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Average Formulas, Average Tricks:-
Suppose there are N numbers, then their average is their sum divided by that is,
Average=(sum / N)
Weighted Average: The average between two sets of numbers is closer to the set with more numbers.
if 3 batsmen scored 25 runs and 2 batsmen scored 35 runs the average of the team won’t be 30. Rather it will be
= (25×3 + 35×2 )/5 = 29 . This is nearer to 25 since more batsmen scored 25 runs.
Average =total of data/No. of data
If the value of each item is increase by x, then the average of the group will also increase by x.
If the value of each item is decreased by y, then the average of the group of items will also decrease by y.
If the value of each item is multiplied by the same value m, then the average of the group or items will also get multiplied by m.
If the value of each item is multiplied by the same value n, then the average of the group or items will also get divided by n.
If we know only the average of the two groups individually, we cannot find out the average of the combined group of items.
Average of x natural no’s = (x+1)/2
Average of even No’s = (x+1)
Average of odd No’s = x
Change in the value of a Quantity and its effect on the Average
When one/more than one quantity are removed but replaced with same no. of quantities of different value,
Change in the no. of quantities and its effect on Average
+ = if quantities are Added, – = if quantities are removed
Average Solved Example, Average Short Cuts,Tricks
Ex.1: Find the average of all prime numbers between 30 and 50?
Sol: there are five prime numbers between 30 and 50.
They are 31,37,41,43 and 47.
Therefore the required average=(31+37+41+43+47)/5->199/5-> 39.8.
Ex.2. find the average of first 40 natural numbers?
Sol: sum of first n natural numbers=n(n+1)/2;
So,sum of 40 natural numbers=(40*41)/2 =820.
Therefore the required average=(820/40) =20.5.
Ex.3. find the average of first 20 multiples of 7?
Sol: Required average =7(1+2+3+…….+20)/20 =(7*20*21)/(20*2) =(147/2)=73.5.
Ex.4. the average of four consecutive even numbers is 27. find the largest of these numbers?
Sol: let the numbers be x,x+2,x+4 andx+6. then,
(x+(x+2)+(x+4)+(x+6))/4) = 27
=(4x+12)/4 = 27
=x+3=27 x=24.
Therefore the largest number=(x+6)=24+6=30.
Ex.5. there are two sections A and B of a class consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class?
Sol: total weight of(36+44) students=(36*40+44*35)kg =2980kg.
Therefore weight of the total class=(2980/80)kg =37.25kg.
Ex:6.nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them?
Sol: Let the average expenditure of all nine be Rs.x
Then 12*8+(x+8)=9x or 8x=104 or x=13.
Total money spent = 9x=Rs.(9*13)=Rs.117.
Ex.7: 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.
Sol: Let the third number be x.
Then second number = 3x.
First number=3x/2.
Therefore x+3x+(3x/2)=(44*3) or x=24
So largest number= 2nd number=3x=72.
Ex.8:The average of 25 result is 18.The average of 1st 12 of them is 14 & that of last 12 is 17.Find the 13th result.
Sol: Clearly 13th result=(sum of 25 results)-(sum of 24 results)
=(18*25)-(14*12)+(17*12)
=450-(168+204)
=450-372
=78.
Ex.9:The Average of 11 results is 16, if the average of the 1st 6 results is 58 & that of the last 63. Find the 6th result.
Sol: 6th result = (58*6+63*6-60*11)=66
Ex.10:The average waight of A,B,C is 45 Kg. The avg wgt of A & B be 40Kg & that of B,C be 43Kg. Find the wgt of B.
Sol. Let A,B,c represent their individual wgts.
Then,
A+B+C=(45*3)Kg=135Kg
A+B=(40*2)Kg=80Kg & B+C=(43*2)Kg=86Kg
B=(A+B)+(B+C)-(A+B+C)
=(80+86-135)Kg
=31Kg.
Ex. 11. The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by 3 months. Find the age of the teacher.
