Quantitative Aptitude PYQs

Calculation:

Given: n(Z) = 90 and the number of elements in Y and Z are in the ratio 4 : 5

To find: b = ?

n(Y) = 16 + 18 + 17 + b

n(Z) = 12 + 18 + 17 + c = 90 

 

\(\rm \frac{n(y)}{n(z)}=\frac{16+18+17+b}{12+18+17+c}=\frac{4}{5}\ \Rightarrow \frac{51+b}{90}=\frac{4}{5}\ \Rightarrow (51+b)\times 5=360 \Rightarrow255+5b=360\ \Rightarrow5b=105\ \Rightarrow b=21\)

Hence, option (3) is correct.

Calculation:

Given: n(Z) = 90, 

From above diagram, we get

n(X) = a + 46, n(Y) = b + 51, n(X ∩ Y) = 16 + 18 = 34,  n(Y ∩ Z)  = 17 + 18 =35,  n(X ∩ Z) = 12 + 18 = 30,  n(X ∩ Y ∩ Z) = 18

n(X) + n(Y) + n(Z) - n(X ∩ Y) - n(Y ∩ Z) - n(X ∩ Z) + n(X ∩ Y ∩ Z) = 

a + 46 + b + 51 + 90 - 34 - 35 - 30 + 18

= a + b + 106

Hence, option (4) is correct.

Concept:

Complement of X = U - X, where U is a universal set.

 

Calculation:

We know, n(X) + n(Y) + n(Z) - n(X ∩ Y) - n(Y ∩ Z) - n(X ∩ Z) + n(X ∩ Y ∩ Z) = n(X ∪ Y ∪ Z)

Here U - n(X ∪ Y  ∪ Z) = p, whre U is a universal set 

⇒ U - (a + b + 106) = p

⇒ U = p + (a + b + 106) 

Now the number of elements in the complement of X, U - n(X) = p + (a + b + 106) - (a + 16 + 18 + 12)

= p + b + 60

Hence, option (1) is correct.

Concept:

Let \(\rm \dfrac p q = \rm \dfrac r s\) then componendo and dividend: \(\rm\frac{p+q}{p-q}=\rm\frac{r+s}{r-s}\)⇒\(\rm\frac{p-q}{p+q}=\rm\frac{r-s}{r+s}\)

\(\rm \sin(A+B)=sinAcosB+cosAsinB \ and \rm \sin(A-B)=sinAcosB-cosAsinB\)

\(\rm \sec^2A=1+tan^2A\)

\(\rm cos2A=cos^2A-sin^2A\)

 

Calculation:

Given: \(\rm \frac {\tan 3A}{\tan A} = K,\)

\(⇒ \rm \frac{\frac{\sin 3 A}{\cos 3 A}}{\frac{\sin A}{\cos A}}=k\ ⇒ \frac{\sin 3 A \cdot \cos A}{\cos 3 A \cdot \sin A}=\frac{K}{1}\)

Applying componendo and dividend, 

\(\rm \frac{\sin 3 A \cdot \cos A-cos 3 A \cdot \sin A}{\sin 3 A \cdot \cos A+cos 3 A \cdot \sin A}=\frac{k-1}{k+1}\ \Rightarrow \frac{sin(3A-A)}{sin(3A+A)}=\frac{k-1}{k+1}\ \Rightarrow \frac{sin(2A)}{sin(4A)}=\frac{k-1}{k+1}\ \Rightarrow \frac{2sin(A)cosA}{2sin(2A)cos2A}=\frac{k-1}{k+1}\ \Rightarrow \frac{sin(A)cosA}{2sin(A)cosAcos2A}=\frac{k-1}{k+1}\ \)

