BOUDDHA MERIDIAN
SCHOOL
FIRST TERMINAL MODEL QUESTION – 2016/17
Class: VII Subject: Maths F.M.: 100
Time: 3 hours P.M.: 40
Attempt ALL the questions. Group ‘A’ [40 × 1 = 40]
1. Write 'True' or 'False' for the following statements: a) The set of even number less than 5 is unit set.
b) If A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}, A and B are overlapping sets.
c) The even prime number is 2.
d) The possible factors of 6 are 1, 2, 3 and 6.
e) H.C.F. of 12 and 60 is 60.
f) The square of 13 is 169.
g) The sum of –3x and –5x is –8x.
h) The value of a is 1.
i) The supplement of 0 is 180.
j) If a and b are in linear pair, the value of a + b is 30.
2. Fill in the blanks with correct answers:
a) The cardinal number of null set is _________.
b) If P = {3, 6, 9} and Q = {2, 3, 4}, then P Q is _________.
c) The value of 3 in binary number system is _________.
d) L.C.M. of 9 and 11 is _________.
e) The square of 40 _________.
f) The cube root of 64 is _________.
g) The product of 6xy2 and x2y2 is _________.
h) The square form of m2 + 2mn + n2 is _________.
i) The quotient of 59 54 is _________,
j) If x and 70 are vertically opposite angles, then value of x is _________.
3. Choose the correct answers:
a) If A = {1, 2, 3}, A is a _________ set.
i) unit ii) empty iii) finite iv) infinite
b) The symbol {} represents _________ set.
i) finite ii) null iii) infinite iv) unit
c) The prime factors of 6 are __________.
i) 2 & 3 ii) 1, 2, 3 & 6 iii) 0 & 6 iv) none of above
d) L.C.M. of 2, 3 and 5 is _______.
i) 30 ii) 2 iii) 0 iv) 1
e) The value of is _______.
i) 800 ii) 80 iii) 40 iv) 400
f) The value of is __________.
i) 12 ii) 35 iii) 2 iv) 3
g) The exponential form of 81 is ________.
i) 33 ii) 32 iii) 34 iv) 35
h) The square form of x2 – 6x + 9 is _________.
i) (x – 3)2 ii) (x + 3)2 iii) (x + 9)2 iv) (x – 6)2
i) The product of –3x2 –2x3 –x is __________.
i) 6x6 ii) 6x5 iii) –6x6 iv) 6x5
j) Each angle of an equilateral triangle measures _________.
i) 50 ii) 90 iii) 60 iv) 180
4. Match the following:
Group 'A' Group 'B'
a) i) 1112
b) A B ii) 1
c) 7 iii) 20
d) (32)2 iv) A is a proper subset of B
e) v) Ten billions
f) 1010 vi) 5 102
g) 500 vii) Reflex angle
h) px – x viii)60
i) 300 ix) A B
j) x) 1024
Group ‘B’ [10 × 2 = 20]
5. a) If M = {1, 2, 3, 4, 5, 6, 7, 8} and N = {2, 4, 6, 8, 10}, then find M N with Venn-diagram.
b) Write the possible subsets of the set {1, 2, 3}.
6. a) Convert 97 into binary number system.
b) Find the H.C.F. of 24, 30 and 36.
7. a) Find the cube root of 512.
b) A pair of supplementary angles are in the ratio 3 : 2, find them.
8. a) What should be added to 2pq – 3qr + 6 to get 9pq + 4qr – 3?
b) Simplify:
9. a) Find the value of x, 2x and 3x.
b) Find the value of x, y and z.
Group ‘C’ [10 × 4 = 40] 10.Rewrite these sets in set builder method:
a) A = {0, 1, 2, 3, 4, 5} b) B = {a, e, i, o, u}
11.If M = {1, 2, 3, 4}, N = {3, 4, 5, 6} and U = {1, 2, ..., 10}, list the elements of the following set operations and illustrate each of them in Venn-diagrams:
a) M N b) N – M
12.Find the greatest number of children to whom 32 oranges, 48 bananas and 80 mangoes can be distributed equally. Also, find the shares of each fruit among them.
13.Find the square root of 1225 by division method.
14.What is the smallest number by which 675 multiplied, so that the product is a perfect cube?
15.Find the product of (x – 2) (2x2 + 3x – 1).
16.If = 6, find the value of m3 – .
17.Find the quotient (x2 – x – 20) (x – 5).
