Quantitative Aptitude PYQs

Shortcut TrickWe have \(96 - 64a^3 + \frac{8}{a^6} - \frac{48}{a^3 } - t^3 = 0\)

Only one equation contains two variable 

So, We can solve this question by value putting method 

Let a = 1 we can't take a = 0  since 1/0 = ∞ 

Now, On putting a = 1 in the equation we get 

96 - 64 + 8 - 48 - t³ = 0

⇒ -8 - t³ = 0

⇒ t = -2 

Now, we have to find the value of a2t + 4a3

On putting a = 1 and t = -2 in the above equation, we get 

1² × (-2) + 4 × 1³ = 2

∴ The value of a2t + 4a3 is 2.

 

Alternate Method 

Given:

\(96 - 64a^3 + \frac{8}{a^6} - \frac{48}{a^3 } - t^3 = 0\)

Formula Used:

(a - b)³ = a³ - b³ - 3a²b + 3ab²

Calculation:

We have \(96 - 64a^3 + \frac{8}{a^6} - \frac{48}{a^3 } - t^3 = 0\)

\(⇒ 3.(\frac{2}{a^2}) (4a)^2 - (4a)^3 + (\frac{2}{a^2})^3 - 3.(\frac{2}{a^2})^2.(4a) - t^3 = 0\)

\(⇒ (\frac{2}{a^2})^3 - (4a)^3- 3.(\frac{2}{a^2})^2.(4a) + 3.(\frac{2}{a^2}) (4a)^2- t^3 = 0 \)

\(⇒ [\frac{2}{a^2} -4a]^3- t^3 = 0 \)

\(⇒ t^3 = [\frac{2}{a^2} -4a]^3 \)

\(⇒ t = [\frac{2}{a^2} -4a]\)

Now, We have to find the value of a2t + 4a3

⇒ \(a^2[\frac{2}{a^2} -4a] + 4a^3\)

⇒ 2 - 4a³ + 4a³

⇒ 2

∴ The value of a2t + 4a3 is 2.

Given:

\(A + B = \rm \frac{x^2 - 8}{x + 2} \ \ and \ A - B = \frac{-x^2 + 2x + 4}{x + 2}\)

Calculation:

We have 

\(A + B = \rm \frac{x^2 - 8}{x + 2} \ \ ---(i)\)

\(A - B = \frac{-x^2 + 2x + 4}{x + 2} \space ---(ii)\)

Subtract equation (ii) from equation (i), we get 

\(2B = \rm \frac{x^2 - 8}{x + 2} - \frac{[-x^2 + 2x + 4]}{x + 2}\)

\(2B = \frac{x^2-8+x^2 - 2x - 4}{x + 2}\)

\(2B = \frac{2(x^2-x - 6)}{x + 2}\)

\(B = \frac{(x^2-x - 6)}{x + 2}\)

⇒ B = (x - 3) (x + 2)/ (x + 2)

⇒ B = (x - 3)

Now, we have to check which option is similar to (x - 3)

On solving option (i) we get (x - 2)/(x + 2) ≠ (x - 3)

On solving option (ii) we get (x + 2)/(x - 2) ≠ (x - 3)

On solving option (iii) we get (2x - 1)(x - 3)/(2x - 1) = (x - 3)

∴ The 3^rd option is correct opiton.

Calculation:

P(x) = \(\frac{x^4}{(x^2 - y^2) (x^2 - z^2) }+ \frac{y^4}{(y^2 - x^2) (y^2 - z^2)} + \frac{z^4}{(z^2 - x^2)(z^2 - y^2)}\)

To simplify this expression, put any value except the value in which p(x) become indeterminate.

Put z = 0

⇒ P(x) = \(\frac{x^4}{(x^2 - y^2) x^2 }+ \frac{y^4}{(y^2 - x^2) y^2 } \)

⇒ P(x) = \(\frac{x^2}{(x^2 - y^2) }- \frac{y^2}{(x^2 - y^2) } \)

⇒ P(x) = \(\frac{(x^2-y^2)}{(x^2 - y^2) }\)

⇒ P(x) = 1

Given:

(2ab - b2) : (6a2 - ab) = 1 : 6

Calculation:

We have 

(2ab - b2) : (6a2 - ab) = 1 : 6

⇒ 6 × (2ab - b2) = 1 × (6a2 - ab)

