Transcript document

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
• Permutation and Combination
• Percentage
• Cubes
Numerical Reasoning
• Boats and Streams
• Time and Distance (Trains)
• Data Sufficiency
• Profit and Loss
• Calendar
• Clocks
• Data Interpretation
Problems on Numbers
Division Algorithm:
Dividend = the number to be divided.
Divisor = the number by which it is divided.
Dividend / Divisor = Quotient.
Quotient * Divisor = Dividend.
Quotient * Divisor + Remainder = Dividend.
Problems on Numbers
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
Geometrical Progression:
Tn = arn – 1.
Sn = a(rn – 1)/(r-1);
n/2 *[2a+(n-1)d)]
Basic Formulae
1. ( a+b)2 = a2 + b2 + 2ab
2. (a-b)2 = a2 +b2 -2ab
3. ( a+b)2 - (a – b)2 = 4ab
4. (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
Problem on Numbers
A monkey starts climbing up a tree 20 feet tall.
Each hour it claims 3 feet and slips back 2 feet.
How much time would it take the monkey to
reach the top?
(Satyam)
Problem on Numbers
Solution:
= 17 + 1= 18 hours
Problem on Numbers
The length of the side of a square is represented
by x + 2. the length of the side of an equilateral
triangle is 2x, if the square and the equilateral
triangle have equal perimeter the find the value
of x.
Problem on Numbers
Solution:
Side of the square is x + 2
Side of the triangle 2x
P = 4(x + 2)
= 4x +8
Perimeter = 3*2x
= 6x
4x + 8 = 6x
x=4
Problem on Numbers
On sports day if 30 children were made to stand
in a column, 16 column could be formed if 24
children were made to stand in a column. How
many columns could be formed?
(Satyam)
Problem on Numbers
Solution:
Total no of children = 30 * 16 = 480
No. of column of 24 children each = 480 / 24
= 20
Problem on Numbers
5/9 part of the population in a village are males
if 30% of the males are married what is the
percentage of unmarried females in the total
population?
Problem on Numbers
Solution:
Let population be = x
Males = 5 x/9
Married Men =30%(5x/9)= x/6 (Married females)
Total females = x – 5x/9 = 4x/9
Unmarried females = 4x/9 – x/6 = 5x/18
Percentage = (5x / 18)*100%
x
= 250 / 9%
Problems on Numbers
How many terms of the A.P. 1, 4, 7…. are
needed to give the sum 715 ?
Problems on Numbers
Solution:
Sum of n terms of A.P
Sn = n/2 *[2a+(n-1)d)]
Sn = 715, n =? a = 1 d = 3
715 = n/2 *[2*1+(n-1)3]
715 =n/2 *[2+(n-1)3
1430 = n[2+3n -3]
1430 = n[3n – 1]
1430 = 3n2 -n
3n2 – n – 1430 = 0 Solving the Quadratic Equation,
n =22
Number of terms needed is 22
Problems on Numbers
The sum of the digits of a two-digits number
is 8. if the digits are reversed the number is
increased by 54. Find the number?
Problems on Numbers
Solution:
x+ y = 8 --------------------1
10y+x = 10x+y+54
10y + x - 10x – y = 54
-x + y = 6 ---------------------2
Solve equation 1&2
x= 1, y = 7
Required number = 10*1 +7 =17
Problems on Ages
The ages of two persons differ by 10 years. If
5 years ago, the elder one be 2 times as old as
the younger one, find their present ages.
Problems on Ages
x-y = 10; x = 10 + y
x- 5 = 2(y-5)
y + 10 -5 = 2y -10
y+5 = 2y -10
2y- y = 15
y=15: x = 25
Their present ages are 15 years and 25 years.
Problems on Ages
The present ages of three persons are in the
proportion of 4:7:9. 8 years ago, the sum of
their ages was 56. Find their present ages ?
Problems on Ages
Solution:
Three person’s age ratio = 4:7:9
Sum of their age = 56, after 8 years
their sum of their ages = 80
A’s age = 4/20 *80 = 16
B’s age = 7/20 *80 = 28
C’s age = 9/20 *80 = 36
Their present ages are 16, 28 and 36.
Problems on Ages
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, find
the age of Father.
Problems on Ages
Solution:
Father age = 3(x+y)
F+20 = x+20+y+20
3x+3y+20 = x+y+40
3x +3y = x+y+40 -20
3x-x+3y – y =20
2x+2y = 20
x+y = 10
F = 3*10 =30
The father’s age is 30.
