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Blue Lotus. Aptitude Numerical Reasoning. Numerical Reasoning. Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn Diagram. Numerical Reasoning. Area and Volume Probability Time and Work (Pipes) SI and CI Average - PowerPoint PPT Presentation
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Blue Lotus
Aptitude
Numerical Reasoning
Numerical Reasoning
• Problems on Numbers
• Problems on Ages
• Ratio and Proportion
• Alligation or Mixture
• Chain Rule
• Partnership
• Venn Diagram
• Area and Volume
• Probability
• Time and Work (Pipes)
• SI and CI
• Average
• Permutation and Combination
• Percentage
• Cubes
Numerical Reasoning
• Boats and Streams
• Time and Distance (Trains)
• Data Sufficiency
• Profit and Loss
• Calendar
• Clocks
• Data Interpretation
Numerical Reasoning
Arithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d; Sum of n terms of A.P Sn = n/2 *(a + L) or n/2 *[2a+(n-1)d)]
a = 1st term, n = number of term, d= difference, Tn = nth term
Geometrical Progression: Tn = arn – 1.
Sn = a(rn – 1)/(r-1); a = 1st term , r = 1st term / 2nd term
Problems on Numbers
Basic Formulae
1. ( a+b)2 = a2 + b2 + 2ab2. (a-b)2 = a2 +b2 -2ab3. ( a+b)2 - (a – b)2 = 4ab4. (a+b)2 + (a – b)2 = 2 (a2 +b2)5. (a2 – b2) = (a+b) (a-b)6. (a+b+c)2 =a2 +b2 +c2 + 2(ab +bc+ca)7. (a3 +b3) = ( a+b) (a2 –ab +b2)8. (a3 –b3) = (a-b) (a2 +ab + b2)9. (a3+b3+c3 -3abc) = (a+b+c) (a2+b2+c2-ab-bc-ca) If a+b+c = 0, then (a3+b3+c3) =3abc
Problems on Numbers
A man ate 100 Bananas in 5 days, each day eating 6 more than the previous day. How many bananas did he eat on the first day?
Problems on Numbers
Solution:
First day be x
Then x+6, x+12, x+18, x+24
5x + 60 = 100
x=8
In first day he ate 8 Bananas.
Problems on Numbers
The traffic light at three different road crossings
change after every 48 seconds, 72 seconds, and 108
seconds respectively. If they all change
simultaneously at 10:20:00 hours then at what time
they again change simultaneously?
Problems on Numbers
Solution:
LCM (48,72,108) = 432 seconds
= 7 minutes 12 seconds
The next change will be at 10:27:12 hours
Problems on Numbers
Raja had 85 currency notes in all, some of which
were of Rs. 100 denomination and the remaining
of Rs. 50 denomination. The total amount of all
these currency notes was Rs. 5000. How much
amount did she have in the denomination of Rs.
50?
Problems on Numbers
Solution:
Let x be the number of Rs. 50 notes
50x + 100 (85-x) = 5000
Solving the above equation,
Amount = Rs. 3500
Problems on Numbers
Mr. Raja is on tour and he has Rs. 360 for his
expenses. If he exceeds his tour by 4 days, he
must cut down his daily expenses by Rs. 3. For
how many days is Mr. Raja on tour?
Problems on Numbers
Solution: Let x be the number of days
360 / x – 360 / (x+4) = 3
Solving the above equation,
The number of days of the tour x = 20 days
Problems on Numbers
If the numerator of a fraction is increased by
25% and the denominator is decreased by 20%
the new value is 5/4. What is the original value?
Problems on Numbers
Solution:
NR = x + 25x/100 DR = y -20y/100
(5x/4)/(4y/5) = 5/4
Solving the equation,
The fraction is 4/5
Problem on Numbers
There is a circular pizza with negligible
thickness that is cut into X pieces by 4 straight
line cuts. What is the maximum and minimum
value of x respectively?
Problem on Numbers
Answer:
Maximum = 11.
Minimum = 5.
Problem on Numbers
A man was engaged on a job for 30 days on
condition that he would get a wage of Rs. 10 for
the day he works but he has to pay a fine of Rs.
2 for each day of his absence. If he gets Rs. 216
at the end. Find the number of days he was
absent?
Problem on Numbers
Solution:
10x – 2(30 –x) = 216
x = 23
Absent for = (30 – 23) = 7 days
Problems on Numbers
I bought a car with a peculiar 5-digit number
license plate which on reversing could still be
read. On reversing the value is increased by
78633. What is the original number if all the
digits were different from 0 - 9?
Problems on Numbers
Solution:
Only 0,1,6,8,9 can be reversed. Hence the
number is 89601. It’s reverse is 10968 and
difference is 78633
Problems on Numbers
Naval collected 8, spiders and beetles in a little
box. When he counted the legs he found that
they were altogether 54. How many beetles and
how many spiders did he collect?
Problems on Numbers
Solution:
Spiders have 8 legs and beetles have 6 legs.
S + B = 8, 8S + 6B = 54.
Solving the above equations
The number of spiders = S = 3
The number of beetles = B = 5
Father’s age is three times the sum of the
ages of his two children, but twenty years
hence his age will be equal to sum of their
ages ?
Problems on Ages
Solution:
Father age = 3(x+y)F+20 = x+y+403x+3y+20 = x+y+403x-x+3y – y =202x+2y = 20 x +y = 10F = 3*10 =30 The father’s age is 30.
Problems on Ages
Problems on Ages
A man’s age is 125% of what it was 10 years
ago, but 83 1/3% of what it will be after ten
years. What is his present age?
Problems on Ages
Solution:Let the present age be x years,Then, 125% of (x-10) = x83 1/3% of (x+10) = x125% of (x-10) = 83 1/3 % of (x+10)5/4 (x-10) = 5/6 (x+10)5x/12 = 250 / 12 x = 50 years
Problem on Ages
My age three years hence multiplied by 3 and
from that subtracted three times my age three
years ago will give you my exact age. Find my
age?
Problem on Ages
Solution:
(x+3)3 – 3(x-3) = x
x =18
Problem on Ages
A boy asks his father “ What is the age of grand
father? Father replied “ He is x years old in x2
years” and also said “ We are talking about 20 th
century” What is the year of birth of grand
father?
Problem on Ages
Solution:
20th century means 1900 – 2000 year
Perfect square 44 * 44 = 1936 (present)
Year of birth = 1936 - 44 = 1892
Problem on Ages
A wizard named Nepo says “ I am only three
times my son's age. My father is 40 years more
than twice my age. Together the three of us are a
mere 1240 years old. How old is Nepo?
Problem on Ages
Solution:
N = 3S
F = 40 + 2N
F + N + S = 1240
Solve 3 equations
Age of Nepo = 360 years
Problems on Ages
Joe’s father will be twice his age 6 years from
now. His mother was twice his age 2 years
before. If Joe will be 24 two years from now,
what is the difference between his father’s &
mothers age? (TCS)
Problems on AgesSolution:F + 6 = 2(J +6)M-2 = 2(J-2)Joe is now 22, F = 2(22) +12 – 6 = 50M = 2(22) – 4+2 = 42Difference = 50 -42 =8
Problems on Ages
One year ago the ratio of Baskaran’s and
Saraswati’s age was 6 : 7 respectively. Four
years hence, this ratio would become 7 : 8. How
old is Saraswathi?
Problems on Ages
Solution:
One year ago,
Baskaran’s age = 6x Saraswathi’s age = 7x
Four years hence
[(6x + 1) + 4] / [(7x+1) + 4] = 7 / 8
Simplifying x =5
Saraswathi’s present age is 7x +1 = 36 years
Puzzle
Can you make 120 by using 5 zeros?
