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Numbers
Ratio, Proportion, Variation
Averages,Mixtures and Alligation
Time,Speed and Distance
Time and Work
Percentages
Profit and Loss
Simple & Compound Interest
Fractions
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Progression
Permutation and Combination
Probability
Geometry and Mensuration
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Boats & Streams
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P3Home > General Placement
Preparation > Courses > Quantitative > Time and Work > Time
and Work-Key Notes
Time and Work-Key Notes
Time and Work-Key Notes-aptitude-questions-answers-basics
Time and Work Aptitude basics, practice questions, answers and
explanations
Prepare for companies tests and interviews
Time & Work
(M1D1HI)/ W1 = (M2D2H2)/W2
Where M1 & M2 represents no of labourers; D1 & D2 represents no of
days; H1 & H2 represents no of hours; W1&W2 represents work done.
If there are 2 persons A & B such that A can do work in ‘a’ days and B can
do work in ‘b’days. Such that ‘a’ is a multiple of ‘b’, then, time taken by
them to complete the work together = Bigger no/Sum of ratios
Eg: A can do work in 9 days, B can do work in 18 days. In how many days
they will complete the work together.
Bigger no=18, Ratio=9:18=1:2
No of days = 18/(1 + 2)
= 6 days
If ‘a’ is not a multiple of ‘b’, then time taken by A&B to complete the work
together
= (LCM)/(Sum of ratios)
Eg: A can do a piece of work 30 days. B can do work in 45 days. In how
many days they will complete the work together?
LCM = 90, Ratio= 30:45=2:3
No of days= 90/(2 + 3) = 90/5 = 18
If there are 3 persons A, B & C whose time taken a,b,c days respectively, to
complete a certain work. Time taken by them to complete the work = LCM
of (a, b, c)/(LCM/a + LCM/b + LCM/c)
Note: For 3 persons the common format is
Step1: Find the LCM
Step2: Find the individual share of work i.e. LCM/a, LCM/b, LCM/c.
Step3: Rest methods depend on the question i.e. follow the question
Eg: A, B and C can do a work in 15,20,45 days respectively. In how many
days they can complete the work together.
LCM=180
No of days= [180/ (180/15 + 180/20 + 180/45)
= [180/ (12+9+4)]
= [180/25]
= 7.2 days
Pipes & Cisterns
If there are 2 pipes A & B such that A (inlet pipe) & B (outlet pipe). Such
that A can fill tank in ‘a’ minutes and B can empty the tank in ‘b’ minutes ,
then time taken to fill the tank if both are operated together = Bigger
no/Difference of ratios
Eg: A can fill tank in 9 minutes, B can empty the tank in 18 minutes.. In
what time the tank be filled, if both pipes work simultaneously?
Bigger no=18, Ratio=9:18=1:2
Time taken to fill the tank = 18/(2 - 1)
= 18 minutes
If ‘a’ is not a multiple of ‘b’, then time taken by A&B to fill the tank.
= (LCM)/(Difference of ratios)
Eg.: An inlet pipe can fill the tank in 30 minutes. B an outlet pipe can
empty the tank in 45 minutes. In what time the tank be filled if both pipes
work simultaneously?
Time taken to fill the tank= LCM = 90
= Ratio= 30:45=2:3
= 90/(3 - 2) = 90/1 = 90 minutes
If there are 3 pipes A, B & C, in which A, B are inlet pipes which takes
a,b,minutes respectively to fill the tank and C an outlet pipe which takes c
minutes to empty the tank
Time taken by them to fill the tank, if all of them are operated together.
= LCM of abc/ (LCM/a + LCM/b - LCM/c)
Eg: A, B two inlet pipes takes 15,18 minutes to fill the tank and C an oulet
pipe takes 45 minutes to empty the tank respectively. In what time the
tank be filled if all of them are operated together?
LCM=90
No of days= [90/(90/15 + 90/18 - 90/45)
= [90/(6+5-2)]
= [90/9]
= 10 minutes
Note: In case of division of money with respect to share of each person’s
work then share of A = bc/ab+bc+ac
In case of division of money with respect to share of each person’s work
then share of B = ac/ab+bc+ac
In case of division of money with respect to share of each person’s work
then share of C = ab/ab+bc+ac
Same as Share of A:(LCM/a)/ (LCM/a + LCM/b + LCM/c)
Share of B:(LCM/b)/ (LCM/a + LCM/b + LCM/c)
Share of C:(LCM/c)/ (LCM/a + LCM/b + LCM/c)
Eg: A,B,C can do a work in 15,20,45 days respectively. They get Rs 500 for
their work. What is the share of A?
LCM = 180
Share of A = (LCM/a x Total amount)/LCM/a + LCM/b + LCM/c
= (180/15)/(180/15 +180/20 + 180/45)
= (12/25) * 500
= Rs.240
patterns.
Exercise questions
1. Two workers A and B manufactured a batch of identical parts. A worked
for 2 hours and B worked for 5 hours and they did half the job. Then they
worked together for another 3 hours and they had to do (1/20)th of the
job. How much time does B take to complete the job, if he worked alone?
A) 24 hours
B) 12 hours
C) 15 hours
D) 30 hours
2. Pipe A can fill a tank in 'a' hours. On account of a leak at the bottom of
the tank it takes thrice as long to fill the tank. How long will the leak at the
bottom of the tank take to empty a full tank, when pipe A is kept closed?
A) (3/2)a hours
B) (2/3)a
C) (4/3)a
D) (3/4)a
3. A and B working together can finish a job in T days. If A works alone and
completes the job, he will take T + 5 days. If B works alone and completes
the same job, he will take T + 45 days. What is T?
A) 25
B) 60
C) 15
D) None of these
4. A man can do a piece of work in 60 hours. If he takes his son with him
and both work together then the work is finished in 40 hours. How long will
the son take to do the same job, if he worked alone on the job?
A) 0 hours
B) 120 hours
C) 24 hours
D) None of these
5. A, B and C can do a work in 5 days, 10 days and 15 days respectively.
They started together to do the work but after 2 days A and B left. C did the
remaining work (in days)
A) 1
B) 3
C) 5
D) 4
6. X alone can do a piece of work in 15 days and Y alone can do it in 10
days. X and Y undertook to do it for Rs.720. With the help of Z they finished
it in 5 days. How much is paid to Z?
A) Rs.360
B) Rs.120
C) Rs.240
D) Rs.300
7. Ram starts working on a job and works on it for 12 days and completes
40% of the work. To help him complete the work, he employs Ravi and
together they work for another 12 days and the work gets completed. How
much more efficient is Ram than Ravi?
