Formulae:
Factorial Notation:
Let n be positive integer.Then ,factorial n dentoed by n!
is defined as n! = n(n-1)(n-2). . . . . . . .3.2.1
eg:- 5! = (5 * 4* 3 * 2 * 1)
= 120
0! = 1
Permutations:
The different arrangements of a given number of things by
taking some or all at a time,are called permutations.
eg:- All permutations( or arrangements)made with the letters
a,b,c by taking two at a time are (ab,ba,ac,ca,bc,cb)
Numbers of permutations:
Number of all permutations of n things, taken r at a time is
given by nPr = n(n-1)(n-2). . .. . . (n-r+1)
= n! / (n-r)!
An Important Result:
If there are n objects of which p1 are alike of one kind;
p2 are alike of another kind ; p3 are alike of third kind and
so on and pr are alike of rth kind, such that
(p1+p2+. . . . . . . . pr) = n
Then,number of permutations of these n objects is:
n! / (p1!).(p2!). . . . .(pr!)
Combinations:
Each of different groups or selections which can be formed by
taking some or all of a number of objects,is called a combination.
eg:- Suppose we want to select two out of three boys A,B,C .
then ,possible selection are AB,BC & CA.
Note that AB and BA represent the same selection.
Number of Combination:
The number of all combination of n things taken r at atime is:
nCr = n! / (r!)(n-r)!
= n(n-1)(n-2). . . . . . . tor factors / r!
Note: nCn = 1 and nC0 =1
An Important Result:
nCr = nC(n-r)
Problems
1.Evaluate 30!/28!
Sol:- 30!/28! = 30 * 29 * (28!) / (28!)
= 30 * 29 =870
2.Find the value of 60P3
Sol:- 60P3 = 60! / (60 – 3)! = 60! / 57!
= (60 * 59 *58 * (57!) )/ 57!
= 60 * 59 *58
= 205320
3. Find the value of 100C98,50C 50
Sol:- 100C98 = 100C100-98)
= 100 * 99 / 2 *1
= 4950
50C50 = 1
4.How many words can be formed by using all the letters of the
word “DAUGHTR†so that vowels always come together &
vowels are never together?
Sol:-
(i) Given word contains 8 different letters
When the vowels AUE are always together we may suppose
them to form an entity ,treated as one letter
then the letter to be arranged are DAHTR(AUE)
these 6 letters can be arranged in 6p6 = 6!
= 720 ways
The vowels in the group (AUE) may be arranged in 3! = 6 ways
Required number of words = 760 * 6 =4320
(ii)Total number of words formed by using all the letters of
the given words
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 40320
Number of words each having vowels together
= 760 * 6
= 4320
Number of words each having vowels never together
= 40320 – 4320
= 36000
5.In how many ways can a cricket eleven be chosen out of a batch
of 15 players.
Sol:- Required number of ways
= 15C 11 = 15C (15-11)
= 15 C 4
15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1
= 1365
6.In how many a committee of 5 members can be selected from 6men
5 ladies consisting of 3 men and 2 ladies
Sol:- (3 men out of 6) and (2 ladies out of 5) are to be chosen
Required number of ways
=(6C3 * 5C2)
= 200
7.How many 4-letter word with or without meaning can be formed out
of the letters of the word 'LOGARITHMS' if repetition of letters is
not allowed
Sol:- 'LOGARITHMS' contains 10 different letters
Required number of words
= Number of arrangements of 100 letters taking
4 at a time
= 10P4
= 10 * 9 * 8 * 7
= 5040
8.In how many ways can the letter of word 'LEADER' be arranged
Sol:- The word 'LEADER' contains 6 letters namely
1L,2E,1A,1D and 1R
Required number of ways
= 6! / (1!)(2!)(1!)(1!)(1!)
= 6 * 5 * 4 * 3 * 2 *1 / 2 * 1
=360
9.How many arrangements can be made out of the letters of the word
'MATHEMATICS' be arranged so that the vowels always come
together
Sol:- In the word ' MATHEMATICS' we treat vowels
AEAI as one letter thus we have MTHMTCS(AEAI)
now we have to arrange 8 letters out of which M occurs
twice ,T occurs twice & the rest are different
Number of ways of arranging these letters
= 8! / (2!)(2!)
= 10080
now AEAI has 4 letters in which A occurs 2 times and the rest
are different
Number of ways of arranging these letters
= 4! / 2! = 12
Required number of words = (10080 * 12)
= 120960
10.In how many different ways can the letter of the word 'DETAIL' be
arranged in such a way that the vowels occupy only the odd positions
Sol:- These are 6 letters in the given word , out of which
there are 3 vowels and 3 consonants
Let us mark these positions as under
(1)(2) (3) (4)(5)(6)
now 3 vowels can be placed at any of the three places out of 4
marked 1,3,5
Number of ways of arranging the vowels = 3P3 = 3! =6
Also,the 3 consonants can be arranged at the remaining 3 positions
Number of arrangements = 3P3 = 6
Total number of ways = (6 * 6) =36
11.How many 3 digit numbers can be formed from the digits 2,3,5,6,7
and 9 which are divisible by 5 and none of the digits is repeated?
Sol:- Since each desired number is divisible by 5,
so we much have 5 at the unit place. The hundreds place
can now be filled by any of the remaining 4 digits .so, there
4 ways of filling it.
Required number of numbers = (1 * 5 * 4)
= 20
12.In how many ways can 21 books on English and 19 books on Hindi
be placed in a row on a self so that two books on Hindi may not
be together?
Sol:- In order that two books on Hindi are never together,
we must place all these books as under:
X E X E X . . . . . . . . . . X E X
Where E denotes the position of an English and X that of
a Hindi book.
Since there are 21 books on English,the number of places
marked X are therefore 22.
Now, 19 places out of 22 can be chosen in
22 C 19 = 22 C 3 =22 * 21 * 20 / 3 * 2 *1
Hence the required number of ways = 1540
13.Out of 7 constants and 4 vowels how many words of 3 consonants
and 2 vowels can be formed?
Sol:- Number of ways of selecting (3 consonants out of 7) and
(2 vowels out of 4)
= 7C3 * 4C2
= 210
Number of groups each having 3 consonants and 2 vowels = 210
Each group contains 5 letters
Number of ways of arranging 5 letters among themselves
= 5! = (5 * 4 * 3 * 2 * 1)
= 210
Required number of words = (210 * 210)
= 25200
Tuesday 12 November 2013
Permutations and Combinations
Clocks
General Concepts:
The face or dial of a watch is a circle whose circumference is
divided into 60 equal parts,called minute spaces.
A clock has two hands, the Smaller one is called the hour hand
or short hand while the larger one is called the minute hand or
long hand.
Important points:
a) In every 60 minutes, the minute hand gains 55 minutes on the
hour hand
b)In every hour, both the hands coincide once ,i.e 0 degrees.
c)the hands are in the same straight line when they are coincident
or opposite to each other. i.e 0 degrees or 180 degrees.
d)when the two hands are at right angles, they are 15 minute spaces
apart,i.e 90 degrees.
e)when the hands are in the opposite directions,they are 30 minute
spaces apart,i.e 180 degrees.
f)Angle traced by hour hand in 12hrs = 360 degrees.
g)Angle traced by minute hand in 60 min = 360 degrees. If a watch
or a clock indicated 8.15,when the correct time is 8, it is said
to be 15 minutes too fast. On the other hand, if it indicates 7.45,
when the correct time is 8,it is said to be 15 minutes slow.
h)60 min --> 360 degrees
1 min --> 60
i)the hands of a clock coincide in a day or 24 hours is 22 times,
in 12hours 11minutes.
j)the hands of clock are straight in a day is 44 times .
k)the hands of a clock at right angle in a day is 44 times .
l)the hands of a clock in straight line but opposite in direction is
22 times per day
Simple Problems:
Type1:
Find the angle between the hour hand and the minute hand
of a clock when the time is 3.25
solution : In this type of problems the formulae is as follows
30*[hrs-(min/5)]+(min/2)
In the above problem the given data is time is 3.25. that is
applied in the
formulae
30*[3-(25/5)]+(25/2)30*(15-25)/5+25/2
= 30*(-10/5)+25/2
= -300/5+25/2
= -600+(25/2)=-475/10=-47.5
i.e 47 1/20
therefore the required angle is 47 1/20.
Note:The -sign must be neglected.
Another shortcut for type1 is :
The formulae is
6*x-(hrs*60+X)/2
Here x is the given minutes,
so in the given problem the minutes is 25 minutes,
that is applied in the given formulae
6*25-(3*60+25)/2
150-205/2
(300-205)/2=95/2
=47 1/20.
therefore the required angle is 47 1/20.
Type2:
At what time between 2 and 3 o' clock will be the hands of a
clock be together?
Solution : In this type of problems the formulae is
5*x*(12/11)
Here x is replaced by the first interval of given time.
here i.e 2. In the above problem the given data is between
2 and 3 o' clock
5*2*12/11 =10*12/11=120/11=10 10/11min.
Therefore the hands will coincide at 10 10/11 min.past2.
Another shortcut for type2 is:
Here the clocks be together but not opposite
to each other so the angle is 0 degrees. so the formulae is
6*x-(2*60+x)/2=06*x-(120+x)/2=012x-120-x=0
11x=120
x=120/11=10 10/11
therefore the hands will be coincide at 10 10/11 min.past2.
Medium Problems
Type3:
At what time between 4 and 5 o'clock will the hands of a clock
be at rightangle?
Solution : In this type of problems the formulae is
(5*x + or -15)*(12/11)
Here x is replaced by the first interval of given time
here i.e 4
Case 1 : (5*x + 15)*(12/11)
(5*4 +15)*(12/11)
(20+15)*(12/11)
35*12/11=420/11=38 2/11 min.
Therefore they are right angles at 38 2/11 min .past4
Case 2 : (5*x-15)*(12/11)
(5*4-15)*(12/11)
(20-15)*(12/11)
5*12/11=60/11 min=5 5/11min
Therefore they are right angles at 5 5/11 min.past4.
Another shortcut for type 3 is:
Here the given angle is right angle i.e 900.
Case 1 : The formulae is 6*x-(hrs*60+x)/2=Given angle
6*x-(4*60+x)/2=90
6*x-(240+x)/2=90
12x-240-x=180
11x=180+240
11x=420
x=420/11= 38 2/11 min
Therefore they are at right angles at 38 2/11 min. past4.
Case 2 : The formulae is (hrs*60+x)/2-(6*x)=Given angle
(4*60+x)/2-(6*x)=90
(240+x)/2-(6*x)=90
240+x-12x=180
-11x+240=180
240-180=11x
x=60/11= 5 5/11 min
Therefore they art right angles at 5 5/11 min past4.
Type 4:
Find at what time between 8 and 9 o'clock will the hands of a
clock be in the same straight line but not together ?