Puzzle
Solution:
Fact (0! + 0! + 0! + 0! + 0!)
= Fact (5)
= 5!
= 120
Ratio and Proportion
• Ratio: The Relationship between two
variables is ratio.
• Proportion: The relationship between
two ratios is proportion.
Ratio and Proportion
The two ratios are a : b and the sum nos. is x
ax bx -------- and ------- a + b a + b
Similarly for 3 numbers a : b : c
Ratio and Proportion
A sporting goods store ordered an equal
number of white and yellow balls. The tennis
ball company delivered 45 extra white balls
making the ratio of white balls to yellow balls
1/5 : 1/6. How many white tennis balls did the
store originally order for? (TCS Question)
Ratio and Proportion
Solution: Let the number of yellow balls be x
(x + 45) : x = 1/5 : 1/6
Solving the above equation,
The number of white balls originally ordered
would be = 225 balls
Ratio and Proportion
The ratio of the rate of flow of water in pipes
varies inversely as the square of the radius of
the pipes. What is the ratio of the rates of flow
in two pipes of diameters 2 cm and 4 cm
respectively?
Ratio and Proportion
Solution:R1 = 1st PipeR2 = 2nd PipeR1 α 1/r12
R2 α 1/r22
R1: R2 = 1/r12: 1/r22
= 1/12: 1/22
= 1/1: 1 / 4 = 4:1 Ratio of rates of flow is 4:1
Ratio and Proportion
I participated in a race. 1/5th of the
participants were before me and 5/6th of
them behind me. Find the total number of
participants.
(Infosys Question)
Ratio and Proportion
Solution:
Let the total number of participants be x.
(x – 1)/5 + 5 (x-1)/6 = x
Solving the above equation,
The total number of participants would be = 31
Ratio and Proportion
If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7,
find A : D?
Ratio and Proportion
Solution :
(A /B *B/C*C/D) =2/3*4/5*6/7
A : D = 16 : 35
Ratio and Proportion
A bag contains 50 paise, 25 paise and 10 paise
coins in the ratio 5 : 9 : 4, amounting to Rs. 206.
Find the number of coins of each type.
Ratio and Proportion
Solution : The actual ratio would be5 * 50/100 : 9 * 25/100 : 4 * 10/100 = 50 : 45 : 8Value of 50 paise coins=206*50/103=10025paise coins =206*45/103=9010paise coins =206*8/103 =16Deriving from this ratio,Number of 50 paise coins = 100*2=200Number of 25 paise coins = 90*4 =360Number of 10 paise coins = 16*10 =160
Ratio and Proportion
There are 20 poles with a constant distance
between each pole. A car takes 24 seconds to
reach the 12th pole. How much time will it take
to reach the last pole?
Ratio and Proportion
Solution:
11 : 19 = 24 : x
By Solving
x = 41.45 seconds
Ratio and Proportion
Annual incomes of A and B is in the ratio 5 : 4
and their annual expenses bear a ratio of 4 : 3. If
each of them saves Rs. 500 at the end of the year
then find the annual income.
Ratio and Proportion
Solution:
IncomeA : B = 5x : 4x
Expenses A : B = 4y : 3y
5x – 4y = 500 and 4x – 3y = 500
Solving the above equations,
A’s annual income = Rs. 2500
B’s annual income = Rs. 3500
Puzzle
Can you draw
three concentric circles with
a line passing through their center
without lifting hand?
•(Quantity of cheaper / Quantity of costlier)
(C.P. of costlier) – (Mean price)
= --------------------------------------
(Mean price) – (C.P. of cheaper)
Alligation or Mixture
Alligation or Mixture
Cost of Cheaper Cost of costlier c d
Cost of Mixture m
d-m m-c
(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)
A merchant has 100 kg of salt, part of which
he sells at 7% profit and the rest at 17% profit.
He gains 10% on the whole. Find the quantity
sold at 17% profit?
Alligation or Mixture
Solution: 7 17 10 (17-10) (10-7) 7 : 3The quantity of 2nd kind = 3/10 of 100kg = 30kg
Alligation or Mixture
In what ratio two varieties of tea one costing Rs. 27 per kg and the other costing Rs. 32 per kg should be blended to produce a blended variety of tea worth Rs. 30 per kg. How much should be the quantity of second variety of tea, if the first variety is 60 kg?
Alligation or Mixture
Solution:
27 32
30
2 3
Quantity of cheaper tea = 2
Quantity of superior tea 3
Quantity of cheaper tea =2*x/5 = 60 , x=150
Quantity of superior tea = 3 * 150/5 = 90 kg
Alligation or Mixture
Alligation or Mixture
A man buys two cows for Rs. 1350 and sells one
so as to lose 6% and the other so as to gain 7.5%
and on the whole be neither gains nor loss. What
does each cow cost?
Alligation or Mixture
Solution:
-6 7.5
0
7.5 6
Ratio = 5 : 4
Cost of first cow = 1350*5/9=Rs. 750
Cost of second cow = Rs. 600
Three types of tea A,B,C costs Rs. 95/kg, Rs.
100/kg. and Rs 70/kg respectively. How many
kg of each should be blended to produce 100
kg of mixture worth Rs.90/kg given that the
quantities of B and C are equal?
Sathyam Question
Alligation or Mixture
Answer:B+C/2 A85 95
905 5
Ratio is 1:1 so A = 50 , B + C = 50
The quantity would be 50 : 25 : 25
Alligation or Mixture
In what proportion water must be added to
spirit to gain 20% by selling it at the cost price?
Alligation or Mixture
Solution:Profit%=20%Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3
0 10 25/3
5/3 25/3
The ratio is 1: 5
Alligation or Mixture
Chain Rule
Direct Proportion :
A B
A B
Indirect Proportion:
A B
A B
Chain Rule
If 36 men can do a piece of work in 25 days then
in how many days 15 men will do it?
Chain Rule
Solution:
Men Days
36 25
15 x (Indirect)
x = 36 * 25/15
x = 60 days
If 12 carpenters working 6 hours a day can make
460 chairs in 24 days, how many chairs will 18
carpenters make in 36 days, each working 8 hours
a day?
Chain Rule
Solution:
Men Days Hours Chairs
12 24 6 460
18 36 8 x
x/460=18*8*36/12*6*24
x=1380
They will make 1380 chairs.
Chain Rule
Chain Rule
If 1 man can load 1 box in a truck in 9 minutes
and a truck can hold up to 8 boxes. How many
trucks completely are loaded when 18 men
work for 1.5 hours? ( CTS )
Chain Rule
Solution:
Men Box Min Trucks
1 1 9 1
18 8 90 x
x = 18 *1* 90/1*8 * 9 = 22.5
The number of trucks completely loaded = 22
Chain Rule
A garrison of 3300 men has provisions for 32 days when given at a rate of 850 grams per head. At the end of 7 days reinforcement arrives and it was found that now the provisions will last 8 days less when given at the rate of 825 grams per head. How many more men can it feed? (Sathyam question)
Chain Rule
Solution:Men Days Grams3300 17 850x 25 825
x = 3300 * 25 * 850/17 * 825 = 50005000 – 3300 = 1700
It can feed 1700 more men.
Chain RuleIf 3 men or 6 boys can do a piece of work in
10 days, working 7 hours a day. How many
days will it take to complete a piece of work
twice as large with 6 men and 2 boys working
together for 8 hours a day?