A)50%
B) 200%
C) 60%
D)100%
8. A red light flashes 3 times per minute and a green light flashes 5 times in
two minutes at regular intervals. If both lights start flashing at the same
time, how many times do they flash together in each hour?
A) 30
B) 24
C) 20
D) 60
9. A and B can do a piece of work in 21 and 24 days respectively. They
started the work together and after some days A leaves the work and B
completes the remaining work in 9 days. After how many days did A
leave?
A) 5
B) 7
C) 8
D) 6
10. Ram, who is half as efficient as Krish, will take 24 days to complete a
work if he worked alone. If Ram and Krish worked together, how long will
they take to complete the work?
A) 16 days
B) 12 days
C) 8 days
D) 18 days
Answer Key
1.C; 2.A; 3.C; 4.B;5.D; 6.B; 7.D; 8.A; 9.B; 10.C
Key numbers notes
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.
Important formulas
i. ( a + b )( a - b ) = ( a 2 - b2 )
ii. ( a + b ) 2 = ( a2 + b2 + 2 ab )
iii. ( a - b )2 = ( a2 + b2 - 2 ab )
iv. ( a + b + c ) 2 = a2+ b2 + c2 + 2 ( ab + bc + ca )
v. ( a3 + b3 ) = ( a + b )( a2 - ab + b2 )
vi. ( a3 - b3 ) = ( a - b )( a2 + ab + b2)
vii. Sum of natural numbers from 1 to n
viii. Sum of squares of first n natural numbers is =
ix. Sum of cubes of first n natural numbers is A
x. HCF= (HCF of the numerators)/(LCM of the denominators)
xi. LCM= (LCM of the numerators)/HCF of the denominators
xii. Product of two numbers = Product of their H.C.F. and L.C.M
Note: When a number N is raised to any integral power m, the digit in the
unit’s
place of the resulting value can be determined without actually evaluating
the
power. The digits when raised to powers will give values in which the digits
in
the unit’s place follow a cylindrical pattern. Following is the pattern to
calculate
the digit in the unit’s place of any derived power.
HCF models:If N is a composite number such that N = ap . bq . cr ….. where a, b, c are
prime factors of N and p,q,r ….. are positive integers, then
(a) The number of factors of N is given by the expression (p + 1) (q + 1) (r +
1)…
(b) It can be expressed as the product of two factors in 1/2 {(p + 1) (q + 1) (r
+ 1)…..} ways
(c) If N is a perfect square, it can be expressed
(i)as a product of two DIFFERENT factors in 1/2 {(p + 1) (q + 1) (r + 1)….. 1} ways
(ii)as a product of two factors in 1/2 {(p + 1) (q + 1) (r + 1) ….+1} ways
(d) Sum of all factors of N = (ap+1 – 1 / a – 1) . (bq+1 – 1 / b – 1) . (cr+1 – 1 / c –
1)…..
(e) The number of co-primes of N (< N), Ø(N) = N(1 – 1/a) (1 – 1/b) (1 – 1/c)
….
(f) Sum of the numbers in (e) = N/2 . Ø(N)
(g) It can be expressed as a product of two factors in 2n-1, where ‘n’ is the
number of different prime factors of the given number N.
Exercise Questions
1. 117 * 117 + 83 * 83 = ?
a) 20698
b) 20578
c) 21698
d) 21268
2. (1/4)3 + (3/4)3 + 3(1/4)(3/4)(1/4 + 3/4) =?
a) 1/64
b)27/64
c) 49/64
d)1
3. The difference of two numbers is 1365. On dividing the larger number by
the smaller, we get 6 as quotient and 15 as reminder. What is the smaller
number ?
a) 240
Time and Distance- Key Notes
Time and Distance- Key Notes-aptitude-questions-answers-basics
b) 270
c) 295
d) 360
4. The 7th digit of (202)3 is
a) 2
b) 4
c) 8
d) 6
5. H.C.F. of two numbers is 16. Which one of the following can never be
their L.C.M
a) 32
b) 80
c) 64
d) 60
6. What is the remainder when 9 + 92 + 93 + .... + 98 is divided by 6?
a) 3
b) 2
c) 0
d)5
7. The sum of the first 100 natural numbers is divisible by
a) 2, 4 and 8
b) 2 and 4
c)2 only
d)none of these
8. For what value of 'n' will the remainder of 351n and 352n be the same
when divided by 7?
he driven 25% faster on an average he would have reached 4 minutes
earlier than the scheduled time. How far is his office?
A) 24 km

Time, Speed and Distance Aptitude basics, practice questions,
answers and explanations
Prepare for companies tests and interviews
B) 72 km
C) 18 km
D) Data Insufficient
Distance= Speed*time
For a non-uniform motion Average speed= Total distance travelled/Total
time taken
When the body travels at 'u' m/s for t1 seconds and 'v' m/s for t2 seconds,
then
Average speed= (ut1+vt2)/(t1+t2)
When the body travels l distance at 'u' m/s and 'm' distance at 'v' m/s;
Average speed = (mu+lv)/(l+m)
5. A man and a woman 81 miles apart from each other, start traveling
towards each other at the same time. If the man covers 5 miles per hour
to the women's 4 miles per hour, how far will the woman have travelled
when they meet?
A) 27
B) 36
C) 45
D) None of these.
Relative Speed: Speed of a moving body w.r.t. another moving body is
called relative speed.
Speed of A w.r.t. B
(i) When they are moving in same direction; Relative speed of A= A-B
(ii) When they are moving in opposite direction; Relative speed of A= A+B
Key points on Trains
When a train is crossing a pole distance travelled by the train= length of
train
When a train of length l is crossing a bridge of length b; the distance
travelled by train=l+b
When a train of length l is crossing a platform of length p; then distance
travelled by train=l+p
When a train of length l1 is crossing/ overtaking another train l2; then
distance travelled = l1+l2
Exercise questions
1. Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B
traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves
Delhi for Mumbai at 9 P.M. If all three trains meet at the same time
between Mumbai and Delhi, what is the speed of Train C if the distance
between Delhi and Mumbai is 1260 kms?
A) 60 km/hr
B) 90 km/hr
C) 120 km/hr
D) 135 km/hr
6. The speed of a motorboat itself is 20 km/h and the rate of flow of the
river is 4 km/h. Moving with the stream the boat went 120 km. What
distance will the boat cover during the same time going against the
stream?
