Mathematics Mcqs

A. 45 cm
B. 18 cm
C. 90 cm
D. None of these

Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(16)2 * h = (4/3)π (12)3 => h = 9 cm

A. 2 : 5
B. 1 : 5
C. 3 : 5
D. 4 : 5
The volume of the cone = (1/3)πr2h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V1/V2 = r12h1/r22h2 => 1/10 = (1)2h1/(2)2h2
=> h1/h2 = 2/5
i.e. h1 : h2 = 2 : 5

A. 252 m
B. 704 m
C. 352 m
D. 808 m
In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 500 resolutions.
= 500 * 2 * 22/7 * 22.4 = 70400 cm = 704 m

A. Rs. 4800
B. Rs. 3600
C. Rs. 3560
D. Rs. 4530
Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) – (6 * 3) – 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs. 4530

A. 10
B. 100
C. 1000
D. 10000
Along one edge, the number of small cubes that can be cut
= 100/10 = 10
Along each edge 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000

A. 12 π cm
B. 16 π cm
C. 24 π cm
D. 32 π cm
Radius of larger circle
= 2×196−−−√=28cm
Circumference of smaller circle
= (37×28)cm=12cm
Circumference of smaller circle
= 2πr=2π×12= 24πcm

A. 23.57 cm
B. 47.14 cm
C. 84.92 cm
D. 94.94 cm
Let the side of the square be a cm.
Parameter of the rectangle = 2(16 + 14) = 60 cm Parameter of the square = 60 cm
i.e. 4a = 60
A = 15
Diameter of the semicircle = 15 cm
Circimference of the semicircle
= 1/2(∏)(15)
= 1/2(22/7)(15) = 330/14 = 23.57 cm to two decimal places

A. 3.34
B. 2
C. 4.5
D. 5.5
Area of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}2 – ∏[4/2]2
= ∏[2.252 – 22] = ∏(0.25)(4.25) { (a2 – b2 = (a – b)(a + b) }
= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq m

A. 4192 sq m
B. 4304 sq m
C. 4312 sq m
D. 4360 sq m
Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l2 – ∏s2
= ∏{1762/∏2 – 1322/∏2}
= 1762/∏ – 1322/∏
= (176 – 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m

A. 200 sq cm
B. 72 sq cm
C. 162 sq cm
D. Cannot be determined
Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.