1] The average age of students of a class is 15.8 years.The average age of boys in the class is 16.4 years and that of the girls is is 15.4 years.The ratio of the numbers of boys to the girls in the class is
a]2:3 b]3:2 c]3: 5 d] 2:5

let the ratio be a:b
total number of boys  = 16.4*a
total number of girls  = 15.4*b
total number of students in class = 16.4a + 15.4b
the average age of all students is 15.8
the average age of boys is 15.8 a
the average age of girls is 15.8b
16.4a + 15.4b = 15.8a + 15.8b
0.6 a = 0.4 b
a/b = 0.4/0.6 = 2/3

1]A man bought a number of clips at 3 for a rupee and an equal number at 2 for a rupee.At what price per dozen should he sell them to make a profit of 20%
a]12 b]6 c]10 d] 2

Lets assume a dozen of clips were brought at 3 for a rupee and another dozen of clips were brought at 2 for a rupee
total CP for 1 dozen in which the cost is 3 for a rupee= 12/3 =4
total CP for 1 dozen in which the cost is 2 for a rupee= 12/2 = 6

total cp for 2 dozens = 6+4 = 10
percentage profit = 20% on Rs. 10 = 20/100 * 10 = Rs.2
After percentage profit the SP of 2 dozen is
Profit = SP - CP
SP = Profit + CP
SP = 2+ 10 = 12
SP of 1 dozen = 12/2 = 6

1] Find the average of first nine multiples of 3
a]12 b]12.5 c]15 d]18.5

average = 3[1+2+3+4+5+6+7+8+9] / 9 = (3*45) /9 = 15

2] Find the average of first twenty multiples of 7
a]72 b]42.5 c]15.6 d]73.5

3] Find the average of first five multiples of 3
a]12 b]12.5 c]15 d]9

4] Find the average of first five multiples of 4
a]12 b]12.5 c]15 d]18.5

5] Find the average of first ten multiples of 3
a]12 b]16.5 c]15 d]18.5

1]A and B together can do a work in 8 days.If A alone can do it in 12 days,then in how many days can B alone do it?

time for [A + B]'s days 1 work = 1/8
time for A's days 1 work =  1/12
time for B's days 1 work = ?

Time of [A+B]'s 1 day work = Time of A's 1 day work + Time of B's 1 day work
1/8 = 1/12+ B
B = 1/8 - 1/12 = 1/24

B's 1 days work = 24 days

1]A rectangle had width 'a' and length 'b'.If the width is decreased by 20% and the length is incresed by 10%.Then what is the area of the new rectangle in percentage compared to ab?
a]12 b]22 c]32 d]42
Let original length be 'b' and 'a'
original area = ab
new length = [(100+10)/100]  * b= (110/100)* b
new breadth = [(100-20)/100] * a = (80/100)*a
new area = (110/100) * b * (80/100)* a = (88/100) *ab
change area = ab - (88/100) * ab = 12 ab/100
% change  or decrease = 12ab/100 * (1/ab) * 100 = 12

2]If the length of a rectangle is increased by 30% while its breadth is decreased by 20%.What is happens to its area?
a]10% decrease b]10% increase c]4% decrease d]4% increase

3]If the length of a rectangle is increased by 70% while its breadth is decreased by 20%.Then percentage decrease in area will be
a]30%  b]32% c]36% d]34%

4]If the length of a rectangle is increased by 10% while its breadth is decreased by 10%.The area of new rectangle is
a]neither increased or decreased b]1% increase c]1% decrease d]none

5]If the length of a rectangle is doubled while its breadth is halved.Then percentage change in area will be
a]50 b]75 c]no change d]none

1] How many pieces each of length 4.5 m can be cut out from 225 m of wire?
a]50 b]51 c]52 d]53

number of pieces = 225/4.5 = 50

2] How many pieces each of length 3.5 m can be cut out from 15 m of cloth?
a]4.285 b]4.28 c]4.2 d]none

3] How many pieces each of length 85 cm can be cut out from 42.5 m of rod?
a]50 b]51 c]52 d]53

4] How many pieces each of length 3/7 m can be cut out from 2(4/7) m roll of ribbons?
a]6 b]5 c]2 d]3

5] How many scarfs each of length 3.5 sq.m long can be cut out from 15 m area of cloth?
a]25 b]21 c]22 d]23

6] A pile of metal sheets is 39.9 cm in height.If each sheet is 0.7 cm thick,how many sheets are there in the pile?
a]57 b]51 c]52 d]53

1]Bread contains 0.06 part of protein by weight.A patient requires 42 grams of protein per day.How many grams of bread are required by the patient?
a]700 b]600 c]500 d]400
number of grams = 42/ 0.06 = 700

