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# Maths Quiz for KVS and NVS Exams

Q1. Diameter of a wheel is 3 m. The wheel revolves 28 times in a minute. To cover 5,280 km distance, the wheel will take (take π=22/7):
(a) 10 minute
(b) 20 minute
(c) 30 minute
(d) 40 minute

Q2. The difference of the areas of two squares drawn on 2 line segments of different lengths is 32 cm2. Find the length of the greater line segment, if one is longer than the other by 2 cm.
(a) 7 cm
(b) 9 cm
(c) 11 cm
(d) 16 cm

Q3. The perimeters of two squares are 40 cm and 32 cm. The perimeter of a third square whose area is, equal to the difference of the areas of the first two square is:
(a) 24 cm
(b) 42 cm
(c) 40 cm
(d) 20 cm

Q4. Four equal-sized maximum circular plates are cut off from a square paper sheet of area 784 cm2. The circumference of each plate is:
(a) 22 cm
(b) 44 cm
(c) 66 cm
(d) 88 cm
Q5. The area of a square is 196 cm2. Whose side is half the radius of a circle. The circumference of the circle is equal to breadth of a rectangle, if perimeter of the rectangle is 712 cm. What is the length of the rectangle?
(a) 196 cm2
(b) 186 cm2
(c) 180 cm2
(d) 190 cm2

Q6. A lawn is in the form of a rectangle having its breadth and length in the ratio 3 : 4. The area of the lawn is 1/12 hectare. The breadth of the lawn is:
(a) 25 metres
(b) 50 metres
(c) 75 metres
(d) 100 meters

Q7. The diameter of a circular wheel is 7 m. How many revolutions will it make in travelling 22 km?
(a) 100
(b) 400
(c) 500
(d) 1000
Q8. The area of the circumcircle of an equilateral triangle is 3π cm^2. The perimeter of the triangle is:
(a) 3√3 cm
(b) 9 cm
(c) 18 cm
(d) 3 cm

Q9. A playground is in the shape of a rectangle, A sum of Rs. 1000 was spent to make the ground usable at the rate of 25 paisa per m2. The breadth of the ground is 50 m. If the length of the ground is increased by 20 m, what will be the expenditure in rupees at the same rate per m2?
(a) 1,250
(b) 1,000
(c) 1,500
(d) 2,250

Q10. A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. The area of the circle is [Take π=22/7]
(a) 125 cm2
(b) 230 cm2
(c) 550 cm2
(d) 616 cm2
Solutions

S1. Ans.(b)
Sol. Circumference of wheel = 2?r
=2×22/7×3/2=66/7  m
Wheel will take =(5280 × 7)/(28 × 66)=20 m

S2. Ans.(b)
Sol. Let, the side of one of the squares be x cm. Then, side of longer square = (x + 2) cm
Now, ((x+2)^2)-(x^2)=32
Or, (x^2)+4x+4-(x^2)=32
Or, 4x = 28
Or, x = 7 cm
Length of the side of the longer square = 7 + 2 = 9 cm.

S3. Ans.(a)
Sol. Side of the two squares are 40/4=10 cm and 32/4 = 8 cm
Let, the side of the third square be ‘x’ cm
Then, (x^2)=(10^2)-(8^2)=100-64=36
Or, x = 6 cm
∴ Perimeter of the third square = 6 × 4 = 24 cm.

S4. Ans.(b)
Sol. Side of square =√784=28 cm
Radius of each plate = 7 cm
∴ Circumference = 2×22/7×7 = 44 cm.

S5. Ans.(c)
Sol. Area of square (a^2)=196
∴a=√196=14 cm
Radius of a circle = 14 × 2 = 28 cm
∴ Circumference =22/7×2×28=176 cm
Now according to the question b = 176 cm
Also, 2(l + b) = 712
2(l + 176) = 712
l + 176 = 356
∴ l = 356 – 176
∴ l = 180 cm2

S6. Ans.(a)
Sol. 1/12 hectare =1/12×10000 m^2
=2500/3 m^2
∴3x×4x=2500/3
⇒ (x^2)=2500/(3 × 3 × 4)⇒x=50/6
⇒ Width =3x=(3×50/6)=25 m
S7. Ans.(d)
Sol. Distance covered by the wheel in the one revolution =πd=22/7×7= 22 m
∴ Number of revolutions =(22 × 1000)/22=1000

S8. Ans.(b)
∴ Area of circumcircle =π×(Side^2)/3=3π
⇒ (Side^2)=9⇒ Side = 3 cm
∴ Perimeter of the triangle = 3 × 3 = 9 cm

S9. Ans.(a)
Sol. Area of rectangular filed =1000/(1/4)
=4000 m^2
∴ Length =(4000/50) = 80 m
New length of field = (80 + 20) = 100 m
Area = (100 × 50) = 5000 m^2
∴ Required expenditure
= Rs. (5000×1/4) = Rs. 1250
S10. Ans.(d)
Sol. 2?r = 2(18 + 26)
⇒ 2×22/7×r=44×2
⇒ r = 14 cm
∴ Area of circle =πr^2
=22/7×14×14=616 cm^2