\(\begin{aligned} &\rm \Rightarrow \frac{1}{2 \cos 2 A}=\frac{k-1}{k+1}\ &\rm \Rightarrow \frac{1}{2\left(\cos ^{2} A-\sin ^{2} A\right)}=\frac{k-1}{k+1}\ &\rm \Rightarrow \frac{\sec ^{2} A}{2\left(1-\tan ^{2} A\right)}=\frac{k-1}{k+1}\ &\rm \Rightarrow \frac{1+\tan ^{2} A}{1-\tan ^{2} A}=\frac{2 k-2}{k+1}\ &\rm \Rightarrow \frac{1+\tan ^{2} A-1+\tan ^{2} A}{1+\tan ^{2} A+1-\tan ^{2} A}=\frac{2 k-2-k-1}{2 k-2+k+1}\ &\rm \Rightarrow \frac{2 \tan ^{2} A}{2}=\frac{k-3}{3 k-1}\ &\rm \Rightarrow \tan ^{2} A=\frac{k-3}{3 k-1} \end{aligned}\)

Hence, option (2) is correct.

Calculation:

We have, \(\rm \tan^2A=\frac {k - 3}{3k - 1}\)

For real values of tan² A:

\(\rm \frac {K - 3}{3K - 1}\) ≥ 0

At k = 1/3 and 3, \(\rm \frac {K - 3}{3K - 1}\)<0

∴For real values of tan A, K cannot lie between \(\rm \frac 1 3;and;3\)

Hence, option (2) is correct.

Concept:

In triangle ABC, with sides a, b, and c sine rule is given by, \(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)

sin (180 - θ) = sin θ 

Calculation:

1. In triangle ABD, using sine rule

sin θ /AB = sin α / AD

⇒ AD sin θ = AB sin α

So ,this is correct statement. 2. In triangle ABD, using sine rule

sin θ /AB = sin (180-(θ +α )) / BD

⇒ BD sin θ = AB sin (α + θ)

So ,this is also correct statement.

Hence, option (3) is correct.

Concept:

In triangle ABC, with sides a, b, and c sine rule is given by, \(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)

sin (A + B) = sinA.cosB + cosA.sin B

Calculation:

\(\begin{array}{l} \rm In \ \Delta B C D \ \rm B D^{2}=B C^{2}+C D^{2} \ \rm B D^{2}=P^{2}+q^{2} \ \rm B D=\sqrt{P^{2}+q^{2}} \end{array}\)

In triangle ABD, using sine rule,

\( \rm \frac{AB}{\sin \theta} =\frac{B D}{\sin (180-(\theta+\alpha))} \ \Rightarrow\rm \frac{AB}{\sin \theta} =\frac{B D}{\sin (\theta+\alpha)} \ \Rightarrow A B =\frac{B D \sin \theta}{\sin (\theta+\alpha)} \ = \frac{B D^{2} \sin \theta}{{B D}(\sin (\theta+\alpha)} \)

\(\begin{array}{l} \rm= \frac{B D^{2} \sin \theta}{{3 D(\sin \theta \cos \alpha+\cos \theta \sin \alpha)}} \ \rm =\frac{\left(p^{2}+q^{2}\right) \sin \theta}{\sqrt{p^{2}+q^{2}}\left(\sin \theta \frac{q}{\sqrt{p^{2}+q^{2}}}+\cos\theta \frac{p}{\sqrt{p^{2}+q^{2}}}\right)} \end{array}\)

= \(\rm \frac {\left( p^2 + q^2 \right) \sin \theta}{p \cos \theta + q \sin \theta}\)

Hence, option (1) is correct.

Concept:

tan (A - B) = \(\rm \frac{\tan A-\tan B}{1+\tan A \tan B}\)

 

Calculation:

\(\rm \tan θ = \frac {\cos 17^\circ - \sin 17^\circ}{\cos 17^\circ + \sin 17^\circ},\)

RHS = \( \frac {\cos 17^\circ - \sin 17^\circ}{\cos 17^\circ + \sin 17^\circ}\)

Divide by cos 17°

\(=\frac{\frac{\cos 17}{\cos 17}-\frac{\sin 17}{\cos 17}}{\frac{\cos 17}{\cos 17}+\frac{\sin 17}{\cos 17}}\)

\(\begin{aligned} &= \frac{1-\tan 17}{1+\tan 17} \& = \frac{\tan 45^{\circ}-\tan 17}{1+1 \times \tan 17^{\circ}} \ & = \frac{\tan 45-\tan 17}{1+\tan 45 \tan 17} \end{aligned}\)

= tan (45 - 17) = tan (28°) 

Now,

⇒ tan θ = tan (28°) 

∴ θ = 28° 

Hence, option (2) is correct.