1] A man has some hens and cows.If the number of heads be 48 and the number of feet equals 140.Then the number of hens will be
a]22 b]23 c]24 d]26

let the number of hens be x
let the number of cows be y

x+y= 48
2x + 4y = 140 = > x+2y = 70

x+y= 48
x+2y = 70
-------------
y = 22

x+22 = 48
x = 26

2] There are deer and peacocks ina  zoo.By counting the heads they are 80.The number of their legs is 200.How many peacocks are there?
a]60 b]63 c]64 d]66


2] There are lions and parrots in a  zoo.By counting the heads they are 15.The number of their legs is 50.How many lions are there?
a]10 b]13 c]14 d]16

1]A is thrice as good a workman as B and therefore it is able to finish a job in 60 days less than B.Working together they can do it in
a]20 days b]22 (1/2) days c]25 days d]30 days

ratio of time taken by A and B is 1:3
let days taken by A and B is x and 3x
difference = 2x
2x = 60
x = 30
Days taken by A and B is 30 and 90
total time taken if they work together = 1/30 + 1/90 = 22(1/2) days

2]A is 6 times as fast as B and takes 100 days less to complete a work than B.Find the total number of days taken by A and B to complete the work.
a]17 days b]17(1/2) days c]18 days d]19 days

1]The banker's discount on Rs.1800 at 12% per annum is equal to the true discount on Rs.1872 for the same time at the same rate.Find the time
a]1/3 years b]2/3 years c]4/3 years d]5/3 years

SI on Rs.1872
preset worth of Rs.1872 is 1800
SI on Rs1800 at 12%  = [1872-1800] = 72
SI = PTR/100
72 =( 1800* T * 12)/100
T = 1/3 years

1] A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 150 m long, find its speed.
a] 10 kmph b]20 kmph c]30 kmph d]40kmph

Sol. Relative speed = (150/9) m/sec =(150/ 9 x 18/5) kmph = 60 kmph.
:. Speed of goods train = (60 - 50) kmph = 10 kmph.

2] A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.
a] 12 kmph b]22 kmph c]42 kmph d]62 kmph

1] A train 140 m long is running at 60 kmph. In how much time will it pass a platform 260 m long?
a] 98 s b]24 s c]12 s d]14 s

Sol. Speed of the train = (60 x 5/18) m/sec =50/3m/sec.
       Distance covered in passing the platform = (140 + 260) m = 400 m
       :. Time taken = (400 x 3/50) see = 24 sec.

2] How long does a train 110 metres long running at the speed of 72 km/hour take to cross a bridge 132 metres in length.
a] 9.8 s b]12.1 s c]12.42 s d]14.3 s

1]Find the time taken by a train 180 m long, running at 72 kmph,in crossing an electric pole.
a] 9 s b]8 s c]7 s d]6 s

Sol. Speed of the train = (72 x 5/18) m/sec = 20 m/sec.
Distance moved in passing the pole = 180 m.
Required time taken = (180/20) sec = 9 sec.

2]In what time will a train 100 m long cross an electric pole,if its speed be 144 km/hr?
a] 2.5 s b]4.25 s c]5 s d]12.5 s

1]A dairy man pays Rs.6.4 per litre of milk.He adds water and sells the mixture at Rs.8 per litre,thereby making 37.5% profit.The proportion of water to milk recieved by the customers is
a]10:1 b]10:2 c]10:3 d]10:4

required ratio of the water and the milk in the mixture = x:y
CP of x litres of milk = Rs.6.4 x
SP of x litres of milk = Rs. 8(x+y)
profit % = 37.5
gaint % =( sp/cp *100) - 100
37.5 = [{8(x+y) } /64.x ]* 100 - 100
137.5/100 = (8x + 8y )/6.4x
880x - 800x = 800y
80x = 800 y
x/y = 10:1

1] A man sells an article at 5% profit.If he had brought it at 5% less and sold it for Re.1 less,he would have gained 10%.The cost price of the article is what?
a]100 b]200 c]300 d]400

let the cost price of the article be Rs. x
profit = 5%
selling price = x+(5% of x) = 21x/20
cost price = x - (5% of x) = 95/100
If sold for Rs.1 less
SP = [(21x/20) - 1].Then gain % is 10

{(New SP - SP)/ SP} * 100 = 10
{(2x-20)/19x} *100  = 10
2x-20 = 19x /10
20x - 200 = 19x
x = 200

1]A box of light bulbs contains 24 bulbs.A worker replaces 17 bulbs in the shipping department and 13 bulbs in the accounting department.How many boxes of bulbs did the worker use?
a]1 (1/4) b]2 (1/4) c]1 (3/4) d]1 (5/4)

24 bulbs in 1 box
replaced bulbs = 17+13 = 30
number of boxes of bulbs the worker used = 30/24 = 1 (1/4)

