^-1 = 0

So, statement 3 is correct.

∴ The correct option is (4).

Concept:

If A is a matrix of order m × n and B is a matrix of order p × q, then AB is a matrix of order m × q with mq number of elements.

and AB is defined if and only if n = p

Calculation:

If X is a matrix of order 3 × 3, Y is a matrix of order 2 × 3 and Z is a matrix of order 3 × 2,

1.  (ZY)X is a square matrix having 9 entries.

Order of Z = 3 × 2, order of Y = 2 × 3,

⇒ Order of (ZY) = 3 × 3

And order of (ZY) = 3 × 3, order of X = 3 × 3,

⇒ order of (ZY)X = 3 × 3

⇒ (ZY)X is a square matrix having 3 × 3 = 9 entries.

So, statement 1 is true

2.  Y(XZ) is a square matrix having 4 entries.

Order of X = 3 × 3, order of Z = 3 × 2,

⇒ order of (XZ) = 3 × 2

And order of Y = 2 × 3, order of (XZ) = 3 × 2, 

⇒ order of Y(XZ) = 2 × 2

⇒ Y(XZ) is a square matrix having 2 × 2 = 4 entries.

So, statement 2 is true

3. X(YZ) is not defined.

Order of Y = 2 × 3, order of Z = 3 × 2,

⇒ order of YZ = 2 × 2

order of X = 3 × 3, order of YZ = 2 × 2

and 3 ≠ 2

⇒  X(YZ) is not defined.

So, statement 3 is true

∴ The correct answer is option (4).

Concept:

For a quadratic equation ax² + bx + c = 0

Sum of roots = -b/a

and the product of roots = c/a

Calculation:

If the sum of roots = product of roots

⇒ \(\frac{-b}{a}=\frac ca\)

⇒ -b = c

So, there can be infinitely many quadratic equations with -b = c and a ≠ 0

∴ The correct option is (4).

Concept:

Between any two real numbers, there are infinitely many rationals and irrational.

{The numbers which are not rational are irrational}

Real = Rational ∪ Irrational

Explanation:

1. The set of all irrational numbers between \(\sqrt{2}\) and \(\sqrt{5}\) is an infinite set.

There are infinitely many rationals and irrational between any two real numbers.

 \(\sqrt{2}\) and \(\sqrt{5}\) are two irrational numbers and there are infinitely many rationals and irrationals between them.

⇒ The set of all irrational numbers between \(\sqrt{2}\) and \(\sqrt{5}\) is an infinite set.

So, statement 1 is true.

2. The set of all odd integers less than 100 is a finite set.

The set of all odd integers less than 100 is given by

{......,-5 ,-3 ,-1 ,1 ,3 ,5 ,.....,95, 97, 99}

this set is bounded above but not bounded below

So, It is an infinite set, and statement 2 is false.

∴ The correct option is (1).

Concept:

• The sum of first n natural numbers = 1 + 2 + 3 + .. + n = \(n(n+1) \over 2\)
• The series of form a, a + d, a + 2d, ....., a + (n - 1)d is an AP and the sum of the first n terms of an A.P. is \({n \over 2}(a + a_n) \) where a_n is the n^th term.
• A number is prime if it has no factors except 1 and itself

Calculation:

1. 2 + 4 + 6 + ........ + 2n = n2 + n

 2 + 4 + 6 + ........ + 2n is an A.P.

with first term 2, number of terms n and last term 2n 

⇒ Sum of A.P. = \({n \over 2}(2 + 2n) \) = n(n + 1) = n2 + n

So statement 1 is true

2. The expression n2 + n + 41 always gives a prime number for every natural number n

If n = 41

n2 + n + 41 = 412 + 41 + 41= 41(41 + 1 + 1) = 41 × 43

As n2 + n + 41  has factors 41 and 43

⇒ n2 + n + 41 is not prime for n = 41 ∈ N

So, statement 2 is false

∴ The correct answer is option (1)

Concept:

If α and β are the roots of the equation ax² + bx + c = 0,

then sum of roots = α + β = -\(b \over a\)

and product of roots = αβ = \(c \over a\)

Calculation:

If p, q(p > q) are the roots of the quadratic equation x2 + bx + c = 0 

then, p + q = -b  __(i)

and pq = c   __(ii)

