a. 5:4 b. 9:4 c. 4:5 d. 4:9
a. 18000m2 b. 20000m2 c. 19000m2 d. 25000m2
a. Rs.360 b. Rs.810 c. Rs.900 d. Rs.1800
a. 7m b. 7.5m c. 8m d. 8.5m
a. 180 b. 1800 c. 18 d. 18000
a. 176 b. 192 c. 184 d. 162
a. 1536m2 b. 15360m2 c. 153600m2 d. None of these
a. 5m b. 12m c. 13m d. 14.5m
a. Rs.950 b. Rs.1425 c. Rs.1900 d. Rs.712.50
a. 2.6m b. 3.9m c. 4m d. 4.2m
a. 42m b. 84m c. 63m d. 8.75m
a. 3cm b. 23cm c. 4.5cm d. 3/4 cm
a. 4cm b. 5cm c. 6cm d. 7cm
a. 111 cm2 b. 148cm2 c. 154cm2 d. 259cm2
a. 5.25m b. 7m c. 10.5m d. 21m
a. 18 b. 20 c. 25 d. 30
a. 30cm b. 60cm c. 72cm d. 132cm.
a. 675.5m2 b. 780.6m2 c. 785.8m2 d. 850.5m2
a. 1100cm3 b. 1110cm3 c. 1010cm3 d. None of these
a. 288 b. 48 c. 72 d. 864
Answer & Explanations:
Let, each side=a. Then, original area= a2.
New side= 150a/100= 3a/2. New area= 9a2/4.
Required ratio= 9a2/4:a2= 9:4
Length of the diagonal= Distance covered in 3 min. at 4 km/hr.
= (4000/ 60 *3)= 200m.
Therefore, Area of the field= 1/2 * diagonal2
= ½ * 200*200 = 20000 m2
Area= Total cost/ Rate= (1215/135) hectares= (9*10000) sq.m.
Therefore, side of the square= Ö90000=300m.
Perimeter of the field= (300*4)m= 1200m
Cost of fencing= Rs.(1200*3/4)= Rs.900
Length of the carpet= Total cost/ rate/m= 8100/45=180m
Area of the carpet= 180*75/100= 135m2
Breadth of the room = (Area/ length)= 135/18= 7.5m
Area of the hall= 3600*1500
Area of each stone= (60*50)
Therefore, number of stones= (3600*1500/ 60*50)= 1800
Area of the room= 544*374 sq.cm
Size of largest square tile= HCF of 544 cm & 374cm
Area of 1 tile= 34*34 sq.cm
Therefore, number of tiles= (544*374/34*34)= 176
Perimeter= Distance covered in 8 min
= (12000/60 *8)m= 1600m
Let, length= 3x meters and breadth= 2x meters
Then, 2(3x+2x)= 1600 or x= 160
Therefore, length= 480 m and breadth= 320m
Therefore, area= (480*320)m2 = 153600 m2
Let, length= x meters and breadth= y meters
Then xy=60 and (x2+y2) + x= 5
Therefore, x=60 and (x2+y2)= (5y-x)2
Or xy=60 and 24y2-10xy=0.
Therefore, 24y2-10*60=0 or y2= 25 or =5.
Therefore, x= (60/5)m= 12m. So, length of the carpet = 12m
Let the dimensions of former room be x,y and z.
Then, the area of its 4 walls= 2(x+y)*z sq. units.
Dimensions of another room are 2x, 2y and 2z units.
Therefore, area of 4 walls of this room= 2 (2x+2)*2z= 4*[2(x+y)*z]
= 4 (Area of 4 walls of Ist room)
Therefore, required cost= Rs.(475*4)= Rs.1900
Let, height= 2x metres & (length+ breadth)= 5x metres.
Length of paper= (260/2)m= 130m.
Therefore, area of paper= (130*50/100)= 65m2
Area of 4 walls= (65+15)=80m2
2(length+breadth)*height=80.
Therefore, 2*5x*2x=80 or x2=4 or x=2
Therefore, height of the room= 4m
½ (12+8)d=840 or d=84m
Area= 3/4 *(33)2= 273/4.
Therefore height= 273/4 *2/ 33= 9/2= 4.5cm
Let the other sides be x and y. Then,
X2+y2 = 132 = 169. Also, ½ xy= 30 => xy=60.
Therefore, (x+y)= ((x2+y2)+2xy) = (169+120)= 289=17.
(x-y)= ((x2+y2)-2xy)= (169-120)= 49= 7.
Solving x+y=17, x-y=7, we get x=12 and y=5.
Therefore, shorter side = 5cm
2πr-r=37 or (2π-1)r=37.
Or (2* 22/7 -1)r=37 or 37r/7=37 or r=7.
Therefore, Area= πr2= (22/7 *7*7)= 154 cm2
2π(R-r)=60 => 2*22/7 * (R-r)=60.
Therefore, (R-r)= (66*7/44)= 10.5m
Distance moved by toothed wheel in 15 revolutions= (15*2*22/7*25)
Distance moved by smaller wheel in 1 revolution= (2*22/7*15)
Therefore, required number of revolutions= (15*44/7*25*7/ 44*15)= 25
Length of wire= circumference of circle of radius 42cm= (2* 22/7* 42)= 264cm.
Therefore, perimeter of rectangle= 264 cm.
Let, length= 6x cm & breadth= 5x cm.
Therefore, 2 (6x+5x)= 264 or x=12.
Therefore, smaller side= 60 cm
Required area= (63*63 – 4*1/4 *22/7 * 63/2 * 63/2)= 850.5m2
Volume of theater= Volume of lower portion+ volume of upper portion.
Volume of lower portion (Rectangular prism)= lbh= 8*4*25= 800cm3
Volume of upper portion (triangular prism)= (½ bh)h= ½ *8*3*25=300 cm3
Therefore, Total volume= Volume of lower portion+ volume of upper portion
= 800+300 = 1100cm3
d=12, r=6;
Volume of the largest sphere= 4/3πr3
= 4/3 *π*6*6*6= 288πcm3