Quantitative Aptitude PYQs

Given:

x2 + x + 1 = 0    ----(i)

Calculation:

We have x2 + x + 1 = 0

We have to find the value of x199 + x200 + x201

⇒ x199  (1 + x + x2) 

From (i), we get

⇒ x199 × 0

⇒ 0

∴ The value of x199 + x200 + x201  is 0.

Concept:

If a, b and c are three terms in GP then b² = ac and vice-versa

If three terms a, b, c are in HP, then b = \(\rm \frac{2ac}{a + c}\) and vice-versa

When three terms are in AP, the middle one is called the arithmetic mean of the other two.

If a, b and c are three terms in AP then b = \(\rm \frac{a + c}{2}\) and vice-versa

Calculation:

Given x, y, z are in G.P

y² = xz      ....(1)

On multiply both sides by 9, we get

⇒ (3y)2 = 9xz

Taking log on both sides, we get

⇒ 2In 3y = In 3x + In 3z

⇒ ln(3x), ln(3y), ln(3z) are in A.P

So, ln(3x), ln(3y), ln(3z) are in A.P

Statement 1 is correct.

Again taking eq(1)

y2 = xz 

Taking log on both sides, we get

⇒ 2In y = In x + In z

On adding 2xyz on both the sides, we get

⇒ 2(xyz + In(y)) = (xyz + In(x)) + (xyz + In(z))

⇒ xyz + ln(x), xyz + ln(y), xyz + ln(z) are in A.P

\(\rm \frac{1}{xyz + In(x)}, \frac{1}{xyz + In(y)}, \frac{1}{xyz + In(z)}\) are in H.P

It means  xyz + ln(x), xyz + ln(y), xyz + ln(z) are not in HP.

Statement 2 is not correct.

Formula used:

• If a, b and c are three terms in AP then b = \(\rm \frac{a + c}{2}\)
• log_amⁿ = n log_am
• log_a [mn] = log_a m + log_a n
• If logam = logan then m = n

Calculation:

Given, log102,  log10(2x - 1), log10(2x + 3) are in AP

According to the question

2 log10(2x - 1) =  log102 + log10(2x + 3)

⇒ log10(2x - 1)² =  log10 2(2x + 3)

⇒ log₁₀​(2^2x + 1 − 2^x+1) = log₁₀​(2x+1 + 6) 

From the formula used

⇒ 22x + 1− 2x+1 = 2x+1 + 6

⇒ 22x + 1− 2 × 2x+1 = 6

⇒ 2^2x − 2^x+2 − 5 = 0 

Take, 2^x = y > 0 

y² − 4y − 5 = 0 

⇒ (y − 5)(y + 1) = 0 

⇒ y = 5,−1 

∵ y > 0

⇒ y = 5 

⇒ 2^x = 5 

⇒ x = log₂​5

∴ The value of x is log2​5.

Explanation:

We can solve this by filling the places according to the question 

Step 1:

Since the Number should be less than 500 

So, there are only 3 possibilities at 1 st place i.e. (2, 3, 4) 

Step 2:

According to the question

Repetition is not allowed we have to fix one number (let 2) in the 1^st place 

Now, At 2^nd place, there are a total of 6 ways (3, 4, 5, 6, 7, 9)

Step 3:

According to the question

Repetition is not allowed we have to fix one number (let 3) in the 2^nd place 

Now, At 3^rd place, there are a total of 5 ways (4, 5, 6, 7, 9)

∴ The total number of different 3-digit numbers less than 500 = 3 × 6 × 5 = 90.

Formula used:

sum of first n terms of GP formula

S_n = \(\rm \frac{a(r^{n} - 1)}{r - 1}\), \(\rm r\neq 1 \) and r > 1

Calculation:

11 = 1 + 10¹

111 = 1 + 10 + 10²

1111 = 1 + 10 + 10² + 10³

(1111 ..... n-digit) = 1 + 10 + 10² + 10³ + ...... + 10^n-1

 1 + 10 + 102 + 103 + ...... + 10n-1  are in G.P

let P = 1 + 10 + 102 + 103 + ...... + 10n-1

⇒ p = \(\rm \frac{a(r^{n} -1)}{r - 1} = \frac{10^{n} -1}{9}\)

Now, we have to find the value of (9p2 + p)

⇒ 9 × \(\rm \frac{(10^{n} -1)^{2}}{81} + \frac{10^{n} -1}{9}\)

⇒\(\rm \frac{(10^{n} -1)^{2}}{9} + \frac{10^{n} -1}{9}\)

⇒ \(\rm \frac{(10^{2n} \space+\space 1 -\space 2\space ×\space 10^{n} \space+\space 10^{n}\space-1\space)}{9} \)

⇒ \(\rm \frac{(10^{2n} - 10^{n})}{9} \)

⇒ \(\rm \frac{10^{n} (10^{n}- 1)}{9} \)

⇒ 10ⁿ p

∴ The value of 9p2 + p is 10n p.

