Question

APP1/19
APP1/19
136. The sum of 2 times a number and 10 is 32, then the number is
(a) 11
(b) 7
(c) 3.2
(d) none of these
(b) -1
(d) 4
144. If m = 2 then the value of 9-5 m is
(a) 1
(C) - 4
145. 2 x 103 in the simple form is
(a) 200
(c) 20000
137
Among the angles 60°, 90°, 110°, 48° the obtuse angle is
(a) 90°
(b) 48°
(c) 110°
(d) 60°
(b) 2000
(d) 4000
146. Value of 2° + 3° is
(a) O
(d) 2
(b) 1
25
138. The measure of supplement of 125° is
(a) 100°
(C) 65°
(b) 25°
(d) 55°
ilan
ABCD is
147. The area of parallelogram 1/4 cm
o
عاه
7 cmc
139. The complement of the angle
(a) 14 sq. cm
(c) 11 sq. cm
26) 28 sq. cm
(d) none of these
is
1200
(a) 160°
(b) 20°
(c) 70°
(d) 170°
ABC is
148. The area of the triangle 4 cm
в з cm с
140.
In the figure 100
find x.
(a) 12 sq. cm
(c) 7 sq. cm
(b) 6 sq. cm
(d) none of these
(a) 80°
f
ot 100°
(c) 90°
(d) none of these
149. If the area of triangle ABC is 36 cm2 then length of base BC is
?
11001
ww
ans. 05
141. In the figure
where lines 'l' and 'm' are parallel the angle x =
>m
3 cm
(a) 110°
(b) 70°
(c) 10°
(d) 20°
142. The circumference of a circle of diameter 10 cm is (take a = 3.14)
(a) 31.4 cm
(b) 15.7 cm
(c) 250 cm
fe) none of these
323
BD C
(a) 28 cm
(b) 20 cm
(c) 24 cm
(d) none of these
150. The area of a square park whose perimeter is 320 m is
(a) 160 m
61 6400 sq. m
(c) 320 sq.m
(d) 100 sq. m
320
143.
Simplified form of 21b-32 + 7b-20b is
(a) 8b+ 32
(b) 8b - 22
(0) 76 - 32
(d) 80-32
JJ
OO
21b-206-28 6 237
- 486-37
b737
-22-
-23-
99​

Answer 1

D- none of these
Okk

Answer 2

Answer:

11

-1

2000

110 deg

2

Step-by-step explanation:

Let say number =X

2X + 10= 32

2X = 22

X = 11

9-5m

= 9-5×2

=9-10

= -1

2×10^3

=2×1000

=2000

angle greater than 90 deg are obtuse angle

so 110 deg

2° + 3°

= 1 + 1

=2

supplement of 125° is (180-125) =55 deg

The area of a square park whose perimeter is 320 m is

6400 sq m

as perimeter = 4 × side

=>side = 320/4 = 80 m

area = side^2 = 80^2 = 6400 sq m

Simplified form of 21b-32 + 7b-20b is

8b-32