Algebra - Simple Equations - Previous Year CAT/MBA Questions
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Mohan sought feedback from a few of his businessmen friends, who were familiar with both the branches. Here is what they said:
- Businessman 1 : Customers of Connaught place and Panipat are very different.
- Businessman 2 : Customers in Panipat are extremely happy with Ram’s behaviour.
- Businessman 3 : Panipat branch does not use the same quality of ingredients but maintains good hygiene and taste.
- Businessman 4 : Who knows, tomorrow the customers of Panipat might also appreciate what Connaught place customers appreciate today!
If Mohan thinks all these are valid concerns, which of the following actions would be best for the business?
- (a)
Training Kishan to replace Ram in a few months.
- (b)
Not worrying about ingredients as long as business grows.
- (c)
Bringing Ram to Connaught place branch.
- (d)
Naming the Panipat branch as ‘Ram’s’, and changing it back to Mohan’, when needed.
- (e)
Asking Kishan to run the Panipat branch.
Answer: Option D
Text Explanation :
It can be inferred from the businessmen's statements that none of them has any harsh or strong opinion about Ram's approach of not using organic vegetables. Thus, according to the businessmen, substituting organic vegetables at the Panipat branch is not required.
Options 1 and 5 would unnecessarily complicate things as Kishan would take more time to get himself acquainted to the Panipat branch, also it is not guaranteed that it would yield the desired results for Mohan. Eliminate options 1 and 5.
The passage states that giving a free hand to Ram might have long term negative consequences. However, option 2 doesn't take this into consideration. Eliminate option 2.
Ram has already established the Panipat branch and he is the ideal person to carry out future operations there. Bringing him back to Connaught place will complicate things and serve no purpose. Eliminate option 3.
If the name of the Panipat branch is changed to 'Ram's' instead of 'Mohan's', then the identity of the business in Panipat will be linked to Ram. Thus, option 4 satisfies all the opinions expressed by the four businessmen.
Hence, the correct answer is option 4.
Workspace:
After discussing with a few customers, Mohan realised that compromising on the quality of ingredients at Panipat branch may not be good idea but at the same time he also realised that Panipat branch had grown fast. He was contemplating following five actions. Which of the following actions would be the best for the future of his business?
- (a)
Creating awareness campaign for organic vegetables in Panipat.
- (b)
Mohan himself should took after the Panipat branch.
- (c)
Close down the Panipat branch.
- (d)
Send Kishan to Panipat branch and bring Ram to Connaught place permanently.
- (e)
Hire a new person to run the Panipat branch
Answer: Option B
Text Explanation :
If an awareness campaign related to organic vegetables is generated in Panipat it can malign the reputation of the Panipat branch as it was previously not using organic vegetables.
As Mohan is an experienced person and is the one responsible for taking the Connaught Place branch to where it is today, he is the best candidate to run the Panipat branch as he is likely to generate the required sales whilst also maintaining the quality of food.
Closing down the Panipat branch would be a silly idea, as the place is already established and generates revenue.
Sending Kishan to the Panipat branch is a bad idea as he would take more time to get himself acquainted to the place. Mohan has already established the place and is the better person between the two brothers to carry out the operations there.
Option 5 can be eliminated for the same reason presented above to eliminate option 4.
Hence, the correct answer is option 2.
Workspace:
Answer the following question based on the information given below.
MBA entrance examination comprises two types of problems: formula-based problems and application-based problem. From the analysis of past data, Interesting School of Management (ISM) observes that students good at solving application-based problems are entrepreneurial in nature. Coaching institutes for MBA entrance exams train them to spot formula-based problems and answer them correctly, so as to obtain the required overall cut-off percentile. Thus students, in general, shy away from application-based problem and even those with entrepreneurial mind-set target formula-based problems.
Half of a mark is deducted for every wrong answer.
