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D B
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BOUDDHA MERIDIAN SCHOOL
FIRST TERMINAL MODEL QUESTION – 2016/17
Class: X Subject: C. Maths F.M.: 100
Time:3 hours P.M.: 40
Group ‘A’[9 (2 + 2) = 36] 1. a) Simplify:
b) Simplify: (1 – xa – b)–1 + (1 – xb – a)–1
2. a) Simplify:
b) Simplify: + 2– 2
3. a) Solve for x: – = 0
b) Solve for x: 23x – 5ax – 2 = 2x – 2a1 – x
4. a) Avay is 4 years younger than Mousam and sum of their ages is 66 years. Find their ages.
b) Find two positive consecutive even numbers if the sum of their
squares is 244.
5. a) The area of a equilateral triangle is 9cm2 then calculate its perimeter.
b) If the length of AC = 12 cm,
BD = 10 cm, and ABCD is a rhombus, then
find the area of ECD.
6. a) Calculate the volume of the prism according to information given in the diagram.
b) A cylindrical tank contains 4,62,000 litre of water, if its radius of the base is 3.5 m then find its height.
7. a) Find the circumference of the great circle of a
sphere whose volume is cm3.
b) Calculate volume of the cone according to information given in the adjoining diagram.
8. a) In the adjoining circle, O is the centre of a circle. If AOC = 130,find the value of x and y.
b) In the adjoining figure, O is centre of the
circle and BD is a diameter. If ACB = 40o, find the size of ABD.
b) In the given figure, ABD is a tangent at B and BFE = 25o. If BE = ED, find the size of FBA.
Group ‘B’ [16 4 = 64]
10.In a class of 60 students, 15 liked Football only, 20 liked Basketball
only and 5 did not both of the games. Represent the information in a Venn diagram and find the number of students who like both of the games.
11.In a group of student, 20 study Accounts, 21 study Mathematics, 18
study History, 7 study Accounts only, 10 study Mathematics only, 6 study Accounts and Mathematics only and 3 study Mathematics and History only. Represent the above information in a Venn diagram. Also find the total number of students and number of students who study all the subjects.
12.Find the H.C.F. of: x3 – 9x, x4 – 2x3 – 3x2&x3 – 27
13.Find the L.C.M. of: 2a3 16, a2 4a + 4 and a2 3a + 2
14. Simplify:
15.Simplify:
16.If a + b + c = 0, prove that: + + = 1
17.Solve for x:22x +3 – 9 2x + 1 = 0
18.Solve for x:= 6 –
19.A number of two digits exceed four times the sum of its digits by 3. If
the number is increased by 18, the result is equal to the number formed by interchanging its digits. Find the number.
20.The present ages of a mother and her daughter are 35 years and 12 years
respectively. Find, how many years ago the product of their ages was 210.
21.The curved surface area of a cylinder is equal to 23 ofits total surface area. If its total surface area is 924 cm2, find its volume.
22.A metallic sphere of radius 7cm is cut in to two hemispheres. Find the
change in total surface area and volume?
23.If BAC and BOC are inscribed and central angles standing on the
same arc BC, prove that BOC = 2BAC.
24.Verify experimentally the opposite angles of a cyclic quadrilateral are
supplementary. (Two figures are necessary but the radii of the circles must not be less than 3 cm)
25.In the adjoining figure, O is the centre of circle. If AB is a diameter
and AD || OC, prove that arc BC = arc CD.