Concept:

cos (A + B) = cosA. cosB - sinA. sin B

Calculation:

Given: cos 2B = 3 sin² A   ....

(1) and  3sin 2A = 2 sin 2B     .... (2) Divide (1) by (2),

\(\frac{cos(2B)}{2sin(2B)}=\frac{3sin^2(A)}{3sin(2A)}\) .... (3) We know that,

sin 2A = 2 sin A∙ cos A

Hence, equation (3) becomes,

\(\frac{cos(2B)}{2sin(2B)}=\frac{sin^2(A)}{2sin(A)cos(A)}\)

⇒ \(\frac{cos(2B)}{sin(2B)}=\frac{sin(A)}{cos(A)}\)

⇒ cos (2B) cos (A) = sin (2B) sin (A)

⇒ cos (2B) cos (A) - sin (2B) sin (A) = 0

From above concept,

cos (A + 2B) = 0

⇒ (A + 2B) = cos^-1 (0)

⇒ (A + 2B) =  π/2

Concept:

\(\begin{array}{l} \rm \sin 3 x=3 \sin x-4 \sin ^{3} x \ \rm \cos 3 x=4 \cos ^{3} x-3 \cos x \end{array}\)

 

Calculation:

sin 3x + cos 3x + 4 sin³ x - 3 sin x + 3 cos x - 4 cos³ x

= 3 sin x - 4 sin³ x + 4 cos³ x - 3 cos x  + 4 sin³ x - 3 sin x + 3 cos x - 4 cos³ x

= 0

Hence, option (1) is correct.

Concept:

 −1 ≤ cosx ≤ 1

 

Calculation:

Here, y = 2 + cos x

We know,  −1 ≤ cosx ≤ 1

Adding 2, we get

⇒ 2 −1 ≤ 2 + cosx ≤ 2 + 1

⇒ 1 ≤ y ≤ 3

So the value of y-coordinate lies between y = 1 to y = 3

Hence, option (4) is correct.

Concept:

2 sin x. cos x = sin 2x

Calculation:

Let, x = 8 cos 10° ⋅ cos 20° ⋅ cos 40°

Now, multiplying both sides by sin 10°

\(\begin{array}{l} ⇒ x \sin 10=4 \times 2 \sin 10 . \cos 10 . \cos 20 . \cos 40 \ ⇒ x \sin 10=2 \times 2 \sin 20 . \cos 20 . \cos 40 \ ⇒ x \sin 10=2 \times \sin 40 . \cos 40 \ ⇒ x \sin 10=\sin 80 \ ⇒ x \sin 10=\sin (90-10)=\cos 10 \end{array}\)

⇒ x = cos 10/ sin 10

⇒ x = cot 10° 

Hence, option (2) is correct.

Concept:

\(\rm \cos (A)-\cos (B)=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

\(\cos 3θ =4 \cos ^{3} θ-3 \cos θ \)

 

Calculation:

cos 48° - cos 12°  

 \(= \rm-2 \sin \left(\frac{48+12}{2}\right) \sin \left(\frac{48-12}{2}\right)\ =-2\sin 30\sin18\ =-2\times \frac{1}{2}sin 18\ =-\sin 18\)

sin 18°

 \(\begin{array}{l} \Rightarrow θ=18^{\circ} \ \Rightarrow 2 θ=36^{\circ} \ \Rightarrow \sin 2 θ=\sin 36^{\circ} \ \Rightarrow \sin 2 θ=\sin (90-54) =\cos 54\ \Rightarrow \sin 2 θ=\cos 3θ \ \Rightarrow 2 \sin θ \cos θ=4 \cos ^{3} θ-3 \cos θ \ \Rightarrow 2 \sin θ \cos θ=\cos θ\left(4 \cos ^{2} θ-3\right) \end{array}\)