Concept:

Quadratic equation:

ax² + bx + c = 0

roots of equal magnitude and opposite sign then

b = 0, a ≠ 0, c ≠ 0

Calculation:

The given equation 3x2 - (k2 + 5k)x + 3k2 - 5k = 0

Here, a = 3, b = -(k2 + 5k) and c = 3k2 - 5k

According to the concept used

Case I. b = 0

⇒ (k² + 5k) = 0

⇒ k(k + 5) = 0

⇒ k = 0, -5

Case II. c ≠ 0

3k2 - 5k ≠ 0

⇒ k(3k - 5) ≠ 0

⇒ k ≠ 0, k ≠ 5/3.

From Case I and II, we get k = -5

But the question says roots are real

So, k = -5 is not possible.  [at k = -5, D < 0]

∴ The option 4^th is the correct option.

Formula used:

n! = n × (n - 1) × (n - 2) × ....... × 3 × 2 × 1

Calculation:

a_n = n (n)!

a_n = (n + 1 - n)n!

a_n = (n + 1)n! - n!

a_n = (n + 1)! - n!

Hence, 

a₁ = 2! - 1! 

a₂ = 3! - 2! 

---

a₁₀ = 11! - 10!

Now,

 a1 + a2 + a3 +...+ a10 

2! - 1! + 3! - 2! + ......11! - 10! 

= 11! - 1

∴ The value of a1 + a2 + a3 +...+ a10 is 11! - 1.

Formula used:

For any quadratic equation ax² + bx + c = 0,

sum of the roots = \(\rm \frac{-b}{a}\)

product of the roots = \(\rm \frac{c}{a}\)

Calculation:

Given equation 

x² + px + q = 0

Also, given p, q are the roots of the equation.

Sum of roots = -p/1

⇒ p + q = −p  

⇒ 2p + q  = 0   .....(1)

And product of roots = q/1

⇒ pq = q

⇒ p = 1

putting the value of p in eq(1), we get

2(1) + q = 0

⇒ q = -2

∴ q has only one value.

Concept::

• The interchange of any two rows (or columns) of the determinant changes its sign.
• The determinant remains unaltered if its rows are changed into columns and the columns into rows.

 

Calculation:

\(\Delta = \begin{vmatrix} a & b & c\ d & e & f\ g & h & i \end{vmatrix}\)

\(A = \begin{vmatrix} 3d + 5g & 4a + 7g & 6g\ 3e + 5h & 4b + 7h & 6h\ 3f + 5i & 4c + 7i & 6i \end{vmatrix}\)

\(⇒ A = 6\begin{vmatrix} 3d + 5g & 4a + 7g & g\ 3e + 5h & 4b + 7h & h\ 3f + 5i & 4c + 7i & i \end{vmatrix}\)

C₁ → C1 - 5C₃, C₂ → C2 - 7C₃

 \(A = 6\begin{vmatrix} 3d & 4a & g\ 3e & 4b & h\ 3f & 4c& i \end{vmatrix}\)

\(A =6\times 3\times4 \begin{vmatrix} d & a & g\ e & b & h\ f & c& i \end{vmatrix}\)

Using the property of determinants

\(A =-72 \begin{vmatrix} a & b & c\ d & e & f\ g & h& i \end{vmatrix}\)

∴  A = -72Δ

Concept:

• If a, b and c are three terms in GP then b2 = ab and vice-versa
• If three terms a, b, c are in HP, then b = \(\rm \frac{2ac}{a + c}\)  and vice-versa
• When three quantities are in AP, the middle one is called as the arithmetic mean of the other two.
• If a, b and c are three terms in AP then b = \(\rm \frac{a + c}{2}\)  and vice-versa
• If a, b and c are in A.P then 1/a, 1/b, 1/c are in H.P and vice-versa

Calculation:

\(\frac{1}{b+c}, \frac{1}{c+a},\frac{1}{a+b}\) are in HP

(b + c), (c + a) and (a + b) are in A.P

⇒ 2(c + a) = b + c + a + b    

⇒ 2c + 2a = 2b + a + c     -----(i)

⇒ 2c + 2a - 2b - a - c = 0

⇒ 2b = c + a

So, a, b, c are in AP

Now, Let (b + c)2, (c + a)2, (a + b)2 are in GP

(c + a)² = √[(b + c).(a + b)]²

⇒ c² + a² + 2ac = (b + c).(a + b)

⇒ c2 + a2 + 2ac = ab + b² + ac + bc

⇒ c2 + a2 - b2 + ac - ab - bc = 0

From here we are unable to check the relation between a, b and c 

So, Our assumption was wrong 

So, b + c)2, (c + a)2, (a + b)2 are not in GP.