⇒ 12ab - 6b² = 6a² - ab

⇒ 6a2 - 13ab  + 6b2 = 0

⇒ 6a2 - 9ab - 4ab + 6b2 = 0

⇒ 3a (2a - 3b) - 2b (2a - 3b) = 0 

⇒ (2a - 3b) (3a - 2b) = 0 

⇒ (2a - 3b) = 0 or (3a - 2b) = 0 

⇒ 2a = 3b or 3a = 2b

Case:1   2a = 3b

⇒ a : b = 3 :  2

⇒ (a + b) : (a - b) = (3 + 2) : (3 - 2) 

⇒ (a + b) : (a - b) = 5

Case:2   3a = 2b 

⇒ a : b = 2 : 3

⇒ (a + b) : (a - b) = (2 + 3) : (2 - 3) 

⇒ (a + b) : (a - b) = -5

∴ The value of (a + b) : (a - b) is -5 or 5.

Formula used:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 

Calculation:

Given that,

\(\frac{x}{b + c -a } = \frac{y}{b - c - a} = \frac{z }{a + b -c} = k\)

⇒ x = k(b + c - a)  ----(1)

⇒ y = k(b - c - a)  ----(2)

⇒ z = k(a + b - c)  ----(3)

Now, let

P = x2 + y2 + z2 - 2xy - 2yz + 2zx 

As, term zx is positive and other terms xy & yz are negative, then x and z must be negative.

⇒ P = (y - x - z)²

⇒ P = [k(b - c - a) - k(b + c - a) - k(a + b - c)]²

⇒ P = [kb - kc - ka - kb - kc + ka - ka - kb + kc]²

⇒ p = [- ka - kb - kc]²

⇒ P = k²(a + b + c)²

Given:

Ratio of three numbers = 2 : 3 : 5

Sum of squares of the numbers = 1368

Calculation:

Let the numbers be 2x, 3x and 5x 

According to the question:

4x² + 9x² + 25x² = 1368

⇒ x² = 36

⇒ x = 6, - 6

According to the question the numbers are positive 

So, We will not taking negative sign

⇒ 2x = 12, 3x = 18, 5x = 30

∴ Sum of all numbers = 60

Calculation:

Statement:1    \(\frac{a^2 - b^2 }{a^2 + b^2 } > \frac{a - b}{a + b} \) where a > b > 0

Let a = 2 and b = 1

⇒ \(\frac{2^2 - 1^2 }{2^2 + 1^2 } > \frac{2 - 1}{2 + 1} \)

⇒ \(\frac{3 }{5 } > \frac{1}{3} \) which is correct.

Hence, statement 1 is correct.

Statemnt:2   \(\frac{a^3 + b^3 }{a^2 + b^2 } > \frac{a^2 + b^2}{a + b} \) only when a > b > 0

Let a = 2 and b = 1

⇒ \(\frac{2^3 + 1^3 }{2^2 + 1^2 } > \frac{2^2 + 1^2}{2 + 1} \)

⇒ \(\frac{9 }{5 } > \frac{5}{3} \) Which is also correct.

This inequality also holds for a = 1 and b = 2 i.e.

Statement 2 is correct for b > a > 0 as well.

But according to statement 2, inequality will be correct only if a > b > 0

So, statement 2 is incorrect.

**Hence, statement

(1) is correct but (2) statement (2) is incorrect.**

Mistake PointsPlease note the language of  the second statement

"only when a > b > 0"

We have proved thet, statement is correct for 

a = 2 & b = 1 (i.e. a > b > 0) and b = 2 & a = 1 (i.e. b > a > 0)

Hence, due to only word, statement 2 will become incorrect.

