Asb Session 1

8/16/2019 Asb Session 1

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Copyright © Amity University

Analytical Skill Building Semester - V


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Chapter -1

Topic: Quantitative Reasoning

• Contents: – Number System

– Number Theory

– Percentage Method

– Profit and loss

– Time and Distance

– MCQ


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• Number theory is the branch of puremathematics concerned with the properties ofnumbers in general, and integers in particular,

as well as the wider classes of problems thatarise from their study.

• Number theory may be subdivided into severalfields, according to the methods used and the

type of questions investigated.

http://en.wikipedia.org/wiki/Pure_mathematicshttp://en.wikipedia.org/wiki/Pure_mathematicshttp://en.wikipedia.org/wiki/Numberhttp://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/Numberhttp://en.wikipedia.org/wiki/Pure_mathematicshttp://en.wikipedia.org/wiki/Pure_mathematics


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• Number line• In basic mathematics, a number line is a picture of a

straight line on which every point is assumed tocorrespond to a real number and every real number to a

point. Often the integers are shown as specially-markedpoints evenly spaced on the line. Although this imageonly shows the integers from −9 to 9, the line includes allreal numbers, continuing "forever" in each direction, andalso numbers not marked that are between the integers.

It is often used as an aid in teaching simple addition andsubtraction, especially involving negative numbers.


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7/76Copyright © Amity University

• Composite number• A composite number is a positive integer which has a

positive divisor other than one or itself. In other words acomposite number is any positive integer greater thanone that is not a prime number .

• So, if n > 0 is an integer and there are integers 1 < a, b <n such that n = a × b, then n is composite. By definition,every integer greater than one is either a prime number or a composite number. The number one is a unit – it isneither prime nor composite. For example, the integer 14

is a composite number because it can be factored as2 × 7. Likewise, the integers 2 and 3 are not compositenumbers because each of them can only be divided byone and itself.

http://en.wikipedia.org/wiki/Positive_integerhttp://en.wikipedia.org/wiki/Divisorhttp://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/1_(number)http://en.wikipedia.org/wiki/Prime_numberhttp://en.wikipedia.org/wiki/Prime_numberhttp://en.wikipedia.org/wiki/Unit_(ring_theory)http://en.wikipedia.org/wiki/Unit_(ring_theory)http://en.wikipedia.org/wiki/Prime_numberhttp://en.wikipedia.org/wiki/Prime_numberhttp://en.wikipedia.org/wiki/1_(number)http://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/Divisorhttp://en.wikipedia.org/wiki/Positive_integer


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• Integers

• Integers are like whole numbers, but they alsoinclude negative numbers ... but still no

fractions allowed!• So, integers can be negative {-1, -2,-3, -4, -5, …

}, positive {1, 2, 3, 4, 5, … }, or zero {0}

• We can put that all together like this:• Integers = { ..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...}



• Prime number

• A natural number

• 1, 2, 3, 4, 5, 6, ...

• is called a prime or a prime number if it isgreater than 1 and has exactly two divisors, 1and the number itself. Natural numbers greater

than 1 that are not prime are called composite.• Illustration showing that 11 is a prime number

while 12 is not

http://en.wikipedia.org/wiki/Natural_numberhttp://en.wikipedia.org/wiki/Divisorhttp://en.wikipedia.org/wiki/Composite_numberhttp://en.wikipedia.org/wiki/Composite_numberhttp://en.wikipedia.org/wiki/Divisorhttp://en.wikipedia.org/wiki/Natural_number


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• 5 is again prime: none of the numbers 2, 3, or 4 divide 5.Next, 6 is divisible by 2 or 3, since

• 6 = 2 · 3.• Hence, 6 is not prime. The image at the right illustrates

that 12 is not prime: 12 = 3 · 4. More generally, no evennumber bigger than 2 is prime: any such number n hasat least three distinct divisors, namely 1, 2, and n. Thisimplies that n is not prime. Accordingly, the term odd prime refers to any prime number greater than 2. In a

similar vein, all prime numbers bigger than 5, written inthe usual decimal system, end in 1, 3, 7 or 9, since evennumbers are multiples of 2 and numbers ending in 0 or 5are multiples of 5.

http://en.wikipedia.org/wiki/Even_numberhttp://en.wikipedia.org/wiki/Even_numberhttp://en.wikipedia.org/wiki/Decimalhttp://en.wikipedia.org/wiki/Decimalhttp://en.wikipedia.org/wiki/Even_numberhttp://en.wikipedia.org/wiki/Even_number


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• Unit fraction

• A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positiveinteger . A unit fraction is therefore thereciprocal of a positive integer, 1/n.

