Q1. 20% of a number is 7, what is the number?
(a) 35
(b) 40
(c) 21
(d) 28
Q2. What is the volume of the cubical can having dimensions 20 cm × 12 cm × 8 cm
(a) 1920 cm³
(b) 1800 cm³
(c) 2000 cm³
(d) 400 cm³
Q3. 3.12 × 5.8 =
(a) 18096
(b) 18.096
(c) 1809.5
(d) 2.314
Q4. How long will a sum of money take to double, if it invested at 8% p.a. simple interest ?
(a) 12 years
(b) 12.5 years
(c) 25 years
(d) 10 years
Q5. A number N744 is divisible by 11. Find the value of the smallest natural number N ?
(a) 7
(b) 2
(c) 3
(d) 8
Q6. A wire is fold and formed a circle which have area of 616 cm². Find the length of the wire.
(a) 84 cm
(b) 88 cm
(c) 80 cm
(d) 82 cm
Q7. Sum and difference of two numbers is 15 and 1 respectively. Find the numbers.
(a) 6,9
(b)8,7
(c)3,12
(d)11,4
Q8. A piece of cloth costs Rs 35. If the piece were 4 m longer and each meter were to cost Rs 1 lesser, then the total cost would remain unchanged. How long is the piece of cloth?
(a) 10 m
(b) 14 m
(c) 12 m
(d) 8 m
Q9. If P travelled the first half of a journey at 40 km/hr and the remaining distance at 50 km/hr, what is the average speed of his travel?
(a) 44.44 km/hr
(b) 53.33 km/hr
(c) 45 km/hr
(d) 60 km/hr
Q10. Find the LCM of 15, 25 and 29.
(a) 2335
(b) 3337
(c) 2175
(d) 2375
SOLUTIONS
S1. Ans.(a)
Sol.
Number→7/20×100=35
S2. Ans.(a)
Sol.
Dimension of cube → 20 cm × 12 cm × 8 cm
Volume → 1920 cm³
S3. Ans.(b)
Sol.
3.12 × 5.8 = 18.096
S4. Ans.(b)
Sol.
Let principle, and time is P and t
So,
(P×s×t)/100=P
t=100/8=12.5 years
S5. Ans.(a)
Sol.
N744 divisible by 11
By hit and trail method
N = 7
7744/11=704
S6. Ans.(b)
Sol.
Area of circle → 616 cm³
Radius → r
πr² = 616
r = 14
circumference = 2πr = 88 cm
S7. Ans.(b)
Sol.
Let numbers is x and y
x + y = 15 …(i)
x – y = 1 …(ii)
Solving (i) and (ii)
x = 8, y = 7
S8. Ans.(a)
Sol.
Let length = ℓ
Cost per meter=35/l
ATQ,
35/(l+4)=35/l–1
ℓ = 10
S9. Ans.(a)
Sol.
Average speed=(2×40×50)/((40+50) )
=400/9 km⁄hr
=44.44 km/hr
S10. Ans.(c)
Sol.