Sol. Total age of 39 persons = (39 x 15) years
= 585 years.
Average age of 40 persons= 15 yrs 3 months
= 61/4 years.
Total age of 40 persons = (_(61/4 )x 40) years= 610 years.
:. Age of the teacher = (610 – 585) years=25 years.
Ex. 12. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new man.
Sol. Total weight increased =(1.8 x 10) kg =18 kg.
:. Weight of the new man =(53 + 18) kg =71 kg.
Ex. 13. There were 35 students in a hostel. Due to the admission of 7 new students, ;he expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs 1. Wbat was the original expenditure of the mess?
Sol. Let the original average expenditure be Rs. x. Then,
42 (x – 1) – 35x=42 = 7x= 84 = x =12.
Original expenditure = Rs. (35 x 12) =Rs. 420. .
14. A batsman makes a score of 87 runs in the 17th inning and thus increases his avg by 3. Find his average after 17 th inning.
Sol. Let the average after 17th inning = x.
Then, average after 16th inning = (x – 3).
:. 16 (x – 3) + 87 = 17x or x = (87 – 48) = 39.
Ex.15. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km perhour. Find the average speed of the train during the whole journey.
Sol. Required average speed = ((2xy)/(x+y)) km / hr
=(2 x 84 x 56)/(84+56)km/hr
= (2*84*56)/140 km/hr
=67.2 km/hr.
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Question 1 of 31
1. Question
One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
Correct
Let original length = x metres and original breadth = y metres.
Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65y=>New Area =65x∗65y=>New Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100=44%Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65y=>New Area =65x∗65y=>New Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100=44%
Incorrect
Let original length = x metres and original breadth = y metres.
Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65y=>New Area =65x∗65y=>New Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100=44%Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65y=>New Area =65x∗65y=>New Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100=44%
Question 2 of 31
2. Question
The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field ?
Correct
Explanation:
Let breadth =x metres.
Then length =(115x/100)metres.
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?
Correct
Explanation:
We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet
Area to be fenced = 2B + L = 2*34 + 20
= 88 feet
Incorrect
Explanation:
We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet
Area to be fenced = 2B + L = 2*34 + 20
= 88 feet
Question 4 of 31
4. Question
The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .
Correct
Explanation:
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm
Area of third Square = 10*10 – 8*8
= 36 cm
So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm
Incorrect
Explanation:
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm
Area of third Square = 10*10 – 8*8
= 36 cm
So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm
Question 5 of 31
5. Question
The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.
Correct
Explanation:
Let the diagonals of the squares be 2x and 5x.
Then ratio of their areas will be
Area of square=12∗Diagonal212∗2×2:12∗5x24x2:25×2=4:25Area of square=12∗Diagonal212∗2×2:12∗5x24x2:25×2=4:25
Incorrect
Explanation:
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm
Area of third Square = 10*10 – 8*8
= 36 cm
So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm
Question 6 of 31
6. Question
A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m
Question 7 of 31
7. Question
A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is
Correct
Explanation:
Number of bricks =Courtyard area1 brick area=(2500×160020×10)=20000Number of bricks =Courtyard area1 brick area=(2500×160020×10)=20000
Incorrect
Explanation:
Number of bricks =Courtyard area1 brick area=(2500×160020×10)=20000Number of bricks =Courtyard area1 brick area=(2500×160020×10)=20000
Question 8 of 31
8. Question
The length of a rectangle is three times of its width. If the length of the diagonal is
810−−√810 , then find the perimeter of the rectangle.
Correct
Explanation:
Let Breadth = x cm,
then, Length = 3x cm
What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?
Correct
Explanation:
In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344
So required cost will be = 344 * 10 = 3440
Incorrect
Explanation:
In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344
So required cost will be = 344 * 10 = 3440
Question 10 of 31
10. Question
A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is ?
Correct
Explanation:
Let original length = x
and original width = y
Decrease in area will be
The perimeters of 5 squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the area of these square is:
Correct
Clearly first we need to find the areas of the given squares, for that we need its side.
