3 Focus Areas For Quant

You don’t need to pick up textbooks replete with advanced engineering mathematics to solve Quantitative Aptitude in the computer-based CAT. The QA section in the CAT is nothing but the theory learnt in school up to Class XII. You need to focus on as many practice tests as you can. This is the most important focus area while preparing for QA in CBT-CAT. Every mock test should be taken as seriously as the actual CAT. Pace yourself and time yourself for every test that you take. For those who are in college, you can take tests immediately since you are familiar with concepts and theory. However, working professionals tend to fall out of touch with academics. So they need to brush up theorems and formulae to prepare for CBT-CAT.

You can begin with topic tests when you initially start to prepare yourself. Once you are confident, you can begin to attempt full-length QA tests. Make sure to analyse your performance after every test that you attempt. Mock tests are a great source of feedback mechanism. You can use practice tests to judiciously select questions. It is okay if you are particularly weak in one area but you will still be at a risk. Hence it makes sense to improvise your performance in every area so that you can answer every question effortlessly.

CBT-CAT Quantitative Aptitude can be classified into 3 categories:

Algebra and Number Theory- This section provides maximum questions in any CAT QA section- about 60%. You need to focus on Probability, Card problems, Permutations and Combinations, Progressions, Quadratic and Linear/Simultaneous equations, Logarithms, Functions, etc. The most important being Number Theory. These problems are simple in nature. You can master a few tricks from textbooks to solve them. You can refer to Higher Algebra by Hall and Knight for the same.

Geometry, coordinate geometry and mensuration- Every CAT paper has about 4 questions on mensuration, a few questions on coordinate geometry, but Geometry is given maximum weightage. The topics that you need to cover include basic theorems involving circles, triangles and parallel lines. A general question asked in CAT is to find the length of certain sides or the value of certain angles. So topics like similarity of triangles and congruency need to be covered. Refrain from solving IIT-JEE level questions in coordinate geometry. A school-level textbook is enough to prepare for mensuration.

Arithmetic and miscellaneous- For Arithmetic, you need to cover problems on Time, Speed and Distance and Set Theory. An NCERT textbook will serve the purpose. Miscellaneous problems don’t fall under any specific category. They basically test your mathematical aptitude.

Preparing For CAT Quant

Management entrance tests comprise of four areas- Maths, English, Comprehension and Reasoning. Some institutes also have an additional section on business judgment or general knowledge. Previously there were four sections in CAT, now there are three. Preparing for CAT in the right way is very important to crack the test.

Start to prepare as early as you can. This way you get accustomed to the kind of questions asked. Spend time developing your vocabulary and reading skills. An excellent way to do this is to read the newspaper daily. Not only will it update you on the latest events and developments, it will also help you learn new words and improve your language skills.

The preparation for CAT can be divided into three stages. Students should start with the ‘basics’ first. Thoroughly browse each chapter and revise your class 10 lessons. Get an idea of figures. Attempt to solve the tricky, puzzling questions in question banks. Working on exercises on each topic and revising your basics usually takes about three months.

Once you are done with basics, you should undertake sectional tests. These tests comprise of 30-40 sums each, and it contains different sums from each chapter. The key is to time yourself and finish one section in about 40 minutes. Initially, you may get a low score, but you can gradually aim for a higher score.

At this stage, it is crucial to practice CAT-type questions. These questions are not available in guidebooks or textbooks. You can gather information and collect it from various sources. There are question banks available in the market. Some specialized institutes offer notes to prepare for CAT. So you can subscribe to such notes and prepare for CAT.

After you clear the first two stages, you can attempt doing mock-CATs. Mock CAT is a similar CAT-like paper that gives you an insight into the exam and trains you efficiently. It is a similar test modeled on CAT. Allot two hours to yourself to finish the entire paper. Check your full test score as well as your sectional score. This will give you a general idea about CAT and where you stand in your preparation.

CAT is difficult to crack. However the right attitude, hard work, determination and persistence is the key to cracking CAT. Cracking CAT is not rocket science. Regular practice and a positive attitude definitely helps.

