Data Analysis and Sufficiency Test

Result

Data Analysis and Sufficiency Test - 8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

Directions For Questions

Direction: The question below consists of a question and three statements number I, II and III given below it. You have to decide whether the data provided in the statement are sufficient to answer the question.

...view full instructions

A bag contains some number of Diamond of 4 different colors (Red, Pink, blue and yellow). What is the total number of Diamond in the bag?

I. If 5 Pink Diamond from the bag are replaced by Yellow Diamond, Then the ratio of probabilities of picking 2 yellow Diamond at a time to picking 3 Blue Diamond at a time becomes $(\frac{1}{116})$ .

II. The different between probabilities of picking 2 Red Diamonds at a time and probability of picking 2 Blue Diamonds at a time is $(\frac{23}{330})$ .

III. If a Diamond is picked from the bag, then there is 40% chance that it is a red Diamond. Probability of picking 2 Red Diamond from the bag is $(\frac{26}{165})$ .

A
The data in only statement I and II together are sufficient to answer the question

B
The data in only statement I and III together are sufficient to answer the question

C
The data in only statement III alone is sufficient to answer the question

D
The data in at least any two of the 3 statements is necessary to answer the question

Solution

From Statement I:

Let number of Red, Pink, Blue and Yellow Diamond in the bag be P, Q, R and S respectively.

Let total number of diamond in the bag be T.

T = (P + Q + R + S)

After replacing 5 Pink Diamond by Yellow diamond.

Probability of picking 2 Yellow diamond =

$\frac{^{(S + 5)} C_{2}}{^{T} C_{2}} = \frac{(S + 5) \times (S + 4)}{T \times (T - 1)}$

Probability of packing 3 Blue diamond

$= \frac{^{R} C_{2}}{^{T} C_{2}}$

$= \frac{R \times (R - 1) \times (R - 2)}{T \times (T - 1) \times (T - 2)}$

The ratio of probabilities of picking 2 Yellow diamond at a time to picking 3 Blue diamond at a time

$= \frac{\frac{(S + 5) \times (S + 4)}{T \times (T - 1)}}{\frac{R \times (R - 1) \times (R - 2)}{T \times (T - 1) \times (T - 2)}}$

$= \frac{(S + 5) \times (S + 4) \times (T - 2)}{R \times (R - 1) \times (R - 2)} = \frac{1}{116}$

⇒ 116 × (S + 5) × (S + 4) × (T – 2) = (R × (R – 1) × (R – 2)

⇒ 116 × (S + 5) × (S + 4) × (T – 2) = (R × (R – 1) × (R – 2)

So, data in statement I alone is not sufficient to answer the question.

From Statement II:

Let number of Red diamonds and number of Blue diamonds in the bag be P and R respectively.

Lets total number of diamond be T.

Probability of picking 2 Red diamonds = $\frac{^{p} C_{2}}{^{T} C_{2}}$

$= \frac{P \times (P - 1)}{T \times (T - 1)}$

Probability of picking 2 Blue diamonds =

$\frac{^{R} C_{2}}{^{T} C_{2}} = \frac{R \times (R - 1)}{T \times (T - 1)}$

As per given,

$\frac{P \times (P - 1)}{T \times (T - 1)} - \frac{R \times (R - 1)}{T \times (T - 1)} = \frac{23}{330}$

$(P \times (P - 1)) - (R \times (R - 1)) = (\frac{23}{330}) \times T \times (T - 1)$

So, data in statement II alone is not sufficient to answer the question.

Even combining data from statement I and II together is also not sufficient to answer the question.

From Statement III:

Probability of picking a red diamond = $\frac{Number of red diamond}{Total number of diamond}$

$= \frac{40}{100} = \frac{2}{5}$

So, let number of red diamond be 2T and total number of diamond be 5T

Probability of picking 2 red diamond from the bag

$= \frac{^{2 T} C_{2}}{^{5 T} C_{2}}$

$= \frac{2 T \times (2 T - 1)}{5 T \times (5 T - 1)} = \frac{26}{165}$

$\frac{4 T 2 - 2 T}{25 T 2 - 5 T} = \frac{26}{165}$

660T² – 330T = 650T² – 130T

10T2 = 200T

T = 20

Total number of diamonds in the bag is 100.