\( \Rightarrow 2 \sin θ= 4(1-\sin ^{2} θ)-3 \ \Rightarrow 4-4 \sin ^{2} θ-3-2 \sin θ=0\ \Rightarrow -4 \sin ^{2} θ-2 \sin θ+1=0\ \Rightarrow 4 \sin ^{2} θ+2 \sin θ-1=0 \)

\( \Rightarrow \sin θ= \frac{-2 \pm \sqrt{4+16}}{8} \ =\frac{-2 \pm 2 \sqrt{5}}{8} \ \sin θ =\frac{-1 \pm \sqrt{5}}{4} \)

Now since θ is in I quadrant 

sin 18° =  \(\frac{-1 + \sqrt{5}}{4} \)

cos 48° - cos 12°  = - sin 18° = \(\frac{1 - \sqrt{5}}{4} \)

= \(\frac{1 - \sqrt{5}}{4} \)

Hence, option (2) is correct.

Concept:

Pythagoras theorem: In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Sine Rule in triangle ABC, having sides a, b, c:

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}\)

sin A - cos B = \(\rm 2\cos(\frac{A +B}{2})\sin(\frac{A -B}{2})\)

Calculation:

1. We have,

sin \(\rm B = \frac 1 3=\frac{P}{H}\)

Hwence, Right angle triangle can be drawn as,

Here, P = 1

H = 3

B = x

Using Pythagoras theorem,

3² = 1² + x²

x² = 8

\(\therefore x=2\sqrt{2}\)

Now , Cosec C = \(\frac{3}{2\sqrt{2}}\)

Hence Ststement (1) is not correct.

2. Given,

b cos B = c cos C

Using the sine rule,

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}\)

⇒ b = K sin B & c = K sin C

⇒ 2 sin B cos B = 2 sin C cos C

Assume,

K = 2

⇒ 2 sin B cos B = 2 sin C cos C

⇒ sin 2B = sin 2C  [2 sin x cos x = sin 2x]

⇒ sin 2B - sin 2C = 0

Using Formula:

sin C - sin D = 2 cos \((\frac{C+D}{2})\) × sin \((\frac{C-D}{2})\)

⇒ 2 cos (B + C) sin (B - C) = 0

In this case,

Either,

cos (B + C) = 0

Hence, (B + C) = 90° ....

(1) Or,

Or sin (B -C) = 0

B - C = 0

⇒ B = C  .... (2) From equation (1) & (2),

B = C = 45°

So ABC must be issosceles 

Hence, option (2) is correct.

Concept:

• Consider complex number z = x + iy,
• \(\text { Modulus of } z=|z|=\sqrt{x^{2}+y^{2}}\)
• Division of modulus of two complex numbers \(\rm |\frac{z_1}{z_2}|= \frac{|z_1|}{|z_2|}\)
• sin²θ + cos²θ = 1

 

Calculation:

We have to find the value of \(|\frac {\cos θ + i \sin θ}{\cos θ - i \sin θ}|\)

Now, According to the concept of division of modulus

\(⇒ \frac {|\cos θ + i \sin θ|}{|\cos θ - i \sin θ|}\)

\(⇒ \frac{\cos^2θ +sin^2θ }{cos^2θ +(-sin θ )^2}\)

⇒ 1/1 = 1

Hence, option (2) is correct.

