Data Analysis and Sufficiency Test - 7

From I:

Let CP of 1 kg of pure wheat = Rs. 1

Then SP of kg of mixture = Rs. 1 and gain = 25%

CP of 1 kg of mixture = 1 ×$\frac{100}{125}$ = Rs.$\frac{4}{5}$

Ratio of wheat to impurity =$\frac{4}{5}:\frac{1}{5}$ = 4 : 1

Percentages impurity =$\frac{1}{4+1}$ × 100 = 20%

So, statement I alone is sufficient.

From II:

Amount of mixture = 45 kg and amount of impurity is z kg.

Given:

⇒$\frac{5}{2}$ × z = 150% of$\frac{2}{3}$ × 45

⇒ z =$\frac{150\times 2\times 45\times 2}{100\times 3\times 5}$

⇒ z = 18

% impurity =$\frac{18}{45}$ × 100 = 40%

So, statement II alone is sufficient.

From III:

Let total quantity of mixture = 5z and quantity of pure wheat = 3z

Quantity of impurity in the mixture = 5z – 3z = 2z

% impurity =$\frac{2z}{5z}$ × 100 = 40%

So, statement III alone is sufficient.

Quantity A:

Since radius of both sphere and hemisphere will be same, we don't need to calculate actual values. We will therefore calculate only percentage increase.

Total surface area of sphere = 4πr²

Total surface area of two hemispheres = 2 × 3πr² = 6πr²

Percentage increase =$\frac{6{\mathrm{\pi r}}^2-4{\mathrm{\pi r}}^2}{4{\mathrm{\pi r}}^2}$ × 100 = 50%

Quantity B:

Words starting with A = 4! = 24

Words starting with E = 4! = 24

After both these series, words starting with N will come.

Words with N will be in this order,

Words starting with NA = 3! = 6

Again, for S and T, we will have NEAST and NEATS.

So, we have to further add 1 in the order

From here only we can tell, the number at which NEATS will come will be (24 + 24 + 6 + 1) which is more than that in Quantity A.

So, Quantity A < Quantity B

Quantity A:

Let part invested at 8% be x.

Part invested at 12% = 16000 - x

Total interest in three years = [8% of x + 12% of (16000 - x)] × 3

Now,

[8% of x + 12% of (16000 - x)] × 3 = 4680

8x + 192000 - 12x = 156000

4x = 36000

x = 9000

Quantity B:

Let principal be X.

Amount will be 2X

Time = 16 years

Interest = 2X - X = X

Interest =$\frac{\mathrm{P}\times \mathrm{r}\times \mathrm{t}}{100}$

X =$\frac{\mathrm{X}\times \mathrm{r}\times 16}{100}$

r =$\frac{100}{16}$ = 6.25%

Now, Principal = Rs. 8000

Rate = 12.5%

Interest = 12.5% of 8000 = 1000

Amount = Rs. 9000

So, Quantity A = Quantity B or no relationship can be established.

Quantity A:

Suppose dealer’s cost price is Rs. 1000 for 1000 gram good.

Since he is using 20% less weight

⇒ Actual Cost price = Rs. 800

As he promises to sell his good at 30% loss

⇒ Actual selling price = 1000 × 0.7 = Rs. 700

⇒ Actual loss =$\frac{800-700}{800}$ = 12.5%

Quantity B:

Suppose dealer’s cost price is Rs. 1000 for 1000 gram good.

Since he is using 25% less weight

⇒ Actual Cost price = Rs. 750

If profit is 40% on Rs. 750 as given in the question

⇒ Selling price = 750 × 1.4 = Rs. 1050

Profit at which the shopkeeper promises to sell =$\frac{1050-1000}{1000}$ = 5%

⇒ x = 5%

∴ Quantity A > Quantity B

Quantity A:

Rate of interest is 4% for the first years, 5% for the second year and 6% for the third year,

$\Rightarrow \mathrm{P}\times \frac{104}{100}\times \frac{105}{100}\times \frac{106}{100}=2893.80$

⇒ P = Rs. 2500

Quantity B:

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

$\Rightarrow 2173=\mathrm{P}\left[{\left(1+\frac{5}{100}\right)}^2-1\right]$

⇒ P = 2173 ×$\frac{400}{41}$ = Rs. 21200

⇒ SI = P × R ×$\frac{\mathrm{T}}{100}$ = 21200 × 5 ×$\frac{2}{100}$ = Rs. 2120

∴ Quantity A > Quantity B

Using statement I:

From the diagram:

a + b + c + d + e + f + g = 700      ----(1)

47% of the students take Music

∴ c + d + f + g = 0.47 × 700 = 329      ----(2)

18% of the students take only science

∴ e = 0.18 × 700 = 126

12% of the students only take science and Mathematics

b = 0.12 × 700 = 84

Using equation (2) and values of e and b in equation (1)

a + 84 + 126 + 329 = 700

⇒ a = 161

∴ 161 students select only Mathematics

The data in statement I alone are sufficient to answer the question

Statement II:

a + b + c + d + e + f + g = 700      ----(1)

48% of the students take either two or more of all of the given subjects.

∴ b + c + d + f = 0.48 × 700 = 336      ----(2)

36% of the students do not like Mathematic.

∴ e + f + g = 0.36 × 700 = 252      ----(3)

From these information we cannot find value of a

∴ Statement II alone is not sufficient.

The final quantity of mixture B after adding 5 liters of milk and 10 liters of water = 40 + 15 = 55

So, the quantity of water in the final mixture of B =$\frac{2}{5}$ × 55 = 22 liters

Final quantity of mixture C after adding 3 liters of milk and 6 liters of water = 56 + 9 = 65

So, the quantity of milk in the final mixture of C =$\frac{9}{13}$ × 65 = 45 liters

∴ The quantity of milk in the final mixture of C is more than the quantity of water in final mixture B by (45 – 22) = 23 liters.

The final quantity of mixture D after adding 10 liters of milk and 5 liters of water = 60 + 15 = 75

So, the quantity of water in the final mixture of D =$\frac{2}{5}$ × 75 = 30 liters

Final quantity of mixture E after adding 6 liters of milk and 8 liters of water = 62 + 14 = 76

So, the quantity of water in the final mixture of C =$\frac{10}{19}$ × 76 = 40 liters

The final quantity of mixture A after adding 4 liters of milk and 4 liters of water = 60 + 8 = 68

So, the quantity of milk in the final mixture of B =$\frac{13}{17}$ × 68 = 52 liters

Final quantity of mixture C after adding 3 liters of milk and 6 liters of water = 56 + 9 = 65

So, the quantity of milk in the final mixture of C =$\frac{9}{13}$ × 65 = 45 liters

∴ Required ratio = (30 + 40) : (52 + 45) = 70 : 97

The final quantity of mixture E after adding 6 liters of milk and 8 liters of water = 62 + 14 = 76

The final quantity of mixture G after adding 3 liters of milk and 3 liters of water = 72 + 6 = 78

So, the final mixture G is 2 more than the final mixture E.

∴ Required percentage =$\left(\frac{78-76}{76}\right)\times 100=\frac{100}{38}$ = 2.63% more

The final quantity of mixture B after adding 5 liters of milk and 10 liters of water = 40 + 15 = 55

Now, 20% of the mixture B is sold

Remaining quantity =$\frac{4}{5}$ × 55 = 44 liters

Water content in it =$\frac{2}{5}\times 44=\frac{88}{5}$

∴ Water content in the remaining mixture = 17.6 liters.