Examples are 1/1, 1/2, 1/3, 1/4 etc

http://en.wikipedia.org/wiki/Rational_numberhttp://en.wikipedia.org/wiki/Vulgar_fractionhttp://en.wikipedia.org/wiki/Numeratorhttp://en.wikipedia.org/wiki/1_(number)http://en.wikipedia.org/wiki/Denominatorhttp://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/Reciprocal_(mathematics)http://en.wikipedia.org/wiki/Reciprocal_(mathematics)http://en.wikipedia.org/wiki/Integerhttp://en.wikipedia.org/wiki/Denominatorhttp://en.wikipedia.org/wiki/1_(number)http://en.wikipedia.org/wiki/Numeratorhttp://en.wikipedia.org/wiki/Vulgar_fractionhttp://en.wikipedia.org/wiki/Rational_number


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• Q2 The least multiple of 7, which leaves aremainder of 4, when divided by 6, 9, 15 and 18is:

• A.74 B.94 C.184 D.364• Answer: Option D• Explanation:L.C.M. of 6, 9, 15 and 18 is 90.Let

required number be 90k + 4, which is multiple of

7.Least value of k for which (90k + 4) is divisibleby 7 is k = 4. Required number = (90 x 4) + 4 =364.


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• 5 (change of baseformula).

• The change of base formula is extra-usefulbecause calculators usually give onlyresults (generally speaking, log with nobase specified is taken as log base 10).


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• If log 27 = 1.431, then the value of log 9 is:• A.0.934 B.0.945 C.0.954 D.0.958

• Answer: Option C• Explanation:log 27 = 1.431• log (33 ) = 1.431• 3 log 3 = 1.431• log 3 = 0.477• log 9 = log(32 ) = 2 log 3 = (2 x 0.477) = 0.954.


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• The fundamental concept to remember when performingcalculations with percentages is that the percent symbolcan be treated as being equivalent to the pure numberconstant 1 / 100 = 0.01 , for example 35% of 300 can bewritten as (35/100) × 300 = 105.

• To find the percentage that a single unit represents outof a whole of N units, divide 100% by N. For instance, ifyou have 1250 apples, and you want to find out whatpercentage of these 1250 apples a single applerepresents, 100%/1250 = (100/1250)% provides the

answer of 0.08%. So, if you give away one apple, youhave given away 0.08% of the apples you had. Then, ifinstead you give away 100 apples, you have given away100 × 0.08% = 8% of your 1250 apples.


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• Here are other examples:• What is 200% of 30?• Answer:

• What is 13% of 98?

• Answer:• 3 )60% of all university students are female. There are 2400 female

students. How many students are in the university?• Answer: , therefore .


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4 There are 300 cats in the village, and 75 ofthem are black. What is the percentageof black cats in that village?

• Solution

• and therefore n% = 25%.


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• OPERATING EXPENSES: are fixedexpenses, such as rent, and utilities.

• OPERATING INCOME: is profit after

operating income.• EARNINGS BEFORE TAXES: is income

including other income and expenses, but

before taxes.• INCOME TAXES: are federal, state, andlocal taxes.


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• Several problem types based on these formulas arediscussed in the examples below

• Q1. If a merchant offers a discount of 40% on themarked price of his goods and thus ends up selling atcost price, what was the % mark up?

ANS: If the merchant offers a discount of 40% on themarked price, then the goods are sold at 60% of themarked price.

• The question further states that when the discountoffered is 40%, the merchant sells at cost price.

• Therefore, selling @ 40% discount = 60% of markedprice (M) = cost price (C)i.e., a mark up 66.66%


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• Q2. A shopkeeper bought an almirah from awholesale dealer for Rs 4500 andsold it for Rs 6000. Find his profit or losspercent.

Answer: Here C.P. of the almirah = Rs 4500S.P. of the almirah = Rs 6000Since S.P. > C.P., there is a profitProfit = S.P. – C.P.