ISM wants more students with entrepreneurial mind-set in the next batch. To achieve this, ISM is considering following proposals:
- Preparing a question paper of two parts, Parts A and Part B of duration of one hour each. Part A and Part B would consist of formula-based problems and application-based problems, respectively. After taking away Part A, Part B would be distributed. The qualifying cut-off percentile would be calculated on the combined scores of two parts.
- Preparing a question paper comprising Part A and Part B. While Part A would comprise formula-based problems, Part B would comprise application-based problems, each having a separate qualifying cut-off percentile.
- Assigning one mark for formula-based problems and two marks for application based problems as an incentive for attempting application-based problems.
- Allotting one mark for formula-based problems and three marks for application-based problem, without mentioning this is the question paper.
Which of the following proposal (or combination of proposals) is likely to identify students with best entrepreneurial mind-set?
- (a)
II
- (b)
I & II
- (c)
I & III
- (d)
II & III
- (e)
II & IV
Answer: Option D
Text Explanation :
If the percentile for formula based questions and application based questions would be combined, students might maximise their score merely by attempting formula based questions. This is undesirable.
We can rule out proposal I.
Statement II is quite strong a stance, and would force students to attempt questions from both sections. However, students may attempt close to zero application based questions.
Statement III gives additional incentive to attempt ‘application-based problems’. Another advantage is that it might avoid the extreme outcomes of students attempting very few questions from ‘application-based problems’. A combination of proposals II and III would be ideal.
As it is already mentioned in the passage that even students with an entrepreneurial mindset avoid application based questions in order to score better, the college will not be able to identify students with an entrepreneurial mindset if they decide to not disclose the higher weightage of marks allotted to application based questions. Also, it is an unfair course of action. Hence, statement IV is not valid.
Hence, the correct answer is option 4.
Workspace:
ISM conducts a common entrance examination every year. This year, the question paper would comprise 60 questions with an equal mix of formula-based problems and application-based problems. All questions would carry equal marks. Balaji is appearing for the examination. Before, appearing for the examination he gets the following information from coaching institutes:
- Application-oriented problems take more time to solve in an examination hall.
- Chances of silly mistakes would be low in application-based problems.
- ISM would assist the students with bank loans to start a new venture.
- Options are generally confusing for formula-based problems.
- ‘Practice makes a man perfect’ can apply only to formula-based problems.
- Students get very good campus jobs.
Based on above information, which of the following options would help him to be better prepared for the examination?
- (a)
I &II
- (b)
I, II & V
- (c)
II, III & VI
- (d)
IV, V & VI
- (e)
I, II, IV & V
Answer: Option E
Text Explanation :
If Balaji already knows that solving application based problems takes more time, he can focus more on formula based problems and achieve the required cut-off percentile.
Similarly, if Balaji knows that the chances of making silly mistakes in application based questions is low, he can plan to attempt those questions towards the end of the exam, where the chances of making silly mistakes is higher due to stress.
Statements III and VI are unrelated to the issue presented in the question.
Knowing that options for formula based questions are tricky, Balaji can avoid attempting those questions.
By knowing that the popular idiom “practice makes a man perfect” is applicable to formula based questions, he can decide whether to attempt such questions or to avoid them, based on his capability.
Thus, only statements I, II, IV and V are valid for the question presented.
Hence, the correct answer is option 5.
Workspace:
Answer the questions on the basis of information given in the following case.
The Disciplinary Committee of Nation Political Party (NPP) is meeting today to decide on the future of two of their party members, Mr. Loyal and his son Mr. Prodigal. Mr. Prodigal is the prime accused in the brutal murder of Mr. Victim, an opposition party leader. Mr. Prodigal is in police custody and his appeal for bail has got rejected. Mr. Loyal claims that his son is innocent and Mr. Victim’s death was the result of internal rivalry in the opposition party. Though Mr. Loyal is not accused in this case, his weakness for his son is well known. The media is blaming him for influencing key witnesses to protect his son. Severe criticism of his father-son duo, both by the media and some social activists, is damaging the image of the party. However, Mr. Loyal has significant followers within the party and is considered an asset to the party. Any harsh decision against Mr. Loyal would adversely affect the future of NPP and could even lead to a split in the party. This would benefit the opposition.