5 Ways To Score Higher In Quant and DI

Quantitative ability and data interpretation are often considered the more difficult and challenging sections of the CAT. If you want to ace the test, these logic-based sections need to be given attention. Practicing and evaluating your ability to calculate faster online will increase speed and help achieve accuracy, if done everyday. Quantitative reasoning ability needs to be honed through regular online testing and analysis. Comparative intelligence sets you apart from the lakhs of students answering the CAT. Students are registering themselves for online test programmes to stay ahead.CAT 2009 2010

Here are five tips to help you ace the CAT:
1. What methods should I use to answer?
• Eliminate the options as you go along. This method known as the ‘Elimination Mehtod’ is usually the fastest way to reach the answer.
• Substitute some values in the question like you do with algebra problems.
• Go from the question to the answer. This method known as ‘Direct Method’ is least effective.
• Use the Direct method and Elimination method in tandom.

2. How do I select the right questions to answer?
Sometimes, it is even possible to skip almost half of the questions and still crack the test. Select questions with easy statements and dissimilar answer choices. This skill can be developed by analyzing Mock CATs regularly. If CAT 09 is linear and not adaptive, then this skill of selecting the right questions to answer is vital.

3. How do I improve my calculating speed?
Students spend an average of 20 per cent of their time on calculation in the Quant and DI section and the rest is spent on understanding the concept of the question. Spend at least 15 minutes everyday daily on training your addition, subtraction, multiplication and division skills. Refresh your ability to recall tables, squares and cubes up to 30, and multiplication between common numbers like 13 x 12, 18 x 24.

4. How do I ensure accuracy of my answers?
The time provided does not permit students to solve all questions. Ensure that you solve a reasonable number of questions at a high accuracy rate rather than a large number of the questions in your paper at a low rate of accuracy. Make the most of the time allotted.

5. How do I navigate tricky questions?
Questions can be crafted in such a way that they use language to trick the reader. Read carefully and try to understand what it is really being asked.

6. How do I gear-up for a change in the Quant-DI Pattern?
Be prepared for any change in the pattern of CAT 2009 . There maybe five or four options per question; it maybe be 1/4th negative or 1/3rd negative for a wrong answer; the number of questions may decrease or increase; the sections may have sub-sections and further sub-sub-sections with varying marks per question. All these probabilities should be addressed and discussed with your online trainers or teacher so that you are equipped to deal with them.

Quant - Number System Ex. 1

Exercise1 + Answers + Explanations

Question:
The sum of the first 100 numbers, 1 to 100 is divisible by

(1) 2, 4 and 8
(2) 2 and 4
(3) 2 only
(4) None of these

Correct Answer - (3)

Explanation:
The sum of the first 100 natural numbers is given by (n(n + 1))/2 = (100(101))/2 = 50(101).
101 is an odd number and 50 is divisible by 2. Hence, 50 (101) will be divisible by 2.

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Question:
What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?

(1) 176
(2) 187
(3) 54043
(4) 748

Correct Answer - (2)

Explanation:
The marbles used to tile the floor are square marbles. Therefore, the length of the marble = width of the marble.As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74. 5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.
The HCF of 578 and 374 = 34.

Hence, the side of the square is 34.

The number of such square marbles required = = 187 marbles.

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What is the remainder when 9^1 + 9^2 + 9^3 + …. + 9^8 is divided by 6?

(1) 3
(2) 2
(3) 0
(4) 5

Correct Answer - (3)

Explanation:
6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave a remainder of 0. Or in other words it is perfectly divisible by 6.
On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3. Therefore, when each of the 8 powers of 9 listed above are divided by 6, each of them will leave a remainder of 3.

The total value of the remainder = 3 + 3 + …. + 3 (8 remainders) = 24. 24 is divisible by 6. Hence, it will leave no remainder.

Hence the final remainder when the expression 9^1 + 9^2 + 9^3 + ….. + 9^8 is divided by 6 will be equal to ‘0′.

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Question:
What is the reminder when 91 + 92 + 93 + …… + 99 is divided by 6?