Solution : In this type of problems the formulae is
(5*x-30)*12/11
x is replaced by the first interval of given time Here i.e 8
(5*8-30)*12/11
(40-30)*12/11
10*12/11=120/11 min=10 10/11 min.
Therefore the hands will be in the same straight line but not
together at 10 10/11 min.past 8.
Another shortcut for type 4 is:
Here the hands of a clock be in the same
straight line but not together the angle is 180 degrees.
The formulae is (hrs*60+x)/2-(6*x)=Given angle
(8*60+x)/2-6*x=180
(480+x)/2-(6*x)=180
480+x-12*x=360
11x=480-360
x=120/11=10 10/11 min.
therefore the hands will be in the same straight line but not
together at 10 10/11 min. past8.
Type 5:
At what time between 5 and 6 o’ clock are the hands of a
3 minutes apart ?
Solution : In this type of problems the formuae is
(5*x+ or - t)*12/11
Here x is replaced by the first interval of given time here xis 5.
t is spaces apart
Case 1 : (5*x+t)*12/11
(5*5+3)*12/11
28*12/11 = 336/11=31 5/11 min
therefore the hands will be 3 min .apart at 31 5/11 min.past5.
Case 2 : (5*x-t)*12/11
(5*5-3)*12/11
(25-3)*12/11=24 min
therefore the hands wi be 3 in apart at 24 min past 5.
Typicalproblems
problems:
A watch which gains uniformly ,is 5 min,slow at 8 o'clock in
the morning on sunday and it is 5 min.48 sec.fast at 8 p.m on
following sunday. when was it correct?
Solution :
Time from 8 am on sunday to 8 p.m on following sunday = 7 days
12 hours = 180 hours
the watch gains (5+(5 4/5))min .or 54/5 min. in 180 hours
Now 54/5 minare gained in 180 hours.
Therefore 5 minutes are gained in(180*5/54*5)hours=83 hours20 min.
=3 days11hrs20min.
therefore watch is correct at 3 days 11 hours 20 minutes after 8 a.m
of sunday
therefore it wil be correct at 20 min.past 7 p.m on wednesday
Calenders
Important Facts and Formulae:
a) Odd Days : The number of days more than the complete number
of weeks in a given period is number of odd days during that
period.
b) Leap year : Every year which is divisible by 4 is called a
leap year. Thus each one of he year 1992,1996,2004,2008,..etc,
is a eap year.
Every 4th century is a leap year but no other century is a
leap year thus each one of 400,800,1200,1600,2000,etc is a
leap year. None of the 1900,2010,2020,2100,etc is a leap
year. An year which is not a leap year is called Ordinary year.
c)An ordinary year has 365 days.
d) A leap year has 366 days.
e)Counting of odd days :
i)1 ordinary year = 365 days =52 weeks+1 day,Therefore An
ordinary year has 1 Odd day.
ii)One leap year = 366 days =52 weeks+2 days, Therefore a leap
year has 2 Odd days.
iii) 100 years = 76 ordinary years+ 24 leap years
= [(76*52) weeks+76 days]+[(24*52)weeks+48 days]
= 5200 weeks+124days=[5217 weeks+5 days]
therefore 100 years contain 5 odd days
iv)200 years contain 10(1week+3days), i.e 3 odd days
v)300 years contain 15(2 weeks+1 day), i.e 1 odd day
vi)400 years contain (20+1), i.e 3 weeks,so 0 Odd days
similarly each one of 800,1200,1600,etc contains 0 odd days.
Note:(7n+m) odd days , where m less than or equal to 7
is equivalent to m odd days ,thus ,8 odd days = 1 odd day etc.
f) Some codes o remember the months and weeks:
a) Week
Sunday - 1
Monday - 2
Tuesday - 3
Wednesday - 4
Thursday - 5
Friday - 6
Saturday - 0
b) Month
jan - 1 july - 0
feb - 4 Aug - 3
Mar - 4 Sep - 6
Apr - 0 Oct - 1
May - 2 Nov - 4
june - 5 Dec - 6
Simple problems:
Shortcuts : This shortcut must be applied only starting
with 19 series.
Example:
What day of the week on 17th june , 1998?
Solution : 5 -> the given month code(august)
17 -> the given date
98->(19 th century after years)
24-> ((47/4) = 11 i.e how many leap years
--------
total = 144 ((144/7) = 20 and the remainder is 4)
therefore in the above week table the no 4 code
represents wednesday
so the required day is wednesday.
Problem 1:
The first republic day of the India was celebrated on 26th
January,1950. It was
Solution : 01
26
50
12
----------
total = 89 ((89/7) = 12 and the remainder is 5)
therefore in the above week table represents the number 5
as thursday, so the required day was Thursday.
Problem 2:
Find on which day 15th august1947 ?
Solution :
03
15
47
11
----------
total = 76
Then (76)/7 = 6 odd days
6 indicates friday in the above week table.
Therefore required day is friday.
Problem 3:
Find on which day jan 26th 1956 ?
Solution :
01
26
56
14
-1 (-1 indicates leapyear(i.e 1956),so 1 reduce from the total)
---------
total = 96
Then (96)/7 = 5 odd days
5 indicates thursday in the above week table
Therefore our required day is Thursday.
Problem 4:
Today is friday after 62 days,it will be :
Solution : Each day of the week is repeated after 7 days.
so, after 63 days,it will be friday. Hence ,after 62 days,
it will be thursday.
Therefore the required day is thursday.
Problem 5:
Find the day of the week on 25th december,1995?
Solution :
06
25
95
23
---------
total = 149
Then (149)/7=(23)=2 odd days
Therefore the required day is "Monday".
Medium Problems
Problem 1:
jan 1, 1995 was a sunday.what day of the week lies on
jan 1,1996?
Solution :
01
01
96
24
-1(since 1996 was leap year)
---------
total = 121
Then (121)/7 = (17) = 2 odd days
Therefore our required day wasMonday.
Problem 2:
On 8th feb,1995 it wednesday. The day of the week on
8th feb,1994 was?
Solution :
04
08
94
23
---------
total = 129
Then (129)/7 = (18) = 3 odd days
Therefore the required day is Tuesday.
Problem 3:
may 6,1993 was thursday.what day of the week was on
may 6,1992 ?
Solution :
02
06
92
23
-1
----------
total = 122
Then (122)/7 = (17) = 3 odd days
Therefore the required day is Tuesday
Problem 4:
jan 1, 1992 was wednesday. What day of the week was
on jan 1,1993 ?
Solution :
01
01
93
23
----------
total = 118
Then (118)/7 = (16) = 6 odd days
Therefore the required day is Friday.
Problem 5:
January 1,2004 was a thursday,what day of the week lies
on jan ,2005?
solution :
The year 2004 being a leap year, it has 2 odd days. so,
first day of the 2005 will be 2 days beyond thursday and
so it will be saturday
therefore the required day is Thursday.
Problem 6:
On 8th march,2005,wednesday falls what day of the week was
it on 8th march,2004?
Solution : the year 2004 being a leap year,it has 2 odd days.
so, the day on8th march,2005 will be two days beyond the day
on 8th march,2004.but 8th march,2005 is wednesday. so,
8th march,2004 is monday.
Therefore the required day is Monday.
Problem 7:
what was the day of the week on 19th september ,1986 ?
Solution :
06
19
86
21
---------
total = 132
Then ((132/7 = 18 and the remainder is 6)
In the above week table represents the number 6 is friday.
Therefore the required day is Friday.
Typical problems
Problem 1:
On what dates of october,1994 did monday fall ?
Solution : 01
01
94
23
-------
total = 119
Then (119)/7 = (17) = 0 odd days
so the day is saturday
Therefore in october first the day is saturday.so,
the monday fell on 3rd october 1994.During october 1994,
monday fell on 3rd ,10th,17th and 24th.
Problem 2:
How many days are there from 2nd january 1995 to
15 th march,1995 ?
Solution : Jan + Feb + March
30 + 28 + 15 = 73 days
Problem 3:
The year next to 1996 having the same calendar as that
of 1996 is ?
Solution : Starting with 1996 , we go on countig the
number of odd days till the sum is divisible by 7.
Year 1996 1997 1998 1999 2000
odd days 2 1 1 1 2
2 + 1 + 1 + 1 + 2 = 7 odd days i.e odd day.
Therefore calendar for 2001 will be the same as
that of 1995.
Problem 4:
The calendar for 1990 is same as for :
Solution:
count the number of days 1990 onwards to get
0 odd day.
Year 1990 1991 1992 1993 1994 1995
oddd days 1 1 2 1 1 1
1 + 1 + 2 + 1 + 1 + 1 = 7 or 0 odd days
Therefore calendar for 1990 is the same as for the
year 1996.
Problem 5:
The day on 5th march of year is the same day on what
date of the same year?
Solution:
In the given monthly code table represents the march
code and november code both are same.that means any
date in march is the same day of week as the
corresponding date in november of that year, so the
same day falls on 5th november.
Areas
Important Facts and Formulae:
Results On Triangle
1.Sum of the angles of a triangle is 180 degrees.
2.The sum of any two sides of a triangle is greater
than third side.
3.Pythagoras Theorem:
In a right angled triangle (Hypotenuse)2 = (Base)2 +(Height)2
4.The line joining the mid point of a side of a triangle
to the opposite vertex is called the MEDIAN.
5.The point where the three medians of a triangle meet,
is called CENTROID. The centroid divides each of the
medians in the ratio 2:1
6.In an isosceles triangle, the altitude from the
vertex bisects the base
7.The median of a triangle divides it into two triangles
of the same area.
8.The area of the triangle formed by joining the mid points
of the sides of a given triangle is one-fourth of the area
of the given triangle.
Results On Quadrilaterals
1.The diagonals of a Parallelogram bisect each other.
2.Each diagonal of a Parallelogram divides it into two
triangles of the same area.
3.The diagonals of a Rectangle are equal and bisect
each other
4.The diagonals of a Square are equal and bisect each
other at right angles.
5.The diagonals of a Rhombus are unequal and bisect
each other at right angles.
6.A Parallelogram and a Rectangle on the same base
and between the same parallels are equal in area.
7.Of all he parallelogram of given sides the parallelogram
which is a rectangle has the greatest area.
Formulae
1.Area of a RECTANGLE = length * breadth
Length = (Area/Breadth) and Breadth = (Area/Length)
2.Perimeter of a RECTANGLE = 2(Length + Breadth)
3.Area of a SQUARE = (side)2 = ½ ( Diagonal)2
4.Area of four walls of a room = 2(length + breadth) * height
5.Area of a TRIANGLE = ½ * base * height
6.Area of a TRIANGLE = √[s * (s-a) * (s-b) * (s-c)],
where a,b,c are the sides of the triangle and s = 1/2(a+b+c)
7.Area of EQUILATERAL TRIANGLE = √(3/4)* (side)2
8.Radius of in circle of an EQUILATERAL TRIANGLE of
side a = r / 2√3
9.Radius of circumcircle of an EQUILATERAL TRIANGLE
of side a = r / √3
10.Radius of incircle of a triangle of area ∆ and
semi perimeter S = ∆ / s
11.Area of a PARALLELOGRAM = (base * height)
12.Area of RHOMBUS = 1/2 (product of diagonals)
13.Area of TRAPEZIUM =
=1/2 * (sum of parallel sides)* (distance between them)
14.Area of a CIRCLE =  r2 where r is the radius
15.Circumference of a CIRCLE = 2r
16.Length of an arc = 2 rø / 360, where ø is central angle
17.Area of a SECTOR = ½ (arc * r) = r2ø / 360
18.Area of a SEMICIRCLE = r2 / 2
19.Circumference of a SEMICIRCLE = r
Simple Problems
1.One side of a rectangular field is 15m and one of its diagonal
is 17m. Find the area of field?
