Maths - Lesson 2
P% Increase
(1 + P%)
Increasing Numerator , Decreasing Denominator
*Resultant will be more* Ex 3/8 < 5/6
Quadratic Equation
ax² + bx + c = 0 x₁ = [-b + √(b² - 4ac) ] / 2a x₂ = [-b - √(b² - 4ac) ] / 2a x₁ and x₂ are called Roots of quadratic Equations
Q. The product of the roots of the equation mx² + 6x + (2m -1)= 0 is -1 . Then m is
c/a = -1 -> ( 2m - 1) / m = -1 -> 3m = 1 -> m = 1/3
Q. $5783 is divided among Sherry , Berry and Cherry in such a way that if $28 , $37 and $18 be deducted from their respective shares , they have money in 4:6:9 ratio . Find Sherry's Share .
Assume that three values are 4x , 6x and 9x then (4x+ $28) + (6x+$37) + (9x + $18 ) = $5783 19x + $83 = $5783 19x = $5783 - $83 19x = $5700 x = $300 Sherrie's Share 4*$300 + $28 = $1228
STD Application - Train - Stationary Object
Case I - Train Crossing a stationary object without Length Speed of Train * Time = Length of Train Case II - Train Crossing a Stationary object with Length L Speed of Train * Time = (Length of Train + Length of Object )
Q
Firs Quartile is 25% , so here 25% of 40 = 10 Students Hence 10th Student score comes in First Quartile which is 1500 - 1600
Vieta's Formula
If ax² + bx + c = 0 has roots as x₁ and x₂ then x₁ + x₂ = -b/a x₁x₂ = c/a The equation can be written as x² - (x₁ +x₂)x + x₁x₂ = 0
B. J. Upton of the Tampa Bay Rays hit 78 RBIs in 2012; this is the 90th percentile value on this chart. How many players hit between 56 and 78 RBIs? 14 22 28 34 42
Q₃ = 56 = 75% Q₄ = 139 TBR hit 78 which is 90% So 90 -75 = 15% Players scored between 56 and 78 - i.e 15% of 280 = 42
Depreciation of Value
V₁ = V₀ ( 1 - r/100)∧t
Ranges where the Ends are Excluded
shown as (1, 2 ) where 1 and 2 are excluded
Attached
Ans D
Statistics
1. Range = Max - Min 2. Standard Deviation is the difference of each point on list from Median and it cant be Negative as its distance. 3. If all the numbers on a list is Identical then the Standard Deviation(SD) = 0 4. If all the numbers on a list are the same distance from the mean , SD = that distance . 5. Lots of points close to the mean -> Small SD; Lots of points far from the mean -> Large SD 6. List ± K doesnt change SD 7. List * K -> K*(SD
Attached
Ans 98%
Q
Ans B ( I and II Only)
Q. The cost of Manufacture of an article is made up of 4 components A , B , C , D which have a ratio of 3:4:5:6 resp . If there are respective changes in the cost of +10% , -20% , -30% and +40% then what would be the percentage change in cost.
Assume Cost of each component is 30 , 40 , 50 and 60 . Now as per change its 33 , 32, 35 and 84 Hence % Change in cost = (3-8-15+24)/180 = 4/180 = ~2.4%
Q. What is the sum of all the multiples of 20 from 160 to 840 inclusive
First Number = 160/20 = 8 Last Number = 840/20 = 42 Total Number of terms ( Incl. of first and Last) = (42 - 8) + 1 = 34 + 1 = 35 Terms Sum = (160 + 840) * 35/2 = 1000 * 17.5 = 17500
Changing Denominator , Same Numerator
Indirectly Proportional to Denominator a/b > a/c if c>a Ex: 4/5 > 4/7
Q
Lets Say the Rice production by Indochina is 100 then Vietnam will be 28 Now if Vietnam increase by 25% i.e 7 then Total Increase in Indochina will also be 7. So Indochina will increase by 7%
Q. A Camp has 30 girls whose heights have an Average of 130 cm and a standard deviation of 4 Cm . Suppose 4 more girls join the Camp . Which set of heights for these 4 girls would most Increase the standard deviation of all the girls at the camp? a) {126, 126, 134, 134} b) {127, 129 , 131 ,133} c) {129, 130 , 130 , 131} d) {140, 140 , 140 , 140}
Lets say SD = 4 cm Average = 130 cm Number of Girls = 30
Median
Note - Changing Highest or Lowest Numbers doesn't impact Median . Case I - When N is Odd ex - { 1, 2 , 3, 4, 5} Here median is 3 ( Middle Term) Case II - When N is Even ex - { 1, 2 , 3, 4, 5, 6} Here middle terms are {3,4} and hence median is (3 + 4)/2 = 3.5 ex - { 11, 22 , 31, 43, 54, 66} Here middle terms are {31,43} and hence median is (31 + 43)/2 = 37
Geometric Progression
The n-th term of a geometric sequence with initial value a and common ratio r is given by Nth term = ar^{n-1}. Sum = [a(1 - r^n)]/(1 - r) Sum for Infinite Terms = a /[1-r]
Series of Percentage Change
*Multiply the multiplier* An amount raised by 40% (1.4) and then decreased by 14% (0.86) , whats the difference at the end x * 1.4 * 0.86 = New Amount x * 1.204 = New Amount That means 1 + 0.204 i.e 20.4 % increase
Q. If the difference between CI and SI on a certain sum of money is Rs 72 at 12% per year for 2 years . Find the Amount
1st Year -> Simple Interest = I Compound Interest = I 2nd Year -> Simple Interest = 2I Compound Interest = 2I + Interest on I (Accrued on 1st Year) Hence after two years CI - SI = 72 2I + Interest on I -2I = 72 Interest on I = 72 12% of I = 72 I = 7200/12 = 600 Rs So Principal is 12% of Principal = 600 Principal = 5000Rs
Average Speed
2S₁S₂/(S₁ + S₂) Or Total Distance / Total Time i.e (D₁ + D₂) / (D₁/S₁ + D₂/S₂)
Compare fractions by Cross Multiplication
7/11 ?? 5/8 --> 7*8 ?? 11*5 --> 56 > 55 Hence 7/11 > 5/8
PCG - Application 5 - Effect of Change in Denominator
Calculate the percentage Change between Ratios A1 = 10/20 , A2 = 10/25 Calculate the Percentage change in the denominator in *Reverse Fashion* i.e The required percentage change in denominator should be from 25 to 20 (25-20)/25 i.e 20% Drop
Set Theory
Check Illustration 5 on Page 633 , Arun Sharma
Weighted Average - Approach 2 - Proportion Distance( Only involves two groups)
Group 1 : N₁ number of Elements , P₁ is the proportion of N₁ to whole group ( N₁ + N₂) and A₁ is the Average . Lets say D₁ is the distance from A₁ to Overall Average . Group 2 : N₂ number of Elements , P₂ is the proportion of N₂ to whole group ( N₁ + N₂) and A₂ is the Average . Lets Say D₂ is the distance from A₂ to Overall Average . P₁/P₂ = D₂/D₁ or P₁ α 1/ D₁ and P₂ α 1/D₂ Lets Assume that P₁ > P₂ then D₁ < D₂ i.e Overall Average will be closer to heavy weights and farther to the light weight .
