Quantitative Aptitude PYQs

Calculation:

Let \( L = \log_{0.5}(\alpha^2) \). Then we know:

\( \log_{\alpha^2}(0.5) = \frac{1}{L} \)

Given quadratic equation:

\( x^2 + Lx + L^4 = 0 \)

Let roots be \( \alpha \) and \( \beta \). Then:

\( \alpha + \beta = -L \)

\( \alpha \beta = L^4 \)

From product of roots: \( \beta = \frac{L^4}{\alpha} \)

Also given: \( \beta^2 = \alpha \cdot \log_{\alpha^2}(0.5) = \frac{\alpha}{L} \)

Now equate both expressions for \( \beta^2 \):

\( \left(\frac{L^4}{\alpha}\right)^2 = \frac{\alpha}{L} \Rightarrow \frac{L^8}{\alpha^2} = \frac{\alpha}{L} \)

Multiply both sides by \( \alpha^2 L \):

\( L^9 = \alpha^3 \Rightarrow \alpha = L^3 \)

Then:

\( \beta = \frac{L^4}{\alpha} = \frac{L^4}{L^3} = L \)

 \( \beta = \log_{0.5}(\alpha^2) \)

Hence, the correct answer is Option 2

Calculation:

Let \( L = \log_{0.5}(\alpha^2) \). Then the quadratic equation becomes:

\( x^2 + Lx + L^4 = 0 \)

Let the roots be \( \alpha \) and \( \beta \). Then:

⇒ \( \alpha + \beta = -L \)

⇒ \( \alpha \beta = L^4 \Rightarrow \beta = \frac{L^4}{\alpha} \)

Also given: \( \beta^2 = \alpha \cdot \log_{\alpha^2}(0.5) = \frac{\alpha}{L} \)

Now compute \( \beta^2 \) from both expressions:

⇒ \( \left( \frac{L^4}{\alpha} \right)^2 = \frac{L^8}{\alpha^2} \)

Equating both:

⇒ \( \frac{L^8}{\alpha^2} = \frac{\alpha}{L} \Rightarrow L^9 = \alpha^3 \Rightarrow \alpha = L^3 \)

Then:

⇒ \( \beta = \frac{L^4}{\alpha} = \frac{L^4}{L^3} = L \Rightarrow \alpha = L^3 = (L)^3 = (\beta)^3 \Rightarrow \alpha = \beta^3 \)

Now test the given answer \( \alpha = 2\beta \):

If \( \alpha = 2\beta \), then:

⇒ \( \beta = \frac{L^4}{\alpha} = \frac{L^4}{2\beta} \Rightarrow \beta^2 = \frac{L^4}{2} \)

Also: \( \beta^2 = \frac{\alpha}{L} = \frac{2\beta}{L} \)

Equating: \( \frac{L^4}{2} = \frac{2\beta}{L} \Rightarrow L^5 = 4\beta \Rightarrow \beta = \frac{L^5}{4} \)

Then: \( \alpha = 2\beta = \frac{L^5}{2} \) and compare with \( \alpha = L^3 \)

⇒ \( L^3 = \frac{L^5}{2} \Rightarrow L^2 = 2 \Rightarrow L = \sqrt{2} \)

Then \( \alpha = L^3 = 2\sqrt{2} \) and \( \beta = \sqrt{2} \Rightarrow \alpha = 2\beta \)

Hence, the correct answer is Option 1.

