(1) ___________ is a binomial.
(a) 2x2
+ 2x + 3 (b) 2x + y + z (c) x + 5 (d) 2x
(2) The numerical coefficient of –5x2
y
4
z
3
is ___________.
(a) 2 (b) 3 (c) 4 (d) -5
(3) 5x 0 (–2x) = ___________.
(a) 502x (b) 502x2
(c) -502x2
(d) 0
(4) (–2x) (3x) (4x) = ____________.
(a) 24x3
(b) –24x3
(c) 24x (d) –24x
(5) (x + 5)2
=____________.
(a) x
2
+ 25 (b) x
2
– 25 (c) x
2
+ 10x + 25 (d) x
2
– 10x +25
(6) 2x2
(3x – 2y) = ____________.
(a) 6x3
– 4x2
y (b) 6x2
– 4x2
y (c) 6x3
+ 4x2
y (d) –12x3
y
(7) 2p2
q and ___________ are like terms.
(a) 2pq2
(b) –2pq2
(c) –5p2
q (d) 3p2
q
2
(8) In the term
, the coefficient of 'x' is ____________.
(a)
(b)
(c)
(d)
(9) (5a – 4b) – (6b – 2a) = ____________.
(a) 3a – 2b (b) 7a – 10b (c) 7a + 10b (d) 3a + 2b
(10) (3x – 1)(3x + 1) = ____________.
(a) 6x2
– 1 (b) 6x2
+ 1 (c) 9x2
– 1 (d) 9x2
+ 1 (11) √
= ___________.
(a) 6 (b) 8 (c) 12 (d) 18
(12) √
√
= ___________.
(a) 9 (b) 1 (c) 3 (d) 4
(13) √
= ___________.
(a)
(b)
(c)
(d)
(14) Square root of 324 is ___________.
(a) 18 (b) 28 (c) 14 (d) 162
(15) √ = ___________.
(a) 20 (b) 40 (c) 160 (d) 80
(16) √ = ___________.
(a) 5 (b) 15 (c) 25 (d) 35
(17) 2
2
= ___________.
(a) 4 (b) 8 (c) 6 (d) 2
(18) 3
2
= ___________.
(a) 6 (b) 8 (c) 9 (d) 27
(19) 172
= ___________.
(a) 17 (b) 34 (c) 324 (d) 289
(20) 1
2
+ 32
= ___________.
(a) 9 (b) 0 (c) 10 (d) 4
(21) 2
3
= ___________.
(a) 8 (b) 6 (c) 4 (d) 2
(22) 6
3
= ___________.
(a) 6 (b) 18 (c) 36 (d) 216
(23) 0
3
+ 13
= ___________.
(a) 0 (b) 1 (c) 3 (d) 13
(24) 2
3
– 2
2
= ___________.
(a) 2 (b) 4 (c) 0 (d) 1(25) 1 –
= ___________.
(a)
(b) (
) (c) (
) (d)
(26)
= ___________
(a) 0 (b)
(c) not defined (d)
(27) (
) (
) = ___________
(a) 1 (b) (–1) (c)
(d)
(28) The value of 7x + 1 = 4x + 7 is ___________.
(a) 2 (b) (–2) (c) 4 (d) (–4)
(29)
, then x = ___________.
(a) 6 (b) 2 (c) 12 (d) 3
(30) If 3x – 1 = 8, then x = ___________.
(a) 3 (b)
(c) –3 (d) 8
(31) If x – 3 = 7, then x = ___________.
(a) 7 (b) 10 (c) 3 (d) 4
(32) The measure of diagonals of a ___________ is equal.
(a) parallelogram (b) trapezium (c) rhombus (d) square
(33) In which quadrilateral, all angles are of 900
?
(a) parallelogram (b) kite (c) square (d) Rhombus
(34) Sum of all the exterior angle of polygon is ___________.
(a) 1800
(b) 900
(c) 3600
(d) 5200
(35) 103
= ___________.
(a) 10 (b) 100 (c) 1000 (d) 10000
Answers
Answer:
1. Use Euclid’s division algorithm to find the HCF of:
i. 135 and 225
ii. 196 and 38220
iii. 867 and 225
Solutions:
i. 135 and 225
As you can see, from the question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,
225 = 135 × 1 + 90
Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,
135 = 90 × 1 + 45
Again, 45 ≠ 0, repeating the above step for 45, we get,
90 = 45 × 2 + 0
The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.
Hence, the HCF of 225 and 135 is 45.
ii. 196 and 38220
In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,
38220 = 196 × 195 + 0
We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.
Hence, the HCF of 196 and 38220 is 196.
iii. 867 and 225
As we know, 867 is greater than 225. Let us apply now Euclid’s division algorithm on 867, to get,
867 = 225 × 3 + 102
Remainder 102 ≠ 0, therefore taking 225 as divisor and applying the division lemma method, we get,
225 = 102 × 2 + 51
Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,
102 = 51 × 2 + 0
The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,225) = HCF(225,102) = HCF(102,51) = 51.
Hence, the HCF of 867 and 225 is 51.
Step-by-step explanation:
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