Which of the following actions would adversely affect both NPP and Mr. Loyal, the most?
- (a)
Take no action against Mr. Loyal.
- (b)
Suspend Mr. Prodigal from the party with immediate effect.
- (c)
Expel Mr. Loyal from the party with immediate effect.
- (d)
Ban Mr. Loyal from entering party premises till completion of court proceedings.
- (e)
Initiate an internal inquiry to find the truth.
Answer: Option C
Text Explanation :
Taking no action against Mr. Loyal would not affect either party as he is not directly involved in the murder case.
Suspending Mr. Prodigal from the party would upset Mr. Loyal as it is common knowledge that the latter has a soft spot for Mr. Prodigal, and it might (remote possibility) lead to Mr. Loyal separating from the party.
Expelling Mr. Loyal from the party would obviously affect him as his political repute might be affected. The NPP party will also be affected as it is stated in the passage that Mr. Loyal's followers would part along with him. This entire situation would benefit the opposition party.
Banning Mr. Loyal from party premises till the court case is cleared will affect both parties, but the situation will not be as grave as compared to option 3.
Initiating an internal inquiry may not affect any of the parties as the information of an internal inquiry can be kept away from Mr. Loyal.
Thus, the strongest possibility of both the parties getting adversely affected is presented in option 3. Option 4 is close but can be eliminated on the basis that the effect of Mr. Prodigal's suspension from the party would not directly affect Mr. Loyal, who may still continue with the party.
Hence, the correct answer is option 3.
Workspace:
At the Disciplinary Committee meeting, members came up with the following suggestions. Which of the following suggestions would harm the party, the least?
- (a)
Maintain status-quo
- (b)
Expel Mr. Prodigal from the party with immediate effect to maintain party’s clean image
- (c)
Initiate an internal inquiry to find the truth.
- (d)
Suspend Mr. Prodigal from the party with immediate effect but announce that he will be taken back if the court declares him innocent.
- (e)
Suspend both Mr. Loyal and Mr. Prodigal from the party with immediate effect.
Answer: Option D
Text Explanation :
Maintaining a status-quo, i.e. taking no action at all, would convey a wrong message to the public and the media; it can have a negative impact on the party's future.
Expelling Mr. Prodigal is certain to upset Mr. Loyal, as the fact that he has a weakness for his son is common knowledge.
If an internal inquiry is set against Mr. Prodigal, it may upset Mr. Loyal. The party cannot afford the risk of losing Mr. Loyal as he is an asset to the party.
The safest bet for the party would be to temporarily dismiss Mr. Prodigal, and take him back if he is acquitted of the murder charge. This course of action would not only help the party gain some brownie points with the public and media, but will also not upset Mr. Loyal as much as the other alternatives can.
Option 5 if implemented will definitely affect the party in an adverse manner, as Mr. Loyal is an asset to them.
Hence, the correct answer is option 4.
Workspace:
Mr. Opportunist, a veteran member of NPP, stakes his claims to be nominated as an NPP candidate in the upcoming election. Mr. Opportunist presented the following arguments in favour of his candidature to the NPP Executive Committee.
- Mr. Loyal’s candidature in the upcoming election will adversely impact NPP’s chances. Hence, the party should not nominate him.
- The party should call a press conference to disown Mr. Loyal. This would enhance the party’s image.
- The party would not be able to take any strong disciplinary action against Mr. Loyal, if he gets re-elected.
- I have a lot of goodwill and significant followers in the constituency,
- None of my close relatives are into active politics.
Which of the following combinations would best strengthen the claim of Mr. Opportunist?
- (a)
I & III
- (b)
I &IV
- (c)
II & III
- (d)
III & V
- (e)
IV & V
Answer: Option B
Text Explanation :
Argument I rules out the party's intention of nominating Mr. Loyal in the upcoming election making way for Mr. Opportunist. Hence, it makes for a strong argument supporting the claim of Mr. Opportunist. Hence, eliminate options 3, 4 and 5.