Chain Rule
Solution:One man’s work = 2 boys’ work
So, 6 men and 2 boys = 6 (2) + 2 = 14 boysBoys Hours Days Work6 7 10 114 8 x 2
x = 6 * 7 * 10 * 2/14 * 8They will finish the work in 7 ½ days.
Types:
• A invested Rs.x and B invested Rs.y then
A : B = x : y
• A invested Rs.X and after 3 months B invested
Rs.Y then the share is
• A : B = X * 12 : Y * 9
Partnership
Partnership
A and B are two partners in a business. A
contributes Rs. 1200 for 5 months and B Rs. 750
for 4 months. If total profit is Rs. 450, find the
respective shares.
Partnership
Solution:
Ratio is1200 * 5 : 750 * 4 = 2 : 1
Profit share of A = 2/3 * 450 = Rs. 300
Profit share of B = Rs. 150
Partnership
A and B invest in a business in the ratio 3 : 2. If
5% of the total profit goes to charity and A’s
share is Rs. 855, what is the total profit %?
Partnership
Solution:
Let the total profit be Rs. 100
After paying charity A’s share = 3/5 *95 = 57
If A’s share is Rs. 57, the total profit is 100
If A’s share is Rs. 855, the total profit is
100 * 855/57 = Rs. 1500
The total profit = Rs. 1500
Partnership
A,B,C entered into a partnership by making an
investment in the ratio of 3 : 5 : 7. After a year
C invested another Rs. 337600 while A
withdrew Rs. 45600. The ratio of investments
then changed into 24: 59 : 167. How much did A
invest initially?
Partnership
Solution:
Let the investments of A, B, and C be 3x, 5x, 7x
(3x – 45600) : 5x : (7x + 337600) = 24 : 59 : 167
(3x – 45600)/5x = 24/59
x = 47200
Initial investment of A = 47200 * 3 = Rs. 141600
A,B,C started a business each investing Rs.
20000. After 5 months A withdrew Rs. 5000, B
withdrew Rs. 4000 and C invested Rs. 6000
more. At the end of the year a total profit of Rs.
69900 was recorded. Find the share of each.
(Satyam Question)
Partnership
Solution:
(20000*5+15000*7) : (20000*5+16000*7) :
(20000*5 + 26000*7)
The ratio is 205 : 212: 282
A’s share = 205/699 * 69900 = Rs. 20500
B’s Share = Rs. 21200
C’s share = Rs. 28200
Partnership
A starts business with a capital of Rs. 1200. B
and C join with some investments after 3 and 6
months respectively. If at the end of a year the
profit is divided in the ratio 2 : 3 : 5 respectively
what is B’s investment in the business?
Partnership
Solution:
(1200 * 12) : ( x * 9) : (y * 6) = 2 : 3 : 5
(1200 * 12)/2 = 9x/3
x = 2400
B’s investment = Rs. 2400
Partnership
Partnership
A and B enter into partnership for a year. A
contributes Rs.1500 and B Rs. 2000. After 4
months, they admit C who contributes Rs. 2250.
If B withdraws his contribution after 9 months,
find their profit share at the end of the year? (In
the ratio)
Partnership
Solution:
A: B: C = 1500*12: 2000*9: 2250*8
= 18000: 18000: 18000
= 1: 1: 1
Profit share at the end of the year = 1: 1: 1
• If A can do a piece of work in n days, then A’s 1
day’s work = 1/n
• If A is thrice as B, then:
Ratio of work done by A and B = 3:1
Ratio of times taken by A and B= 1:3
Time and Work
Pipes and Cisterns
• P1 fills in x hrs. Then part filled in 1 hr is 1/x
• P2 empties in y hrs. Then part emptied in 1 hr
is 1/y
• P1 and P2 both working simultaneously which fills in x
hrs and empties in y hrs resp ( y>x) then net part filled is 1/x – 1/y
• P1 can fill a tank in X hours and P2 can empty the full
tank in y hours( where x>y), then on opening both pipes, the net part empties in hour 1/y -1/x
Pipes and Cisterns
A can do a piece of work in 8 days. A undertook
to do it for Rs. 320. With the help of B, he
finishes the work in 6 days. Find B’s share.
Time and Work
Solution:
(A + B)’s 1 day work = 1/6
A’s 1 day work = 1/8
B’s 1 day work = 1/24
Ratio of their work = 3 : 1
B gets ¼ * 320 = Rs. 80
Time and Work
5 men or 8 women do equal amount of work in a day.
A job requires 3 men and 5 women to finish the job
in 10 days. How many women require to finish the
job in 14 days?
Satyam Question
Time and Work
Solution:
5 men = 8 women, so 1 man = 8/5 women
3m + 5w = 3 (8/5) w + 5 w = 49/5 w
Days Women
10 49/5
14 x
The number of women required = 7
Time and Work
A can do a piece of work in 30 days, and B in 50
days and C in 40 days. If A is assisted by B on
one day and by C on the next day alternatively
work will be completed in ?
Time and Work
Solution:
A = 1/30, B = 1/50, C = 1/40
A+B = 1/30+1/50 = 8/150
A+C = 1/30+1/40 = 7/120
2 days work done by A & B = 8/150 + 7/120) = 67/600
16 days work done = 67 * 8/600 = 536/600 = 67/75
Work left = 1-67/75 = 8/75
17th day A & B are working = 8/75 – 8/150 = 4/75
18th day A&C are working 4/75 = 120/7* 4/75 = 32/35
They will finish the work in 17 + 32/35 = 17 32/35 days
Time and Work
Time and Work
Two men undertake to do a piece of work for
Rs.1400. First man alone can do this work in
7 days while the second man alone can do
this work in 8 days. If they working together
complete this work in 3 days with the help of
a boy, how should money be divided?
Time and Work
Solution:Wages of the first man for 3 days= work done by him in 3 days * Rs. 1400= 3/7 * 1400 = Rs. 600Wage of second man for 3 days = work done by him in 3 days * Rs. 1400 = 3 / 8 *1400 = Rs. 525Wages of the boy for 3 days = Rs. 1400 – Rs. (600 + 525) = Rs. 275 There shares will be 600, 525, 275.
Time and Work
2 men and 3 boys can do a piece of work in 10
days while 3 men and 2 boys can do the same
work in 8 days. In how many days can 2 men
and 1 boy do the work?
Satyam Question
Time and Work
Answer:
1 man’s work = x
1 boy’s work = y
2x + 3y = 1/10 and 3x + 2y = 1/8
Solving the above equations,
The required number of days = 12½ days
Pipes A and B running together can fill a cistern
in 6 minutes. If B takes 5 minutes more than A
to fill the cistern then find the time in which A
and B will fill the cistern separately.
Time and Work (Pipes)
Answer:
1/x + 1/(x+5) = 1/6
Solving the above equation,
Pipe A will fill the cistern in 10 minutes.
Pipe B will fill the cistern in 15 minutes.
Time and Work (Pipes)
A cistern has two taps which fill it in 12 minutes
and 15 minutes respectively. There is also a
waste pipe in the cistern. When all the pipes are
opened, the empty cistern is filled in 20 minutes.
How long will a waste pipe take to empty a full
cistern ?
Time and Work (Pipes)
Solution:
All the tap work together = 1/12 + 1/15 - 1/20
= 5/60 + 4/60 – 3/60
= 6/60
= 1/10
The waste pipe can empty the cistern in 10 minutes.
Time and Work (Pipes)
One fill pipe A is 3 times faster than second
fill pipe B and takes 32 minutes less than the
fill pipe B. when will the cistern be full if both
pipes are opened together ?