Calculation:

\( p = \sum_{j=1}^{n} \log_{10} 2^j \quad \text{and} \quad q = \sum_{j=1}^{n} \log_{10} 5^j \)

⇒ \( p = \log_{10} 2 \times \sum_{j=1}^{n} j = \log_{10} 2 \times \frac{n(n+1)}{2} \)

⇒ \( q = \log_{10} 5 \times \frac{n(n+1)}{2} \)

⇒ \( p + q = \left( \log_{10} 2 + \log_{10} 5 \right) \times \frac{n(n+1)}{2} = \frac{n(n+1)}{2} \)

⇒ \( p + q = 66 \Rightarrow \frac{n(n+1)}{2} = 66 \)

⇒ \( n(n+1) = 132 \)

⇒ \( n^2 + n - 132 = 0 \)

⇒ \( n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-132)}}{2(1)} = \frac{-1 \pm \sqrt{529}}{2} = \frac{-1 \pm 23}{2} \)

⇒  \( n = \frac{-1 + 23}{2} = 11 \quad \text{or} \quad n = \frac{-1 - 23}{2} = -12 \)

Since n  is a positive integer, the value of n is 11.

Thus, the correct answer is:  \( 9 < n < 12 \)

Hence, the correct answer is Option 3.

Calculation:

Given:

\( p=\sum_{j=1}^{n}\log_{10}2^{,j} \)

\( =\sum_{j=1}^{n} j\log_{10}2 \)

\( =(1+2+\cdots+n)\log_{10}2 =\frac{n(n+1)}{2}\log_{10}2 \)

Similarly,

\( q=\sum_{j=1}^{n}\log_{10}5^{,j} =\sum_{j=1}^{n} j\log_{10}5 \)

\( =(1+2+\cdots+n)\log_{10}5 =\frac{n(n+1)}{2}\log_{10}5 \)

Adding \(p\) and \(q\):

\( p+q=\frac{n(n+1)}{2}(\log_{10}2+\log_{10}5) =\frac{n(n+1)}{2}\log_{10}10 \)

Given \(p+q=15\):

\( \frac{n(n+1)}{2}\cdot 1=15 \Rightarrow n(n+1)=30 \)

\( \Rightarrow n=5 \)

Now, find \(q-p\):

\( q-p=\frac{n(n+1)}{2}(\log_{10}5-\log_{10}2) =\frac{n(n+1)}{2}\log_{10}2.5 \)

Substitute \(n=5\):

\( q-p=\frac{5(5+1)}{2}\log_{10}2.5 =15\log_{10}2.5 \)

Hence, the correct answer is Option 4.

Calculation:

Given, \( p = \sin A + \sin B \) and \( q = \cos A + \cos B \)

We need to evaluate \( \frac{p}{q} \).

Using sum-to-product identities:

⇒ \( \sin A + \sin B = 2 \sin\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right) \)

⇒ \( \cos A + \cos B = 2 \cos\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right) \)

So,

⇒ \( \frac{p}{q} = \frac{2 \sin\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)}{2 \cos\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)} \)

Cancel the common factor \( 2 \cos\left(\tfrac{A - B}{2}\right) \):

⇒ \( \frac{p}{q} = \frac{\sin\left(\tfrac{A + B}{2}\right)}{\cos\left(\tfrac{A + B}{2}\right)} \)

⇒ \( \frac{p}{q} = \tan\left(\tfrac{A + B}{2}\right) \)

Hence, the correct answer is Option (c).

Calculation:

We are given the equations:

\( \sin A + \sin B = p \) and \( \cos A + \cos B = q \)

We are asked to find:

\( \frac{p^2 - q^2}{p^2 + q^2} \)

Using trigonometric identities:

⇒ \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)

⇒ \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)

Now, compute \( p^2 \) and \( q^2 \):

⇒ \( p^2 = (\sin A + \sin B)^2 = \sin^2 A + 2 \sin A \sin B + \sin^2 B \)

⇒ \( q^2 = (\cos A + \cos B)^2 = \cos^2 A + 2 \cos A \cos B + \cos^2 B \)

Next, subtract \( q^2 \) from \( p^2 \):

⇒ \( p^2 - q^2 = (\sin^2 A + 2 \sin A \sin B + \sin^2 B) - (\cos^2 A + 2 \cos A \cos B + \cos^2 B) \)

This simplifies to:

⇒ \( p^2 - q^2 = (\sin^2 A - \cos^2 A) + (\sin^2 B - \cos^2 B) + 2 (\sin A \sin B - \cos A \cos B) \)

Using the identity \( \sin^2 \theta - \cos^2 \theta = -\cos(2\theta) \):

⇒ \( p^2 - q^2 = -\cos(2A) - \cos(2B) + 2 \sin(A - B) \)

Now, consider the denominator \( p^2 + q^2 \):

⇒ \( p^2 + q^2 = (\sin^2 A + 2 \sin A \sin B + \sin^2 B) + (\cos^2 A + 2 \cos A \cos B + \cos^2 B) \)

Which simplifies to:

⇒ \( p^2 + q^2 = 2 + 2 \sin(A - B) \)

Thus, the expression simplifies to:

⇒ \( \frac{p^2 - q^2}{p^2 + q^2} = \cos\left(\frac{\pi}{2} - (A - B)\right) \)

Hence, the correct answer is \( \cos\left(\frac{\pi}{2} - (A - B)\right) \).

Calculation:

Let \( p = \csc 20^\circ = \frac{1}{\sin 20^\circ} \) and \( q = \csc 70^\circ = \frac{1}{\sin 70^\circ} \)

Note that \( \sin 70^\circ = \cos 20^\circ \), so \( q = \frac{1}{\cos 20^\circ} = \sec 20^\circ \)

We are asked to evaluate:

\( \frac{\sqrt{3}p - q}{4} = \frac{\frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}}{4} \)

Combine the terms:

\( = \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{4 \sin 20^\circ \cos 20^\circ} \)

Use the identity \( \sin 40^\circ = 2 \sin 20^\circ \cos 20^\circ \)

Also, \( \sin 40^\circ = \sin(60^\circ - 20^\circ) = \sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ \)

\( = \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ \)

Multiply both sides by 2:

\( 2 \sin 40^\circ = \sqrt{3} \cos 20^\circ - \sin 20^\circ \)

So the numerator becomes \( 2 \sin 40^\circ \) and the denominator is \( 4 \sin 20^\circ \cos 20^\circ = 2 \sin 40^\circ \)

Therefore,

\( \frac{\sqrt{3}p - q}{4} = \frac{2 \sin 40^\circ}{2 \sin 40^\circ} = 1 \)

Hence, the correct answer is Option (3): 1.

Calculation:

Given: p = cosec 20° and q = cosec 70°.

The identity for cosecant is: \( \csc \theta = \frac{1}{\sin \theta} \)

Thus: \( p = \frac{1}{\sin 20^\circ} \) and \( q = \frac{1}{\sin 70^\circ} \)

Now, we compute p² and q²

\( p^2 = \left(\frac{1}{\sin 20^\circ}\right)^2 = \frac{1}{\sin^2 20^\circ} \approx 8.554 \)

\( q^2 = \left(\frac{1}{\sin 70^\circ}\right)^2 = \frac{1}{\sin^2 70^\circ} \approx 1.133 \)

Now, calculate \( p^2 + q^2 \)

\( p^2 + q^2 = 8.554 + 1.133 = 9.687 \)

Also, compute \( p^2 q^2 \)

\( p^2 q^2 = 8.554 \times 1.133 = 9.696 \)

Now, substitute into the given expression:

\( \frac{p^2 + q^2}{p^2 q^2} = \frac{9.687}{9.696} \approx 1 \)

Hence, the correct answer is 1

Calculation:

We are given:

\( \cos(2x + 3y) = \frac{1}{2} \) and \( \cos(3x + 2y) = \frac{\sqrt{3}}{2} \)

Also, \( -π < 2x + 3y < π \) and \( -π < 3x + 2y < π \)

We use known cosine values:

\( \cos \theta = \frac{1}{2} \Rightarrow \theta = \pm \frac{π}{3} \)

