Showing posts with label quant. Show all posts
Showing posts with label quant. Show all posts

Wednesday, June 27, 2012

DATA SUFFICIENCY questions

Mark A) If the question can be answered with statement I alone but not statement II alone, or can be answered with statement II alone but not statement I alone
Mark B) If the question cannot be answered with statement I alone or with statement II alone, but can be answered if both statements are used together
Mark C) If the question can be answered with either statement alone
Mark D) If the question cannot be answered with the information provided 

1. A point P is identified as P(m,n). What is the ratio AP:BP given that the points A and B are identified as A(5,-4) and B(1,6)?

Stmt 1. m = 3. Not sufficient.
Stmt 2. n = 2.5 Not sufficient.

Either statement alone is not sufficient. Both put together, we can answer the questions. Ans B

2. A point P is identified as P(m,n). What is the ratio AP:BP given that the points A and B are identified as A(5,-4) and B(5,6)?

Stmt 1. m = 5. Not sufficient. It tells us that all points lie on the line x = 5, but this is not enough
Stmt 2. n = 3 Not sufficient.

Both put together, this is enough.

In the above question, would n = 1 have been sufficient? Think about that.

3. A survey of 100 people tried to find the number of people who can write with both their left and right hands. What is the maximum number of people who could write left-handed and right-handed?

Stmt 1. 50 people can write only with their left hand. 40 people can write only with their right hand. Sufficient: Maximum of 10 people could write left-handed and right-handed.
Stmt 2. 50 people can write with their left hand. 40 people can write with their right hand. Sufficient: Maximum of 40 people could write left-handed and right-handed.

Answer Choice C

4. What is the slope of a line?

Stmt 1: The line makes 135 degrees with the negative direction of x - axis. Sufficient: The line makes 45 degrees with positive x axis. This should be enough.
Stmt 2: The line makes an isosceles right triangle with the coodinate axes and the product of the intercepts is negative. Sufficient: Either both intercepts are positive and equal or negative and equal. Slope = -1
Answer Choice C 


Sunday, June 24, 2012

Linear Races and Circular races

Linear Races


Linear Races :- when we have straight tracks for races.


Terms used to define linear races & their actual meanings:-
A gives B a start of 10 meters : B starts a race 10 meters ahead of A.

A gives B a start of 10 seconds : B starts 10 seconds before A.

A beats B by 10 meters : When A reaches finishing line, B is 10 meters behind.

A beats B by 10 seconds : B takes 10 seconds more than A to finish race.

A beats B by 10 meters or 10 seconds: B takes 10 seconds to cover 10 meters. Speed of B is 1 m/s.

Beat time : difference between time take by loser & winner.

Winner's distance: Length of the race track.

Time take by winner =  time taken by loser - beat time.

Dead Heat : Tie


Q1. X beats Y by 60 meters or by 12 seconds in a 2 Km race.Find Speed of X, Speed of Y, Time taken by X & time taken by Y.

Solution:-

Speed of Y = 60 / 12 = 5 m/s

Time taken by Y = 2000 / 5 = 400 minutes

Time taken by X = 400 - 60 = 340 minutes
Speed of X = 2000 / 340 = 5.88 m/s


Q2. In a 1000 m race, Neeta beats Geeta by 50 m & Seeta by 100 m. By what distance will Geeta beat Seeta ?
(1) 48.36 m (2) 50 m(3) 52.64 m (4) 51.28 m (5) 52.36 m


Solution :-
Distance covered by Neeta : Geeta in winning time by Neeta is = 1000 : 950 = 20:19.

Since time taken is same, ratio of their speed is also 20:19
Distance covered by Neeta : Seeta in winning time by Neeta is = 1000 : 900 = 20:18.
Since time taken is same, ratio of their speed is also 20:18
Hence Ratio of speeds of Geeta to Seeta would be 19:18.

Distance covered by Seeta when Geeta covers 1000 m = 1000 * 18/19 = 947.36
So Geeta beats Seeta by 1000 - 947.36 = 52.64 m. Hence answer option 3.


Q3. M beats N by 30 m or 5 seconds. Which of the following statements is/are true ?
I. Speed of N can not be found.
II. Speed of M can be found.
III. Distance covered by N can be found.
IV. N takes 6 more seconds to meet M.