Squaring equation (i),

⇒ (p + q)² = (-b)²

⇒ p² + q² + 2pq = b²

⇒ p2 + q2 + 2c = b2  (From (i))

⇒ p2 + q2 = b2 - 2c  __(iii)

Take equation (iii) - 11(ii),

⇒ p2 + q2 − 11pq = b2 - 2c - 11c = 0

⇒ b² = 13c  __(iv)

Now (p - q)² =  p2 + q2 - 2pq

Put values from (ii) and (iii),

⇒ (p - q)2 =  b2 - 2c  - 2c = b2 - 4c

⇒ (p - q)2 =  13c - 4c {from (iv)}

⇒ (p - q)2 =  9c

⇒ p - q = 3√c

∴ The correct answer is option (1).

Concept:

Sine Rule:

In a triangle Δ ABC, where a is the side opposite to A, b is the side opposite to B, c is the side opposite to C and where R is the circum-radius:

\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)

The right-angled triangle follow the Pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

The hypotenuse is the greatest side of the right-angled triangle

Solution:

Given, Length of sides of polygon a = 1

Number of sides of polygon n = 12

 

The right-angled triangle follow the Pythagoras theorem,

\(\frac{sin 15}{0.5}=\frac{sin 75}{R}\)

\(\Rightarrow \frac{\sqrt3-1}{(2\sqrt2)0.5}=\frac{\sqrt6+\sqrt2}{4R}\)

\(\Rightarrow \frac{2(\sqrt3-1)}{(2\sqrt2)}=\frac{\sqrt6+\sqrt2}{4R}\)

\(\Rightarrow \frac{(\sqrt3-1)}{(\sqrt2)}=\frac{\sqrt6+\sqrt2}{4R}\)

\(\Rightarrow R=\frac{\sqrt6+\sqrt2}{4}\times \frac{\sqrt2}{\sqrt3-1}\)

\(\Rightarrow D = 2R=\frac{\sqrt6+\sqrt2}{4}\times \frac{2\sqrt2}{\sqrt3-1}\) ...... (Diameter D = 2R)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{2}\times \frac{\sqrt2}{\sqrt3-1}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt2 \sqrt2}\times \frac{\sqrt2}{\sqrt3-1}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt2}\times \frac{1}{\sqrt3-1}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2} \times \frac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}\)

\(\Rightarrow D =\frac{6+2\sqrt{12}+2}{4}\)

\(\Rightarrow D =\frac{8+2\sqrt{12}}{4}\)

\(\Rightarrow D =\frac{8+2\sqrt{4 \times3}}{4}\)

\(\Rightarrow D =2+\sqrt3\)

Concept:

The range of a function is the set of all possible outputs of the function.

Calculation:

Given  A = {7, 8, 9, 10, 11, 12, 13; 14, 15, 16} 

and f ∶ A → N be defined by f(x) = the highest prime factor of x.

Range of f = f(A)

⇒ Range of f = {f(7), f(8), f(9), f(10), f(11), f(12), f(13); f(14), f(15), f(16)}

⇒ Range of f = {f(7), f(2³), f(3²), f(2 × 5), f(11), f(2² × 3), f(13); f(2 × 7), f(3 × 5), f(2⁴)}  

⇒ Range of f = {7, 2, 3, 5, 11, 3, 13; 7, 5, 2}  

⇒ Range of f = { 2, 3, 5, 7,11,13}

⇒ |Range of f| = 6

∴ The correct answer is option (3).

Explanation:

Given, R = {(x, y): x, y ∈ N and x2 = y3}.

1. (x, x) ∈ R for all x ∈ N

 If (x,x) ∈ R ⇒ x2 = x3

⇒ x²(x - 1) = 0 

⇒ (x, x) ∈ R for x = 0, 1 only

So, statement 1 is not correct.

2. (x, y) ∈ R ⇒ (y, x) ∈ R

 let x = 8 and y = 4

⇒ x² = 64 = 4³ = y³

⇒ (x,y) ∈ R 

but y² = 16 ≠ 8³ ⇒ (y,x) ∉ R

So, statement 2 is not correct.

