solutions / cbse 10th mathematics sample paper
TRANSCRIPT
Solutions / CBSE 10th Mathematics Sample Paper
Section – A
1.We have ax by a b ... i
bx ay a b ... ii
Multiplying (i) by a and (ii) by b, we get 2 2a x aby a ab ... iii
2 2b x aby ab b ... iv (1 mark)Adding (iii) and (iv), we get
2 2 2 2a b x a b x 1 (½ mark)
Substituting x = 1 in (i), we geta 1 by a b
by b y 1 (½ mark)
OR
Let the father’s age be x years and son’s age be y years. Now according to problem
x 3y 3 x 3y 3 …(i) (½ mark) and …(ii) (½ mark) Subtracting (i) from (ii), we get
y 10 (½ mark)Substituting y = 10 in (i), we get
x 3 10 3 x 33 (½ mark) Hence, father’s age = 33 years, son’s age = 10 years
2. We have HCF = , LCM = and . We Know
(½ mark)
(1 mark)
(½ mark)2. We have
; a 0 , b 0, x 0
(½ mark)
(½ mark)
(½ mark) (½ mark)
3.Let a be the first term and d be the common difference of an A.P., then 7a a 7 1 d 7a a 6d [Using an = a + (n – 1)d]
and 11a a 11 1 d 11a a 10d (½ mark) According to the problem
7 117a 11a
7 a 6d 11 a 10d (½ mark)
4 a 17d 0
a 18 1 d 0 (½ mark)
18a 0 (½ mark)
5. Cash price of the electric iron = Rs. 440 Cash down payment = Rs. 200 Balance to be paid = Rs. (440 – 200) = Rs. 240 (½ mark) Instalment paid = Rs. 244 Total price charged under the instalment plan = Rs. (200 + 244) = Rs. 444 Therefore, interest charged = Rs. 444 – Rs. 440 = Rs. 4 (½ mark)
Therefore, Rs. 4 is the interest charged on Rs. 240 for 1 month i.e. year
Int. 100rate %
Pr incipal Time
(½ mark)
4 1001
24012
4 10020
20
(½ mark)
Hence, rate 20% per annum
6.
A
B CD
In and , we have(Given)
And [Common] [By AA criterion of similarity] (½ mark)
(½ mark)
(½mark)
(½ mark)
ORS R
Q
M
P O
40°65°
50°
[Angle in semi circle]
In triangle QRP, QRP QPR PQR 180
90 QPR 65 180
QPR 25
PQRS is a cyclic quadratic. QRS QPS 180
90 PRS 25 40 180 PRS 25 (1 mark)
PMQR is also a cyclic quadratic. MQR RPM 180
50 65 25 QPM 180
QPM 40 (1 mark)7. Total number of outcomes = 17
(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 Number of favourable outcomes = 9 P(an odd number) = 9/17 (1 mark)
(ii) Numbers divisible by 3 or 5 are 3, 5, 6, 10, 12, 15 Number of favourable outcomes = 6 P(a number divisible by 3 or 5) =6/17 (1 mark)
Section – B
8. We have y = x
x 0 1 4 y 0 1 4
y = 2x
x 0 1 2 y 0 2 4
and x + y = 6 y = 6 – x
x 0 1 3 y 6 5 3
(1 mark)
y
y
xxO
y = 2
x
(0,6
)(1
,5)
(1,1)
(0,0)
(4,4)B (2,4
) y = x
(1,2
)
x + y = 6
(3,3)
A
(1½ mark)
From figure, the three straight lines intersect at O(0, 0), A(3, 3) and B(2, 4). Vertices of the required triangle OAB are (0, 0), (3, 3) and (2, 4).
