MATHEMATICS
Previous year question paper with solutions for MATHEMATICS Mar-2019
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Question paper 1
Part A
Questions from 1 to 8 carry 1 mark each.
1. Complete the prime factor tree:
diagram
Answer:
35
2. The graph of y=p (x) is given. Find the number of zeroes of p (x)
diagram
(a) 0 (b) 1
(c) 2 (d) 3
Answer:
(b) 1
3. Write first four term of the A.P. when the first term a=4 and common difference d=-3
Answer:
a= 4 d= -3
Find fur terms are
a, a+d ,a+2d, a+3d
4, 4-3, 4-6,4-9
4, 1, -2 ,-5
4. which point lies on the x-axis from the following :
(a) (1,1) (b) (2,0)
(c) (0,3) (d) (-4,-2)
Answer:
(b) (2,0)
5. The hypotenuse is the ______ side in right triangle.
Answer:
Bigger
6. Write the formula for finfing the area of the sector of a circle with angle θ.
Answer:
7. The formula for finding the surface area of the sphere is
Answer:
False
8. Probability of an event E+Probability of the event 'not E'=1
Answer:
True
Part-B
Questions from 9 to 16 carry 2 marks each.
9. Express 5005 as a product of its prime factors.
Answer:
5005 = 5×7×11×13
It is prime factorization of 5005
10. Find the zeroes of the quadratic polynomial 3x2-x-4 and verify the relationship between the zeroes and the coefficients.
Answer:
Polynomial is given by
Then its zeroes are
x(3x-4)+1(3x-4)=0
(x+1)(3x-4)=0
X+1=0 , 3x-4=0
X=-1 ,
General quadratic equation is
Comparing with
A=3 , b=-1 C= -4
Sum of zeroes =
Product of zeroes =
(-1)
verified
11. The coach of a team buys 3 bats and 6 balls for Rs. 3900. Later, he buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically.
Answer:
. Let the cost of one bat = x
Let the cost of one ball = y
Algebraically Representation is
12. Check whether the equation x (2x+3) =x2 +1 is quadratic equation.
Answer:
Equation is given
Yes it is quadratic ? highest power of x is 2
13. PQR is triangle right angled at P and M is a point on QR such that . Show that PM2=QM.MR.
Diagram
Answer:
Let
Similarly
In
(AAA Similarity)
We know that corresponding sides of similar triangles are proportional.
14. In given figure, if TP and TQ are the two tangents to a circle with centre O so that , find angle
Diagram
Answer:
Given that TQ and TP are two tangents of the circle.
We know that radius is perpendicular to tangent, therefore, OP⊥TP and OQ ⊥TQ
In quadrilateral POQT,
15. The following table gives the literacy rate (in percentage) of 35 cities. Find the man literacy rate.
Literacy rate (in %)
45-55
55-65
65-75
75-85
85-95
Number of Cities
3
10
11
8
3
Answer:
h class internal
h = 10
a assumed Mean
a = 70
Literacy rate (in%)
No. of cities
45 – 55
3
50
-20
-2
-6
55 – 65
10
60
-10
-1
-10
65 – 75
11
70
0
0
0
75 – 85
8
80
10
1
8
85 – 95
3
90
20
2
6
Totol
Mean () = a +
Mean () = 69.43%
16. One card is drawn from a well-shuffled deck of 52 cards. find the probability of getting :
(i) a spade
(ii) the queen of diamond.
Answer:
Total no. of cards = 52
(i) a spare
Total no. of spare =13
Probability =
(ii) The Queen of diamond.
No. of queen of diamond=1
Probability =
Part-C
Questions from17 to 24 carry 4 marks each.
17. The sum of the reciprocals of Rehman's ages,(in years) 3 years ago and 5 years from now . Find his present age.
Answer:
Let present age of Rehman = x
from required information we have given that
3(2x+2)
6x + 6 =
x(x-7)+(x-7)=0
(x+3) (x-7)=0
X= -3 x = 7
Rejected
Present age is 7 years
18. How many terms of the A.P.: 9,17,25,........ must be taken to give a sum of 636 ?
Answer:
Given that
a = 9 d = 8
N = ?
Formula ,
636 =
2 × 636 = n [18+8n-8]
1272 = n [8n+10]
1272 =
4n(n-12) +53 (n-12) = 0
(4n+53) (n-12)=0
4n+53=0 n-12=0
n= n=12 Ans
Rejected
? 12th terms whose sum is 636
19. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the-segment joining the point of contract at the centre.