A) 80 km
B) 180 km
C) 60 km
D) 100 km
2. Two trains A and B start simultaneously from stations X and Y towards
each other respectively. After meeting at a point between X and Y, train A
reaches station Y in 9 hours and train B reaches station X in 4 hours from
the time they have met each other. If the speed of train A is 36 km/hr, what
is the speed of train B?
A) 24 km/hr
B) 54 km/hr
C) 81 km/hr
D) 16 km/hr
3. A man moves from A to B at the rate of 4 km/hr. Had he moved at the
rate of 3.67 km/hr, he would have taken 3 hours more to reach the
destination. What is the distance between A and B?
A) 33 kms
B) 132 kms
C) 36 kms
D) 144 kms
7. I travel the first part of my journey at 40 kmph and the second part at
60 kmph and cover the to
4hours 5hours 6hours
8. tes later than usual. What is his usual time?
A) 30 min
B) 60 min
C) 70 min
D)50 min
9. A passenger train covers the distance between stations X and Y, 50
minutes faster than a goods train. Find this distance if the average speed
of the passenger train is 60 kmph and that of goods train is 20 kmph.
A) 20 kms
B) 25 kms
C) 45 kms
D) 40 kms
10. Ram covers a part of the journey at 20 kmph and the balance at 70
kmph taking total of 8 hours to cover the distance of 400 km. How many
hours has been driving at 20 kmph?
A) 2 hours
B) 3 hours 20 minutes
C) hours 40 minutes
D)3 hours 12 minutes
Answer Key
1.C; 2.B; 3.B; 4.C; 5.B; 6.A; 7.C; 8.B; 9.B; 10.D
4. A man driving his bike at 24 kmph reaches his office 5 minutes late. Had
Percentage- Key Notes
Percentage- Key Notes-aptitude-questions-answers-basics
Percentages Aptitude basics, practice questions, answers and
explanations
5. A trader buys goods at a 19% Amount on the label price. If he wants to
make a profit of 20% after allowing a Amount of 10%, by what % should
his marked price be greater than the original label price?
A) +8%
B) -3.8%
Prepare for companies tests and interviews
Percentage= (Sum of quantities)/(Number of quantities)
Percentage increase by x%= ((x+100)/100)*Initial
Percentage decrease by x%= ((100-x)/100)*Initial
Some common percentage conversions
Exercise questions
1. A trader makes a profit equal to the selling price of 75 articles when he
sold 100 of the articles. What % profit did he make in the transaction?
A) 33.33%
B) 75%
C) 300%
D) 150%
2. A merchant buys two articles for Rs.600. He sells one of them at a profit
of 22% and the other at a loss of 8% and makes no profit or loss in the end.
What is the selling price of the article that he sold at a loss?
A) Rs. 404.80
B) Rs. 440
C) Rs. 536.80
D) Rs. 160
3.A trader professes to sell his goods at a loss of 8% but weights 900 grams
in place of a kg weight. Find his real loss or gain percent.
A) 2% loss
B) 2.22% gain
C) 2% gain
D) None of these
4. Rajiv sold an article for Rs.56 which cost him Rs.x. If he had gained x% on
his outlay, what was his cost?
A) Rs. 40
B) Rs. 45
C) Rs. 36
D)Rs. 28
Permutation and Combination- Key Notes-aptitude-questions-answersbasics
 Permutation and Combination Aptitude basics, practice
questions, answers and explanations
Prepare for companies tests and interviews
nP
r
= n!/ (n-r)!
C) +33.33%
D) None of these
6. If apples are bought at the rate of 30 for a rupee. How many apples
must be sold for a rupee so as to gain 20%?
A) 28
B) 25
C) 20
D) 22
7. Two merchants sell, each an article for Rs.1000. If Merchant A
computes his profit on cost price, while Merchant B computes his profit
on selling price, they end up making profits of 25% respectively. By how
much is the profit made by Merchant B greater than that of Merchant A?
A) Rs.66.67
B) Rs. 50
C) Rs.125
D) Rs.200
8. A merchant marks his goods in such a way that the profit on sale of 50
articles is equal to the selling price of 25 articles. What is his profit
margin?
A) 25%
B) 50%
C) 100%
D) 66.67%
9. A merchant marks his goods up by 75% above his cost price. What is the
maximum % Amount that he can offer so that he ends up selling at no
profit or loss?
A) 75%
B) 46.67%
C) 300%
D) 42.85%
10. The price of a T.V. is increased 30% before budget and in budget 20%
is also increased. Then total increase in price will be
A) 50%
B) 56%
C) 55%
D) 59%
Answer Key
1.C; 2.A; 3.B; 4.A; 5.A; 6.B; 7.B; 7.B; 8.C; 9.D; 10.B
so that the relative position of the vowels and consonants remain the
same as in the word EDUCATION?
A) 9!/4
B) 9!/(4!*5!)
C) 4!*5!
D) None of these
nC
r
nC
r
= n!/r!(n-r)!
= nC(n-r)
nC = nPr/r!
r
0!=1
1!=1
nP =n!
n
nP =n
1
nC =n
1
nC =1
n
Exercise questions
1. How many words can be formed by re-arranging the letters of the word
ASCENT such that A and T occupy the first and last position respectively?
A)5!
B)4!
C)6! - 2!
D)6! / 2!
2. There are 2 brothers among a group of 20 persons. In how many ways
can the group be arranged around a circle so that there is exactly one
person between the two brothers?
A) 2 * 19!
B)18! * 18
C) 19! * 18
D)2 * 18!
3. There are 12 yes or no questions. How many ways can these be
answered?
A) 1024
B) 2048
C) 4096
D)144
4. How many ways can 4 prizes be given away to 3 boys, if each boy is
eligible for all the prizes?
A) 256
B) 12
C) 81
D) None of these
5. A team of 8 students goes on an excursion, in two cars, of which one can
seat 5 and the other only 4. In how many ways can they travel?
A) 9
B)26
C)126
D) 3920
6. How many numbers are there between 100 and 1000 such that at least
one of their digits is 6?
A) 648
B) 258
C) 654
D)252
7. How many ways can 10 letters be posted in 5 post boxes, if each of the
post boxes can take more than 10 letters?
