Quantitative Aptitude PYQs

For hollow sphere: r₁ = 6/2 = 3 cm and r₂ = 4/2 = 2 cm

Volume of the hollow sphere = 4/3π (r₁³ - r₂³)

⇒ 4π/3 × (27 - 8) = 76π/3

Since it is melted into a cone of base diameter 8 cm:

Volume of the cone = 76π/3

⇒ π/3 × 4² × h = 76π/3

⇒ h = 76/16 = 4.75 cm

For solid metallic cylinder: height = 10 cm and radius = 6 cm

Volume of the cylinder = π × 36 × 10 = 360π

Flat surface area = 2π × 36 = 72π

This cylinder is melted to make 2 cones with ratio of volumes 1 : 2

Hence,

Volume of smaller cone = 360 π/3 = 120π

Volume of larger cone = (360 π/3) × 2 = 240π

Suppose radiuses of smaller and larger cones are r and R, respectively;

Since the height of both cones is same as 10 cm;

Now,

π/3 × r² × 10 = 120 π

⇒ r = 6 cm

And

π/3 × R² × 10 = 240 π

⇒ R = 6√2 cm

Now,

Flat surface area of 2 cones = πr² + πR² = π × 36 + π × 36 × 2 = 108π

Hence,

Percentage increase = [108π - 72π]/72π = 50%

We know that:

(112, 15, 113) is a Pythagoras triplet with one number 15:

⇒ 113² = 112² + 15²

Hence,

Maximum perimeter = 112 + 113 + 15 = 240 cm

Given:

Length of rod = 24 feet

Formula used:

The surface area of each face of a cube with side a = a²

Calculation:

Since a skeleton cube has 12 sides:

Length of each side of the skeleton cube

= 24/12 = 2 feet

Hence, the area of one of the faces of the largest cube

= 2² = 4 square feet

Given:

EB = 25 cm and FC = 16 cm

Since CF and CP are tangents to the circle from the same point C, CP = CF = 16 cm

Similarly, BP = BE = 25 cm

BC = 16 + 25 = 41 cm

CQ is perpendicular to BA:

QB = 25 - 16 = 9 cm

In ΔBCQ:

CQ = √(41² - 9²) = 40 cm

Hence, diameter of the circle = 40 cm

Given:

Radii of three copper spheres = 3 cm, 4 cm, and 5 cm.

Formula used:

The volume of sphere = 4/3(πr³)

Calculation:

Sum of volumes of 3 copper spheres = Volume of new sphere.

⇒ 4/3(πr3) = 4π/3 × (27 + 64 + 125)

⇒ 4/3(πr3) = 4π/3 × 216

⇒ r3 = 6³

⇒ The radius of the larger sphere would be 6 cm.

Given:

Volume of a hemisphere is 155232 cm³

Formula Used:

Volume of Hemisphere = 2π/3 × r3

Calculation:

2π/3 × r³ = 155232

⇒ π × r³ = 232848

⇒ 22/7 × r³ = 232848

⇒ r³ = 74088

⇒ r = 42 cm

We can say that the bucket is in the form of frustum:

r = 3 cm and R = 6 cm, h = 7 cm

Now,

Volume of the frustum = π/3(r² + R² + rR)h

⇒ π/3(3² + 6² + 18) × 7

⇒ 22/21 × (3² + 6² + 18) × 7

⇒ 22 × (9 + 36 + 18)

⇒ 22/3 × 63

⇒ 462 cm³

Given:

Height  = 8 cm

The radius of its base = 6 cm

Formula used:

Slant height l = \(\sqrt {{r^2} + {h^2}} \) 

Total surface area of cone = πr2 + πrl

Calculation:

Height = 8 cm and radius = 6 cm

Slant height l = \(\sqrt {{8^2} + {6^2}} \) = 10 cm

Total surface area of cone = πr² + πrl

⇒ π (36 + 60)

∴  96π cm²

Given:

Six cubes, each with 12 cm edge are joined end to end.