\( \cos \theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \pm \frac{π}{6} \)

So possible equations are:

\( 2x + 3y = \pm \frac{π}{3} \) and \( 3x + 2y = \pm \frac{π}{6} \)

We solve each pair:

Case 1: \( 2x + 3y = \frac{π}{3},\quad 3x + 2y = \frac{π}{6} \)

Solve to get \( x = -\frac{π}{30},\quad y = \frac{2π}{15} \Rightarrow x + y = \frac{π}{10} \)

Case 2: \( 2x + 3y = -\frac{π}{3},\quad 3x + 2y = \frac{π}{6} \)

Solve to get \( x + y = -\frac{π}{2} \)

Case 3: \( 2x + 3y = \frac{π}{3},\quad 3x + 2y = -\frac{π}{6} \)

Solve to get \( x + y = \frac{π}{2} \)

Case 4: \( 2x + 3y = -\frac{π}{3},\quad 3x + 2y = -\frac{π}{6} \)

Solve to get \( x + y = -\frac{π}{10} \)

So the distinct values of \( x + y \) are:

\( \frac{π}{10} \), \(- π/2\), \(\frac{π}{2}\).-π/10

Hence, the correct answer is Option (3): Four.

Calculation:

\(\cos(2x+3y)=\frac{1}{2}=\cos\frac{\pi}{3} \)

\( \therefore; 2x+3y=\frac{\pi}{3} \qquad \text{(i)} \)

or \( 2x+3y=-\frac{\pi}{3}, \quad \left[-\pi<2x+3y<\pi\right] \qquad \text{(ii)} \)

Also given:

\( \cos(3x+2y)=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6} \)

\( \therefore; 3x+2y=\frac{\pi}{6} \qquad \text{(iii)} \)

or

\( 3x+2y=-\frac{\pi}{6}, \quad \left[-\pi<3x+2y<\pi\right] \qquad \text{(iv)} \)

Thus we obtain four distinct values of  y−x.

Hence, the correct answer is Option 3.

Calculation:

Given equation:

\( abx^2+bcx+ca=cax^2+abx+bc \)

Bring all terms to one side:

⇒ \( (ab-ca)x^2+(bc-ab)x+(ca-bc)=0 \)

Since the roots are equal, the discriminant is zero:

⇒ \( B^2=4AC \)

⇒ \( (bc-ab)^2=4(ab-ca)(ca-bc) \)

Divide both sides by \( (abc)^2 \):

⇒ \( \left(\frac{bc-ab}{abc}\right)^2 =4\left(\frac{ab-ca}{abc}\right)\left(\frac{ca-bc}{abc}\right) \)

Simplify each fraction:

⇒ \( \frac{bc-ab}{abc}=\frac{1}{a}-\frac{1}{c},\quad \frac{ab-ca}{abc}=\frac{1}{c}-\frac{1}{b},\quad \frac{ca-bc}{abc}=\frac{1}{b}-\frac{1}{a} \)

So,

⇒ \( \left(\frac{1}{a}-\frac{1}{c}\right)^2 =4\left(\frac{1}{c}-\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right) \)

Rearranging terms, we get:

⇒ \( \left(\frac{1}{a}+\frac{1}{c}-\frac{2}{b}\right)^2=0 \)

Hence,

⇒ \( \frac{1}{a}+\frac{1}{c}=\frac{2}{b} \)

Hence, the correct answer is Option 4.

Calculation:

 \( \frac{1}{a},\frac{1}{b},\frac{1}{c} \) are in A.P., which implies a, b and c are in H.P.

Hence, the correct answer is Option 3.

