Data Analysis and Sufficiency Test - 4

Quantity I:

Total aces and Jacks = 4 + 4 = 8

Remaining number of cards in a pack = 44

Total number of kings = 4

Required probability =$\frac{4}{44}$

$=\frac{1}{11}$

Quantity II:

Number of pink balloons = 25

Number of Red Balloons = 30

Total number of balloons in a box = 25 + 30 = 55

Number of ways that 2 balloons are drawn = ⁵⁵C₂ = 1485

Number of ways that both balloons are of same color = ²⁵C₂ + ³⁰C₂ = 300 + 435 = 735

Required probability =$\frac{735}{1485}=\frac{49}{99}$

So, we can observe Quantity I < Quantity II.

Formula:

Average =$\frac{\mathrm{Total}\mathrm{women}}{\mathrm{Number}\mathrm{of}\mathrm{cities}}$

Calculations:

Women >> BMW >> Jaipur

Percent of Women who have BMW car in Jaipur = 30%

Number of Women who have BMW in Jaipur = 200 ×$\frac{30}{100}$ = 60

Women >> BMW >> Udaipur

Percent of Women who have BMW car in Udaipur = 40%

Number of Women who have BMW in Udaipur = 150 ×$\frac{40}{100}$ = 60

Total Number of Women who have BMW in Udaipur and Jaipur = 60 + 60 =120

Required Average of Women =$\frac{\mathrm{Required}\mathrm{Women}}{\mathrm{Numbers}\mathrm{of}\mathrm{Cities}}$

⇒$\frac{120}{2}$ = 60

∴ The average of women who have BMW cars in the respective cities is 60.

Men >> Audi >> Jaipur

Percent of men who have Audi car in Jaipur = 60%

Number of men who have Audi in Jaipur = 125 ×$\frac{60}{100}$ = 75

Men >> Audi >> Udaipur

Percent of men who have Audi car in Udaipur = 20%

Number of men who have Audi in Udaipur = 125 ×$\frac{20}{100}$ = 25

⇒ Required Difference = 75 - 25 = 50

∴ The difference between the men who have the Audi cars in cities Jaipur and Udaipur is 50.

Formula:

Required Degree =$\frac{\mathrm{Men}\mathrm{who}\mathrm{have}\mathrm{BMW}\mathrm{in}\mathrm{Delhi}\mathrm{and}\mathrm{Mumbai}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cars}}$ × 360°

Calculations:

Number of men who have BMW in Delhi = 175 ×$\frac{80}{100}$ = 140

Number of men who have BMW in Mumbai = 275 ×$\frac{40}{100}$ = 110

Total number of men who have BMW car in Delhi and Mumbai cities = 140 + 110 = 250

Total number of cars in all the cities = 200 + 125 + 175 + 250 + 275 + 300 + 150 + 125 = 1600

Degree value of men who have BMW cars in respective cities =$\frac{\mathrm{Required}\mathrm{Men}\mathrm{who}\mathrm{have}\mathrm{BMW}\mathrm{in}\mathrm{given}\mathrm{cities}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cars}}\times 360°$

⇒$\frac{250}{1600}$ × 360° = 56.25°

∴ The degree value of Required men who have BMW cars in given cities is 56.25°.

Percent of Women who have Audi cars in Jaipur = 40%

Number of Women who have Audi cars in Jaipur = 125 ×$\frac{40}{100}$ = 50

Percent of Women who have Audi cars in Delhi = 20%

Number of Women who have Audi cars in Delhi = 250 ×$\frac{20}{100}$ = 50

Percent of Women who have Audi cars in Mumbai = 60%

Number of Women who have Audi cars in Mumbai = 300 ×$\frac{60}{100}$ = 180

Percent of Women who have Audi cars in Udaipur = 80%

Number of Women who have Audi cars in Udaipur = 125 ×$\frac{80}{100}$ = 100

∴ Mumbai’s women have Highest number of cars in four cities.