= Rs 6000 – Rs 4500= Rs 1500Profit % = 1500/4500 = 33.33% .


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• Q3. A trader professes to sell his goods at a loss of 8%but weights 900 grams in place of a kg weight. Find hisreal loss or gain per cent ?

Answer: The trader professes to sell his goods at a loss

of 8%. Therefore, Selling Price = (100 - 8)% of CostPriceor SP = 0.92CPBut, when he uses weights that measure only 900 gramswhile he claims to measure 1 kg.Hence, CP of 900gms = 0.90 * Original CPSo, he is selling goods worth 0.90CP at 0.92CPTherefore, he makes a profit of 0.02 CP on his cost of0.9 CPprofit %= 2.22%


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• 1. A car travels at 30km/hr for the first 2 hrs & then 40km/h for the next 2hrs.Find the ratio of distance traveledS1/S2=V1/V2=3/4

• 2. Two cars leave simultaneously from points A & B (100km apart)& theymeet at a point 40 km from A. What is Va/Vb?

• T is constant so V1/V2=S1/S2=40/60=4/6

• 3. A train meets with an accidient and moves at 3/4th its original speed. Dueto this , it is 20 mins late. Find the original time for the journey beyond thepoint of accident?

• Method1 : Think about 2 diff. Situations , 1st with accident and another w/oaccident , then S is constant

• So V1/V2=T1/T2 =>V1/[3/4)*V2]=(T1+20)/T1• =>4/3=(T1+20)/T1 =>T1=60

• Method 2: Velocity decreases by 25% so time will increase by 33.3%• 33.3%=20 mins =>100%=60 mins • CONVERSION: 1km/hr=1000m/h=1000/3600m/sec=5/18m/sec


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Q3Excluding stoppages, the speed of a bus is 54kmph and including stoppages, it is 45 kmph.For how many minutes does the bus stop perhour?

• A.9 B.10 C.12 D.20• Answer: Option B• Explanation:

• Due to stoppages, it covers 9 km less.• Time taken to cover 9 km = 9/54x 60 min= 10

min.


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• Train Problems

The basic equation in train problem is the same S=VTThe following things need to be kept in mind whilesolving the train related problems.

• When the train is crossing a moving object , the speedhas to be taken as the relative speed of the train withrespect to the object.The distance to be covered when crossing an object,

whenever trains crosses an object will be equal to:Length of the train + Length of the object


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• 2. A man's speed with the current is 15 km/hrand the speed of the current is 2.5 km/hr. Theman's speed against the current is:

• A.8.5 km/hr B. 9 km/hr

• C.10 km/hr D. 12.5 km/hr • Answer: Option C• Explanation: • Man's rate in still water = (15 - 2.5) km/hr = 12.5

km/hr.• Man's rate against the current = (12.5 - 2.5)

km/hr = 10 km/hr.


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• Circular motion• The circular motion is rotation along a circle: a circular

path or a circular orbit. It can be uniform, that is, withconstant angular rate of rotation, or non-uniform, that is,

with a changing rate of rotation. The rotation around afixed axis of a three-dimensional body involves circularmotion of its parts. The equations describing circularmotion of an object do not take size or geometry intoaccount, rather, the motion of a point mass in a plane is

assumed. In practice, the center of mass of a body canbe considered to undergo circular motion.

http://en.wikipedia.org/wiki/Rotationhttp://en.wikipedia.org/wiki/Circlehttp://en.wikipedia.org/wiki/Orbithttp://en.wikipedia.org/wiki/Uniform_circular_motionhttp://en.wikipedia.org/wiki/Non-uniform_circular_motionhttp://en.wikipedia.org/wiki/Rotation_around_a_fixed_axishttp://en.wikipedia.org/wiki/Rotation_around_a_fixed_axishttp://en.wikipedia.org/wiki/Center_of_masshttp://en.wikipedia.org/wiki/Center_of_masshttp://en.wikipedia.org/wiki/Rotation_around_a_fixed_axishttp://en.wikipedia.org/wiki/Rotation_around_a_fixed_axishttp://en.wikipedia.org/wiki/Non-uniform_circular_motionhttp://en.wikipedia.org/wiki/Non-uniform_circular_motionhttp://en.wikipedia.org/wiki/Non-uniform_circular_motionhttp://en.wikipedia.org/wiki/Uniform_circular_motionhttp://en.wikipedia.org/wiki/Orbithttp://en.wikipedia.org/wiki/Circlehttp://en.wikipedia.org/wiki/Rotation


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• First meeting time = Circumference /Relative velocity

• First Meeting at starting point The firstmeeting at the starting point will occurafter a time that is obtained by the LCM ofthe times that each of the bodies takes the

complete one full round.


