Data Analysis and Sufficiency Test - 1

Quantity I:

Sample space when three coins are tossed = {HHH, HHT, HTH, THH, TTT, THT, TTH, HTT}

Number of outcomes = 8

Required probability = $\frac{7}{8}$ (= 0.875)

Quantity II:

The probability that at least 1 bag is a school bag

= 1 – (Probability that no bag is school bag)

The probability that no bag is schoolbag = Probability that both bags are backpack

The probability that both bag are backpack $=\frac{{}^2{\mathrm{C}}_2}{{}^5{\mathrm{C}}_2}=\frac{1}{10}$

Probability that at least 1 bag is schoolbag $=1-\frac{1}{10}=\frac{9}{10}$

∴ The probability that at least one bag is a school bag is $\frac{9}{10}$ (= 0.9)

So, we can observe Quantity I < Quantity II.

Quantity I:

Let the price of a kurta and price of a pant be Rs. 3a and Rs. 2a respectively.

Total price of 15 kurtas and 10 pants = 390 × (15 + 10) = Rs. 9750

⇒ 15 × 3a + 10 × 2a = 9750

⇒ a = 150

The price of a kurta = 3 × 150 = Rs. 450

The price of a pant = 2 × 150 = Rs. 300

Required difference = 450 – 300 = Rs. 150

Quantity II: Rs. 150

So, we can observe Quantity I = Quantity II.

Quantity I:

Price of shoe = Rs.450 and price of a bag = Rs. 400

Ratio of price of a bag and price of a shoe = 400 : 450 = 8 : 9

Ratio of number of bags and number of shoes = 5 : 6

Compounded ratio = (5 : 6) and (8 : 9)

= (5 × 8) : (6 × 9)

= 20 : 27

The price of bags = $\frac{20}{47}$ × 2350 = Rs. 1000

Quantity II: Rs. 1090

So, we can observe Quantity I < Quantity II.

Quantity I:

Let salary of Ram be Rs. A.

Savings of Ram = A × $\frac{40}{100}$ = Rs. $\frac{2\mathrm{A}}{5}$

Amount spent on Rent = $\frac{3\mathrm{A}}{5}\times \frac{30}{100}$ = Rs. $\frac{9\mathrm{A}}{50}$

Amount spent on food = $\frac{21\mathrm{A}}{50}\times \frac{25}{100}=\frac{21\mathrm{A}}{200}$

Amount spent on shopping = $\frac{63\mathrm{A}}{200}\times \frac{80}{100}=\frac{63\mathrm{A}}{250}$

Amount remaining = 1 – $\frac{63A}{250}$ = 1122

⇒ A = 1500

Amount spent on Shopping = $\frac{63}{250}$ × 1500 = Rs. 378

Required percentage = $\frac{378}{1500}$ × 100 = 25.2%

Quantity II:

⇒ 12% of 300 + 65% of 700 = 34% of 250 + ?% of 4000 + 10

$\Rightarrow \frac{12}{100}\times 300+\frac{65}{100}\times 700=\frac{34}{100}\times 250+\frac{?}{100}\times 4000+10$

⇒ 36 + 455 = 85 + 40 × ? + 10

⇒ 491 = 95 + 40 × ?

⇒ ? = 9.9%

So, we can observe Quantity I > Quantity II.

Quantity I:

$\Rightarrow \frac{1}{\mathrm{A}}+\frac{1}{\mathrm{B}}=\frac{1}{12}$

$\Rightarrow \frac{1}{\mathrm{A}}+\frac{1}{\mathrm{C}}=\frac{1}{10}$

$\Rightarrow \frac{1}{\mathrm{A}}+\frac{1}{\mathrm{B}}+\frac{1}{\mathrm{C}}=\frac{3}{20}$

$\Rightarrow \frac{1}{\mathrm{C}}=\frac{3}{20}-\frac{1}{12}=\frac{1}{15}$

$\Rightarrow \frac{1}{\mathrm{A}}=\frac{1}{10}-\frac{1}{15}=\frac{1}{30}$

$\Rightarrow \frac{1}{\mathrm{B}}=\frac{1}{12}-\frac{1}{30}=\frac{1}{20}$

Ratio of their work efficiency,

$=\frac{1}{30}=:\frac{1}{20}:\frac{1}{15}$

= 2 : 3 : 4

Required difference $=\frac{4}{9}\times 8100-\frac{2}{9}\times 8100$

= Rs. 1800

Quantity II:

The marked price of an item = Rs. 2750

Discount percentage = 10%

The selling price of an item = 2750 × $\frac{80}{100}$ = Rs. 2200

Cost price of an item = 2200 × $\frac{100}{110}$ = Rs. 2000

Required difference = 2200 – 2000 = Rs. 200

So, we can observe Quantity I > Quantity II.

Quantity I:

Side of a square = $\frac{56}{4}$ = 14 cm

Radius of a circle = 7 cm

Area of a circle = $\frac{22}{7}$ × 7 × 7 = 154 sq.cm

Area of a rectangle = 154 sq.cm

Length of a rectangle = 7 + 4 = 11 cm

Breadth of a rectangle = $\frac{154}{11}$ = 14 cm

Perimeter of a rectangle = 2(11 + 14) = 50 cm

Quantity II: 50 cm

So, we can observe Quantity I = Quantity II

Quantity I:

Let price of a mango and an apple be Rs.A and Rs.B respectively.

⇒ 12A + 16B = $\frac{180}{7}$ × 28 = 720

⇒ 3A + 4B = 180      ----(i)

Also, we have:

⇒ 10A – 5B = 50

⇒ 2A – B = 10      ----(ii)

Multiply equation (ii) by 4,

Equation (ii) ⇒ 8A – 4B = 40      ----(iii)

Adding equations (i) and (iii), we get:

11A = 220

⇒ A = $\frac{220}{11}$ = 20      ----(iv)      

Substituting the value of A into equation (ii), we get:

(2 × 20) – B = 10

⇒ B = 40 – 10 = 30      ----(v)

So, the ratio between the price of an apple and that of a mango is:

$\frac{\mathrm{B}}{\mathrm{A}}=\frac{30}{20}=3:2$

Quantity II: $\frac{3}{8}$

So, we can observe Quantity I > Quantity II.

Quantity I:

Speed of the train = 54 kmph = 15 m/s

Let the length of the train be A m.

$\Rightarrow \frac{200+\mathrm{A}}{15}=30$

⇒ A = 250 m

Let the speed of running man be M m/s.

Relative speed = 15 + A

$\Rightarrow \frac{250}{15+\mathrm{A}}=12.5$

⇒ A = 5 m/s

Quantity II: 5 m/s

So, we can observe Quantity I = Quantity II.

Quantity I:

Let the salary of A and B be Rs.A and Rs.B respectively.

⇒ A – B = 5000

$\Rightarrow \frac{15\mathrm{A}}{100}+\frac{30\mathrm{B}}{100}=5250$

⇒ A + 2B = 35000

Solving,

A = Rs. 15000 and B = Rs. 10000

Average expenditure of A and B = $\frac{\left(15000+10000\right)-5250}{2}$

= Rs. 9875

Quantity II: Rs. 9860

So, we can observe Quantity I > Quantity II.

Quantity I:

Pipe P's 1 hour's work = $\frac{1}{10}$

Pipe Q's 1 hour's work = $\frac{1}{15}$

Pipe (P + Q)’s 2 hours work = $\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$

To fill the tank completely it will take 12 hours.

Quantity II: 15 hours

So, we can observe Quantity I < Quantity II.