Time and Work (Pipes)
Solution:B = xA = 3xB = x (1/3) = x/3x – 32 =x/33x – 96 = x3x – x = 962x = 96x = 96/2 = 48B = 48 -32 = 16A and B together = 1/16 + 1/48 = 1/12
The cistern will be filled in 12 minutes when both the pipes are opened together.
Time and Work (Pipes)
Time and Work (Pipes)
Two pipes A and B can fill a tank in 20 min. and
40 min. respectively. If both the pipes are opened
simultaneously, after how much time A should
be closed so that the tank is full in 10 minutes?
Time and Work (Pipes)
Solution:Let B be closed after x minutes. Then,Part filled by (A+B) in x min + part filled by A in( 10 – x min)=1x (1/20 +1/40) + (10 – x) * 1/20 = 1x (3/40) + 10 – x/20 = 13x/40 + 10 –x/ 20 = 13x + 20 – 2x = 403x – 2x = 40 -20x = 20 min. A must be closed after 20 minutes.
Time and Work (Pipes)
Two taps A and B can fill a tank in 6 hours and
4 hours respectively. If they are opened on alternate
hours and if pipe A is opened first, in how many
hours, the tank shall be full?
Time and Work (Pipes)
Solution:A’s work in hour = 1/6, B’s work in 1 hour = 1/4(A + B)’ s 2 hr work = 1/6+1/4 = 5/12(A + B)’ s 4 hr work = 10/12 = 5/6Remaining Part = 1- 5/6 = 1/6Now, it is A’s turn and 1/6 part is filled by A in
1 hour.Total time = 4+1 =5
Area and Volume
Cube:
• Let each edge of the cube be of length a. then,
• Volume = a3cubic units
• Surface area= 6a2 sq.units.
• Diagonal = √3 a units.
Cylinder:
• Let each of base = r and height ( or length) = h.
• Volume = πr2h
• Surface area = 2 πr h sq. units
• Total Surface Area = 2 πr ( h+ r) units.
Area and Volume
Cone:
• Let radius of base = r and height=h, then
• Slant height, l = √h2 +r2 units
• Volume = 1/3 πr2h cubic units
• Curved surface area = πr l sq.units
• Total surface area = πr (l +r)
Area and Volume
Sphere:
• Let the radius of the sphere be r. then,
• Volume = 4/3 πr3
• Surface area = 4 π r2sq.units
Area and Volume
Circle: A= π r 2
Circumference = 2 π r
Square: A= a 2
Perimeter = 4a
Rectangle: A= l x b
Perimeter= 2( l + b)
Area and Volume
Triangle:
A = 1/2*base*height
Equilateral = √3/4*(side)2
Area of the Scalene Triangle
S = (a+b+c)/ 2
A = √ s*(s-a) * (s-b)* (s-c)
Area and Volume
What is the cost of planting the field in the form
of the triangle whose base is 2.8 m and height
3.2 m at the rate of Rs.100 / m2
Area and Volume
Solution:
Area of triangular field = ½ * 3.2 * 2.8 m2
= 4.48 m2
Cost = Rs.100 * 4.48
= Rs.448..
Area and Volume
Area and Volume
Find the length of the longest pole that can be
placed in a room 14 m long, 12 m broad, and 8
m high.
Area and Volume
Solution:
Length of the longest pole = Length of the diagonal of the room
= √(142 + 122 + 82)
= √ 404 = 20.09 m
Area of a rhombus is 850 cm2. If one of its
diagonal is 34 cm. Find the length of the other
diagonal.
Area and Volume
Solution:
850 = ½ * d1 * d2
= ½ * 34 * d2
= 17 d2
d2 = 850 / 17
= 50 cm
Second diagonal = 50cm
Area and Volume
A grocer is storing small cereal boxes in large
cartons that measure 25 inches by 42 inches by 60
inches. If the measurement of each small cereal
box is 7 inches by 6 inches by 5 inches then what
is maximum number of small cereal boxes that can
be placed in each large carton ?
Area and Volume
Solution:
No. of Boxes = 25*42*60 / 7*6*5 = 300
300 boxes of cereal box can be placed.
Area and Volume
Area and Volume
If the radius of a circle is diminished by 10%,
what is the change in area in percentage?
Area and Volume
Solution:
= x + y + xy/100
= -10 - 10 + 10*10/100
= -19%
Diminished area = 19%.
Area and Volume
A circular wire of radius 42 cm is bent in the
form of a rectangle whose sides are in the ratio
of 6:5. Find the smaller side of the rectangle?
Area and Volume
Solution:
length of wire = 2 πr = (22/7*14*14)cm
= 264cm
Perimeter of Rectangle = 2(6x+5x) cm
= 22xcm
22x =264 x = 12 cm
Smaller side = (5*12) cm = 60 cm
Area and Volume
A farmer owns a square field with a pole in one
of the corners to which he tied his cow with a
rope whose length is about 10 m. What is the
area available for the cow to graze?
(Caritor Question)
Area and Volume
Answer:
Area = 78.5 m2
Area and Volume
If the length of a rectangle is reduced by 20%
and breadth is increased by 20%. What is the
percentage change in the area?
(Infosys Question)
Area and Volume
Solution:
x + y + (xy/100)%
= - 20 + 20 – 400/100
= -4
The area would decrease by 4%
Area and Volume
A power unit is there by the bank of the river of
750 m width. A cable is made from power unit to
power a plant opposite to that of the river and
1500 m away from the power unit. The cost of
the cable below water is Rs. 15 per meter and
cost of cable on the bank is Rs. 12 per meter.
Find the total cost. (TCS Question)
Area and Volume
Answer:
Cable under water is 750m, on the bank 750m
(750 * 15) + (750 * 12) = 20250
Total cost = Rs. 20250
• Probability:
P(E) = n(E) / n (S)
• Addition theorem on probability:
n(AUB) = n(A) + n(B) - n(AB)
• Mutually Exclusive:
P(AUB) = P(A) + P(B) Where P(A B)=0
• Independent Events:
P(AB) = P(A) * P(B)
Probability
There are 6 pairs of gloves of different sizes. In how
many ways can you choose one for the left hand and
one for the right hand such that they are not of the
same pair?
(Caritor Question)
Probability
Solution:
The number of possibilities are 6 * 6 = 36
They should not be from the same pair = 36 - 6
=30
30 ways.
Probability
A speaks the truth 80% of the times, B speaks
the truth 60% of the times. What is the
probability that they tell the truth at the same
time?
(Wipro Question)
Probability
Answer :
P(Both tell truth) = P(A) * P (B)
= 0.8 * 0.6
P = 0.48
Probability
A group consists of equal number of men and
women. Of them 10% of men and 45% of women
are unemployed. If a person is randomly selected
from the group find the probability for the selected
person to be an employee.
(Satyam Question)
Probability
Solution:
Let the number of men is 100 and women be 100
Employed men and women = (100-10)+(100-45)
= 145
Probability = 145 / 200 = 29 / 40
Probability
Probability
The probability of an event A occurring is 0.5 and that of B is 0.3. If A and B are mutually exclusive events. Find the probability that neither A nor B occurs?
Probability
Solution:
It is Mutually exclusive events P(A n B)=0
Probability = 1 – ( P(A) + P (B) – P(A n B) )
= 1 – (0.5 + 0.3 – 0)
= 0.2
Simple Interest = PNR / 100
Amount A = P + PNR / 100
When Interest is Compound annually:
Amount = P (1 + R / 100)n
C.I = A-P
Simple / Compound
Interest
• Half-yearly C.I.:
Amount = P (1+(R/2)/100)2n
• Quarterly C.I. :
Amount = P (1+(R/4)/100)4n
Simple / Compound Interest
Simple/compound interest
Difference between C.I and S.I for 2 years
= P*(R/100)2.