Number of men who have BMW in Jaipur = 200 ×$\frac{70}{100}$ = 140

Number of Women who have Audi in Jaipur = 125 ×$\frac{40}{100}$ = 50

So, ratio of men who have BMW car in Jaipur with the women who have Audi car in Jaipur

= 140 ∶ 50

= 14 ∶ 5

Given:

Distance travelled = 1000 km

Time taken to travel upstream = 40 hours + Time taken to travel downstream

Speed of wind = 50% more

Calculations:

Let the speed of the stream = X km per hour

The speed of the ship in still water = (100 + 200)% of X

= 300% of X

= 3 × X km per hour

According to the question,

$\Rightarrow \frac{1000}{3\mathrm{X}-\mathrm{X}}-\frac{1000}{3\mathrm{X}+\mathrm{X}}=40$

$\Rightarrow \frac{1000}{2\mathrm{X}}-\frac{1000}{4\mathrm{X}}=40$

⇒ 1000 × 2 = 8X × 40

⇒ X = 6.25 km/hr

Quantity I:

The required time =$\frac{1000}{3\mathrm{X}-\mathrm{X}}$

$=\frac{1000}{2\mathrm{X}}$

$=\frac{1000}{2\times 6.25}$

∴ Required time = 80 hours

⇒ New speed of stream = 1.5x

Quantity II:

W hen the speed of stream was increased by 50% then the new speed of the stream = 150% of x

New speed of stream = 1.5 × 6.25

New speed of stream = 9.375 km/hr

The speed of the ship in downstream = 3x + x = 4x, where x = New speed of stream

⇒ 9.375 × 4 = 37.5 km/hr

The required time =$\frac{2000}{37.5}$ ≈ 53 hours

∴ Quantity I > Quantity II

Given:

Let P(eating a rasgulla) be x.

Calculation:

P(eating a rasgulla) =$\frac{\mathrm{Number}\mathrm{of}\mathrm{rasgulla}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{sweets}}$

To find the probability of eating a rasgulla

we need the number of rasgulla and total number of sweets.

As none of the statement mentioned the total number of sweets.

So, we cannot find the probability of eating a rasgulla.

∴ All the statements together are not sufficient to answer the question.

Given:

Let Rate of Interest = R%

From Statement I:

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

Compound Interest (C.I.) = Amount – Principal

Given, Amount earned at principal Rs. 1000 for 3 years in C.I. is equal to Simple interest earned on principal Rs. 6655 for two years.

⇒ Amount At C.I. for 3 yrs. = S.I. for 2 yrs.

$\Rightarrow 1000\times {\left(1+\frac{\mathrm{R}}{100}\right)}^3=\frac{6655\times \mathrm{R}\times 2}{100}$

Only R is variable in above equation so we can easily find out the value of R.

⇒ R =10%

So, Statement I is sufficient to answer.

From Statement II:

Given, If total Interest earned in 37 years on on Rs.3000 at simple interest is Rs. 1110.

S.I. =$\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$

⇒ 1110 =$\frac{3000\times \mathrm{R}\times 37}{100}$

⇒ R =10 %

So, Statement II is sufficient to answer.

From Statement III:

Since only Difference between compound interest in 4th year and 3rd year is given.

There is no mention of Principal(P).

That leaves us with to variables R and P.

So, it is not sufficient to answer.

So, Statement I and II are sufficient to answer the question.

Given:

Let the two digit no be AB = 10a+b

where,

a is tens digit and b is ones digit.

The difference between digits = 2

it means either a – b = 2 or b – a = 2

From Statement I:

As Difference between the cubes of two number is given.

That gives us two cases,

⇒ a³ – b³ = 366 or b³ – a³ = 366

Neither cases helps us to reach a conclusion.

From Statement II:

We get that,

⇒ b > a

But not sufficient to answer.

From Statement III:

As difference between the squares of two number is 32.

That gives us two cases,

⇒ a² – b² = 32 or b² – a² = 32

Also from given, we have two cases,

⇒ Either a – b = 2 or b – a = 2

Neither cases helps us to reach a conclusion.

But joining From Statement II and III:

Since b > a

⇒ b – a = 2 and b² – a² = 32

From above two equation we get values of a and b as 7 and 9.

∴ Statement II and III are sufficient to answer the question.