Q. A can run around a circular track in 4 minutes . B Can do the same in 7 Minutes , if they start running together from the same starting point , when will they meet for the first time at the point diametrically opposite to the starting point .
Here Distance is constant. Hence Speed α 1/ Time Speed of A / Speed of B = Time of B / Time of A -> 7/4 If the ratio of speed is 7/4 then they will meet at 3 locations on the circle and if you divide circle in 3 parts it wont be diametrically opposite to the starting point . Hence they will never meet at this point .
Q. A train crosses a man traveling in another train in the opposite direction in 8 Sec . However the train requires 25 Second to cross same man if the trains are traveling in the same direction . If the length of the first train is 200 mts and second train is 160 mts . Find the speed of first train .
Here first train is referencing man and not the second train . hence we should ignore the Length given for second train . Lets Say S₁ = Speed of Train A S₂ = Speed of Train B (S₁ + S₂)*8 = 200 -> S₁ + S₂ = 25 - Eqn 1 (S₁ - S₂)*25 = 200 -> S₁ + S₂ = 8 - Eqn 2 Solving Eq 1 and 2 2S₁ = 33 -> S₁ = 16.5 m/s
Rounding
If the Number in tenth place is 0 - 4 then we *Round Down* . If the Number in tenth place is 5 - 9 then we *Round up* Trick - Look only for number at tenth place and follow above logic . The Subsequent decimal after tenth place is IRRELEVANT . Ex - 7.12 Rounded Off to nearest Integer is 7 7.499999 Rounded off to nearest Integer 7 7.5000001 Rounded off to nearest Integer 8 65536 Rounded to nearest Thousand Place is 66000 as 5 was present in the Hundredth place ( One place RIGHT to the Thousand Place ) 1492 Rounded to nearest Ten is 1490 13456 Rounded to nearest hundred is 13500
PCG - Application 3 - Product Constancy Application
Lets Say x = Price of Commodity in Rs/Kg y = Quantity xy = Expenditure If Price increases by say 25% , by how many percentage should Quantity decrease to maintain Same Expenditure . New Price = 1.25x To maintain same expenditure , Quantity should be multiplied by some number to get (1.25x)*(?y) = xy ? = 1/1.25 ? = 100/125 = 4/5 = 0.8 There Quantity should be multiplied by 0.8 or decrease by 20% to maintain same Expenditure .
Quartiles
Lets say List is {24, 25, 26, 27 , 30 , 32 , 40 , 44, 50 , 52, 55 , 57} Then Median will be = (32+ 40)/2 = 36 Median can also be called as Quartile 2. Lower List - L₁ = {24, 25, 26, 27 , 30 , 32 } Median of L₁ = (26+27)/2 = 53/2 = 26.5 which is also called as Quartile 1 . Upper List - L₁ = {40 , 44, 50 , 52, 55 , 57} Median of L₁ = (50+52)/2 = 102/2 = 51 which is also called as Quartile 3 . Hence Q₁ = 26.5 , Q₂ or Median = 36 , Q₃ = 51 and Mean = (24 + 25+ 26 + 27 + 30 +32 +40 +44 + 50 + 52 + 55 + 57)/2 =462/12 = 38.5
Normal Distribution
M = Median S = Standard Deviation M + S = 34% or S = 34% M + 2S = 34% + 13.5% M + 3S = 34% + 13.5% + 2.5% M - S = 34% M - 2S = 34% + 13.5% M - 3S = 34% + 13.5% + 2.5%
Boxplots
Minimum is the Start Q₁ = 25% Q₂/Median = 50% Q₃ = 75% Q₄ = Maximum These 5 Lines describe Box Plots. Interquartile Range - IQR = Q₃ - Q₁ i.e the Box where 50% People Land
Q. Car P & Q are approaching each other on the same highway . Car P is moving at 49 mph and Car Q is at 61 mph . At 2:00 PM they are approaching each other and 121 miles apart . Eventually they pass each other . At what clock time are they moving away from each other and 44 miles apart .