Diagram
Prove that the summ of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Answer:
20. Find the coordinates of the point which divides the join of (-1,7) and (4,-3) in the ratio 2:3.
Diagram
Answer:
x = 1
y = 3
c(1,3)
OR
20. Find the value of y for which the distance between the points P (2,-3) and Q (10,y) is 10 units.
Answer:
P(2,-3) and Q(10,y)
distance between P and Q =10 units
by distance formula
=
100 =
y(y+9)-3(y+9)=0
(y-3) (y+9)=0
Y=3 y=-9 Rejected
21. Match the follwing:
(i) sin (90o-A) (a) sin A
(ii) cos 00 (b) 0
(iii) sin 00 (c) 1
(iv) cos (90o-A) (d) cos A
Answer:
Match the following
(i)
(ii)
(iii)
22. A kite is flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is . Find the length of the string, assuming that there is no slack in the string.
Diagram
Answer:
23. Draw a circle of radius 3 cm. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Answer:
Simply Draw
24. In a circle of radius 21 cm, an arc subtends an angle of at the centre. find
Diagram
(i) the length of the arc
(ii) area of the sector formed by the arc.
Answer:
. Radius of circle = 21cm
(i) The length of arc
=
(ii) Area of the sector formed by the arc
Part-D
Questions 25 to 28 carries 6 marks each.
25. Check graphically whether the pair of equations x+3y=6 and 2x-3y=12 is consistent. If so, solve them graphically.
Answer:
Given equations are
x + 3y = 6
2x - 3y = 12
x+3y=6
X
0
6
3
y
2
0
1
2x-3y=12
x
6
0
y
0
-4
x = 6 , y = 0
is solution of above equations.
OR
25. 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Answer:
Let the no. of days taken by 1 woman to complete the work = x
Let the no. of days taken by 1 man to complete the work = y
So, the work done by 1 man in 1 day =
The work done by 1 man in 1 day =
According to first condition
According to the second condition,
From equation 1 and 2
Y = 36
X = 18
Y = 36 Put in equation (2)
x = 18
26. If a line is drawn parllel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio, Prove it.
Answer:
We need to prove that
Let us join BE and CD and then draw DM ⊥AB and EN⊥AB
Now Area of
So ar(ADE)=
ar(BDE)=
ar(ADE)=
ar(DEC)=12 EC×DM
We know that and DEC are on the same base DE and between the same Parallels BC and DE.
So ……………………..(3)
Hence Proved
OR
26. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that .
Answer:
We need to prove that
Let
We know that TP=TQ
So, TPQ is an isosceles tringle
We know
So
27. A solid pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1cm3 of iron has approximately 8g mass. (use π = 3.14)
Diagram
Answer:
Radius of larger cylinder () = 12cm
Height of larger cylinder () = 220cm
Radius of larger cylinder () = 8cm
Height of larger cylinder () = 60cm
Volume of pole =volume of large cylinder +volume of smaller cylinder
=
=
Mass of 1 of iron = 8g
of iron
= 111532.8 ×8 g
= 892262.4 g = 892.262 kg
OR
27. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
Diagram
Answer:
Radius of hemispherical part =
= radius of cylindrical part(r)=3.5m
Height of cylindrical part (h)=10m
The total surface area of article
= CSA of cylindrical part + CSA of two Hemispherical part
=
=
=
= 119
Hence, the total surface area of the article is 374
28. A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained:
Height (in cm)
No. of girls
Less than 140
4
Less than 145
11
Less than 150
29
Less than 155
40
Less than 160
46
Less than 165
51
Find the median height.
Answer:
Height
NO. of girls
Less than 140
4
Less than 145
11
Less than 150
29
Less than 155
40
Less than 160
46
Less than 165
51
Now
This observation lies in the class 145-150 then
L = 145
C.f = 11
F = 18
H = 5
Median = l+
Median = 145+
= 145+
Median height of the girls is 149.03cm
OR
28. The following data gives the information on the observed lifetime (in hours) of 225 electrical components:
Lifetimes
(in hours)
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
10
35
52
61
38
29
Determine the moodel lifetimes of the componts.
Answer:
Lifetime
Frequency
0-20
10
20-40
35
40-60
52
60-80
61
80-100
38
100-120
29
In above maximum class frequency is 61;
Belonging to class interval 60-80
Lower class limit (l) of Model class = 60
Frequency () of Model class = 61
Frequency () of class preceding the Modal class = 52
Frequency () of class succeeding the Modal class = 38
Class size (h)=20
Mode =
=
= 60+ 5.625= 65.625