Side of sqaure = Perimeter/4
So sides are,
(244),(324),(404),(764),(804)=6,8,10,19,20Area of new square will be =[62+82+102+192+202]=36+64+100+361+400=961cm2Side of new Sqaure =961−−−√=31cmRequired perimeter =(4×31)=124cm(244),(324),(404),(764),(804)=6,8,10,19,20Area of new square will be =[62+82+102+192+202]=36+64+100+361+400=961cm2Side of new Sqaure =961=31cmRequired perimeter =(4×31)=124cm
Incorrect
Clearly first we need to find the areas of the given squares, for that we need its side.
Side of sqaure = Perimeter/4
So sides are,
(244),(324),(404),(764),(804)=6,8,10,19,20Area of new square will be =[62+82+102+192+202]=36+64+100+361+400=961cm2Side of new Sqaure =961−−−√=31cmRequired perimeter =(4×31)=124cm(244),(324),(404),(764),(804)=6,8,10,19,20Area of new square will be =[62+82+102+192+202]=36+64+100+361+400=961cm2Side of new Sqaure =961=31cmRequired perimeter =(4×31)=124cm
Question 12 of 31
12. Question
50 square stone slabs of equal size were needed to cover a floor area of 72 sq.m. Find the length of each stone slab.
Correct
Explanation:
Area of each slab =
7250m2=1.44m2Length of each slab =1.44−−−−√=1.2m=120cm7250m2=1.44m2Length of each slab =1.44=1.2m=120cm
Incorrect
Explanation:
Area of each slab =
7250m2=1.44m2Length of each slab =1.44−−−−√=1.2m=120cm7250m2=1.44m2Length of each slab =1.44=1.2m=120cm
Question 13 of 31
13. Question
What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?
Correct
In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,
Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm
Required number of tiles =
Area of floorArea of tile=(1517×90241×41)=814Area of floorArea of tile=(1517×90241×41)=814
Incorrect
In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,
Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm
Required number of tiles =
Area of floorArea of tile=(1517×90241×41)=814Area of floorArea of tile=(1517×90241×41)=814
Question 14 of 31
14. Question
The area of a square is 69696 cm square. What will be its diagonal ?
Correct
If area is given then we can easily find side of a square as,
The difference of the areas of two squares drawn on two line segments in 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2 cm.
Correct
Explanation:
Let the lengths of the line segments be x and x+2 cm
then,
The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :
Correct
Explanation:
Area of triangle, A1 = 12∗base∗height=12∗15∗12=90cm2Area of second triangle =2∗A1=180cm212∗20∗height=180=>height=18cmArea of triangle, A1 = 12∗base∗height=12∗15∗12=90cm2Area of second triangle =2∗A1=180cm212∗20∗height=180=>height=18cm
Incorrect
Explanation:
Area of triangle, A1 = 12∗base∗height=12∗15∗12=90cm2Area of second triangle =2∗A1=180cm212∗20∗height=180=>height=18cmArea of triangle, A1 = 12∗base∗height=12∗15∗12=90cm2Area of second triangle =2∗A1=180cm212∗20∗height=180=>height=18cm
Question 18 of 31
18. Question
The sides of a triangle are in the ratio of
12:13:1412:13:14
If the perimeter is 52 cm, then find the length of the smallest side.
Correct
Ratio of sides =12:13:14=6:4:3Perimeter=52cmSo sides are =(52∗613)cm,(52∗413)cm,(52∗313)cmRatio of sides =12:13:14=6:4:3Perimeter=52cmSo sides are =(52∗613)cm,(52∗413)cm,(52∗313)cm
a = 24 cm, b = 16 cm and c = 12 cm
Length of the smallest side = 12 cm
Incorrect
Ratio of sides =12:13:14=6:4:3Perimeter=52cmSo sides are =(52∗613)cm,(52∗413)cm,(52∗313)cmRatio of sides =12:13:14=6:4:3Perimeter=52cmSo sides are =(52∗613)cm,(52∗413)cm,(52∗313)cm
a = 24 cm, b = 16 cm and c = 12 cm
Length of the smallest side = 12 cm
Question 19 of 31
19. Question
The height of an equilateral triangle is 10 cm. find its area.