Sol: Other side = √[(17*17) – (15*15)] = √(289-225) = 8m
Area = 15 * 8 =120 sq. m
2.A lawn is in the form of a rectangle having its sides in the
ratio 2:3 The area of the lawn is 1/6 hectares. Find the length
and breadth of the lawn.
Sol: let length = 2x meters and breadth = 3x mt
Now area = (1/6 * 1000)sq m = 5000/3 sq m
2x * 3x = 5000/3 =>x * x =2500 / 9
x = 50/3
length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m
3.Find the cost of carpeting a room 13m long and 9m broad with
a carpet 75cm wide at the rate of Rs 12.40 per sq meter
Sol: Area of the carpet = Area of the room = 13* 9 =117 sq m
length of the carpet = (Area/width) = 117 * (4/3) = 156 m
Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40
4.The length of a rectangle is twice its breadth if its length
is decreased by 5cm and breadth is increased by 5cm, the area
of the rectangle is increased by 75 sq cm. Find the length of
the rectangle.
Sol: let length = 2x and breadth = x then
(2x-5) (x+5) – (2x*x)=75
5x-25 = 75 => x=20
length of the rectangle = 40 cm
5.In measuring the sides of a rectangle, one side is taken 5%
in excess and the other 4% in deficit. Find the error percent
in the area, calculate from the those measurements.
Sol: let x and y be the sides of the rectangle then
correct area = (105/100 * x) * (96 / 100 *y)
=(504/500 xy) – xy = 4/500 xy
Error% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8%
6.A room is half as long again as it is broad. The cost of
carpeting the room at Rs 5 per sq m is Rs 2.70 and the cost of
papering the four walls at Rs 10 per sq m is Rs 1720. If a door
and 2 windows occupy 8 sq cm. Find the dimensions of the room?
Sol: let breadth=x mt ,length= 3x/2 mt and height=h mt
Area of the floor = (total cost of carpeting /rate)
= 270/5 sq m = 54 sq m
x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m
so breadth = 6m and length=3/2*6 = 9m
now papered area = 1720 /10 = 172 sq m
Area of one door and 2 windows =8 sq m
total area of 4 walls = 172+8 = 180 sq m
2(9+6)*h = 180 => h=180/30 = 6m
7.The altitude drawn to the base of an isosceles triangle is 8cm
and the perimeter is 32cm. Find the area of the triangle?
Sol: let ABC be the isosceles triangle, the AD be the altitude
let AB = AC=x then BC= 32-2x
since in an isoceles triange the altitude bisects the base so
BD=DC=16-x
in ∆ADC,(AC) 2 = (AD) 2 + (DC) 2
x*x=(8*8) + (16-x)*(16-x)
32x =320 => x = 10
BC = 32-2x = 32-20 = 12 cm
Hence, required area = ½ * BC * AD
= ½ * 12 * 10 = 60 sq cm
8.If each side of a square is increased by 25%, find the
percentage change in its area?
Sol: let each side of the square be a , then area = a * a
New side = 125a / 100 = 5a / 4
New area =(5a * 5a)/(4*4) = (25a²/16) – a²
= 9a²/16
Increase %= 9a²/16 * 1/a² * 100%
= 56.25%
9.Find the area of a Rhombus one side of which measures 20cm
and one diagonal 24cm.
Sol: Let other diagonal = 2x cm
since diagonals of a rhombus bisect each other at right angles,
we have
20² = 12² + x² => x = √[20² -12²]= √256 = 16cm
so the diagonal = 32 cm
Area of rhombus = ½ * product of diagonals
= ½ * 24 * 32
= 384 sq cm
10. The area of a circular field is 13.86 hectares. Find the cost
of fencing it at the rate of Rs. 4.40 per meter.
Sol: Area = 13.86 * 10000 sq m = 138600 sq m
r²= 138600 => r² = 138600 * 7/22 => 210 m
circumference = 2r = 2 * 22/7 * 210m = 1320 m
cost of fencing = Rs 1320 * 4.40 = Rs. 5808
Medium Problems:
11.Find the ratio of the areas of the incircle and circumcircle of
a square.
Sol: let the side of the square be x, then its diagonal = √2 x
radius of incircle = x/2 and
radius of circmcircle =√2 x /2 = x/√2
required ratio = x²/4 : x²/2 = ¼ : ½ = 1:2
12.If the radius of a circle is decreased by 50% , find the
percentage decrease in its area.
Sol: let original radius = r and new radius = 50/100 r = r/2
original area = r² and new area = (r/2)²
decrease in area = 3 r²/4 * 1/ r² * 100 = 75%
13.Two concentric circles form a ring. The inner and outer
circumference of the ring are 352/7 m and 528/7m respectively.
Find the width of the ring.
sol: let the inner and outer radii be r and R meters
then, 2r = 352/7 => r = 352/7 * 7/22 * ½ = 8m
2R = 528/7 => R= 528/7 * 7/22 * ½ = 12m
width of the ring = R-r = 12-8 = 4m
14.If the diagonal of a rectangle is 17cm long and its perimeter
is 46 cm. Find the area of the rectangle.
sol: let length = x and breadth = y then
2(x+y) = 46 => x+y = 23
x²+y² = 17² = 289
now (x+y)² = 23² =>x²+y²+2xy= 529
289+ 2xy = 529 => xy = 120
area =xy=120 sq. cm
15.A rectangular grassy plot 110m by 65cm has a gravel path .5cm
wide all round it on the inside. Find the cost of gravelling the
path at 80 paise per sq.mt
sol: area of theplot = 110 * 65 = 7150 sq m
area of the plot excluding the path = (110-5)* (65-5) = 6300 sq m
area of the path = 7150- 6300 =850 sq m
cost of gravelling the path = 850 * 80/100 = 680 Rs
16. The perimeters of ttwo squares are 40cm and 32 cm. Find the
perimeter of a third square whose area is equal to the difference
of the areas of the two squares.
sol: side of first square = 40/4 =10cm
side of second square = 32/4 = 8cm
area of third squre = 10² – 8² = 36 sq cm
side of third square = √36 = 6 cm
required perimeter = 6*4 = 24cm
17. A room 5m 44cm long and 3m 74cm broad is to be paved with squre
tiles. Find the least number of squre tiles required to cover the
floor.
sol: area of the room = 544 * 374 sq cm
size of largest square tile = H.C.F of 544cm and 374cm= 34cm
area of 1 tile = 34*34 sq cm
no. of tiles required = (544*374) / (34 * 34) = 176
18. The diagonals of two squares are in the ratio of 2:5. Find
the ratio of their areas.
sol: let the diagonals of the squares be 2x and 5x respectively
ratio of their areas = ½ * (2x)² : ½*(5x)² = 4:25
19.If each side of a square is increased by 25%. Find the percentage
change in its area.
sol: let each side of the square be a then area = a ²
new side = 125a/100 = 5a/4
new area = (5a/4)² = 25/16 a²
increase in area = (25/16)a² - a² = (9/16)a²
increase % = (9/16)a² * (1/a²) * 100 = 56.25%
20.The base of triangular field os three times its altitude. If the
cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18.
Find its base and height.
sol:
area of the field = total cost/ rate = 333.18 /24.68 = 13.5 hectares
=> = 13.5 * 10000 = 135000 sq m
let the altitude = x mt and base = 3x mt
then ½ *3x * x = 135000 => x² = 90000 => x = 300
base= 900m and altitude = 300m
21.In two triangles the ratio of the areas is 4:3 and the ratio of
their heights is 3:4. Find the ratio of their bases?
Sol:
let the bases of the two triangles be x &y and their heights
be 3h and 4h respectively.
(1/2*x*3h)/(1/2*y*4h) =4/3 => x/y = 4/3 *4/3 = 16/9
22.Find the length of a rope by which a cow must be tethered in order
that it may be able to graze an area of 9856 sq meters.
Sol:
clearly the cow will graze a circular field of area 9856 sq m and
radius equal to the length of the rope.
Let the length of the rope be r mts
then r²=9856 => r²=9856*7/22 = 3136 => r=56m
23.The diameter of the driving wheel of a bus is 140cm. How many
revolutions per minute must the wheel make inorder to keep a speed of
66 kmph?
Sol: Distance to be covered in 1min = (66*1000)/60 m =1100m
diameter = 140cm => radius = r =0.7m
circumference of the wheel = 2*22/7*0.7 = 4.4m
no of revolutions per minute = 1100/4.4 = 250
24.The inner circumference of a circular race track, 14m wide is 440m.
Find the radius of the outer circle.
Sol: let inner radius be r meters.
Then 2r =440 => r=440*7/22*1/2 = 70m
radius of outer circle = 70+4 =84m
25.A sector of 120 degrees, cut out from a circle, has an area of
66/7 sq cm. Find the radius of the circle.
Sol: let the radius of the circle be r cm. Then
r²ø/360 =66/7=> 22/7*r²*120/360 = 66/7 =>r² = 66/7 *7/22*3 =9
radius = 3cm
26.The length of the room is 5.5m and width is 3.75m. Find the cost
of paving the floor by slabs at the rate of Rs.800 per sq meter.
Sol: l=5.5m w=3.75m
area of the floor = 5.5 * 3.75 = 20.625 sq m
cost of paving = 800 *20.625 =Rs. 16500
27.A rectangular plot measuring 90 meters by 50 meters is to be
enclosed by wire fencing. If the poles of the fence are kept 5 meters
apart. How many poles will be needed?
Sol: perimeter of the plot = 2(90+50) = 280m
no of poles =280/5 =56m
28.The length of a rectangular plot is 20 meters more than its breadth.
If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is
the length of the plot in meter?
Sol: let breadth =x then length = x+20
perimeter = 5300/26.50 =200m
2(x+20+x) =200 => 4x+40 =200
x = 40 and length = 40+20 = 60m
29.A rectangular field is to be fenced on three sides leaving a side of
20 feet uncovered. If the area of the field is 680 sq feet, how many
feet of fencing will be required?
Sol: l=20feet and l*b=680 => b= 680/20 = 34feet
length of fencing = l+2b = 20+68 =88 feet
30.A rectangular paper when folded into two congruent parts had a
perimeter of 34cm foer each part folded along one set of sides and
the same is 38cm. When folded along the other set of sides. What is
the area of the paper?