∴ Only the 1^st option is correct.

Formula used:

A = \(\rm \begin{bmatrix} a & b\ c & d \end{bmatrix}\)

det(A) = ad - cb

Identity matrix: 

An identity matrix is a given square matrix of any order which contains on its main diagonal elements with a value of one, while the rest of the matrix elements are equal to zero.

\(\rm I=\begin{bmatrix} 1 & 0\ 0 & 1 \end{bmatrix}​​\)

Calculation:

Given that,

 

\(\rm A^2=\begin{bmatrix} 1 & 2a\ 0 & 1 \end{bmatrix}\)

\(\rm A^3 = A^2\times A=\begin{bmatrix} 1 & 3a\ 0 & 1 \end{bmatrix}\)

Therefore, the value of A100 - A50 - 2A25

⇒ \(\begin{bmatrix} 1 & 100a \ 0 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 50a \ 0 & 1 \end{bmatrix}-2\begin{bmatrix} 1 & 25a \ 0 & 1 \end{bmatrix}\)

⇒ \(\begin{bmatrix} 1 & 100a \ 0 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 50a \ 0 & 1 \end{bmatrix}-2\begin{bmatrix} 1 & 25a \ 0 & 1 \end{bmatrix}\)

⇒ \(\begin{bmatrix} 1 & 50a \ 0 & 1 \end{bmatrix}-\begin{bmatrix} 2 & 50a \ 0 & 2 \end{bmatrix}\)

⇒ \(-2\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\)

∴ A100 - A50 - 2A25  = -2I

Alternate Method\()​

\(\rm A=\begin{bmatrix} 1 & a \ 0 & 1 \end{bmatrix}\)​​​⇒ |A|  = 1 - 0 = 1

 

⇒ A100 - A50 - 2A25

⇒  (1)100 - 150 - 2(1)25

⇒  1 - 1 - 2

⇒  -2

⇒  -2I

Formula used:

\(\rm A = \begin{bmatrix} a & b & c\ d & e & f\ g & h & i \end{bmatrix}\)

det(A) = a(ei - hf) - b(di - gf) + c(dh - eg)

Calculation:

\(\begin{vmatrix} a & -b & a - b - c\ -a & b & -a + b - c\ -a & -b & -a - b + c \end{vmatrix} = kabc \)

R₂ → R₂ + R₁

\(\begin{vmatrix} a & -b & a - b - c\ 0 & 0 & - 2c\ -a & -b & -a - b + c \end{vmatrix} = kabc \)

⇒ a[-(-2c)(-b)] - (-b)[-(-2c)(-a)]

⇒ - 2abc - 2abc + (a - b - c ) × 0 = kabc

⇒ - 4abc = kabc

⇒ k = - 4

∴ The value of k is - 4

Formula used:

Powers of iota.

i² = -1, i³ = -i, i⁴ = 1,

i⁵ = ( i . i4 ) = i,

i⁶ = i² . i4  = -1,

i⁷ = -i...... and so on.

 ⇒ i1 + i2  + i3 + i4 = i -1 - i + 1 = 0

Calculation:

\(\rm \sum\limits_{n=1}^{8n +7} i^n\)

⇒ i¹ + i²  + i³ + i⁴ + i⁵ + i⁶ + i⁷ + ..... + i⁸ⁿ⁺⁷ 

⇒  i1 + i2  + i3 + i4 + i5 + i6 + i7 + ..... +  i8n +  i8n+1 + i8n+2 + .... +  i8n+6  + i8n+7  

⇒ (i -1 - i + 1) +  (i -1 - i + 1) + .... +  (i8n+1 + i8n+2 +  i8n+3 + i8n+4 ) +  (i8n+5 +  i8n+6  + i8n+7)

⇒ 0 + i8n+5 +  i8n+6  + i8n+7 = i1 + i2  + i3 =  i -1 - i = -1

∴ The value of  \(\rm \sum\limits_{n=1}^{8n +7} i^n\) is - 1.

Formula used:

z = a + ib

\(\rm \overline{z} = a - ib\)

Equation of circle: (x - h)² + (y - k)² = r²

Where (h,k) is the coordinates of center of the circle and r is the radius. 

Calculation:

we have \(z \overline{z} + |z|^2 + 4(z + \overline{z})-48 = 0\)

(x + iy)(x - iy) + x² + y² + 4(x + iy + x - iy) - 48 = 0

⇒ x² + y² +  x2 + y2 + 4 × 2x - 48 = 0

⇒ 2(x² + y²) + 8x - 48 = 0

⇒ x² + y² + 4x - 24 = 0

⇒ (x + 2)² + (y - 0)² = \(\rm (\sqrt{28})^{2}\)

Since, the above equation is similar to (x - h)2 + (y - k)2 = r2

∴ The given equation represent a circle.

