Q1. The distance between two places is 12 km. A map scale is 1 : 25000. The distance between the two places on the map, (in cm) is
(a) 24
(b) 36
(c) 48
(d) 60
Sol.
Required difference = (12 km)/25000=1200000/25000
Sol.
Required difference = (12 km)/25000=1200000/25000
= 48 cm
Q2. When half of a number is increased by 15, the result is 39. The sum of digits of the original number is
(a) 6
(b) 7
(c) 9
(d) 12
Sol.
Let number be X
Let number be X
According to the question,
x/2+15=39⇒x=48
∴ Required sum = 4 + 8 = 12
Q3. If for ∆ABC and ∆DEF the correspondence CAB↔EDF gives a congruence, then which of the following is not true?
(a) AC=DE
(b) AB=EF
(c) ∠A=∠D
(d) ∠B=∠F
Sol.
In, ∆ABC and ∆DEF, AB=DF
Q4. If 2/3 x=0.6 and 0.02 y=1, then the value of x+y^(-1) is
(a) 0.92
(b) 1.1
(c) 49.1
(d) 50.9
Sol.
x=0.6×3/2=0.9 and y=1/0.02⇒y^(-1)=0.02
x=0.6×3/2=0.9 and y=1/0.02⇒y^(-1)=0.02
Now, x+y^(-1)=0.9+0.02=0.92
Q5. A square and a circle have equal perimeters. The ratio of the area of the square to the area of the circle is
(a) 1 : 1
(b) 1 : 4
(c) π : 2
(d) π : 4
Sol.
Let side of the square be a.
Let side of the square be a.
Radius of the circle be r.
According to the question.
4 a=2πr⇒2a=πr⇒a=πr/2
∴ Required ratio = a^2/(πr^2 )=(πr⁄2)^2/(πr^2 )
=(π^2 r^2)/(4×πr^2 )=π/4=π:4
Q6. ABCD is a square with AB = (x+16) cm and BC=(3x) cm. The perimeter (〖in cm〗^2 ) is
(a) 16
(b) 24
(c) 32
(d) 96
Sol.
According to the question,
According to the question,
AB=BC,x+16=3x,2x=16
⇒x=8 cm
AB=x+16=8+16=24 cm
∴ Perimeter of ABCD =4×24=96 sq cm
Q7. The mean of 10 numbers is 0. If 72 and -12 are included in these numbers, the new mean will be
(a) 0
(b) 5
(c) 6
(d) 60
Sol.
New mean =(10 × 0 + 72 – 12)/(10 + 2)=5
New mean =(10 × 0 + 72 – 12)/(10 + 2)=5
Q8. The circumference of the base of a right circular cylinder is 44 cm and its height is 15 cm. The volume (〖in cm〗^3 ) of the cylinder is (use π=22/7)
(a) 770
(b) 1155
(c) 1540
(d) 2310
Sol.
Radius of cylinder =44/2π=44/(2 × 22/7)=7 cm
Radius of cylinder =44/2π=44/(2 × 22/7)=7 cm
∴ Volume of cylinder
=πr^2 h=22/7×7×7×15=2310 〖cm〗^3
Q9. If p=3^2000+3^(-2000) and q=3^2000-3^(-2000), then the value of p^2-q^2 is
(a) 1
(b) 2
(c) 3
(d) 4
(d) 4
Sol.
p^2-q^2=(p+q)(p-q)
p^2-q^2=(p+q)(p-q)
=[(3^2000+3^(-2000) )+(3^2000-3^(-2000) )]×[(3^2000+3^(-2000) )-(3^2000-3^(-2000) )]
=2×3^2000×2×3^(-2000)=4
Q10. The value of √(16√(8√4) ) is
(a) 16
(b) 8
(c) 8∛2
(d) 16√2
Sol.
√(16√(8√4) ) =√(16√(8×2)) =√(16×4)=√64=8