Sol: when folded along the breadth
we have 2(l/2 +b) = 34 or l+2b = 34...........(1)
when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2)
from 1 &2 we get l=14 and b=10
Area of the paper = 14*10 = 140 sq cm
31.A took 15 seconds to cross a rectangular field diagonally walking at
the rate of 52 m/min and B took the same time to cross the same field
along its sides walking at the rate of 68m/min. The area of the field is?
Sol: length of the diagonal = 52*15/60 =13m
sum of length and breadth = 68*15/60 = 17m
√(l²+b²)=13 or l+b = 17
area =lb = ½ (2lb) = ½[(l+b)² – (l²+b²)] = ½[17² -169]
=1/2*120 = 60 sq meter
32 . A rectangular lawn 55m by 35m has two roads each 4m wide running in
the middle of it. One parallel to the length and the other parallel to
breadth. The cost of graveling the roads at 75 paise per sq meter is
sol: area of cross roads = 55*4 +35*4-4*4 = 344sq m
cost of graveling = 344 *75/100 =Rs. 258
33.The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.
How much will it cost to lay a three meter wide pavement along the
fencing inside the field @ Rs. 50 per sq m
sol: perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126-6 = 120m
area of the pavement = (126*126)-(120*120)=246â€*6 sq m
cost of pavement = 246*6*50 = Rs. 73800
34.Amanwalked diagonally across a square plot. Approximately what was
the percent saved by not walking along the edges?
Sol: let the side of the square be x meters
length of two sides = 2x meters
diagonal = √2 x = 1.414x m
saving on 2x meters = .59x m
saving % = 0.59x /2x *100%
= 30% (approx)
36.A man walking at the speed of 4 kmph crosses a square field
diagonally in 3 meters.The area of the field is
sol: speed of the man = 4*5/18 m/sec = 10/9 m/sec
time taken = 3*60 sec = 180 sec
length of diagonal = speed * time = 10/9 * 180 = 200m
Area of the field = ½ *(dioagonal)²
= ½ * 200*200 sq m = 20000sq m
37.A square and rectangle have equal areas. If their perimeters
are p and q respectively. Then
sol: A square and a rectangle with equal areas will satisfy the
relation p < q
38.If the perimeters of a square and a rectangle are the same,
then the area a & b enclosed by them would satisfy the condition:
sol: Take a square of side 4cm and a rectangle having l=6cm and
b=2cm
then perimeter of square = perimeter of rectangle
area of square = 16 sq cm
area of rectangle = 12 sq cm
Hence a >b
39.An error of 2% in excess is made while measuring the side of a
square. The percentage of error in the calculated area of the
square is
sol: 100cm is read as 102 cm
a = 100*100 sq cm and b = 102 *102 sq cm
then a-b = 404 sq cm
percentage error = 404/(100*100) = 4.04%
40.A tank is 25m long 12m wide and 6m deep. The cost of plastering
its walls and bottom at 75 paise per sq m is
sol: area to be plastered = [2(l+b)*h]+(l*b)
= 2(25+12)*6 + (25*12)= 744 sq m
cost of plastering = Rs . 744*75/100 = Rs. 5581
41.The dimensions of a room are 10m*7m*5m. There are 2 doors and 3
windows in the room. The dimensions of the doors are 1m*3m. One
window is of size 2m*1.5m and the other 2 windows are of size 1m*1.5m.
The cost of painting the walls at Rs. 3 per sq m is
sol: Area of 4 walls = 2(l+b)*h
=2(10+7)*5 = 170 sq m
Area of 2 doors and 3 windows = 2(1*3)+(2*1.5)+2(1*1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m
cost of painting = Rs. 158 *3 = Rs. 474
42.The base of a triangle of 15cm and height is 12cm. The height of
another triangle of double the area having the base 20cm is
sol: a = ½ *15*12 = 90 sq cm
b = 2a = 2 * 90 = ½ * 20 *h => h= 18cm
43.The sides of a triangle are in the ratio of ½:1/3:1/4. If the
perimeter is 52cm, then the length of the smallest side is
sol: ratio of sides = ½ :1/3 :1/4 = 6:4:3
perimeter = 52 cm, so sides are 52*6/13 =24cm
52*4/13 = 16cm
52 *3/13 = 12cm
length of smallest side = 12cm
44.The height of an equilateral triangle is 10cm. Its area is
sol: a² = (a/2)² +(10)²
a² – a²/4 = 100 =>3a² = 100*4
area = √3/4 *a² = √3/4*400/3 = 100/√3 sq cm
45.From a point in the interior of an equilateral triangle, the
perpendicular distance of the sides are √3 cm, 2√3cm and
5√3cm. The perimeter of the triangle is
sol: let each side of the triangle be ‘a’ cm
then area(AOB) +area(BOC)+area(AOC) = area(ABC)
½ * a *√3 +1/2 *a *2√3 +1/2 * a*5√3 = √3/4 a ²
a/2√3(1+2+5) = √3/4 a ² => a=16
perimeter = 3*16 = 48cm
Complex Probems:
1.If the area of a square with side a s equal to the area of a
triangle with base a, then the altitude of the triangle is
sol: area of a square with side a = a ² sq unts
area of a triangle with base a = ½ * a*h sq unts
a ² =1/2 *a *h => h = 2a
altitude of the triangle is 2a
2.An equilateral triangle is described on the diagonal of a
square. What is the ratio of the area of the triangle to that of
the square?
Sol: area of a square = a ² sq cm
length of the diagonal = √2a cm
area of equilateral triangle with side √2a
= √3/4 * (√2a) ²
required ratio = √3a² : a ² = √3 : 2
3.The ratio of bases of two triangles is x:y and that of their
areas is a:b. Then the ratio of their corresponding altitudes
wll be
sol: a/b =(½ * x*H) /(1/2 * y * h)
bxH = ayh =>H/h =ay/bx
Hence H:h = ay:bx
4 .A parallelogram has sides 30m and 14m and one of its diagonals
is 40m long. Then its area is
sol: let ABCD be the given parallelogram
area of parallelogram ABCD = 2* (area of triangle ABC)
now a = 30m, b = 14m and c = 40m
s = ½(30+14+40) = 42m
Area of triangle ABC = √[ s(s-a)(s-b)(s-c)
= √(42*12*28*2 = 168sq m
area of parallelogram ABCD = 2 *168 =336 sq m
5.If a parallelogram with area p, a triangle with area R and a
triangle with area T are all constructed on the same base and all
have the same altitude, then which of the following statements
is false?
Sol: let each have base = b and height = h
then p = b*h, R = b*h and T = ½ * b*h
so P = R, P = 2T and T = ½ R are all correct statements
6.If the diagonals of a rhombus are 24cm and 10cm the area
and the perimeter of the rhombus are respectively.
Sol: area = ½*diagonal 1 *diagonal 2= ½ * 24 * 10= 120 sq cm
½ * diagonal 1 = ½ * 24 = 12cm
½ * diagonal 2 = ½ *10 =5 cm
side of a rhombus = (12) ² + (5) ² = 169 => AB = 13cm
7.If a square and a rhombus stand on the same base, then the ratio
of the areas of the square and the rhombus is:
sol: A square and a rhombus on the same base are equal in area
8.The area of a field in the shape of a trapezium measures
1440sq m. The perpendicular distance between its parallel sides
is 24cm. If the ratio of the sides is 5:3, the length of the
longer parallel side is:
sol: area of field =1/2 *(5x+3x) *24 = 96x sq m
96x = 1440 => x = 1440 /96 = 15
hence, the length of longer parallel side = 5x = 75m
9.The area of a circle of radius 5 is numerically what percent its
circumference?
Sol: required percentage = (5)²/(2*5) *100 = 250%
10.A man runs round a circular field of radius 50m at the speed of
12m/hr. What is the time taken by the man to take twenty rounds of
the field?
Sol: speed = 12 k/h = 12 * 5/18 = 10/3 m/s
distance covered = 20 * 2*22/7*50 = 44000/7m
time taken = distance /speed = 44000/7 * 3/10 = 220/7min
11.A cow s tethered in the middle of a field with a 14feet long
rope.If the cow grazes 100 sq feet per day, then approximately
what time will be taken by the cow to graze the whole field?
Sol: area of the field grazed = 22/7 * 14 * 14 = 616 sq feet
12.A wire can be bent in the form of a circle of radius 56cm.
If it is bent in the form of a square, then its area will be
sol: length of wire = 2 r = 2 *22/7 *56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88*88 = 7744sq cm
13.The no of revolutions a wheel of diameter 40cm makes in
traveling a distance of 176m is
sol:
distance covered in 1 revolution = 2 r = 2 *22/7 *20
= 880/7 cm
required no of revolutions = 17600 *7/880 = 140
14.The wheel of a motorcycle 70cm in diameter makes 40
revolutions in every 10sec.What is the speed of motorcycle
n km/hr?
Sol: distance covered in 10sec = 2 *22/7 *35/100 *40 =88m
distance covered in 1 sec =88/10m = 8.8m
speed =8.8m/s = 8.8 * 18/5 *k/h = 31.68 k/h
15.Wheels of diameters 7cm and 14cm start rolling simultaneously
from x & y which are 1980 cm apart towards each other in opposite
directions. Both of them make the same number of revolutions per
second. If both of them meet after 10seconds.The speed of the
smaller wheel is
sol: let each wheel make x revolutions per sec. Then
(2 *7/2 *x)+(2 * 7*x)*10 = 1980
(22/7 *7 * x) + (2 * 22/7 *7 *x) = 198
66x = 198 => x = 3
distance moved by smaller wheel in 3 revolutions
= 2 *22/7 *7/2 *3 = 66cm
speed of smaller wheel = 66/3 m/s = 22m/s
16.A circular swimming pool is surrounded by a concrete wall
4ft wide. If the area of the concrete wall surrounding the pool
is 11/25 that of the pool, then the radius of the pool is?
Sol: let the radius of the pool be R ft
radius of the pool including the wall = (R+4)ft
area of the concrete wall =  [(R+4)2 - R2 ]
=> = [R+4+R][R+4-R]
= 8(R+2) sq feet
8(R+2) = 11/25  R2 => 11 R2 = 200 (R+2)
Radius of the pool R = 20ft
17.A semicircular shaped window has diameter of 63cm. Its
perimeter equals
sol: perimeter of window = r +2r = [22/7 * 63/2 +63] = 99+63
= 162 cm
18.Three circles of radius 3.5cm are placed in such a way that
each circle touches the other two. The area of the portion
enclosed by the circles is
sol:
required area = (area of an equilateral triangle of side 7 cm)
- (3 * area of sector with Ø = 6o degrees and r = 3.5cm)
= ( √ ¾ * 7 * 7) – (3* 22/7 *3.5 *3.5*60/360 ) sq cm
= 49√3/4 – 11*0.5*3.5 sq cm = 1.967 sq cm
19. Four circular cardboard pieces, each of radius 7cm are placed
in such a way that each piece touches two other pieces. The area
of the space encosed by the four pieces is
sol: required area = 14*14 – (4 * ¼ * 22/7 * 7 *7) sq cm
= 196 – 154 = 42 sq cm
Compound Intrest
Important Facts and Formulae:
Compound Interest:
Sometimes it so happens that the borrower and the lender
gree to fix up a certain unit of time ,say yearly or
half-yearly or quarterly to settle the previous account.
