Search Previous Year Questions

Given:

Profit on article X = 15%

Selling price of article X = ₹34,500

Loss on article Y = 10%

No overall gain or loss (Net profit = ₹0)

Formula used:

Cost price (CP) = $\frac{\text{Selling Price (SP)}}{1+\frac{\text{Profit\%}}{100}}$

Cost price (CP) = $\frac{\text{Selling Price (SP)}}{1-\frac{\text{Loss\%}}{100}}$

Calculations:

For article X:

CP of article X = $\frac{34500}{1+\frac{15}{100}}$

⇒ CP of article X = $\frac{34500}{1.15}$

⇒ CP of article X = ₹30,000

For article Y:

Total cost price of X and Y = Total selling price of X and Y

⇒ CP of X + CP of Y = SP of X + SP of Y

⇒ ₹30,000 + CP of Y = ₹34,500 + SP of Y

⇒ CP of Y = SP of Y - ₹4,500

CP of article Y = $\frac{\text{SP of Y}}{1-\frac{10}{100}}$

⇒ SP of Y - ₹4,500 = $\frac{\text{SP of Y}}{0.9}$

⇒ 0.9(SP of Y - ₹4,500) = SP of Y

⇒ 0.9SP of Y - ₹4,050 = SP of Y

⇒ SP of Y - 0.9SP of Y = ₹4,050

⇒ 0.1SP of Y = ₹4,050

⇒ SP of Y = ₹4,050 × 10

⇒ SP of Y = ₹40,500

∴ The correct answer is option (2).

Given:

Quantity = 100 quintals

Total = 10 metric tonnes

Conversion:

1 quintal = 100 kg

1 metric tonne = 1000 kg

So,

100 quintals = 100 × 100 = 10,000 kg

10 metric tonnes = 10 × 1000 = 10,000 kg

Calculations:

⇒ (10000 / 10000) × 100 = 100%

∴ The correct answer is: 100%

Given:

Radius of the circle (r) = 2 cm

The circle is inscribed in an equilateral triangle.

Formula used:

Area of an equilateral triangle = $\dfrac{\sqrt{3}}{4}a^2$

Where, a = side of the equilateral triangle

Relation between radius (r) and side (a) of an equilateral triangle:

$\text{r} = \dfrac{\sqrt{3}}{6}a$

Calculation:

$\text{r} = \dfrac{\sqrt{3}}{6}a$

⇒ 2 = $\dfrac{\sqrt{3}}{6}a$

⇒ $\text{a} = \dfrac{2 \times 6}{\sqrt{3}} = \dfrac{12}{\sqrt{3}} = 4\sqrt{3}$

Area of the triangle:

$\text{Area} = \dfrac{\sqrt{3}}{4}a^2$

⇒ $\text{Area} = \dfrac{\sqrt{3}}{4}(4\sqrt{3})^2$

⇒ $\text{Area} = \dfrac{\sqrt{3}}{4} \times 16 \times 3$

⇒ $\text{Area} = 12\sqrt{3}$

∴ The correct answer is option (1).

Given:

The sides of a triangle are k, 1.5k, and 2.25k.

Formula used:

The sum of the squares of medians of a triangle is given by:

$\text{Sum of squares of medians} = \frac{3}{4} (a^2 + b^2 + c^2)$

Where, a, b, and c are the sides of the triangle.

Calculation:

Let a = k, b = 1.5k = 3k/2, c = 2.25k = 9k/4.