1]A single discount of 10%,20% and 40% is equal to a single discount of

a]50% b]56.8% c]60% d]70.28%
Net SP = [100-40%] of [100-20%] of [100-10%] of Rs.100
            = 60% of 80% of 90% of Rs.100
            = 60/100 * 80/100 * 90/100 * 100
            = 43.2
Required discount = 100 - 43.2 = 56.8

2]A single discount of 10%,20% and 5% is equal to a single discount of
a]31.6% b]31.5% c]31.4% d]31.3%

3]A single discount of 10%,12% and 15% is equal to a single discount of
a]50.2% b]67.32% c]60.8% d]70.28%

4]A single discount of 30% and 10% is equal to a single discount of
a]50% b]56% c]63% d]70%

5]A single discount of 20% and 30% is equal to a single discount of
a]50% b]56% c]60% d]70%

6]Two successive discounts of 20% and 15% are qual to a single discount of
a]50% b]68% c]60% d]70%

5]Two successive discounts of 20% and 10% are qual to a single discount of
a]50% b]56% c]60% d]72%

1]By selling a book for Rs.144 a shopkeeper loses 10%.At what price should he sell it inorder to gain 15%?
a]Rs.172 b]Rs.180 c]Rs.184 d]Rs.190

SP with CP as Rs.100 = 90 as there is 10 % loss
New SP with CP as Rs.100 = 115 as he needs 15% gain
SP with CP as Rs.100 : actual SP =New SP with CP as Rs.100 : New SP
 90 : 144 = 115 : x
90 x = 114 * 115
x = 184

2]By selling a watch for Rs.144 a shopkeeper loses 10%.At what price should he sell it inorder to gain 10%?
a]Rs.172 b]Rs.180 c]Rs.176 d]Rs.190

3] There would be 10% loss if rice is sold at Rs.5.50 per kg.At what price per kg should it be sold to earn a profit of 20%
a]Rs.7.20 b]Rs.7.02 c]Rs.7.33 d]Rs.6

4]A fruit seller sells mangoes at the rate of Rs.9/kg and thereby loses 20%.At what price per kg,he should have sold them to make a profit of 5%?
a]Rs.11.82 b]Rs.11.84 c]Rs.11.81 d]Rs.11.83

 5]When a plot is sold for Rs.18700 the owner loses 15%.At what price must the plot be sold in order to gain 15%?
a]Rs.2100 b]Rs.22500 c]Rs.25300 d]Rs.25800

1]How long does a train 110 mts long running at the speed of 132 km/hr take to cross a bridge 165 mts in length?
a]7.5 s b]7.4 s c]7.3 s d]7.2 s

solution
speed of train = d/t = (132*1000)/3600 m/s = 110/3 m/s
total distance covered = (110+165) = 275 m
time taken = [110/3] /275 = 7.5 s

2]How long does a train 110 mts long running at the speed of 72 km/hr take to cross a bridge 132 mts in length?
a]12.1 s b]12.4 s c]12.3 s d]12.2 s

3]A train is moving at a speed of 45 km/hr.If the lenght of the train is 360 mts.How long will it take to cross a bridge 140 mts in length?
a]40 s b]70 s c]30 s d] 20 s

4]How long does a train 110 mts long running at the speed of 36 km/hr take to cross a bridge 132 mts in length?
a]24.2 s b]24.4 s c]24.3 s d]24.2 s

1] A fruit seller loses 10% by selling mangoes at 26 for Rs.5.How many should he sell for a rupee to gain 17.4%?
a] 33.8 b]33.7 c]33.6 d]33.5

let the SP of 26 mangoes be Rs.X
If 10% money lost remaining is Rs.90

90:1 = 117 : x
90x = 117
x = 1.3
if for Rs. 1.3 mangoes sold is 26
For Re.1 mangoes sold = y

1.3 * y = 26 * 1
y = 33.8 Rs

2. By selling 45 lemons for Rs. 40, a man loses 20%. How many should he sell for Rs. 24 to gain 20% in the transaction?
a] 19 b] 18 c] 24 d] 22

1] A worker makes a basket in 2/3 of an hour.If he works for 7(1/2) hours,how many baskets can be made?
a]11(1/4) b]11(1/2) c]11(1/3) d]11(1/5)

[7(1/2) ]/ (2/3) = (15/2) / (2/3 ) = 45/4 = 11(1/4)

A company has 6435 bars of soap.If the company has sold 20% of its stock.How many bars of soap did it sell?
a]1287 b]1286 c]1285 d]1284

solution
20 % of 6435 = (20/100) * 6435 = 1287

1] A train can travel 20% faster than car.Both start from point A at the same time and reach point B 75 kms away from A at the same time.On the way however the train lost about 12.5 minutes while stopping at the stations.The speed of the car is

Solution
Let the speed of the car is x km/hr
speed of the train is x + 20% of x = (6x) / 5 km/hr