Problems on Ages
Jalia is twice older than Qurban. When Jalia was
4 years younger, Qurban was 3 years older the
difference between their ages is 12 what is the
sum of their ages?
Problems on Ages
Solution:
J = 2Q
1
(J – 4) – (Q +3) = 12
2
Solve 1 &2
Sum of their ages = 19 + 38 = 57
Problems on Ages
Ten years ago, Chandrawathi’s mother was 4
times older than her daughter. After 10 years the
mother will be twice older than daughter. What
is the present age of Chandrawathi?
Problems on Ages
Solution:
Let Chandrawathi’s age 10 years ago x
Her mother’s age 10 years ago 4x
(4x + 10 + 10) = 2(x+10+10)
x = 10
Chandrawathi is x +10 = 20 years
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
Concentration of three wines A, B, and C are
10, 20, and 30 percent respectively. They are
mixed in the ratio 2 : 3 : x resulting in a 23%
concentration solution. Find x.
(Caritor Question)
Ratio and Proportion
Answer :
10 : 20 : 30
2 : 3:x
Multiplying
20 : 60 : 30x
20 + 60 + 30x = 230
x=5
Ratio and Proportion
Ajay, Aman, Suman and Geetha rented a
house and agreed to share the rent as follow
Ajay : Aman = 8:15
Aman : Suman = 5:8
Suman: geetha = 4:5
The part of rent paid by Suman will be ?
Ratio and Proportion
Solution:
Ajay : Aman = 8 : 15
In next ratio Aman is 5 to make that 15 multiply by 3 that is ratio
of
Aman : Suman = 15:24
Suman and Geetha contribution is 4:5 but Suman previous
contribution is 24 to make Suman contribution in 24 multiply by
6
Suman: Geetha = 4*6:5*6 = 24:30
Ratio of Ajay:Aman:Suman:Geetha = 8:15:24:30
Suman ratio = 24/77
Ratio and Proportion
60 kg of an alloy A is mixed with 100 kg of alloy
B. If alloy A has lead and tin in the ratio 3 : 2
and alloy B has tin and copper in the ratio 1 : 4,
then what would be the amount of tin in the new
alloy?
Ratio and Proportion
Solution :
The amount of tin in both alloys
= (60*2/5) + (100*1/5)
= 44 kg
Ratio and Proportion
In a class composed of x girls and y boys what
part of the class is composed of girls?
(Satyam Question)
Ratio and Proportion
Answer:
Ratio = x : (x+y) = x / x+y
Ratio and Proportion
In a factory ,the ratio of male workers to
female workers was 5:3. If the number of
female workers was less by 40.What was the
total number of workers in the factory.
Ratio and Proportion
Solution:
Let the number of males is 5x and
females is 3x and Total is 8x
5x – 3x = 40
x = 20
Total number of workers in the factory = 8x
=8*20 = 160
Ratio and Proportion
A Mixture contains milk and water in the ratio
5:1.On adding 5 liters of water, the ratio of
milk to water becomes 5:2. What is the
quantity of milk in the original mixture ?
Ratio and Proportion
Solution:
Let quantity of milk be 5x, and water be x then,
5x
= 5
x+5
2
Hence x=5
Quantity of milk = 5x5 = 25 liters.
Alligation or Mixture
•(Quantity of cheaper / Quantity of costlier)
(C.P. of costlier) – (Mean price)
= -------------------------------------(Mean price) – (C.P. of cheaper)
Alligation or Mixture
Cost of Cheaper
c
Cost of costlier
d
Cost of Mixture
m
d-m
m-c
(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)
Alligation or Mixture
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
:
3
The ratio is 7:3
The quantity of 2nd kind = 3/10 of 100kg
= 30kg
Alligation or Mixture
A certain type of mixture is prepared by
mixing brand A at Rs. 9/kg with brand B at Rs.
4/kg. If the mixture is worth Rs. 7/kg how
many kg of brand A are needed to make 40 kgs
of the mixture?
Satyam Question
Alligation or Mixture
Solution:
B
4
A
9
7
2
3
Brand A required = 3*40 / 5 = 24kg
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 he neither gains nor
loses. How much does each cow cost?
Alligation or Mixture
Solution:
-6%
7.5%
0
7.5
6
Ratio is 5 : 4
The cost of first cow
The cost of second cow
= 5*1350/9 =Rs. 750
= Rs. 600
Alligation or Mixture
How much water be added to 14 liters of milk
worth Rs. 5.4 per liter so that the value of the
mixture may be Rs. 4.20 per liter?