In such cases ,the amount after the first unit of time
becomes the principal for the 2nd unit ,the amount after
second unit becomes the principal for the 3rd unit and so
on. After a specified period ,the difference between the
amount and the money borrowed is called Compound Interest
for that period.
Formulae:
Let principal=p,Rate=R% per annum Time=nyears
1.When interest is compounded Annually,
Amount=P[1+(R/100)]n
2.When interest is compounded Halfyearly,
Amount=P[1+((R/2)100)]2n
3.When interest is compounded Quaterly,
Amount=P[1+((R/4)100)]4n
4.When interest is compounded Annually,but time in fractions
say 3 2/5 yrs Amount=P[1+(R/100)]3[1+((2R/5)/100)]
5.When rates are different for different years R1%,R2%,R3%
for 1st ,2nd ,3rd yrs respectively
Amount=P[1+(R1/100)][1+(R2/100)][1+(R3/100)]
6.Present Worth of Rs.X due n years hence is given by
Present Worth=X/[1+(R/100)]n
Simple Problems
1.Find CI on Rs.6250 at 16% per annum for 2yrs ,compounded
annually.
Sol: Rate R=16,n=2,Principle=Rs.6250
Method1:
Amount=P[1+(R/100)]n
=6250[1+(16/100)]2
=Rs.8410
C.I=Amount-P
=8410-6250
=Rs.2160
Method2:
Iyear------------------6250+1000
\\Interest for 1st yr on 6250
II yr---------------6250+1000+160
\\Interest for I1yr on 1000
C.I.=1000+1000+160
=Rs.2160
2.Find C.I on Rs.16000 at 20% per annum for 9 months
compounded quaterly
Sol:
MethodI:
R=20%
12months------------------------20%
=> 3 months------------------------5%
For 9 months,there are '3' 3months
--------16000+800
--------16000+800+40
--------16000+800+40+10+2
=>Rs.2522
MethodII: Amount=P[1+(R/100)]n
=16000[1+(5/100)]3
=Rs.18522
C.I=18522-16000
=Rs.2522
Complex Problems
1.The difference between C.I and S.I. on a certain sum
at 10% per annum for 2 yrs is Rs.631.find the sum
Sol:
MethodI:
NOTE:
a) For 2 yrs -------->sum=(1002D/R2)
b) For 3 yrs -------->sum=(1003D/R2(300+R))
Sum=1002*631/102
=Rs.63100
MethodII:
Let the sum be Rs.X,Then
C.I.=X[1+(10/100)]2-X
S.I=(X*10*2)/100
=X/5
C.I-S.I.=21X/100-X/5
=X/100
X/100=631
X=Rs.63100
2.If C.I on a certain sum for 2 yrs at 12% per annum is
Rs.1590. What would be S.I?
sol:
C.I.=Amount-Principle
Let P be X
C.I=X[1+(12/100)]2-X
=>784X/625-X=1590
=>X=Rs.6250
S.I=(6250*12*2)/100=Rs.1500
3.A sum of money amounts to Rs.6690 after 3 yrs and to
Rs.10035b after 6 yrs on C.I .find the sum
sol:
For 3 yrs,
Amount=P[1+(R/100)]3=6690-----------------------(1)
For 6 yrs,
Amount=P[1+(R/100)]6=10035----------------------(2)
(1)/(2)------------[1+(R/100)]3=10035/6690
=3/2
[1+(R/100)]3=3/2-----------------(3)
substitue (3) in (1)
p*(3/2)=6690
=>p=Rs.4460
sum=Rs.4460
4.A sum of money doubles itself at C.I in 15yrs.In how many
yrs will it become 8 times?
sol: Compound Interest for 15yrs p[1+(R/100)]15
p[1+(R/100)]15=2P
=>p[1+(R/100)]n=8P
=>[1+(R/100)]n=8
=>[1+(R/100)]n=23
=>[1+(R/100)]n=[1+(R/100)]15*3
since [1+(R/100)] =2
n=45yrs
5.The amount of Rs.7500 at C.I at 4% per annum for 2yrs is
sol:
Iyear------------------7500+300(300------Interest on 7500)
IIyear ----------------7500+300+12(12------------4% interest
on 300)
Amount=7500+300+300+12
=Rs.8112
6.The difference between C.I and S.I on a sum of money for
2 yrs at 121/2% per annum is Rs.150.the sum is
sol:
Sum=1002D/R2=( 1002*150) /(25/2)2=Rs.9600
7.If the S.I on sum of money at 15% per annum for 3yrs is Rs.1200,
the C.I on the same sum for the same period at same rate is------
sol:
S.I=1200
P*T*R/100=1200
P*3*5/100=1200
=>P=Rs.8000
C.I for Rs.8000 at 5% for 3 yrs is-------------8000+400
-----8000+400+20
-------------8000+400+20+20+1
C.I =400+400+20+400+20+20+1
=Rs.1261
Simple Intrest
Important Facts and Formulae:
Principal or Sum:- The money borrowed or lent out for a
certain period is called Principal or the Sum.
Interest:- Extra money paid for using others money is
called Interest.
Simple Interest:- If the interest on a sum borrowed for
a certain period is reckoned uniformly,then it is called
Simple Interest.
Formulae:
Principal = P
Rate = R% per annum
Time = T years. Then,
(i)Simple Interest(S.I)= (P*T*R)/100
(ii) Principal(P) = (100*S.I)/(R*T)
Rate(R) = (100*S.I)/(P*T)
Time(T) = (100*S.I)/(P*R)
Simple Problems
1.Find S.I on Rs68000 at 16 2/3% per annum for 9months.
Sol:- P=68000
R=50/3% p.a
T=9/12 years=4/3 years
S.I=(P*R*T)/100
=(68000*(50/3)*(3/4)*(1/100))
=Rs 8500
Note:If months are given we have to converted into
years by dividing 12 ie., no.of months/12=years
2.Find S.I on Rs3000 at 18% per annum for the period from
4th Feb to 18th April 1995
Sol:- Time=(24+31+18)days
=73 days
=73/365=1/5 years
P= Rs 3000
R= 18% p.a
S.I = (P*R*T)/100
=(3000*18*1/5*1/100)
=Rs 108
Remark:- The day on which money is deposited is not
counted while the day on which money is withdrawn is
counted.
3. In how many years will a sum of money becomes triple
at 10% per annum.
Sol:- Let principal =P
S.I = 2P
S.I = (P*T*R)/100
2P = (P*T*10)/100
T = 20 years
Note:
(1) Total amount = Principal + S.I
(2) If sum of money becomes double means Total amount
or Sum
= Principal + S.I
= P + P = 2P
Medium Problems
1.A sum at Simple interest at 13 1/2% per annum amounts
to Rs 2502.50 after 4 years.Find the sum.
Sol:- Let Sum be x. then,
S.I = (P*T*R)/100
= ((x*4*27)/(100*2))
= 27x/100
Amount = (x+(27x)/100)
= 77x/50
77x/50 = 2502.50
x = (2502.50*50)/77
= 1625
Sum = 1625
2. A some of money becomes double of itself in 4 years
in 12 years it will become how many times at the same
rate.
Sol:- 4 yrs - - - - - - - - - P
12 yrs - - - - - - - - - ?
(12/4)* P =3P
Amount or Sum = P+3P = 4 times
3. A Sum was put at S.I at a certain rate for 3 years.
Had it been put at 2% higher rate ,it would have
fetched Rs 360 more .Find the Sum.
Sol:- Let Sum =P
original rate = R
T = 3 years
If 2% is more than the original rate ,it would have
fetched 360 more ie., R+2
(P*(R+2)*3/100) - (P*R*3)/100 = 360
3PR+ 6P-3PR = 36000
6P = 36000
P = 6000
Sum = 6000.
4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest
rate is increased by 3%, it would amount to how much?
Sol:- S.I = 920 - 800 = 120
Rate = (100*120)/(800*3) = 5%
New Rate = 5 + 3 = 8%
Principal = 800
Time = 3 yrs
S.I = (800*8*3)/100 = 192
New Amount = 800 + 192
= 992
5. Prabhat took a certain amount as a loan from bank at
the rate of 8% p.a S.I and gave the same amount to Ashish
as a loan at the rate of 12% p.a . If at the end of 12 yrs,
he made a profit of Rs. 320 in the deal,What was the
original amount?
Sol:- Let the original amount be Rs x.
T = 12
R1 = 8%
R2 = 12%
Profit = 320
P = x
(P*T*R2)/100 - (P*T*R1)/100 =320
(x*12*12)/100 - (x*8*12)/100 = 320
x = 2000/3
x = Rs.666.67
6. Simple Interest on a certail sum at a certain rate is
9/16 of the sum . if the number representing rate percent
and time in years be equal ,then the rate is.
Sol:- Let Sum = x .Then,
S.I = 9x/16
Let time = n years & rate = n%
n = 100 * 9x/16 * 1/x * 1/n
n * n = 900/16
n = 30/4 = 7 1/2%
Complex Problems
1. A certain sum of money amounts t 1680 in 3yrs & it
becomes 1920 in 7 yrs .What is the sum.
Sol:- 3 yrs - - - - - - - - - - - - - 1680
7 yrs - - - - - - - - - - - - - 1920
then, 4 yrs - - - - - - - - - - - - - 240
1 yr - - - - - - - - - - - - - ?
(1/4) * 240 = 60
S.I in 3 yrs = 3*60 = 18012
Sum = Amount - S.I
= 1680 - 180
= 1500
we get the same amount if we take S.I in 7 yrs
I.e., 7*60 =420
Sum = Amount - S.I
= 1920 - 420
= 1500
2. A Person takes a loan of Rs 200 at 5% simple Interest.
He returns Rs.100 at the end of 1 yr. In order to clear
his dues at the end of 2yrs ,he would pay:
Sol:- Amount to be paid
= Rs(100 + (200*5*1)/100 + (100*5*1)/100)
= Rs 115
3. A Man borrowed Rs 24000 from two money lenders.For one
loan, he paid 15% per annum and for other 18% per annum.
At the end of one year,he paid Rs.4050.How much did he
borrowed at each rate?
Sol:- Let the Sum at 15% be Rs.x
& then at 18% be Rs (24000-x)
P1 = x R1 = 15
P2 = (24000-x) R2 = 18
At the end of ine year T = 1
(P1*T*R1)/100 + (P2*T*R2)/100 = 4050
(x*1*15)/100 + ((24000-x)*1*18)/100 = 4050
15x + 432000 - 18x = 405000
x = 9000
Money borrowed at 15% = 9000
Money borrowed at 18% = (24000 - 9000)
= 15000
4.What annual instalment will discharge a debt of Rs. 1092
due in 3 years at 12% Simple Interest ?