Formula used:

(a + b)² = a² + b² + 2ab

(a + ib) (a - ib) = a² - i²b² = a² + b²

i² = -1

Concept Used:

if y = x² then we can say that x is the square root of y

Calculation:

We have the algebric expression \(\rm 2a+2\sqrt{a^2 + b^2}\)

⇒ {(a + ib) + (a - ib)} + \(\rm 2\sqrt{(a + ib)(a - ib)}\)

⇒ \(\rm {(\sqrt{(a + ib)})^{2} + ((\sqrt{(a - ib)})^{2} + 2\sqrt{(a + ib)(a - ib)}}\)

⇒ \(\rm (\sqrt{a + ib} + \sqrt{a - ib})^{2}\)

∴ The square root of  \(\rm 2a+2\sqrt{a^2 + b^2}\) is  \(\rm (\sqrt{a + ib} + \sqrt{a - ib})\).

Formula used:

For any quadratic equation ax2 + bx + c = 0,

when α and β are the roots then

α + β = \(\rm \frac{-b}{a}\)

αβ = \(\rm \frac{c}{a}\)

(sinθ + cos θ)2 = sin2θ + cos2θ + 2 sin θ cos θ

sin2θ + cos2θ = 1

Calculation:

According to the question

sin θ and cos θ are the roots of the equation ax² + bx + c = 0,

sin θ + cos θ = \(\rm \frac{-b}{a}\)     ----(i)

sin θ . cos θ = \(\rm \frac{c}{a}\)     ----(ii)

(sinθ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ    ----(iii)

Form (i), (ii) and (iii), we get 

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

⇒ b² = a² + 2ac

⇒ a² - b² + 2ac = 0

∴ The correct relation between a, b and c is a2 - b2 + 2ac = 0.

Formula used:

• C(n,r) = \(\rm \frac{n!}{r!(n-r)!}\)
• If a, b and c are in A.P then 2b = a + c

Calculation:

According to the question C(n, 4), C(n, 5) and C(n, 6) are in AP

2 C(n, 5) = C(n, 4) + C(n, 6) 

⇒ 2 × \(\rm \frac{n!}{5!(n-5)!}\) = \(\rm \frac{n!}{4!(n-4)!}\) + \(\rm \frac{n!}{6!(n-6)!}\)

⇒ \(\rm \frac{1}{60(n-5)(n - 6)!} = \frac{1}{24(n-4)(n - 5)(n - 6)!} + \frac{1}{720(n - 6)!}\)

Multiply by 24(n - 5)(n - 6) on both side

⇒ \(\rm \frac{1}{5} - \frac{1}{2(n-4)} = \frac{n - 5}{60}\)

Again, multiply 60(n - 4) on both side

⇒ 12(n - 4) = 30 + (n - 5)(n - 4)

⇒ 12n - 48 = 30 + n² - 9n + 20

⇒ n² - 21n + 98 = 0

⇒ n = 14, 7

∴ The value of n is 7.

Concept:

If n is a positive integer and r is a whole number, such that r < n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time.

It can be represented as ⁿP_r = \(\frac{n!}{(n-r)!}\).

The combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” 

ⁿC_r = \(\frac{n!}{r!(n-r)!}\) ,  when n < r

Where n = distinct object to choose from

C = Combination

r = spaces to fill

Calculation:

Vowels = 2

Consonants = 5

Total Alphabets = 7

Since 4 letter words must include 2 vowels, we don't need to select them, and the rest of the 2 letters will be taken from 5 consonants.

Number of ways of selecting 2 letters from 5 consonants = ⁵C₂ = 10

Arrangement of all 4 letters will be given by 4! = 24 ways

Total number of arrangements = 5C2 × 4! = 10 × 24 = 240 ways

∴ The total number of words that can be formed is 240.

Explanation:

We only need 2 points to form straight lines

We have 20 points so the number of straight lines drawn is ²⁰C₂ ​= 190

 

We only need 3 points to form a triangle

We have 20 points, so number of triangles drawn is ²⁰C₃ = 1140

∴ Statement 1 is incorrect and Statement 2 is correct.

Formula used:

If a and b are real numbers and n is a positive integer, then

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ an - 1 b¹ + .........  ⁿCn bⁿ  for \(\rm 0 \leq r \leq n\)

Total number of terms in (a + b)ⁿ = n + 1

 

Calculation:

We have to find the total number of terms in \(\rm \left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+2\right)^{21}\) where a ≠ 0, b ≠ 0

⇒ \([{\rm \left(\frac{a}{b}+\frac{b}{a}\right)^{2}}]^{21}\)

⇒ \({\rm \left(\frac{a}{b}+\frac{b}{a}\right)^{42}}\)

∴ The total number of terms in the expression is 42 + 1 = 43.