Mix Test - 6

Given,

Principal \((P)=\) Rs. \(16000\)

Time \((n)=9\) months \(=3\) quarters

Rate \((r)=20 \%\) per annum \(=5 \%\) per quarter

As we know,

Amount \(=P\times\left(1+\frac{r}{100}\right)^{n}\)

\(\therefore\) Amount \(=\left[16000 \times\left(1+\frac{5}{100}\right)^{3}\right]\) 

\(=\left[16000 \times\left(\frac{105}{100}\right)^{3}\right]\) 

\(=16000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\)

\(=\) Rs. \(18522\)

Hence, the correct option is (B).

Given,

\(k\) is an odd number

Let \(k=1\)

\(\left(8^{2 k}+5^{2 k}\right)\)

\(=\left(8^{2 \times 1}+5^{2\times 1}\right)\)

\(=8^{2}+5^{2}\)

\(=64+25\)

\(=89\)
Hence, the correct option is (A).

Given:

The radius of spherical balls is 8 cm.

The size of the box 80 cm × 48 cm × 16 cm

The diameter of the spherical ball = 16 cm

Number of spherical balls that can be adjusted along with the length = $\frac{80}{16}$ = 5

Number of spherical balls that can be adjusted along with the breadth = $\frac{48}{16}$ = 3

Number of spherical balls that can be adjusted along with the height = $\frac{16}{16}$ = 1

There will be only one layer.

Total number of spherical balls = 5 × 3 × 1 = 15

∴ The required result will be 15.

Hence, the correct option is (A).

Let the value of coins be \(8 x: 4 x: 3 x\)

So,

Number of coins \(8 x \times 1: 4 x \times 2: 3 x \times 4\)

\(=8 x: 8 x: 12 x\)

\(\therefore\) Total coins:

\(\Rightarrow 8 x+8 x+12 x=28 x\)

\(28 x=280\) (according to the question)

\(x=\frac{280}{28}=10\)

\(\therefore\) No. of \(50P\) coins are \(=8 x=8 \times 10=80\)

Hence, the correct option is (A).

We know that:

The volume of the cuboid = Length \(\times\) breadth \(\times\) height

The volume of the cube \(=\) Side \(^{3}\)

The number of cubes = \(\frac{\text{Volume of the cuboid}}{\text{The volume of the cube}}\)

Total volume to rectangular box \(=15 \times 12.5 \times 7.5\)

Volume of 1 box \(=2.5 \times 2.5 \times 2.5\)

Number of cubes that can fit \(=\frac{(15 \times 12.5 \times 7.5)}{(2.5 \times 2.5 \times 2.5)}=90\)

\(\therefore\) The number of the cubes is 90

Hence, the correct option is (C).

Let the number be \(x\).

According to the question,

\(40 \%\) of \(x+60=50 \%\) of \(x\)

\(\Rightarrow \frac{40}{100}\times x+60=\frac{50}{100}\times x\)

\(\Rightarrow \frac{40x}{100}+60=\frac{50x}{100}\)

\(\Rightarrow 60=\frac{50}{100} x-\frac{40}{100} x\)

\(\Rightarrow 60=\frac{10 x}{100}\)

\(\Rightarrow x=\frac{60 \times 100}{10}\)

\(\Rightarrow x=600\)

So, the number is \(600\).

Hence, the correct option is (D).

Given:

The selling price of the article = 864

The loss percentage of article = 28

Formula:

Loss% = $\frac{(\mathrm{Cost} \mathrm{price}-\mathrm{Selling} \mathrm{price})}{\mathrm{Cost} \mathrm{price}}\times 100$

Let be assume cost price of article is C.

$\Rightarrow \mathrm{C}\times \frac{(100-28)}{100}=864$

$\Rightarrow \mathrm{C}=\frac{(864\times 100)}{72}=12\times 100=1200$

⇒ The selling price at 20% profit = $1200\times \left(1+\frac{20}{100}\right)$

$=1200\times \frac{6}{5}=1440$

∴ The required result will be 1440.

Hence, the correct option is (C).

Given:

The cost price of pen is 80% of selling price.

Formula:

Profit = Selling price – Cost Price

Profit percentage = $\left(\frac{\mathrm{Profit}}{\mathrm{Cost} \mathrm{Price}}\right)\times 100$

Let the Selling price of pen be 100 x.