Sol:- Let each instalment be Rs x
(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092
28x/25 + 31x/25 + x =1092
(28x +31x + 25x) = (1092 * 25)
84x = 1092 * 25
x = (1092*25)/84 = 325
Each instalement = 325
5.If x,y,z are three sums of money such that y is the simple
interest on x,z is the simple interest on y for the same
time and at the same rate of interest ,then we have:
Sol:- y is simple interest on x, means
y = (x*R*T)/100
RT = 100y/x
z is simple interest on y,
z = (y*R*T)/100
RT = 100z/y
100y/x = 100z/y
y * y = xz
6.A Sum of Rs.1550 was lent partly at 5% and partly at 5%
and partly at 8% p.a Simple interest .The total interest
received after 3 years was Rs.300.The ratio of the money
lent at 5% to that lent at 8% is:
Sol:- Let the Sum at 5% be Rs x
at 8% be Rs(1550-x)
(x*5*3)/100 + ((1500-x)*8*3)/100 = 300
15x + 1500 * 24 - 24x = 30000
x = 800
Money at 5%/ Money at 8% = 800/(1550 - 800)
= 800/750 = 16/15
7. A Man invests a certain sum of money at 6% p.a Simple
interest and another sum at 7% p.a Simple interest. His
income from interest after 2 years was Rs 354 .one
fourth of the first sum is equal to one fifth of the
second sum.The total sum invested was:
Sol:- Let the sums be x & y
R1 = 6 R2 = 7
T = 2
(P1*R1*T)/100 + (P2*R2*T)/100 = 354
(x * 6 * 2)/100 + (y * 7 * 2)/100 = 354
6x + 7y = 17700 ———(1)
also one fourth of the first sum is equal to one
fifth of the second sum
x/4 = y/5 => 5x - 4y = 0 —— (2)
By solving 1 & 2 we get,
x = 1200 y = 1500
Total sum = 1200 +1500
= 2700
8. Rs 2189 are divided into three parts such that their
amounts after 1,2& 3 years respectively may be equal,
the rate of S.I being 4% p.a in all cases. The Smallest
part is:
Sol:- Let these parts be x,y and[2189-(x+y)] then,
(x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100
4x/100 = 8y/100
x = 2y
By substituting values
(2y*1*4)/100 = (2189-3y)*3*4/100
44y = 2189 *12
y = 597
Smallest Part = 597
9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and
the remainder at 10%.If his annual income is Rs.561. The
capital is:
Sol:- Let the capital be Rs.x
Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)
+ (5x/12 * 10/100 * 1) = 561
7x/300 + x/50 + x/24 = 561
51x = 561 * 600
x = 6600
Alligation or Mixtures
Important Facts and Formulae:
1.Allegation:It is the rule that enables us to find the
ratio in which two of more ingredients at the given price
must be mixed to produce a mixture of a desired price.
2.Mean Price:The cost price of a unit quantity of the
mixture is called the mean price.
3.Rule of Allegation:If two ingredients are mixed then
Quantity of Cheaper / Quantity of Dearer =
(C.P of Dearer – Mean Price) /(Mean Price–C.P of Cheaper).
C.P of a unit quantity of cheaper(c)
C.P of unit quantity of dearer(d)
Mean Price(m)
(d-m) (m-c)
Cheaper quantity:Dearer quantity = (d-m):(m-c)
4.Suppose a container contains x units of liquid from which
y units are taken out and replaced by water. After n
operations the quantity of pure liquid = x (1 – y/x)n units.
Simple Problems
1.In what ratio must rice at Rs 9.30 per Kg be mixed with rice
at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?
Solution:
C.P of 1 Kg rice of 1st kind 930 p
C.P of 1 Kg rice of 2n d kind 1080p
Mean Price 1000p
80 70
Required ratio=80:70 = 8:7
2.How much water must be added to 60 liters of milk at
11/2 liters for Rs 20 so as to have a mixture worth
Rs 10 2/3 a liter?
Solution: C.P of 1 lit of milk = 20*2/3 = 40/3
C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3
Mean Price 32/3
8/3 32/3
Ratio of water and milk =8/3 : 32/3 = 1:4
Quantity of water to be added to 60 lit of milk
=1/4*60=15 liters.
3.In what ratio must water to be mixed with milk to gain
20% by selling the mixture at cost price?
Solution:Let the C.P of milk be Re 1 per liter
Then S.P of 1 liter of mixture = Re.1
Gain obtained =20%.
Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6
C.P of 1 liter of water 0 C.P of 1 liter of milk1
Mean Price 5/6
1/6 5/6
Ratio of water and milk =1/6 : 5/6 = 1:5.
4.In what ratio must a grocer mix two varieties of pulses
costing Rs 15 and Rs 20 per Kg respectively so as to get
a mixture worth Rs 16.50 per Kg?
Solution:
Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of
2nd kind 20
Mean Price Rs 16.50
3.50 1.50
Required ratio =3.50 : 1.50 = 35:15 = 7:3.
5. 4Kg s of rice at Rs 5 per Kg is mixed with 8 Kg of rice
at Rs 6 per Kg .Find the average price of the mixture?
Solution:
rice of 5 Rs per Kg rice of 6 Rs per Kg
Average price Aw
6-Aw Aw-5
(6-Aw)/(Aw-5) = 4/8 =1/2
12-2Aw = Aw-5
3Aw = 17
Aw = 5.66 per Kg.
6.5Kg of rice at Rs 6 per Kg is mixed with 4 Kg of rice to
get a mixture costing Rs 7 per Kg. Find the price
of the costlier rice?
Solution: Using the cross method:
rice at Rs 6 per Kg rice at Rs x per Kg
Mean price Rs 7 per Kg
5 4
x-7:1=5:4
4x-28 = 5
4x=33=>x=Rs 8.25.
Therefore price of costlier rice is Rs 8.25 per Kg
Medium Problems
1.A butler stole wine from a butt of sherry which contained
40% of spirit and he replaced,what he had stolen by wine
containing only 16% spirit. The butt was then of 24%
strength only. How much of the butt did he steal?
Solution:
Wine containing 40%spirit Wine containing 16% spirit
Wine containing 24% spirit
8 16
They must be mixed in the ratio of =1:2.
Thus 1/3 of the butt of sherry was left and hence the
butler drew out 2/3 of the butt.
2.The average weekly salary per head of the entire staff
of a factory consisting of supervisors and the laborers
is Rs 60.The average salary per head of the supervisors
is Rs 400 and that of the laborers is Rs 56.Given that
the number of supervisors is 12.Find the number of
laborers in the factory.
Solution:
Average salary of laborer Rs 56 Average salary of
supervisors Rs 400
Average salary of entire staff Rs 60
340 4
Number of laborer / Number of Supervisors = 340 / 4=85/1
Thus,if the number of supervisors is 1,number of
laborers =85.
Therefore if the number of supervisors is 12 number of
laborers 85*12=1020.
3.The cost of type 1 rice is Rs 15 per Kg and type 2 rice
is Rs 20 per Kg. If both type1 and type 2 are mixed in the
ratio of 2:3,then the price per Kg of the mixed variety
of rice is?
Solution:Let the price of the mixed variety be Rs x per Kg.
Cost of 1 Kg of type 1 rice Rs 15 Cost of 1 Kg of type 2
rice Rs 20
Mean Price Rs x
20-x x-15
(20-x) /( x-15) = 2/3
=> 60-3x = 2x-30
5x = 90=>x=18.
4.In what ratio must a grocer mix two varieties of tea worth
Rs 60 a Kg and Rs 65 a Kg so that by selling the mixture
at Rs 68.20 a Kg he may gain 10%?
Solution:S.P of 1 Kg of the mixture = Rs 68.20,gain =10%
S.P of 1 Kg of the mixture = Rs (100/110*68.20)=Rs 62.
Cost of 1 Kg tea of 1st kind 60 Cost of 1 Kg tea of 2nd
kind 65
Mean Price Rs 62
3 2
Required ratio =3:2.
5.A dishonest milkman professes to sell his milk at cost price
but he mixes t with water and there by gains 25% .The
percentage of water in the mixture is?
Solution:Let C. P of 1 liter milk be Re 1.
Then S.P of 1 liter mixture=Re 1. Gain=25%
C.P of 1 liter mixture =Re(100/125*1) = Re 4/5.
C.P of 1 liter milk Re 1 C.P of 1 liter of water 0
Mean Price 4/5
4/5 1/5
Ratio of milk to water =4/5 : 1/5 = 4:1
Hence percentage of water n the mixture=1/5*100=20%.
12.A merchant has 1000Kg of sugar,part of which he sells
at 8% profit and the rest at 18% profit. He gains 14% on the
whole. The quantity sold at 18% profit is?
Solution:
Profit on 1st part 8% Profit on 2nd part 18%
Mean Profit 14%
4 6
Ratio of 1st and 2nd parts =4:6 =2:3.
Quantity of 2nd ind =3/5*1000Kg =600 Kg.
6.A jar full of whiskey contains 40% alcohol. A part of
this whiskey is replaced by another containing 19% alcohol
and now the percentage of alcohol was found to be 26%.
The quantity of whiskey replaced is?
Solution:
Strength of first jar 40% Strength of 2nd jar 19%
Mean Strength 26%
7 14
So,ratio of 1st and 2nd quantities =7:14 =1:2
Therefore required quantity replaced =2/3.
7.A container contains 40lit of milk. From this container
4 lit of milk was taken out and replaced by water.
This process was repeated further two times.
How much milk is now contained by the container?
Solution:Amount of milk left after 3 operations = 40(1-4/40)3lit
=(40*9/10*9/10*9/10)
= 29.16 lit
Complex Problems
1.Tea worth Rs 126 per Kg are mixed with a third variety in
the ratio 1:1:2. If the mixture is worth Rs 153 per Kg ,
the price of the third variety per Kg will be?
Solution:
Since First and second varieties are mixed in equal proportions
so their average price =Rs (126+135)/2 = 130.50.
So the mixture is formed by mixing two varieties ,one at
Rs 130.50 per Kg and the other at say Rs x per Kg in the
ratio 2:2 i e,1:1 we have to find x.
Costof 1Kg tea of 1st kind RS 130.50 Costof 1Kg tea of 2n d
kind Rs x.
Mean Price Rs 153
x-153 22.50
(x=153)/22.5 = 1 =>x-153 = 22.5
x = 175.50.
Price of the third variety =Rs 175.50 per Kg.
2.The milk and water in two vessels A and B are in the ratio 4:3
and 2:3 respectively. In what ratio the liquids in both the
vessels be mixed to obtain a new mixture in vessel c
consisting half milk and half water?
Solution:Let the C.P of milk be Re 1 per liter.
