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# Maths Questions for DSSSB Exam : 7th August 2018 (Solutions)

Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB Exam & STET Exam.
Q1. If diameter of a circle is increased by 12.5%, find the percentage increase in its perimeter.
(a) 12.5%
(b) 144.25%
(c) 26.56%
(d) None of these
Q2. The sides of a triangle area in the ratio 12: 15: 20. If its perimeter be 94 cm, find the length of smallest side of the triangle.
(a) 18 cm
(b) 22.5 cm
(c) 24 cm
(d) 27 cm
Q3. The side of a square is 5 cm which is 13 cm less than the diameter of a circle. What is the approximate area of the circle?
(a) 245 sq. cm
(b) 255 sq. cm
(c) 235 sq. cm
(d) 265 sq. cm
Q4. Find the area of a square inscribed in a circle of radius 4 cm.
(a) 50 sq. cm
(b) 47 sq. cm
(c) 32 sq. cm
(d) 38 sq. cm
Q5. The area of an equilateral triangle is √243/2 sq. cm. Find the length of its side.
(a) 3 cm
(b) 3√3 cm
(c) 9 cm
(d) 3√2 cm
Q6. The inner circumference of a circular race track of 7 m wide is 440 m. Find the radius of the outer circle.
(a) 57 m
(b) 68 m
(c) 77 m
(d) 69 m
Q7. A cylinder is of the height 8 m and has base radius 8 m. The maximum length of the rod that can be placed in it is
(a) 8√5 m
(b) 9√5 m
(c) 8√3 m
(d) 8 (2π + 1) m
Q8. If the sides of a triangle area 50 m, 78 m and 112 m, then the perpendicular distance from the opposite angle on the side 112 m is
(a) 10 m
(b) 30 m
(c) 5 m
(d) 20 m
Q9. How many balls of radii 1 cm can be made by melting a cube of side 22 cm?
(a) 5324
(b) 2662
(c) 2541
(d) 1347
Q10. If the area of trapezium is 250 m², then the distance between the parallel sides of lengths 15 m and 10 m, is
(a) 18 m
(b) 20 m
(c) 22 m
(d) 30 m
Solutions
S1. Ans.(a)
Sol. Percentage change in perimeter is same as percentage change in diameter of the circle.
∴ Required percentage change = 12.5%

S2. Ans.(c)
Sol. Let sides of the triangle are 12x, 15x and 20x.
Then, 12x + 15x + 20x = 94
⇒ 47x = 94
⇒ x = 2
Thus, the smallest side = 12x
= 12 × 2 = 24 cm

S3. Ans.(b)
Sol. Diameter of the circle = 5 + 13 = 18 cm
∴ Radius of circle = 18/2 = 9 cm
∴ Area of circle = π(9)² = 22/7 × 81
= 255 sq cm