1. Statement I is true.
2. Statement I, II & III are true.
3. Statement I & III are true.
4. All statements are true.
5. All statements are false.

Solution:-
By given data only speed of N can be found. To find other parameters we need length of track. Hence no statement is true. So answer option 5. 

Q4. A gives B a start of 15 seconds. A can run at speed of 80 mps & B can run at speed of 40 mps. In how much time will A  meet B on straight track after B starts race?
(1) 10 s (2) 12 s(3) 15 s  (4) 20 s (5) 30 s

Solution:-
Distance covered by B in early 15 seconds = 40 * 15 = 600 m
Relative speed of A & B = 80 - 40 = 40 mps
Time taken after A begin = 600 / 40 = 15 seconds.
Total time after B begins race = 15 + 15 = 30 seconds.Hence answer option 5. 

Q5. In above question, what would have been speed of A so that it would meet B after 2 mins after B begins his race ? (All other data remains same)
(1) 42.52 m/s(2) 45 m/s (3) 45.15 m/s (4) 45.71 m/s (5) 50 m/s

Solution:-
Relative Distance = 600 meters.
Time = 120 - 15 = 105 seconds
Relative Speed = 600 / 105 = 5.71 m/s
Speed of A = 40 + 5.71 = 45.71 m/s

Circular races


Circular Races : Circular races are on circular tracks where one can meet other person more than once.

When two persons A & B starts from same point at same time on a circular track then we can find

I. after how much time they meet for first time :- they meet for first time when one covers one more lap than other person. Relative distance would be length of track & using relative speed, time taken can be found.


II. After how much time they will meet for first time at starting point : this can be find out by taking LCM of time taken  by individual to cover one lap.


Q1. Two person X & Y start from the same point and move along a circular track of 60 m. Speed of X is 5 m/s & speed of Y is 7 m/s. After how much time will they meet for the first time ?

(1) 30 s (2) 15 s (3) 12 s (4) 16 s (5) Can not be determined


Solution:-

Since we don't know the directions of X & Y we can not determined answer. It is possible that they are running in same direction or they might be running in opposite direction. Hence answer option 5.



Q2. Two friends Raj & Rahul start a race on circular track of 240 m from same point in same direction at same time. Speed of the Raj is 20 m/s & that of Rahul is 25 m/s. After how much time will they meet for first time ?

(1) 10 s (2) 12 s (3) 24 s (4) 36 s (5) 48 s




Solution:-

Since same direction is same their related speed is : 25 - 20 = 5 m/s

Related distance to meet for first time is one lap of track = 240 m

Time taken : 240 / 5 = 48 seconds.




Q3. In above question, what would be time taken if they are running in opposite direction ?

(1) 3.33 s (2) 5.33 s (3) 8.33 s (4) 10 s (5) 12 s





Solution:-

Since opposite direction is same their related speed is : 25 + 20 = 45 m/s

Related distance to meet for first time is one lap of track = 240 m

Time taken : 240 / 45 = 5.33 seconds.



Q4. Two friends Raj & Rahul start a race on circular track of 500 m from same point in same direction at same time. Speed of the Raj is 20 m/s & that of Rahul is 25 m/s. After how much time will they meet for first time at starting point?

(1) 20 s (2) 25 s (3) 100 s (4) 200 s (5) 500 s


Solution:- 

Time taken to meet at starting point would be when both complete laps at same time. That is LCM of their time taken to complete track.

Time taken by Raj to complete track = 500/20 = 25 s

Time taken by Rahul to complete track = 500/25 = 20 s

Time taken to meet for first time at starting point = LCM(20,25) = 100 s.


Q5. In above question, what would be time taken to meet for first time if they are moving along circular track in opposite direction ?

(1) 20 s (2) 25 s (3) 100 s (4) 200 s (5) 500 s


Solution:- 

Time taken to meet at starting point would be when both complete laps at same time. That is LCM of their time taken to complete track.

Time taken by Raj to complete track = 500/20 = 25 s

Time taken by Rahul to complete track = 500/25 = 20 s

Time taken to meet for first time at starting point = LCM(20,25) = 100 s.


Meeting at starting point in circular races is independent of direction.



when more than 2 people are running in circular track. For e.g. 3 persons X, Y & Z.

I. after how much time they meet for first time :- It can be found by determining the time taken between two people & then between three.
II. After how much time they will meet for first time at starting point : LCM of time taken by X, Y & Z.

