Search Previous Year Questions

Given:

The total price of shares = Rs. 2500000

Price of one share = Rs. 100

Concept used:

Number of shares = Total price of shares/Price of one share

Calculation:

The total price of shares = Rs. 2500000

Price of one share = Rs. 100

Number of shares = Total price of shares/Price of one share

⇒ Number of shares = 2500000/100 = Rs. 25000

∴ The number of shares circulated by the company is 25000.

Given:

Total number of cards = 52

Formula used:

Probability = Number of favorable outcome/ Total number of outcome

Calculation:

Total number of the spade in a standard deck of card = 13

Total number of aces in a standard deck of card = 4

Number of an ace in spade = 1

Number of favorable outcome = 13 + 4 – 1 = 16

Total number of outcome = 52

Probability = Number of favorable outcome/ total number of outcome

⇒ Probability = 16/52 = 4/13

∴ The probability that the card is a spade or an ace is 4/13.

Given:

ABCD is a cyclic quadrilateral

∠ADC = 4y + 20°

∠ABC = 4x°

∠BAD = 3y – 5°

∠BCD = 7x + 5°

Concept used:

Sum of the opposite angle of a cyclic quadrilateral is 180°.

Calculation:

ABCD is a cyclic quadrilateral

∴ ∠ADC + ∠ABC = 180°

⇒ 4y + 20° + 4x° = 180°

⇒ x + y + 5° = 45°

⇒ x + y = 40°        ……(1)

And,

∠BAD + ∠BCD = 180°

⇒ 3y – 5° + 7x + 5° = 180°

⇒ 7x + 3y = 180°     …….(2)

Multiply 3 in equation (1) and subtract it in equation (2).

7x + 3y – 3x – 3y = 180° - 120°

⇒ 4x = 60°

⇒ x = 15°

Put value of x in equation (1)

x + y = 40°

⇒ 15° + y = 40°

⇒ y = 25°

∴ The ordered pair (x, y) is (15°, 25°).

Given:

25, 26, 27, 23, 27, 26, 24, x, 27, 26, 25, 25

Concept used:

The mode is the value that appears most often in a set of data values.

Calculation:

Following data is arranged in ascending order:

23, 24, 25, 25, 25, 26, 26, 26, 27, 27, 27, x.

The number of repetitions in data.

23 and 24 - 1 time

25 and 26 - 3 times

27 - 3 times

If Mode = 27

i.e., 27 appears most often (4 times) in given sets of data values.

∴ x = 27

Given:

x = 2, y = 1, z = -3

Formula used:

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

Calculation:

x + y + z = 2 + 1 + (-3) = 3 – 3 = 0

If x + y + z = 0,

Then x3 + y3 + z3 – 3xyz = 0

∴ x3 + y3 + z3 – 3xyz = 0.

Given:

Jatin buys an article for 10% less than its value

sells it for 10% more than its value

Formula used:

Profit% = [(S.P – C.P)/C.P] × 100

Calculation:

Let original value be Rs. 10

He bought it at = Rs. 10 × (90/100) = Rs. 90

He sells it at = Rs. 10 × (110/100) = Rs. 110

Gain = Rs. 110 – Rs. 90 = Rs. 20

Gain% = (20/90) × 100%

⇒ 200/9% = \(22\frac{2}{9}\)%

∴ The gain percentage is \(22\frac{2}{9}\)%

Basic quadratic equation is ax² + bx + c  where sum of roots =-b/a and product of roots = c/a

⇒ Here two roots be α and β which are 2x and 3x.

⇒ α × β = c/a  

⇒ α × β = 54

⇒ 6x² = 54

⇒ x = 3 so α = 6 and β = 9

⇒ Now sum of roots = α + β = -b/a 

⇒ p = 15

Given:  

The ratio of the heights of the two right circular cones and their diameters is given.

Formula used:

The volume of cone = \(\frac{1}{3}\pi {r^2}h\)

Calculation:

Here we have two cones with two different radius and height.

Let the radius and the height of the first cone be r₁ and h₁. Similarly the radius and the height of the second cone be r₂ and h₂.

Volume of the first cone be \(\frac{1}{3}\pi {r_1}^2{h_1}\) and the volume of the second cone be \(\frac{1}{3}\pi {r_2}^2{h_2}\)

⇒ r₁ = 2 and r₂ = 2.5, h₁ = 25 and h₂ = 64

⇒ placing the above value in the equation \(\frac{{\frac{1}{3}\pi {r_1}^2{h_1}}}{{\frac{1}{3}\pi {r_2}^2{h_2}}}\)

⇒ \(\frac{{\pi {r_1}^2{h_1}}}{{\pi {r_2}^2{h_2}}}\) = \(\frac{{4 \times 25}}{{6.25 \times 64}}\) = \(\frac{1}{4}\)

∴ Required ratio = 1: 4

Given: 

A = {BELOW}, B = {WOOL}

Concept used:

If a set has “n” elements, then the number of subset of the given set is 2ⁿ

Calculation:

⇒ C = A – B

⇒ C = {B, E} taking out the common letters.

The subsets of C = 2² = 4

∴ The number of subsets of C is 4

Given:

Diagonals BD = 24 and AC = 10.

Concept used: 

Rhombus has all four sides equal and their diagonals bisect each other at 90° 

Let both the diagonals meet at O,

OD = OB = 12 and AO = OC = 5

In Δ OCD use the Pythagoras theorem.

In Δ OCD, OD² + OC² = DC² ,

12² + 5² = DC²

⇒ DC = 13

The perimeter of the Rhombus is 4a where a is the sides of the rhombus.

∴ Perimeter = 4 × 13 = 52 cm.