Alligation or Mixture
Solution: The cost of water is 0.
W
M
0
540
420
120
420
Ratio is 4 : 14
The amount of water added for 14 liters is 4 liters.
Alligation or Mixture
There are 65 students in a class, 39 rupees are
distributed among them so that each boy gets
80 paise and girl gets 30 paise. Find the number
of boys and girls in that class.
Alligation or Mixture
Solution: “Money per boy or girl” is considered.
Per student = 3900/65 = 60 paise.
Girls
Boys
30
80
60
20
30
Girls : Boys = 2 : 3
Number of boys = 39
Number of girls = 26
Chain Rule
Direct Proportion :
A
B
A
B
Indirect Proportion:
A
B
A
B
Chain Rule
A man completes 5/8 of a job in 10 days. At
this ratio how many more days will it take for
him to finish the job?
Chain Rule
Answer:
Jobs
5/8
3/8
Days
10
x
5x / 8 = (10*3) / 8 (Direct )
The number of days to complete = 6
Chain Rule
A certain number of men can finish a piece of
work in 100 days. If however there were 10
men less it will take 10 days more for the work
to be finished. How many men were there
originally?
(Satyam Question)
Chain Rule
Solution:
Men
x
(x-10)
Days
100
110
(indirect )
100x = 110(x – 10 )
The number of men present originally = 110
Chain Rule
15 men take 21 days of 8 hours each to do a
piece of work. How many days of 6 hours each
would it take for 21 women if 3 women do as
much work as 2 men?
(Satyam Question)
Chain Rule
Solution:
3 women = 2 men
Men Days Hours
15
21
8
14
x
6
x = 15*8
21
14*6S
x = 30
The number of days required is 30.
Chain Rule
6 cats kill 6 rats in 6 minutes. How many cats
will be needed to kill 100 rats in 50 minutes?
(Satyam question)
Chain Rule
Solution:
Cats
6
x
Rats
6
100
Minutes
6
50
x/6 = (100*6)/(6*50)
The number of cats needed would be 12.
Chain Rule
39 persons can repair a road in 12 days, working
5 hours a day. In how many days will 30
persons working 6 hours a day, complete the
same work?
Chain Rule
Solution : Men
days
hours
39
12
5
30
x
6
More hours less days ( inverse proportion )
Less men more days ( inverse proportion )
x
5
39
5 * 39 * 12
 ------- = ---- * ----  x = --------------12
6
30
6 * 30
 x = 13 days.
Partnership
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
A sum of money is divided among A, B, C such
that for each rupee A gets, B gets 65 paise, and C
gets 35 paise. If C’s share is Rs. 560, what is the
sum?
(TCS Question)
Partnership
Solution :
A
:
B
:
C
100
:
65
:
35
Total = 100 + 65 +35 = 200
C’s share = 560
(35*x)/200 = 560
X = 3200
The sum invested is Rs. 3200
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:
A and B = 3 : 2
Let Profit be x
x – 5%
x- 5x/100 = 95x/100
A’s share is = 3/5 * 95x/100 = 855
19x/100 =285
19x = 28500
x = 26500/19 = 1500
Total profit is Rs.1500
Partnership
A,B,C subscribe Rs. 50,000 for a business. A
subscribes Rs. 4000 more than B and B
subscribes Rs. 5000 more than C. Out of a total
profit of Rs. 35,000 how much does A receive?
(TCS Question)
Partnership
Answer :
Ratio A
B
C
9000 + x : 5000 + x : x
14000+3x = 50000
On solving we get x = 12000
Ratio A B
C
21 : 17 :12 Total = 50
A receives = (21* 35000)/50 = Rs. 14700
Partnership
A began a business with Rs. 450 and B jointed
with Rs. 300. When did B join if the profit at the
end of the year was divided between them in the
ratio 2 : 1?
(Caritor Question)
Partnership
Answer :
Ratio
A
:
B
450x12 : 300 (12 – x )
5400 / ( 300( 12 – x )) = 2 / 1
On solving x = 3
B invested money after 3 months
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 is 1: 1: 1
Time and Work
• 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
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
Pipes and Cisterns
• 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
Time and Work
10 men can complete a piece of work in 15 days
and 15 women can complete the same work in 12
days. If all the 10 men and 15 women work
together, in how many days will the work get
completed ?
