Data Analysis and Sufficiency Test - 2

Percentage of male employees who are using Maruti car in company Q than the male employee who are using Maruti car in company P = \((\frac{4800}{12,000})\) × 100 = 40%

Total female who are not using any car = 8,000 + 600 + 7,500 + 3,600 = 19,700

Average of female who are not using any car = $\frac{19,700}{4}$ = 4,925

∴ Average of female who are not using any car is 4,925.

The males who are using Hyundai car in company P more than female who are using Mahindra car in company P = 6,000 – 5,000 = 1,000

More percent = $\frac{1,000}{5,000}$ × 100 = 20%

∴ The male employee who are using Hyundai car in company P are 20% more than female employee who are using Mahindra car in P.

The males who are using Hyundai car in company Q less than female who are using Ford car in company S = 16,320 – 14,400 = 1,920

Less % = $\frac{1,920}{16,320}$ × 100 = 11.76%

∴ The male employee who are using Hyundai car in company P are 11.76% less than female employee who are using Ford car in S.

The difference of male and female employee who are not using any car in company Q

⇒ 4,800 – 600 = 4,200

∴ The difference of male and female employee who are not using any car in company Q is 4,200.

From statement I:

Ratio of efficiency of A, B and C is 5 : 6 : 9

A takes 8 days more than C to finish the work

⇒ (9 – 5) units = 8 days

⇒ 4 units = 8 days

⇒ 1 unit = 2 days

∴ A will complete the work in = 5 × 2 = 10 days

From statement II:

A and B working together can complete the work in 30 days

$\Rightarrow \frac{1}{\mathrm{A}}+\frac{1}{\mathrm{B}}=\frac{1}{30}$

B is twice efficient than A.

Ratio of efficiency of A and B is 1 : 2.

⇒ (1 + 2) units = 30 days

⇒ 3 units = 30 days

⇒ 1 unit = 10 days

∴ A will complete the work in = 10 days

∴ Data in statement I alone or in statement II alone is sufficient to answer the question.

Formula:

L_t + L_p = Speed of the Train × Time taken to cross the Platform

⇒ Length of Train = L_t

⇒ Length of Platform = L_p

Calculation:

Statement I:

Length of the Train A = L_t = 100 m

Length of the Platform (L_p) = 2 × L_t = 2 × 100 = 200 m

Speed of Train A = S_ta

Time Taken to Cross the platform = 10 Sec

L_t + L_p = Speed of the Train × Time taken to cross the Platform

L_t + 2 × L_t = S_ta × 10

100 + 200 = S_ta × 10

300 = S_ta × 10

S_ta = 30 m/s

Time Taken by Train B = 20 s

Length of the Platform = 200 m

Length of the train B = L_b

Speed of Train B = S_b

L_b + L_p = Speed of the Train × Time taken to cross the Platform

L_t + 200 = S_b × 20

Length of the Train B is not given neither speed is given two variable needs two equation.

Statement II:

Speed of Train A = S_a

Speed of Train B = S_b

Sum of the speed of Train A and B is 200 m/s

S_a + S_b = 200 m/s      ----(i)

Difference between the speed of Train A and Train B is 50 m/s

S_a – S_b = 50 m/s      ----(ii)

Dividing (i) and (ii)

$\frac{{\mathrm{S}}_{\mathrm{a}}+{\mathrm{S}}_{\mathrm{b}}}{{\mathrm{S}}_{\mathrm{a}}-{\mathrm{S}}_{\mathrm{b}}}=\frac{200}{50}$

$\frac{{\mathrm{S}}_{\mathrm{a}}+{\mathrm{S}}_{\mathrm{b}}}{ {\mathrm{S}}_{\mathrm{a}}-{\mathrm{S}}_{\mathrm{b}}}=\frac{4}{1}$

Using Componendo and Dividendo

$\frac{{\mathrm{S}}_{\mathrm{a}}}{{\mathrm{S}}_{\mathrm{b}}}=\frac{5}{3}$

Sa : Sb = 5 : 3

∴ The statement II alone is sufficient to answer the question, but the statement I alone is not sufficient.

Formula:

Simple interest $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$

Calculations:

From question:

Interest earned by Pritesh Singh = $\frac{10,000\times \mathrm{x}\times 10}{100}$ = Rs. 1000x

Statement I: Interest earned by Sachin = $\frac{20,000\times 15\times 10}{100}$ = Rs. 30,000

Statement II: Interest earned by Sachin = 2 × Interest earned by Pritesh.

From statement I and II:

⇒ 30,000 = 2 × 1,000x

⇒ 30,000 = 2,000x

⇒ x = $\frac{30,000}{2,000}$ = 15

From statement I and II we can find the value of x.

∴ Both the statements I and II together are needed to answer the question.

Formula:

Volume of sphere $\mathrm{=}\frac{4}{3}{\mathrm{\pi r}}^3$

Volume of hemisphere $=\frac{2}{3}{\mathrm{\pi r}}^3$

Surface area of sphere = 4πr²

Surface area of hemisphere = 3πr²

Calculation:

From statement I:

Let radius of sphere and hemisphere be r and R respectively

Volume of sphere = $\frac{1}{4}$ × Volume of hemisphere

Radius of sphere = r

Volume of sphere = $\frac{4}{3}{\mathrm{\pi r}}^3$ And, radius of hemisphere = R

So, volume of hemisphere $=\frac{2}{3}{\mathrm{\pi R}}^3$

Now, $\frac{4}{3}{\mathrm{\pi r}}^3=\left(\frac{1}{4}\right)\frac{2}{3}{\mathrm{\pi R}}^3$

$\Rightarrow 4{\mathrm{r}}^3=\frac{{\mathrm{R}}^3}{2}$

As we can see that value of R and r can’t be determined

From statement II:

Radius of sphere = r

Radius of hemisphere = R

Surface area of sphere = 4πr²

Surface area of hemisphere = 3πR²

⇒ 4πr² × 3 = 3πR²

⇒ 12πr² = 3πR²

As we can see that from statement II also, we can’t find the value of r and R.

∴ Neither statement I nor statement II is sufficient to answer the question.

Given:

Rahul wants to invests Rs.4,000 in two Edugorilla package RBI grade B course and SBI PO course

RBI grade B course = 10% p.a.

SBI PO course = 15% p.a.

Calculations:

From questions: Let Rahul invests x amount in RBI grade B course then in SBI PO course = (4,000 – x)

Interest earned at RBI grade B course $=\frac{\mathrm{x}\times 10\times \mathrm{T}}{100}$

Interest earned at SBI PO course $=\frac{\left(4,000-\mathrm{x}\right)\times 15\times \mathrm{T}}{100}$

From statement I:

$\Rightarrow \frac{\mathrm{x}\times 10\times \mathrm{T}}{\left(4,000-\mathrm{x}\right)\times 15\times \mathrm{T}}=\frac{1}{6}$

$\Rightarrow \frac{10\mathrm{x}}{\left(4,000-\mathrm{x}\right)\times 15}=\frac{1}{6}$

$\Rightarrow \frac{2\mathrm{x}}{\left(4,000-\mathrm{x}\right)\times 3}=\frac{1}{6}$

⇒ x = 800

⇒ RBI grade B course = x = 800

⇒ SBI PO course = 4,000 – 800 = 3,200

From statement II:

Rahul invests 2,500 in SBI PO course

Rahul invests in RBI course = 4,000 – 2,500 = 1,500

From statement I and II we can’t find the time.

∴ Neither statement I nor statement II is sufficient to answer the question.