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Mathematics Miscellaneous Quiz

Q1. The area of a rectangular football field is 24200 sq. m. It is half as broad as it is long. What is

the approx. minimum distance a man will cover if he wishes to go from one corner to the opposite

one?

(a) 283 m

(b) 246 m

(c) 576 m

(d) 289 m

Q2. A ladder is resting with one end in contact with the top of a wall of height 60 m and the other

end on the ground is at a distance of 11 m from the wall. The length of the ladder is :

(a) 61 m

(b) 71 m

(c) 87 m

(d) None of these

Q3. The area of a minor sector subtending the central angle at the centre 𝟒𝟎° is 8.25 𝐜𝐦𝟐. What is

the area of the remaining part (i.e., major sector) of the circle?

(a) 82.5 cm2

(b) 74.25 cm2

(c) 66 cm2

(d) None of these

Q4. If a piece of wire 25 cm long is bent into an arc of a circle

subtending an angle of 75° at the centre, then the radius of the

circle (in cm) is:

(a) π

120

(b) 60

π

(c) 60 π

(d) None of these

Q5. If the circumference of a circle is 4.4 m, then the area of the

circle (in m²) is :

(a) 49/π

(b) 49π

(c) 4.9π

(d) None of these

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Q6. A circular wire of radius 4.2 m is cut and bent in the form of a rectangle whose longer side is

20% more than its shorter side. The longer side of the rectangle is :

(a) 7.2 m

(b) 72 cm

(c) 8 m

(d) None of these

Q7. The difference between the circumference and the diameter of the circle is 15 m. What is the

area of the circle?

(a) 225 m2

(b) 165 m2

(c) 156 m2

(d) None of these

Q8. If the driving wheel of a bicycle makes 560 revolutions in travelling 1.1 km. Find the diameter

of the wheel:

(a) 31.5 cm

(b) 30.5 cm

(c) 62.5 cm

(d) None of these

Q9. A 5m wide lawn is cultivated all along the outside of a rectangular plot measuring 90m × 40m.

The total area of the lawn is :

(a) 1441 m2

(b) 1400 m2

(c) 2600 m2

(d) 420 m2

Q10. A room is half as long again as it is wide. The cost of carpeting it at 62 paise per square metre

is Rs. 2916.48. Find the cost of white washing the ceiling at 30 paise per metre :

(a) Rs. 2211.5

(b) Rs. 1114.2

(c) Rs. 1411.2

(d) can’t be determined

Q11. Find the length of the wire required to fence a square field

6 times having its area 5 hectares:

(a) 2400√3m

(b) 1600 m

(c) 2400√5 m

(d) 2400 m

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Q12. Find the cost of paving a courtyard 316.8 m × 65 m with stones measuring 1.3 m × 1.1 m at

Rs. 0.5 per stone :

(a) Rs. 1440

(b) Rs. 7200

(c) Rs. 72,000

(d) None of these

Q13. The perimeters of two squares are 40 cm and 32 cm. The perimeter of a third square whose

area is the difference of the areas of the two squares is

(a) 24 cm

(b) 42 cm

(c) 40 cm

(d) 20 cm

Q14. If the diagonals of two squares are in the ratio of 2: 5, their areas will be in the ratio of

(a) √2 ∶ √5

(b) 2 ∶ 5

(c) 4 : 25

(d) 4 : 5

Q15. The diagonal of a square A is (𝐚 + 𝐛). The diagonal of a square whose area is twice the area of

square A, is

(a) 2(a + b)

(b) 2(a + b)2

(c) √2(a + b)

(d) √2(a − b)

Q16. The length and perimeter of a rectangle are in the ratio 5 : 18. Then, length and breadth will

be in the ratio

(a) 4 : 3

(b) 3 : 5

(c) 5 : 4

(d) 4 : 7

Q17. Through each vertex of a triangle, a line parallel to the

opposite side is drawn. The ratio of the perimeter of new

triangle, thus formed, with that of the original triangle is

(a) 3 : 2

(b) 4 : 1

(c) 2 : 1

(d) 2 : 3

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Q18. If the perimeter of a right angled isosceles triangle is (𝟒√𝟐 + 𝟒) cm, the length of the

hypotenuse is

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 10 cm

Q19. The base and altitude of a right angled triangle are 12 cm and 5 cm, respectively. The

perpendicular distance of its hypotenuse from the opposite vertex is

(a) 44

13 cm

(b) 48

13 cm

(c) 5 cm

(d) 7 cm

Q20. From a point in the interior of an equilateral triangle the perpendicular distances of the

sides are √𝟑 cm, 𝟐√𝟑 cm and 𝟓√𝟑 cm. The perimeter (in cm) of the triangle is

(a) 64

(b) 32

(c) 48

(d) 24

Q21. Three circles P, Q and R touch each other as shown below. The radius of each of the circle P

and Q is (√𝟐 + 𝟏) cm, while that of the smaller circle is 1 cm. The perimeter of the shaded region is

(a)

π

4(2√2 − 1) cm

(b) π

2(2 + √2) cm

(c) π

2(2 + √1) cm

(d) π

4(2 − 1) cm

Q22. In a triangle ABC, AB + BC = 12 cm, BC + CA = 14 cm and CA

+ AB = 18 cm. Find the radius of the circle (in cm) which has the

same perimeter as the triangle.