Time and Work
Solution:
10 men = 15 days means 1day work = 1/15
15 men = 12 days means 1 day work = 1/12
10 men + 15 women = 1/15 + 1/12
= 4+5/60 = 9/60
= 3/20
= 6 2/3
The work will be completed in 6 2/3 days.
Time and Work
A and B can finish a piece of work in 30 days, B
and C in 40 days, while C and A in 60 days .In
how many days A, B and C together can do the
work ?
Time and Work
Solution:
A + B = 30 days = 1/30
B + C = 40 days = 1/40
C+ A = 60 days = 1/60
All work together
A+B+C+B+C+A = 1/30 +1/40 +1/60
2(A+B+C) = 1/30+1/40+1/60
= 4+3+2 /120 = 9/120
= 9/240 = 3/80
= 26 2/3
A, B and C can finish the work in 26 2/3 days
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. In how many
days alternatively work will be completed?
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
Work done by A & B and A & C = ( 8/150 + 7/120)
= 67/600
For 16 days , work done = 67x8 / 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 = 120 / 7 * 4 / 75 = 32/35
They will finish the work in 17 32/35 days
Time and Work
A can do a piece of work in 12 days. B is 60%
more efficient than A. Find the number of days
B takes to do the same piece of work.
Time and Work
Answer :
Work Days
100
12
160
x
On solving , x =( 100*12) / 160 = 7 ½ Days
B will take 7½ days.
Time and Work
Seven men can complete a work in 12 days.
They started the work and after 5 days, two
men left. In how many days will the work be
complete the remaining work.
Time and Work
Solution:
(7*12) men can complete the work in 1 day
1 man’s 1 day’s work = 1/84
7 men’s 5 day’s work = 1/12 *5 = 5/12
Remaining work = (1-5/12) = 7/12
5 men’s 1 day’s work =5*( 1/84 )= 5/84
5/84 work is done by them in 1 day
7/12 work is done by them in 84/5 *7/12 = 49 / 5
= 9 4/5 days
Time and Work (Pipes)
Pipe A can fill a cistern in 20 minutes and pipe B
in 30 minutes and pipe C can empty the same in
40 minutes. If all of them work together, find the
time taken to fill the tank.
(Satyam Question)
Time and Work (Pipes)
Answer:
The time taken is 17 1/7 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)
A cistern is provided by two taps A and B. A
can fill it in 20 minutes and B in 25 minutes.
Both taps are kept open for 5 minutes and then
the second is turned off. How many minutes
will be taken to fill the tank ?
Time and Work (Pipes)
Solution:
In one day (A +B) can do = 1/20 + 1/25 = 9/100
In 5 days (A +B ) can do = 5x9/100 = 9/20
Work left
= 1 – 9 /20 =11/20
A alone can do the remaining = ( 11/20) x 20
= 11
The cistern will be filled in 11 minutes.
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 by A in ( 10 – x min) =1
x (1/20 +1/40) + (10 – x) * 1/20 = 1
x (3/40) + (10 – x) /20 = 1
3x/40 + (10 –x) / 20 = 1
3x + 20 – 2x = 40
3x – 2x = 40 -20
x = 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 = 5/6
Remaining Part = 1- 5/6 = 1/6
Now, 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.
Area and Volume
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 = ½ * 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 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
If the radius of a circle is diminished by 10%,
what is the change in area in percentage?
Area and Volume
Solution:
x = - 10 , y = - 10
= x + y +xy/100 %
= -10 - 10 – (10*10/100) %
= - 19%
Area changed = 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
If the length of a rectangle is increased by 30%
decreased by 20%, then what percent of its area
would increase?
(TCS Question)
Area and Volume
Solution:
x = 30 y = - 20 (Decreased )
Using formula = x + y + (xy/100)%
The increase would be 4%
Area and Volume
The sides of the right triangular field containing
the right angle are x and x + 10 and its area is
5500 m2. Find the equation to determine its area.
(Caritor Question)
Area and Volume
Answer:
The equation is 1 / 2*x(x+10) = 5500
x2 + 10x – 11000 = 0
Area and Volume
The length and breadth of a rectangular plot are
in the ratio 7 : 5. If the length is reduced by 5 m
and breadth is increased by 2 m then the area is
reduced by 65 m2. Find the length and breadth of
the rectangular plot.