5 / (6x) * 75 + 25 / (2* 60) = 75/x
 375 / 6x + 5/24 = 75/x
375 / 6x - 75/x= - 5/24
-[80/6x] = - 5/24
  x = 60

1] A starts 3 minutes after B for a place 4.5 km distant. B on reaching his destination immediately returns and after walking a km meets A.If A can cover 1km in 18 mins what is B's speed?
a]6 km/hr b]7 km/hr c]8 km/hr d] 5 km/hr

A covers 3.5 km before meeting B.So time taken for it is =  [18 * 3.5] + 3 = 66 mins
B covers 5.5 km [11/2 km] before meeting A in 66 mins i.e., [66/60] hours = time taken

B's speed = [d/t] = [ 11/2 ]/ [11/10] = 5 km/hr

1] The sum of 6th and 15th elements of an AP is equal to the sum of space 7th,10th and 12th elements of the same progression.Which element of the series whould be equal to zero?
a]8th b]9th c] 11th d] 13th

solution
(a+5d) + (a+7d) = (a+6d) + (a+9d) + (a+11d)
2a+19d = 3a + 26d
a = -7d

T[8] = a+7d = -7d +7d = 0

1] A worker makes a basket in [2/3] of an hour.If he works for 7(1/2) hours,how many baskets will he make?
a]11(1/4) b]11(3/4) c]11(5/4) d] 11(6/4)

solution
{7(1/2)}/ (2/3) = (15/2) / (2/3) = 11(1/4)

1] Find the area of the traingle given the base and the corresponding altitude are 55mm and 34mm.
a] 931 b]932 c]933 d]935

area of the traingle = (1/2) * base * height = (1/2) *55*34 = 935 sq mm

2] Find the area of the triangle given the base and the corresponding altitude are 3.5mm and 2.4mm.
a] 4.1 b]4.2 c]4.3 d]4.5

3] Find the area of the triangle given the base and the corresponding altitude are 16.7mm and 10.9mm.
a] 91.015 b]91.014 c]91.014 d]91.013

a] The average of 30 results is 20 and the average of other 20 results is 30. What is the average of all the results?
a] 24 b] 48 c] 25 d] 50

Total numbers for in average of 30 results = 30*20 = 600
Total numbers for in average of 20 results = 20*30 = 600
Total numbers = 30+20

average of all the results = 600+600 /500 = 24

1] A mixture contains milk and water in the ratio 5:1.On adding 5 litres of water,the ratio of milk and water becomes 5:2.The quantity of milk in the mixture is
a]16 litres b]25 litres c]32.5 litres d]22.75 litres

5x/(x+5) = 5/2
10x = 5x+25
5x = 25
x = 5

quantity of milk = 5x = 25 litres

2] A mixture contains alcohol and water in the ratio 4:3.On adding 5 litres of water,the ratio of alcohol and water becomes 4:5.The quantity of alcohol in the mixture is
a]16 litres b]10 litres c]2.5 litres d]25 litres

1] A man can row downstream at 14 kmph and upstream at 9 kmph.Man's rate in still water is
a]5 kmph b]23 kmph c] 11.5 kmph d] none

rate in still water is = (1/2)(14+9) = 23/2 = 11.5 kmph

2] A man can row upstream at 7 kmph and downstream at 10 kmph.Man's rate in still water is
a]5 kmph b]23 kmph c] 8.5 kmph d] none

3] If a man can swim downstream at 6 kmph and upstream at 2 kmph.Man's rate in still water is
a]5 kmph b]23 kmph c] 4 kmph d] none

4] A man can row upstream at 11 kmph and upstream at 16 kmph.Man's rate in still water is
a]5 kmph b]23 kmph c] 13.5 kmph d] none

5] In one hour a boat goes 11km along the stream and 5 kmph against the stream.The speead of the boat in still water is
a]5 kmph b]23 kmph c] 11.5 kmph d] none

1]A man can row upstream at 11 kmph and downstream at 8 kmph.The speed of the stream is
a]3kmph b]9.5 kmph c]1.5 kmph d]6 kmph

speed of stream= (1/2) (11-8) = 3/2 = 1.5 kmph

2]A man can row upstream at 8 kmph and downstream at 13 kmph.The speed of the stream is
a]3kmph b]9.5 kmph c]2.5 kmph d]6 kmph

3]A man can row upstream at 6 kmph and downstream at 4 kmph.The speed of the stream is
a]1 kmph b]9.5 kmph c]1.5 kmph d]6 kmph

4]A man can row upstream at 14 kmph and downstream at 9 kmph.The rate of the current is
a]3kmph b]9.5 kmph c]2.5 kmph d]6 kmph

5]A man can row upstream at 11 kmph and downstream at 16 kmph.The speed of the stream is
a]3kmph b]9.5 kmph c]2.5 kmph d]6 kmph