Argument III simply rephrases the argument stated in option II; it is a weak line of reasoning. So, eliminate option 1.
Argument IV is a strong reason as to why it would be a good idea to nominate Mr. Opportunist.
Hence, the correct answer is option 2.
Workspace:
The Disciplinary Committee has decided to suspend Mr. Loyal from the party because they felt he was influencing the judicial process. However, Mr. Loyal feels that the committee is biased and he is being framed. Now, election has been announced. The last time. Mr. Loyal had won with a majority on account of his good work. Which of the following options is most likely to resurrect Mr. Loyal’s immediate political career?
- (a)
The main opposition party has invited Mr. Loyal to join the party and contest the election Chance of winning is high.
- (b)
Not participation in the campaign and instructing his followers to stay away from the campaigning process.
- (c)
Ask his followers to support the NPP nominated candidate and display his loyalty to NPP.
- (d)
Mr. Loyal should contest as an independent candidate. But because of a split in votes, his chances of winning would be low.
- (e)
Influence the nomination process through his followers within NPP, to get one of his close associates nominated.
Answer: Option A
Text Explanation :
All the options except 1 seem incongruous with respect to the information provided in the passage.
Option 1 is most feasible and will give a boost to Mr. Loyal's political career. Joining another party would do no harm to his career, given his popularity among voters as well as the support of opposition party. So, eliminate options 2, 3, 4 and 5.
Hence, the correct answer is option 1.
Workspace:
Innovative Institute of Business (IIB) has decided to be the first green campus in India. IIB Administration has advised all campus residents to reduce carbon footprints. IIB faculty members did a brainstorming and came up with the following suggestions:
- Replacing electricity source for street lights with solar panels.
- Replacing the existing buildings with environment friendly buildings.
- Organizing a seminar on ‘Towards a Sustainable Future’ involving all students, staff, and experts from around the country.
- Introducing a compulsory course on sustainability to increase awareness among students.
- Conducting an initial energy audit to explore where IIB can reduce carbon footprints.
Which of the following options would be the most preferred sequence of actions to reduce carbon footprints on campus?
- (a)
II, IV, V
- (b)
IV, V, III
- (c)
V, I, II
- (d)
V, I, III
- (e)
V, III, I
Answer: Option C
Text Explanation :
From the given suggestions, the foremost step should be to locate the places where IIB can start implementing its measures to curb carbon footprints which is stated in suggestion V.
Organizing a seminar to create awareness regarding a sustainable future is an indirect means of increasing awareness. Similarly, introducing a compulsory course is also not an effective way of educating the students.
The next logical step should therefore be to implement an actual strategy to reduce carbon footprints. Solar panel operated lights and environmentally-friendly buildings are practical and effective ways to reduce carbon footprints.
Thus, rule out options 1, 2, 4 and 5.
Option 3 is the best answer as it contains concrete actions.
Hence, the correct answer is option 3.
Workspace:
In an examination, two types of questions are asked: one mark questions and two marks questions. For each wrong answer, of one mark question, the deduction is 1/4 of a mark and for each wrong answer, of two marks question, the deduction is 1/3 of a mark. Moreover, 1/2 of a mark is deducted for any unanswered question. The question paper has 10 one mark questions and 10 two marks questions. In the examination, students got all possible marks between 25 and 30 and every student had different marks. What would be the rank of a student, who scores a total of 27.5 marks?
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
None of theabove
Answer: Option A
Text Explanation :
Maximum marks = 30
Marks scored if exactly 29 questions were answered correctly.
Case (1): One question is incorrectly answered.
Case (1a): A question of 1 mark is answered incorrectly.
Total marks = 30 – 1 – 0.25 = 28.75
Case (1b): A question of 2 marks is answered incorrectly.
Total marks = 30 – 2 – 0.33 = 27.67
Case (2): One question is not answered.