Relative Speed = 110 miles per hour Gap = 121 + 44 = 165 miles Time = Distance / Speed Time = 165/110 = 1.5 hours So Answer = 2 + 1.5 = 3:30 PM
Q. Four runners A , B , C and D are running around a circular track with speeds in the ratio 1:2:3:4 resp. They start together at the same point , What is the number of unique possible locations of meeting .
Relative Speed of A:B ->Speed of B - Speed of A = 2-1 = 1 Relative Speed of A:C ->Speed of C - Speed of A = 3-1 = 2 Relative Speed of A:D ->Speed of D - Speed of A = 4-1 = 3 Relative Speed of B:C ->Speed of C - Speed of B = 3-2 = 1 Relative Speed of B:D ->Speed of D - Speed of B = 4-2 = 2 Relative Speed of C:D ->Speed of C - Speed of D = 4-3 = 1 Unique possibility is 4
Speed Time and Distance
Speed = Distance/Time
STD Case II : When Time is Constant
Speed α Distance Scenario 1 - A Car travels at 30Km/hr for first 2 hours and then travels at 40 km/hr for next 2 hours . Find the ratio of the distance traveled. at Constant Time -> S₁ / S₂ = D₁ / D₂ -> D₁ / D₂ = 3/4 Scenario 2 - Two cars move simultaneously from Point A and Point B towards each other . The Speed of Car A is 20 m/s and 25 m/s resp . Find the meeting point if total distance is 900 Km. at Constant Time -> S₁ / S₂ = D₁ / D₂ -> D₁ / D₂ = 4/5 Hence D₁ = 400 Km and D₂ = 500 Km
Percentiles
The percentile of a given value is determined by the percentage of the values that are smaller than that value. For example, a test score that 160 out of 170 can be 78th percentile and score of 167 out of 170 can be 95th percentile . 1.The percentile of the lowest score is the 0th percentile. 2. The percentile of the highest score is 99th Percentile .( There's no such thing as 100th percentile)
Q: *Application for DI* - Pg 184 Arun Sharma Year 1: Scooters - 23.47% Total Sales - 17342.34 Cr Year 2: Scooters - 26.55% Total Sales - 19443.56 Cr Find the change in Scooter Percentage from Year 1 to Year 2 .
These can be approx to Year 1 : Scooters Sale - 23.47% of 17342.34 - 0.2347*17342.34 Cr -> 23 * 173 Year 2 :Scooters Sale-26.55% of 19443.56 - 0.2655*19443.56 Cr -> 26 * 194 Now 23 to 26 is 1.13% Rise and 173 ti 194 is 1.12% Rise Total Change - 1.13*1.12 = 1.2656 i.e 26.5% Increase Overall
Time Required to meet first time on Circular Track
Time required to meet first time = Track Length/ (Relative Speed) Case I - Same Direction - Circle : 200mts , A :2m/s , B : 5m/sec Then to meet first time. 200/(5-2) = 200/3 = 66.66 seconds Case II - Opposite Direction - Circle : 200mts , A :2m/s , B : 5m/sec Then to meet first time. 200/(5+2) = 200/7 = 28.57 seconds
STD Case I : When Speed is Constant
Time α Distance Scenario 1 - A Car moves for 2 hours at a speed of 25 Kmph and another car takes 3 hours at the same speed . So whats the distance covered by second car . At Constant Speed -> t₁ / t₂ = D₁/D₂ = 2/3 D₁ = 25*2 = 50 Km Hence D₂ = (3/2)*50 = 75 Km
Q
Total Income = 650 + 180 = 830 M Total Students = ~3000 Mean = 830 * 10⁶ / 3 * 10³ = ~276000 Closest option is E
Weighted Average - Approach 1 - Basic formula
Weighted Average = Sum total of all groups /(Total Number of Elements in all group together) Weighted Average =(n₁A₁ + n₂A₂)/(n₁ + n₂)
Decreasing Numerator , Decreasing Denominator *Subtraction - Repels *
a) If we *Subtract Same number to Numerator/Denominator then the result will move Away from 1* For Ex - 5/6 (0.83) is subtracted by 2 Numerator/Denominator 3/4 (0.75) then the result moves away from 2 . ii) If we *Subtract Different number to Numerator/Denominator then the result will move Away from that number* For Ex - 5/6 (0.83) is subtracted by 1/2(0.5) Numerator/Denominator 4/4 (1) then the result moves away from 2 .
Odd Function
f(x) = -f(-x)
Even Function
f(x) = f(-x)
Relation between Means
A * H = G² Where A is A.P Mean = (a + b) /2 H is Harmonic Mean = 2ab/(a+b) G is Geometric Mean = (ab)½
Correlation
Correlation is about the General pattern of all the points Together . +ive Correlation will go in upward direction but it may have points going up and down . The General pattern representing all dots together is Correlation and doesn't depend on movement from point to point. -ive Correlation will go in upward direction but it may have points going up and down . The General pattern representing all dots together is Correlation and doesn't depend on movement from point to point.
Dividendo - If a/b = c/d then (a-b)/b = (c-d)/d
True Ex - 10/2 = 50/10 then 8/2 = 40/10
Geometric Mean
G = (ab)½ where a and b are two quantities
Ranges where Ends are included
shown as [1 ,2 ] where 1 and 2 are included
Q. Solve the inequality 1/x < 1
(1/x) -1 < 0 (1-x)/x < 0 or (x -1 )/x > 0 This can only happen when both numerator and denominator are of same sign Case I - Both are positive x - 1 > 0 and x > 0 hence x > 1 Case II - Both are negative -(x-1)> 0 and -x > 0 -> (x-1)< 0 and x < 0 -> Hence x < -1 Answer : ( -∞ , -1) U (1 , ∞)
Harmonic Mean
H = 2ab /[a+b]
Componendo and Dividendo If a/b = c/d then (a + b) /( a - b) = ( c+ d) /( c -d )
True if 10/2 = 50/10 then 12/8 = 60/40
Conversion between kmph and m/s
1 km/h = 5/18 m/s 1 m/s = 18/5 km/h 1 miles/hour = 1.6 km/hr -> 1.6 * 5/18 m/s -> 4/9 m/s
Compound Interest
Amount = P( 1 + r/100)∧n where time = n years r = rate per annum P = Principal Case - When compound Interest is reckoned half yearly Amount = P( 1 + (r/2)/100)∧2n Case - When compound interest is reckoned quarterly Amount = P(1 + (r/4)/100)∧4n
Q. We have unlimited supplies of 20% HCL Solution and 50% HCL Solution .Now we combine x liters of first with y liters of second to produce 7 Liters of a 40% HCL Solution . What does X Equal.