6]A man can row upstream at 8 kmph and downstream at12 kmph.The rate of current is
a]2 kmph b]9.5 kmph c]1.5 kmph d]6 kmph

1] The fathers age was 5 times his son's age 5 years ago and will be 3 times son's age after 2 years. The ratio of their present ages is
a] 5:2 b]5:3 c]10:3 d]11:5

let son's age be 5 years ago be x and father's age 5x
After 5 years [i.e. present age] ,sons and father's age x+5 and 5x+5
After 2 years
3(x+5+2) = 5x+5+2
3x+15+6 = 5x+7
         14  =  2x
             x= 7 years
present age = x+5 = 12 and 5x+5 = 40
ratio = 40:12 = 20:6 = 10:3

2] The age of a father 10 years ago was thrice the son's age.10 years hence the fathers age will be twice that of his son.The ratio of their present ages is
a] 8:5 b]7:3 c]5:2 d]9:5

3] 1year ago a father was four times as old as his son.In 6 years timehis age exceeds twice his son's age by 9 years.Ratio present ages is
a] 13:4 b]12:5 c]11:3 d]9:2

4] 1year ago pramila was four times as old as her daughter.In 6 years time pramila's age exceeds her daughter's age by 9 years.Ratio present ages is
a] 13:4 b]12:5 c]11:3 d]9:2

1] The speed of the boat in still water is 10 kmph.If it can travel 24 km down stream and 14 km in the upstream in the equal time,indicate the speed of the flow of the stream.
a]38/100  b]100/38 c ]100+38  d] 100-38

solution:
let speed of flow of water = x
speed of boat downstream =  10+x
speed of boat upstream =  10-x
speed = distance travelled/time => d= st
distance downstream = 24 km
 distance travelled downstream= 24/(10+x)
distance upstream = 14
distance travelled upstream = 14/(10-x)
 since its given that distance covered upstream and down stream is in equal time
 24/(10+x) = 14/(10-x)
240-24x = 140x + 14x
x = 100/38

2] The speed of the boat in still water is 10 kmph.If it can travel 26 km down stream and 14 km in the upstream in th same time,indicate the speed of the flow of the stream.
a]8 b]3 c ]10 d] 2

1]If a:b = 5:9 and b:c = 4:7,find a:b:c
a] 20,30,60 b]20,36,60 c]20,30,63 d]20,36,63

a:b = 5:9
b:c = 4:7
b:c = [4* (9/4)] : [ 7*(9/4)]
      = 9: (63/4)
a:b:c = 5:9: (63/4) = 20: 36: 63

2]If a:b = 5:7 and b:c = 6:11,find a:b:c
a] 20,42,60 b]20,36,77 c]20,30,63 d]30,42,77

3]If a:b = 4:6 and b:c = 8:10,find a:b:c
a] 12,30,60 b]12,36,60 c]8,30,63 d]8,12,15

4]If a:b = 3:5 and b:c = 4:7,find a:b:c
a] 20,35,60 b]12,36,60 c]20,35,63 d]12,20,35

1] Three numbers are in the ratio 1:2:3 and their HCF is 12.The numbers are
a]12,24,36  b]12,26,36 c] 12,24,38 d] 14,24,36

let the required numbers be x,2x and 3x
the HCF is 12
so x= 12
the numbers are x = 12 , 2x = 24, 3x = 36

1] Two numbers are in the ratio of 15: 11.If their HCF is 13,find the number
a]190,140  b]191,141  c]192,143  d]195,143

let the required numbers be 15x and 11x
Then HCF is x
So x= 13
The numbers are 15*13 = 195 and 11*13 = 143

2] Two whole numbers are in the ratio of 15: 11.If their HCF is 7,find the number
a]100,77  b]105,141  c]105,77  d]195,143

2] A boat can be rowed 9 km upstream or 18 km downstream in a period of 3 hours.What is the speed of the boat in still water.
a] 2.5 kmph  b] 4.5 kmph  c] 6 kmph d] 8 kmph

solution
rate downstream = (9/3) = 3 kmph
rate upstream = (18/3) = 6 kmph
speed in still water = (1/2) (3+6) = 4.5 kmph

1] A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream,it takes 4 hours.What is the speed of the boat in still water.
a] 2  b] 4  c] 6 d] 8

solution
rate downstream = (16/2) = 8 kmph
rate upstream = (16/4) = 4 kmph
speed in still water is (1/2) (8+4) = 6 kmph

1] A man rows upstream 16 km and downstream 28 km taking 5 hours each time. the velocity of the current is
a] 2.4 km/hr  b] 1.2 km/hr  c] 3.6 km/hr  d] 1.8 km/hr

solution
 speed upstream = (16/5) = 3.2 km/hr
 speed downstream = (28/5) = 5.6 km/hr
velocity of current = (1/2)(3.2~ 5.6) = 1.2 km/hr