Case (2a): A question of 1 mark is not answered.
Total marks = 30 – 1 – 0.5 = 28.5
Case (2b): A question of 2 marks is not answered.
Total marks = 30 – 2 – 0.5 = 27.5
If these marks are arranged in descending order, (27.5) comes in the fifth place.
i.e., the rank of a student, who scores a total of 27.5 marks, would be 5.
Hence, option (a).
Workspace:
x, 17, 3x – y2 – 2, and 3x + y2 – 30, are four consecutive terms of an increasing arithmetic sequence. The sum of the four number is divisible by:
- (a)
2
- (b)
3
- (c)
5
- (d)
7
- (e)
11
Answer: Option A
Text Explanation :
Since it is an A.P.,
Therefore, 17 – x = 3x + y2 – 30 – 3x + y2 + 2
Or, x + 2y2 = 45 … (i)
Again, 17 − x = 3x − y2 − 2 −17
Or, 4x – y2 = 36 … (ii)
Solving equations (i) and (ii),
we get x = 13.
Sum of the A.P. = x + 17 + 3x – y2 – 2 + 3x + y2 – 30
= 7x – 15 = 7(13) – 15 = 76
Out of the given options, 76 is only divisible by 2.
Hence, option (a).
Workspace:
In quadrilateral PQRS, PQ = 5 units, QR = 17 units, RS = 5 units, and PS = 9 units. The length of the diagonal QS can be:
- (a)
> 10 and < 12
- (b)
> 12 and < 14
- (c)
> 14 and < 16
- (d)
> 16 and < 18
- (e)
cannot be determined
Answer: Option B
Text Explanation :
Let QS = x, we get the following figure
In any triangle, the sum of any two sides must be greater than the third side. Similarly, the difference between any two sides must be smaller than the third side. Hence,
In ∆QRS,
x + 5 > 17
⇒ x > 12 …(i)
In ∆PQS,
x < 9 + 5
⇒ x < 14 …(ii)
Combining (i) and (ii), we get
12 < x < 14
Hence, option (b).
Workspace:
Consider the formula,
- (a)
increases
- (b)
decreases
- (c)
increases and then decreases
- (d)
decreases and then increases
- (e)
cannot be determined
Answer: Option A
Text Explanation :
Where K1 and K2 are constants
⇒ 1/S decreases when ω increases.
⇒ S increases when ω increases.
Hence, option (a).
Workspace:
Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student can get an A grade in the course if the average of her scores is more than or equal to 90.Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for the last quiz and he realizes that he will score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. How many quizzes did Prof. Suman take?
- (a)
6
- (b)
7
- (c)
8
- (d)
9
- (e)
None of these
Answer: Option D
Text Explanation :
Total exams given = x
Score of previous exams = y
Hence, option (d).
Workspace:
A polynomial “ax3 + bx2 + cx + d” intersects x-axis at 1 and –1, and y-axis at 2. The value of b is:
- (a)
-2
- (b)
0
- (c)
1
- (d)
2
- (e)
Cannot be determined
Answer: Option A
Text Explanation :
ax3 + bx2 + cx + d intersects x axis at 1 & –1
∴ a + b + c + d = 0
–a + b – c + d = 0
∴ 2(b + d) = 0
∴ b + d = 0
ax3 + bx2 + cx + d intersects y axis at 2
0 + d = 2
∴ d = 2
From (1) & (2), b = –2
Hence, option (a).
Workspace:
The sum of the possible values of X in the equation |X + 7| + |X – 8| = 16 is:
- (a)
0
- (b)
1
- (c)
2
- (d)
3
- (e)
None of the above
Answer: Option B
Text Explanation :
|X + 7| + |X – 8| = 16
For X ≥ 8, |X + 7| = X + 7 and |X – 8| = X – 8
∴ |X – 7| + |X – 8| = X + 7 + X – 8 = 2X – 1
2X – 1 = 16
For -7 ≤ x < 8,
∴ |X + 7| = X + 7 & |X – 8| = 8 – X
∴ |X + 7| = |X – 8| = X + 7 + 8 – X = 15 ≠ RHS
0 ≤ x < 8 is not possible.