Data in statement III alone is sufficient to answer the question.
Question 2
1 / -0.25

Directions For Questions

Direction: The following question has three statements. Study the question and the statements to decide which of the statement(s) is/are necessary to answer the question.

...view full instructions

A committee having 2 Hindi teachers and 2 English teachers is to be formed from some ‘Hindi teachers’ and ‘English teachers’. The probability of doing so is $\frac{3}{5}$ . Find the total number of Hindi and English teachers.

Statement I: Number of ways to choose 2 teachers out of total teachers is 15.

Statement II: Number of ways to choose 2 Hindi teachers out of total Hindi teachers is 3.

Statement III: Number of English teachers is 3.

A
Statement I alone is sufficient

B
Statement II alone is sufficient

C
Statement III alone is sufficient

D
Each of the statements alone is sufficient

Solution

Let the number of Hindi and English teachers is ‘H’ and ‘E’ respectively

A committee having 2 Hindi teachers and 2 English teachers is to be formed from H ‘Hindi teachers’ and E ‘English teachers’. The probability of doing so is $\frac{3}{5}$

We need to find total number of Hindi and English teachers i.e. (H + E)

Statement I:

Number of ways to choose 2 teachers out of total teachers is 15

Suppose (H + E) = x

⇒ x (x - 1) = 30

Now, Product of two consecutive numbers is given as 30 (6 × 5) so, x = 6

⇒ (H + E) = 6

Statement I alone is sufficient to answer the question

Statement II:

Number of ways to choose 2 Hindi teachers out of total Hindi teachers is given by 3

⇒ ^HC₂ = 3

⇒ H (H - 1) = 6

Now, Product of two consecutive numbers is given as 6 (3 × 2) so H = 3

From equation 1, we can find the value of E also

Statement (II) alone is sufficient to answer the question

Statement III:

Number of English teachers is 3

As, E = 3, value of H can be found from equation (1)

Statement III alone is sufficient.

∴ Each of the statements alone is sufficient to answer the question.
Question 3
1 / -0.25

Directions For Questions

Direction: The following question has three statements. Study the question and the statements to decide which of the statement(s) is/are necessary to answer the question.

...view full instructions

Mohan travel from Delhi to Agra from train and he notices that with stoppages, he covers the total distance in 52.8 hours. He wants to calculate speed of the train without stoppages. Find the speed of the train excluding stoppages.

Statement I∶ Distance between Delhi and Agra is 500 km.

Statement II∶ The first stop lasts 2 minutes, second 4 minutes and others like this 6,8….100 minutes.

Statement III∶ The train can cross the platform of stations between Delhi and Agra in 10 seconds each.

A
Only statement I is sufficient

B
Statement I and statement II are sufficient

C
Statement I, statement II and statement III are sufficient

D
Only statement III is sufficient

Solution

Statement I:

Let time without stoppages be x

Speed = $\frac{500}{x}$

Statement II:

= $\frac{2 + 4 + 6 + 8 \dots . + 100}{60}$ hours

= 50 × $\frac{51}{60}$

= 42.5 hours

X = 52.8 – 42.5

= 10 hours

Required speed = $\frac{500}{10}$

= 50 kmph

Statement III:

The data for train crossing a platform is not required here.

∴ Statement I and statement II are sufficient to answer the question.
Question 4
1 / -0.25

Directions For Questions

Direction: In the following question, two statements are numbered as A and B. On solving these statements, we get quantities A and B respectively. Solve both quantities and choose the correct option.

...view full instructions

Quantity A: A bag contains 50 slips. Numbers are written on each slip from 51 to 100.

One slip is drawn randomly from the bag. What is the probability, it is a square number?