(Caritor Question)
Area and Volume
Answer:
Difference in areas = 65
(7x *5x) – (7x – 5) (5x + 2) = 65
x=5
The length of the rectangle = 7x = 7*5 = 35 m
The breadth of the rectangle = 5x = 5*5= 25 m
Probability
•
Probability:
P(E) = n( E) / n( S)
•
Addition theorem on probability:
n(AUB) = n(A) + n(B) - n(AB)
•
Mutually Exclusive:
P(AUB) = P(A) + P(B)
•
Independent Events:
P(AB) = P(A) * P(B)
Probability
There are 19 red balls and 1 black ball. Ten balls
are placed in one jar and remaining in one jar.
What is probability of getting black ball in right
jar ?
(Infosys -2008)
Probability
Answer:
Probability is 1/2.
Probability
Sam and Jessica are invited to a dance party. If
there are 7 men and 7 women in total at the
dance and 1 woman and 1 man are chosen to
lead the dance, what is the probability that Sam
and Jessica will not chosen to lead the dance ?
Probability
Solution:
Probability of selecting Sam and Jessica
= 7C1 * 7C1 = 49
n(s) = 49
n (є) = 1 ( selecting Sam and Jessica)
P (є) = n (є) / n(s)
= 1/49
Not selecting Sam and Jessica = 1- 1/49
The probability = 48/49
Probability
There are 5 distinct pairs of white socks and 5
pairs of black socks in a cupboard. In the dark,
how many socks do I have to pull out to ensure
that I have at least one correct pair of white
socks?
TCS Question
Probability
Answer:
12 times
Probability
In a simultaneous throw of two coins, what is
the probability of getting at least one head?
TCS Question
Probability
Answer :
Probability = 3/4
Probability
Three unbiased coins are tossed. What is the
probability of getting at least 2 heads?
Probability
Answer :
The probability is 4/8 = 1/2
Probability
A can solve 90% of the problems given in a book
and B can solve 70%. What is the probability
that at least one of them will solve a problem
selected at random from the book?
Probability
Answer:
A does not solve the problem = 1- 90/100 = 1/10
B does not solve the problem = 1- 70/100 = 3/10
Either A or B to solve = 1 – (1/10)*(3/10)
= 1- 3/100
= 97/100
Probability
Two unbiased Dice are thrown. Find the
probability that neither a doublet nor a total of
10 will appear ?
Probability
Solution:
A= Doublet (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
B =(6,4),(4,6),(5,5)
P(A)= 6 / 36 , P(B) = 3 / 36 , P(A n B) =1/36
Probability = 1 – P( Au B)
= 1 – ( P(A) + P(B) - P(AnB) )
= 7/9
Simple / Compound Interest
Simple Interest = PNR / 100
Total amount A = P + PNR / 100
When Interest is Compound annually:
Amount = P (1 + R / 100)n
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
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 }
Simple / Compound Interest
The simple interest on a certain sum is 16 / 25
of the sum. Find the rate percent and time if
both are numerically equal.
Sathyam Question
Simple / Compound Interest
Solution:
R = N and I = 16/25 P
R = I * 100 / P*N
R*N = 64 where R = N
R=8
The rate of interest is 8%
Simple / Compound Interest
A sum of S.I. at 13½% per annum amounts to
Rs. 2502.5 after 4 years. Find the sum.
Sathyam Question
Simple / Compound Interest
Answer:
A = P + PNR / 100
A = P (1 +NR/100)
The principal is Rs. 1625
Simple/compound interest
The difference between the compound and
simple interest on a certain sum for 2 years at the
rate of 8% per annum is Rs.80,What is the sum?
Wipro Question
Simple/Compound interest
Answer :
C.I – S.I = P(R/100)2
80 = P*(8*8/100*100)
The sum is Rs.12,500
Simple/Compound interest
If a sum of money compounded annually
amounts of thrice itself in 3 years, in how many
years will it become 9 times itself?
CTS Question
Simple/Compound interest
Solution:
A = 3P and find n when A = 9P
The number of years required = 6 years
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
Simple/Compound interest
What will be the C.I on Rs. 15625 for 2½ years
at 4% per annum?
Simple/Compound interest
Solution:
A = P(1+R/100)n
A = 15625 ((1+4/100)2 (1+4*1/2/100))
= 17238
CI = A –P = 17238 - 15625
Compound interest = Rs. 1613
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
The average weight of 5 persons sitting in
boat is 38 Kg .If the average weight of the
boat and the persons sitting in the boat is
52Kg,What is the weight of the boat?