Q6. If X, Y & Z are starts their race by moving along a circular track of length 120 m from same point at same time in same direction. Find the time taken for them to meet for first time if speed of X is 2m/s, Y is 3m/s & that of Z is 5 m/s.
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s

Solution:-
All three will meet only when X meets Y.
relative distance = 120 m
relative speed of X & Y = 3 - 2 = 1 m/s
time taken for them to meet for first time = 120 s


All three will meet only when X meets Z.
relative distance = 120 m
relative speed of X & Z = 5 - 2 = 3 m/s
time taken for them to meet for first time = 120 /3 =  40 s

All three will meet for first time when X meets Y & Z together for first time : LCM (120,40)= 120 s


Q7. in above question, what time they will meet for first time at starting point ?
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s


Solution:- 

Time taken to meet at starting point would be when all complete laps at same time. That is LCM of their time taken to complete track.

Time taken by X to complete track = 120/2 = 60 s

Time taken by Y to complete track = 120/3 = 40 s
Time taken by Z to complete track = 120/5 = 24 s

Time taken to meet for first time at starting point = LCM(60,40,24) = 120 s.
It is independent of directions. Let them run in any direction. You don't worry whenever we are finding their first meet at starting point.


Q8. If X, Y & Z are starts their race by moving along a circular track of length 120 m from same point at same time. Y & Z are running in same direction while X is running in opposite direction. Find the time taken for them to meet for first time if speed of X is 2m/s, Y is 3m/s & that of Z is 5 m/s.
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s

Solution:-
All three will meet only when X meets Y.
relative distance = 120 m
relative speed of X & Y = 3 + 2 = 5 m/s
time taken for them to meet for first time = 120/5 = 24 s


All three will meet only when Y meets Z.
relative distance = 120 m
relative speed of X & Z = 5 - 3 = 2 m/s
time taken for them to meet for first time = 120 /2 =  60 s

All three will meet for first time when Y meets X & Z together for first time : LCM (24,60)= 120 s.

Q9. If A overtakes B for the first time in the middle of 6th lap. Find ratio of speed of A to B. We know they started their race from same point at same time.
(1) 6:5 (2) 11:9 (3) 5:6 (4) 9:11 (5) Can not be determined.

Solution:-
Overtakes means same direction.
A overtakes B for first time when he covers 5.5 laps. Same time A would cover 4.5 laps.
Ratio of speeds = ratio of distance covered = 5.5 : 4.5 = 55:45 = 11:9

Time and distance

TIME & DISTANCE :
  • Distance = Speed * Time
  • 1 km/hr = 5/18 m/sec
  • 1 m/sec = 18/5 km/hr
  • Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is 2xy/(x+y) kmph
  • Likewise, travelling 3 equal distance with speeds x,y and z, average speed = 2xyz / (xy+yz+zx)
  • Time taken after meeting (same distance):-
    if 2 persons starts journey from A & B and move towards each other and meet at C, after meeting they reach opposite points in x & y hrs respectively.

    then speed before they meet is  √y : √x
    Q. Seeta start her journey from Mumbai towards Pune at speed of  20 kmph & her sister Geeta starts her journey from Pune towards Mumbai at speed of 30 kmph. They meet at Lonavala. What is ratio of time taken by them to reach their destinations after they meet ?
    (1) 9:4 (2) 2:3 (3) 3:2 (4) 1:1 (5) Data insufficient
    Solution :-
    Speed before they meet is x:y i.e. 20:30
    time after they meet will be y2:x2 = 900 : 400 = 9:4. Hence answer option 1.
     

