Mathematics 10 Past Papers 2015 MCQs Tests Federal Board

Federal Board, FBISE Mathematics 10 MCQs Tests of 2015 Past Papers are available on this page. There is only one Mathematics objective paper available on the FBISE website that is an overseas paper. MCQs Test of only Overseas paper is given below. Chapter-wise MCQs Tests of Mathematics 10 are available HERE.

Past Paper 2015 (Overseas)
Practice MCQs Test

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Class 10 Mathematics
Past Paper 2015 (Overseas)
Practice MCQs Test

1 / 15

The length of each side of a regular pentagon is 5cm. Its perimeter is:

10cm

20cm

25cm

15cm

2 / 15

If R={(0,2),(2,3),(3,3),(3,4)} then Domain of R is:

{0,3,4}

{0,2,3}

{2,3,4}

{0,2,4}

3 / 15

The semi-circumference and the diameter of a circle both subtend a central angle of:

90°

360°

270°

180°

4 / 15

The measures that are used to determine the degree or extent of variation in a data set are called measures of:

Variance

Central tendency

Dispersion

Average

5 / 15

Tangents drawn at the end points of the diameter of a circle are:

Parallel

Collinear

Non parallel

Perpendicular

6 / 15

If b-4ac<0 , then roots of equation ax²+bx+c=0 are:

Imaginary

Equal

Unequal

Rational

7 / 15

An arc subtends a central angle of 40° then the corresponding chord will subtend a central angle of:

40°

20°

60°

80°

8 / 15

The opposite angles of any quadrilateral inscribed in a circle are:

Acute angles

Supplementary angles

Complementary angles

Obtuse angles

9 / 15

A∩A'=

∅

A

Aͨ^C

U

10 / 15

A fraction in which the degree of the numerator is equal to or greater than the degree of denominator is called:

Irrational fraction

Rational fraction

Proper fraction

Improper fraction

11 / 15

A line which has two points in common with a circle is called:

Secant of a circle

Tangent of a circle

Sine of a circle

Cosine of a circle

12 / 15

The angles subtended by an arc at the circumference of a circle are:

Un-equal

Obtuse

Equal

Acute

13 / 15

The solution set of 5x²=15x is:

{3,-3}

{0,-3}

{0}

{0,3}

14 / 15

If u/v=v/w=k then:

u=w² k

u=wk²

u=vk²

u=v² k

15 / 15

15° is equal to:

π/4 radians

π/12 radians

π/3 radians

π/6 radians

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