-7 ≤ x < 0
∴ |X + 7| = 7 + X & |X – 8| = -X + 8
16 = |X + 7| + |X – 8| = 7 + X – X + 8 = 15
∴ -7 ≤ x < 0 is not possible.
Now, x < -8 |X + 7| = - 7 – x |X – 8| = -X + 8
∴ |X + 7| + |X – 8| = - 7 – X – X + 8 = 1 – 2X = 16
⇒ 2X = -15 ⇒ X = -7.5
Hence, option (b).
Workspace:
There are two windows on the wall of a building that need repairs. A ladder 30 m long is placed against a wall such that it just reaches the first window which is 26 m high. The foot of the ladder is at point A. After the first window is fixed, the foot of the ladder is pushed backwards to point B so that the ladder can reach the second window. The angle made by the ladder with the ground is reduced by half, as a result of pushing the ladder. The distance between points A and B is
- (a)
< 9 m
- (b)
≥ 9 m and < 9.5 m
- (c)
≥ 9.5 m and < 10 m
- (d)
≥ 10 m and < 10.5 m
- (e)
≥ 10.5 m
Answer: Option E
Text Explanation :
The following figure represents the length and the position of the ladder,
By Pythagoras Theorem,
The approximate distance between A and B can be given as,
Hence, option (e)
Workspace:
Amitabh picks a random integer between 1 and 999, doubles it and gives the result to Sashi. Each time Sashi gets a number from Amitabh, he adds 50 to the number, and gives the result back to Amitabh, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let ‘x’ be the smallest initial number that results in a win for Amitabh. The sum of the digits of ‘x’ is:
- (a)
3
- (b)
5
- (c)
7
- (d)
9
- (e)
None of these
Answer: Option C
Text Explanation :
Let the smallest number be ‘X’
Amitabh has to win and X is the least possible number in the range 1 -999 ∴ step 4 has to be the last step.
⇒ 16X + 750 > 1000
The least possible value of X = 16
Sum of the digit = 1 + 6 = 7
Hence, option (c).
Workspace:
Consider four natural numbers: x, y, x + y, and x – y. Two statements are provided below:
- All four numbers are prime numbers.
- The arithmetic mean of the numbers is greater than 4.
Which of the following statements would be sufficient to determine the sum of the four numbers?
- (a)
Statement I.
- (b)
Statement II.
- (c)
Statement I and Statement II.
- (d)
Neither Statement I nor Statement II.
- (e)
Either Statement I or Statement II.
Answer: Option A
Text Explanation :
Considering Statement I:
Since this is the only possible solution, statement I is sufficient.
Considering Statement II:
Since no unique solution is possible, statement II is not sufficient.
Hence, option (a).
Workspace:
Triangle ABC is a right angled triangle. D and E are mid points of AB and BC respectively. Read the following statements.
- AE = 19
- CD = 22
- Angle B is a right angle.
Which of the following statements would be sufficient to determine the length of AC?
- (a)
Statement I and Statement II.
- (b)
Statement I and Statement III.
- (c)
Statement II and III.
- (d)
Statement III alone.
- (e)
All three statements.
Answer: Option E
Text Explanation :
Option 1: If we take statements I and II, we have 2 medians of a right angled triangle, without knowing which angle is a right angle.
Option 2: If we take statements I and III, we have 1 median of a right angled triangle, and B is the right angle.
Option 3: If we take only statements II and III, we have 1 medians of a right angled triangle, and B is the right angle.
Option 4: If we only know that B is a right angle, we cannot determine the length of AC.
Option 5: If we have all 3 statements, then we have 2 medians of a right angled triangle, and we know B is the right angle. Hence, we can find AC.
Hence, option (e).