Difference between C.I and S.I for 3 years
= P{(R/100)3+ 3(R/100)2 }
If the compound interest on a certain sum for
two years is Rs.60.60 and simple interest is
Rs. 60, then the rate of interest per annum is ?
Simple / Compound Interest
Solution:
SI for 2 years
60 = P*2*R/100
PR/ 100 = 30
C.I – S.I = P (R/100)2
P(R/100 * R/100) = 0.60
30 * R/100 =0.6 = 2%
The rate of interest is 2%
Simple / Compound Interest
Find in what time a given sum of money will
quadruple itself at simple interest at the rate
of one paise per rupee per month ?
Simple / Compound Interest
Solution:
4P after n years
4P = P+ P*n*12/100
4 = 1+ 12n/100
n = 25 years
The period is 25 years.
Simple / Compound Interest
If Rs. 1000 is invested at 5% simple interest if
the interest is added to the principal every ten
years, the amount will become Rs. 2000 after
how many years ?
Simple / Compound Interest
Solution:Amount = Principle + interest for 10 years = 1000 +( 1000*10*5/100 ) = 15001500 after 10 years2000 after n years500 = 1500*5*n/100 n = 6 2/3yearTotal 10 + 6 2/3 = 16 2/3 Total time is 16 2/3 years
Simple / Compound Interest
Simple/compound interest
A sum of money amounts to Rs. 6690 after 3
years and to Rs. 10035 after 6 years on C.I. Find
the sum.
(Satyam Question)
Simple/Compound interest
Solution:
P(1 + R/100)3 = 6690 -----------1
P(1 + R/100)6 = 10035 -----------2
Dividing (1 + R/100)3 = 10035/6690 = 3/2
Substitute in equation 1 then P(3/2)=6690
P = 6690 * 2/3 = 4460
The sum is Rs.4460
Simple/Compound interest
What will be the difference between S.I and C.I
on a sum of Rs. 4500 put for 2 years at 5% per
annum?
Simple/Compound interest
Solution:
C.I – S.I = P (R/100)2
Difference = Rs. 11.25
Average
• Average is a simple way of representing an
entire group in a single value.
• “Average” of a group is defined as:
X = (Sum of items) / (No of items)
Average
A batsman makes a score of 90 runs in the 17th
innings and then increases his average by 3. Find
his average after 17th innings.
(Infosys Question)
Average
Solution:
Let the average of 17th innings be x
Average after the 16th innings = x-3
16(x-3) + 90 = 17x
x = 42
Average
The average of 11 observations is 60. If the
average of 1st five observations is 58 and that
of last five is 56, find sixth observation?
Average
Solution:
5 observations average = 58
Sum = 58*5 = 290
Last 5 observation average = 56
Sum = 56*5 = 280
Total sum of 10 numbers = 570 (290 + 280)
Total sum of 11 numbers = 660 (11*60)
6th number = 90 (660 – 570)
Average
The average of age of 30 students is 9 years. If
the age of their teacher included, it becomes 10
years. Find the age of the teacher?
Average
Solution:
30 students total age = 30*9=270
Including the teacher’s age = 31*10=310
Difference = 310 – 270 = 40 years
Average
Of the three numbers second is twice the first
and is also thrice the third. If the average of the
three numbers is 44, find the largest number.
Average
Answer:
Let the first number be x
Average = (x + 2x + 2x/3)/3 = 44
Largest number is 72
Average
In a coconut groove, (x+2) trees yield 60 nuts
per year, x trees yield 120 nuts per year and (x-
2) trees yield 180 nuts per year. If the average
yield per year per tree be 100 find x.
Average
Answer:
[(x+2)*60 + (x*120) + (x-2) *180] =100
[x+2+x+x-2]
By simplifying x = 4
Average
In a cricket team of 11 boys one player weighing
42 kg is injured and his place is taken by another
player. If the average weight of the team is
increased by 100 g as a result of this, then what
is the weight of the new player?
Average
Solution:
Average weight of 11 boys is increased by 0.1 kg
The total increase in weight = 0.1 * 11 = 1.1 kg
Weight of the new boy = 42 + 1.1 = 43.1 kg
Permutations and Combinations
• Factorial Notation: n! = n(n-1)(n-2)….3.2.1 • Number of Permutations: n! / (n-r)!
• Combinations: n! / r!(n –r)!
• The number of Combinations of ‘n’ things taken
‘r’ at a time in which ‘p’ particular things will
always occur is n-pCr-p
• The number of Combinations of ‘n’ things taken
‘r’ at a time in which ‘p’ particular things never
occur is n-pCr
Permutations and Combinations
A foot race will be held on Saturday. How many
different arrangements of medal winners are
possible if medals will be for first, second and
third place, if there are 10 runners in the race …
Permutations and Combinations
Solution:
n = 10
r = 3
n P r = n!/(n-r)!
= 10! / (10-3)!
= 10! / 7!
= 8*9*10
= 720
Number of ways is 720.
Permutations and Combinations
To fill a number of vacancies, an employer must
hire 3 programmers from among 6 applicants,
and two managers from 4 applicants. What is
total number of ways in which she can make her
selection ?
Permutations and Combinations
Solution:
It is selection so use combination formula
Programmers and managers = 6C3 * 4C2
= 20 * 6 = 120
Total number of ways = 120 ways.
Permutations and Combinations
Permutations and Combinations
A man has 7 friends. In how many ways can
he invite one or more of them to a party?
Permutations and Combinations
Solution:
In this problem, the person is going to select his friends for party, he can select one or more person, so addition
= 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7
= 127
Number of ways is 127
Permutations and Combinations
If nP5 = 20 nP3 then what is the value of n?
Permutations and Combinations
Solution:
n!/(n-5)! = 20 * n!/(n-3)!
n = 8
Permutations and Combinations
Find the number of different 8 letter words
formed from the letters of the word EQUATION
if each word is to start with a vowel
Permutations and Combinations
Solution:
For the words beginning with a vowel, the first
letter can be any one of the 5 vowels, the
remaining 7 places can be filled by
7P7 = 5040
The number of words = 5 * 5040 = 25200
Permutations and Combinations
In how many different ways can the letters of the
word TRAINER be arranged so that the vowels
always come together?
Permutations and Combinations
Solution:
A,I,E can be arranged in 3! Ways
(5! * 3!) / 2! = 360 ways
Percentage
• By a certain Percent, we mean that many
hundredths.
• Thus, x Percent means x hundredths, written
as x%
•Finding out of Hundred.
If Length is increased by X% and Breadth is decreased by Y% What is the percentage Increase or Decrease in Area of the rectangle?
Formula: X+Y+ XY/100 %
Decrease 20% means -20
Percentage
Percentage
Two numbers are respectively 30% & 40%
less than a third number. What is second
number as a percentage of first?
Percentage
Solution:Let 3rd number be x.1st number = x – 30% of x = x – 30x/100 = 70x/ 100 = 7x/102nd number = x – 40% of x = x – 40x/100 = 60x/ 100 = 6x/10Suppose 2nd number = y% of 1st n umber6x / 10 = y/100 * 7x /10 y = 600 / 7 y = 85 5/7 85 5/7%
Percentage
After having spent 35% of the money on
machinery, 40% on raw material, 10% on
staff, a person is left with Rs.60,000. The
total amount of money spent on machinery
and the raw material is?