(½ mark)
9. We have
(1 mark)
(1 mark)
(1 mark)
10. Let x be the larger number, then square of the smaller number will be 4x(1 mark)
2x 4x 45 0 x 9 x 5 0 (½ mark)
x = 9 or x = –5 (½ mark) The numbers are natural numbers
x = 9 (½ mark) Square of smaller number = 4 × 9 = 36 Smaller number = (½ mark) Hence, larger number = 9 and smaller number = 6
11. Let a be the first term and d be the common difference of the A.P. and (½ mark)
and (½ mark) and
and (½ mark)
(1 mark)
(½ mark)
22S 902
12. Let each instalment be Rs. x.r = Rate of interest = 5% per annum CI.
n = Time = 2 years For first instalment:
A = Rs.x, time = 1 year, r = 5% per annum and P = P11
15
x P 1100
nr
Using A P 1100
(½ mark)
120
P Rs. x21
For second instalment:A = Rs. x, time = 2 years, r = 5% per annum and P = P2
(½ mark)
2
220
P Rs. x21
Total sum borrowed = Rs. 4100 [Given]
Therefore,
(1mark)
20 20x 1 Rs.4100
21 21
(½ mark)
(½ mark)Hence, annual instalment is Rs. 2,205.
13.
(1 mark)
(1 mark)
= 1 + 1 – 2 + 3 (1 mark)= 3
OR
LHS = (½ mark)
(½ mark)
(½ mark)
(½ mark)
(½ mark)
(½ mark)
= RHS
14. Radius of solid metallic sphere
Volume of the sphere = (1 mark)
Radius of the base of the cone =
Volume of the cone = (1 mark)
Let n be the number of cones so formed.
n 3.0625 1543.5
(1 mark)
15.A
B C
PQ I
L5 cm
4 cm 3 cm
(2 marks)Steps of construction: (1 mark)1. Construct the ABC in which BC = 5 cm, CA = 3 cm and AB = 4 cm.2. Bisectand and . Let bisectors of these angles be BP and CQ respectively.
Let these intersect at I. 3. Draw the perpendicular IL on BC.4. Taking I as centre and IL as radius, draw the circle.
This is the required incircle of the ABC. Radius IL = 1 cm 16.
A(3, –4) P(p, –2) Q B(1, 2)
1 2
5, q
3
Let AB be the line-segment. Clearly P divides AB in the ratio of 1 : 2 internally. (½ mark)
[Using section formula]
(1 mark)
Now Q is mid-point of PB (½ mark)
[Using mid-point formula]
q = 0 (1 mark)
OR
A(x , y )1 1
C(x , y )3 3B(x , y )2 2 D
1
2
Let D be the mid-point of BC,
(½ mark)
Let G (x, y) be the centroid of the triangle. Since, G divides AD in the ratio 2:1. (½ mark)
(1 mark)
and (1 mark)
Hence, the co-ordinates of the centroid of the triangle are
.
17.
A
T
C D B
AT is a tangent and TB is a chord of the circle. ATB = TCB ... (i) [Angles in the alternate segments] (½ mark) TD is the bisector of BTC.
... (ii) (½ mark)Adding (i) and (ii), we get
... (iii) (½ mark) As is the exterior angle of
... (iv) (½ mark) From (iii) and (iv), we get
Hence, is an isosceles triangle (1 marks)
18. Let the co-ordinates of point on the Y-axis is P(0, y). (½ mark) Also let A and B denote the points . Since AP = BP,
(½ mark)
(½ mark)
(½ mark) (½ mark)(½ mark)
Hence, the point on Y-axis is (0, -2). .
19. The following table gives the share of each item as a component of 360°.
Items Expenditure (in percent) Share as a component of 360°
Food Entertainment Other expenditure Savings
40 25 20 15
040360 144
100
025
360 90100
020
360 72100
015
360 54100
Total
100 360°
(1½ mark)
Pie Chart:
Food144°
Entertainment90°
Other expenditure
72°
Savings54°
(1½ mark)
Section – C
20.
hO
Q
P
Cloud
P´
R
H
y
x
(1 mark)
According to figure, the cloud is at P and the image of the cloud P is at with respect to the lake.