2] A man rows upstream 30 km and downstream 18 km taking 5 hours each time. What is the velocity of the current?
a] 2.4 km/hr  b] 1.2 km/hr  c] 3.6 km/hr  d] 1.8 km/hr

3] A man rows upstream 12 km and downstream 37 km taking 5 hours each time. What is the velocity of the current?
a] 4.5 km/hr  b] 2.5 km/hr  c] 3 km/hr  d] 4 km/hr

1] The sum of two numbers is 25 and their difference is 13.Find their product
a]111  b]112  c]113  d]114

let the numbers be x and y
x + y = 25
x - y = 13
-------------
2x  = 38
x = 19

x - y = 13
19 - y = 13
y = 6

product of two numbers x * y = 19 * 6 = 114

2] the difference between two numbers is 114 and their sum is 20.what is their product?
a]51  b]52  c]53  d]54

3] If the sum of two numbers is 97 and their difference is 37.Their product is
a]2011  b]1120  c]1013  d]2010

1] The simple interest on a certain sum of money at 5% per annum for 3 years is Rs.300.Find the sum.
a]2000  b]1000  c]1500  d]2500

SI = PTR/100
300 = (P*3*5)/100
P = 2000

2] The simple interest on a certain sum of money at 4% per annum for 3 years is Rs.48.Find the sum.
a]400 b]100  c]500  d]200
3] The simple interest on a certain sum of money at 4% per annum for 5 years is Rs.800.Find the sum.
a]4000  b]1000  c]1500  d]2500
4] The simple interest on a certain sum of money at 3(3/4)% per annum for 2(1/3) years is Rs.210.Find the sum.
a]2400  b]1000  c]1500  d]2500

1]Subtracting 40% of a number from the number,we get the result as 30.The number is
a]50  b]100  c]150  d]200

let the number be x
x - 40% of x = 30
[60/100] x    = 30
                  x = 50

1]Aniket started a business investing Rs.16000.Sanket joined him after 4 months by investing Rs.24000.What would be Sanket's share in the profit of Rs.20000 at the end of the year.
a] 1:1  b]1:2  c]1:3  d]1:4

solution
 16000*12 : 24000*8
16*12        :  8*24
1                :     1

These problems feature two or more fractions

1] Two third of one fourth of a number is 30.The number is
a] 180 b]108  c]144  d]none

solution
2/3 of 1/4 of x = 30
                     x = 180

2] If one fourth of two fifths of a number is 36.What is the number ?
a] 180 b]190 c]250  d]360

3] Two third of three fifth of a number is 14.What is the number?
a] 180 b]108  c]35 d]none

4 ]one fourth of one fourth of a number is 30.The number is
a] 180 b]108  c]144  d]576

5] If one third of one fourth of a number is 12.What is the number ?
a] 180 b]190 c]250  d]144

6] If one fourth of two fifths of a number is 36.What is the number ?
a] 180 b]190 c]250  d]360

7] If three fourth of one fifth of a number is 60.What is the number ?
a] 180 b]190 c]250  d]400

8] If one third of one fourth of a number is 12.What is the number ?
a] 180 b]190 c]250  d]144

9] If three fourth of one fifth of two third a number is 30.What is the number ?
a] 180 b]190 c]250  d]300

In an election 4000 votes were cast.One of the candidates got 40% votes.He was defeated by how m any votes?
a]2400  b]1600  c]800  d]600

Solution:
candidate who lost got 40% of 4000= 40/100 * 4000 = 1600
difference of votes = 2400-1600 = 800 votes

If the price of the kerosene was raised by 10%,find by how much percent a householder must reduce his consumption of kerosene so that not to increase his expenditure?

a]28%  b]20%  c]10.08%  d]10%

Raised price of kerosene is 110/100 of the original price
hence consumption should become 100/110 of the original consumption so as to keep the expenditure same
reduction = [1-100/10] of original consumption = 10/100 of original consumption= [10/100] * 100 = 9.09%

1]A towel was 50 cm broad and 100 cm long.When bleached,it was found to have lost 20 percent of its length and 10 percent breadth.The percentage decrease in area is
a] 28 %  b] 20%  c]  10.08%  d] 10%


original area = 100* 50
reduced length = 80% of 100 = 80
reduced breadth  = 90% of 50= 45
reduced area = 80*45
reduction in area = 500 - 3600 = 1400
% reduction = [400/5000] 100 = 28 %

 In an election between two candidates, A and B, A secured 56% of the votes and won by 48000 votes. Find the total number of votes polled if 20% of the votes were declared invalid.
a. 500000        b. 400000                    c. 600000        d. None

Valid Votes:
A got 56%  =>  B got 44%
Difference = 12% = 48000
So, 100% = 400000. These are valid votes.
But valid votes are only 80% of total votes.
So, 80% of total votes  = 400000  =>  total votes = 500000