(a) 5

2

(b) 7

2

(c) 9

2

(d) 11

2

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Q23. A circle is inscribed in a square whose length of the diagonal is 12√𝟐 cm. An equilateral

triangle is inscribed in that circle. The length of the side of the triangle is

(a) 4√3 cm

(b) 8√3 cm

(c) 6√3 cm

(d) 11√3 cm

Q24. The ratio of the areas of a regular hexagon and an equilateral triangle having same

perimeter is

(a) 2 : 3

(b) 6 : 1

(c) 3 : 2

(d) 1 : 6

Q25. Three circles of diameter 10 cm each, are bound together by a rubber band, as shown in the

figure. The length of the rubber band (in cm), if it is stretched as shown, is

(a) 30

(b) 30 + 10π

(c) 10π

(d) 77

Q26. A fraction becomes 𝟏

𝟔 when 4 is subtracted from its numerator and 1 is added to its

denominator. If 2 and 1 are respectively added to its numerator and the denominator, it becomes 𝟏

𝟑. Then, the LCM of the numerator and denominator of the fraction, must be

(a) 14

(b) 350

(c) 5

(d) 70

Q27. Five bells begin to toll together and toll respectively at

intervals of 6, 7, 8, 9 and 12 seconds. After how many seconds

will they toll together again ?

(a) 204

(b) 504

(c) 304

(d) 514

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Q28. Let x be the least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each

case but when divided by 9 leaves remainder 0. The sum of digits of x is

(a) 24

(b) 21

(c) 22

(d) 18

Q29. A number between 1000 and 2000 which when divided by 30, 36 and 80 gives a remainder

11 in each case is

(a) 1152

(b) 1451

(c) 1641

(d) 1712

Q30. A General of an Army wants to create a formation of square from 36562 army men. After

arrangement, he found some army men remained unused. Find the number of men unused? (a) 36

(b) 65

(c) 81

(d) 97

Solutions

S1. Ans.(b)

Sol. l ∶ b = 2 ∶ 1

2x2 = 24200

Diagonal = √(110)2 + (220)2 = 110√12 + 22

= 110 × √5 = 110 × 2.236

= 246 m

S2. Ans.(a)

Sol. Using Pythagoras theorem –

x2 = (60)2 + (11)2

x = 61 m

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S3. Ans.(c)

Sol. ATQ,

Angle =40

360=

1

9

So, 360−40

360=

8

9

Area of major sector = 8 × 8.25 = 66 cm2

S4. Ans.(b)

Sol. ATQ-

2π × R ×75

360= 25

R =60

π

S5. Ans.(d)

Sol. ATQ,

2πr = 4.4 ⇒ r =7

10m

πr2 = 0.49 πm2

S6. Ans.(a)

Sol. ATQ,

2πr = 2 ×22

7× 4.2 = 26.4 meter

26.4 = 2(6x + 5x)

6x = 7.2 m

S7. Ans.(d)

Sol. ATQ,

(2πr − 2r) = 15

2r (22

7− 1) = 15 ⇒ r =

7

2

πr2 =22

7×

7

2×

7

2= 38.5 m2

S8. Ans.(c)

Sol. ATQ.

πD × 560 = 1.1 × 1000

= πD × 560 = 1100

∴ D =110 × 7

56 × 22

=5

8m ⇒

5 × 100

8⇒ 62.5 cm

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S9. Ans.(b)

Sol. Total area of the lawn = (90 + 40 + 10) × 10 = 1400

S10. Ans.(c)

Sol. Area of floor = Area of roof

and from options, 30 almost 50% of 62

so you need not to solve just look out for the appropriate option this 1411.2 is almost 50% of 2916.48

and rest of the options are not satisfactory.

S11. Ans.(c)

Sol. 5 hectares = 50000 m2

so, area = a2 = 50000

a =100√5 m

so, perimeter = 400√5

required wire = 400√5 × 6=2400√5 m

S12. Ans.(b)

Sol. No. of stones =Area of courtyard

Area of one stone=

316.8m×65m

1.3m×1.1m= 14400

Cost = Rate × number of stones

= (0.5 × 14400) = 7200

S13. Ans.(a)

Sol. Let sides of squares are a, b and c.

Given a =40

4= 10 cm

b = 32

4= 8 cm

ATQ —

c2 = (10)2 − (8)2

= 36

∴ c = 6 cm

So perimeter a third square = 6 × 4 = 24 cm

S14. Ans.(c)

Sol. Ratio of areas = [diagonal1

diagonal2]

2

= (2

5)

2

⇒4

25

S15. Ans.(c)

Sol. ATQ

2 [1

2× (a + b)2] =

1

2× (d)2

d2 = 2(a + b)2

d = √2(a + b)

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S16. Ans.(c)

Sol. Length

Perimetr=

5

18 ⇒ perimetre = 2(ℓ + b)

∴ b =(Perimeter−2ℓ)

2

So Length (ℓ)

Breath(b)=

5(18−5×2)

2

⇒5×2

18⇒

5

4

S17. Ans.(c)

Sol. According to mid point theorem CA ∥ PQ.

