Set II - Delhi
Q. 8. What is the angle subtended at the centreof a circle of radius 7 cm, by an arc of length 11 cm?
Ans . 90°
Sol .
Q. 9. What is the common difference of an AP whose nth term is 3 + 5n? 1
Sol .
nth term, an = 3 + 5n
When n = 1, a1 = 3 + 5(1) = 8
When n = 2, a2 = 3 + 5(2) = 3 + 10 = 13
Common difference, d = a2 – a1 = 13 – 8 = 5
Q. 10 . What is the probability that two friends have different birthdays? 1
Sol .
Total no. of days in a year = 365
SECTION B
Questions number 11 to 15 carry 2 marks each.
Q. 11. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x).
Sol . Dividend = Divisor X Quotient + Remainder
Q. 12. If tan (2A) = cot(A – 21°), where 2A is an acute angle, then find the value of A.
Sol :
Q. 13. If A(–2, 4), B(0, 0) and C(4, 2) are the vertices of a Δ ABC, then find the length of median through the vertexA.
Ans. 
Sol .
Q. 14. If sin 3A = cos(A – 10°), then find the value of A, where 3A is an acute angle. 2
Sol . sin 3A = cos(A – 10°) 
Q. 15. Find the values of x for which the distance between the points P(4, – 5) and Q(12, x) is 10 units. 2
Sol .
PQ = 10 …(Given)
PQ2 = 100 …(Squaring both sides)
(12 – 4)2 + [x – (– 5)]2 = 100
82 + (x + 5)2 = 100
(x + 5)2 = 100 – 64 = 36
x + 5 = ± 6
x + 5 = 6 or x + 5 = – 6
x = 6 – 5 = 1 or x = – 6 – 5 = – 11
PQ2 = 100 …(Squaring both sides)
(12 – 4)2 + [x – (– 5)]2 = 100
82 + (x + 5)2 = 100
(x + 5)2 = 100 – 64 = 36
x + 5 = ± 6
x + 5 = 6 or x + 5 = – 6
x = 6 – 5 = 1 or x = – 6 – 5 = – 11
SECTION C
Questions number 16 to 25 carry 3 marks each.
Q. 16. Prove that 
is an irrational number.
is an irrational number.Sol.
Let us assume, to the contrary, that
is rational That is, we can find coprimes a and b 
such that
is rational That is, we can find coprimes a and b such that 
is rational, and so
is rational But this contradicts the fact that 
is irrational. This contradiction has arisen because of our incorrect assumption that 
is rational. So, we conclude that 
is irrational.Q. 17. Check graphically whether the pair of equations 3x + 5y = 15 and x – y = 5 is consistent. Also find the coordinates of the points where the graphs of the equations meet the y-axis.
Sol .
| x | 0 | -5 | 5 |
| y | 3 | 6 | 0 |
| x | 5 | 7 | 3 |
| y | 0 | 2 | -2 |
By plotting the points and joining them, the lines intersect at A(5, 0)
x = 5, y = 0
The pair of given equation is consistent.
The equation 3x + 5y = 15 meets the y-axis at
B(0, 3)The equation x – y = 5 meets the y-axis at C(0, – 5)
x = 5, y = 0The pair of given equation is consistent.The equation 3x + 5y = 15 meets the y-axis at
B(0, 3)The equation x – y = 5 meets the y-axis at C(0, – 5)
Or
Places A and B are 160 km apart on a highway. One car starts from A and another from B at the
same time. If the cars travel in the same direction at different speeds, they meet in 8 hours, but if
they travel towards each other, they meet in 2 hours. What are the speeds of the two cars?
same time. If the cars travel in the same direction at different speeds, they meet in 8 hours, but if
they travel towards each other, they meet in 2 hours. What are the speeds of the two cars?
Sol .
Let the speed of 1st car starting from place A = x km/hr
and the speed of 2nd car starting from place B = y km/hr
When two cars travel in the same direction
Distance covered by 1st car in 8 hours
= 160 km + Distance covered by 2nd car
8x = 8y + 160 …( Q Distance = Speed
time)
x = y + 20 …(i) (Dividing both sides by 8)
When two cars travel towards each other
2x + 2y = 160 …(
D = S
T)
x + y = 80 …(Dividing both sides by 2)
y + 20 + y = 80 …[From (i)]
2y = 80 – 20 = 60
From (i), x = 30 + 20 = 50
Speed of car starting from place A = 50 km/hr
and speed of car starting from place B = 30 km/hr.
and the speed of 2nd car starting from place B = y km/hr
When two cars travel in the same direction
Distance covered by 1st car in 8 hours
= 160 km + Distance covered by 2nd car
8x = 8y + 160 …( Q Distance = Speed
time)x = y + 20 …(i) (Dividing both sides by 8)
When two cars travel towards each other
2x + 2y = 160 …(
D = ST)x + y = 80 …(Dividing both sides by 2)
y + 20 + y = 80 …[From (i)]
2y = 80 – 20 = 60
From (i), x = 30 + 20 = 50
Speed of car starting from place A = 50 km/hrand speed of car starting from place B = 30 km/hr.
Q. 18. Find the roots of the equation :
Sol .
Q. 19. Find the sum of all three digit numbers which are divisible by 7.
Sol. To find : 105 + 112 + 119 + … + 994
Ist term, a = 105
Common difference, d = 112 – 105 = 7
an = 994
a + (n – 1)d = 994
105 + (n – 1)7 = 994
(n – 1)7 = 994 – 105 = 889
Ist term, a = 105
Common difference, d = 112 – 105 = 7
an = 994
a + (n – 1)d = 994
105 + (n – 1)7 = 994
(n – 1)7 = 994 – 105 = 889
Q. 20. Prove that :
Sol. See Q. 20, 2008(I OD).