(1) 0
(2) 3
(3) 4
(4) None of these

Correct Answer - (2)

Explanation:
Any number that is divisible by ‘6′ will be a number that is divisible by both ‘2′ and ‘3′. i.e. the number should an even number and should be divisible by three (sum of its digits should add to a multiple of ‘3′). Any number that is divisible by ‘9′ is also divisible by ‘3′ but unless it is an even number it will not be divisibly by ‘6′. In the above case, ‘9′ is an odd number. Any power of ‘9′ which is an odd number will be an odd number which is divisible by ‘9′.

Therefore, each of the terms 91, 92 etc are all divisibly by ‘9′ and hence by ‘3′ but are odd numbers.

Any multiple of ‘3′ which is odd when divided by ‘6′ will leave a reminder of ‘3′. For example 27 is a multiple of ‘3′ which is odd. 27/6 will leave a reminder of ‘3′. Or take 45 which again is a multiple of ‘3′ which is odd. 45/6 will also leave a reminder of ‘3′.

Each of the individual terms of the given expression 91 + 92 + 93 + …… + 99 when divided by 6 will leave a reminder of ‘3′. There are 9 such terms. The sum of all the reminders will therefore be equal to 9*3 = 27.

27/6 will leave a reminder of ‘3′.

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Question:
Find the value of 1.1! + 2.2! + 3.3! + ……+n.n!

(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
Correct Answer - (3)

Explanation:
1.1! can be written as (2!-1!), likewise 2.2! = (3!-2!) , 3.3! = (4!-3!) and n.n! = (n+1)!-n!.
Thus by summing it up we get the result as given below.
1.1! + 2.2! + 3.3! + ……+n.n! = (2!-1!) + (3!-2!) + (4!-3!) +……+ (n+1)!-n! = (n+1)!-1.

Quant - Number System Ex. 2

Exercise2 + Answers + Explanations

Question:
‘a’ and ‘b’ are the lengths of the base and height of a right angled triangle whose hypotenuse is ‘h’. If the values of ‘a’ and ‘b’ are positive integers, which of the following cannot be a value of the square of the hypotenuse?

(1) 13
(2) 23
(3) 37
(4) 41

Correct Answer - (2)

Explanation:
The value of the square of the hypotenuse = h2 = a2 + b2

As the problem states that ‘a’ and ‘b’ are positive integers, the values of a2 and b2 will have to be perfect squares. Hence we need to find out that value amongst the four answer choices which cannot be expressed as the sum of two perfect squares.

Choice 1 is 13. 13 = 9 + 4 = 32 + 22. Therefore, Choice 1 is not the answer as it is a possible value of h2

Choice 2 is 23. 23 cannot be expressed as the sum two numbers, each of which in turn happen to be perfect squares. Therefore, Choice 2 is the answer.

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Question:
Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

(1) 599
(2) 1021
(3) 263
(4) Cannot be determined

Correct Answer - (1)

Explanation:
Two numbers when divided by a common divisor, if they leave remainders of x and y and when their sum is divided by the same divisor leaves a remainder of z, the divisor is given by x + y - z.

In this case, x and y are 431 and 379 and z = 211. Hence the divisor is 431 + 379 - 211 = 599.

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Question:
What is the least number that should be multiplied to 100! to make it perfectly divisible by 350?

(1) 144
(2) 72
(3) 108
(4) 216

Correct Answer - (2)

Explanation:
100! has 348 as the greatest power of 3 that can divide it. Similarly, the greatest power of 2 that can divide 100! is 297. 297 = 448 * 21.
Therefore, the largest power of 12 that can divide 100! is 48.
Therefore, for 350 to be included in 100!, 100! needs to be multiplied by 32 * 23 = 9 * 8 = 72.

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Question:
A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11, viz 88?

(1) 3
(2) 21
(3) 59
(4) 68

Correct Answer - (3)

Explanation:
When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7.
Therefore, d1r2 + r1 = 8*7 + 3 = 59.

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Question:
What is the total number of different divisors including 1 and the number that can divide the number 6400?

(1) 24
(2) 27
(3) 27
(4) 68

Correct Answer - (2)

Explanation:
To find the number of divisors for a number, express then number as the product of the prime numbers like ax * by * cz .

In this case, 6400 can be expressed as 64*100 = 26 * 4 * 25 = 28 * 52.