Equation 1 - X + Y = 7 Equation 2 - 0.2X + 0.5Y = 2.8 Solving Eqn 1 and 2 03.X = 0.7 -> X = 7/3
PCG - Application 4 - Compare two numbers
Ex - B's Salary is 25% more than A . By what percent is A's Salary less than B's Salary Lets Say A's Salary is x then B's Salary is 1.25x % A's Salary less than B = (1.25x - 1x)/1.25 =20%
Minima of Q.E
If the quadratic is in the form y = ax² + bx + c and *a is Positive Integer* then k = c - b²/4a will be the minima of Q.E
Relative Speed -Two bodies moving in the opposite direction . Their relative speed will be S₁ + S₂.
In this case case of bodies moving to and fro between two points A and B starting from opposite ends of the path. Distance Covered at 1st meeting = D Distance Covered at 2nd meeting = D + 2D Distance Covered at 3rd meeting = D + 2*(3-1)D Distance Covered at 4th meeting = D + 2*(4-1)D Distance Covered at10th Meeting = D + 2*(10-1)D Distance Covered at nth Meeting = D +2D(n-1)
Q. Car X and Y are travelling from A to B on the same route at constant speed . Car X is initially behind Car Y , but car X's speed is 1.25 times car Y's speed . Car X passes Car Y at 1:30 PM . At 3:15 Car X reaches B and at that moment Car y is still 35 Miles away from B . What is the speed of Car X
Speed of Car X = 1.25 *Speed Of Car Y At 1:30 PM - Car X and Y are at start point Speed of Car X - Speed of Car Y = Gap / Time (1.25y - y) = 35/1.75 0.25y = 20 -> y = 80 miles/hour x = 1.25*80 = 100 miles/hour Another Way to Solve this - At 1:30 as Time is constant Speed of Car x / Speed of Car Y = Distance of Car X /Distance of Car Y 1.25Y/Y = D/(D - 35) D = 35*5 Total Time = 105 Minutes = 105/60 Hours Speed of Car X = (35*5/105)*60 = 100 Km/hr
Componendo - If a/b = c/d then (a+b)/b = (c+d)/d
True Ex - if 1/2 = 3/6 then 3/2 = 9/6
Average of AP or Any Equally Spaced Number
Consider the AP - 2, 6, 10 , 14, 18 , 22 Average is - (2+6+10+14+18+22)/6 = 12 You can also calculate average by adding Corresponding terms i.e 1st and 6th Term = (2 + 22)/2 = 12 i.e 2nd and 5th Term = (6+18)/2 = 12 i.e 3rd and 4th Term = (10 + 14) /2 = 12
Percentage Change Graphics - Application 1 - Successive Change
A's Salary is increase by 20%(1.2) and then decrease by 20%(0.8) , What is the net percentage change in A's Salary . 1.2*0.8*x -> 0.96x Hence A's Salary has gone down by 4%
Q. A certain sum of money amounts to Rs 704 in 2 years and Rs 800 in 5 years . Find the principal
Amount = 704 in 2 Years Amount = 800 in 5 years Interest earned in 3 years = 800 - 704 = 96 Rs Interest earned each year = 32 Rs Principal = 704 - 32 -32 = Rs 640
PCG - Application 6 - Calculate Ratio Changes
Calculate the percentage Change between Ratios A1 = 10/20 , A2 = 15/25 *Numerator Effect* - Calculate in the normal order . Change from 10 to 15 (15-10)/10 - 50% Increase - 1.5 *Denominator Effect* - Calculate in Reverse Order Change from 25 to 20 - 20% Decrease - 0.8 Hence total Change - 1.5*0.8 = 1.20 = 20% Increase
How Many times they will meet on Circular track
Calculate when they met first Time and then find who completes race first . Case I - Same Direction - Race: 1800 m, Circle : 200mts , A :2m/s , B : 5m/sec They will meet first time at 200/(5-2) = 200/3 = 66.66 seconds They will keep on meeting every 66.66 seconds until one completes the race . Looking at the speed obviously 5m/sec will win the race . So time required by B (5m/sec) to complete the race is 1800 /5 = 360 second Number of times they will meet = 360/66.66 = (360/200)*3 = 6.4 i.e 6 Times they will meet Case II- Opposite Direction Race: 1800 m, Circle : 200mts , A :2m/s , B : 5m/sec They will meet first time at 200/(5+2) = 200/7 = 28.57 seconds They will keep on meeting every 28.57 seconds until one completes the race . Looking at the speed obviously 5m/sec will win the race . So time required by B (5m/sec) to complete the race is 1800 /5 = 360 second Number of times they will meet = 360/28.57 =12 times they will meet
Q.At Didymus Corp , there are just two classes of Employees : Silver and Gold . The Average Salary of Gold Employees is $56000 higher then that of Silver Employees . If there are 120 Silver and 160 Gold Employees then the Average Salary of Company is how much higher than the Average Salary for the Silver Employees .