Given:

Work done by (3n - 1) men in (2n + 1) days = x

Work done by (3n + 1) men in (4n - 3) days = y

x : y = 6 : 11

Formula used:

(m₁ × d₁)/w₁ = (m2 × d₂)/w₂

m = number of men

d = number of days

w = work done

Calculation:

Let x be 6a and y be 11a

According to the question:

(3n - 1)(2n + 1)/6a = (3n + 1)(4n - 3)/11a

⇒ 11(6n² + n - 1) = 6(12n² - 5n - 3)

⇒ 66n² + 11n - 11 = 72n² - 30n - 18

⇒ 6n² - 41n - 7 = 0

⇒ (6n + 1)(n - 7) = 0

⇒ n = - 1/6, 7

∴ n = 7

Given:

\(\frac{ay - bx}{c}= \frac{cx - az}{b}= \frac{bz - cy}{a} \)

Calculation:

In the above equation multiple by c, b, and a in the consecutive equations

\(⇒ \frac{ay - bx}{c} \times \frac {c}{c} = \frac{cx - az}{b} \times \frac {b}{b}= \frac{bz - cy}{a} \times \frac {a}{a}\)

\(⇒ \frac{acy - bcx}{c^2} = \frac{bcx - abz}{b^2} = \frac{abz - acy}{a^2}\)

\(⇒ \frac{acy - bcx}{c^2} = \frac{bcx - abz}{b^2} = \frac{abz - acy}{a^2} = k (let) ----(i)\)

On equating equations with k, we get 

\(⇒ {acy - bcx} = c^2 k \space \space ----(ii)\)

\(⇒ {bcx - abz} = b^2 k \space \space ----(iii)\)

\(⇒ {abz - acy} = a^2 k \space \space ----(iv)\)

On adding equation (ii), (iii) and (iv), we get 

⇒ 0 = k(a2 + b2 + c2)

⇒ k = 0    ----(v)

From equation (i) and (v), we get

acy = bcx

⇒ x/y = a/b    ----(vi)

bcx = abz

⇒ x/z = a/c    ----(vii)

abz = acy

⇒ y/z = b/c     ----(viii)

From (vi), (vii) and (viii), we get

x = ar, y = br and z = cr  [let r be the ratio]

Now, we have to check the given conditions

Condition 1. is true from equation (vi)

Condition 2. is \(\frac{x + y+ z}{a + b + c} = \frac{z}{c}\)

⇒ (ar + br + cr)/(a + b + c) = r   [LHS]

⇒ cr/c = r  [RHS]

Condition 2. is also true 

∴ Both the conditions 1 and 2 are true.

Given:

Area of the rectangular garden = 375 m²

65 m wire can fence 3 sides of the garden

Formula used:

Area of a rectangle = lb

l = length of the rectangle

b = breadth of the rectangle

Calculation:

We have, lb = 375      ----(i)

According to the question,

There are two possibilities 

2b + l = 65      ----(ii)

or 2l + b = 65      ----(iii)

Solving (i) and (ii), we get 

⇒ 2b + 375/b = 65

⇒ 2b² - 65b + 375 = 0

⇒ (2b - 15) (b - 25) = 0

⇒ b = 7.5, 25 and l = 50, 15

Pair (50, 7.5) is possible here, as length > breadth

So, The perimeter of rectangle = 2 (50 + 7.5) = 115m

Solving equation (i) and (iii), we get 

⇒ l = 7.5, 25 and b = 50, 15

Pair (25, 15) is possible here, as length > breadth

So, The perimeter of rectangle = 2 (25 + 15) = 80m

∴ Form option we can say that 80m is the perimeter of the garden.

Given:

One root of ax2 - 4ax + 15 = 0 is 3/2

Calculation:

If 3/2 is root of ax2 - 4ax + 15 = 0; then it will satisfy the equation

⇒ 9a/4 - 6a + 15 = 0

⇒ -15a/4 = -15

⇒ a = 4

Equation becomes 4x2 - 16x + 15 = 0

⇒ 4x2 - 10x - 6x + 15 = 0

⇒ (2x - 5)(2x - 3) = 0

⇒ x = 3/2, 5/2

Now, Sum of squares of roots = (3/2)² + (5/2)²

⇒ 34/4 = 17/2

∴ The value of sum of square of roots is 17/2.