Having done that, the way to find the number of divisors is by multiplying the indices of each of the prime numbers after incrementing the indices by 1.
i.e. the number of divisors = (x+1)(y+1)(z+1).
In this case, (8 + 1)(2 + 1) = 9*3 = 27.

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Quant - Number System Ex. 3

Exercise3 + Answers + Explanations

Question:
When 26854 and 27584 are divided by a certain two digit prime number, the remainder obtained is 47. Which of the following choices is a possible value of the divisor?

(1) 61
(2) 71
(3) 73
(4) 89

Correct Answer - (3)

Explanation:
When both 26854 and 27584 are divided by a divisor, the remainder obtained happens to be the same value 47.

Which means that the difference between 26854 and 27584 is a multiple of the divisor. Else the remainders would have been different.

=> 27584 - 26854 = 730

Now we need to find out a two digit prime number, whose multiple will be 730. 730 can be written as 73 * 10, where 73 is a prime number and greater than 47.
Hence, the two digit prime divisor is 73.

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Question:
How many times will the digit ‘0′ appear between 1 and 10,000?

(1) 4000
(2) 4003
(3) 2893
(4) 3892
Correct Answer - (3)

Explanation:

In two digit numbers, ‘0′ appears 9 times.
In 3-digit numbers, either 1 ‘0′ or 2 ‘0s’ can appear.
When one ‘0′ appears: The first digit of a 3-digit number can be any of the 9 digits other than ‘0′. The second digit can be any of the ‘9′ digits and the third digit be ‘0′.Hence there will 9*9 = 81 such combinations. The ‘0′ can either be the 2nd digit or the third digit, hence there are 2 arrangements possible.Therefore, there are a total of 81 * 2 = 162 three digit numbers with ‘0′ as one of the digits.

When two ‘0s’ appear: The first digit of the number can be any of the 9 digits. The second and third digit are 0s.
Hence there are 9 such numbers and 0 appears 9*2 = 18 times. Hence, ‘0′ appears 18 + 162 = 180 times in three digit numbers. In 4-digit numbers, 0 might be either 1 or 2 or 3 of the digits.

When one ‘0′ appears: The remaining three digits can be any of the 9 digits.i.e. 93 combinations = 729. The ‘0′ can either be the second or third or fourth digit. Hence 3*729 numbers = 2187.

When two 0s appear: The remaining two digits can be any of the 9 digits.Hence 92 = 81 possible combinations. The two 0s can be in two of the three positions = 3*81 = 243 such numbers and 2*243 = 486 number of times 0 will be appear.

When three 0s appear: There are 9 such numbers and 0 appears thrice in each. Hence 9*3 = 27 times.
Hence the number of 0s in 4-digit numbers = 2187 + 486 + 27 = 2700 times.
There is one 5-digit number - 10,000 and 0 appears four times in this number.
Hence the total number of times 0 appears between 1 and 10,000 = 9 + 180 + 2700 + 4 = 2893.

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Question:
What number should be subtracted from x^3 + 4x^2 – 7x + 12 if it is to be perfectly divisible by x + 3?

(1) 42
(2) 39
(3) 13
(4) None of these
Correct Answer - (1)

Explanation:
According to remainder theorem when, then the remainder is f(-a). In this case, as x + 3 divides x3 + 4×2 – 7x + 12 – k perfectly (k being the number to be subtracted), the remainder is 0 when the value of x is substituted by –3.

=> (-3)3 + 4(-3)2 – 7(-3) + 12 – k = 0
=> -27 + 36 + 21 + 12 = k
=> k = 42

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Question:
What is the value of M and N respectively? If M39048458N is divisible by 8 and 11; Where M and N are single digit integers?

(1) 7, 8
(2) 8, 6
(3) 6, 4
(4) 5, 4

Correct Answer - (3)

Explanation:
If the last three digits of a number is divisible by 8, then the number is divisible by 8 (test of divisibility by 8).
Here, last three digits 58N is divisible by 8 if N = 4. (Since 584 is divisible by 8.)

For divisibility by 11. If the digits at odd and even places of a given number are equal or differ by a number divisible by 11, then the given number is divisible by 11.

