Mathematics Mock Test Papers
1. Without using trigonometirc tables, evaluate:
Solution:-
= -6
2. Evaluate:
Solution:-
3. Evaluate:
Solution:-
= 0
4. A man is standing on the deck of a ship, which is 8m above water level. He observes the angle of elevation of the top of a hill as 600 and angle of depression of the base of the hill as 300. Calculate the distance of the hill from the ship and the height of the hill.
Solution: - Let B be man, D the base of the hill, x be the distance of hill from the ship and h + 8 be the height of the hill.
In
tan 600 = AC/BC
In
tan 300 = CD/BC
Height of the hill = h + 8 = 24 + 8 = 32m
Distance of the hill from the ship =
5. A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 600. When he moves 40m away from the bank, he finds the angle of elevation to be 300. Find the height of the tree and the width of the river.
Solution:- Let height of the tree be y and width of the river be x. CD = 40m
In
tan 300 = AB/BD
In
tan 600 =AB/BC
Putting value of y from (ii) to (i)
Height of the tree =
Width of the river = 20m.
6. From a window (h metres high above the ground) of a house in a street, the angle of elevation and depression. of the top and the foot of another house on the opposite side of the street are respectively. Show that the height of the opposite house is
.
Solution :- Let W be the window AB the house and WP the width of the street.
In
In
Height of the =
7. From the top of a tower 100 m high, the angles of depression of the top and bottom of a pole standing on the same plane as the tower are observed to be 300 and 450 respectively. Find the height of the pole.
Solution:- Let AB be the tower and CD be the pole.
Let CD = h. AB = 100m
In
In
Height of the pole = 42.26
8. The angles of elevation and depression of the top and bottom of a light house from the top of a building, 60m high, are 300 and 600 respectively. Find
(i) The difference between the heights of the light house and building.
(ii) Distance between the light house and the building.
Solution:- Let AB be the building and CE be the light house.
AD = BC =
tan 300 = DE/AD
(i) CE - CD = DE = 20m
(ii) Distance = AD = BC = 34.64m
9 . An unbiased dice is tossed.
- Write the sample space of the experiment
- Find the probability of getting a number greater than 4.
- Find the probability of getting a prime number.
Solution:-
- Sample space = {1, 2, 3, 4, 5, 6}
n(s) = 6
- E = event of getting a number greater than 4
= {5, 6}
n (E) = 2
P (> 4) = Probability of a number greater than 4
= n(E)/n(S) = 2/6 = 1/3
- E = Event of getting a prime number
= {2, 3, 5}
n (E) = 3
P(Prime number) = Probability of a prime number
10. Out of 400 bulbs in a box, 15 bulbs a defective. One ball is taken out at random from the box. Find the probability that the drawn bulb is not defective.
Solution:- Total number of bulbs = 400
Total number of defective bulb = 15
Total number of non-defective bulbs = 400-15
= 385
P (not defective bulb) = 385/400
= 77/80
11. Find the probability of getting 53 Fridays in a leap year.
Solution:- No. of days in a leap year = 366
366 days = 52 weeks and 2 days.
A leap year must has 52 Fridays
The remaining two days can be
(a) Sunday an Monday
(b) Monday and Tuesday
(c) Tuesday and Wednesday
(d) Wednesday and Thursday
(e) Thursday and Friday
(f) Friday and Saturday
(g) Saturday and Sunday
Out of 7 case, 2 cases have Friday
P (53 Friday) = 2/7
12 . Three unbiased coins are tossed simultaneously. What is the probability of getting exactly two heads?
Solution: - When three coins are tossed simultaneously, the sample space is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
n(S) = 8
E = Set of cases favorable to the event
= {HHT, HTH, THH}
n(E) = 3
P (exactly two heads) =
13. A dice is thrown twice. Find the probability of getting (a) doublets (b) prime number on each die.
Solutions: - Sample space =
S = { (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
n (S) = 36
(i) E = Events getting doublet
= {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n (E) = 6
P(doublet) =
(ii) E = Events getting prime number on each die.
= {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}
n(E) = 9
P (getting prime number on each die)
= n(E)/n(S) = 9/36 = 1/4
14 . A bag contains 5 red balls, 8 White balls, 4 green balls and 7 black balls. A ball is drawn at random from the bag. Fine the probability that it is. (i) Black (ii) not green.
Solution:- Red balls = 5
White balls = 8
Green balls = 4
Black balls = 7
Total balls = 24
(i) P (Black balls) = 7/24
(ii) P (not a green ball) = 1- P (green ball)
= 1 - 4/24 = 20/24
= 5/6
15. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Solution:- P (neither an ace nor a king)
= 1 – p (either an ace or a king)
= 1 – 8/52 {no. of ace = 4}
{no. of king = 4}
Total = 8
= (52 – 8)/52
= 44/52
= 11/13
16. Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.
Solution: - Using Distance Formula
Diagonal AC = diagonal BD
A, B, C and D are vertices of a square
17. Determine the ratio in which the point P (m, 6) divides the join of A(-4, 3) and B(2, 8). Also find the value of m.
Solution:-
Let AP : PB = K : 1
Co-ordinates of P are
Burt co-ordinates of P are given (m, 6)
18. The coordinates of the mid-point of the line joining the points (2p + 2, 3) and (4, 2q + 1) are (2p, 2q). Find the value of p and q.
Solution:-
Let the given points be A (2p + 2, 3), B(4, 2q + 1) and C (2p, 2a)
Co-ordinates of C are
19. Find the ratio in which the line-segment joining the points (6, 4) and (1, -7) is divided by x-axis.
Solution:-
Let the two points A (6, 4) and B (1, -7), be divided by x-axis at C (x, 0)
Let AC : CB = K : 1
Co-ordinates of C are
As C lies on x-axis
20. Prove that coordinates of the centroid of a triangle ABC, with vertices (x1 y1), (x2, y2) and (x3, y3) are given by
Solution:-
Let the vertices of triangles be A(x1, y1), B(x2, y2) and C(x3, y3)
Let D be the mid-point of BC.
Co-ordinates of D are
Let centroid of triangle ABC be G.
Co-ordinates of G are
21. Find the value of m for which the points with co-ordinates (3, 5), (m, 6) and (1/2, 15/2) are collinear.
Solution :- Let the three points be A (3, 5) B (m, 6) and (1/2, 15,2)
As points A, B and care collinear
22. In the adjoining fig AD is the bisector of If BD = 4cm, DC = 3cm and AB = 6cm, determine AC.
Solution:- In is the bisector of
23. In the adjoining fig, AD is bisector of If AB = 5.6cm, AC = 4cm, DC = 3cm, find BC.
Solution:- In is the bisector of
24. In an equilateral triangle PQR, the side QR is trisected at S. prove that
Solution:-
Given:- In an equilateral is trisected at S.
To Prove:-
Construction:- is drawn
Proof:- QD = DR = QR/2 ------------(i)
Side QR is trisected at S(given)
In is acute
25. In the given figure, ABC is right angled triangle with the AB = 6cm and AC = 8cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r, the radius of the inscribed circle.
Solution:- In right
BC2 = AB2 + AC2 [By Pathagoras theorem]
= 62 + 82
= 36 + 64
= 100
= 1/2 X 6 X 8 = 24cm2
= 24/12
= 2cm
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