Mix Test - 2

Given:

The equations x - y = 6 and \(\frac{x}{3}\) + \(\frac{y}{2}\) = 12

Calculations:

The equations x - y = 6 and \(\frac{x}{3}\) + \(\frac{y}{2}\) = 12

The equations x – y = 6

Then x = y + 6

Put the value of x in the equation \(\frac{x}{3}\) + \(\frac{y}{2}\) = 12

Then \(\frac{(y+6)}{3}+\frac{y}{2}=12\)

2y + 12 + 3y = 72

⇒ 5y = 60

⇒ y = 12

⇒ x = 12 + 6 = 18

\(\therefore\) The value of x and y is 18 and 12

Hence, the correct option is (B).

Let, 'n' be the number of cylinders that can be formed;

\(\therefore\) Volume of spherical ball \(=\mathrm{n} \times\) Volume of one cylinder

\(\Rightarrow \frac{4}{3} \pi r^{3}=\mathrm{n} \times \pi r^{2} h\)

\(\Rightarrow \frac{4}{3} \times \frac{22}{7} \times 9 \times 9 \times 9=\mathrm{n} \times \frac{22}{7} \times 0.6 \times 0.6 \times 1\)

\(\Rightarrow \mathrm{n}=2700\)

Hence, the correct option is (D).

Let the price of $1$ unit of the commodity $=Rs.x$

If the Price is decreased by $40\%$

Price of $1$ unit of the commodity becomes $=Rs.(1-0.40)x=Rs.0.60x$

If consumption is increased by $30\%$:

$1.30$ numbers of units consumed in place of $1$ unit

Price of $1.30$ unit of the commodity becomes $=Rs.(1.30\times 0.60x)=Rs.0.78x$

$\therefore$ Percentage decrease in expenditure $=\left[\frac{(x-0.78x)}{x}\right]\times 100 =0.22\times 100\%=22\%$

Hence, the correct option is (A).

Given:

\(\sqrt{10+\sqrt{25+\sqrt{121}}}\)

\(=\sqrt{10+\sqrt{25+11}}\)

\(=\sqrt{10+6}=\sqrt{16}=4\)

Hence, the correct option is (D).

Formula:

Amount $=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

$\Rightarrow 51840=25000\times {\left(1+\frac{20}{100}\right)}^T$

$\Rightarrow \frac{51840}{25000}={\left(1+\frac{1}{5}\right)}^{\mathrm{T}}$

$\Rightarrow \frac{1296}{625}={\left(\frac{6}{5}\right)}^{\mathrm{T}}$

$\Rightarrow {\left(\frac{6}{5}\right)}^4={\left(\frac{6}{5}\right)}^{\mathrm{T}}$

∴ T = 4 years

Hence, the correct option is (C).

Ratio of three iron balls \(=x: 2 x: 3 x\)

We know that,

Volume of sphere \(=\frac{4}{3} \times \pi r^{3}\)

According to the question,

\(\frac{4}{3} \times \pi \times\left[x^{3}+(2 x)^{3}+(3 x)^{3}\right]=\frac{4}{3} \times \pi \times 6^{3}\)

\(\Rightarrow x^{3}+8 x^{3}+27 x^{3}=216\)

\(\Rightarrow 36 x^{3}=216\)

\(\Rightarrow x^{3}=6\)

\(\Rightarrow x=\sqrt[3] 6\)

\(\therefore\) Diameter of the smallest ball is \(=2 \sqrt[3]{6}\)

Hence, the correct option is (C).

Let the number be $x$.

If the number is increased by $10\%$, then

The new number will be $=x+10\%$ of $x=x+0.1x=1.1x$

Again, the number is decreased by $20\%$, then

The new number will be $=1.1x-20\%$ of $1.1x=0.88x$

Percentage change $=(x-0.88x)\times \frac{100}{x}=12\%$ decrease

Hence, the correct option is (B).

Let the length and breadth of the rectangle be 5x and 4x respectively.

The perimeter of the rectangle = 54 cm

Perimeter of the rectangle = 2 × (l + b)

⇒ 2 × (5x + 4x) = 54

⇒ 9x = 27

⇒ x = 3 cm

Now, Length of the rectangle = 5 × 3 = 15 cm

Breadth of the rectangle = 4 × 3 = 12 cm

Area of the rectangle = 15 × 12

Area of the rectangle = 180 cm²

Hence, the correct option is (A).

Let the base of the parallelogram be \(x\).

So, the base of the triangle \(=(\frac{5 }{7})x\)

Let the height of the parallelogram be \(\mathrm{P}\).

Let the height of the triangle be \(\mathrm{T}\).

Hence, we obtain the following:

The area of the triangle \(=\frac{1}{2} \times [(\frac{5}{7}) x] \times 1\)

The area of the parallelogram \(=x \times P\)

Now, both areas are equal so, we get:

\(\frac{1}{2} \times[(\frac{5}{7})x] \times \mathrm{T}=x \times \mathrm{P}\)

\(\Rightarrow \frac{\mathrm{T}}{\mathrm{P}}=\frac{14}{5}\)

∴ The required ratio of the respective heights of the triangle and the parallelogram is 14 ∶ 5.

Hence, the correct option is (C).

Given,

\(\sqrt{0.01} \times \sqrt[3]{0.008}-0.02\)

\(=\sqrt{(0.1)^{2}} \times \sqrt[3]{(0.2)^{3}}-0.02\)

\(=0.1 \times 0.2-0.02\)

\(=0.02-0.02\)

\(=0\)

Hence, the correct option is (A).