Therefore, (M+9+4+4+8)-(3+0+8+5+N)=(M+5) should be divisible by 11 => when M = 6.

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Question:
How many zeros contained in 100!?

(1) 100
(2) 24
(3) 97
(4) Cannot be determined

Correct Answer - (2)

Explanation:
If the number N = 2×1.3×2.5×3 ….. pxq. It is clear that each pair of prime factors 2 and 5 will generate one zero in the number N, because 10 = 2X5.

For finding out the number of zeros in the number it is sufficient to find out how many two’s and five’s appear in the expansion of each product N.

No. of five’s = No. of two’s = Hence it is evident that there are only 24 pairs of primes 2 and 5 and therefore 100! ends in 24 zeros.

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Quant - Number System Ex. 4

Exercise4 + Answers + Explanation

Question:
When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?

(1) 11
(2) 17
(3) 13
(4) 23

Correct Answer - (3)

Explanation:
Let the divisor be d.
When 242 is divided by the divisor, let the quotient be ‘x’ and we know that the remainder is 8. Therefore, 242 = xd + 8

Similarly, let y be the quotient when 698 is divided by d.
Then, 698 = yd + 9.

242 + 698 = 940 = xd + yd + 8 + 9
940 = xd + yd + 17
As xd and yd are divisible by d, the remainder when 940 is divided by d should have been 17.
However, as the question states that the remainder is 4, it would be possible only when leaves a remainder of 4.
If the remainder obtained is 4 when 17 is divided by d, then d has to be 13.

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Question:
What is the total number of different divisors of the number 7200?

(1) 20
(2) 4
(3) 54
(4) 32

Correct Answer - (3)

Explanation:
Express 7200 as a product of prime numbers. We get 7200 = 25 * 32 * 52.
The number of different divisors of 7200 including 1 and 7200 are (5 + 1)(2 + 1)(2 + 1) = 6 * 3 * 3 = 54

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Question:
When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

(1) 10
(2) 7
(3) 192
(4) None of these

Correct Answer - (2)

Explanation:
Let the number be ‘a’.
When ‘a’ is divided by 36, let the quotient be ‘q’ and we know the remainder is 19
=> and remainder is 19
=> a = 36q + 19

now, a is divided by 12

=> 36q is perfectly divided by 12
Therefore, remainder = 7

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Question:
A person starts multiplying consecutive positive integers from 20. How many numbers should he multiply before the will have result that will end with 3 zeroes?

(1) 11
(2) 10
(3) 6
(4) 5

Correct Answer - (3)

Explanation:
A number will end in 3 zeroes when it is multiplied by 3 10s. To get a 10, one needs a 5 and a 2.
Therefore, this person should multiply till he encounters three 5s and three 2s.
20 has one 5 (5 * 4) and 25 has two 5s (5 * 5).
20 has two 2s (5 * 2 * 2) and 22 has one 2 (11 * 2).

Therefore, he has to multiply till 25 to get three 5s and three 2s, that will make three 10s. So, he has to multiply from 20 to 25 i.e. 6 numbers.

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Question:
How many four digit numbers exist which can be formed by using the digits 2, 3, 5 and 7 once only such that they are divisible by 25?

(1) 4! - 3!
(2) 4
(3) 8
(4) 6

Correct Answer - (2)

Explanation:
The numbers that are divisible by 25 are the numbers whose last two digits are ‘00′, ‘25′, ‘50′ and ‘75′.
However as ‘0′ is not one of the four digits given in the problem, only numbers ending in ‘25′ and ‘75′ need to be considered.

Four digit number whose last two digits are ‘25′ using 3 and 7 as the other two digits are 3725 and 7325.
Similarly, four digit numbers whose last two digits are ‘75′ using 2 and 3 as the other two digits are 2375 and 3275.

Hence there are four 4-digit numbers that exist.

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Quant - Time, Speed & Distance

Questions along with respective explanations.

Q.1. Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260 kms?

(1) 60 km/hr (2) 90 km/hr (3) 120 km/hr (4) 135 km/hr

Correct Answer - (3)

Solution:
All three trains meet at the same time between Delhi and Mumbai. Which means Train A and Train B are at the same point at that time. This will happen when Train B is overtaking Train A.

