Q1.
A sum on compound interest becomes three times in 4 years then with the same interest rate , the sum will become 27 time in ________
years.
(a) 12 (b) 10 (c) 8 (d) 15
Solution.
(x1)1/n1 = (x2)1/n2
(3)1/4 = (27)1/n2
(3)1/4 = (3)3/n2
1/4 = 3/x2
x2 = 12
Solution
Q2.
Mr A invested Rs. 3500 for 6 years Rs. 4000 for 4 years at the same rate of simple interest. He received a total simple interest
of Rs 2700. Find the rate.
(a) 7.3% (b) 8% (c) 8.2% (d) 9%
Solution.
SI =
3500 x 6 x R/100
+
4000 x 4 x R/100
2700 = 210R + 160R
2700 = 370R
R =
2700/370
= 7.3%
Solution
Q3.
The point p(x,y) lies in the 2nd quadrant, which is greater x or y.
(a) y (b) x (c) x=y (d) None
Solution.
(a) y
Note : ordinate (y) is positive and abscissa is negative .
Solution
Q4.
In the given figure, If x is greater than y by (1/10)th of a right angle . find the value of x.
(a) 950 (b) 850 (c) 650 (d) 750
Solution.
x = 1/9 x 900 + y
x = 100 + y
x + y = 180
100 + y + y = 180
2y = 170
y = 850
So, x = 950
Solution
Q5.
In the given figure, a ball is thrown and lands in the interior of the circle what is the probability that the ball will land in the unshaded region
(a) 19/14π (b) 0 (c) 1 (d) 48/25π
Solution.
2r = AC = 5
r = 5/2
Area of the circle = π25/4 = 25π/4 cm
2 Area of the rectangle = 12 cm
2
P(Ball land in the unshaded part) =
Area of unshaded part /Whole area (Area of the circle)
12 /25π/4
48 /25π
Solution
Q6.
(i) Circles having the same centre are called concentric circles.
(ii) The circles having same radii are congruent.
which statement is right .
(a) only (i) (b) only (ii) (c) both (d) none of these
Solution.
(c) both
Solution
Q7.
An arc of a circle is of length 10 π cm and the sector it bounds has an area of 40 π cm2. Find the radius of the circle.
(a) 5 cm (b) 40 cm (c) 10 cm (d) 8 cm
Solution.
θ/360o x 2 π r / θ/360o x π r2
=
10 π / 40 π
2 / r
=
1 / 4
r = 8 cm
Solution
Q8.
In the triangle , the side opposite to the larger angle is longer.
(a) True (b) False (c) None of these
Solution.
(a) True
Solution
Q9.
Factorize x4 + 6x2 + 5
(a) (x2 + 3 - 2x) (b) (x2 + 1) (x2 + 5)
(c) (x2 + 1) (d) (x2 + 6)
Solution.
(x2 + 3)2
= x4 + 6x2 + 9 - 4
= (x2 + 3)2 - (2)2
(x2 + 3 - 2) (x2 + 3 + 2)
= (x2 + 1)(x2 + 5)
Solution
Q10.
If (x- 1)3 = 64, find the value of x.
Solution.
(x - 1)3 = 43
x - 1 = 4
x = 5
So, 53 = 125
Solution