[1]
[1]
Marking Schemes
Group ‘A’[9 (2 + 2) = 36] 1. a)
= [1]
= [1]
b) (1 – xa – b)–1 + (1 – xb – a)–1
= + [1]
= = 1 [1]
2. a)
= [1]
=
= [1]
b) + 2– 2
= + 2 – 2
= 4 + 6 – 2 5 [1]
= 0 [1]
3. a) – = 0
=
Cubing on both sides,
ax – b = a – bx
ax + bx = a + b [1]
x = 1
Checking & Conclusion [1]
b) 23x – 5ax – 2 = 2x – 2a1 – x Equating corresponding power,
3x – 5 = x – 2 &x – 2 = 1 – x [1]
x = &x =
Hence, x = [1]
4. a) Suppose Avay's age = 'x' years Mousam's age = 'y' years
x<y x = y – 4 ...(i)
x + y = 66 ...(ii)
Hence Ajay's age = x = 31 years
& Mousam's age = y = 35 years
b) Let us suppose required consecutive even numbers be x&x + 2.
x2 + (x + 2)2 = 244 [1]
x = – 12 (Rejected) &x = 10 When x = 10, x + 2 = 12
Hence 10 & 12 [1]
5. a) As we know that,
Area of equilateral triangle (A) = 9=
a = = 6 cm [1]
Perimeter (P) = 3a = 18 cm [1]
b) 1st diagonal of rhombus (d
1) = AC = 12 cm
2nd diagonal of rhombus (d
2) = BD = 10 cm
Area of rhombus ABCD (A )=(d1d2) [1]
= (12 10) = 60 cm2
ECD =1
2ABCD = 1
2 X 60 = 30cm2 [1]
6. a) Suppose perpendicular (p) = 8 cm
hypotenuse(h) = 10 cm
base (b) = ? By Pythagoras theorem,
h = p2 + b2
b = = 6 cm [1]
Volume (V) = (pb) h
= (8 6) 30
b) Volume of cylindrical tank (V) = 4,62,000l
= 462 m3
Radius (r) = 3.5 m Height (h) = ?
By formula, V = r2h
462 = (3.5)2h [1]
h = 14 m [1]
7. a) Volume of sphere (V) =cm3
Circumference of great circle (C) = ? By formula,
V = r3 = r3
r = = 3 cm [1]
Again,
C = 2r = 6 cm [1]
b) Radius (r) = = 7 cm
Slanting height (l) = 25 cm Volume (V) = ?
Vertical height (h) = ? As we know that,
l2 = r2 + h2
h = = 24 cm [1]
By formula,
V = r2h = 8624 cm3 [1]
8. a) i) AOC (Obtuse) + AOC (Reflex) = 360 [ ...]
AOC (Reflex) = 360 – 120 = 240
ii) x = 240 = 120 with reason [1]
iii) y = x = 120 with reason [1]
b) Join C & D, ACD = 90 with reason
i) or, BAC + ACD = 90
or, 40 + ACD = 90
ACD = 50 [1]
ii) ABD = 50 [1]
9. a) i) XOZ (Reflex) = 2XYZ
= 240 with reason [1]
ii) XOZ (Obtuse) = 360 – 240 with reason [1]
a = 120
b) i) EFB =EBD = 25 with reason
EBD = EDB = 25 [1]
ii) ABF = BEF = 25 + 25 = 50 [1]
Group ‘B’ [16 4 = 64]
10. For listing given information properly in set form [1]
n(U) = 60, no(F) = 15, no(B) = 20 , n(FB) = 5
n(U) = no(F) + no(B) + n(F∩B) + n(FB) [1]
or, 60 = 15 + 20 + n(F∩B) + 5
or, n(F∩B) = 20 [1]
Representing in a Venn-diagram [1]
11. For listing given information properly in set form [1]
n(A) = 20, n(M) = 21, n(H) = 18, no(A) = 7, no(M) = 10, no(A∩M)
= 6, no(H∩M) = 3
Drawing in Venn Diagram [1]
n(M) = no(M) + no(A∩M) + no(H∩M) + n(A∩M∩H) [1]
or, 21 = 10 + 6 + 3 + n(A∩M∩H)
or, n(A∩M∩H) = 2 [1]
12.1st expression =x3 – 9x = x (x2 – 9)
2nd expression =x4 – 2x3 – 3x2 = x2 (x2 – 2x – 3)
= x2 (x – 3) (x + 1) [1]
3rd expression = x3 – 27 = x3 – 33
= (x – 3) (x2 + 3x + 9) [1]
H.C.F. = (x – 3) [1]
13.1st expression = 2a3 – 16 = 2 (a3 – 8) = 2 (a3 – 23)
= 2 (a – 2)(a2 + 2a + 4) [1]
2nd expression = 4a2 – 4a + 4
= (a – 2) (a – 2) [1]
3rd expression = a2 – 3a + 2
= (a – 2) (a – 2) [1]
L.C.M. = 2 (a – 2)(a – 2) (a – 1)(a2 + 2a + 4)
= 2 (a – 1)(a – 2)(a3 – 8) [1]
14.