A reduction of 10% in price of sugar enables a housewife to buy 5 kg more for Rs. 300/-. Find the reduced price per kg of sugar.
a. 5/-                b. 4.5/-                         c. 6/-                d. None

Total money = Rs. 300.
Saving of the lady = 10% of 300 = 30/-
With 30/- she bought 5 kg sugar => each kg costs Rs. 6/-

 From a 20lt solution of alt and water with 20% salt, 2lt of water is evaporated. Find the new % concentration of salt.
a. 20%             b. 23%                         c. 25%             d. None


In 20lt, salt = 20% => 4 lt.
New volume = 18 lt (2 lt evaporated)
So, new % = 4/18 X 100 = 22.22%

 In a list of weights of candidates appearing for police selections, the weight of A is marked as 58 kg instead of 46.4 kg. Find the percentage of correction required.
a. 30                b. 20                            c. 24                d. None



% correction = (58-46.4)/58 X 100 = 20%

A person spends 20% of his income on rent, 20% of the rest on food, 10% of the remaining on clothes and 10% on groceries. If he is left with Rs. 9520/- find his income.
a. 10000/-                    b. 15000/-                    c. 20000/-        d. None


Three successive decreases of 20%, 20% and 10% => 0.8 X 0.8 X 0.9 = 0.576
Again 10% decrease => 0.576 – 0.1 = 0.476.
So, 0.476 x = 9520 => x = 20000.

 A shopkeeper offers three successive discounts of 10%, 20% and 30% to a customer. If the actual price of the item is Rs. 10000, find the price the custome has to pay to the shopkeeper.
a. 5040/-                      b. 4000/-                      c. 6000/-          d. None
Total discount = 0.9 X 0.8 X 0.7 = 0.504 of actual price.
So, price = 0.504 X 10000 = 5040.

 If  10lt solution of water and alcohol containing 10% alcohol is to be made 20% alcohol solution, find the volume of alcohol to be added.
a. 1 lt                            b. 1.25 lt                      c. 1.5 lt             d. 2 lt


In 10 lt, alcohol is 10% = 1 lt.
Let x lt alcohol is added.
So, (1+x)/(10+x) = 20% = 1/5  =>  x = 1.25 lt.

A is twice B and B is 200% more than C. By what percent is A more than C?
a. 400                          b. 600                          c. 500              d. 200

A = 2B and B = 3C (ince 200% more)
A = 6C  =>  500 % more.

 In an examination, a student secures 40% and fails by 10 marks. If he scored 50%, he would pass by 15 marks. Find the minimum marks required to pass the exam.
a. 250                          b. 100                          c. 110              d. 125

50% of max marks – 40% of max marks = 25
 max marks = 250
Pass marks = 40% of max + 10 => 100 + 10 = 110.

If A is 20% taller than B, by what percent is B shorter than A?
a. 20%                         b. 25%                         c. 16.66%        d. None

A = 1.2 B
=> B = A/1.2
 => 0.8333A => 16.66%.
(OR) Decrease from 1.2 to 1 => 16.66%.

 

The population of a town increases at a rate of 10% for every year. If the present population is 12100, find the population two years ago.
a. 11000                      b. 9800                        c. 10000          d. 10120


1.1 X 1.1 X x = 12100 => x = 10000.

 A solution of salt and water contains 15% salt. If 30 lt water is evaporated from the solution the concentration becomes 20% salt. Find the original volume of the liquid before water evaporated.
a. 100 lt                        b. 120 lt                       c. 200 lt            d. None


let 'x' be the quantity of solution of salt and water
Salt content before evaporation of water= 15% of x = 0.15x (x = volume of solution)
Salt content after evaporation of water = 20% of [x-30] = 1/5  (since 30 lt evaporated)
equating both of them

(15/100) * x = (20/100) * (x-30)
3x/20 = 1/5 * (x - 30)
3x/20 - 1/5x = - 30/5
15x- 20x = - (6*20*5)
- 5x = -(6*20*5)
x = 120 litres

 If 240 lt of oil is poured into a tank, it is still 20% empty. How much more oil is to be poured to fill the tank?
a. 300 lt                        b. 60 lt                         c. 120 lt            d. None


22.20% empty
 => 80 % full = 240 lt
 => 20% = 60 lt

Finding increase in percentage difference by comparing two different values of percentages

A and B were hired for the same salary. A got two 40% hikes whereas B got a 90% hike. What is the percentage difference in the hikes thay got?
a. 16%                         b. 6%                           c. 10%             d. 8%

Solution
A => 1.4 X 1.4 = 1.96
B => 1.9     
1.96 -1.9=0.06
=> 6% difference.