3. (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R

 If (x, y) ∈ R and (y, z) ∈ R

⇒ x2 = y3 and y2 = z3}

⇒ x2 = (z^3/2)3 

⇒ x2 = (z3/2)3 ≠ z^9/2

So, statement 3 is not correct.

∴ The correct answer is option (4).

Concept:

A ∪ B = {x| x ∈ A or x ∈ B}

A ∩ B = {x| x ∈ A and x ∈ B}

Explanation:

1. A ∩ B = A ∩ C ⇒ B = C

 If A = {1,2,3,4,5}, B = {1,2,p,q} and C = {1,2,m,n}

Here, A ∩ B = A ∩ C = {1,2}

But B ≠ C

So statement 1 is false.

2. A ∪ B = A ∪ C ⇒ B = C

 If A = {1,2,3,4}, B = {1,2,p,q} and C = {p,q}

Here,A ∪ B = A ∪ C = {1,2,3,4,p,q}

But B ≠ C

So statement 2 is false.

∴ The correct option is (4).

Concept:

If z = x + iy 

⇒ Modulus of z = \( |z| = \sqrt {x^2 +y^2}\)

Calculation:

Given, 

z = \(\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ}\) where i = \(\sqrt{−1}\)

Multiply numerator and denominator by (1 + i sin θ),

\(⇒ z =\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ} \times \frac{1+\text{i}\sin θ}{1+\text{i}\sin θ} \),

\(⇒ z =\frac{(1+\text{i}\sin θ)^2}{1^2−(\text{i}\sin θ)^2} \)

\(⇒ z =\frac{(1+2{i}\sin θ - \sin^2\theta )}{1+\sin^2 θ} \)

\(⇒ z =\frac{(1 - \sin^2\theta )}{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}\)

\(⇒ z = \frac{ \cos^2\theta }{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}\)

\(⇒ |z| =\sqrt{(\frac{ \cos^2\theta }{1+\sin^2 θ})^2 +(\frac{2\sin θ}{1+\sin^2 θ})^2}\)

\(⇒ |z| ={1 \over 1+ \sin^2 \theta }\sqrt{\cos^4 \theta + 4\sin^2 \theta}\)

\(⇒ |z| = {1 \over 1+ \sin^2 \theta }\sqrt{(1 -\sin^2 \theta)^2 + 4\sin^2 \theta}\)

\(⇒ |z| ={1 \over 1+ \sin^2 \theta }\sqrt{1 -2\sin^2 \theta + \sin^4 \theta + 4\sin^2 \theta}\)

\(⇒ |z| ={1 \over 1+ \sin^2 \theta }\sqrt{1 +2\sin^2 \theta + \sin^4 \theta }\)

\(⇒ |z| ={1+ \sin^2 \theta \over 1+ \sin^2 \theta } = 1\)

∴ The correct answer is option (1).

Concept:

If z = x + iy is purely real, then y = 0

Calculation:

Given, 

z = \(\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ}\) where i = \(\sqrt{−1}\)

Multiply numerator and denominator by (1 + i sin θ) 

\(⇒ z =\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ} \times \frac{1+\text{i}\sin θ}{1+\text{i}\sin θ} \)

\(⇒ z =\frac{(1+\text{i}\sin θ)^2}{1^2−(\text{i}\sin θ)^2} \)

\(⇒ z =\frac{(1+2{i}\sin θ - \sin^2\theta )}{1+\sin^2 θ} \)

\(⇒ z =\frac{(1 - \sin^2\theta )}{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}\)

Since z is purely real,

⇒ \(\frac{2\sin θ}{1+\sin^2 θ} = 0\)

⇒ sin θ = 0

⇒ θ = nπ 

∴ The correct answer is option (3).

Concept:

If z = x + iy is purely imaginary, then x = 0

Calculation:

Given, 

z = \(\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ}\) where i = \(\sqrt{−1}\)

Multiply numerator and denominator by (1 + i sin θ), 

\(⇒ z =\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ} \times \frac{1+\text{i}\sin θ}{1+\text{i}\sin θ} \),

\(⇒ z =\frac{(1+\text{i}\sin θ)^2}{1^2−(\text{i}\sin θ)^2} \)

\(⇒ z =\frac{(1+2{i}\sin θ - \sin^2\theta )}{1+\sin^2 θ} \)

\(⇒ z =\frac{(1 - \sin^2\theta )}{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}\)

\(⇒ z = \frac{ \cos^2\theta }{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}\)

Since z is purely imaginary,

⇒ \(\frac{\cos^2 θ}{1+\sin^2 θ} = 0\)

⇒ cos²θ = 0

⇒ θ =  \(\frac{(2n+1)\pi}{2}\)

∴ The correct answer is option (2).