A) 510
B) 105
C) 10P5
D) 10C5
8. In how many ways can the letters of the word EDUCATION be rearranged
Boats and Streams
Aptitude basics, practice questions, answers and explanations
Prepare for companies tests and interviews
Boats and Streams
The water of a stream, usually, keeps flowing at a certain speed, in a
particular direction. This speed is called the current of the stream. A boat
develops speed because of its engine power. The speed with which it
9. In how many ways can 15 people be seated around two round tables
with seating capacities of 7 and 8 people?
A) 15!/(8!)
B) 7!*8!
C) (15C8)*6!*7!
D)2*(15C7)*6!*7!
10. If the letters of the word CHASM are rearranged to form 5 letter words
such that none of the word repeat and the results arranged in ascending
order as in a dictionary what is the rank of the word CHASM?
A) 24
B)31
C) 32
D)30
11. How many words of 4 consonants and 3 vowels can be made from 12
consonants and 4 vowels, if all the letters are different?
A) 16C7 * 7!
B) 12C4 * 4C3 * 7!
C) 12C3 * 4C4
D) 12C4 * 4C3
12. In how many ways can 5 letters be posted in 3 post boxes, if any
number of letters can be posted in all of the three post boxes?
A) 5C3
B) 5P3
C) 53
D)35
13. How many number of times will the digit '7' be written when listing
the integers from 1 to 1000?
A) 271
B) 300
C) 252
D)304
14. There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either
with a red or a green ball in such a way that at least 1 box contains a green
ball and the boxes containing green balls are consecutively numbered. The
total number of ways in which this can be done is
A) 5
B) 21
C) 33
D) 60
15. What is the value of 1*1! + 2*2! + 3!*3! + ............ n*n!, where n!
means n factorial or n(n-A(n-2)...1
A) n(n-A(n-A!))
B) (n+A!)/(n(n-A))
C) (n+A! - n!) D) (n + A! - 1!)
Answers
1.B; 2.D; 3.C; 4.C; 5.C; 6.D; 7.A; 8.C; 9.C; 10.C; 11.B; 12.D; 13.B; 14.B; 15.D
stream.
The speed of the boat in still water (in km/hr)is:
a)3
b)5
c)8
d)9
Solution: Speed in still water=1/2(11+5)kmph=8kmph
3. Speed of a boat in still water is 16km/h. If it can travel 20km
travels when there is no current is called speed of boat in still water. When
the boat moves in the direction of the current is said to be with the stream/
current or downstream. When the boat moves in the direction opposite to
that of the current, it is said to be against the stream is called upstream.
Eg:-If the speed of a boat in still water is ‘u’km/hr and the speed of the
stream is ‘v’km/hr then:
* Speed downstream=(u+v)km/hr
* Speed upstream = (u-v)km/hr
If the speed downstream is u km/hr and the speed upstream is v km/hr,
then:
* Speed of boat in still water = ½(u+v)km/hr
* Speed (Rate) of stream = ½ (u-v)km/hr
Examples
a) A man can row a boat 12 km/h with the stream and 8km/h against the
stream.
Find his speed in still water.
a) 2km/hr
b) 4km/hr
c) 8km/hr
d) 10km/hr
Solution: Speed of boat in still water = ½(u+v) km/hr = ½ (12+8)=10km/hr
b) A man can row a boat 27km/h with the stream and 11km/h against the
stream.
Find speed of stream
a) 2km/hr
b)4km/hr
c)8km/hr
d)10km/hr
Solution: Speed (Rate) of stream = ½ (u-v) km/hr = ½ (27-11)=8km/hr
c) A boat running downstream covers a distance of 16km in 2 hours while
for covering the same distance upstream, it takes 4 hours. What is the
speed of the boat in still water?
a)4km/hr
b)6km/hr
c)8km/hr
d) None of these
Rate of downstream=(16/2 ) kmph=8kmph
Rate of upstream =(16/4) kmph=4kmph
Therefore Speed in still water=1/2(8+4) kmph=6kmph
Note: If ratio of downstream and upstream speeds of a boat is ‘a:b.’
Then ratio of time taken= b:a
Speed of stream=a-b/a+b *Speed in still water
Speed in still water =a+b/a-b *Speed of stream
Exercise Questions
1. A man rows downstream 32 km and 14km upstream. If he takes 6 hours
to cover each distance, then the velocity (in kmph) of the current is:
a)1/2
b)1
c)1and ½
d)2
Solution: Rate downstream=(32/6)kmph; Rate upstream=(14/6)kmph
Velocity of current=1/2(32/6-14/6)kmph=3/2kmph=1.5kmph
2. In one hour, a boat goes 11km along the stream and 5km against the
The set of all the 'things' currently under discussion is called the universal
set (or sometimes, simply the universe). It is denoted by U. The universal
set doesn’t contain everything in the whole universe. On the contrary, it
restricts us to just those things that are relevant at a particular time. For
example, if in a given situation we’re talking about numeric values –
quantities, sizes, times, weights, or whatever – the universal set will be a
suitable set of numbers (see below). In another context, the universal set
may be {alphabetic characters} or {all living people}, etc.
2.Empty set
downstream in the same time as it can travel 12 km upstream, the rate of
stream is.
a)1km/h
b)2km/h
c)4km/h
d)5km/h
Solution: Speed downstream: Speed upstream=20:12=5:3
Speed of current=5-3/5+3*16=4km/h
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Sets and Union- Keynotes
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A set can be defined as a collection of things that are brought together
because they obey a certain rule. These 'things' may be anything you like:
numbers, people, shapes, cities, bits of text ..., literally anything. The key
fact about the 'rule' they all obey is that it must be well-defined. In other
words, it enables us to say for sure whether or not a given 'thing' belongs
to the collection. If the 'things' we're talking about are English words, for
example, a well-defined rule might be: '... has 5 or more letters'. A rule
which is not well-defined (and therefore couldn't be used to define a set)
might be: '... is hard to spell'
Elements
A 'thing' that belongs to a given set is called an element of that set.
For example: Henry VIII is an element of the set of Kings of England
Notation
Curly brackets {...... } are used to stand for the phrase 'the set of ...'. These
braces can be used in various ways.
For example: We may list the elements of a set: { − 3, − 2, − 1,0,1,2,3}.
We may describe the elements of a set: { integers between − 3 and 3
inclusive}.
We may use an identifier (the letter x for example) to represent a typical
element, a | symbol to stand for the phrase 'such that', and then the rule
or rules that the identifier must obey:
{x | x is an integer and | x | < 4} or {x|x ϵ Z, |x| <4 }
The last way of writing a set - called set comprehension notation - can be
generalized as:
x | P(x), where P(x) is a statement (technically a propositional function)
about x and the set is the collection of all elements x for which P is true.
