Q1.
If a:b = 4:3, and a:c = 5:6 . find b:c.
(a)5/8 (b)4/3 (c)2/3 (d)6/5
Solution.
b/c = (b/a) x (a/c)
= 3/4 x 5/6
b/c = 5/8
Solution
Q2.
Product of (21).(22)(23) is divisible by which set of numbers.
(a)6,3 (b)4,8 (c)4,5 (d)11,8
Solution.
(6,3)
Note: we know that,the product of three consecutive positive integer is divisible by 6.
Solution
Q3.
In a triangle ABC ED ∥ CB and
AD/DB
= 5/7
. AE = 5 cm find EC.
(a) 8 (b) 9 (c) 6 (d) 7
Solution.
By thale's theorem
let EC = x
AD/DB
=
AE/EC
5/7
=
5/x
x = 7 cm
Solution
Q4.
Find the area of triangle which is formed by points A(0,0), B(0,4) and C(6,0) .
(a) 12 unit (b) 24 unit (c) 6 unit (d) 8 unit
Solution.
(a) 12 unit
Area of triangle = 1/2 x base x height
Solution
Q5.
The mean of x and y is 8, and mean of x, y and z is 10, find the value of z.
(a) 18 (b) 14 (c) 20 (d) 16
Solution.
(x+y)/2 = 8
x + y = 16
(x + y + z)/3 = 10
x + y + z = 30
z= 14
Solution
Q6.
In a right-angled triangle sum of two perpendicular sides is 7 cm and sum of square of these perpendicular side is 25.The area of triangle is __________ cm2 .
(a) 8 cm2 (b) 12 cm2 (c) 4 cm2 (d) 6 cm2
Solution.
x + y = 7
x 2 + y 2 = 25
(x + y)2 = x2 + y2 + 2xy
49 = 25 + 2xy
2xy = 24
xy = 12
Area of triangle = 1/2xy = 6 cm2
Solution
Q7.
Find the volume of a cone having radius of the base as 30 cm and its slant height as 50 cm
(use π = 3.14)
(a) 49160 cm3 (b) 9980 cm3 (c) 1100 cm3 (d) 37680cm3
Solution.
r = 30 cm
l = 50 cm
h = √l2 - r2 = 40 cm
Volume of cone is = 1/3πr2h
= 37680cm3
Solution
Q8.
Total surface area of a cuboid is 65 cm2 and its lateral surface area is 35cm2. find the area of its base?
(a) 18cm2 (b) 25cm2 (c) 15cm2 (d) 20cm2
Solution.
Total surface area of cuboid = 2(Area of base) + lateral surface area .
65 = 2(Area of base) + 35
Area of base = 15 cm2
Solution
Q9.
If the 8th term of an A.P. is 31 and the 15th term is 16 more than the 11th term, find a and d.
(a) 4,7 (b) 3,4 (c) 3,2 (d) 3,1
Solution.
T8 = 31 and T15 = 16 + T11
a + 14d = 16 + a + 10d
d = 4
a + 7d = 31
a + 7 x 4 = 31
a = 31 - 28 = 3
Solution
Q10.
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is.
(a) 5n (b) 2n2
(c) 4n + 3 (d) 4n + 6
Solution.
Tn = Sn - Sn-1
Tn = 2n2 + 5n - {2(n - 1)2 + 5(n-1)}
T n = 2n2 + 5n - {2n2 + 2 - 4n + 5n - 5}
Tn = 2n2 + 5n - 2n2
+ 3 - n
Tn = 4n + 3
Solution