Difference of Average = $56000 Proportion of Silver = 120/280 = 3/7 Proportion of Gold = 160/280 = 4/7 Distance of Silver/Distance of Gold = 4/3 hence Divide the line in 8 parts , with each part weighting $8000 (56000/7) . Distance of Silver from Average = 4* One Part = 4*8000 = 32000
Changing Numerator , Same Denominator
Directly Proportional to Numerator a/b > c/b if a > c Ex : 2/3 < 4/3
PCG - Application 2- Product Change
Ex 1 :A product 10 * 10 * 10 becomes 11 * 12 * 13 10 to 11 -> 10% increase - 1.1 10 to 12 -> 20% Increase - 1.2 10 to 13 -> 30% Increase - 1.3 Total Change - 1.1 * 1.2 * 1.3 = 1.716 i.e 71.6% Increase
Maxima of Q.E
If the quadratic is in the form y = ax² + bx + c and *a is Negative Integer* then k = c - b²/4a will be the maxima of Q.E
Q. Due to a 25% price hike in the price of Rice , A person is able to purchase 20 kg less of rice for Rs 400 . Find the initial price .
Initial Expenditure = xy New Expenditure = (1.25x)*(0.8)y Now as per question - Person purchased 20Kg of Less rice . i.e 0.2y = 20 . Hence y =100 Kg We know that xy = 400 if y = 100 then x = 4 . So Price of rice before increase was 4 then it rose to 5( 25 % Increase)
Q. Two numbers are in the ratio P:Q . When 1 is added to both the numerator and denominator , the ratio gets changed to R/S . Again when 1 is added to both the Numerator and Denominator , it becomes 1/2. Find the sum of P and Q . a)3 b)4 c)5 d)6
Lets Say P + Q = 3 The possible values of P/Q are 1/2 or 2/1 , If we add 2 to numerator and denominator then the resultant is 3/5 , 4/2 which is not equal to 1/2 . Hence its not correct . Lets say P + Q = 4 The possible values of P/Q are 1/3 , 3/1 ( 2/2 is not possible ) . If we add 2 to numerator and denominator then the resultant is 3/5 , 5/3. Hence not correct . Let say P + Q = 5 The possible values of P/Q are 2/3 , 3/2 , 1/4 , 4/1 . If we add 2 to numerator and denominator then 4/6 , 5/4, 3/6 i.e 1/2 Hence this is the correct Answer .
Q. The ROI for first three years is 6% , for next 4 years 7% and for the period beyond 7 years is 7.5% annually . If a man lent out Rs 1200 for 11 years , find the total interest earned by him .
Total Interest = (1200*3*6/100) + (1200*4*7/100) + (1200*4*7.5/100) = 12(18 + 28 + 30) = 12 * 76 = 912
Q. In order to maximize its profits , AMS Corporate defines a function . Its unit sales price is $700 and the function representing the cost of production = 300 + 2p², where p is the total units produced or sold . Find the most profitable production level . Assume that everything is necessarily sold.
Total Selling Price = $700p Cost Price = 300 + 2p² Profit = Selling Price - Cost Price 700p - 2p² -300 = 0 -> -2p² + 700p - 300 = 0 -> p² -350p + 150 = 0 30775 To Calculate Maxima x = -b/2a = 350/2 = 175 is the Level y = c - b²/4a = 150 - 122500/4 = -30475 is the value at that level
Q. Find the Sum of an AP of 17 Terms , whose 3rd term is 8 and 8th term in 28 .
T₃ : a + (3-1)d = 8 -> a + 2d = 8 T₈ : a + (8-1)d = 28 -> a + 7d = 28 Solving above equation 5d = 20 -> d = 4 . Hence the common difference is 4. Now T₈ and T ₁₀ are corresponding term s. T₁₀ = 28 + 4 +4 = 36 Average is = (36 + 28 )/2 = 64/2 = 32 Hence Sum = 32*17 = 320 + 224 = 544
STD Application - Train - Moving Object
Case I - Train Crossing a Moving Object without Length In Opposite Direction : (Speed of Train + Speed of Object)*Time = Length of Train In Same Direction : (Speed of Train - Speed of Object)*Time = Length of Train Case II -Train Crossing a Moving Object with Length In Opposite Direction : (Speed of Train + Speed of Object)*Time = (Length of Train + Length of Object ) In Same Direction : (Speed of Train - Speed of Object)*Time = (Length of Train + Length of Object )
Q. A shopkeeper allows a rebate of 25% to the buyer . He sells only smuggled goods and as a bribe he pays 10% of the cost of the Article . If his cost price is $2500 then find what should be the marked price if he desires to make a profit of 9.09%.
Cost Price = $2500 Bribe + Cost Price = $250 + $2500 = $2750 Market Price = 0.75* Selling Price 0.75* SP = $2750 * 1.0909 SP = $4000
discriminant
D = b² - 4ac is called the "discriminant" If D < 0 then the quadratic equation has no real solutions (it has 2 complex solutions). If D = 0 then the quadratic equation has 1 solution that is given by the formula: x₁ = [-b + √(b² - 4ac) ] / 2a = -b/2a x₂ = [-b - √(b² - 4ac) ] / 2a = -b/2a If D > 0 then the quadratic equation has 2 distinct solutions x₁ = [-b + √(b² - 4ac) ] / 2a x₂ = [-b - √(b² - 4ac) ] / 2a
Increasing Numerator , Increasing Denominator *Multiplication*
Doesn't Change any thing 4/12 = 24/72
Q. On a test in a class of more than 40 students , the scores had mean = median = 81 . Two absent students then took the test ; they received grades of 83 and 47 . What are the new mean and median
Old Median = 81 New Median = Even after adding 2 Students the Median will not change . hence New Median = 81 . Please note that adding 1 will change the median as then it will be the middle number . Old Mean = 81 New Mean = Will be less than 81 as 47 is really low value . New Mean = Sum of 42 Numbers / 42 =[ Sum of 40 Numbers + (83+47) ] /42 =[ Old Mean * 40 + 130 ] /42 = 81*40 + 130 /42 = 80.23
Q. A runs at 3 m/s and B runs at 2 m/s , The race is of total 1800 mts and one round is 200 mts. Calculate if they run in same direction .