The symbol ϵ is used as follows:
ϵ means 'is an element of ...'. For example: dog ϵ {quadrupeds}
ɇ means 'is not an element of ...'. For example:
Washigton DC ɇ {European capital cities}
A set can be finite: {British citizens}or infinite: {7, 14, 21, 28, 35, …. }.
Sets will usually be denoted using upper case letters: A, B, ...
Elements will usually be denoted using lower case letters: x, y, ...
Some Special Sets
1.Universal Set
P contains $, ;, &, ..., so A ⊂ P
But Q and F are just different ways of saying the same thing, so Q = F.
The use of ⊂ and ⊆ is clearly analogous to the use of < and ≤ when
comparing two numbers.
Note: Every set is a subset of the universal set, and the empty set is a
subset of every set.
5.Disjoint
Two sets are said to be disjoint if they have no elements in common.
The set containing no elements at all is called the null set, or empty set. It is
denoted by a pair of empty braces: { } or by the symbol f. It may seem odd
to define a set that contains no elements. Bear in mind, however, that one
may be looking for solutions to a problem where it isn't clear at the outset
whether or not such solutions even exist. If it turns out that there isn't a
solution, then the set of solutions is empty.
For example:
If U = {words in the English language} then {words with more than 50
letters}= f .
If U = {whole numbers} then {x|x2 = 10} = f .
Operations on the empty set
Operations performed on the empty set (as a set of things to be operated
upon) can also be confusing. (Such operations are nullary operations.) For
example, the sum of the elements of the empty set is zero, but the product
of the elements of the empty set is one (see empty product). This may
seem odd, since there are no elements of the empty set, so how could it
matter whether they are added or multiplied (since “they” do not exist)?
Ultimately, the results of these operations say more about the operation in
question than about the empty set. For instance, notice that zero is the
identity element for addition, and one is the identity element for
multiplication.
3.Equality
Two sets A and B are said to be equal if and only if they have exactly the
same elements. In this case, we simply write:
A=B
Note two further facts about equal sets:
The order in which elements are listed does not matter.
If an element is listed more than once, any repeat occurrences are ignored.
So, for example, the following sets are all equal:
{1, 2, 3} = {3, 2, 1} = {1, 1, 2, 3, 2, 2}
(You may wonder why one would ever come to write a set like {1, 1, 2, 3, 2,
2}. You may recall that when we defined the empty set we noted that there
may be no solutions to a particular problem - hence the need for an empty
set. Well, here we may be trying several different approaches to solving a
problem, some of which in fact lead us to the same solution. When we
come to consider the distinct solutions, however, any such repetitions
would be ignored.)
4.Subsets
If all the elements of a set A are also elements of a set B, then we say that A
is a subset of B, and we write: A ⊆ B
For example: If T = {2, 4, 6, 8, 10} and E = {even integers}, then T ⊆ E
If A = {alphanumeric characters} and P = {printable characters}, then A ⊆ P
If Q = {quadrilaterals} and F = {plane figures bounded by four straight lines},
then Q ⊆ F
Notice that A ⊆ B does not imply that B must necessarily contain extra
elements that are not in A; the two sets could be equal – as indeed Q and F
are above. However, if, in addition, B does contain at least one element
that isn’t in A, then we say that A is a proper subset of B. In such a case we
would write: A ⊂ B
In the examples above:
E contains 12, 14, ... , so T ⊂ E
All whole numbers, positive, negative and zero form the set of integers. It is
sometimes denoted by Z. So Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}
In blackboard bold, it looks like this: Z
3.Real numbers
If we expand the set of integers to include all decimal numbers, we form
the set of real numbers. The set of reals is sometimes denoted by R.
A real number may have a finite number of digits after the decimal point
(e.g. 3.625), or an infinite number of decimal digits. In the case of an
infinite number of digits, these digits may:
recur; e.g. 8.127127127...
For example: If A = {even numbers} and B = {1, 3, 5, 11, 19}, then A and B
are disjoint.
Operations on Sets
1.Intersection
The intersection of two sets A and B, written A ∩ B, is the set of elements
that are in A and in B.
(Note that in symbolic logic, a similar symbol,^, is used to connect two
logical propositions with the AND operator.)
For example, if A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, then A ∩ B = {2, 4}.
We can say, then, that we have combined two sets to form a third set
using the operation of intersection.
2.Union
In a similar way we can define the union of two sets as follows:
The union of two sets A and B, written A ∪ B, is the set of elements that
are in A or in B (or both).
(Again, in logic a similar symbol,V, is used to connect two propositions
with the OR operator.)
So, for example, {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}.
You'll see, then, that in order to get into the intersection, an element must
answer 'Yes' to both questions, whereas to get into the union, either
answer may be 'Yes'.
The ∪ symbol looks like the first letter of 'Union' and like a cup that will
hold a lot of items. The ∩ symbol looks like a spilled cup that won't hold a
lot of items, or possibly the letter 'n', for intersection. Take care not to
confuse the two.
3.Difference
The difference of two sets A and B (also known as the set-theoretic
difference of A and B, or the relative complement of B in A) is the set of
elements that are in A but not in B.
This is written A - B, or sometimes A \ B.
For example, if A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, then A - B = {1, 3}.
4.Complement
The set of elements that are not in a set A is called the complement of A.
It is written A′ (or sometimes AC, or Â). Clearly, this is the set of elements
that answer 'No' to the question Are you in A?.
For example, if U = N and A = {odd numbers}, then A′ = {even numbers}.
Notice the spelling of the word complement: its literal meaning is 'a
complementary item or items'; in other words, 'that which completes'. So
if we already have the elements of A, the complement of A is the set that
completes the universal set.
5.Cardinality
The cardinality of a finite set A, written | A | (sometimes #(A) or n(A)), is
the number of (distinct) elements in A. So, for example:
If A = {lower case letters of the alphabet}, | A | = 26.
Some special sets of numbers
Several sets are used so often, they are given special symbols.
1.The natural numbers
The 'counting' numbers (or whole numbers) starting at 1, are called the
natural numbers. This set is sometimes denoted by N. So N = {0, 1, 2, 3,
...}.
Note that, when we write this set by hand, we can't write in bold type so
we write an N in blackboard bold font: N
2.Integers
Step 1: The recurring number is 23
Step 2: 23/99 [the number 23 is of length 2 so we have added two nines]
Step 3: Reducing it to lowest terms : 23/99 [it can not be reduced further].