Finding percentage increase,Given the change is the quantity/number over a span of time

The population of a town doubled every 5 years from 1960 to 1975. What is the percentage increase in population in this period?
a. 800                          b. 400              c. 700              d. 600

Solution :From 1960 to 1975, in 15 years population doubled every 5 yrs => three times
So, 2 X 2 X 2 = 8 times => 700% more.

In a test of 80 questions, Jyothsna answered 75% of the first 60 questions correctly. What % of the remaining questions she has to answer correctly so that she can secure an overall percentage of 80 in the test?
a. 80%                         b. 90%             c. 85%             D. 95%                                              
Solution
[(75% X 60) + (x% X 20)] / 80 = 80%  =>  x = 95.   (since required is 80%)
(OR)  60 out of 80 is 3/4. So,  (3/4 X 75)  +  (1/4 X x)  = 80  =>  x =95.

Given the reminder of a number when it is divided by a certain divisor.Finding the reminder if the same number is divided by another divisor

A number when divided by 32 gives a the reminder 29.This number when divided by 8 will leave the reminder
a] 3 b] 5 c] 7 d] 29                                                                                                                     Ans 5

let the number be x,
When it is divided by 32,let the quotient be k and reminder 29
32) x (k
      ---
      29

therefore x = 32k  + 29
                  = [(8 * 4 * k) + (8*6) - 5]
5 is the reminder

Finding percentage increase in area of a circle when the radius is increased by a certain percent

1] If radius of a circle is increased by 100%,then the area is increased by
a] 100%  b] 200%  c]300%  d]400%                                                                                  

Solution
let the original radius be R cm

New radius = {(100+100)/100}R cm = (200/100) R = 2R cm

original area = pi * R^2  cm^2

New area =  pi * (2R)^2  cm^2

change in area = pi *4* R^2 -  pi  * R^2 = 3 * pi * R^2

% increase in area =  3 * pi * R^2 * [1/ pi  * R^2 ] * 100 =300 %                       
                          
Practice questions

2] If radius of a circle is increased by 6%,then the area is increased by
a] 12.36%  b] 12.35%  c]12.34%  d]12.33%         

*3] If radius of a circle is doubled,how many times of the area of original circle will it become now?
a] 100%  b] 200%  c]300%  d]400%       
 
4] If radius of a circle is increased by 1%,what is the increased percent in the area 
a] 1%  b] 2%  c]1.1%  d]2.01%         

    

Solving for unknown numbers

1]Three fourth of a number is 60 more than its one third.The number is
a] 84  b]108  c]144  d]none

Solution
Let the number be 'x'
(3/4) x = 60+(1/3) x
(3/4) x - (1/3) x = 60
(9x-4x)/12 = 60
5x = 12*60
x = 144



2]If one seventh of a number exceeds its eleventh part by 100.The number is
a] 770  b]1100 c]1825  d]1925


3]A number exceeds its four seventh by 18. What is the number ?
a]42   b]24  c]144  d]none


4]A number exceeds its one third by 50. What is the number?
a] 70  b]75  c]144  d]none

5]A number exceeds its three fifth by 14. What is the number?
a] 70  b]35c]144  d]45

  Friends,Math-o-Phobics and Apti enthusiasts..this is a blog dedicated to Mathematical Aptitude.
I have always been weak in Mathematics or should i say i suffered from matho-phobia that completely ruined my otherwise excellent acedemic records starting from my class 7th Karnataka State Public Exams upto my grad exams.Whatever explainations i may give,at the end of the day am just weak in MATH,thats all!
  Being a capricorn,the perfection seeker i am,entire life it gnawed me that i sucked at math! No matter how much i tried i couldnt get out of that phobia and nor did my folks helped me in getting rid of it.It resulted in just pass marks in the plus two and ruined my entire life coz i couldnt go according to my nerdy plans of getting into the course i wanted and the college i wanted! As a result,i hated my college life,my course and ended up in a career i didnt want.Life became an entire mess.i blame everything on my MATH Result even today and will keep doing it till i die.I had math in Grad too which i made sure to pass with good marks out of fear of not getting graduation! That's a different story altogether.
  Many things changed but the yearning to master the math always remained.And hence i kept it as one of the MANY GOALS of my life[capricorns suck i know!].I was hell bound to learn it and master it,no matter how long it took.My perseverance for 8 years has paid and finally am out of Math-o-phobia coz today i am fairly good at aptitude as i easily crack it in the entrance exams and all.
  This blog is for those people who feel mathematical aptitude is an iron nut.Its just a concoction of questions with solutions and practice questions.Solve them and feel confident for entrance exams as there is no escaping this section in any entrance exam at any level.Its a necessary evil.Why not make it a friend by giving some time to it once in a while?
Moral of the story=Most of us suffer from Math-o-phobia,consciously subconsciously.Elders need to pay attention towards it.Given proper guidance and practice time,we can come out of it..!

Thanks for stopping by
Happy math-ing!
Komal :)