Quantity B: $\frac{2}{25}$

A
Quantity A ≥ Quantity B

B
Quantity A ≤ Quantity B

C
Quantity A < Quantity B

D
Quantity A > Quantity B

Solution

Quantity A:

P (E) = $\frac{Number of favorable outcomes}{Total number of outcomes}$

Let E is an event of a probability.

Number of favorable outcomes = 3 i.e. (64, 81,100)

∴ P (E) = $\frac{3}{50}$

Quantity B: $\frac{2}{25}$

∴ Quantity A < Quantity B
Question 5
1 / -0.25

Direction: The following question has two statements. Study the question and the statements to decide which of the statement(s) is/are necessary to answer the question.

Find the time taken by 8 men, 4 women and 6 kids together to complete the piece of work.

Statement I: 6 men and 4 women together can complete the work in 12 days and 1 man and 8 women together can complete the work in 17 days.

Statement II: The ratio of efficiencies of a woman and a kid is 5 : 2.

A
Both Statements I and II together are not sufficient to answer the question.

B
Statements I and II together are required to answer the question.

C
Either statement I or statement II alone is sufficient to answer the question.

D
Statement II alone is sufficient to answer the question.

Solution

Statement I:

6 men and 4 women together can complete the work in 12 days and 1 man and 8 women together can complete the work in 17 days.

Suppose the efficiencies of man and woman are ‘m’ and ‘w’ respectively.

So,

(6 × m + 4 × w) × 12 = (1 × m + 8 × w) × 17

⇒ 72m + 48w = 17m + 136w

⇒ 55m = 88w

⇒ m : w = 8 : 5

So, ratio of efficiencies of a man and a woman = 8 : 5

Combining statement II:

Ratio of efficiencies of a woman and a kid = 5 : 2

Ratio of efficiencies of a man, a woman and a kid = 8 : 5 : 2

Suppose the work done by men, women and kid be 8x, 5x and 2x respectively

Now,

Suppose the time taken by 8 men, 4 women and 6 kids together is T days

(6 × 8x + 4 × 5x) × 12 = (8 × 8x + 4 × 5x + 6 × 2x) × T

⇒ (48x + 20x) × 12 = (64x + 20x + 12x) × T

⇒ 68x × 12 = 96x × T

⇒ T = 8.5

Time taken by 8 men, 4 women and 6 kids together = 8.5 days

∴ Statements I and II together are required to answer the question.
Question 6
1 / -0.25

Directions For Questions

Direction: Read the data carefully and answer the following question.

In a city during 2005 six engineering University A, B, C, D, E, and F Having 84000 students in three branches CSE, ME and ECE. CSE Students from C is 15% of total students from CSE and total student from University D is 12% Of total students in the city. ME student from University A is 9% of total ME students and total students from university F is 20% more than total students from D. Ratio of CSE, ME and ECE students in the city is 4 : 5 : 3 and the total students from University B is 15120. ECE from University D is $\frac{3}{5}$ th of CSE Student from C and ratio of students from CSE, ME and ECE from university F 3 : 5 : 4. CSE students from University D is 25% more than ECE students from university A and ME student from University E is 24% of total ME students and ratio of CSE, ME and ECE student from University A is 13 : 10 : 12. Total number of ECE student from University A is 18% of total ECE student And ratio of number of students from CSE and ECE from University B is 3 : 2. ECE students from university E Is 25% more than ECE students from F and ratio of ME students from University B to C is 2 : 3.

...view full instructions

It is assumed that in 2018, because of JIO, $40 \%$ of the total population of India will buy products online. If in 2018, the population of India was increased by $5 \%$ over previous year then in 2018 , total how many people will buy product in India?

A
231 million

B
243 million

C
239 million

D
233 million

Solution

Let the total population of India in $2016=x$ million then

$20 \%$ of $x$ million $=100$ millions

$x=100 \times 5=500$ millions

The population of India in $2017=110 \%$ of $500=550$ million

The population of India in $2018=105 \%$ of 550 million $=577.5$ million

In 2018, because of JIO, $40 \%$ of the total population of India will buy product online $=40 \%$ of 577.5 $=\frac{40 \times 577.5}{100}=231$ million
Question 7
1 / -0.25

Directions For Questions

Direction: Read the data carefully and answer the following question.