Train A starts 3 hours before Train B. Therefore, by the time Train B leaves Mumbai, Train A has covered 3 * 60 = 180 kms.

The relative speed between Train A and Train B = 90 - 60 = 30 kmph. Therefore, Train B will overtake Train A in = 6 hours from the time Train B leaves Mumbai. That is at 3 A.M, Train B will overtake Train A. The point between Mumbai and Delhi at which Train B overtakes Train A will be 6*90=540 kms from Mumbai.

Train C will also be at that point at 3 A.M while Train B is overtaking Train A. And Train C would have travelled 1260-540 = 720 kms in these 6 hours. Therefore, the speed of Train C = 120 km/hr.

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Q.2 A train travels at an average speed of 90 km/hr without any stoppages. However, its average speed decrease to 60km/hr on account of stoppages. On an average, how many minutes per hour does the train stop?

(1) 12 minutes (2) 18 minutes (3) 24 minutes (4) 20 minutes

Correct Answer - (4)

Solution:
If it travelled at 90 km / hr, it would have crossed 90 kms in an hour. However, it covered only 60 kms due to stoppages.

The distance it covered decreased by 1/3 or it covered only 2/3rd of the distance that it can cover for which the traveling time would have been 2/3rd of an hour. The remaining 1/3rd of an hour was spent in stoppages. Therefore, the train stops on an average for 20 minutes every hour.
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Q.3 A man goes from city A to city B situated 60 kms apart by a boat. His onward journey was with the stream while the return journey was an upstream journey. It took him four and half hours to complete the round trip. If the speed of the stream is 10 km/hr, how long did it take him to complete the onward journey?
(1) 3 hours (2) 3.5 hours (3) 2.25 hours (4) 1.5 hours

Correct Answer - (4)

Solution:
The average speed for the round trip = km/hr
Let the speed during the onward journey be ‘D’ km/hr. Let the speed of the boat in still water be ‘B’ km/hr.
Therefore, D = B + S => D = B + 10 (As the speed of the stream is 10 km/hr).
Let the speed during the return journey be ‘U’ km/hr.
Therefore, U = B - S = B - 10

As the distance between A and B is the same as the distance between B and A, the average speed is given by the formula =
=> => 3B2 - 300 = 80B.
=> 3B2 - 80B - 300 = 0 => 3B2 - 90B + 10B - 300 = 0
=> 3B(B - 30)+10(B - 30) = 0
=> (B-30)(3B+10) = 0
=> B = 30 or B = -10/3

As speed is a positive quantity, B = 30.
Therefore, D = 30 + 10 = 40 km/hr and U = 30 - 10 = 20 km/hr.

His onward journey was done at a speed of 40 km/hr. The distance covered was 60 kms.
Therefore, the time taken for the onward journey = 1.5 hours

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Q. 4 The speed of a motor boat itself is 20 km/h and the rate of flow of the river is 4 km/h. Moving with the stream the boat went 120 km. What distance will the boat cover during the same time going against the stream?
(1) 80 km (2) 180 km (3) 60 km (4) 100 km

Correct Answer - (1)

Solution:
Let the distance to be covered by the boat when it is travelling against the stream be x.
The boat goes down the river at a speed of 20 + 4 = 24 km/h and up the river at a speed of 20 – 4 = 16 km/h.
Since the time taken is same 120/24 = x/16
Therefore, x = 80 km.

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Q. 5 A passenger train covers the distance between stations X and Y, 50 minutes faster than a goods train. Find this distance if the average speed of the passenger train is 60 kmph and that of goods train is 20 kmph.
(1) 20 kms (2) 25 kms (3) 45 kms (4) 40 kms
Correct Answer - (2)

Solution:
Let ‘d’ be the distance between the stations X and Y.
Time taken by the passenger train to cover the distance ‘d’ = d/60 hour
Time taken by the goods train to cover the distance ‘d’ = d/20 hour
Time difference between these two trains is given by 50 minutes or 50/60 hour

i.e., d/20 –d/60 = 50/60

d = 25kms.