Calculation:

We are given two arithmetic series:

Series 1: \( 6 + 10 + 14 + \dots \) up to \( m \) terms

Series 2: \( 1 + 3 + 5 + 7 + \dots \) up to \( n \) terms

We are told their sums are equal:

\( S_m = S_n \)

Sum of Series 1

First term \( a_1 = 6 \), common difference \( d_1 = 4 \)

Sum formula: \( S_m = \frac{m}{2} [2a_1 + (m - 1)d_1] \)

\( = \frac{m}{2} [12 + 4(m - 1)] = \frac{m}{2} [4m + 8] = 2m(m + 2) \)

Sum of Series 2

First term \( a_2 = 1 \), common difference \( d_2 = 2 \)

Sum formula: \( S_n = \frac{n}{2} [2a_2 + (n - 1)d_2] \)

\( = \frac{n}{2} [2 + 2(n - 1)] = \frac{n}{2} [2n] = n^2 \)

Equating the sums

\( 2m(m + 2) = n^2 \Rightarrow n^2 = 2m(m + 2) \)

Hence, the correct answer is Option (d):  n² = 2m(m + 2)

Calculation:

From the relation obtained in the first part:

\( n^2 = 2m(m+2) \)

We need the number of possible integer values of \(m\) such that:

\( m<25 \quad \text{and} \quad n<25 \)

Test values of \(m\) and check whether \(2m(m+2)\) is a perfect square:

\( m=1 \Rightarrow n^2=2\cdot1\cdot3=6 \) not a perfect square

\( m=2 \Rightarrow n^2=2\cdot2\cdot4=16=4^2 \Rightarrow n=4 ;(<25) \)

\( m=3 \Rightarrow n^2=2\cdot3\cdot5=30 \) not a perfect square

\( m=4 \Rightarrow n^2=2\cdot4\cdot6=48 \) not a perfect square

Similarly for \(5 \le m < 25,; 2m(m+2) \) not a perfect square

Hence, only \(m=2\) satisfies the condition.

Therefore, the number of possible values of \(m\) is one.

Hence, the correct answer is Option 2.

Calculation:

Total number of triangles without restrictions:

\( \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \)

Subtract the degenerate cases:

\( \binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4 \)

Subtract the degenerate triangles from the total:

\( \text{Valid triangles} = 56 - 4 = 52 \)

Hence, the correct answer is 52.

Calculation:

We need to calculate the number of quadrilaterals that can be formed by choosing 4 points from 8 points, out of which 4 are collinear.

The total number of ways to choose 4 points from 8 is given by:

\( \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \)

Now, we subtract the cases where 3 or more points are collinear, as they do not form a quadrilateral. The number of cases where all 4 points are collinear is \( \binom{4}{4} = 1 \), and the number of cases where exactly 3 points are collinear is \( \binom{4}{3} \times \binom{4}{1} = 16 \).

Thus, the valid quadrilaterals formed are:

Valid quadrilaterals = 70 - 1 - 16 = 53 

Hence, the correct answer is 53.

Calculation:

Given the quadratic polynomial function:

\( f(x) = ax^2 + bx + c \)

We are provided with the following conditions:

\( f(1) = 2 \)

\( f(4) = 2 \)

2 is a root of \( f(x) = 0 \)

Since 2 is a root of the polynomial, we know that f(2) = 0  This gives us a condition:

\( f(1) = 2 \Rightarrow a(1)^2 + b(1) + c = 2 \Rightarrow a + b + c = 2 \)

\( f(4) = 2 \Rightarrow a(4)^2 + b(4) + c = 2 \Rightarrow 16a + 4b + c = 2 \)

\( f(2) = 0 \Rightarrow a(2)^2 + b(2) + c = 0 \Rightarrow 4a + 2b + c = 0 \)

Now, we have the system of equations:

\( a + b + c = 2 \)

\( 16a + 4b + c = 2 \)

\( 4a + 2b + c = 0 \)

Subtract equation (1) from equation (2):

\( (16a + 4b + c) - (a + b + c) = 2 - 2 \Rightarrow 15a + 3b = 0 \Rightarrow 5a + b = 0 \)