If the area of a square with the side a is equal to the area of a triangle with base a, then the altitude of the triangle is.
Correct
Explanation:
We know area of square =a2Area of triangle =12∗a∗h=>12∗a∗h=a2=>h=2aWe know area of square =a2Area of triangle =12∗a∗h=>12∗a∗h=a2=>h=2a
Incorrect
Explanation:
We know area of square =a2Area of triangle =12∗a∗h=>12∗a∗h=a2=>h=2aWe know area of square =a2Area of triangle =12∗a∗h=>12∗a∗h=a2=>h=2a
Question 21 of 31
21. Question
What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle ?
Correct
Explanation:
As far as questions of Area or Volume and Surface area are concerned, it is all about formulas and very little logic. So its a sincere advice to get all formulas remembered before solving these questions.
Lets solve this,
Area of rectangle =l∗bArea of triangle =12l∗bRatio =l∗b:12l∗b=1:12=2:1Area of rectangle =l∗bArea of triangle =12l∗bRatio =l∗b:12l∗b=1:12=2:1
Incorrect
Explanation:
As far as questions of Area or Volume and Surface area are concerned, it is all about formulas and very little logic. So its a sincere advice to get all formulas remembered before solving these questions.
Lets solve this,
Area of rectangle =l∗bArea of triangle =12l∗bRatio =l∗b:12l∗b=1:12=2:1Area of rectangle =l∗bArea of triangle =12l∗bRatio =l∗b:12l∗b=1:12=2:1
Question 22 of 31
22. Question
The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:
Correct
Option D
Explanation:
We know the product of diagonals is 1/2*(product of diagonals)
A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.
Correct
Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
Area of triangle =12∗b∗h1Area of rectangle =b∗h2As per question 12∗b∗h1=b∗h212∗b∗h1=b∗100h1=100∗2=200mArea of triangle =12∗b∗h1Area of rectangle =b∗h2As per question 12∗b∗h1=b∗h212∗b∗h1=b∗100h1=100∗2=200m
Incorrect
Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
Area of triangle =12∗b∗h1Area of rectangle =b∗h2As per question 12∗b∗h1=b∗h212∗b∗h1=b∗100h1=100∗2=200mArea of triangle =12∗b∗h1Area of rectangle =b∗h2As per question 12∗b∗h1=b∗h212∗b∗h1=b∗100h1=100∗2=200m
Question 24 of 31
24. Question
Find the circumference of a circle, whose area is 24.64 meter sqaure
Correct
Explanation:
Area of Square =π∗r2=>π∗r2=24.64=>r2=24.6422∗7=>r2=7.84=>r=7.84−−−−√=>r=2.8Circumference =2π∗r=2∗227∗2.8=17.60mArea of Square =π∗r2=>π∗r2=24.64=>r2=24.6422∗7=>r2=7.84=>r=7.84=>r=2.8Circumference =2π∗r=2∗227∗2.8=17.60m
Incorrect
Explanation:
Area of Square =π∗r2=>π∗r2=24.64=>r2=24.6422∗7=>r2=7.84=>r=7.84−−−−√=>r=2.8Circumference =2π∗r=2∗227∗2.8=17.60mArea of Square =π∗r2=>π∗r2=24.64=>r2=24.6422∗7=>r2=7.84=>r=7.84=>r=2.8Circumference =2π∗r=2∗227∗2.8=17.60m
Question 25 of 31
25. Question
The wheel of a motorcycle, 70 cm in diameter makes 40 revolutions in every 10 seconds. What is the speed of the motorcycle in km/hr
Correct
Explanation:
In this type of question, we will first calculate the distance covered in given time.
