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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.=x∗115×100=460×2=(460×100/115)x2=400x=20=x∗115×100=460×2=(460×100/115)x2=400x=20
Incorrect
Explanation:
Let breadth =x metres.
Then length =(115x/100)metres.=x∗115×100=460×2=(460×100/115)x2=400x=20=x∗115×100=460×2=(460×100/115)x2=400x=20

Question 3 of 31
3. Question
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 feetArea to be fenced = 2B + L = 2*34 + 20
= 88 feetIncorrect
Explanation:
We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feetArea 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 cmArea of third Square = 10*10 – 8*8
= 36 cmSo side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cmIncorrect
Explanation:
We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cmArea of third Square = 10*10 – 8*8
= 36 cmSo 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 beArea 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 cmArea of third Square = 10*10 – 8*8
= 36 cmSo 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.
Correct
From the question, 2b+l = 30
=> l = 302bArea=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0Area=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0
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 mIncorrect
From the question, 2b+l = 30
=> l = 302bArea=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0Area=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0
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)=20000Incorrect
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 cmx2+(3x)2=(810−−√)2=>10×2=640=>x=8×2+(3x)2=(810)2=>10×2=640=>x=8
Incorrect
Explanation:
Let Breadth = x cm,
then, Length = 3x cmx2+(3x)2=(810−−√)2=>10×2=640=>x=8×2+(3x)2=(810)2=>10×2=640=>x=8

Question 9 of 31
9. Question
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 = 344So 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 = 344So 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=xy−(80×100×90y100)=(xy−1825xy)=725xyDecrease = (7xy25xy×100)%=28%=xy−(80×100×90y100)=(xy−1825xy)=725xyDecrease = (7xy25xy×100)%=28%
Incorrect
Explanation:
Let original length = x
and original width = y
Decrease in area will be=xy−(80×100×90y100)=(xy−1825xy)=725xyDecrease = (7xy25xy×100)%=28%=xy−(80×100×90y100)=(xy−1825xy)=725xyDecrease = (7xy25xy×100)%=28%

Question 11 of 31
11. Question
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 cmRequired 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 cmRequired 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,
Side=69696−−−−−√=264cmwe know diagonal =2√×side=2√×264=1.414×264=373.296cmSide=69696=264cmwe know diagonal =2×side=2×264=1.414×264=373.296cm
Incorrect
If area is given then we can easily find side of a square as,
Side=69696−−−−−√=264cmwe know diagonal =2√×side=2√×264=1.414×264=373.296cmSide=69696=264cmwe know diagonal =2×side=2×264=1.414×264=373.296cm

Question 15 of 31
15. Question
If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters is :
Correct
Explanation:
a2b2=2252561516< =>4a4b=4∗154∗16=1516=15:16a2b2=2252561516< =>4a4b=4∗154∗16=1516=15:16Incorrect
Explanation:
a2b2=2252561516< =>4a4b=4∗154∗16=1516=15:16a2b2=2252561516< =>4a4b=4∗154∗16=1516=15:16 
Question 16 of 31
16. Question
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,(x+2)2−x2=32×2+4x+4−x2=324x=28x=7cm(x+2)2−x2=32×2+4x+4−x2=324x=28x=7cm
Incorrect
Explanation:
Let the lengths of the line segments be x and x+2 cm
then,(x+2)2−x2=32×2+4x+4−x2=324x=28x=7cm(x+2)2−x2=32×2+4x+4−x2=324x=28x=7cm

Question 17 of 31
17. Question
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=18cmIncorrect
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 cmIncorrect
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.
Correct
Explanation:
Let each side be a cm, then(a2)2+102=a2< =>(a2−a24)=100< =>3a24=100a2=4003Area=3√4∗a2=(3√4∗4003)cm2=1003√cm2(a2)2+102=a2< =>(a2−a24)=100< =>3a24=100a2=4003Area=34∗a2=(34∗4003)cm2=1003cm2
Incorrect
Explanation:
Let each side be a cm, then(a2)2+102=a2< =>(a2−a24)=100< =>3a24=100a2=4003Area=3√4∗a2=(3√4∗4003)cm2=1003√cm2(a2)2+102=a2< =>(a2−a24)=100< =>3a24=100a2=4003Area=34∗a2=(34∗4003)cm2=1003cm2

Question 20 of 31
20. Question
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=2aIncorrect
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)Let one diagonal be d1 and d2
So as per question12∗d1∗d2=15012∗10∗d2=150d2=1505=3012∗d1∗d2=15012∗10∗d2=150d2=1505=30
Incorrect
Option D
Explanation:
We know the product of diagonals is 1/2*(product of diagonals)Let one diagonal be d1 and d2
So as per question12∗d1∗d2=15012∗10∗d2=150d2=1505=3012∗d1∗d2=15012∗10∗d2=150d2=1505=30

Question 23 of 31
23. Question
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), thenArea 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), thenArea 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=17.60mIncorrect
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 * CircumferenceSo 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 be40 * Circumference =40 * 2*\pi*r =40∗2∗227∗35=8800cm=8800100m=88mDistance covered in 1 sec =8810=8.8mSpeed=8.8m/s=8.8∗185=31.68km/hr40 * Circumference =40 * 2*\pi*r =40∗2∗227∗35=8800cm=8800100m=88mDistance covered in 1 sec =8810=8.8mSpeed=8.8m/s=8.8∗185=31.68km/hr
Incorrect
Explanation:
In this type of question, we will first calculate the distance covered in given time.
Distance covered will be, Number of revolutions * CircumferenceSo 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 be40 * Circumference =40 * 2*\pi*r =40∗2∗227∗35=8800cm=8800100m=88mDistance covered in 1 sec =8810=8.8mSpeed=8.8m/s=8.8∗185=31.68km/hr40 * Circumference =40 * 2*\pi*r =40∗2∗227∗35=8800cm=8800100m=88mDistance covered in 1 sec =8810=8.8mSpeed=8.8m/s=8.8∗185=31.68km/hr

Question 26 of 31
26. Question
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 newarea 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 newarea 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