Monday, May 30, 2011

Quant… Basic Formulae

Consolidated some of the basic formula.
ALGEBRA :
1. Sum of first n natural numbers = n(n+1)/2
2. Sum of the squares of first n natural numbers = n(n+1)(2n+1)/6
3. Sum of the cubes of first n natural numbers = [n(n+1)/2]2
4. Sum of first n natural odd numbers = n2
5. Average = (Sum of items)/Number of items
Arithmetic Progression (A.P.):
An A.P. is of the form a, a+d, a+2d, a+3d, …
where a is called the ‘first term’ and d is called the ‘common difference’
1. nth term of an A.P. tn = a + (n-1)d
2. Sum of the first n terms of an A.P. Sn = n/2[2a+(n-1)d] or Sn = n/2(first term + last term)
Geometrical Progression (G.P.):
A G.P. is of the form a, ar, ar2, ar3, …
where a is called the ‘first term’ and r is called the ‘common ratio’.
1. nth term of a G.P. tn = arn-1
2. Sum of the first n terms in a G.P. Sn = a|1-rn|/|1-r|
Permutations and Combinations :
1. nPr = n!/(n-r)!
2. nPn = n!
3. nP1 = n
1. nCr = n!/(r! (n-r)!)
2. nC1 = n
3. nC0 = 1 = nCn
4. nCr = nCn-r
5. nCr = nPr/r!
Number of diagonals in a geometric figure of n sides = nC2-n

Tests of Divisibility :

1. A number is divisible by 2 if it is an even number.
2. A number is divisible by 3 if the sum of the digits is divisible by 3.
3. A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
4. A number is divisible by 5 if the units digit is either 5 or 0.
5. A number is divisible by 6 if the number is divisible by both 2 and 3.
6. A number is divisible by 8 if the number formed by the last three digits is divisible by 8.
7. A number is divisible by 9 if the sum of the digits is divisible by 9.
8. A number is divisible by 10 if the units digit is 0.
9. A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places, is divisible by 11.
H.C.F and L.C.M :
H.C.F stands for Highest Common Factor. The other names for H.C.F are Greatest Common Divisor (G.C.D) and Greatest Common Measure (G.C.M).
The H.C.F. of two or more numbers is the greatest number that divides each one of them exactly.
The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
Two numbers are said to be co-prime if their H.C.F. is 1.
H.C.F. of fractions = H.C.F. of numerators/L.C.M of denominators
L.C.M. of fractions = G.C.D. of numerators/H.C.F of denominators
Product of two numbers = Product of their H.C.F. and L.C.M.

PERCENTAGES :