Percentage
Solution:Let total salary =100%Salary = 100Spending: Machinery + Raw material + staff = 35+40+10 = 85Remaining percentage = 100 – 85 = 1515 = 60000100 = ? By chain rule, we get 400000In this 400000 , 75% for machinery & raw material = 4, 00,000* 75/100
=Rs 3 00000
Percentage
If the number is 20% more than the other,
how much percent is the second number less
than the first?
Percentage
Solution:
Let x =20
= x / (100+x) *100
=( 20 /120 )*100
=16 2/3
The percentage is 16 2/3
Percentage
Solution:Let capacity of the tank be 100 liters. Then,Initially: A type petrol = 100 litersAfter 1st operation:A = 100/2 = 50 liters, B = 50 litersAfter 2nd operation:A = 50 / 2+50 = 75 liters, B = 50/2 = 25 litersAfter 3rd operation:A = 75/2 37.5 liters, B = 25/2 +50 = 62.5 litersRequired Percentage = 37.5%
Percentage
Answer:
Increase % in the area of the new rectangle
= x + y + (xy/100)%
= 20 + 10 + (200/100)
= 32%
Increase 32%
Percentage
If A’s income is 40% less than B’s income, then
how much percent is B’s income more than A’s
income?
Percentage
Answer:
B’s income = 40/(100-40) * 100%
= 66 2/3%
Percentage
Ramesh loses 20% of his pocket money. After spending 25% of the remainder he has Rs. 480 left. What was his pocket money?
Percentage
Answer:
x (1-20/100) (1-25/100) = 480
Solving the equation, x = 800
Pocket money = Rs. 800
Boats and streams
•Up stream – against the stream
•Down stream – along the stream
•u = speed of the boat in still water
•v = speed of stream
•Down stream speed (a)= u+v km / hr
•Up stream speed (b)=u-v km / hr
•u = ½(a+b) km/hr
•V = ½(a-b) km / hr
A boat’s crew rowed down a stream from A to B
and up again in 7 ½ hours. If the stream flows at
3km/hr and speed of boat in still water is 5
km/hr. , find the distance from A to B ?
Boats and streams
Solution: Down Stream = Sp. of the boat + Sp. of the stream = 5 +3 =8Up Stream = Sp. of the boat – Sp. of the stream = 5-3 = 2Let distance be xDistance/Speed = Time X/8 + x/2 = 7 ½ X/8 +4x/8 = 15/2 5x / 8 = 15/2 5x = 15/2 * 8 x =12
Boats and streams
Boats and Streams
A boat goes 40 km upstream in 8 hours and 36
km downstream in 6 hours. Find the speed of the
boat in still water in km/hr?
Boats and Streams
Solution:
Speed of the boat in upstream = 40/8 = 5 km/hr
Speed of the boat in downstream = 36/6 = 6 km/hr
Speed of the boat in still water = 5+6/2 = 5.5 km/hr
Boats and Streams
A man rows to place 48km distant and back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. Find the rate of the stream?
Boats and Streams
Solution:Down stream 4km in x hours. Then,Speed of Downstream = 4/x km/hr, Speed of Upstream = 3/x km/hr48/ (4/x) + 48/(3/x) = 14 , x = 1/2Speed of Down stream = 8, Speed of upstream = 6Rate of the stream = ½ (8-6) km/hr = 1 km/hr
Boats and Streams
In a stream running at 2 km/h a motor boat goes 10 km upstream and back again to the starting point in 55 minutes. Find the speed of the motor boat in still water.
Boats and Streams
Solution:
Let the speed be x km/h
10/(x+2) + 10/(x-2) = 55/60
Solving the above equation,
x = 22 km/h
Boats and Streams
A swimmer can swim a certain distance in the direction of current in 5 hours and return the same distance in 7 hours. If the stream flows at the rate of 1 km/h, find the speed of the swimmer in still water.
Boats and Streams
Solution:
Let the speed of the swimmer in still water be x
Downstream speed = (x+1) * 5
Upstream speed = (x-1) * 7
5(x+1) = 7(x-1)
Solving the equation, x = 6
Speed of the swimmer = 6 km/h
Time and Distance
•Speed:- Distance covered per unit time is
called speed.
Speed = distance/time
•Distance = speed*time
•Time = distance/speed
• Distance covered α Time (direct variation).
• Distance covered α speed (direct variation).
• Time α 1/speed (inverse variation).
Time and Distance
• Speed from km/hr to m/sec - ( * 5/18).
• Speed from m/sec to km/h, - ( * 18/5).
• Average Speed:-
Average speed = Total distance traveled
Total time taken
Time and Distance
The jogging track in a sports complex is 726m
in circumference. Suresh and his wife start from
the same point and walk in opposite direction at
4.5km/hr and 3.75km/hr respectively. They will
meet for the first time in how many minutes ?
Time and Distance
Solution:
Suresh speed m/hr = 4.5*5/18 *60*60 = 4500
His wife speed m/hr = 3.75*5/18*60*60 = 3750
Meeting time
4500/60 x +3750/60x = 756 m
x = 5.28 m
They will meet after 5.28minutes
Time and Distance
Ram travels from P to Q at 10km/hr and returns at
15km/hr. Shyam travels P to Q and returns at
12.5km/hr. If he takes 12 minutes less than ram
then what is the distance between P and Q ?
Time and Distance
Solution:
Average Speed of Ram = 2(10*15)/25 = 12km/hr
Average Speed of Shyam = 2(12.5 *12.5)/25 = 12.5
Ram’s speed in m/sec = 12*5/18 = 10/3
Shyam’s speed in m/sec = 12.5*5/18 = 625/36
(x/ 10/3) – (x/625/36) = 12*60
3x/10 – 30x/625 = 720
= 30
Distance between P and Q is 30 km
Time and Distance
If I walk 30 miles/hr I reach 1 hour before and if I walk
20 miles/hr I reach 1 hour late. Find the distance between
two points and if the exact time of reaching destination is
11 a.m. then find the speed with which I walk?
Infosys Question
Time and Distance
Solution:
x/30-x/20 = 2
By Solving the equation,
The total distance between two points = 120 miles
TheAverage speed required = 24 miles/hr
Time and Distance
Time and Distance
By walking at ¾ of his usual speed, a man
reaches office 20 minutes later than usual.
Find his usual time?
Time and Distance
Solution:
Usual time = Numerator * late time
= 3*20
= 60
Time and Distance
A car travels a certain distance taking 7 hours in forward journey. During the return journey it increased the speed by 12 km/hr and takes the time of 5 hours. What is the distance traveled?
(Satyam Question)
Time and Distance
Solution:
Let speed be x , D = S * T =7x
7x = (x+12) 5
Solving the above equation,
x = 30
D = 7 * 30 = 210
Actual Distance = 210 km
Time and Distance (Trains)
A train starts from Delhi to Madurai and at the
same time another train starts from Madurai to
Delhi after passing each other they complete
their journeys in 9 and 16 hours, respectively.
At what speed does second train travels if first
train travels at 160 km/hr ?
Time and Distance (Trains)
Solution:
Let x be the speed of the second train
S1 / S2 = √T2/T1
160/x = √16/9
160/x = 4/3
x = 120
The speed of second train is 120km/hr.
Time and Distance (Trains)
How long will a train 100 m long traveling at
72 km/hr take to overtake another train 200 m
long traveling at 54 km/h?
(Satyam Question)
Time and Distance (Trains)
Solution:
They are traveling in the same direction. Hence the Relative speed is x -y
72 – 54 = 18 km/h
18 * 5/18 = 5 m/sec
To travel 5 m it takes 1 second
To travel 300m it will take300/5 = 60 seconds
It will take 60 seconds or 1 minute.