Mixed Recurring to Fractions:
If N= 0.abcbcbc…. Then N = abc - a / 990 = Repeated & non-repeated digits
- Non repeated digits / As many 9's as repeated digits followed by as many
zero as non - repeated digits
Eg: 0.25757..... = 257 - 2 / 990 = 255 / 990 = 17 / 60.
... or they may not recur; e.g. 3.141592653...
In blackboard bold: R
4.Rational numbers
Those real numbers whose decimal digits are finite in number, or which
recur, are called rational numbers. The set of rationals is sometimes
denoted by the letter Q.
A rational number can always be written as exact fraction p/q; where p and
q are integers. If q equals 1, the fraction is just the integer p. Note that q
may NOT equal zero as the value is then undefined.
For example: 0.5, -17, 2/17, 82.01, 3.282828... are all rational numbers.
In blackboard bold: Q
5.Irrational numbers
If a number can't be represented exactly by a fraction p/q, it is said to be
irrational.
Examples include: √2, √3.
Fractions
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Simple or Vulgar Fraction
A number expressed with numerator and denominator. Say I have 3 of 10
apples then I will express it as 3/10. The total is written below a horizontal
or diagonal line, and the number of parts comprising the fraction
(numerator) is written above. Such fractions are called vulgar fractions or
simple fractions. Eg:[ 3/4 ]
Decimal Fraction
Expressing the fraction in decimal values (denominator a power of 10) is
called decimal fraction. 1/2 is expressed as 0.5 in decimal fraction. Eg:[
0.45773 ]
Converting a decimal to vulgar fraction:
Step 1: Calculate the total numbers after decimal point.
Step 2: Remove the decimal point from the number.
Step 3: Put 1 under the denominator and annex it with "0" as many as the
total in step a.
Step 4: Reduce the fraction to its lowest terms.
Example: Consider 0.44
Step 1: Total number after decimal point is 2
Step 2 and 3: 44/100
Step 4: Reducing it to lowest terms : 44/100 = 22/50 = 11/25
Converting a recurring decimal to vulgar fraction
A decimal with recurring value is called recurring decimal.
E.g: 2/9 will give 0.22222222...... where 2 is recurring number.
Method:
Step 1: Separate the recurring number from the decimal fraction.
Step 2: Annex denominator with "9" as many times as the length of the
recurring number.
Step 3: Reduce the fraction to its lowest terms.
Example: Consider 0.2323232323
Sn= a/1-r+ dr (1-rn-1)/ (1-r)2 - [a+(n-1)d]rn/ 1-r
The sum to infinity, S= a/ 1-r + dr (1-rn-1)/(1-r)2 ; r<1
Exponential Series
Progression
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Important Equations and Formula
Sum of first n natural numbers= n(n+1)/2
Sum of the squares of first n natural numbers= (n(n+1)(2n+1))/6
Sum of the cubes of first n natural numbers= [n (n+1)/2]2
Sum of first n natural odd numbers= n2
Average = Sum of items/ Number of items
Arithmetic Progression (AP): An AP 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'.
nth term of an AP; tn=a+(n-1)d
Sum of the first n terms of an AP; Sn= n/2 [2a+(n-1)d] or Sn= n/2 (first
term+last term)
Geometrical Progression (GP):
A GP is of the form a, ar, ar2, ar3......... where a is called the 'first term' and
r is called the 'common ratio'.
nth term of a GP; tn= arn-1
Sum of the first n terms in a GP; Sn= a(1-rn)/1-r
Sum of infinite series of progression; S= a/(1-r)
Geometric mean of two number a and b is given as GM= sqrt(ab)
Harmonic Progression (HP)
If a1,a2,a3,...................an are in AP, then 1/a1, 1/a2, 1/a3, ........1/an, are in
HP
nth term of this HP, tn =1/(1/a1+(n-1)(a1-a2/a1a2) ) nth term of this HP
from the end, tn=1/ (1/a1-(n-1)(a1-a2/a1a2))
If a and b are two non-zero numbers and H is harmonic mean of a and b
then a, H, b from HP and then H=2ab/(a+b)
Arithmetico-Geometric series
A series having terms a, (a+d)r, (a+2d)r2,...... etc is an ArithmeticoGeometric series where a is the first term, d is the common difference of
the Arithmetic part of the series and r is the common ratio of the
Geometric part of the series.
The nth term tn= [a+(n-1)d]rn-1
The sum of the series to n terms is
ex = 1+x/1!+x2/2!+x3/3!+..........
(e is an irrational number)
coefficient of xn= 1/n!; Tn+1=xx/n!
e-x = 1-x/1!+x2/2!- x3/3!+..........
Logarithmic Series
loge (1+x)= x-x2/2+x3/3+x4/4+........
(-1<x
1)
loge (1-x)= -x-x2/2-x3/3- x4/4 -........
(-1x<
1)
Key notes on Compound interest
Compound Interest: (Amount - Principal)
Amount= P* (1+R/100)n
loge (1+x)/(1-x)= 2-(x+x3/3+x5/5+.........)
(-1<x 1)
Simple and Compound Interest- Key Notes
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Important formula and equations
Principal: The money borrowed or lent out for a certain period is called the
principal or the sum.
Interest: Extra money paid for using other's money is called interest.
Simple Interest (SI): If the interest on a sum borrowed for certain period is
reckoned uniformly, then it is called simple interest.
Let Principal= P, Rate= R% per annum (p.a) and Time= T years. Then
(i) Simple Interest= (P*R*T)/100
(ii) P = (100*SI)/(R*T); R= (100*SI)/(P*T) and T= (100*SI)/(P*R)
Key notes on Simple Interest
A sum of money becomes n times itself in T years at simple interest, then
the rate of interest is
Rate= 100(n-l)%
T
If a sum of money becomes n times in T years at SI then it will be m times
of itself in ..... years
Required time= (m-l)*T years
(n-l)
If SI on a sum of money is 1/xth of the principal and the time T is equal to
the rate percent R, then
Rate= Time=
A certain sum is at SI at a certain rate for T years. And if it had been put at
R1 % higher rate, then it would fetch Rs.x more, then the
Principal= x*100
T*R1
The annual payment that will discharge a debt of Rs.P due in T years at the
arte of interest R% per annum is Annual payment =
100P
100T+RT(T-1)
2
Let the rate of interest for first 1 years is r1% per annum, for the next t2
years is r2 % per annum and for the period beyond that is r3 %. Suppose all
together the simple interest for t3 years is Rs.I. Then
Principal=100*I
t1r1+t2r2+(t3-t1-t2)r3
The simple interest on a certain sum of money at r1 % per annum for t1
years is Rs.m. The interest on the same sum for t2 years at r2 % per annum
is n.