Alternate Method We have given complex numbers,

z = \(\rm \frac {\cos θ + i \sin θ}{\cos θ - i \sin θ},\)

Multiply in the numerator and denominator by (cos θ + i sin θ)

⇒ z = \(\rm \frac {\cos θ + i \sin θ}{\cos θ - i \sin θ}\times \rm \frac {\cos θ + i \sin θ}{\cos θ + i \sin θ}\)

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

We know that,

sin²A + cos²A = 1

and, i² = -1

Hence, ⇒ z = \(\rm \frac {(\cos^2 θ + i^2 \sin^2 θ+2i\ cosθ \ sinθ )}{\cos^2 θ + \sin^2 θ}\)

or, z = (cos²θ - sin²θ) + 2i sin θ cos θ ....

(1) Using formula,

cos 2θ = (cos2θ - sin2θ)

and, sin 2θ = 2sin θ cos θ

Hence, equation (1) becomes,

z = cos 2θ + i sin 2θ

Hence, |z| = \(\sqrt{(cos^22\theta+sin^22\theta)}\) = 1

Concept:

A proper subset is one that contains few elements of the original set.

A superset is one which contain all the element including elements of the original set.

 

Calculation:

The proper subsets of {1, 2, 3, 4} and supperset of set {3} are

{1, 3}, {2, 3}, {3, 4}, {1, 2, 3}, {2, 3, 4}, {1, 3, 4}, {1, 2, 3, 4}

Hence, option (3) is correct.

Concept:

If A is the arithmetic mean of numbers a and b and is given by ⇔ \(\rm AM = \dfrac{a+b}{2}\)

If G is the geometric mean of the numbers a and b and is given by  ⇔ \(\rm GM = \sqrt{ab}\)

If H is the Harmonic mean of numbers a and b and is given by ⇔ \(\rm HM = \dfrac{2ab}{a+b}\)

Relation between AM, GM and HM

G² = AH

AM ≥ GM ≥ HM

 

Calculation: 

\(\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\ \rm q&\rm r&\rm p\ \rm r&\rm p&\rm q \end{array}} \right|,\)

\(\begin{array}{l} =\rm p\left(q r-p^{2}\right)-q\left(q^{2}-p r\right)+r\left(p q-r^{2}\right) \ =\rm p q r-p^{3}-q^{3}+p q r+p q r-r^{3} \ =\rm3 p q r-\left(p^{3}+q^{3}+r^{3}\right) \end{array}\)

As we know AM ≥ GM

Now, AM of p³, q³ and r³ = \(\rm \dfrac{p^3+q^3+r^3}{3}\)

Now, GM of p³, q³ and r³ = \(\rm ({p^3\times q^3 \times r^3})^{\frac{1}{3}} = pqr\)

Therefore, \(\rm \dfrac{p^3+q^3+r^3}{3}\) > pqr

⇒ \(\rm p^{3}+q^{3}+r^{3} > 3pqr\)

\(\Rightarrow \rm3 p q r-(p^{3}+q^{3}+r^{3}) < 0\)

∴ D < 0

Hence, option (2) is correct.

Concept:

(1 + x)ⁿ  = ⁿΣ_r-0 nC_r . x^r = [C₀ + C₁ x + C₂ x² + … C_n x_n]

\(\large{\rm n_{C_{n-r}}=n_{C_{r}}}\)

Calculation:

\(\begin{aligned} &(1+\mathrm{x})^{9}={ }^{9} \mathrm{C}_{0}+{ }^{9} \mathrm{C}_{1} \mathrm{x}+{ }^{9} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots . .^{9} \mathrm{C}_{9} \mathrm{x}^{9}\ &\text { Substituting } \mathrm{x}=1, \text { we get }\ &2^{9}={ }^{9} \mathrm{C}_{0}+{ }^{9} \mathrm{C}_{1}+{ }^{9} \mathrm{C}_{2}+\ldots .^{9} \mathrm{C}_{9}----(1)\end{aligned} \)