Time and Distance (Trains)
Rajee starts her journey from Delhi to Bhopal and
simultaneously Sheela starts from Bhopal to Delhi.
After crossing each other they finish their remaining
journey in 5 4/9 hours and 9 hours respectively.
What is Sheela’s speed if Rajee’s speed is 36 km/hr.
Time and Distance (Trains)
Answer:
Rajee’s speed = Square root of (t2/t1)
Sheela’s speed
The speed of Sheela = 28 km/hr
Time and Distance (Trains)
Excluding stoppages, the speed of a train is
54 km/hr and including stoppages it is 45
km/hr. For how many minutes does the train
stop per hour?
Time and Distance (Trains)
Solution:
x
= _____________
x/45 + x/36
Take L.C.M and find the x value
x = 12
It stops 12 minutes/hr
Time and Distance (Trains)
A train 130 m long passes a bridge in 21 seconds
moving with a speed of 90 km/hr. Find the
length of the bridge?
Time and Distance (Trains)
Solution:Let the length of the bridge be xSpeed of the train = (length of the train + length
of the bridge) / Time taken90*5/18 = (130 +x )/21Length of the bridge = 395 m
• Gain =(S.P.)-(C.P.)
• Loss =(C.P.)-(S.P.)
• Loss or gain is always reckoned on C.P.
• Gain% = [(Gain*100)/C.P.]
• Loss% = [(Loss*100)/C.P.]
• S.P. = ((100 + Gain%)/100)C.P.
• S.P. = ((100 – Loss%)/100)C.P.
Profit and Loss
Sundeep buys two CDs for Rs.380 and sells one
at a loss of 22% and the other at a gain of 12%.
If both the CDs are sold at the same price, then
the cost price of two CDs is ?
Profit and Loss
Solution:CDs =Rs.3801st CD = x -22x/1002nd CD = y + 12y/100SP1 = SP2x – 22x/100 = y – 12y/100 x = 56y/39 x + y = 380 56y/39 + y = 380 y = 156 x = 224 Cost of the two CDs are Rs. 224 and Rs.156
Profit and Loss
Mr. Ravi buys a cooler for Rs. 4500. For how
much should he sell so that there is a gain of 8%
Profit and Loss
Solution:
S.P = (100+8%/100)/CP
= (108/100) * 4500 = 4860
He should sell it for Rs. 4860.
Profit and Loss
Profit and Loss
If the manufacturer gains 20%, the wholesale dealer 25%, and the retailer 35%, then find the cost of production if the retail price is Rs. 1265.
Profit and Loss
Solution: Let C.P be x
Gain 35% of 25% of 20% of x = 1265
135/100 * 125/100 * 120/100 * x = 1265
x = Rs.624.7
Profit and Loss
Manoj sells a shirt to Yogesh at a profit of 15% and Yogesh sells it to Suresh at a loss of 10%. Find the resultant profit or loss.
Profit and Loss
Solution:
Resultant profit = (x + y + xy/100)
= 15 – 10 – (150/100)
= 3½%
The resultant profit is 3.5%
Profit and Loss
Mr. Verma sold his scooter for Rs. 10500 at a gain of 5%. Find the cost price of the scooter.
Profit and Loss
Solution:
C.P = (100/(100+P%))*S.P
= (100/(100+5) )* 10500
= 10000
Cost price = Rs. 10000
Profit and Loss
A man bought a horse and a cart. If he sold the horse at 10% loss and the cart at 20% gain, he would not lose anything, but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. What is the cost of one horse and one cart?
Satyam Question
Profit and Loss
Solution : C.P of horse = x; C.P of cart = y
10x/100 = 20y/100
5x/100 = 5y/100 + 10
Solving the above two equations,
C.P of horse = x = Rs. 400
C.P of cart = y = Rs. 200
Profit and Loss
A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs. 10.50 less he would have gained 30%. Find the cost price of the article.
Satyam Question
Profit and Loss
Solution:
First C.P = x, S.P = 125% of x = 5x/4
Second C.P = 80x/100 = 4x/5
S.P = 130/100 * 4x/5 = 26x/5
5x/4 - 26x/25 = 10.50
x = 50
The cost price is Rs. 50
CalendarOdd days:
0 = Sunday
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
CalendarMonth code: Ordinary year
J = 0 F = 3
M = 3 A = 6
M = 1 J = 4
J = 6 A = 2
S = 5 O = 0
N = 3 D = 5
Month code for leap year after Feb. add 1.
Calendar
Ordinary year = (A + B + C + D )-2
-----------------------take remainder
7
Leap year = (A + B + C + D) – 3
------------------------- take remainder
7
What is the day of the week on 30/09/2007?
Calendar
Solution:
A = 2007 / 7 = 5
B = 2007 / 4 = 501 / 7 = 4
C = 30 / 7 = 2
D = 5
( A + B + C + D )-2
= -----------------------
7
= ( 5 + 4 + 2 + 5) -2
----------------------- = 14/7 = 0 = Sunday
7
Calendar
What was the day of the week on 13th May,
1984?
Calendar
Solution:
A = 1984 / 7 = 3
B = 1984 / 4 = 496 / 7 = 6
C = 13 / 7 = 6
D = 2
( A + B + C + D) -3
= -----------------------
7
= 14/7= 0, Sunday.
Calendar
Calendar
On what dates of April 2005 did Sunday fall?
CalendarSolution: You should find for 1st April 2005 and then you find the Sundays
date.A = 2005 / 7 = 3B = 2005 / 4 = 501 / 7 = 4C = 1 / 7 = 1D = 6 (A + B + C + D) -2 = ----------------------- 7 3 + 4 + 1 + 6 -2 ----------------------- = 12 / 7 = 5 = Friday. 7 1st is Friday means Sunday falls on 3, 10, 17, 24
Calendar
What was the day on 5th January 1986?
CalendarSolution:
A = 1986 / 7 = 5B = 1986 / 4 = 496/7 = 6C = 5 / 7 = 5D = 0 (A + B + C + D) -2 = ----------------------- 7 5 + 6 + 5 + 0-2 = ----------------------- = 14 / 7 = Sunday 7
Clock:
•In every minute, the minute hand gains 55 minutes on the hour hand
•In every hour both the hands coincide once. = (11m/2) – 30h (hour hand to min hand)
= 30h – (11m/2) (min hand to hour hand)
•If you get answer in minus, you have to subtract your
answer with 360 o
Clocks
Clocks
Find the angle between the minute hand and
hour hand of a clock when the time is 7:20.
Solution:
= 30h – (11m/2)
= 30 (7) – 11 20/2
= 210 – 110
= 100
Angle between 7: 20 is 100o
Clocks
Clocks
How many times in a day, the hands of a
clock are straight?
Clocks
Solution:
In 12 hours, the hands coincide or are in opposite direction 22 times a day.
In 24 hours, the hands coincide or are in opposite direction 44 times a day.
Clocks
How many times do the hands of a clock
coincide in a day?
ClocksSolution:
In 12 hours, the hands coincide or are in opposite direction 11 times a day.
In 24 hours, the hands coincide or are in opposite direction 22 times a day.
Clocks
At what time between 7 and 8 o’clock will the
hands of a clock be in the same straight line but,
not together?
Clocks
Solution: h = 7
= 30h – 11m/2
180 = 30 * 7 – 11 m/2
On simplifying we get ,
5 5/11 min past 7
Clocks
At what time between 5 and 6 o’clock will the
hands of a clock be at right angles?