Average
Soluton:
Average weight of 5 persons = 38 Kg
Total weight of the 5 persons =38*5=190
Total weight including the boat weight is
=52*6=312
Weight of the Boat = 312-190= 122 Kg
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
Total sum of 11 numbers = 660
6th number =90
(290 + 280)
(11*60)
(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
Diff= 310-270=40 years
Average
What is the average of even numbers from 1
to 81?
Average
Solution:
Average = (last even + 2) / 2
= (80 + 2 ) / 2
= 41
Average
The Average of 50 numbers is 38. If two
numbers namely 45 and 55 are discarded, What
is the average of the remaining numbers?
Average
• Solution:
Total of 50 numbers = 50*38=1900
Total of 48 numbers = 1900-(45+55)=1800
Average
=1800/ 48 = 37.5
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)!
Permutations and Combinations
• 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
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 also the person 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
There are 5 gentlemen and four ladies to dine at
a round table. In how many ways can they seat
themselves so that no two ladies are together?
Permutations and Combinations
Solution:
5! = 120 ways
Permutations and Combinations
In a chess board there are 9 vertical and 9
horizontal lines. Find the number of rectangles
formed in the chess board.
Permutations and Combinations
Solution:
9C2 * 9C2 = 1296
Permutations and Combinations
In how many ways can a cricket team of 11
players be selected out of 16 players , if one
particular player is to be excluded?
Permutations and Combinations
Solution:
If one particular player is to be excluded, then
selection is to be made of 11 players out of 15.
15C11= 15!/( 11!*4!)=1365 ways
Percentage
• By a certain Percent, we mean that many
hundredths.
• Thus, x Percent means x hundredths, written
as x%
Percentage
•If Length is increased by X% and Breadth is
decreased by Y% What is the Percentage
•Finding out of Hundred.
Increase or Decrease in Area of the rectangle?
Formula: X+Y+ XY/100 %
Decrease 20% means -20
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/10
2nd number = x – 40% of x = x – 40x/100 = 60x/ 100
= 6x/10
Suppose 2nd number = y% of 1st number
6x / 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
= 100
Spending:
Machinery + Raw material + staff = 35+40+10 = 85
Remaining = 100 – 85 = 15
15 = 60000
100 = ?
( By Chain rule)
= 4, 00,000
In this 4, 00,000 75% for machinery and raw material
= 4, 00,000* 75/100 = 3, 00,000
Rs. 3, 00,000
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
An empty fuel tank of a car was filled with A
type petrol. When the tank was half empty, it
was filled with B type petrol. Again when the
tank was half empty, it was filled with A type
petrol. When the tank was half – empty again,
it was filled with B type petrol. What is the
percentage of A type petrol at present in the
tank?
Percentage
Solution:
Let capacity of the tank be 100 liters. Then,
Initially: A type petrol = 100 liters
After 1st operation:
A = 100/2 = 50 liters, B = 50 liters
After 2nd operation:
A = 50 / 2+50 = 75 liters, B = 50/2 = 25 liters
After 3rd operation:
A = 75 / 2 = 37.5 liters, B = 25/2 +50 = 62.5 liters
Required Percentage = 37.5%
Percentage
Find the percentage increase in the area of a
rectangle whose length is increased by 20% and
breadth is increased by 10%
Percentage
Answer:
Percentage of Area Change=( X +Y+ XY/100)%
=20+10+20*10/100
=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:
Percentage = R*100%/(100-R)
= (40*100)/ (100-40)
=66 2/3%
Percentage
Fresh grapes contain 90% water by weight while
dried grapes contain 20% water by weight. What
is the weight of dry grapes available from 20 kg
of fresh grapes?
Percentage
Solution:
Fresh grapes contain 10% pulp
20 kg fresh grapes contain 2 kg pulp
Dry grapes contain 80% pulp.
2 kg pulp would contain = 2/0.8 =20/8 = 2.5kg
Hence 2.5 kg Dry grapes.
Percentage
If A is 20% of C and B is 25% of C then what
percentage is A of B?
Percentage
Answer:
Given A is 20% of C and B is 25% of C
A of B is = (20*100%)/25 = 80%
A is 80% of B
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
Boats and Streams
A boat is rowed down a river 40 km in 5 hours
and up a river 21 km in 7 hours. Find the speed
of the boat and the river.