Given:

Product of the digits of a 2-digit number = 8

If 63 added to the number, its digits interchange

Calculation:

Let the number be 10x + y

According to the question

xy = 8    ----(i)

Also, 10x + y + 63 = 10y + x

⇒ 9y - 9x = 63 

⇒ y = x + 7      ----(ii)

From equation (i) and (ii), we have

⇒ 8/x = x + 7

⇒ x² + 7x - 8 = 0

⇒ (x + 8) (x - 1) = 0

⇒ x = 1, - 8; negative term rejected

⇒ y = 8

Hence, the number is 18

∴ Sum of the digits = 1 + 8 = 9

Given:

Speed of motor boat in still water(b) = 30 kmph

Total downstream distance = 60 km

Total upstream distance = 60 km

Total time 9/2 hours

Formula used:

d = b + w

u = b - w

w = speed of stream, d = downstream speed

u = upstream speed

Calculation:

According to the question:

60/(30 + w) + 60/(30 - w) = 9/2

⇒ 3600 = 9/2(900 - w²)

⇒ (900 - w2) = 800

⇒ w2 = 100

⇒ w = 10

∴ Speed of the stream = 10 kmph

Given:

Present age of father  = Sum of present ages of 4 children

10 year hence,

Sum of ages of children = 1.6 × age of the father

Calculation:

Let the present age of father be f and present sum of four children be c

According to the question

f = c    ----(i) and 

1.6(f + 10) = c + 40    ----(ii)

Form (i) and (ii), we get

⇒ 1.6f + 16 = f + 40

⇒ f = 40

∴ Present age of father = 40 years

Given:

Sum of numerator and denominator = 10

If numerator increased by 3 and denominator decreased by 1, fraction becomes 1

Calculation:

Let n be the numerator and (10 - n) be the denominator

⇒ Fraction is n/(10 - n)

According to the question:

(n + 3)/(10 - n - 1) = 1

⇒ n + 3 = 9 - n

⇒ n = 3

⇒ Denominator = 10 - n = 7

∴ Difference between numerator and denominator = 7 - 3 = 4

Given:

Equations 7x + ky = 27 and kx + 7y = 19 have unique solution

Formula used:

a₁x + b₁​y + c₁ ​= 0 

a₂x + b₂y + c₂​ = 0

If both equations have unique solution,

then a₁/a₂ ≠ b₁/b₂

Calculation:

For equations to have unique solution:

7/k ≠ k/7

⇒ k² ≠ 49

∴  k ≠ 7

Given:

Principal(P) doubles itself in 5 years when compounded annually

Formula used:

Amount(A) = P[1 + (R/100)]^T

R = Rate of interest

T = time period

Calculation:

According to the question:

2P = P[1 + (R/100)]⁵

⇒ 2 = [1 + (R/100)]5

⇒ 2^1/5 = 1 + (R/100) ----(i)

Let the required time period be T years

⇒ 4P = P[1 + (R/100)]^T ----(ii)

From (i) and (ii), we get

4 = [2^1/5]^T

⇒ 2^1/5 = 4^1/T

⇒ 21/5 = 22/T

⇒ 2/T = 1/5

⇒ T = 10

∴ It will become 4 times in 10 years

Formula Used:

The angle between the hour hand and minutes hand

θ = (30H - 11 M/2)

Calculation:

H = 3 and angle = 0°

When minute & hour hands coincide, then θ = 0

⇒ 0 = 30 × 3 - 11 M/2

⇒ 0 = 90 - 11 M/2

⇒ 90 = 11 M/2

⇒ 180 = 11 M

⇒ M = \(16\frac{4}{11}\)

⇒ M = \(16\frac{4}{11}\) Minutes

∴  Both hour hand and minute hand will coincide past 3 O’clock between 16 to 17 minutes.

Given:

Sheela can stitch a suit in 2 days

Meena can stitch a suit in 3/2 days

Formula used:

Work = Efficiencey × Time

Calculation:

Efficiency of Sheela = 1/2 

Efficiency of Meena = 2/3

Combined efficiency = 1/2 + 2/3 = 7/6 suits per day

Now, They can stitch 7/6 suit in 1 day 

So they can stitch 1 suit in 6/7 days 

So they can stitch 35 suits in 35 × 6/7 = 30 days 

∴ Days taken for stitching 35 suits = 30 days

Given:

2x - 3y - 7 = 0

Formula used:

(a - b)³ = a³ - 3a²b + 3ab² - b³

Calculation:

2x - 3y = 7

By cubing both sides

⇒ 8x³ - 3. (2x)² (3y) + 3 (2x) (3y)² - 27y³ = 343

⇒ 8x3 - 36x2y + 54xy2 - 27y3 = 340 + 3

⇒ 8x3 - 36x2y + 54xy2 - 27y3 - 340 = 3

∴ The value of 8x3 - 36x2y + 54xy2 - 27y3 - 340 is 3.