High School Math Problems with Solutions

HIGH SCHOOL MATH PROBLEMS WITH SOLUTIONS

Problem 1 :

Simplify the following

Solution :

L.H.S

= (1 + tan²A) / (1+cot²A)

sec²A - tan²A = 1

sec²A = 1 + tan²A

cosec²A - cot²A = 1

cosec²A = 1 + cot²A

= sec²A / cosec²A

= (1/cos²A) / (1/sin²A)

= sin²A/cos²A

= tan²A

Problem 2 :

From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (take √3 = 1.732)

Solution :

In triangle BCD,

tanθ = Opposite side / Adjacent side

tan 30 = BC/CD

1/√3 = 10/CD

CD = 10√3 ------(1)

In triangle ACD,

tan 45 = AC/CD

1 = (10 + x) /CD

CD = 10 + x ------(2)

(1) = (2)

10√3 = 10 + x

x = 10(√3 - 1)

x = 10(10.732 - 1)

x = 10(0.732)

x = 7.32

The height of the flag staff is 7.32 m.

Problem 3 :

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower is 30°. Find the height of the tower.

Solution :

Height of the tower = AB, Distance between the foot of the tower and point of observation = BC.

tanθ = Opposite side / Adjacent side

tan θ = AB/BC

tan 30 = AB/30

1/√3 = AB/30

AB = 30/√3

AB = (30/√3) ⋅ (√3/√3)

AB = 30√3/3

AB = 10√3

Problem 4 :

The coefficient of variation of a series is 69% and its standard deviation is 15. Find the mean.

Solution :

Coefficient of variation = (σ/x̅) ⋅ 100%

69/100 = (15/x̅) ⋅ 100%

0.69 = 15/x̅

x̅ = 15/0.69

x̅ = 21.73

So, the required mean is 21.73.

Problem 5 :

If the two dice are thrown. what is the probability of getting the same number in both the dice.

Solution :

When two dice are thrown, the sample space

n(S) = 36

Let A be the event of getting doublet.

A = { (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) }

n(A) = 6

p(A) = n(A)/n(S)

p(A) = 6/36

Problem 6 :

2 cubes each of volume 64 cm³ are joined end to end. Determine the surface area of the resulting cuboid.

Solution :

Volume of one cuboid = 64 cm³

a³ = 64

Side length of cube is 4 cm.

Side length of cuboid = 4 + 4 = 8 cm

breadth = 4 cm and height = 4 cm

Surface area of cuboid = 2(lb + bh + hl)

= 2(8⋅4 + 4⋅4 + 4⋅8)

= 2(32 + 16 + 32)

= 2(80)

= 160 cm²

Problem 7 :

A metallic sphere of radius 4.2 cm is melted and recast into a shape of a cylinder of radius 6 cm. Determine the height of the cylinder.

Solution :

Radius of sphere = 4.2 cm

Radius of cylinder = 6

Volume of sphere = Volume of cylinder

(4/3) πr³ = πr² h

(4/3) 4.2³ = 6² h

h = 98.784/36

h = 2.744

So, the height of the cylinder is 2.744 cm.

Problem 8 :

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Determine the curved surface area of the frustum.

Solution :

Slant height (l) = 4 cm

Circumference of larger circle = 18

2πR = 18

R = 18/2π

R = 9/π

2πr = 6

r = 6/2π

r = 3/π

Surface area of the frustum cone = πl (R + r)

= π ⋅ 6(9/π + 3/π)

= 6(9+3)