$a^2 + b^2 + c^2 = k^2 + \frac{9}{4}k^2 + \frac{81}{16}k^2$

$a^2 + b^2 + c^2 = \frac{16}{16}k^2 + \frac{36}{16}k^2 + \frac{81}{16}k^2 = \frac{133}{16}k^2$

Sum of squares of medians = $\frac{3}{4} \times \frac{133}{16}k^2$

Sum of squares of medians = $\frac{3 \times 133}{4 \times 16}k^2 = \frac{399}{64}k^2$

Therefore, the sum of the squares of the medians is: $\frac{399}{64}k^2$

Given:

2s = a + b + c ⇒ s = (a + b + c)/2

s(s − a)(s − b)(s − c) × $\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]$

Let us take:

a = b = c = 2 ⇒ a + b + c = 6 ⇒ 2s = 6 ⇒ s = 3

Then: s − a = s − b = s − c = 1

Now evaluate:

s(s − a)(s − b)(s − c) × $\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]$

= 3 × 1 × 1 × 1 × [1/1 + 1/1 + 1/1 − 1/3]

= 3 × (3 − 1/3) = 3 × (8/3) = 8

Check which option gives 8:

abc = 2 × 2 × 2 = 8 (Match)

2abc = 16 (Not matching)

4abc = 32 (Not matching)

ab + bc + ca = 12 (Not matching)

Therefore, the correct answer is option 1.

Given:

Principal (P) = ₹10,000

Rate of Interest (r) = 4% per annum

Time (t) = 1 year

Compounding frequency = Quarterly (every 3 months), so n = 4

Formula used:

A = P$(1+\frac{r}{n×100})^{n×t}$

Calculation:

A = 10,000$(1+\frac{4}{4×100})^{4×1}$

⇒ A = 10,000$(1+\frac{1}{100})^4$

⇒ A = 10,000$(1+0.01)^4$

⇒ A = 10,000$(1.01)^4$

⇒ A = 10,000 × 1.04060401

⇒ A ≈ ₹10,406

∴ The correct answer is option (1).

Given:

p = 0.099999...

Calculation:

Let p = $0.0{\overline9}$ = x. Multiply both sides by 10:

10x = 0.9999

100x = 9.99999...

100x − 10x = 9.99999... − 0.99999...

90x = 9

x = 0.1

So, x = 1. Therefore, p = 1.

p = 0.1 into the expression 70p² + 43p − 5:

70p² + 43p − 5 = 70(0.1)² + 43(0.1) − 5

= 0.70 + 4.3 − 5

= 0

∴ The value of 70p² + 43p − 5 is 0.

Given:

Find the remainder when 2¹⁰¹ is divided by 101

Formula used:

By Fermat's Little Theorem:

If p is a prime and a is not divisible by p, then:

a^p−1 ≡ 1 (mod p)

So, 2¹⁰⁰ ≡ 1 (mod 101)

Calculations:

2¹⁰¹ = 2 × 2¹⁰⁰

⇒ 2¹⁰¹ ≡ 2 × 1 ≡ 2 (mod 101)

∴ The remainder is 2.

Given:

The quadratic equation is: x² + p·x + q = 0

Roots of the equation are: p ≠ 0 and q ≠ 0 (same as the coefficients)

Formula used:

Let α and β be the roots of the equation:

α + β = −p

α · β = q

But it's given that roots are p and q themselve

p + q = −p  ...(i)

p × q = q  ...(ii)

Calculations:

From (ii): p × q = q

Divide both sides by q (since q ≠ 0):

⇒ p = 1

Substitute p = 1 into (i):

⇒ 1 + q = −1 ⇒ q = −2

Now find: p² + q²

= 1² + (−2)² = 1 + 4 = 5

∴ The correct answer is option 4.

Given:

The equations: x² + px + q = 0 and x² + qx + p = 0 (p ≠ q) have a common root.

Formula used:

If the equations have a common root, let the common root be α. Substituting α in both equations, we have:

α² + pα + q = 0

α² + qα + p = 0

By subtracting these equations, we eliminate α² and solve for p and q.

Calculation:

From the two equations:

α² + pα + q = 0

α² + qα + p = 0

Subtract the second equation from the first:

⇒ (pα + q) - (qα + p) = 0

⇒ pα - qα + q - p = 0

⇒ α(p - q) = p - q

If p ≠ q, we can cancel (p - q):

⇒ α = 1

Substitute α = 1 in the first equation:

1² + p(1) + q = 0

⇒ 1 + p + q = 0

⇒ p + q = -1

∴ The correct answer is option (1).