Then the sum= (m-n)*100
r1t1-r2t2
C) 7200
D)9600
5. A man invests Rs.5000 for 3 years at 5% p.a. compound interest
reckoned yearly. Income tax at the rate of 20% on the interest earned is
When the interest is compounded K times a year, Amount= P( 1 + R /
K*100)kt
When the interest is paid half yearly, say at r%per annum compound
interest, then the amount after t years is given by:
P( 1 + R / 2*100)2t
Similarly, if the interest is paid quarterly, say at r% per annum compound
interest, then the amount due after t years is given by:
P( 1 + r / 4 * 100)4t
Under the method of equated instalments, the value of each instalment is
the same.
Equal Annual Instalment under
(a) Simple Interest, x = 2P(100 + nr)
n[200 + (n - 1)r]
(b) Compound Interest, x = Pr / 100[1 – (100/100 + r) n ]
Exercise questions
1.A father left a will of Rs.35 lakhs between his two daughters aged 8.5
and 16 such that they may get equal amounts when each of them reach
the age of 21 years. The original amount of Rs.35 lakhs has been
instructed to be invested at 10% p.a. simple interest. How much did the
elder daughter get at the time of the will?
A) Rs.17.5 lakhs
B)Rs.21 lakhs
C) Rs.15 lakhs
D) Rs. 20 lakhs
2.What will Rs.1500 amount to in three years if it is invested in 20% p.a.
compound interest, interest being compounded annually?
A) 2400
B) 2592
C) 2678
D)2540
3. If a sum of money grows to 144/121 times when invested for two years
in a scheme where interest is compounded annually, how long will the
same sum of money take to treble if invested at the same rate of interest
in a scheme where interest is computed using simple interest method?
A) 9 years
B) 22 years
C) 18 years
D)33 years
4. The population of a town was 3600 three years back. It is 4800 right
now. What will be the population three years down the line, if the rate of
growth of population has been constant over the years and has been
compounding annually?
A) 6000
B) 6400
deducted at the end of each year. Find the amount at the end of the third
year.
A) 5624.32
B)5630.50
C)5788.125
D)5627.20
6. The difference between the compound interest and the simple interest
on a certain sum at 12% p.a. for two years is Rs.90. What will be the value
of the amount at the end of 3 years?
A) 9000
B) 6250
C) 8530.80
D)8780.80
7. Vijay invested Rs.50,000 partly at 10% and partly at 15%. His total
income after a year was Rs.7000. How much did he invest at the rate of
10%?
A) Rs.40,000
B)Rs.40,000
C)Rs.12,000
D)Rs.20,000
8. A sum of money invested for a certain number of years at 8% p.a. simple
interest grows to Rs.180. The same sum of money invested for the same
number of years at 4% p.a. simple interest grows to Rs.120 only. For how
many years was the sum invested?
A) 25 years
B) 40 years
C) 33 years and 4 months
D)Cannot be determined
9. How long will it take for a sum of money to grow from Rs.1250 to
Rs.10,000, if it is invested at 12.5% p.a simple interest?
A) 8 years
B) 64 years
C) 72 years
D)56 years
10. Rs.5887 is divided between Shyam and Ram, such that Shyam's share at
the end of 9 years is equal to Ram's share at the end of 11
years, compounded annually at the rate of 5%. Find the share of Shyam.
A) 2088
B) 2000
C) 3087
D)None of these
Answer Key
1.B; 2.B; 3.B; 4.B; 5.A; 6.D; 7.B; 8.A; 9.D; 10.C
Geometry and Mensuration-Key Points
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Where , a = length of two equal side , b= length of base of isosceles
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Geometry and Mensuration
Mensuration: Mensuration is the branch of mathematics which deals with
the study of Geometric shapes , Their area , Volume and different
parameters in geometric objects.
Some important mensuration formulas are:
1.Area of rectangle (A) = length(l) * Breath(b);
triangle.
10.Area of trapezium (A) =(a+b)/2
Where , “a” and “b” are the length of parallel sides.
2.Perimeter of a rectangle (P) = 2 * (Length(l) + Breath (b))
3.Area of a square (A) = Length (l) * Length (l)
4.Perimeter of a square (P) = 4 * Length (l)
11.Perimeter of a trapezium (P) = sum of all sides
5.Area of a parallelogram(A) = Length(l) * Height(h)
12.Area f rhombus (A) = Product of diagonals / 2
13.Perimeter of a rhombus (P) = 4 * l
where l = length of a side
14.Area of quadrilateral (A) = 1/2 * Diagonal * (Sum of offsets)
6.Perimeter of a parallelogram (P) = 2 * (length(l) + Breadth(b))
7.Area of a triangle (A) = (Base(b) * Height(b)) / 2
15. Area of a Kite (A) = 1/2 * product of it’s diagonals
16. Perimeter of a Kite (A) = 2 * Sum on non-adjacent sides
17. Area of a Circle (A) =
, where , r= radius of the circle
And for a triangle with sides measuring “a” , “b” and “c” , Perimeter =
a+b+c
and s = semi perimeter = perimeter / 2 = (a+b+c)/2
And also . Area of triangle,
This formulas is also knows as “Hero’s formula”.
8.Area of triangle(A)
=
18. Circumference of a Circle =
r= radius of circle
d= diameter of circle
19. Total surface area of
cuboid = 2(lb+bh+lh)
where , l= length , b=breadth , h=height
20. Total surface area of
cuboid = 6l2
where , l= length
22. length of diagonal of cuboid =
9. Area of isosceles triangle =
22. length of diagonal of cube =
38. Volume of hollow cylinder =
23. Volume of cuboid = l * b * h
24. Volume of cube = l * l* l
where , R = radius of cylinder , r= radius of hollow , h = height of cylinder
39. Surface area of a right square pyramid =
25. Area of base of a cone =
Where , a = length of base , b= length of equal side ;
of the isosceles triangle forming the slanting face.
