Pawan Rao Maths Book PDF Download

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Pawan Rao Maths Book PDF Download 

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     Related Topics

  1. सरलीकरण (Specification)
  2. आसन्न मान (Approximate Value)
  3. समीकरणों की समानता (Equality of Equations)
  4. घातांक एवं करणी (Indices and Surds)
  5. लघुत्तम समापवर्तक एवं महत्तम समापवर्तक (LCM and HCF)
  6. संख्या पद्धति (Number System)
  7. प्रतिशतता (Percentage)
  8. लाभ, हानि और बट्टा (Profit, Loss and Discount)
  9. साधारण ब्याज (Simple Interest)
  10. चक्रवृद्धि ब्याज (Computed Interest)
  11. अनुपात एवं समानुपात (Ratio and Proportion)
  12. साझेदारी (Partnership)
  13. मिश्रण (Mixture)
  14. मिस्र समानुपात (Mixture)
  15. मिश्र समानुपात (Mixed Proportion)
  16. उम्र सम्बन्धी प्रश्न (Problems on Age)
  17. समय तथा काम (Time and Work)
  18. नल तथा टंकी (Pipes and Cistern)
  19. औसत (Average)
  20. समय तथा दूरी (Time and Distance)
  21. रेलगाङी से सम्बंधित प्रश्न (Problems Based on Trains)
  22. नाव एवं धारा (Bots and Streams)
  23. क्षेत्रफल (Area)
  24. पृष्ठ क्षेत्रफल तथा आयतन (Surface Area and Volume)
  25. आंकणो की पर्याप्तता (Data Sufficiency)
  26. संख्या श्रेणी (Number Series)
  27. क्रमचय एवं संचय (Permutation and Combination)
  28. प्रायिका सिद्धांत (Probability Theory)
  29. एकघातीय एवं द्विघातीय समीकरण (Linear & Quadratic Equations)
  30. ग्राफ (Graph)
  31. सारणीकरण (Tabulation)
  32. लघुगणक (Logarithms)
  33. श्रेणी (Progression)

 

Pawan Rao maths book PDF : Download Now 2021

Math Question And Answer

Q.2: The perimeters of two similar triangles ABC and PQR are 156 cm and 46.8 cm respectively. If BC = 19.5 cm and QR = x cm, then the value of x is:
(A) 3.76 cm
(B) 5.85 cm
(C) 4.29 cm
(D) 6.75 cm
(SSC CHSL August 2021)

Q.3: The radius of a sphere is 9 cm. It is melted and drawn into a wire of radius 0.3 cm. The length of the wire is:
(A) 112 m
(B) 108 m
(C) 118 m
(D) 106 m
(SSC CHSL August 2021)

Ans : (B) 108 m
Volume of sphere = Wire volume
\frac {4}{3} \pi r^3 = \pir2h
\frac 43 \times \pi \times 9^3 = \pix 0.3 x 0.3 x h
h =\frac {4\times9\times9\times9}{3\times0.3\times0.3}
h = 10800 cm = 108 m

Q.4: One side of a rectangular field is 39 m and its diagonal is 89 m. What is the area of the field ?
(A) 3120 m2
(B) 2100 m2
(C) 2160 m2
(D) 3140 m2
(SSC CHSL August 2021)

Ans : (A) 3120 m2
Side = 39
Diagonal = 89
Second side = \sqrt{(89)^2 - (39)^2}
= \sqrt{7921-1521}
Second Side = \sqrt{6400} =80
Area of field = First Side x Second side (l x b)
= 80 x 39 = 3120 m2

Q.5: The base of a triangle to the perimeter fo a square whose diagonal is 7\sqrt2 cm and its height is equal to the side of square whose area is 169 cm2 . The area (in cm2 ) of the triangle is:
(A) 130
(B) 182
(C) 175
(D) 156
(SSC CHSL August 2021)

Ans : (B) 182
Diagonal of square 7\sqrt2
a\sqrt2=7\sqrt2
a = 7
Perimeter of Square = 4a = 28
Perimeter of square = Base of Triangle = 28
Area of Square = 169
Side of Square = 13
Height of Triangle = Side of square = 13
Area of Triangle = \frac12 x b x h
\frac12 \times28\times13 =182