Distance covered will be, Number of revolutions * Circumference
So we will be having distance and time, from which we can calculate the speed. So let solve.
Radius of wheel = 70/2 = 35 cm
Distance covered in 40 revolutions will be
Explanation:
In this type of question, we will first calculate the distance covered in given time.
Distance covered will be, Number of revolutions * Circumference
So we will be having distance and time, from which we can calculate the speed. So let solve.
Radius of wheel = 70/2 = 35 cm
Distance covered in 40 revolutions will be
If the radius of a circle is diminished by 10%, then the area is diminished by
Correct
Let the original radius be R cm. New radius = 2R
Area=πR2New Area =π2R2=4πR2Increase in area =(4πR2−πR2)=3πR2Increase percent =3πR2πR2∗100=300%Area=πR2New Area =π2R2=4πR2Increase in area =(4πR2−πR2)=3πR2Increase percent =3πR2πR2∗100=300%
Incorrect
Let the original radius be R cm. New radius = 2R
Area=πR2New Area =π2R2=4πR2Increase in area =(4πR2−πR2)=3πR2Increase percent =3πR2πR2∗100=300%Area=πR2New Area =π2R2=4πR2Increase in area =(4πR2−πR2)=3πR2Increase percent =3πR2πR2∗100=300%
Question 27 of 31
27. Question
If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?
Correct
2πR1=4π=>R1=22πR2=8π=>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π2πR1=4π=>R1=22πR2=8π=>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π
Incorrect
2πR1=4π=>R1=22πR2=8π=>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π2πR1=4π=>R1=22πR2=8π=>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π
Question 28 of 31
28. Question
The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
Correct
Explanation:
Perimeter = Distance covered in 8 min. =
12000
x 8
m = 1600 m.
60
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m^{2} = 153600 m^{2}.
Incorrect
Explanation:
Perimeter = Distance covered in 8 min. =
12000
x 8
m = 1600 m.
60
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m^{2} = 153600 m^{2}.
Question 29 of 31
29. Question
The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
Correct
Explanation:
Perimeter = Distance covered in 8 min. =
12000
x 8
m = 1600 m.
60
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m^{2} = 153600 m^{2}.
Incorrect
Explanation:
Perimeter = Distance covered in 8 min. =
12000
x 8
m = 1600 m.
60
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m^{2} = 153600 m^{2}.
Question 30 of 31
30. Question
An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:
Correct
Explanation:
100 cm is read as 102 cm.
A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2} – A_{1}) = [(102)^{2} – (100)^{2}]
= (102 + 100) x (102 – 100)
= 404 cm^{2}.
Percentage error =
404
x 100
%
= 4.04%
100 x 100
Incorrect
Explanation:
100 cm is read as 102 cm.
A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2} – A_{1}) = [(102)^{2} – (100)^{2}]
= (102 + 100) x (102 – 100)
= 404 cm^{2}.
Percentage error =
404
x 100
%
= 4.04%
100 x 100
Question 31 of 31
31. Question
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
Correct
Explanation:
Let original length = x metres and original breadth = y metres.
Original area = (xy) m^{2}.
New length =
120
x
m
=
6
x
m.
100
5
New breadth =
120
y
m
=
6
y
m.
100
5
New Area =
6
x x
6
y
m^{2}
=
36
xy
m^{2}.
5
5
25
The difference between the original area = xy and new-area 36/25 xy is
= (36/25)xy – xy
= xy(36/25 – 1)
= xy(11/25) or (11/25)xy
Increase % =
11
xy x
1
x 100
%
= 44%.
25
xy
Incorrect
Explanation:
Let original length = x metres and original breadth = y metres.
Original area = (xy) m^{2}.
New length =
120
x
m
=
6
x
m.
100
5
New breadth =
120
y
m
=
6
y
m.
100
5
New Area =
6
x x
6
y
m^{2}
=
36
xy
m^{2}.
5
5
25
The difference between the original area = xy and new-area 36/25 xy is
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