26. Curved surface area of a cone =C
Where , r = radius of base , l = slanting height of cone
40. Volume of a right square pyramid =
27. Total surface area of a cone =
28. Volume of right circular cone =
Where , r = radius of base of cone , h= height of the cone (perpendicular to
base)
29. Surface area of triangular prism = (P * height) + (2 * area of triangle)
Where , p = perimeter of base
30. Surface area of polygonal prism = (Perimeter of base * height ) + (Area
of polygonal
base * 2)
41. Area of a regular hexagon =
42. area of equilateral triangle =
43. Curved surface area of a Frustums =
44. Total surface area of a Frustums =
31. Lateral surface area of prism = Perimeter of base * height
32. Volume of Triangular prism = Area of the triangular base * height
45. Curved surface area of a Hemisphere =
33. Curved surface area of a cylinder =
Where , r = radius of base , h = height of cylinder
46. Total surface area of a Hemisphere =
34. Total surface area of a cylinder =
47. Volume of a Hemisphere =
35. Volume of a cylinder =
36. Surface area of sphere =
48. Area of sector of a circle =
where , r= radius of sphere , d= diameter of sphere
37. Volume of a sphere =
a) 243 cm2
Exercise questions
1. What is the area of an equilateral triangle of side 16cm ?
b) 64 3 cm2
c) 363 cm2
d)323 cm2
where , \theta = measure of angle of the sector , r= radius of the sector
3. Triplicate ratio of a : b is a3: b3
4. Sub-triplicate ratio of a : b is cuberoot of (a) : cuberoot of (b)
If a/b = c/d; then a=bc/d
2. Consider the following figure. <A=x+30; <D= x-40. Find <B?
a. 125o
b. 55o
c. 155o
d. 122o
3. Find the area of a square, the product of whose diagonals is 66 cm2
a) 30 cm2
b) 33 cm2
c) 36 cm2
d) 42 cm2
4. A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1
cubic centimeter cubes, how many 1 cubic centimeter cubes will have
exactly one of their sides painted?
a) 9
b) 61
c) 98
d) 54
5. Find the area of a trapezium whose parallel sides are 20 cm and 18 cm
long, and the distance between them is 15 cm.
a) 225 cm2
b)275 cm2
c) 285 cm2
d)315 cm2
6. Examine the figure. <ADB=25o; Find <OBC:
a. 115o
b. 25o
c. 50o
d. 65o
7. The sector of a circle has radius of 21 cm and central angle 1350. Find its
perimeter.
a) 91.5 cm
b)93.5 cm
c) 94.5 cm
d) 92.5 cm
8. The volumes of two cones are in the ratio of 1 :10 and the radii of the
cones are in the ratio of 1 : 2, what is the ratio of their vertical heights?
a) 2 : 5
b) 1 : 5
c) 3 : 5
4) 4 : 5
Answer Key
1.c; 2.b; 3.b; 4.d; 5.c; 6.a; 7.a; 8.a
Ratio,Proportion and Variation- Key Points
Ratio,Proportion and Variation- Key Points-aptitude-questions-answersbasics
Ratio, Proportion, Variation Aptitude basics, practice questions, answers
and explanations
Prepare for companies tests and interviews
Ratio:
1. Duplicate ratio of a : b is a2 : b2
2. Sub-duplicate ratio of a : b is sqrt of(a) : sqrt of (b)
5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is the total amount?
a. Rs. 14000
b. Rs. 15000
c. Rs. 20000
d. None of these
8. If Rs. 782 be divided into three parts, proportional to ½:2/3:3/4, then the
(i) (a+b)/b = (c+d)/d
(ii) (a-b)/b = (c-d)/d
(iii) a/c = b/d
(iv) b/a = d/c
(v) (a+b)/(a-b) = (c+d)/(c-d)
If a/b=c/d; then simplest possible value of a=c, b=d
(vi) In the ratio a : b, if a > b, then (a + x / b + x) < a / b (x > 0)
(vii) In the ratio a : b, if a < b, then (a + x / b + x) > a / b (x > 0)
(viii) In the ratio a : b, if a = b, then (a + x / b + x) = a / b (x > 0)
Proportion & Variation:
If a is directly proportional to b; then a=kb
If a is inversely proportional to b; then a=k/b
If a is directly proportional to b and inversely proportional to c, then a is
directly proportional to b/c
=> a= kb/c
Exercise Questions
1. Ram, Sham and Suresh start business investing in the ratio 1/2 : 1/3:
1/6. The time for which each of them invested their money was in the
ratio 8:6:12 respectively. If they get profit of Rs.18000 from the business,
then how much share of profit will Ram get?
a. Rs.4000
b. Rs.6000
c. Rs.8000
d. Rs. 10000
2. The ratio of the number of boys and girls in a college is 7 : 8. If the
percentage increase in the number of boys and girls be 20% and 10%
respectively, what will be the new ratio?
a. 8 : 9
b. 17 : 18
c. 21 : 22
d. Cannot be determined
3. p,q and r are three positive numbers and Q=(p+q+r)/2; If (Q-p):(Q-q):(Qr) = 2:5:7, then find the ratio of p,q and r ?
a. 4:3:7
b. 12:9:7
c.9:7:4
d. 4:3:2
4.A and B together have Rs. 1210. If 4/15 of A's amount is equal to 2/5 of
B's amount, how much amount does B have?
a. Rs. 460
b. Rs. 484
c. Rs. 550
d. Rs. 664
5. Two numbers are respectively 20% and 50% more than a third
number.The ratio of the two numbers is
a. 2 : 5
b. 3 : 5
c. 4 : 5
d. 6 : 7
6. The ratio of the cost prices of two articles A and B is 4:5.The articles are
sold at a profit with their selling prices being in the ratio 5:6.If the profit
on article A is half of its cost price, find the ratio of the profits on the
articles A and B?
a. 7:10
b. 9:11
c. 5: 9
d. 10:11
7. A sum of money is to be distributed among A, B, C, D in the proportion of
first part is:
a. Rs. 182
b. Rs. 190
c. Rs. 196
d. Rs. 204
9.A bag contains 50 paisa, 20 paisa and 10 paisa coins in the ratio 5:3:1.If
the total amount in the bag is 640 Rs,find the difference in the amounts
contributed by 50 paisa and 20 paisa coins.
a. Rs.300
b. Rs.400
c. Rs.380
d. None of these
10. The speed of an engine is proportional to the square root of the
number of wagons attached to it. Without any wagons attached to it the
speed of the engine is 60km/hr. With 16 wagons attached to it the speed of
the engine is 40km/hr; find the maximum number of wagons that can be
attached so that the train moves.
a. 144
b. 145
c. 142
d. 143
Answer Key:
1.c; 2.c; 3.b; 4.b; 5.c; 6.d; 7.a; 8.d; 9.c; 10.d