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• VERY SHORT ANSWER TYPE QUESTIONS :• Q.1 Give an example of motion in which

displacement and distances are not equal butspeed remains constant during the motion.

• Sol. Uniform circular motion.• Q.2 Can an object be accelerated without

speeding up or slowing down?• Sol. Yes. When an object is in uniform

circular motion, its speed remains constantbut there is an acceleration towards centre ofcircular path.


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• Q.3 Write S.I. unit and dimensions of angularvelocity.

• Sol. Radian per second (rad s.1) Dimensions

[T.1]• Q.4 What do you mean by uniform angular

velocity.

• Sol. If a body covers equal angles intervals oftime then the angular velocity of the body iscalled uniform angular velocity.


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• Q.5 Define average angular acceleration.• Sol. Average angular acceleration is defined as the ratio of the

change in angular velocity to the time taken by the body toundergo this change

• Q.6 Write the S.I. unit and dimensions of angular acceleration.• Sol. Unit radian sec.2. Dimensions [M0L0T.2]

• Q.7 What is the direction of velocity vector of a particle inuniform circular motion.

• Sol. The direction of velocity vector of a particle in uniform

circular motion is along the tangent to the circular path.


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• Q.8 Uniform circular motion is an example ofaccelerated motion. Give reason.

• Sol. Yes. uniform circular motion is acceleratedmotion. In this motion the direction of velocitychanges continuously.

• Q.9 What provides the centripetal force to satelliterevolving around the earth?

• Sol. The centripetal force of the satellite revolving aroundthe earth is provided by the gravitational force betweenthe earth and satellite.

• Q.10 Write the relation between, the linear velocity (v),angular velocity (w) and radius of the circular path (r).

• Sol. v = rw


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• Clock •

For clock problems consider the clock as acircular track of 60km.

• Min. hand moves at the speed of 60km/hr (thinkmin. hand as a point on the track) and hour handmoves at 5km/hr and second hand at the speedof 3600 km/hr.Relative speed between hr hand and mins hand= 55


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• Q1 An accurate clock shows 8 o'clock in themorning. Through how may degrees will thehour hand rotate when the clock shows 2 o'clockin the afternoon?

• A. 144º B. 150º• C. 168º D. 180º• Answer: Option D• Explanation:

• Angle traced by the hour hand in 6 hours =360/12 x 6 º = 180º.•


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• Multiple Choice questions• Q1….. are like whole numbers, but they also include negative

numbers

• A Integers

• B Prime• C whole numbers

• D None

• Q2 Which number is prime

• A 6 B 5 C 9 D 12


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• Q3 LCM stands for• A Largest common multiple • B Least common multiple

• C Lower common mode • D Least common mode• Q4 If either a or b is 0, LCM(a, b) is

defined to be• A zero B One C half D one and half


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Q5 Find the highest common factor of 14 and 28.

A 21 B 28 C 14 D 12

Q6 Natural numbers are the set of all non

fractional numbers from A 1 to infinity.

B 0 to infinity

C 0 to 1D Greater than 1


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• Q7 Number theory is the branch of pure mathematics concernedwith the properties of numbers in general, and ………in particular .

• A integers

• B Prime

• C Natural

• D Whole

Q8 A composite number is a ……..which has a positivedivisor other than one or itself

A positive integer B negative integerC integer D prime integer


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• 9 a rational number is any number that can be expressed as thequotient or fraction a/b of two integers, with the denominator b

A equal to zero

B not equal to zero

C not equal to one

D equal to one

10 Distance is ……..proportional to Velocitywhen time is constant

A indirectly B inversely C directly D none


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• Key• 1-A• 2- B• 3- B

• 4-A• 5-C• 6-A• 7-a

• 8-a• 9-b• 10-C


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Thank You

Documents

Asb Session 1