Q1.
Find two rational number between 1/5 and 1/3.
(a) 1/14 and 4/10 (b) 3/10 and 2/12 (c) 4/15 and 3/10 (d) 2/15 and 5/10
Solution.
Let q1 and q2are two required rational numbers then,
q1 = 1/2(1/5 + 1/3)
= 1/2(3+5/15) = 1/2
(8/15) = 4/15
Similarly,
q2 = 1/2(4/15 + 1/3)
= 1/2 (12 + 15/45)
= 1/2 (27/45) = 1/2 (3/5) = 3/10
Hence, two rational numbers between 1/5 and 1/3 are 4/15 and 3/10.
{Formula Apply : Find rational numbers between a and b
, (a+b)/2}.
Solution
Q2.
The expresion 26n - 32n has a factor
(a)11 (b)9 (c)7 (d)6
Solution.
(23)2n - (3)2n = 82n - 32n
2n is an even integer {xn - an is
exactly divisible by x+a when n is even.}
Solution
Q3.
If 12 bananas are bought for Rs.11 and 11 bananas are sold for Rs.12. find gain or loss percent .
Solution.
C.P of 12 bananas = Rs.11
C.P of 1 banana = 11/12
S.P of 11 bananas = Rs.12
S.P of 1 bananas = 12/11
S.P > C.P
gain = 12/11 - 11/12
= (144 - 121/132) = 23/132
gain % = (23/132) x (12/11) x 100
= 2300/121
Solution
Q4.
Simplify
231 x 231 - 144 x 144
/
231 + 144
(a) 144 (b) 87 (c) 41 (d) 16
Answer.
(b) 87
Answer
Q5.
Find the H.C.F of 20x2 y9 z, 16x5 y3, 12x3 y10
(a) 4x2 y3 (b) 2xy (c) 2 (d) 1
Solution.
H.C.F of 20, 16, 12 = 4
Highest power of x common to all = x2
Highest power of y common to all = y3
z is not common so omit it
HCF is 4x2 y3
Solution
Q6.
Find an angle which is 360 more than its supplement.
(a) 980 (b) 1300 (c) 1080
(d) 1200
Solution.
let both angles are x and y
x + y = 180
x - y = 36
(c) 1080
Solution
Q7.
The area of a square is 121 cm2. If the same wire is bent in the form of circle, find the area of the circle.
(a) 170 cm2 (b) 164 cm2 (c) 154 cm2 (d) 180 cm2
Solution.
Side of a square is 11,
So perimeter of square is 44
2πr = 44
2 x 22/7 x r = 44
r = 7 cm
Area of the circle = πr2 = 154
Solution
Q8.
The edges of a cuboid are in the ratio 3:2:1 and its Total surface area is 44 cm2, Its volume is ________ cm3
(a) 12√2 (b) 12 (c) 18 (d) 10√2
Solution.
2(6x2 + 2x2 + 3x2) = 44
11x2 = 22
x2 = 2
x = √2
v = 12√2
Solution
Q9.
If tanθ + cotθ = 2, find the value of tan2θ + cot2θ.
(a) 4 (b) 1 (c) 3 (d) 2
Solution.
(tanθ + cotθ)2 = 4
tan2θ + cot2θ + 2.tanθ.cotθ = 4
tan2θ + cot2θ = 2
Solution
Q10.
If (x- 1)3 = 64, find the value of x3.
(a)115 (b) 100 (c) 152 (d) 125
Solution.
(x - 1)3 = 43
x - 1 = 4
x = 5
So, 53 = 125
Solution