PR P'R Let OP = x, OQ = y and PR = P’R = H In right triangle OPQ,
...(i) (1 mark)
In right triangle ,
… (ii) (1 mark)
Subtracting (ii) and (iii), we get
… (iii) (½ mark)
Again, in right triangle OPQ,
… (iv) (½ mark)
Dividing (v) and (iv), we get
(½ mark)
(½ mark)
OR
45°
O Ax
60°
Q
P
4,0
00 m
(1 mark) Let P and Q be the positions of the two aeroplanes.In right-angled ,
tan60° =
... (i) (1½ mark)
In right-angled ,
tan45° =
[From (i)] (1½ mark)
Distance between two aeroplanes
= PQ = AP – AQ
(1 mark)21. Part – I
A
B D C
P
Q S RGiven:
To prove:
Construction: Draw and (½ mark)
Proof:
... (i) (½ mark)
In and
... (ii) (1 mark)
But
[From (ii)] ... (iii) (½ mark)
[From (i) and (iii)] (½ mark)
Hence, (½ mark)
Part – II
(½ mark)
(½ mark)
(½ mark)
22. Part – I
A
Q
B
M
P
O
S T
Given: ST is a tangent to the circle with centre O. P and Q are points on
major and minor arc AB respectively.To prove: and Construction: Draw a diameter AM and join MB. (½ mark)Proof: [Angle in a semicircle]
MAT 90
In ,... (i)
Also ... (ii) (1 mark)From (i) and (ii),
(½ mark)Also [Angle on the same segment AB]
(½ mark)AQBP is a cyclic quadrilateral.
(½ mark)
(½ mark) Hence, and
Part – II
A
B
C
S T
35°75°
Using part I,Since angles in the alternate segments are equal.
and (½ mark) and (1 mark)
ORPart – I
B
CA
D
1
2O
Given: ABCD is a cyclic quadrilateral inscribed in a circle with centre O.To prove: A + C = 180º and B + D = 180° Construction: Join AO and CO (½ mark)Proof: Arc ABC subtends angle 1 at the centre and subtends ADC at point D(Remaining part of the circle). ADC = ½ (1) … (i)Again arc ADC subtends 2 at the centre and subtends ABC at point B(Remaining part of the circle). ABC = ½ (2) ... (ii) (1½ mark)Adding (i) and (ii), we getADC + ABC = ½ (1 + 2) (½ mark) ADC + ABC = ½ (360°) = 180° (½ mark)B + D = 180°Similarly, A + C = 180° (½ mark)
Part – II
x + 1 05 y + 5
4 y – 4
2 x + 4
D
A
B
C
Applying this property on given angles of circle, we get2x + 4 + 4y – 4 = 180°and x + 10 + 5y + 5 = 180° x + 2y = 90° … (iii) and x + 5y = 165° … (iv) (½ mark)Subtracting (iii) from (i), we get3y = 75° y = 25° (½ mark)Substituting y = 25° in (iii), we getx + 2 × 25° = 90° x = 40° (½ mark)Thus, values of x and y are 40° and 25° respectively.
23. Class Mid-value xi Frequency fi fixi
0-10 5 8 4010-20 15 10 15020-30 25 9 22530-40 35 12 42040-50 45 11 495
fi = 50 fixi = 1330 (3 mark)
(1 mark)
(1 mark)
24.
24 m
12 m
21 m
5 m
16 m
O
AB
Radius of both conical and cylindrical part = r = 12 m, height of conical part = 16 m, height of cylindrical part (h) = 5 m.Lateral surface area of the tent … (i) (1 mark)
O
1 6 m
A 1 2 m B
2 2 212 16 2 144 256 400
(1 mark)Lateral surface area of tent
(1 mark)
(1 mark)
The cost of the cloth
(1 mark)
25. Annual income of Usha = Rs. 4,10,000Deduction on (P.M.’s Relief Fund + charitable society)
= Rs. 30,000 + Rs. 20,000
= Rs. 30,000 + Rs. 10, 000 = Rs. 40, 000 (1 mark)Total savings (PPF + LIC + NSCs)
= Rs. (60,000 + 4 × 4500 + 30,000)= Rs. 10,8,000
But 100% exemption for savings upto Rs. 1,00,000 (1 mark)Taxable income = Rs. (410000 – 40000 – 100000)
= Rs. 2,70,000 (1 mark)
Total income tax = Rs. [13000 + (270000 – 250000) ]
= Rs. [13000 + 6000]= Rs. 19,000 (1 mark)
Education Cess @ 2% = = Rs. 380 (½ mark)
Net tax payable = Rs. (19000 + 380) = Rs. 19,380 (½ mark)