Milk in 1 liter mixture of A = 4/7 liter.
Milk in 1 liter mixture of B = 2/5 liter.
Milk in 1 liter mixture of C = 1/2 liter.
C.P of 1 liter mixture in A=Re 4/7
C.P of 1 liter mixture in B=Re 2/5.
Mean Price = Re 1/2.
By rule of allegation we have:
C.P of 1 liter mixture in A C.P of 1 liter mixture in B
4/7 2/5
Mean Price ½
1/10 1/14
Required ratio = 1/10 : 1/14 = 7:5.
3.How many Kg s of wheat costing him Rs 1.20,Rs 1.44
and Rs 1.74 per Kg so that the mixture may be worth
Rs 1.41 per Kg?
Solution:
Step1:Mix wheat of first and third kind to get a mixture
worth Rs 1.41 per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of
3rd kind 174p
Mean Price 141p
33 21
They must be mixed in the ratio =33:21 = 11:7
Step2:Mix wheats of 1st and 2n d kind to obtain a mixture
worth of 1.41.per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 2n d
kind 144p
Mean Price 141p
3 21
They must be mixed in the ratio = 3:21=1:7.
Thus,Quantity of 2n d kind of wheat / Quantity of
3rd kind of wheat = 7/1*11/7= 11/1
Quantities of wheat of 1st :2n d:3rd = 11:77:7.
4.Two vessels A and B contain spirit and water mixed in
the ratio 5:2 and 7:6 respectively. Find the ratio n which
these mixture be mixed to obtain a new mixture in vessel
c containing spirit and water in the ratio 8:5?
Solution:Let the C.P of spirit be Re 1 per liter.
Spirit in 1 liter mix of A = 5/7 liter.
C.P of 1 liter mix in A =5/7.
Spirit in 1 liter mix of B = 7/13 liter.
C.P of 1 liter mix in B =7/13.
Spirit in 1 liter mix of C = 8/13 liter.
C.P of 1 liter mix in C =8/13.
C.P of 1 liter mixture in A 5/7 C.P of 1 liter mixture
in B 7/13
Mean Price 8/13
1/13 9/91
Therefore required ratio = 1/13 : 9/91 = 7:9.
5.A milk vendor has 2 cans of milk .The first contains 5% water
and the rest milk. The second contains 50% water. How much milk
should he mix from each of the container so as to get 12 liters
of milk such that the ratio of water to milk is 3:5?
Solution:Let cost of 1 liter milk be Re 1.
Milk in 1 liter mixture in 1st can = 3/4 lit.
C.P of 1 liter mixture in 1st can =Re 3/4
Milk in 1 liter mixture in 2n d can = 1/2 lit.
C.P of 1 liter mixture in 2n d can =Re 1/2
Milk in 1 liter final mixture = 5/8 lit.
Mean Price = Re 5/8.
C.P of 1 lt mix in 1st Re3/4 C.P of 1 lt mix in 2nd Re1/2
Mean Price 5/8
1/8 1/8
There ratio of two mixtures =1/8 :1/8 = 1:1.
So,quantity of mixture taken from each can=1/2*12
= 6 liters.
6.One quantity of wheat at Rs 9.30 p
er Kg are mixed
with another quality at a certain rate in the ratio 8:7.
If the mixture so formed be worth Rs 10 per Kg ,what is
the rate per Kg of the second quality of wheat?
Solution:Let the rate of second quality be Rs x per Kg.
C.P of 1Kg wheat of 1st 980p C.P of 1 Kg wheat of 2nd 100x p
Mean Price 1000p
100x-1000p 70 p
(100x-1000) / 70 = 8/7
700x -7000 = 560
700x = 7560 =>x = Rs 10.80.
Therefore the rate of second quality is Rs10.80
7.8lit are drawn from a wine and is then filled with water.
This operation is performed three more times.The ratio of
the quantity of wine now left in cask to that of the water
is 16:81. How much wine did the cask hold originally?
Solution:
Let the quantity of the wine in the cask originally be
x liters.
Then quantity of wine left in cask after
4 operations = x(1- 8/x)4lit.
Therefore x((1-(8/x))4)/x = 16/81.
(1- 8/x)4=(2/3) 4
(x- 8)/x=2/3
3x-24 =2x
x=24.
8.A can contains a mixture of two liquids A and B in the
ratio 7:5 when 9 liters of mixture are drawn off and the
can is filled with B,the ratio of A and B becomes 7:9.
How many liters of liquid A was contained by the can initially?
Solution:
Suppose the can initially contains 7x and 5x liters
of mixtures A and B respectively .
Quantity of A in mixture left = (7x- (7/12)*9 )lit
= 7x - (21/4) liters.
Quantity of B in mixture left = 5x - 5/12*9
= 5x - (15/4) liters
Therefore (7x – 21/4)/ (5x – 15/4+9)=7/9
(28x-21)/(20x +21)= 7/9
(252x -189)= 140x +147
112x = 336
=> x=3.
So the can contains 21 liters of A.
9.A vessel is filled with liquid,3 parts of which are water
and 5 parts syrup. How much of the mixture must be drawn off
and replaced with water so that the mixture may be
half water and half syrup?
Solution:
Suppose the vessal initially contains 8 liters of liquid.
Let x liters of this liquid be replaced with water
then quantity of water in new mixture
= 3-(3x/8)+x liters.
Quantity of syrup in new mixture = 5 - 5x/8 liters.
Therefore 3 - 3x/8 +x = 5 - 5x/8
5x+24 = 40-5x
10x = 16.
x= 8/5.
So part of the mixture replaced = 8/5*1/8 =1/5.
Boats and Streams
Important facts:
1)In water, the direction along the stream is called down stream.
2)Direction against the stream is called upstream.
3)The speed of boat in still water is U km/hr and the speed of
stream is V km/hr then
speed down stream =U + V km/hr
speed up stream = U – V km/hr
Formulae:
If the speed down stream is A km/hr and the speed up stream is
B km/hr then speed in still water = ½(A+B) km/hr
rate of stream =1/2(A-B) km/hr
Problems:
1. In one hour a boat goes 11 km long the stream and 5 km
against the stream. The speed of the boat in still water is?
Sol:
Speed in still water = ½ ( 11+5) km/hr= 8 kmph
2.A man can row 18 kmph in still water. It takes him thrice
as long as row up as to row down the river. find the rate
of stream.
Sol:
Let man's rate up stream be xkmph
then, in still water =1/2[3x+x]=2x kmph
so, 2x= 18, x=9
rate upstream =9kmph
rate downstream =27 kmph
rate of stream = ½ [27-9]
= 9kmph
3.A man can row 71/2kmph in still watre . if in a river
running at 1.5 km an hour, if takes him 50 min to row to
place and back. how far off is the place?
Sol: speed down stream =7.5+1.5=9kmph
speed upstream =7.5-1.5=6kmph
let the required distence x km. then ,
x/9+x/6=50/60 = 2x+3x= 5/6*18
5x=15, x=3
Hence, the required distence is 3 km
4.A man can row 3 quarters of a km aganist the stream is
111/4 min. the speed of the man in still water is ?
Sol: rate upstream = 750/625 m/sec =10/9 m/sec
rate downstream =750/450 m/sec = 5/3 m/sec
rate in still water =1/2[10/9+5/3] = 25/18 m/sec
= 25/18*18/5=5 kmph
5.A boat can travel with a speed of 13 kmph in still water.
if the speed of stream is 4 kmph,find the time taken by
the boat to go 68 km downstream?
Sol: Speed down stream = 13+4= 17 kmph
time taken to travel 68km downstream =68/17 hrs
= 4 hrs
6.A boat takes 90 min less to travel 36 miles downstream then
to travel the same distence upstream. if the speed of the
boat in still water is 10 mph . The speed of the stream is :
Sol: Let the speed of the stream be x mph .
then, speed downstream = [10+x]mph
speed upstream =[10-x] mph
36/[10+x] - 36/[10-x] = 90/60 =72x*60= 90[100-x2]
(x+50)(x-2) =0
x=2 kmph
7.At his usual rowing rate, Rahul 12 miles down stream in a
certain river in 6 hrs less than it takes him to travel the
same distence upstream. but if he could double his usual
rowing rate for his 24 miles roundthe down stream 12 miles
would then take only one hour less than the up stream 12 miles.
what is the speed of the current in miles per hours?
Sol: Let the speed in still water be x mph and the speed of
the curren be y mph.
then, speed upstream = (x-y)
speed downstream =(x+y)
12/(x-y) - 12/(x+y) = 6
6(x2 – y2) m= 2xy => x2 – y2 =4y -(1)
and 12/(2x-y) - 12/(2x+y) =1 => 4x2 – y2 = 24y
x2= ( 24y + y2)/4 -->(2)
from 1 and 2 we have
4y+ y2 =( 24y+y2)/4
y=8/3 mph
y= 22/3 mph
8.There is a road beside a river. two friends started from
a place A, moved to a temple situated at another place B
and then returned to A again. one of them moves on a cycle
at a speed of 12 kmph, while the other sails on a boat at a
speed of 10 kmph . if the river flows at the speedof 4 kmph,
which of the two friends will return to place A first ?
Sol: Clearly, The cyclist moves both ways at a speed of 12 kmph
so, average speed of the cyclist = 12 kmph
the boat sailor moves downstream = (10+4) = 14 kmph
upstream =(10-4) = 6 kmph
So, average speed of the boat sailor =[ 2*14*6]/[14+6] kmph
=42/5 kmph =8.4 kmph
Since, the average speed of the cyclist is greater, he will
return to A first.
9.A boat takes 19 hrs for travelling downstream from point A to
point B. and coming back to a point C midway between A and B.
if the velocity of the sream is 4 kmph. and the speed of the
boat in still water is 14 kmph. what is the distence between
A and B?
Sol:
speed downstream =14+4 =18 kmph
speed upstream = 14 -4 = 10 kmph
let the distence between A and B be x km. then,
x/18 + (x/2)/10 = 19
x/18 + x/20 =19
19x/180 =19 =>x = 180km
Hence, the distence between A and B bw 180 km
Trains
General Concept:
(1) Time taken by a train x mt long in passing a signal post
or a pole or a standing man = time taken by the train to cover x mt
(2) Time taken by a train x mt long in passing a stationary
object of length y mt = time taken by the train to cover x+y mt
(3) Suppose two trains or two bodies are moving in the same
direction at u kmph and v kmph such that u > v then their
relative speed is u-v kmph
(4)If two trains of length x km and y km are moving in opposite
diredtions at u kmph and vmph,then time taken by the train to
cross each other = (x+y)/(u+v) hr
(5) Suppose two trains or two bdies are moving in opposite direction
at u kmph and v kmph then,their relative speed = (u+v) kmph
(6)If two train start at the same time from 2 points A & B towards
each other and after crossing they take a & b hours in reaching B & A
respectively then A's speed : B's speed = (b^1/2 : a^1/2 )
Problems
(1)Find the time taken by a train 180m long,running at 72kmph in
crossing an electric pole
Solution:
Speed of the train =72*5/18m/s =20 m/s
Distance move din passing the pole = 180m
Requiredtime = 180/20 = 9 seconds
(2)A train 140 m long running at 60kmph.In how much time will it
pass a platform 260m long.