Same Direction :- A's Speed = 3m/s B's Speed = 2m/s Relative Speed = 3m/s - 2m/s = 1m/s Time required to meet at Starting Point = LCM of Time required by A and by B LCM of (200/3 , 200/2) = 200 seconds Distance Covered by A = 200*3m/s = 600m Distance Covered by B = 200*2m/s = 400m Time required to meet first time = Track Length/ (Relative Speed) = 200/1 = 200se Distance Covered by A = 200*3m/s = 600 Distance Covered by B = 200*2m/s = 400m Hence they are meeting first time at the start line .
Q. Three containers A , B and C are having mixtures of Milk and water in the ratio 1:5 , 3:5 and 5:7 resp. If the capacities of the containers are in the ration 5:4:5 , find the ration of milk to water if the mixtures of all the three containers are mixed together .
Total 14 Liters (5L+ 4L+ 5L) Milk in 14 L - ( (5/6)*1 +(4/8)*3 + (5/12)*5) Water in 14 L- ((5/6)*5 +(4/8)*5 + (5/12)*7) (5/6 + 3/2 + 25/12)/ (25/6 +20/8 +35/12) 0.833 + 1.5 + 2.08 / 4.166 + 2.5 + 2. 4.3/8.7 = 0.494
Arithmetic Progression
n = Total Number of Terms a = First Term Sum = [n(a+L)]/2 Nth Term = L = [a + (n -1)d] Sum = Total Number of Terms * Average of the AP
P% Decrease
(1 - P%)
Mean
Mean = ( Sum of N Entries) / N
Simple Interest
Simple Interest = ( P * R * T )/ 100 Amount = Principal + Simple Interest
Decreasing Numerator , Increasing Denominator
*Resultant will be Less* Ex 8/6 < 6/7
If 10 Persons can clean 10 Floors by 10 Mops in 10 Days . In how many days can 8 Person Clean 8 floors by 8 Mops .
10 Men in 10 Days - 100 Man Days are required to clean the 10 Floors . 1 Floor will require 10 Man Days . 8 Floor require 80 Man Days . 8 Person can clean in 10 Days .
Q. Three Quantities A, B , C are such that AB = KC where K is constant . When A is kept constant , B varies directly as C . When B is constant A varies directly with C and When C is constant A varies inversely with B . Initially A was at 5 and A:B:C was 1:3:5 Find the value of A when B equals 9 at C Constant .
AB = KC Hence 5B = KC A:B:C = 1:3:5 as A is 5 , B is 15 and C is 25 -> 5*15 = K*25 -> K = 3 When C is constant AB =KC -> A*9 = 3*25 A =8.33
Q. Two Swimmers with speed of 5 Laps/10 minutes and 4 laps/10 minutes are swimming . Where would they meet in 9th Lap.
As the time is constant Distance α Speed Speed of A = 5 Laps/10 min Speed of B = 4 Laps/10 min Hence Distance of A : Distance of B = 5/4 At 9th Lap - A would have covered 5 Laps and B would have covered 4 Laps and hence they would meet at start .
Relative Speed
Case I - Two Bodies moving in Opp Direction at Speed S₁ and S₂ , Relative Speed will be S₁ + S₂ Case II - Two Bodies moving in same Direction , Relative Speed |S₁ -S₂|
Where they will meet on Circular Track
Case I- Same Direction Race:1800 m, Circle:200m , A:2m/s , B:5m/s The ratio of Speed is 2:5 and as they are moving in Same Direction they will meet (5-2) , 3 location on circle i.e Divide circle in 3 equal parts and they will meet there . Each meeting point will be 200/3 = 66.66 seconds apart . First Location on Circle where they will meet : A will cover = 2 m/s* 66.66s = 132 mts ~ B will cover = 5 m/s* 66.66s = 333 mts Hence First Location is = 132mts i.e 132mts on Circle Second Location on Circle where they will meet : A will cover = 2 m/s* 66.66s * 2 = 264 mts ~ B will cover = 5 m/s* 66.66s * 2 = 666.66 mts Hence Second Location is = 200 + 64mts i.e 64mts on Circle Third Location on Circle where they will meet : A will cover = 2 m/s* 66.66s * 3= 399.96 mts ~ B will cover = 5 m/s* 66.66s * 3 = 999.99mts Hence Third Location is = ~400mts i.e 200mts on Circle Therefore , A and B will meet after every 66.66 Seconds at 132mts , 64 mts from the starting line and on the Starting line . These are 3 location where they keep on meeting until A finishes race. Case II- Opposite Direction Race:1800 m, Circle:200m , A:2m/s , B:5m/s The ratio of Speed is 2:5 hence they will meet (5+2) , 7 location on circle i.e Divide circle in 7 equal parts and they will meet there . Each meeting point will be 200/7 = 28.57 seconds apart . First Location on Circle where they will meet : A will cover = 2 m/s* 28.57s = ~57mts B will cover = 5 m/s* 28.57s = ~150mts Hence First Location is = ~57mts Second Location on Circle where they will meet : A will cover = 2 m/s* 28.57s * 2 = 114mts ~ B will cover = 5 m/s* 28.57s * 2 = 285mts Hence Second Location is = 114mts Third Location on Circle where they will meet : A will cover = 2 m/s* 28.57*3= 171.42 mts B will cover = 5 m/s* 28.57*3 = ~428mts Hence Third Location is = ~171.42mts Therefore , A and B will meet after every 28.57 seconds at 57mts ,114mts , 171.42mts , 28mts , 85mts , 142mts from from the starting line and on the Starting line . These are 7 location where they keep on meeting until A finishes race.