Formula used:

Surface area of cuboid = 2 × (lb + bh + hl)

Calculation:

Length of cuboid = 12 × 6 = 72 cm

Breadth of cuboid = 12 cm

Height = 12 cm

Now, surface area of cuboid = 2 × (lb + bh + hl)

⇒ 2 × (72 × 12 + 12 × 12 + 12 × 72) cm²

⇒ 2 × (864 + 144 + 864) cm²

⇒ 2 × 1872 = 3744 cm²

∴ The surface area of resulting cuboid is 3744 cm²

x + 1/x = 26/5

⇒ x + 1/x = 5 + 1/5

Hence, x = 5 or 1/5

Hence, two such numbers are possible.

Statement 1: If p is relatively prime to each of q and r, then p is relatively prime to the product qr.

Suppose p = 8, q = 15 and r = 7

Here, p is relatively prime to each of q and r.

Now, product qr = 105

Here, p i.e. 8 is relatively prime to qr i.e. 105 also.

Hence, statement 1 is correct.

Statement 2: If p divides the product qr and if p divides q, then p must divide r.

Suppose p = 4, q = 8 and r = 7

Here, qr = 56

Here, p divides the product qr and p divides q also.

But p does not divide r here.

Hence, statement 2 is not correct.

Calculation:

Suppose the present ages of Radha and Rani are X and Y, respectively;

According to the question:

(X - 5) = 3(Y - 5)

⇒ 3Y - X = 10

⇒ X = 3Y - 10      -----(1)

And

(X - 1) = 2(Y - 1)

⇒ 2Y - X = 1

From equation (1)

⇒ 2Y - (3Y - 10) = 1

⇒ -Y + 10 = 1

⇒ Y = 9

From equation (1)

⇒ X = 3 × 9 - 10 = 17

∴ The age difference between Radha and Rani = 17 - 9 = 8

Suppose he buys x apples and y oranges:

5x + 7y = 500

Possible non-zero values of (x, y) pair:

(2, 70), (9, 65), (16, 60), (23, 55), (30, 50), (37, 45), (44, 40), (51, 35), (58, 30), (65, 25), (72, 20), (79, 15), (86, 10), (93, 5)

Hence, total 14 ways are possible.

Alternate Method

5x + 7y = 500

for x and y to be integers y should be divisible by 5.

Let y = 5a

So 5x + 35a = 500

x + 7a = 100

Minimum value of x = 1

So 7a = 99, a = 14( to be integer)

Given:

y ∝ 1/√x

Or

y = k/√x

Given x = 36 when y = 36;

⇒ 36 = k/6

⇒ k = 216

Now when y = 54:

54 = 216/√x

⇒ √x = 4

⇒ x = 16

Concept used:

(a + b)² = a² + 2ab + b²

Calculations:

16 + 6√7

⇒ 7 + 9 + 6√7

= (√7)² + 3² + 2 × 3 × √7

⇒ (3 + √7)²

∴ The square root is (3 + √7)

Suppose:

X = 7²⁵

Taking log :

log₁₀X = 25 log₁₀7

log₁₀X = 25 × 0.845

log₁₀X = 21.125

X = (10)^21.125

Hence, it will have 22 digits.

Y = 8²³ = 2⁶⁹

Taking log :

log₁₀Y = 69 log₁₀2

log₁₀Y = 69 × 0.301

log₁₀Y = 20.769

Y = (10)^20.769

Hence, it will have 21 digits.

Z = 9²⁰ = 3⁴⁰

Taking log :

log₁₀Z = 40 log₁₀3

log₁₀Z = 40 × 0.477

log₁₀Z = 19.08

Z = (10)^19.08

Hence, it will have 20 digits.

x = 14a + 7

And

x = 15a + 5

Value of x in both cases will be equal to 35 when a = 2

Hence, 30 < x < 40

1/Y = 1/15 – 1/20

⇒ 1/Y = 1/60

⇒ Y = 60

Hence, if only Y is opened, it will take 60 minutes to drain out the full tank.

Concept used:

Irrational numbers are real numbers that cannot be represented as a simple fraction. It cannot be expressed in the form of a ratio, such as p/q, where p and q are integers, q ≠ 0. 

Calculation:

1. √75 is an irrational number.

2. • 4x/5 < - 7/8 will be true for x > 1

3. \(\frac{{x - 2}}{x} < 1\), it is not true for x = - 1

4. 4.232323 can be expressed in the form p/q where p and q are integers : true

So, statements 2 and 4 are correct

∴ Required answer is Option 4.