Or
Using Geometry, find the value of sin 60°.
Sol.
Δ ABC be an equilateral, Let each side be 2a. Since each angle in equilateral Δ is 60°
Q. 21. Show that the points (1, 1), (–1, 5), (7, 9) and (9, 5) taken in that order, are the vertices of a rectangle.
Sol. Let A(1, 1), B(– 1, 5), C(7, 9) and D(9, 5) Using Distance formula
Diagonal AC = Daigonal BD = 10
∴ Diagonals are equal …(ii)
From (i) and (ii), we get
Points A, B, C and D are the vertices of a rectangle.
∴ Diagonals are equal …(ii)
From (i) and (ii), we get
Points A, B, C and D are the vertices of a rectangle.
Q. 22. Find the area of the triangle formed by joining the mid-points of the sides of the trianglewhose vertices are (2, 2), (4, 4) and (2, 6).
Sol. Let A(2, 2), B(4, 4), C(2, 6)
Mid-point of AB,
Mid-point of AB,
Q. 23. In Fig. 4, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8cm and OA = 5 cm.Find the length of OC.
Sol. Through O, draw
PQ || BC
So that P lies on AB
and Q lies on DC.
PQ || BC
PQ || BC
So that P lies on AB
and Q lies on DC.
PQ || BC
Q. 24. Find the roots of the equation.
Sol .
Q. 25. Find the sum of all three digit numbers which are divisible by 9. 3
Sol. To find : 108 + 117 + 126 + … + 999
1st term, a = 108
Common difference, d = 117 – 108 = 9
an = 999 a + (n – 1)d = 999
108 + (n – 1).9 = 999
(n – 1)9 = 999 – 108 = 891
1st term, a = 108
Common difference, d = 117 – 108 = 9
an = 999 a + (n – 1)d = 999 108 + (n – 1).9 = 999 (n – 1)9 = 999 – 108 = 891
SECTION D
Questions number 26 to 30 carry 6 marks each.
Q. 26. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Sol. Let shorter side, AB = x m then Diagonal, AC = (x + 60) m and longer side,
BC = (x + 30) m In rt. Δ ABC,
BC = (x + 30) m In rt. Δ ABC,
Q. 27. A boy, who’s eye level is 1.3 m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at any instant is 60°. After 2 seconds, the angle of elevation reduces to 30°. If the speed of wind at that moment is 29 3 m/s, then find the height of the balloon from the ground.
Sol. Let AP be the boy, C and E are two positions
of the balloon. Speed of wind
Distance travelled by the balloon in 2 seconds,
CE = BD =
Let AB = x m and BC = DE = y m
of the balloon. Speed of wind
Distance travelled by the balloon in 2 seconds,
CE = BD =
Let AB = x m and BC = DE = y m
Or
A statue, 1.5 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 45° and from the same point the angle of elevation of the top of the pedestal is 30°. Find the height of the pedestal. 
Sol. Let BC = h m
be the pedestal.
CD be the statue.
Distance from the
point be AB = x m
In rt.,
be the pedestal.
CD be the statue.
Distance from the
point be AB = x m
In rt.,
Q. 28 . Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Using the above, do the following .
If D is a point on the side AC of 
such that
AD : DC = 2 : 3, and E is a point on BC such that
DE || AB; then find the ratio of areas of
and 
.
such thatAD : DC = 2 : 3, and E is a point on BC such that
DE || AB; then find the ratio of areas of
and .Sol. Part I: See Theorem 2,
Page (xxxi).
Part II:
Let AD = 2k, CD = 3k
Then AC = 2k + 3k = 5k
Page (xxxi).
Part II:
Let AD = 2k, CD = 3k
Then AC = 2k + 3k = 5k
Q. 29. The lengths of 40 leaves of a plant are measured correct up to the nearest millimetre and the data is as under . 6
| Length (in mm) | Number of leaves |
| 118 – 126 | 4 |
| 126 – 134 | 5 |
| 134 – 142 | 10 |
| 142 – 150 | 12 |
| 150 – 158 | 4 |
| 158 – 166 | 5 |
Find the mean and median length of the leaves.
Sol
| Length (in mm) fi | Number of leaves | xi |  | Fidi’ | c.f | |
| 118 – 126 | 4 | 122 | -2 | -8 -5 | -13 | 4 |
| 126 – 134 | 5 | 130 | -1 | 9 | ||
| 134 – 142 | 10 | 138 | 0 | 0 | 19 | |
| 142 – 150 | 12 | 146 | 1 | 12 8 15 | 35 | 31 |
| 150 – 158 | 4 | 154 | 2 | 35 | ||
| 158 – 166 | 5 | 162 | 3 | 40 | ||
 |  | |||||
Q. 30. A motor boat whose speed in still water is 16 km/hr, takes 2 hours more to go 60 km upstream than to return to the same spot. Find the speed of the stream. 6
Sol. Let the speed of the stream = x km/hr
Then the speed of motor boat going upstream = (16 – x) km/hr
and the speed of motor boat going downstream = (16 + x) km/hr
According to the question
Then the speed of motor boat going upstream = (16 – x) km/hr
and the speed of motor boat going downstream = (16 + x) km/hr
According to the question
No comments:
Post a Comment