18.In the adjoining figure, find the sizes of unknown angles:
19.Find the sizes of unknown angles in the given figure:
[1]
[1]
[2]
[1]
[1]
Marking Schemes
Class: VII Subject: Maths F.M.: 100
Time: 3 hours P.M.: 40
Group ‘A’
1. a) F b) T c) T d) T e) F
f) T g) T h) T i) T j) F
2. a) 0 b) {3} c) 112 d) 99 e) 1600
f) 4 g) 6x3y4 h) (m + n)2 i) 55 j) 70
3. a) iii) b) ii) c) i) d) i) e) ii)
f) ii) g) iii) h) i) i) iii) j) iii)
4. a) ix) b) iv) c) i) d) x) e) iii)
f) v) g) vi) h) ii) i) vii) j) viii)
Group ‘B’ [10 × 2 = 20]
5. a) M N = {1, 2, 3, 4, 5, 6, 7, 8, 10} [1]
For diagram [1] b) The possible factors are:
{1, 2}, {1, 3}, {2, 3}, {}, {1, 2, 3}
6. a) 2 97 2 48 1 2 24 0 2 12 0
2 6 0 [1]
2 3 0 2 1 1 0 1
97 = 11000012 [1] b) 24 = 2 2 2 3
30 = 2 3 5 [1]
36 = 2 3 3 2
H.C.F. = 2 3 = 6 [1]
7. a) 2 512
2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1
512 = 2 2 2 2 2 2 2 2 2
= [1]
= 2 2 2 = 8
b) Let 3x and 2x be the supplementary angles. or, 3x + 2x = 180
or, 5x = 180
or, x = or, x = 36
2x = 2 36 = 72
3x = 3 36 = 108
8. a) Here, the required expression to be subtracted is: 2pq – 3qr + 6
9pq + 2qr – 3 (–) (–) (+) – 7pq – 7qr + 9
b)
= xab – ca xbc – ab xca – bc
= = x0
[1]
[1]
[1]
[1]
[1]
[2]
[2]
[2]
= 1
9. a) We know,
2x + 3x + x = 180 [Angles at a point of straight line] or, 6x = 180
or, x =
x = 30
2x = 2 30 = 60 3x = 3 30 = 90
b) We know,
z = 155 [V.O.A.] [0.5]
Again,
or, x + 150 = 180 [Straight angle] or, x = 180 – 150
or, x = 30
x = y = 30 [V.O.A.] [0.5]
3x = 3 30 = 90
Group ‘C’ [10 × 4 = 40]
10. a) A = {x : x is a whole number less than 6} [2] b) B = {x : x is an English vowel} [2]
11.M N = {1, 2, 3, 4, 5, 6} [1]
Diagram [1]
N – M = {5, 6} [1]
Diagram [1]
12.Here, the required greatest number of children is the H.C.F. of 60, 90 and 120.
2 60 3 90 2 120
3 30 3 30 2 60
2 10 2 10 3 30
5 5 5 5 2 10
1 1 5 5
1 60 = 2 2 3 5
90 = 2 3 3 5 120 = 2 2 2 3 5
H.C.F. = 2 3 5 = 30
Again, to find the shares of each fruit, 60 30 = 2
90 30 = 3 120 30 = 4
The required greatest number of children is 30.
Each student shares 2 bananas, 3 apples, and 4 mangoes.
13.Let us find the prime factors of 1225. 5 1225
5 245 7 49 7 7
1
1225 = 5 5 7 7
= 5 7 [2]
= 35
14.
[1]
[1.5]
[1.5]
Ans.
Ans.
3 9 3 3 1
6755 = 3 3 3 5 5 = 33 52
5 is the required smallest number. [1]
15. (x – 2) (2x2 + 3x – 1)
= x (2x2 + 3x – 1) – 2 (2x2 + 3x – 1) [1] = 2x3 + 3x2 – x – 4x2 – 6x + 2 [2] = 2x3 – x2 – 7x + 2 [1]
16.We have, m3 – = + 3 m = 63 + 3 6 = 216 + 18 = 234
17.We have given, (x2 – x – 20) (x – 5) x – 5 ) x2 – x – 20 ( x + 4
x2 – 5x (–) (+)
+ 4x – 20 + 4x – 20 (–) (+)
x + 2 Ans.
18.We have given,
y = 80 [V.O.A.] [1]
Again, z + 125 = 180 [Angles at a point of straight line] or, z = 180 – 125
or, z = 55
Again, x + 120 = 180 [Straight angles] or, x = 180 – 120
or, x = 60 x = 60
y = 60
z = 55
19.We have given, w = 110 [V.O.A.]
x = w = 110 [Being alternate angles] y = x = 110 [Being V.O.A.]
y + z = 180 [Sum of pair of co-interior] or, 110 + z = 180
or, z = 180 – 110 = 70 x = 110
y = 110
z = 70
***THE END***