Q. A certain sum of money trebles in 8 years , In how many years it will be 5 times .
In 8 years Interest earned is 2P (i.e Total Amount = P + 2P = 3P) Hence In 16 years Interest earned will be 4P (i.e Total Amount = P + 4P = 5P)
Harmonic Progression
In General If a, b, c , d are in AP then 1/a , 1/b , 1/c and 1/d are in HP
Point to note
In general , while solving through the options , the student should use the principal of starting with the middle (in terms of value), more convenient options . Also whenever it is not mentioned whether we have to assume CI or SI we should always assume SI .
Relative Speed - Two bodies moving in the same direction . Their relative speed will be S₁ - S₂.
In the case when bodies are moving from point A to B and then coming back to A. When they will meet first time distance covered will be 2D So for nth meeting Total distance covered by them is = 2nD
Time Required to meet at Starting Point on Circular Track
LCM of Time Required by A and by B to cover one circle . Case I - Same Direction - Circle : 200mts , A :2m/s , B : 5m/sec Then to meet at starting point. LCM ( 200/2 , 200/3) = LCM of 200 and 200 / HCF of 2 and 3 = 200/1 = 200 Seconds Case II - Opposite Direction - Circle : 200mts , A :2m/s , B : 5m/sec Then to meet at starting point. LCM ( 200/2 , 200/3) = LCM of 200 and 200 / HCF of 2 and 3 = 200/1 = 200 Seconds
Q. The ratio of water and milk in a 30 Liter mixture is 7:3 . Find the quantity of water to be added to the mixture in order to make this ratio 6:1. a) 30 b) 32 c) 33 d) 35
Lets say its a) 30 . then total volume of water is 21 + 30 = 51 and Milk will be 9 Liters . 51/9 is not 6/1 hence this option is not correct . Lets say 32 then Total volume of water is 21 + 32 = 53 and Milk is 9 . 53/9 is not 6/1 hence this option is not correct . Lets say 33 then Total volume of water is 21 + 33 = 54 and 54/9 is 6/1 hence this is correct option .
Q. What is the condition for one root of the quadratic equation ax² + bx + c = 0 to be twice the other
Lets say one root is g and other is 2g g + 2g = -b/a and 2g² = c/a 3g = -b/a and 2g² = c/a i.e g = -b/3a 3(-b/3a)² = c/a 2b² = 9ac
Q. For the first leg of a trip , Fred Traveled A miles at p . During the second Leg , he traveled at a slower speed . There were only two legs in the trip . The entire trip took T hours , and the average Speed for the entire Trip was V . In terms of A , p , T and V what was the average speed of the Second Leg of the Trip . a) (VT - A)/T b) (V - p)/pT c) (V - A/T)/(1 - A/pT) d) (V - A/T)/(V - p)
Lets say that first Leg was of 0 Miles . so if A = 0 , the whole trip is Second Leg and Average Speed of whole trip will be equal to Average Trip of Second Leg . Average Speed of Second Leg = Average Speed = V Putting A = 0 in the options a) (VT - A)/T -> (VT - 0)/T -> V b) (V - p)/pT -> Not True c) (V - A/T)/(1 - A/pT) -> V d) (V - A/T)/(V - p) -> Not True So two options left a) and c) Now lets say A = 20 Miles , p = 40 m/h , T = 2 hours and V= 30 m/h , Total Distance = 60 Average Speed of Second Leg = Distance/Time = 40/1.5 = 400/15 = 26.6~ Putting these value in a) and c) a) (VT - A)/T -> (30*2 - 20)/2 = 20 c) (V - A/T)/(1 - A/pT) -> (30 - 20/2)/(1-20/80) = 80/3 = 26.6 - Correct Answer
Q. A Shopkeeper sells two items at the same price . If he sells one of them at a profit of 10% and one at loss of 10% . Find the percentage profit/loss.
Lets say the selling price for both the product is $10 Cost price of 1st prod with 10% Lost: x*0.9 = $10 -> x = 10/0.9 = 11.11111 Cost price of 2nd prod with 10% Prod: x*1.1 = $10 -> x = 10/1.1 = 9.090909 Hence Total Cost Price = 20.020202 Total Selling Price = 20 % of Profit loss -> (20 - 20.020202)/20.02 -> 0.02020/20.02 -> 1% LOSS Another Way - He sold one at 1.1x and another at 0.9x so total change is 0.99x i.e Combined loss of 1%
Q. Find the value of √(6 + (√6 + (√6 ....
Lets say y = √6 + (√6 .... then y = √(6 +y ) y² -y - 6 = 0 (y-3)(y+2) = 0 y = 3 or y = - 2 -2 is not admissible as Square root can never have -ive values for Real roots . Hence 3 is Answer .