Clocks
Solution: h = 5
90 = 30 * 5 – 11m/2
Solving
10 10/11 minutes past 5
Clocks
Find the angle between the two hands of a clock
at 15 minutes past 4 o’clock
Clocks
Solution:
Angle = 30h – 11m/2
= 30*4 – 11*15 / 2
The angle is 37.5o
Clocks
At what time between 5 and 6 o’clock are the
hands of a clock together?
Clocks
Solution: h = 5
O = 30 * 5 – 11m/2
m = 27 3/11
Solving
27 3/11 minutes past 5
In interpretation of data, a chart or a graph is
given. Some questions are given below this chart
or graph with some probable answers. The
candidate has to choose the correct answer from
the given probable answers.
Data Interpretation
• 1. The following table gives the distribution of students according to professional
courses:
__________________________________________________________________
Courses Faculty
___________________________________
Commerce Science Total
Boys girls Boys girls
___________________________________________________________
• Part time management 30 10 50 10 100
• C. A. only 150 8 16 6 180
• Costing only 90 10 37 3 140
• C. A. and Costing 70 2 7 1 80
__________________________________________________________________
• On the basis of the above table, answer the following questions:
The percentage of all science students over
Commerce students in all courses is
approximately:
(a) 20.5 (b) 49.4
(c) 61.3 (d) 35.1
Data Interpretation
Answer:
Percentage of science students over commerce
students in all courses = 35.1%
Data Interpretation
What is the average number of girls in all
courses ?
(a) 15 (b) 12.5
(c) 16 (d) 11
Data Interpretation
Answer:
Average number of girls in all courses = 50 / 4
= 12.5
Data Interpretation
What is the percentage of boys in all courses
over the total students?
(a) 90 (b) 80
(c) 70 (d) 76
Data Interpretation
Answer:
Percentage of boys over all students
= (450 x 100) / 500
= 90%
Data Interpretation
Data Sufficiency
Find given data is sufficient to solve the problem or not.
A.If statement I alone is sufficient but statement II alone is not sufficient
B.If statement II alone is sufficient but statement I alone is not sufficient
C.If both statements together are sufficient but neither of statement alone is sufficient.
D.If both together are not sufficient
Data Sufficiency
What is John’s age?
I. In 15 years will be twice as old as Dias would be
II. Dias was born 5 years ago. (Wipro)
Data Sufficiency
Answer:
c) If both statements together are sufficient but neither of statement alone is sufficient.
Data Sufficiency
What is the distance from city A to city C in kms?
I. City A is 90 kms from city B.
II.City B is 30 kms from city C
Data Sufficiency
Answer:
d) If both together are not sufficient
Data Sufficiency
If A, B, C are real numbers, Is A = C?
I. A – B = B – C
II. A – 2C = C – 2B
Data Sufficiency
Answer:
D . If both together are not sufficient
Data Sufficiency
What is the 30th term of a given sequence?
I. The first two term of the sequence are 1, ½
II. The common difference is -1/2
Data Sufficiency
Answer:
A. If statement I alone is sufficient but statement II alone is not sufficient
Data Sufficiency
Was Avinash early, on time or late for work?
I. He thought his watch was 10 minute fast.
II. Actually his watch was 5 minutes slow.
Data Sufficiency
Answer:
D. If both together are not sufficient
Data Sufficiency
What is the value of A if A is an integer?
I. A4 = 1
II. A3 + 1 = 0
Data Sufficiency
Answer:
B. If statement II alone is sufficient but statement I alone is not sufficient
Cubes
A cube object 3”*3”*3” is painted with green
in all the outer surfaces. If the cube is cut into
cubes of 1”*1”*1”, how many 1” cubes will
have at least one surface painted?
Cubes
Answer:
3*3*3 = 27
All the outer surface are painted with colour.
26 One inch cubes are painted at least one surface.
Cubes
A cube of 12 mm is painted on all its sides. If
it is made up of small cubes of size 3mm, and if
the big cube is split into those small cubes, the
number of cubes that remain unpainted is
Cubes
Answer:
= 8
Cubes
A cube of side 5 cm is divided into 125 cubes of equal size. It is painted on all 6 sides.
1. How many cubes are coloured on only one side?
2. How many cubes are coloured on only two side?
3. How many cubes are coloured on only three side?
4. How many cubes are not coloured?
Cubes
Answer:
1. 54
2. 36
3. 8
4. 27
Cubes
A cube of 4 cm is divided into 64 cubes of
equal size. One side and its opposite side is
coloured painted with orange. A side adjacent
to this and opposite side is coloured red. A side
adjacent to this and opposite side is coloured
green? Cont..
Cubes
1. How many cubes are coloured with Red alone?
2. How many cubes are coloured orange and Red alone?
3. How many cubes are coloured with three different colours?
4. How many cubes are not coloured?
5. How many cubes are coloured green and Red alone?
Cubes
Answer:
1. 8
2. 8
3. 8
4. 8
5. 8
CubesA 10*10*10 cube is split into small cubes of equal
size 2*2*2 each. A side and adjacent to it is coloured Pink. A side adjacent to Pink and opposite side is coloured Blue. The remaining sides are coloured yellow.
1. Find the no. of cubes not coloured?2. Find the no. of cubes coloured blue alone?3. Find the no. of cubes coloured blue & pink &
yellow?4. Find the no. of cubes coloured blue & pink ?5. Find the no. of cubes coloured yellow & pink ?
Cubes
Answer:
1. 27
2. 18
3. 4
4. 12
5. 12
Venn Diagram
If X and Y are two sets such that X u Y has 18
elements, X has 8 elements, and Y has 15
elements, how many element does X n Y have?
Venn Diagram
Solution:
We are given n (X uY) = 18, n (X) = 8, n (Y) =15. using the formula.
n( X n Y) = n (X) + n (Y) - n ( X u Y)
n( X n Y) = 8 + 15 – 18
n( X n Y) = 5
Venn Diagram
If S and T are two sets such that S has
21elemnets, T has 32 elements, and S n T has
11 elements, how many element elements does
S u T have?
Venn Diagram
Answer:
n (s) = 21, n (T) = 32, n ( S n T) = 11,
n (S u T) = ?
n (S u T) = n (S) + n( T) – n (S n T)
= 21 + 32 – 11 = 42
Venn Diagram
If A and B are two sets such that A has 40
elements, A u B has 60 elements and A n B
has 10 elements, how many element elements
does B have?
Venn Diagram
Answer:
n ( A) = 40, n ( n B) = 60 and n ( A n B) = 10,
n ( A u B) = n ( A) + n (B) – n ( A n B)
60 = 40 + n (B) – 10
n (B) = 30
Venn Diagram
In a group of 1000 people, there are 750 people
who can speak Hindi and 400 who can speak
English. How many can Speak Hindi only?
Answer:
n( H u E) = 1000, n (H) = 750, n (E) = 400,
n( H u E) = n (H) + n (E) – n( H n E)
1000 = 750 +400 – n ( H n E)
n ( H n E) = 1150 – 100 = 150
No. of people can speak Hindi only
_
= n ( H n E) = n ( H) – n( H n E)
= 750 – 150 = 600
Venn Diagram
In a class of 100 students, the number of students passed in English only is 46, in maths only is 46, in commerce only is 58. the number who passes in English and Maths is 16, Maths and commerce is 24 and English and commerce is 26, and the number who passed in all the subject is 7. find the number of the students who failed in all the subjects.
Venn Diagram
Solution:
No. of students who passed in one or more subjects
= 11+ 9 + 13 + 17 + 15 + 19 + 7 = 91
No of students who failed in all the subjects
= 100 -91 = 9
Venn Diagram
In a group of 15, 7 have studied Latin, 8 have
studied Greek, and 3 have not studied either.
How many of these studied both Latin and
Greek?