Quantitative Aptitude Test - 9

The logic of the series can be explained as,

(15)² + 2 = 227

(16)² + 3 = 259

(17)² + 5 = 294

(18)² + 7 = 331

(19)² + 11 = 372

(20)² + 13 = 413

∴ The value of ? is 413.

Cost price = 480 Rs

Profit = 480 × 10% = 48

S. P. = 480 + 48 = 528 Rs

Discount % = 4%

Discount is always given on the marked price,

So

x × 96% = 528

x = 550 Rs

Given:

Total journey distance \(=480 \) km

Ideal average speed \(= 80\) kmph

Planned or time taken in ideal condition \(=\frac{480}{80}=6\) hour \(=360\) minutes

For actual time,

Distance in which average speed of \(80\) kmph is maintained,

Distance covered before break \(=62.5 \%\) of \(480=(\frac{5}{ 8}) \times 480=300\) km

Time taken to cover this distance \(=\frac{300 }{80}=3.75\) hour \(=225\) minutes

After the break,

Distance need to be covered \(=480-300=180 \) km

Time required to cover the distance \(=\frac{180}{ 60}=3\) hour \(=180\) minutes

Actual time of journey \(=225+180+30=435\) minutes

\(\therefore\) Difference in time \(=435-360=75 \) min

For this to be divisible by 11, the sum of numbers at even places - the sum of numbers at odd places must be zero or divisible by 11

⇒ (4 + 1) - (3 + 0) = 5 - 3 = 2

instead of 2, there must be 0 so that it will be divisible by 11.

∴ 2 must be subtracted from the number to make it divisible by 11.

Let, the efficiency of the \(3^{\text {rd }}\) son \(=x\)

Total work \(=72\) units (LCM of \(24,36\) and \(3\) )

Efficiency of Sumit \(=\frac{72}{ 3}=24\)

Efficiency of the \(1^{\text {st }}\) son \(=\frac{72}{ 24}=3\)

Efficiency of the \(2^{\text {nd }}\) son \(=\frac{72}{ 36}=2\)

Since, Sumit can do twice the work as much as done by all the \(3\) sons together

\(\Rightarrow 24=2 \times(3+2+x)\)

\(\Rightarrow x=7\)

Ratio in which the amount will be distributed among the \(3\) sons \(=3: 2: 7\)

\(\therefore\) Share of the \(3^{\text {rd }}\) son \(=600 \times \frac{7}{ 12}=\) Rs. \(350 \)

Given:

Product of the age of Tania and Ronita = 12 years

Let the age of Tania be x years

Age of Ronita = (x + 4) years

According to the question,

⇒ (x) × (x + 4) = 12

⇒ x² + 4x = 12

⇒ x² + 4x – 12 = 0

⇒ x² + (6 – 2) x – 12 = 0

⇒ x² + 6x – 2x – 12 = 0

⇒ x(x + 6) – 2(x + 6) = 0

⇒ (x + 6) (x − 2) = 0

⇒ x = −6, 2

Present age of Ronita = (2 + 4) = 6 years

Age of Ronita after 5 years = (6 + 5) = 11 years

∴ The age of Ronita after 5 years is 11 years.

Total number of balls \(=4+5+6=15\)

Let \(\mathrm{S}\) be the sample space.

\(\mathrm{n}(\mathrm{S})=\) Total number of ways of drawing 3 balls out of \(15={ }^{15} \mathrm{C}_{3}\)

Let\(\mathrm{E}=\) Event of drawing 3 balls, all of them are yellow.

\(\mathrm{n}(\mathrm{E})=\) Number of ways of drawing 3 balls, all of them are yellow

= Number of ways of drawing 3 balls from the total \(5={ }^{5} \mathrm{C}_{3}\)

( \(\because\) there are 5 yellow balls in the total balls)

\(\mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{5} \mathrm{C}_{3}}{{ }^{15} \mathrm{C}_{3}}\)\(=\frac{{ }^{5} \mathrm{C}_{3}}{{ }^{15} \mathrm{C}_{3}}\)

\(=\frac{\left(\frac{5 \times 4}{2 \times 1}\right)}{\left(\frac{15 \times 14 \times 13}{3 \times 2 \times 1}\right)}=\frac{2}{91}\)

From the table given aboveNumber of boys goes on trip = 360

∴Number of boys goes on trip = 360

From the table given above

Number of boys in school is 500 and number of girls in school is 400.

∴ Total number of student in school = 500 + 400 = 900

From the table given above

The sum of number of boys goes on trip and number of girls not goes on trip = 360 + 100 = 460

The sum of number of girls goes on trip and number of boys not goes on trip = 300 + 140 = 440

∴ Required ratio = 460 ∶ 440 = 23 ∶ 22

So, the required ratio is 23 ∶ 22.