Solution:
Distance travelled =140 + 260 m =400 m,
speed = 60 * 5/18 = 50//3 m
time=400*3 / 50 = 24 Seconds
(3)A man is standing on a railway bridge which is 180 m.He finds
that a train crosses the bridge in 20 seconds but himself in
8 sec. Find the length of the train and its sppeed
Solution:
i)D=180+x
T = 20 seconds
S= 180+x / 20 ------------ 1
ii)D=x
T=8 seconds
D=ST
x=8S ------------- 2
Substitute 2 in 1
S=180 + 8 S / 20
S=15 m/s
Length of the train,x is 8 *15 = 120 m
(4)A train 150m long is running with a speed of 68 mphIn wht
time will it pass a man who is running at a speed of 8kmph in
the same direction in which the train is going
Solution:
Relative Speed = 68-8=60kmph*5/18 = 50/3 m/s
time= 150 * 3 /50 =9sec
5)A train 220m long is running with a speed of 59 k mph /..In
what time will it pass a man who is running at 7 kmph in the
direction opposite to that in which train is going.
Solution:
Relative Speed = 59+7=66kmph*5/18 = 55/3 m/s
time= 220/55 * 3 =12sec
(6)Two trains 137m and 163m in length are running towards each
other on parallel lines,one at the rate of 42kmph & another at
48 mph.In wht time will they be clear of each other from the
moment they meet.
Solution:
Relative speed =42+48 = 90 *5/18 = 25m/s
time taken by the train to pass each other = time taken to cover
(137+163)m at 25 m/s
= 300 /25 s =12 s
(7)A train running at 54 kmph takes 20 sec to pass a platform.
Next it takes 12 sec to pass a man walking at 6kmph in the same
direction in which the train is going.Find length of the train
and length of platform
Solution:
Relative speed w.r.t man = 54-6=48kmph
the length of the train is 48 * 5/18 * 12 =160m
time taken to pass platform =20 sec
Speed of the train = 54 * 5/18 =15m/s
160+x =20 *15
x=140m
length of the platform is 140m
(8)A man sitting in a train which is travelling at 50mph observes
that a goods train travelling in opposite irection takes 9 sec
to pass him .If the goos train is 150m long fin its speed
Solution:
Relative speed =150/9 m/s =60 mph
speed of the train = 60-50 =10kmph
(9)Two trains are moving in the sam e direction at 65kmph and
45kmph. The faster train crosses a man in slower train in18sec.the
length of the faster train is
Solution:
Relative speed =65-45 kmph = 50/9 m/s
Distancce covered in18 s =50/9 * 18 = 100m
the length of the train is 100m
(10)Atrain overtakes two persons who are walking in the same
direction in which the train is going at the rate of 2kmph an
4kmph and passes them completely in 9 sec an 10 sec respectively.
The length of train is
Solution:
2kmph = 5/9 m/s
4 mph =10/9 m/s
Let the length of the trainbe x meters and its speed is y m/s
then x / (y- 5/9) = 9 and x / (y- 10/9) = 10
9y-5 =x and 10(9y-10)=9x
9y-x=5 and 90y-9x=100
on solving we get x=50,lenght of trains
(11) Two stations A & B are 110 km apart on a straight line.
One train starts from A at 7am and travels towards B at 20kmph.
Another train starts from B at 8am an travels toward A at a speed
of 25kmph.At what time will they meet
Solution:
Suppose the train meet x hr after 7am
Distance covered by A in x hr=20x km
20x+25(x-1) = 110
45x=135
x=3
So they meet at 10 am
(12)A traintravelling at 48kmph completely crosses another train
having half its length an travelling inopposite direction at 42kmph
in12 sec.It also passes a railway platform in 45sec.the length of
platform is
Solution:
Let the length of the first train be x mt
then,the length of second train is x/2 mt
relative speed = 48+42 kmph =90 * 5/18 m/s = 25m/s
(x+ x/2)/25 =12
x=200
Length of the train is 200m
Let the length of the platform be y mt
speed f the first train = 48*5/18 m/s = 40/3 m/s
200+y * 3/40 = 45
y=400m
(13)The length of a running trsain in 30% more than the length of
another train B runnng in the opposite direction.To find out the
speed of trtain B,which of the following information given in the
statements P & Q is sufficient
P : The speed of train A is 80 kmph
Q : They too 90 sec to cross each other
(a) Either P & Q is sufficient
(b)Both P & Q are not sufficient
(c)only Q is sufficient
(d)Both P & Q are neeed
Ans: B
Solution:
Let the length of th e train A be x mt
Length of the train B = 130/100 x mt =13x/10 mt
Let the speed of B be y mph,speed of the train A=80mph
relative speed= y+80 * 5/18 m/s
time taken by the trains t cross each other is gven by
90 = (x + 13x/10)/ (5y+400 / 18)
to find y,clearly xis also needed
so,both P & Q are not sufficient
(14)The speed of a train A,100m long is 40% more than then the speed
of another train B,180m long running in opposite direction.To fin out
the speed of B,which of the information given in statements P & Q is
sufficient
P :The two trains crossed each other in 6 seconds
Q : The difference between the spee of the trains is 26kmph
(a)Only P is sufficient
(b)Only Q is sufficient
(c)Both P & Q are needed
(d)Both P & Q are not sufficient
Ans : A
Solution:
Let speed of B be x kmph
then,speed of A =140x/100 kmph =7x/5 mph
relative speed = x + 7x/5 =2x/3 m/s
time taken to cross each other = (100+180)*3/2x s =420/x s
now,420/x = 6
x=70 mph
thus,only P is sufficient
(15)The train running at certain speed crosses astationary enginein
20 seconds.to find out the sped of the train,which of the following
information is necessary
(a)Only the length of the train
(b)only the length of the engine
(c)Either the length of the train or length of engine
(d)Both the length of the train or length of engine
Ans : D
Solution:
Since the sum of lengths of the tran and the engine is needed,
so both the length must be known
Time and Distance
Formulae:
I)Speed = Distance/Time
II)Time = Distance/speed
III) Distance = speed*time
IV) 1km/hr = 5/18 m/s
V)1 m/s = 18/5 Km/hr
VI)If the ratio of the speed of A and B is a:b,then the ratio of
the time taken by them to cover the same distance is 1/a : 1/b
or b:a
VII) suppose a man covers a 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
1)A person covers a certain distance at 7kmph .How many meters
does he cover in 2 minutes.
Solution::
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m
2)If a man runs at 3m/s. How many km does he run in 1hr 40min
Solution::
speed of the man = 3*18/5 kmph
= 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
3)Walking at the rate of 4knph a man covers certain distance
in 2hr 45 min. Running at a speed of 16.5 kmph the man will
cover the same distance in.
Solution::
Distance=Speed* time
4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
Complex Problems
1)A train covers a distance in 50 min ,if it runs at a speed
of 48kmph on an average.The speed at which the train must run
to reduce the time of journey to 40min will be.
Solution::
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph
2)Vikas can cover a distance in 1hr 24min by covering 2/3 of
the distance at 4 kmph and the rest at 5kmph.the total
distance is?
Solution::
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.
Solution::
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
4)A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
Solution::
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x --------------1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2
divide 1 & 2 equations
by solving we get x = 40
5)Excluding stoppages,the speed of the bus is 54kmph and
including stoppages,it is 45kmph.for how many min does the bus
stop per hr.
Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
6)Two boys starting from the same place walk at a rate of
5kmph and 5.5kmph respectively.wht time will they take to be
8.5km apart, if they walk in the same direction
Solution::
The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
7)2 trains starting at the same time from 2 stations 200km
apart and going in opposite direction cross each other ata
distance of 110km from one of the stations.what is the ratio of
their speeds.
Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9
8)Two trains start from A & B and travel towards each other at
speed of 50kmph and 60kmph resp. At the time of the meeting the
second train has traveled 120km more than the first.the distance
between them.
Solution::
Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the
theft is discovered at 3pm and the owner sets off in another car
at 75kmph when will he overtake the thief
Solution::
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
10)In covering distance,the speed of A & B are in the ratio
of 3:4.A takes 30min more than B to reach the destion.The time
taken by A to reach the destinstion is.
Solution::
Ratio of speed = 3:4
Ratio of time = 4:3
let A takes 4x hrs,B takes 3x hrs
then 4x-3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
11)A motorist covers a distance of 39km in 45min by moving at a
speed of xkmph for the first 15min.then moving at double the
speed for the next 20 min and then again moving at his original
speed for the rest of the journey .then x=?
Solution::
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph
12)A & B are two towns.Mr.Fara covers the distance from A t0 B
on cycle at 17kmph and returns to A by a tonga running at a
uniform speed of 8kmph.his average speed during the whole
journey is.
Solution::
When same distance is covered with different speeds,then the
average speed = 2xy / x+y
=10.88kmph
13)A car covers 4 successive 3km stretches at speed of
10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.
Solution::
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.
The average speed is.
Solution::
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
15)A bullock cart has to cover a distance of 80km in 10hrs.
If it covers half of the journeyin 3/5th time.wht should be its
speed to cover the remaining distance in the time left.
Solution::
Time left = 10 - 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph
16)The ratio between the speeds of the A& B is 2:3 an
therefore A takes 10 min more than the time taken by B to reach
the destination.If A had walked at double the speed ,he would
have covered the distance in ?
Solution::
Ratio of speed = 2:3
Ratio of time = 3:2
A takes 10 min more
3x-2x = 10 min
A's time=30 min
--->A covers the distance in 30 min ,if its speed is x
-> He will cover the same distance in 15 min,if its speed
doubles (i.e 2x)
17)A is twice as fast as B and B is thrice as fast as C is.
The journey covered by B in?
Solution::
speed's ratio
a : b = 2: 1
b : c = 3:1
Time's ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )
18)A man performed 3/5 of the total journey by ratio 17/20 by
bus and the remaining 65km on foot.wht is his total journey.
Solution::
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km
19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .
Another train N leaves Delhi at 7am and reaches Meerut at 1030am
At what time do the 2 trains cross one another
Solution::
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min
20)A man takes 5hr 45min in walking to certain place and riding
back. He would have gained 2hrs by riding both ways.The time he
would take to walk both ways is?
Solution::
Let x be the speed of walked
Let y be the speed of ride
Let D be the distance
Then D/x + D/y = 23/4 hr -------1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 --------2
substitute 2 in 1
D/x + 15/8 = 23/4
D/x = 23/4 -15/8 =46-15/8 =31/8
Time taken for walk one way is 31/8 hr
time taken to walk to and fro is 2*31/8 = 31/4 hr
=7 hr 45 min
Subscribe to:
Posts (Atom)