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# Maths Questions for DSSSB & KVS/NVS Exams

Q1. A circular wire of diameter 112 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 9 : 7. The smaller side of the rectangle is
(a) 77 cm
(b) 97 cm
(c) 67 cm
(d) 84 cm

Q2. If the perimeter of an equilateral triangle be 18 cm, then the length of each median is
(a) 3√2 cm
(b) 2√3 cm
(c) 3√3 cm
(d) 2√2 cm

Q3. Two equal maximum sized circular plates are cut off from a circular paper sheet of circumference 352 cm. Then the circumference of each circular plate is
(a) 176 cm
(b) 150 cm
(c) 165 cm
(d) 180 cm

Q4. The inradius of an equilateral triangle is √3 cm, then the perimeter of that triangle is
(a) 18 cm
(b) 15 cm
(c) 12 cm
(d) 6 cm

Q5. The difference between the circumference and diameter of a circle is 150 m. The radius of that circle is (Take π=22/7)
(a) 25 metre
(b) 35 metre
(c) 30 metre
(d) 40 metre

Q6. The perimeters of a circle, a square and an equilateral triangle are same and their areas are C, S and T respectively. Which of the following statement is true?
(a) C = S = T
(b) C > S > T
(c) C < S < T
(d) S < C < T

Q7. A horse takes 2 1/2 seconds to complete a round of a circular field. If the speed of the horse was 66 m/sec, then the radius of the field is [Given, π=22/7]
(a) 25.62 m
(b) 26.52 m
(c) 25.26 m
(d) 26.25 m

Q8. The diameter of the front wheel of an engine is 2x cm and that of rear wheel is 2y cm. To cover the same distance, find the number of times the rear wheel will revolve when the front wheel revolves ‘n’ times.
(a) n/xy times
(b) yn/x times
(c) nx/y times
(d) xy/n times

Q9. A bicycle wheel has a diameter (including the tyre) of 56 cm. The number of times the wheel will rotate to cover a distance of 2.2 km is (Assume π=22/7)
(a) 625
(b) 1250
(c) 1875
(d) 2500

Q10. If the altitude of an equilateral triangle is 12√3 cm, then its area would be;
(a) 36√3  cm^2
(b) 144√3  cm^2
(c) 72 cm^2
(d) 12 cm^2
Solutions

S1. Ans.(a)
Sol. circumference of circle = π× diameter
=(22/7)×112 = 352 cm
∴ perimeter of rectangle = 352
2(l + b) = 352
l + b =(352/2)=176
∴ smaller side =(7/16)×176
= 77 cm

S2. Ans.(c)
Sol. perimeter of equilateral triangle = 18 cm
3 × side = 18 cm
side =(18/3) = 6 cm
length of median =√3/2 side
=(√3/2)×6
=3√3 cm

S4. Ans.(a)

Sol. Inradius of equilateral triangle =side/(2√3)
√3=side/(2√3)
side = 6 cm
perimeter of an equilateral triangle = 3 × 6 = 18 cm

S7. Ans.(d)
Sol. Distance covered in 1 revolution = Circumference of circular field = 2πr
Distance = speed × time
= 66 m/s ×(5/2) s = 165m
∴ 2πr = 165
2×(22/7)×r=165
r=(165 ×7)/(2 ×22) = 26.25 m.

S8. Ans.(c)
Sol. Circumference of front wheel × no. of its revolutions = circumference of rear wheel × no. of its revolutions
2πx × n = 2πy × m (let ‘m’ is the revolution of rear wheel)
m=(nx/y)

S9. Ans.(b)
Sol. Distance to be covered in one revolution = Circumference of wheel = π× diameter
=(22/7)×56 = 176 cm
Total distance = 2.2 km
= (2.2 × 1000 × 100) cm
= 22,0000 cm
∴ Number of revolutions
=220000/176=1250