Q. A runs at 3 m/s and B runs at 2 m/s , The race is of total 1800 mts and one round is 200 mts. Calculate if they run in opposite direction . a) First Meeting at Starting point b) How many times will they meet c) When they will meet d) Where on tracks will they meet
Opposite Direction :- A's Speed = 3m/s B's Speed = 2m/s Relative Speed = 3m/s + 2m/s = 5 m/s Time required to meet at Starting Point = LCM of Time required by A and by B Point = LCM of (200/3 , 200/2) = 200 seconds Distance Covered by A = 200*3m/s = 600m Distance Covered by B = 200*2m/s = 400m Time required to meet first time = Track Length/ (Relative Speed) = 200/5 = 40 Seconds Distance Covered by A = 40*3m/s = 120m Distance Covered by B = 40*2m/s = 80m Time required to meet Second Time = 40*2 = 80 s Time required to meet Third Time = 40*3 = 60 s Total Number of meetings : A will finish race in 1800/3 = 600seconds Every 40 Seconds A and B meets hence they will meet 600/40 = 15 Times exactly . Where they will meet on the circle : The ratio of Speed is 3:2 hence they will meet (3+2) , 5 location on circle i.e Divide circle in 5 equal parts and they will meet there .
Q. On a ferry there are 50 cars and 10 Trucks . The Cars have an Average mass of 1200 Kg and the trucks have an average mass of 3000 Kg . What is the average mass of all 60 Vehicles on the ferry .
Solving with Approach 1 - Weighted Av = (n₁A₁ + n₂A₂)/(n₁ + n₂) = (50*1200 + 10*3000)/(50+10) = 1500 Kgs Solving with Approach 2 - Car Ratio = 50/(10 + 50) =5/6 Truck Ratio = 10/(10 +50) = 1/6 Now Car Ratio > Truck Ratio D of Car < D of Truck where D denotes the distance of individual Average to the Overall Average . Car Ratio/Truck Ratio = 5/1 ∴ Car Distance/truck Distance =1/5 Total Distance between individual Average = 3000 -1200 = 1800 Kg Dividing it in (5+1) 6 Parts = 1800/6 = 300 . Hence Each part is 300 Part . Car Distance from Average will be 1*300 = 300Kgs Average Weight of Car is 1200 Kgs Overall Average Weight = 1200+300 = 1500 Kgs Truck Distance from Average will be 5*300 = 1500 Kgs Average Weight of Truck is 3000 Kgs Overall Average Weight = 3000 - 1500 = 1500 Kgs
STD Case III - When Distance is Constant
Speed α 1/ Time Scenario 1 - A Train meets with an Accident and moves at 3/4 its original speed . Due to this , it is 20 Minutes Late .Find the time for the journey beyond the point of Accident. (Point to Note - Accident happened before Start and not on the way ) Lets Say Original Time -> t₀ Original Speed -> S₀ New Time -> t₁ = t₀ + 20 New Speed -> S₁ As the distance is constant S₀/S₁ = t₁/t₀ As per question S₁ = S₀(3/4) S₀/S₁ = 4/3 t₁/t₀ = 4/3 -> t₁ = (4/3)t₀ Also t₀ + 20 = t₁ -> t₀ + 20 = 4/3t₀ Hence t₀ = 60 Minute =1 Hour Scenario 2 - A Man travels from his house at 5Km/Hr and reached his office 20 Minutes late . If his speed had been 7.5 Km/h , he would have reached his office 12 minutes early . Find the distance from his house to his office . Lets Say t₀ = Time when he reach office at right time thats not early or late t₁ = t₀ + 20 and S₁ = 5 km/hr t₂ = t₀ - 12 and S₂ = 7.5 km/hr As the Distance is Constant S₁/S₂ = t₂/t₁ = 5/7.5 = 2/3 So (t₀ - 12)/(t₀ + 20) = 2/3 -> 3t₀ - 36 = 2t₀ + 40 t₀ = 76 Mins hence t₁ = 76 + 20 = 96 Minutes Distance from home = 5 * 96/60 = 8 Km
Vertex of Quadratic Equation
Vertex is the middle point of the quadratic equation or Parabola . ax² + bx + c a) you know the roots of equation for example (-4, 0) and (5,0) then Average of X values (-4+5)/2 = 0.5 will be the X value of Vertex . Y value can be found by inserting x into quadratic equation . b) Vertex is (h,k) where h = -b/2a and k = c - (b²/4a)
Mean and Median
When you have a set of consecutive numbers (integers, evens, odds, multiples), the mean is equal to the median. To locate the median, find the average of the endpoints. For Any Arithmetic Sequence the Mean = Median For any Equally Spaced Sequence Mean = Median
Increasing Numerator , Increasing Denominator *Addition - Attracts *
a) If we add *Same number to Numerator and Denominator then the resultant fraction is closer to 1 * than the original fraction . Ex - 1/3 is 0.33 and 2/4 is 0.5 i.e Closer to 1 and the resultant increased Ex - 4/2 is 2 and 6/4 is 1.5 i.e Closer to 1 and the resultant decreased b) If we add *different number to Numerator and Denominator , Then the resultant factor will be close to that Number * . For Ex - 1/3 is 0.33 and we add 2/5 (0.4) to Numerator/Denominator . The result 3/8 (0.36) will be close to 2/5 . For Ex - 6/5 is 1.2 and we add 2/5 (0.4) to Numerator/Denominator . The result 8/10 (0.8) will be close to 2/5(0.4)
Q. The Ratios of A's Salary to B's Salary is 2:3 . The ratio of B's Salary to C's Salary is 4:5 . The ratio of C's Salary to D's Salary is 6:11 and the ratio of D's Salary to E's is 12:17 . find the ratio of A's Salary to E's Salary .
a/b = 2/3 b/c = 4/5 c/d = 6/11 d/e = 12/17 A:B:C:D:E = N₁N₂N₃N₄ : D₁N₂N₃N₄ : D₁D₂N₃N₄ : D₁D₂D₃N₄: D₁D₂D₃D₄ A:B:C:D:E = 2*4*6*12: 3*4*6*12 : 3*5*6*12: 3*5*11*12 : 3*5*11*17 A:E = (2*4*6*12)/(3*5*11*17)=(16*12)/(55*17)
