SSC CGL 2020 objective Ques (4 results)
SSC CGL 20201)The length, breadth and height of a cuboidal box are in the ratio 7 : 5 : 3 and its whole surface area is 27832 sq cm. Its volume is :
A)208120 \(cm^3\)
B)280120
\(cm^3\)
C)288100 \(cm^3\)
D)288120
288120 \(cm^3\)
l = 7 k cm, b = 5 k cm. h = 3 k cm. TSA = 27832sq.cm. ⇒2 (lb + bh + hl) = 27832 ⇒ k = 14
l = 98cm, b= 70cm, h = 42cm
volume of cuboidal box =( 98 x 70 x 42) cu. cm = 288120 cu cm
SSC CGL 20202)The volumes of spheres A and B are in the ratio 125 : 64. If the sum of radii of A and B is 36 cm,then the surface area (in \(cm^2\)) of A is:
\(1600\pi\)
Let, radius of sphereA = R cm; Radus of sphere B = r cm; so \({{4\over3}\pi{R}^3}\over{{4\over3}\pi{r}^3}\) = \(125\over64\);
\({R^3\over r^3 }=({5\over4})^3\) ; \({R\over r} = {5\over4} = 5: 4\); and R + r = 36 cm; R = \({5\over9}\times36 = 20 cm\);
Surface area of sphere A = \(4\pi{R}^2= 4\pi\times(20)^2=1600\pi {cm}^2\)
SSC CGL 20203)Find the height of a cuboid whose volume is 330 \(cm^3\) and base area is 15 \(cm^2\) .
22 cm
Volume of cuboid = Base area x height
330 = 15 x height
Height = 330/15 = 22 cm
SSC CGL 20204)If the radius of a right circular cylinder is decreased by 10%, and the heightis increased by 20%, then the percentage increase/decrease in its volume is :
decrease by 2.8%
Volume of right circular cylinder =\( \pi r^2h\);
Radius of a right circular cylinder is decreased by 10%, and the height is increased by 20% so,
r1 =\( r \times 90/100 = 0.9r\) ;
h1 = \(h \times 120/100 = 1.2h\) ;
Volume of new right circular cylinder =\(\pi r1^2h1 = \pi (0.9r)^2(1.2h) = 0.972(\pi r^2h)\) ;
Decrements in volume =\( \pi r^2h - 0.972(\pi r^2h) = 0.028(\pi r^2h)\) ;
Percentage Decrements in volume = \(\frac{0.028(\pi r^2h)}{(\pi r^2h)} \times 100\) = 2.8%