As we know, ⁿC_r = ⁿC_n-r

So,

\({ }^{9} \mathrm{C}_{0}+{ }^{9} \mathrm{C}_{1}+ +{ }^{9} \mathrm{C}_{2} +{ }^{9} \mathrm{C}_{3}+ ^{9} \mathrm{C}_{4} ={ }^{9} \mathrm{C}_{9}+{ }^{9} \mathrm{C}_{8}+{ }^{9} \mathrm{C}_{7}+{ }^{9} \mathrm{C}_{6}+{ }^{9} \mathrm{C}_{5} \)     ----(2)

Now, From (1) and (2), we get,

\(2^{9}=2\left[{ }^{9} \mathrm{C}_{9}+{ }^{9} \mathrm{C}_{8}+{ }^{9} \mathrm{C}_{7}+9 \mathrm{C}_{6}+{ }^{9} \mathrm{C}_{5}\right]\)

⇒ \(\left[{ }^{9} \mathrm{C}_{9}+{ }^{9} \mathrm{C}_{8}+{ }^{9} \mathrm{C}_{7}+9 \mathrm{C}_{6}+{ }^{9} \mathrm{C}_{5}\right] = 2^{8}\)

⇒ 256

Hence, option (1) is correct.

Concept:

If A is non-singular then A^-1 exist.

\(\rm A^{-1} = \dfrac{1}{|A|}.adj.A\)

 

Calculations:

Given: the matrix A is a non singular matrix of order 3.

Since A is non-singular A^-1 exist.

\(\rm A^{-1} = \dfrac{1}{|A|}.adj.A\)                      ....(1)

Pre-multiply equation (1) by A

\(\rm AA^{-1} = \dfrac{A}{|A|}.adj.A\)

⇒ I |A| = A.(adj.A)

⇒A.(Adj. A) = |A|I                       ....(2)

Now, post multiply equation (1) by A

\(\rm A^{-1}A =\dfrac{1}{|A|} (adj.A)A\)

⇒ I|A| = (adj.A)A

⇒ (adj.A)A = |A|I                     ....(3)

From (2) and (3), we have

The statement A (adj A) = (adj A) A is correct.

 

Also, A is a non-singular matrix.

Hence, A^-1 exist

AA^-1= I

A (\(\rm \dfrac{1}{|A|}.adj.A\)) = I

A (adj A) = |A| I

take determinants Both side

| A (adj A) | = |  |A| I

We know that if A be an n-rowed square matrix  and ' k'  be any scalar then| K A | = Kⁿ |A|

Now

| A | | adj A| = | A |ⁿ | I_n |

| A | | adj A| = | A |ⁿ

| adj A| = | A |^n-1

The statement |adj A| = |A| is not correct.

Concept:

The equation of line is:

y = mx + c

where m is the slope of the line.

In the square, the diagonals are perpendicular to each other.

Calculations:

Given,

3x + 2y = 5 is the equation of one diagonal of the square.

Consider m₁ be the slope of one diagonal. and m₂ be the slope of other diagonal.

⇒ 2y = - 3x +5

⇒ \(\rm y = \dfrac{-3}{2} x+ \dfrac{5}{2} \)

⇒ m₁ = \(\dfrac{-3}{2}\)

In square, the diagonals are perpendicular to each other.

⇒ m₁.m₂ = -1

⇒m₂ = \(\rm\dfrac{2}{3}\)

Equation of other diagonal is 

 y = m₂x + c

⇒ y = \(\rm\dfrac{2}{3}\) x + c....(1)

Since, the other diagonal is passing through the point (1, -1). It must satisfy the equation of line.

Put y = -1 and x = 1 in above equation

⇒ -1 = \(\rm\dfrac{2}{3}\) + c

⇒ c = \(\rm\dfrac{-5}{3}\)

 Put the value of c in equation (1).

⇒y = \(\rm\dfrac{2}{3}\) x - \(\rm\dfrac{5}{3}\) 

⇒ 3y = 2x - 5

⇒ 2x - 3y = 5 is the equation of other diagonal.

The point (1, -1) is one of the vertices of a square. If 3x + 2y = 5 is the equation of one diagonal of the square, then the equation of the other diagonal is 2x - 3y = 5