In a city during 2005 six engineering University A, B, C, D, E, and F Having 84000 students in three branches CSE, ME and ECE. CSE Students from C is 15% of total students from CSE and total student from University D is 12% Of total students in the city. ME student from University A is 9% of total ME students and total students from university F is 20% more than total students from D. Ratio of CSE, ME and ECE students in the city is 4 : 5 : 3 and the total students from University B is 15120. ECE from University D is $\frac{3}{5}$ th of CSE Student from C and ratio of students from CSE, ME and ECE from university F 3 : 5 : 4. CSE students from University D is 25% more than ECE students from university A and ME student from University E is 24% of total ME students and ratio of CSE, ME and ECE student from University A is 13 : 10 : 12. Total number of ECE student from University A is 18% of total ECE student And ratio of number of students from CSE and ECE from University B is 3 : 2. ECE students from university E Is 25% more than ECE students from F and ratio of ME students from University B to C is 2 : 3.

...view full instructions

Total number students in university A is approximately what percent less than total students of University C taken all three branches together?

A
24.8%

B
31.2%

C
20.8%

D
29.4%

Solution

Students from university A = (4095 + 3150 + 3780) = 11025

Students from university C = (4200 + 9345 + 2075) = 15617

Required percent = $\frac{15617 - 11025}{15617}$ × 100 = 29.4%
Question 8
1 / -0.25

Directions For Questions

Direction: Read the data carefully and answer the following question.

In a city during 2005 six engineering University A, B, C, D, E, and F Having 84000 students in three branches CSE, ME and ECE. CSE Students from C is 15% of total students from CSE and total student from University D is 12% Of total students in the city. ME student from University A is 9% of total ME students and total students from university F is 20% more than total students from D. Ratio of CSE, ME and ECE students in the city is 4 : 5 : 3 and the total students from University B is 15120. ECE from University D is $\frac{3}{5}$ th of CSE Student from C and ratio of students from CSE, ME and ECE from university F 3 : 5 : 4. CSE students from University D is 25% more than ECE students from university A and ME student from University E is 24% of total ME students and ratio of CSE, ME and ECE student from University A is 13 : 10 : 12. Total number of ECE student from University A is 18% of total ECE student And ratio of number of students from CSE and ECE from University B is 3 : 2. ECE students from university E Is 25% more than ECE students from F and ratio of ME students from University B to C is 2 : 3.

...view full instructions

If 75% of total students from B successfully completed their project And the ratio Of male to female students who completed their project from university B is 5 : 7, then what is difference between number of female student from CSE to number of female students from ME who has completed this project successfully in University B, If ratio of total students in B who has completed their project from CSE, ME and ECE are 6 : 7 : 5 and ratio of male students from given three branches in B, who has completed this project are 8 : 11 : 6 respectively?

A
78

B
55

C
63

D
81

Solution

Students successfully completed the project = 15120 × $\frac{75}{100}$ = 11340

Male students successfully completed the project = 11340 × $\frac{5}{12}$ = 4725

CSE student completed the project successfully = 11340 × $\frac{6}{18}$ = 3780

Mail CSE students successfully completed their project = 4725 × $\frac{8}{25}$ = 1512

Female successfully completed the project from CSC = 3780 - 1512 = 2268

ME student completed the project successfully = 11340 × $\frac{7}{18}$ = 4410

Mail ME student completed their project successfully = 4725 × $\frac{11}{28}$ = 2079

Female students successfully completed third project = 4410 – 4079 = 2331

Required difference = 2268 - 2331 = 63
Question 9
1 / -0.25

Directions For Questions

Direction: Read the data carefully and answer the following question.

