Mix Test - 9

As given, from a pack of 52 cards, 1 card is drawn at random.

Total number of cases,

\(P(B)=52\) 

Total face cards,

\(P(A)=16\) [favourable cases] 

So, the probability of a face card drawn \(= \frac{P(A)}{P(B)}\)

\(=\frac{16}{52}\)

\(=\frac{4}{13}\)

Hence, the correct option is (A).

Given,

Distance traveled by the wheel in 120 revolutions = 1584 m

Distance traveled in 1 revolution = perimeter of the wheel

The perimeter of the wheel = 2πr or πd

where r = radius of the wheel and d = diameter of the wheel

Distance traveled in one revolution = πd

Given that, distance traveled in 120 revolutions = 1584 m

⇒ 120 × πd = 1584

⇒ 120 × (\(\frac{22}{7}\)) × d = 1584

⇒ (\(\frac{22}{7}\)) × d = 13.2

⇒ d = 0.6 × 7

⇒ d = 4.2

∴ The diameter of thee wheel is 4.2 m.

Hence, the correct option is (A).

\(21 \%\) of Rs. \(250\)

\(=21 \%\times250 \)

\(=\frac{21}{100}\times250 \)

\(=\frac{21}{2}\times5 \)

\(=\frac{105}{2} \)

\(=\) Rs. \(52.50\)

Hence, the correct option is (A).

Given:

Time = 2 years, Simple Interest = 500, rate = 10%

Formula:

Simple Interest = $\frac{\mathrm{Principal}\times \mathrm{Rate}\times \mathrm{Time}}{100}$

Compound Interest = $\mathrm{Principal}\left[{\left(1 + \frac{\mathrm{Rate}}{100}\right)}^{\mathrm{t}}-1\right]$

Calculation:

Let the principal be ‘P’.

Simple Interest = $\frac{\mathrm{Principal}\times \mathrm{Rate}\times \mathrm{Time}}{100}$

⇒ 500 = $\frac{\mathrm{Principal}\times 10\times 2}{100}$

Principal = 2500

Compound Interest = $\mathrm{Principal}\left[{\left(1+\frac{\mathrm{Rate}}{100}\right)}^{\mathrm{t}}–1\right]$

$=2500\left[{\left(1+\frac{10}{100}\right)}^2–1\right]$

= 525

∴ The compound Interest is Rs. 525.

Hence, the correct option is (A).

Given:

\(\mathrm{SP}=\) Rs. \(x \)

\(\text {Loss }=28 \% \)

\(\text {SP }=\) Rs. \(y \)

\(\text {Profit }=12 \%\)

Let the cost price be Rs. \(100\)

If television is sold at Rs. \(x\), then

\(\text {Loss } \%=28 \% \)

As we know,

\(\text {Loss } \%=\left(\frac{\text { Loss }}{\mathrm{CP}}\right) \times 100 \)

\(28=\left(\frac{\text { Loss }}{{100}}\right) \times 100 \)

\(\Rightarrow \text { Loss }=28 \)

\(\text {SP }=\text { CP }-\text { Loss } \)

\(\text {SP }=100-28=\) Rs. \(72 \)

\(\Rightarrow x=\) Rs. \(72\)

Now,

If television is sold at Rs. \(y\), then Profit \(\%=12 \%\)

\(12=\left(\frac{\text{Profit}}{100}\right) \times 100\)

Profit \(=\) Rs. \(12\)

\(\mathrm{SP}=\mathrm{CP}+\) Profit

\(\mathrm{SP}=100+12=\) Rs. \(112\)

\(\Rightarrow {y}=\) Rs. \(112\)

We have to find \(\frac{y}{x}\)

\(\frac{y}{x}=\frac{112}{72}=\frac{14}{9}\)

\(\therefore\) The ratio of \({y}\) to \({x}\) is \(14: 9\).

Hence, the correct option is (D).

From the table we can say that,

Total stock available of Analog watch = 6000 + 4500 + 5000 + 3000 = 18500

Total stock available of Digital watch = 7000 + 5000 + 4000 + 2500 = 18500

Total stock available of Automatic watch = 5500 + 3500 + 6500 + 3500 = 19000

∴ Required average = $\frac{18500 +18500+19000}{3}=18666.67$

Hence, the correct option is (C).