And CB ∥ RQ.

∴ all lines are equal to each other.

So perimeter of ∆ ABC = 1

2 Perimeter of ∆PQR

S18. Ans.(a)

Sol. Let equal sides of ∆ = a cm

So, ATQ,

2a + √a2 + a2 = 4√2 + 4

2a + a√2 = 4√2 + 4

a(√2 + 2) = 4(1 + √2)

a√2(1 + √2) = 4(1 + √2)

a√2 = 4

a =4

√2⇒ 2√2.

∴ Hypotenuse = side × √2

= 2√2 × √2.

= 4 cm

S19. Ans.(b)

Sol. Lets the perpendicular distance from the vertex to hypotenuse

= x unit

So ATQ 1

2× 5 × 12 =

1

2× x × 13 (hypotenuse)

x =60

13= 4

8

13 cm

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S20. Ans.(c)

Sol. Let ∆ PQR is an equilateral ∆ and its side = x

So, ATQ, √3

4x2 =

1

2× x × √3 +

1

2× x × 2√3 +

1

2× x × 5√3

√3

4x2 =

1

2x[√3 + 2√3 + 5√3]

√3x = 8√3 × 2

x = 16

∴ Perimetre of ∆PQR = 3 × 16 = 48 unit

S21. Ans.(b)

Sol. In ∆ PQR

PQ = √2 + 1 + √2 + 1 = 2(√2 + 1)

PR = √2 + 1 + 1 = √2 + 2

∵ PR2 + RQ2 = PQ2

∴ ∠PRQ = 90°, ∠RPQ = 45°

Thus, required perimeter = ℓ1 + ℓ2 + ℓ3

= 2π(√2 + 1)45

360°+ 2π(√2 + 1)

45

360+ 2π × 1 ×

90

360

= 2π ×1

8× 2(√2 + 1) +

π

2 =

π

2× (2 + √2)

S22. Ans.(b)

Sol. Given AB + BC = 12

BC + CA = 14

CA + AB = 18

∴ 2(AB + BC + CA) = 44

AB + BC + CA = 22

ATQ,

2πr = 22

2πr = 22

∴ r = 11 ×7

22⇒

7

2 cm

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S23. Ans.(c) Sol. Given,

diagonal of □ = 12 √2

∴ side = 12 cm Radius of circle = 6 cm

Circumradius of a ∆ = side

√3

6 =(a)

√3

∴ a = 6√3 cm S24. Ans.(c) Sol. If perimeter are equal of hexagon and equilateral ∆ then side of ∆ will be double of hexagon. Then,

Area of Hexagon

Area of equilateral∆=

6 ×√3

4. a2

√3

4. (2a)2

=6. a2

4. a2=

3

2

∴ Required ratio = 3 : 2 S25. Ans.(b) Sol. Length = 3 × 10 + 2πr = 30 + 2π × 5 = (30 + 10π) cm S26. Ans.(d)

Sol. Let fraction is x

y

∴ x − 4

y + 1 =

1

6 (given)

⇒ cross multiply the equation ⇒ 6x – 24 = y + 1 6x – y – 25 = 0 …….(i)

Again, x + 2

y + 1=

1

3 GIVEN, ⇒ 3x + 6 = y + 1

3x – y + 5 = 0 From equation (i) and (ii) 6x − y = 25 3x − y = −5

x = 10

∴ y = 35

∴ x

y =

10

35

Fraction = 10

35

Numerator = 10 Denominator = 35 LCM (numerator, denominator) ⇒ 2 × 5 × 7 = 70

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S27. Ans.(b)

Sol. LCM of (6, 7, 8, 9, 12)

LCM = 3 × 2 × 7 × 4 × 3 = 504

They will toll after every 504

S28. Ans.(d)

Sol. LCM of 5, 6, 7 and 8 = 840 840n + 3

9 =

810n+30n + 3

9 =

810n

9+

30n + 3

9

⇒ 30n + 3

9

⇒ Take n = 2

⇒ 30(2) + 3

⇒ 63

9 = Remainder = 0

∴ Number is 840n + 3

⇒ 840 (2) +3 [n = 2]

⇒ 1683

Sum of digit = 1+6+8+3 = 18

S29. Ans.(b)

Sol. LCM of 30, 36, 80 = 720

Number = 720 × K + 11

Put K = 2

Then number = 720 × 2 + 11 = 1440 + 11 ⇒ 1451

Put K = 3

Then number = 720 × 3 + 11 = 2160 + 11 ⇒ 2171 > 2000

S30. Ans.(c)

Sol. According to question,

So After making 191 group of square 81 Men will be left.