1. If A is R% more than B, then B is less than A by R / (100+R) * 100
2. If A is R% less than B, then B is more than A by R / (100-R) * 100
3. If the price of a commodity increases by R%, then reduction in consumption, not to increase the expenditure is : R/(100+R)*100
4. If the price of a commodity decreases by R%, then the increase in consumption, not to decrease the expenditure is : R/(100-R)*100
PROFIT & LOSS :
1. Gain = Selling Price(S.P.) – Cost Price(C.P)
2. Loss = C.P. – S.P.
3. Gain % = Gain * 100 / C.P.
4. Loss % = Loss * 100 / C.P.
5. S.P. = (100+Gain%)/100*C.P.
6. S.P. = (100-Loss%)/100*C.P.
Short cut Methods:
1. By selling an article for Rs. X, a man loses l%. At what price should he sell it to gain y%? (or)
A man lost l% by selling an article for Rs. X. What percent shall he gain or lose by selling it for Rs. Y?
(100 – loss%) : 1st S.P. = (100 + gain%) : 2nd S.P.
2. A man sold two articles for Rs. X each. On one he gains y% while on the other he loses y%. How much does he gain or lose in the whole transaction?
In such a question, there is always a lose. The selling price is immaterial.
Formula: Loss % =
3. A discount dealer professes to sell his goods at cost price but uses a weight of 960 gms. For a kg weight. Find his gain percent.
Formula: Gain % =
RATIO & PROPORTIONS:
1. The ratio a : b represents a fraction a/b. a is called antecedent and b is called consequent.
2. The equality of two different ratios is called proportion.
3. If a : b = c : d then a, b, c, d are in proportion. This is represented by a : b :: c : d.
4. In a : b = c : d, then we have a* d = b * c.
5. If a/b = c/d then ( a + b ) / ( a – b ) = ( d + c ) / ( d – c ).
TIME & WORK :
1. If A can do a piece of work in n days, then A’s 1 day’s work = 1/n
2. If A and B work together for n days, then (A+B)’s 1 days’s work = 1/n
3. If A is twice as good workman as B, then ratio of work done by A and B = 2:1
PIPES & CISTERNS :
1. If a pipe can fill a tank in x hours, then part of tank filled in one hour = 1/x
2. If a pipe can empty a full tank in y hours, then part emptied in one hour = 1/y
3. If a pipe can fill a tank in x hours, and another pipe can empty the full tank in y hours, then on opening both the pipes,
the net part filled in 1 hour = (1/x-1/y) if y>x
the net part emptied in 1 hour = (1/y-1/x) if x>y
TIME & DISTANCE :
1. Distance = Speed * Time
2. 1 km/hr = 5/18 m/sec
3. 1 m/sec = 18/5 km/hr
4. Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is 2xy/(x+y) kmph.
PROBLEMS ON TRAINS :
1. Time taken by a train x metres long in passing a signal post or a pole or a standing man is equal to the time taken by the train to cover x metres.
2. Time taken by a train x metres long in passing a stationary object of length y metres is equal to the time taken by the train to cover x+y metres.
3. Suppose two trains are moving in the same direction at u kmph and v kmph such that u>v, then their relative speed = u-v kmph.
4. If two trains of length x km and y km are moving in the same direction at u kmph and v kmph, where u>v, then time taken by the faster train to cross the slower train = (x+y)/(u-v) hours.
5. Suppose two trains are moving in opposite directions at u kmph and v kmph. Then, their relative speed = (u+v) kmph.
6. If two trains of length x km and y km are moving in the opposite directions at u kmph and v kmph, then time taken by the trains to cross each other = (x+y)/(u+v)hours.
7. If two trains start at the same time from two points A and B towards each other and after crossing they take a and b hours in reaching B and A respectively, then A’s speed : B’s speed = (√b : √
SIMPLE & COMPOUND INTERESTS :
Let P be the principal, R be the interest rate percent per annum, and N be the time period.
1. Simple Interest = (P*N*R)/100
2. Compound Interest = P(1 + R/100)N – P
3. Amount = Principal + Interest
LOGORITHMS :
If am = x , then m = logax.
Properties :
1. log xx = 1
2. log x1 = 0
3. log a(xy) = log ax + log ay
4. log a(x/y) = log ax – log ay
5. log ax = 1/log xa
6. log a(xp) = p(log ax)
7. log ax = log bx/log ba
Note : Logarithms for base 1 does not exist.
AREA & PERIMETER :
Shape Area Perimeter
Circle ∏ (Radius)2 2∏(Radius)
Square (side)2 4(side)
Rectangle length*breadth 2(length+breadth)
1. Area of a triangle = 1/2*Base*Height or
2. Area of a triangle = √ (s(s-(s-b)(s-c)) where a,b,c are the lengths of the sides and s = (a+b+c)/2
3. Area of a parallelogram = Base * Height
4. Area of a rhombus = 1/2(Product of diagonals)
5. Area of a trapezium = 1/2(Sum of parallel sides)(distance between the parallel sides)
6. Area of a quadrilateral = 1/2(diagonal)(Sum of sides)
7. Area of a regular hexagon = 6(√3/4)(side)2
8. Area of a ring = ∏(R2-r2) where R and r are the outer and inner radii of the ring.
VOLUME & SURFACE AREA :
Cube :
Let a be the length of each edge. Then,
1. Volume of the cube = a3 cubic units
2. Surface Area = 6a2 square units
3. Diagonal = √ 3 a units
Cuboid :
Let l be the length, b be the breadth and h be the height of a cuboid. Then
1. Volume = lbh cu units
2. Surface Area = 2(lb+bh+lh) sq units
3. Diagonal = √ (l2+b2+h2)
Cylinder :

Let radius of the base be r and height of the cylinder be h. Then,
1. Volume = ∏r2h cu units
2. Curved Surface Area = 2∏rh sq units
3. Total Surface Area = 2∏rh + 2∏r2 sq units
Cone :
Let r be the radius of base, h be the height, and l be the slant height of the cone. Then,
1. l2 = h2 + r2
2. Volume = 1/3(∏r2h) cu units
3. Curved Surface Area = ∏rl sq units
4. Total Surface Area = ∏rl + ∏r2 sq units
Sphere :
Let r be the radius of the sphere. Then,
1. Volume = (4/3)∏r3 cu units
2. Surface Area = 4∏r2 sq units
Hemi-sphere :
Let r be the radius of the hemi-sphere. Then,
1. Volume = (2/3)∏r3 cu units
2. Curved Surface Area = 2∏r2 sq units
3. Total Surface Area = 3∏r2 sq units
Prism :
Volume = (Area of base)(Height