From the table we can say that,

Total stock available of Analog watch = 6000 + 4500 + 5000 + 3000 = 18500

Stock of Analog Titan watches = 6000

∴ Required percentage = $\frac{6000}{18500}$ × 100 = 32.43%

Hence, the correct option is (C).

From the table we can say that,

Total stock of Titan watches = 6000 + 7000 + 5500 = 18500

Total stock of Timex watches = 3000 + 2500 + 3500 = 9000

∴ Required ratio = 9000 : 18500

= 18 : 37

Hence, the correct option is (C).

From the table we can say that,

Total stock of Sonata watches = 4500 + 5000 + 3500 = 13000

Total stock of Fossil watches = 5000 + 4000 + 6500 = 15500

∴ Required percentage = $\frac{15500–13000}{15500}\times 100$

⇒ 16.12% ≈ 16%

Hence, the correct option is (D).

From the table we can say that,

Total stock available of Digital watch = 7000 + 5000 + 4000 + 2500 = 18500

Total stock available of Automatic watch = 5500 + 3500 + 6500 + 3500 = 19000

∴ Required difference = 19000 – 18500 = 500

Hence, the correct option is (B).

The longest rod that can be placed in a room which is 9 meter long, 8 meter broad and 12 meter high is its longest diagonal.

Longest diagonal \(=\sqrt{l^{2}+b^{2}+h^{2}}\)

Where, I= length, b= breadth, h= height

Longest diagonal \(=\sqrt{9^{2}+8^{2}+12^{2}}\)

\(=\sqrt{(81+64+144)}\)

\(=\sqrt{289}\)

= 17 meter

Hence, the correct option is (C). 

Fourth proportion of three numbers a, b, and c $=\frac{\mathrm{b}\times \mathrm{c}}{\mathrm{a}}$

Let the fourth proportion be x.

$\Rightarrow \frac{14}{6}=\frac{28}{\mathrm{x}}$

$\Rightarrow \mathrm{x}=28\times \frac{6}{14}$

⇒ x = 12

∴ The fourth proportion is 12.

Hence, the correct option is (A).

Given,

\(\frac{3}{7}+\left(-\frac{4}{9}\right)+\left(-\frac{11}{7}\right)+\frac{7}{9}\)

Firstly group the rational numbers with same denominators.

\(=\frac{3}{7}+\left(-\frac{11}{7}\right)+\left(-\frac{4}{9}\right)+\frac{7}{9}\)

Now the denominators which are same can be added directly.

\(=\frac{3+(-11)}{7}+\frac{(-4+7)}{9}\)

\(=-\frac{8}{7}+\frac{3}{9}=-\frac{8}{7}+\frac{1}{3}\)

By taking LCM for \(7\) and \(3\) we get, \(21\).

\(=\frac{(-8 \times 3)}{(7 \times 3)}+\frac{(1 \times 7)}{(3 \times 7)}\)

\(=-\frac{24}{21}+\frac{7}{21}\)

Since the denominators are same can be added directly,

\(=\frac{(-24+7)}{21}=-\frac{17}{21}\)

Hence, the correct option is (A).

Please remember that Maximum portability is 1.

So we can get the total probability of non-defective bulbs and subtract it from 1 to get the total probability of defective bulbs.

Total cases of non defective bulbs \(=({ }^{16} C_{2})\)

As we know,

\({ }^{n} C_{r} = \frac{n!}{r!(n-r)!}\)

\(=\frac{16 \times 15}{2 \times 1}\)

\(=120\)

Total cases \(={ }^{20} C_{2}\)

\(=\frac{20 \times 19}{2 \times 1}\)

\(=190\)

Total probability of non-defective bulbs \(=\frac{120}{190}\)

\(=\frac{12}{19}\)

The probability that at least one of these is defective \(=1-\frac{12}{19}\)

\(=\frac{7}{19}\)

Hence, the correct option is (A).

The factor of \(21600=2^{5} \times 3^{3} \times 5^{2}\)

A perfect cube is a number that is obtained by multiplying the same integer three times. Here, 2 comes 5 times and 5 comes 2 times.

To make it a perfect cube, it must be multiplied by \((2 \times 5)\), i.e., 10.

Hence, the correct option is (B).