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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.