Data Analysis and Sufficiency Test - 6

Given:

Income = Rs. 60000

Let amount received by each sister = p

Statement I:

Amount received by Sid’s brother = p + 500

Other expenditure = 16800

Savings =$\frac{60}{100}\times \frac{100}{40}$ × 16800 = 25200

Now, 60000 = 3p + p + 500 + 16800 + 25200

P = Rs. 4375

Percent of money is received by each sister =$\frac{4375}{60000}$ × 100 = 7.29%

Statement II:

Savings = Rs. 25200

Other expenditure =$\frac{40}{100}\times \frac{100}{60}$ × 25200 = Rs. 16800

60000 = 25200 + 16800 + Share of three sister + Share of brother

3p + Share of brother = 18000

Given data is not sufficient to determine the answer.

So, this statement is redundant to fine the answer.

Statement III:

Savings = Rs. 25200

Other expenditure =$\frac{40}{100}\times \frac{100}{60}$ × 25200 = Rs. 16800

60000 = 25200 + 16800 + Share of three sister + Share of brother

4p = 18000

P = Rs. 4500

Percent of money is received by each sister =$\frac{4500}{60000}$ × 100 = 7.5%

Ali = 3 + Bela

Bela = 5 + Chales

Ali = Bela + Chales

(Dyna + 2) : (Ena + 2) = 3 : 4

Dyna =$\frac{3\mathrm{Ena}-2}{4}$

Dyna + Ena + Fany + Geet = 50

7 Ena + 4 Fany + 4 Geet = 202

Statement I:

Geet = Ali + 1

Ali + Chals = 20

Ali + Ali – 8 = 20

Ali = 14 years

Geet = 14 + 1 = 15 years

7Ena + 4Fanny + 4Geet = 202

7Ena + 4Fanny + 4 × 15 = 202

7Ena + 4Fanny = 124

Age of Fanny is not given.

So, this statement is redundant to find the answer.

Statement II:

Dyna + Ena = 17

Dyna = 17 – Ena

Dyna =$\frac{3\mathrm{Ena}-2}{4\mathrm{a}}$

Ena = 10 years

Statement III:

Bela + Dyna = 32

Dyna = 32 – Bela

Dyna =$\frac{3\mathrm{Ena}-2}{4}$

$\frac{3\mathrm{Ena}-2}{4}$= 32 – Bela

Age of Bela is not given.

So, this statement is redundant to find the answer.

$\frac{\mathrm{Height}\mathrm{of}\mathrm{cone}}{\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{cylinder}}=\frac{5}{8}$

⇒ Statement I alone is not sufficient to answer the question

From statement II:

Area of the square = 144 cm²

⇒ Side of the square = 12 cm

Perimeter of the square = 12 × 4 = 48 cm

⇒ Height of the cylinder = 2 × 48 = 96 cm

Statement II alone is not sufficient to answer the question.

From statement III:

Diameter of the cone = 3 × Breadth of the rectangle

Perimeter of the rectangle = 2 × (l + b) = 84 cm

∵ Breadth cannot be determined.

⇒ Statement III alone is not sufficient to answer the question.

If we combine all the statements, we can determine the height of the cone but not the radius or the slant height.

∴ None statement is sufficient to answer.

Quantity A:

Ratio of Selling Price to Cost price is 5 : 4.

Let selling price = 5x

Cost Price = 4x

⇒ Profit = 5x – 4x = x

⇒ Profit percentage =$\left(\frac{\mathrm{x}}{4\mathrm{x}}\right)$ × 100 =25%

∴ The Profit percentage is 25%.

Quantity B: The Discount percentage is 21.25%.

∴ Quantity A > Quantity B

Quantity A:

Given:

Let the speed of boat in upstream = x

then speed of boat in downstream = 2x

speed of boat in still water = b

speed of stream = s

The difference between the speed of boat in still water and speed of stream = 8 km/hr

Calculations:

According to the question:

$\Rightarrow \frac{2\mathrm{x}}{\mathrm{x}}=\frac{\mathrm{b}+\mathrm{s}}{\mathrm{b}-\mathrm{s}}$

By applying Componendo and dividendo

$\Rightarrow \frac{2+1}{2-1}=\frac{2\mathrm{b}}{2\mathrm{s}}$

$\Rightarrow \frac{\mathrm{b}}{\mathrm{s}}=\frac{3}{1}$

So, b – s = 2 unit

As we know, b – s = 8 km/hr

⇒ 2 unit = 8 km/hr

⇒ 1 unit = 4 km/hr

∴ Speed of stream, s = 4 km/hr

Quantity B:

Speed of man =$\frac{10}{9}$ m/s

⇒$\frac{10}{9}$m/s =$\frac{10}{9}\times \frac{18}{5}$km/hr

⇒$\frac{10}{9}$ m/s = 4 km/hr

∴ Quantity A = Quantity B

Quantity A:

The average age of a dance pair = 21 years

⇒ Sum of ages of a dance pair = 42 years

⇒ Sum of ages of a dance pair after 5 years = 52 years ---(i)

⇒ Average of the pair along with their child after 5 years = 21 years

⇒ Sum of ages of pair along with their child after 5 years = (21 × 3) = 63 years ---(ii)

From equation (i) and (ii), we get

⇒ Age of child after 5 years = 63 – 52 = 11 years

⇒ Present Age of child = 11 – 5 = 6 years

The present age of child is 6 years

Quantity B:

The age of a boy is 5 years.

∴ Quantity A > Quantity B

Total number of balls in arena = 7 × 9 = 63

A table has only one ball of each colour.

Number of playing tables = 7

Quantity A:

As Cyan being one of the colours of balls.

Number of cyan balls in the room = 7 balls

Probability of event =$\frac{\mathrm{Favourable}\mathrm{balls}}{\mathrm{Total}\mathrm{balls}}$

⇒ P(picking one cyan balls) =$\frac{7}{63}$

⇒ P(picking one cyan balls) =$\frac{1}{9}$

Quantity B:

As the colour of the cue ball is white, but there is no cue ball in the arena.

⇒ P(picking one white balls) = 0

∴ Quantity A > Quantity B

Quantity A:

Given:

Let Cost Price of an article = 1000

Discount Percentage = z%

Loss percentage = 27.1%

Calculations:

⇒ Loss on an article = 27.1% of 1000 = 271

⇒ Selling Price of an article = 1000 – 271 = 729

According to the question:

⇒ z = 10

∴ The value of z is 10%.

Quantity B: 9%

∴ Quantity A > Quantity B

From the pie chart:

Amount of income of D = 80000 × 0.24 = Rs. 19200

From the bar graph:

Expenditure done on Clothes and Entertainment together = 9000 + 4000 = Rs. 13000

∴ Required difference = 19200 – 13000 = Rs. 6200

From the pie chart:

Total income of A and E together = 80000 × (0.16 + 0.12) = 80000 × 0.28 = Rs. 22400

From the bar graph:

Total expenditure done on Rent and Clothes together= 15000 + 9000 = Rs. 24000

∴ Required ratio = 22400 : 24000 = 14 : 15