= + [1]
= + [1]
= [1]
= [1]
15.
= –
= – [1]
= – [1]
= [1]
=
= [1]
16. + + = 1
= + + [1]
= + + [1]
a + b + c = 0
a + c = – b
– b – c = + a
= + + [1]
=
= 1 [1]
17. 22x +3 – 9 2x + 1 = 0 or, 22x 23– 9 2x + 1 = 0
Let 2x = k and 22x = k2
8k2 – 9k + 1 = 0 [1]
or, 8k2 – 8k – k + 1 = 0 or, 8k (k – 1) – 1 (k – 1) = 0
or, (k – 1) (8k – 1) = 0 [1]
Either, OR
k – 1 = 0 8k – 1 = 0
Replacing 'k' by 2x
2x – 1 = 0 8k – 1 = 0
or, 2x = 1 = 20 or, k =
x = 0 [1] or, 2x = 2–3
x = – 3
Hence x = 0, -3 [1]
18. = 6 –
[1]
or, 5 = 32 –
or, 6 = 42 [1]
Squaring on both sides,
7x = 49
x = 7 [1]
Checking,
When x = 7, = 6 –
= 6 – 1
5 = 5 (True)
Hence, x = 7 [1]
19. 16x + y = 4 (x + y) + 3...(i) [1]
10x + y + 18 = 10y + x...(ii) [1]
Solving equations (i) & (ii),
x = 3 [1]
and y = 3 & required number 10x + y = 35 [1]
20.Let us suppose the product of their ages 210 before 'x' years.
(35 – x)(12 – x) = 210 [1]
x2 – 47x + 210 = 0 [1]
(x – 42)(x – 5) = 0 [1]
Either,
x – 42 = 0 x = 42 (Rejected)
x – 5 = 0 x = 5 [1]
21.TSA = 924 cm2
CSA = 23 of 924 = 616 cm2 [1]
TSA = CSA + 2 . A or, 924 = 616 + 2A
or, 2A = 308
or, A = 154
or, πr2 = 154 or, r2 = 49
or, r = 7 cm [1]
CSA = 2πrh Or, 616 = 2π . 7h
Or, h = 14cm [1]
Volume = πr2h = π. 7.7.14 = 2156 cm3 [1]
22.Volume of Sphere = 4
3πr3 = 4
3π 7.7.7 = 1437.33 cm3 [1]
TSA of Sphere = 4πr2 = 4π7.7 = 616 cm2 [1]
TSA of 1 Hemisphere = 3πr2 = 3π7.7 = 462 cm2
TSA of 2 Hemispheres = 2.462 = 924 cm2 [1]
Change in TSA = 924 – 616 = 308cm2
Hence, the TSA increases by 308cm2 while volume remains same as
same object is made into two halves. [1]
23.Figure + Given + To prove [1]
Construction [1]
i) BOC = BC
& BAC = BC [1]
ii) BAC = BOC [1]
24.For correct figure [1]
Table with correct measurement [1 + 1]
For correct conclusion [1]
25.Given + To prove + Construction [1]
i) DAC = ACO
ii) ACO = CAB
iv) BC=AC [1]
![Table with correct measurement [1 + 1]](https://thumb-us.123doks.com/thumbv2/123dok_us/251947.2020357/6.1262.49.612.55.825/table-with-correct-measurement.webp)
