# Quantitative Aptitude Questions Daily Quiz Day – 19

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335 Dear Aspirants, Our SSC Crackers team is providing a new series of Quantitative Aptitude Questions for Upcoming Exam so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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1) An isosceles triangle with each equal side measuring 5 cm and the base measuring 8 cm has a median drawn to its base. Find its length.

a) 3 cm

b) 4 cm

c) 5 cm

d) 6 cm

2) In a cyclic quadrilateral, two adjacent angles are in the ratio 12:7. If the angle opposite to smaller one of those angles is 110°, what is the measure of the fourth angle?

a) 45°

b) 60°

c) 80°

d) 90°

3) An ice-cream is in the form of an inverted cone surmounted by a hemisphere. The length of the ice-cream is 31 cm while the curved surface area of the scoop is 308 cm2. Find the volume of ice-cream contained in the structure.

a) 1950.67 cm3

b) 1095.67 cm3

c) 1905.67 cm3

d) 1195.67 cm3

4) If the radius of circle is 6 cm and the length of the tangent from a point P is 8 cm, what is the maximum distance of the point P from any point on the circle?

a) 16 cm

b) 10 cm

c) 6 cm

d) 18 cm

5) In the given circle, PQ and RS are two chords produced to intersect externally at a point M. If PM = 15 cm, PQ = 9 cm and SM = 18 cm, find the length of the chord RS.

a) 5 cm

b) 8 cm

c) 15 cm

d) 13 cm

6) In the given figure, angle COV measures 65o and angle DOW measures 80o. Find the measures of exterior angles of the triangle (Angles XCD, ODY and COZ). a) 100o, 115o and 145o respectively

b) 115o, 100o and 145o respectively

c) 100o, 145o and 115o respectively

d) 145o, 115o and 100o respectively

7) Given a metallic pipe which looks like a hollow cylinder. The measure of its internal radius is 4 cm and the external radius is 4.7 cm while the length measures 10 cm. Find the volume of metal used?

a) 165.4 cm3

b) 191.4 cm3

c) 159.2 cm3

d) 200 cm3

8) If the circumference of the given circle is 88 cm, then what is the area of the shaded portion? a) 56 cm2

b) 52 cm2

c) 58 cm2

d) Other than above

9) If a solid cone of volume 27π cm3 is kept inside a hollow cylinder whose radius and height are of that cone, then the volume of water needed to fill the empty space is:

a) 3π cm3

b) 18π cm3

c) 54π cm3

d) 81π cm3

10) By joining the midpoints of which of the following figure, the obtained figure is the same as the given figure?

a) Trapezium

b) Rectangle

c) square

d) Rhombus

Area of isosceles triangle = (b / 4) × √ (4a2 – b2); where ‘b’ stands for base and ‘a’ stands for each of its equal sides.

Thus, area of the given triangle = (8 / 4) × √(4 × 52 – 82 ) = 12 cm2

A median drawn to the base of an isosceles triangle is same as the altitude of that triangle.

Thus, area of the triangle = (1 / 2) × 8 × h = 12

So, h = 3 cm

Let the two adjacent angles be 12k and 7k.

Angle opposite to angle measuring 7k is 110°.

∴ 7k + 110° = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

=> 7k = 70°

∴ k = 10°

The angles are 120° and 70°.

Sum of all angles in a cyclic quadrilateral = 360°

∴ Measure of the fourth angle = 360° – (120° + 70° + 110°)

= 360° – 300° = 60°.  Given, OA = 6 cm and PA = 8 cm.

OA ⊥ PA

∴ OP = √(OA2 + PA2) = √(62 + 82) = 10 cm.

Maximum distance of ‘P’ from any point on the circle will be from the point diametrically opposite to the point where the line joining the centre and ‘P’ meets the circle i.e., BP.

BP = OP + OB = 10 + 6 = 16 cm. Using the property- “The angle between a secant and tangent drawn from the same point on the circle, is equal to the angle subtended in the alternate segment of the circle by the secant”, angles CDO and OCD measure 65o and 80o respectively as angle COV measures 65o and angle DOW measures 80o. Thus, exterior angle COZ measures 145o (measure of exterior angle is equal to the sum of interior opposite angles), angle XCD measures 100o and angle ODY measures 115o.

Volume of the hollow cylinder = πhR2 – πhr2 = π X h (R2 – r2)

Hence the volume is = (22/7) X 10 X (4.72 – 42) = (22/7) X 10 X 6.09 = 191.4 cm3

Let the radius of the circle is r

Circumference= 2 x π x r

88 = 2 x (22/7) x r

r = 14 cm

Area of quadrant = π x 14 x14/4 = (22/7) x 14 x 14/4 = 154 cm2

A(DAOB) = (1/2) x 14 x 14 = 98 cm2

Required area = 154 -98 = 56 cm2

Volume of water needed to fill the empty space = volume of cylinder – volume of cone

= π X r2 X h – 1/3 X π X r2 X h

= 2/3 X π X r2 X h

= 2 X (1/3 X π X r2 X h)

= 2 X 27π cm3

= 54π cm3