In a city during 2005 six engineering University A, B, C, D, E, and F Having 84000 students in three branches CSE, ME and ECE. CSE Students from C is 15% of total students from CSE and total student from University D is 12% Of total students in the city. ME student from University A is 9% of total ME students and total students from university F is 20% more than total students from D. Ratio of CSE, ME and ECE students in the city is 4 : 5 : 3 and the total students from University B is 15120. ECE from University D is $\frac{3}{5}$ th of CSE Student from C and ratio of students from CSE, ME and ECE from university F 3 : 5 : 4. CSE students from University D is 25% more than ECE students from university A and ME student from University E is 24% of total ME students and ratio of CSE, ME and ECE student from University A is 13 : 10 : 12. Total number of ECE student from University A is 18% of total ECE student And ratio of number of students from CSE and ECE from University B is 3 : 2. ECE students from university E Is 25% more than ECE students from F and ratio of ME students from University B to C is 2 : 3.

...view full instructions

If the ratio of the number of males and the number of the female of ME students from University E and F 8 : 7 and 7 : 5 respectively then total male student in ME From E is approximately what percent more/less than total two female students in ME from F?

A
128.2%

B
113.3%

C
121.4%

D
118.5%

Solution

Male ME students from E = 8400 × $\frac{8}{15}$ = 4480

Female ME students from F = 5040 × $\frac{5}{12}$ = 2100

Requires percent = $\frac{4480 - 2100}{2100}$ × 100 = 113.3%
Question 10
1 / -0.25

Directions For Questions

Direction: Read the data carefully and answer the following question.

In a city during 2005 six engineering University A, B, C, D, E, and F Having 84000 students in three branches CSE, ME and ECE. CSE Students from C is 15% of total students from CSE and total student from University D is 12% Of total students in the city. ME student from University A is 9% of total ME students and total students from university F is 20% more than total students from D. Ratio of CSE, ME and ECE students in the city is 4 : 5 : 3 and the total students from University B is 15120. ECE from University D is $\frac{3}{5}$ th of CSE Student from C and ratio of students from CSE, ME and ECE from university F 3 : 5 : 4. CSE students from University D is 25% more than ECE students from university A and ME student from University E is 24% of total ME students and ratio of CSE, ME and ECE student from University A is 13 : 10 : 12. Total number of ECE student from University A is 18% of total ECE student And ratio of number of students from CSE and ECE from University B is 3 : 2. ECE students from university E Is 25% more than ECE students from F and ratio of ME students from University B to C is 2 : 3.

...view full instructions

What is the difference between total CSE students from A, C, D and F to total ECE student from B, C, D and E taken together?

A
2684

B
2728

C
2672

D
2856

Solution

CSE students from A, C, D and F = (4095 + 4200 + 4725 + 3024) = 16044

ECE students from B, C, D and E = (3556 + 2072 + 2520 + 5040) = 13188

Required difference = (16044 – 13188) = 2856

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Data Analysis and Sufficiency Test - 8

Result

Data Analysis and Sufficiency Test - 8

Score

-

Rank

-

SHARING IS CARING

Question 1

The data in only statement I and II together are sufficient to answer the question

The data in only statement I and III together are sufficient to answer the question

The data in only statement III alone is sufficient to answer the question

The data in at least any two of the 3 statements is necessary to answer the question

Solution

Question 2

Statement I alone is sufficient

Statement II alone is sufficient

Statement III alone is sufficient

Each of the statements alone is sufficient

Solution

Question 3

Only statement I is sufficient

Statement I and statement II are sufficient

Statement I, statement II and statement III are sufficient

Only statement III is sufficient

Solution

Question 4

Quantity A ≥ Quantity B

Quantity A ≤ Quantity B

Quantity A < Quantity B

Quantity A > Quantity B

Solution

Question 5

Both Statements I and II together are not sufficient to answer the question.

Statements I and II together are required to answer the question.

Either statement I or statement II alone is sufficient to answer the question.

Statement II alone is sufficient to answer the question.

Solution

Question 6

231 million

243 million

239 million

233 million

Solution

Question 7

24.8%

31.2%

20.8%

29.4%

Solution

Question 8

78

55

63

81

Solution

Question 9

128.2%

113.3%

121.4%

118.5%

Solution

Question 10

2684

2728

2672

2856

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.