Boats and Streams
Solution:
Downstream speed (a) = 40/5 = 8 km/h
Upstream speed
(b)= 21/7 = 3 km/h
Speed of the boat
Speed of the river
=1/2(a+b)= 5.5 km/h
=1/2(a-b) = 2.5 km/h
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 =6km/hr
Speed of the boat in still water = (5+6) / 2
= 5.5 km/hr
Boats and streams
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 = Speed of the boat + Speed of the stream
= 5 +3 =8
Up Stream = Speed of the boat – Speed of the stream
= 5-3 = 2
Let distance be x
Distance/Speed = Time
x/8 + x/2 = 7 ½
x/8 +4x/8 = 15/2
5x / 8 = 15/2
5x = 15/2 * 8
5x = 60
x =12
Boats and Streams
A man rows to place 48 km 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 4 km in x hours. Then,
Speed Downstream = 4 / x km/hr,
Speed Upstream = 3 / x km/hr
48/ (4 / x) + 48/(3 / x) = 14
x=½
Speed of Downstream = 8,Speed of upstream = 6
Rate of the stream = ½ (8-6) km/hr = 1 km/hr
Boats and Streams
A Swimmer can swim a certain distance in the
direction of current in 5 hr and return the same
distance in 7 hr. If the stream flows at the rate of
1 kmph, find the speed of the swimmer in still
water?
Boats and Streams
Answer:
Let the Speed of the Swimmer in still water be x
Distance covered in going = 5(x+1) km
Distance covered in returning = 7(x-1) km
5(x+1) = 7(x-1)
x=6
Speed of the swimmer in still water = 6 kmph.
Time and Distance
•Speed:- Distance covered per unit time is
called Speed.
Speed = Distance / Time
•Distance = Speed * Time
•Time = Distance / Speed
Time and Distance
• 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
If a man walks at the rate of 5 kmph he misses a
train by only 7 minutes. However if he walks at
the rate of 6 kmph he reaches the station 5
minutes before the arrival of the train. Find the
distance covered by him to reach the station.
Sathyam Question
Time and Distance
Solution:
x/5-x/6 = 12/60
( Difference 7+5=12)
Solving the equation,
x=6
The distance covered is 6 km.
Time and Distance
The J and K express from Delhi to Srinagar was
delayed by snowfall for 16 minutes and made up
for the delay on a section of 80 km travelling with
a speed 10 km per hour higher than its normal
speed. Find the original speed of the express?
Time and Distance
Answer:
The train saves 16 minutes by travelling faster over a
section of 80 km
80/x – 80/(x+10) = 16/60
By solving we get x = 50
Time and Distance
If the total distance of a journey is 120 km. If
one goes by 60 kmph and comes back at 40
kmph what is the average speed during the
journey?
Sathyam Question
Time and Distance
Solution:
Average Speed = 2xy / (x + y)
The average speed is 48 kmph.
Time and Distance
By walking at ¾ of his usual aped, 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
If a person walks at 14 km/hr instead of 10
km/hr, he would have walked 20 km more. Find
his actual distance traveled?
Time and Distance
Solution:
Let the distance be x km. Then,
x/10 = (x+20) / 14
14x = 10x +200
4x = 200
x = 50 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)
Two trains 200 m and 150 m long are running
on the parallel rails at the rate of 40 km/hr
and 45 km/hr. In how much time will they
cross each other if they are running in the
same direction?
Sathyam Question
Time and Distance (Trains)
Answer:
Relative speed = x – y = 45 – 40 = 5 km/h
=5*5/18= 25/18 m/s
For 25/18 m it takes 1 sec
For 350 m it takes = 18 * 350 /25 =252
Time taken = 252 seconds
Time and Distance (Trains)
There are 20 poles with a constant distance
between each pole. A train takes 24sec to
reach the 12 pole. How much time will it take
to reach the last pole ?
Time and Distance (Trains)
Solution:
To reach 11 poles it takes 24 sec
For 19 poles it will take x time
Poles
time
11
24
19
x
11x = 19 * 24
x = 19* 24 /11
x = 41.45 sec
It reaches the last pole in 41.45 sec
Profit and Loss
• 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
A man buys an article for Rs. 27.50 and sells it
for Rs. 28.60. Find his gain percentage.
Sathyam Question
Profit and Loss
Answer:
Gain 4%
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
Answer:
C.P = 4500 Profit = 8/100*4500 = 360
S.P = C.P + Profit = 4500 + 360 = 4860
He should sell it for Rs. 4860.
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:
C.P of two CDs =Rs.380
S.P of 1st CD = x – 22 x /100
S.P of 2nd CD = y + 12y /100
SP1 = SP2
x – 22x/100 = y – 12y/100
x = 56/39 y
x + y = 380
56/39y + y = 380
y = 156
x = 224
Cost of the two CDs are Rs. 224 and Rs.156
Profit and Loss
A tradesman fixed his selling price of goods at
30% above the cost price. He sells half the stock
at this price, one quarter of his stock at a
discount of 15% on the original selling price and
rest at a discount of 30% on the original selling
price. Find the Gain Percent altogether?