Workspace:
There are two circles C1 and C2 of radii 3 and 8 units respectively. The common internal tangent, T, touches the circles at points P1 and P2 respectively. The line joining the centers of the circles intersects T at X. The distance of X from the center of the smaller circle is 5 units. What is the length of the line segment P1P2?
- (a)
≤ 13
- (b)
> 13 and ≤ 14
- (c)
> 14 and ≤ 15
- (d)
> 15 and ≤ 16
- (e)
> 16
Answer: Option C
Text Explanation :
Let A, B be the centre of the two circles with radius 3cm, 8cm respectively.
AX = 5cm, AP1 = 3cm
Using Pythagoras theorem, P1X = 4cm
Now ∆ A P1X ≈ ∆ B P2X
⇒ A P1/ B P2 = P1X/ P2X
⇒ 3/8 = 4/ P2X
P2X = 10.66
P1 P2 = P1X + P2X = 14.66
Hence, option (c).
Workspace:
The probability that a randomly chosen positive divisor of 1029 is an integer multiple of 1023 is: a2/b2, then ‘b – a’ would be:
- (a)
8
- (b)
15
- (c)
21
- (d)
23
- (e)
45
Answer: Option D
Text Explanation :
10 = 5 × 2
1029 = 529 × 229
∴ Number of divisors = (29 + 1)(29 + 1) = 30 × 30
We need to find all the divisors of K such that 1029
= K × 1023
K = 106 = 56 × 26
∴ Number of divisors = (6 + 1)(6 + 1) = 7 × 7
The probability that a randomly chosen positive divisor of 1029 is an integer multiple of 1023
∴ a = 7 and b = 30
∴ b – a = 30 – 7 = 23
Hence, option (d).
Workspace:
Circle C1 has a radius of 3 units. The line segment PQ is the only diameter of the circle which is parallel to the X axis. P and Q are points on curves given by the equations y = ax and y = 2ax respectively, where a < 1. The value of a is:
- (a)
- (b)
- (c)
- (d)
1 6
- (e)
None of the above
Answer: Option A
Text Explanation :
Points P and Q have the same Y coordinate but their X coordinates differ by 6 units (Since the diameter = 6 units)
We have two possible cases:
Case 1:
Case 2:
But it is given that a < 1
∴ Case 2 can be discarded
Hence, option (a).
Workspace:
There are two squares S1 and S2 with areas 8 and 9 units, respectively. S1 is inscribed within S2, with one corner of S1 on each side of S2. The corners of the smaller square divides the sides of the bigger square into two segments, one of length ‘a’ and the other of length ‘b’, where, b > a. A possible value of ‘b/a’, is:
- (a)
≥ 5 and < 8
- (b)
≥ 8 and < 11
- (c)
≥ 11 and < 14
- (d)
≥ 14 and < 17
- (e)
> 17
Answer: Option D
Text Explanation :
⇒ a = 0.18, b = (3 – a) = 2.82
∴ b/a = 15.66
Hence, option (d).
Workspace:
Diameter of the base of a water – filled inverted right circular cone is 26 cm. A cylindrical pipe, 5 mm in radius, is attached to the surface of the cone at a point. The perpendicular distance between the point and the base (the top) is 15 cm. The distance from the edge of the base to the point is 17 cm, along the surface. If water flows at the rate of 10 meters per minute through the pipe, how much time would elapse before water stops coming out of the pipe?
- (a)
< 4.5 minutes
- (b)
≥ 4.5 minutes but < 4.8 minutes
- (c)
≥ 4.8 minutes but < 5 minutes
- (d)
≥ 5 minutes but < 5.2 minutes
- (e)
≥ 5.2 minutes
Answer: Option D
Text Explanation :
Applying the concept of similarity,
Volume of water that overflows can be given as,
Since the radius of the hole is 5 mm i.e. 0.5cm,
The volume of water flowing in 1 minute,
(0.52 × 1000)π cm3
Hence, the required time can be given as,
Hence, option (d).
Workspace:
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