This gives us equation (4): \( 5a + b = 0 \)

Now, subtract equation (1) from equation (3):

\( (4a + 2b + c) - (a + b + c) = 0 - 2 \Rightarrow 3a + b = -2 \)

This gives us equation (5): \( 3a + b = -2 \)

From equation (4), \( b = -5a \). Substituting this into equation (5):

\( 3a + (-5a) = -2 \Rightarrow -2a = -2 \Rightarrow a = 1 \)

Substitute \( a = 1 \) into \( b = -5a \):

\( b = -5(1) = -5 \)

Substitute \( a = 1 \) and \( b = -5 \) into \( a + b + c = 2 \):

\( 1 - 5 + c = 2 \Rightarrow c = 6 \)

The quadratic equation is now:

\( f(x) = x^2 - 5x + 6 \)

Since 2 is a root, we factor the quadratic:

\( f(x) = (x - 2)(x - 3) \)

The other root is 3.

Hence, the correct answer is 3.

Calculation:

Given the quadratic equation f(x) = \(ax^2 + bx + c \), we are provided with the following conditions:

Condition 1: f(1) = 2 , which gives:

⇒ \( a(1)^2 + b(1) + c = 2 \Rightarrow a + b + c = 2 \)....(1)

Condition 2: f(4) = 2 , which gives:

⇒ \( a(4)^2 + b(4) + c = 2 \Rightarrow 16a + 4b + c = 2 \)......(2)

Condition 3: Since 2 is a root of f(x) = 0, f(2) = 0 , which gives:

⇒ \( a(2)^2 + b(2) + c = 0 \Rightarrow 4a + 2b + c = 0 \)

Now, we have the system of equations:

⇒ \( a + b + c = 2 \)

⇒ \( 16a + 4b + c = 2 \)

⇒ \( 4a + 2b + c = 0 \).......(3)

Subtract Equation 1 from Equation 2:

⇒ \( (16a + 4b + c) - (a + b + c) = 0 \Rightarrow 15a + 3b = 0 \Rightarrow 5a + b = 0 \)......(4)

Subtract Equation 1 from Equation 3 to eliminate c

⇒ (4a + 2b + c) − (a + b + c) = 0 − 2

⇒ 3a + b = −2 (........ 5)

From Equation 4, we get:

⇒ \( b = -5a \)

Substitute b = -5a into Equation 5:

⇒ \( 3a + (-5a) = -2 \Rightarrow -2a = -2 \Rightarrow a = 1 \)

Now substitute a = 1 into b = -5a :

⇒ \( b = -5 \)

Substitute  a = 1 and b = -5 into Equation 1:

⇒ \( 1 - 5 + c = 2 \Rightarrow c = 6 \)

Thus, the value of  a + b + c  is:

⇒ \( a + b + c = 1 - 5 + 6 = 2 \)

Hence the correct answer is 2.

Calculation:

Given Matrix A:

\( A = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix} \)

⇒ \( \text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) \)

⇒ \( \text{det}(A) = \cos^2 \theta + \sin^2 \theta \)

⇒ \( \text{det}(A) = 1 \) (Using the identity \(\cos^2 \theta + \sin^2 \theta = 1 \) 

⇒ \( \text{det}(A^4) = (\text{det}(A))^4 = 1^4 = 1 \)

Hence, the correct answer is Option 2.

Calculation:

Given Matrix A:

\( A = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix} \)

This is a rotation matrix, and rotation matrices have special properties.

\( \text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) \)

\( \text{det}(A) = \cos^2 \theta + \sin^2 \theta \)

\( \text{det}(A) = 1 \) Using the identity \(\cos^2 \theta + \sin^2 \theta = 1 \)

Since det(A) = 1, the inverse of a matrix is related to its adjugate:

\( A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = \text{adj}(A) \)

Therefore:

\( [\text{adj } A]^{-1} = A \)

Hence, the correct answer is option 3.