Solution:

There are two AP with the different first terms and common difference.

For the first AP A:

Let first term be a and common difference = d

Sum of n terms P = \(\frac{n}{2}[2a+(n-1)d]\)

and n^th term a_n = a + (n - 1)d

For the second AP B:

Let first term be M and common difference = D

Sum of n terms Q = \(\frac{n}{2}[2M+(n-1)D]\)

and n^th term M_n = M + (n - 1)D

It is given that

\(\frac{Sum\ of\ n\ terms\ of\ first\ A.P}{Sum\ of\ n\ terms\ of\ second\ A.P }= \frac{5n+4}{9n+6}\)

⇒ \(\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2M+(n-1)D] }= \frac{5n+4}{9n+6}\)

⇒ \(\frac{[2a+(n-1)d]}{[2M+(n-1)D] }= \frac{5n+4}{9n+6} \)

⇒ \(\frac{a+\left ( \frac{n-1}{2} \right )d}{M+\left ( \frac{n-1}{2} \right )D}= \frac{5n+4}{9n+6} \).....(1)

Now, we need to find the ratio of the first term of A and B

\(\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B}\)

= \(\frac{a}{M}\)

Hence, \(\frac{n-1}{2}=0\Rightarrow n=1\)

Putting n = 1 in equation (1), we get

\(\frac{a}{M}=\frac{5(1)+4}{9(1)+6}\)

⇒ \(\frac{a}{M}=\frac{9}{15}\)

⇒ \(\frac{a}{M}=\frac{3}{5}\)

⇒ \(\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B} =\frac{3}{5}\)

Hence the ratio of the first term of A and B = \(\frac{3}{5}\)

∴ The correct option is (4)

Solution:

There are two AP with the different first terms and common difference.

For the first AP A:

Let first term be a and common difference = d

Sum of n terms P = \(\frac{n}{2}[2a+(n-1)d]\)

and nth term an = a + (n - 1)d

For the second AP B:

Let first term be M and common difference = D

Sum of n terms Q = \(\frac{n}{2}[2M+(n-1)D]\)

and nth term Mn = M + (n - 1)D

It is given that

\(\frac{Sum\ of\ n\ terms\ of\ first\ A.P}{Sum\ of\ n\ terms\ of\ second\ A.P }= \frac{5n+4}{9n+6}\)

⇒ \(\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2M+(n-1)D] }= \frac{5n+4}{9n+6}\)

⇒ \(\frac{[2a+(n-1)d]}{[2M+(n-1)D] }= \frac{5n+4}{9n+6} \)

⇒ \(\frac{a+\left ( \frac{n-1}{2} \right )d}{M+\left ( \frac{n-1}{2} \right )D}= \frac{5n+4}{9n+6} \).....(1)

Now, we need to find the ratio of the 10^th term of A and B

\(\frac{10^{th}\ term\ of\ A}{10^{th}\ term\ of\ B}\)

= \(\frac{a_{10}\ of\ first\ A.P}{M_{10}\ of\ second\ A.P}\)

= \(\frac{a+(10-1)d}{M+(10-1)D}\)

= \(\frac{a+9d}{M+9D}\)                 ---(2)

From equation (1) & (2)

Hence, \(\frac{n-1}{2}=9\Rightarrow n=19\)

Putting n = 19 in equation (1), we get

\(\frac{a+9d}{M+9D}=\frac{5(19)+4}{9(19)+6}\)

 

⇒ \(\frac{a+9d}{M+9D}=\frac{99}{177}\)

⇒ \(\frac{a+9d}{M+9D}=\frac{33}{59}\)

⇒ \(\frac{10^{th}\ term\ of\ A}{10^{th}\ term\ of\ B} =\frac{33}{59}\)

Hence the ratio of the 10^th term of A and B = \(\frac{33}{59}\)

∴ The correct option is (3)