Profit and Loss
Solution:
Let us take C.P of goods = 100
The Marked price
= 130
=1/2*130 + 1/4*0.85*130 + 1/4 *0.7*130
=15.375%
Profit and Loss
Rajesh buys an article with 25% discount on its
Marked Price. He makes a profit of 10% by
selling it at Rs 660. Find the marked price?
Profit and Loss
Solution:
S.P = Rs.660
Profit = 10 %
C.P = 110 / 100*660 = Rs.600
Let x be M.P
75% of x = 600
x = 800
Marked Price = 800
Profit and Loss
A shopkeeper gains the cost of 8 meters of
thread by selling 40 meters of thread. Find
his gain percentage?
Profit and Loss
Answer:
Gain = 20%
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)
The resultant profit is 3.5%
Profit and Loss
Aditya purchases toffees at Rs. 10 per dozen and
sells them at Rs. 12 for every 10 toffees. Find the
gain or loss?
TCS Question
Profit and Loss
Answer:
Profit = 44
Profit and Loss
Anirudh bought 8 lemons for a rupee but sells
only 6 lemons for a rupee. Find his profit
percentage.
CTS Question
Profit and Loss
Answer:
Profit 33 1/3%
Profit and Loss
A radio when sold at a certain price gives a gain
of 20%. What will be the gain if it sold for thrice
the price?
Sathyam Question
Profit and Loss
Solution:
Let C.P be Rs. 100. Hence
S.P = Rs. 120
Gain = S.P – C.P = 360 – 120 = 260
Gain =Rs. 260
Profit and Loss
A grocer purchased 80 kg of rice at Rs. 13.50 per
kg and mixed it with 120 kg rice Rs. 16 per kg.
At what rate per kg should he sell the mixture to
gain 16%?
Sathyam Question
Profit and Loss
Solution:
C.P of 200 kg = (80*13.5 + 120*16) =3000
S.P of 200 kg = 116% of C.P = 3480
The selling price of 1 kg is Rs. 17.40
Profit and Loss
A candidate who gets 20% marks fails by 10
marks but another candidate who gets 42%
marks get 12% more than the passing marks.
Find the maximum marks.
Sathyam Question
Profit and Loss
Answer:
The maximum marks = 100
Odd days:
0 = Sunday
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
Calendar
Calendar
Month 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
Calendar
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
(3 + 6 + 6 + 2) -3
= ----------------------- = 14/7= 0, Sunday.
7
Calendar
On what dates of April 2005 did Sunday fall?
Calendar
Solution:
You should find for 1st April 2005 and then you find the Sundays
date.
A = 2005 / 7 = 3
B = 2005 / 4 = 501 / 7
=4
C=1/7=1
D=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
Calendar
Solution:
A = 1986 / 7 = 5
B = 1986 / 4 = 496/7 = 6
C=5/7=5
D=0
A + B + C + D -2
= ----------------------7
5 + 6 + 5 + 0-2
----------------------- = 14 / 7 = Sunday
7
Clocks
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
Find the angle between the minute hand and
hour hand of a clock when the time is 7:20.
Clocks
Solution:
 = 30h – (11m/2)
= 30 (7) – 11 20/2
= 210 – 110
= 100
Angle between 7: 20 is 100o
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?
Clocks
Solution:
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
Data Interpretation
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.
•
The following table gives the distribution of students according to professional
courses:
1.
__________________________________________________________________
Courses
Faculty
___________________________________
Commerce
Boys
Science
girls
Boys
Total
girls
___________________________________________________________
•
Part time management
•
C. A. only
•
•
30
10
50
10
100
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:
Data Interpretation
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
Answer:
Data Interpretation
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
Answer:
Data Interpretation
Percentage of boys over all students
= (450 x 100) / 500
= 90%
Data Sufficiency
Find given data is sufficient to solve
problem or not.
A.If statement I alone is sufficient
statement II alone is not sufficient
B.If statement II alone is sufficient
statement I alone is not sufficient
C.If both statements together are sufficient
neither of statement alone is sufficient.
D.If both together are not sufficient
the
but
but
but
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
Cubes
A 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?
Venn Diagram
Answer:
3 of them studied both Latin and Greek.