So, the cost price of pen = 80% of 100x = 80x

Profit = 100x - 80x = 20x

Profit % = $\left(\frac{20\mathrm{x}}{80\mathrm{x}}\right)\times 100=\frac{100}{4}=25\%$

∴ The profit percent is 25%.

Hence, the correct option is (A).

Given:

A box contains 3 black balls, 6 red balls, 5 orange balls

Formula:

Probability = $\frac{\mathrm{Total} \mathrm{favourable} \mathrm{events}}{\mathrm{Total} \mathrm{events}}$

Total balls = 14

Favourable cases = 6

Probability of 1st ball being red = $\frac{6}{14}=\frac{3}{7}$

As 1st ball is being thrown, so total balls = 13

Favourable cases = 5

So, Probability = $\frac{5}{13}$

Probability of having both balls as red = $\frac{3}{7}\times \frac{5}{13}=\frac{15}{91}$

Hence, the correct option is (C).

Given: 

\(\frac{4}{7}+0+\left(-\frac{8}{9}\right)+\left(-\frac{13}{7}\right)+\frac{17}{21}\)

Firstly group the rational numbers with same denominators.

\(=\frac{4}{7}+\left(-\frac{13}{7}\right)+\frac{17}{21}+\left(-\frac{8}{9}\right)\)

Now the denominators which are same can be added directly.

\(=\frac{(4-13)}{7}+\frac{17}{21}-\frac{8}{9}\)

\(=-\frac{9}{7}+\frac{17}{21}-\frac{8}{9}\)

By taking LCM for 7, 9 and 21 we get, 63.

\(=\frac{(-9 \times 9)}{(7 \times 9)}+\frac{(17 \times 3)}{(21 \times 3)}-\frac{(8 \times 7)}{(9 \times 7)}\)

\(=-\frac{81}{63}+\frac{51}{63}-\frac{56}{63}\)

Since the denominators are same can be added directly, 

\(=\frac{-81+51-56}{63}=\frac{-86}{63}\)

Hence, the correct option is (A).

Given:

Area of circle = 154 cm²

Calculation:

⇒ π × r² = 154, r = 7 cm

Perimeter of semicircle = 2r + πr

$\Rightarrow \left(2\times 7\right)+\frac{22\times 7}{7}=36$ cm

∴ The required result will be 36 cm.

Hence, the correct option is (C).

Sum of first 222 whole numbers

0+1+2..............221

\(\frac{n(n+2)}{2}\)

\(=\frac{221(221+1)}{2}\)

\(=221 \times 111\)

Unit digit = 1
Hence, the correct option is (C).

Given:

The length of the cube edges is 11 cm.

The largest sphere will be such that the diameter of the sphere will be equal to the edge of the cube.

So, the radius r of the sphere = $\frac{3}{2}$ cm

Volume of the largest sphere = $\left(\frac{4}{3}\right)\times \left(\frac{22}{7}\right)\times {\left(\frac{3}{2}\right)}^3=14.14$ cm3

∴ The required result will be 14.14 cm³.

Hence, the correct option is (B).

Father has three children with atleast one Boy.

Atleast one out of 3 means the number of Boys should atleast be one and it can be up to 3. 

Let's represent boy with B and girl with G.

Then possible outcomes are BBB, BBG, BGG.

And, the favorable outcome i.e. 2 boys and 1 girl is BBG .

P(E) =  $\frac{Number of favorable outcomes}{Number of total Possible outcomes}$

⇒ P(E) = $\frac{1}{3}$

∴ The probability that he has 2 boys and 1 girl is $\frac{1}{3}$.

Hence, the correct option is (B).

Given,

\(=\sqrt[3]{\frac{0.2 \times 0.2 \times 0.2+0.04 \times 0.04 \times 0.04}{0.4 \times 0.4 \times 0.4+0.08 \times 0.08 \times 0.08}}\)

\(=\sqrt[3]{\frac{0.008+0.000064}{0.064+0.000512}}\)

\(=\sqrt[3]{\frac{0.008064}{0.064512}}\)

\(=\sqrt[3]{\frac{8064}{64512}}\)

\(=\sqrt[3]{\frac{1}{8}}\)

\(=\frac{1}{2}\)

\(=0.5\)

Hence, the correct option is (C).