Q.6: If the adjacent sides of a rectangle whose perimeter is 60 cm are in the ratio 3 : 2, then what will be the area of the rectangle ?
(A) 864 cm2
(B) 216 cm2
(C) 60 cm2
(D) 300 cm2
(SSC CHSL August 2021)

Ans : (B) 216 cm2
Ration of side = 3 : 2
Perimeter = 60
Perimeter of rectangle = 2 ( l + b)
2(3n + 2n) = 60
5n = 30, n = 6
l = 3n = 3 x 6 = 18
b = 2n = 2 x 6 = 12
Area = l x b = 18 x 12 = 216 cm2

Q.7: The perimeter of a right angle triangle whose sides that make right angles are 15 cm and 20 cm is:
(A) 60 cm
(B) 40 cm
(C) 70 cm
(D) 50 cm
(SSC CHSL August 2021)

Ans : (A) 60 cm
Diagonal of Right angle triangle
(Third Side ) =\sqrt{(20^2 + 15^2)}
=\sqrt {400 + 225}
Diagonal =\sqrt {625} = 25
Perimeter = Sum of all sides = 15 + 20 + 25= 60 cm

Q.8: The difference between the two perpendicular sides of a right-angled triangle is 17 cm and its area is 84 cm2 . What is the perimeter (in cm ) of the triangle ?
(A) 65
(B) 49
(C) 72
(D) 56
(SSC CHSL August 2021)

Ans : (D) 56
b – h = 17
Area = \frac12x b x h = 84
b x h = 168 = 24 x 7
b = 24
h = 7
Diagonal2 = b2 + h2 = 242 + 72
=> 576 + 49 = 625
Diagonal= 25
Sides of Triangle= 7, 24, 25
Perimeter = 7 + 24 + 25 = 56

Q.9: A solid metallic sphere of radius 10 cm is melted and recast into spheres of radius 2 cm each. How many such spheres can be made ?
(A) 64
(B) 216
(C) 125
(D) 100
(SSC CHSL August 2021)

Ans : (C) 125
Volume of big sphere = n x Volume of small sphere
V =\frac 43\pi r^3
\frac 43 \pi (10)^3 = n x\frac 43\pi (2)^3
10 x 10 x 10 = n x 2 x 2x 2
1000 = n x 8
n = 125

Q.10: The sum of the squares of the sides of a rhombus is 1600 cm2 . What is the side of the rhombus ?
(A) 10 cm
(B) 15 cm
(C) 20 cm
(D) 25 cm
(SSC CHSL August 2021)

Ans : (C) 20 cm
All the sides of rhombus are equal
The sum of the squares of sides = 1600
a2 + a2 + a2 + a2 = 1600
4a2 = 1600
a2 = 400
a = 20

Mensuration Questions for Competitive Exams

Q.11: The volume of a right circular cone is 462 cm3 . If its height is 12 cm, then the area of its base (in cm2 ) is:
(A) 124.5
(B) 103.5
(C) 115.5
(D) 98.5
(SSC CHSL August 2021)

Ans : (C) 115.5
Volume of Cone =\frac13 \pi r^2h
\frac13 \pi r^2h = 462
r2 =\frac{462\times 3}{\pi \times h}
=\frac{462\times 3\times7}{22 \times 12}
r2 =\frac{21\times7}{4}
Area of base =\pi r^2
=\frac {22\times21\times7}{7\times4}
= 115.5

Q.12: How many bricks each measuring 64 cm x 11.25 cm x 6 cm. will be needed to build a wall measuring 8m x 3m x 22.5m ?
(A) 200000
(B) 250000
(C) 67500
(D) 125000
(SSC CHSL August 2021)

Ans : (D) 125000
Volume of wall = V1
Volume of Brick= V2
Number of Bricks =\frac{V1}{V2}
=\frac{8m\times3m\times22.5}{64cm\times11.25cm\times6cm}
=\frac{800\times300\times22.5}{64\times11.25\times6}
= 125000

Q.13: If the radius of a circle is equal to a diagonal of a square whose area is 12 cm2 , then the area of the circle is:
(A) 28\pi cm2
(B) 36\pi cm2
(C) 32\pi cm2
(D) 24\pi cm2
(SSC CHSL August 2021)

Ans : (D) 24\pi cm2
Area of Square = 12
Side of Square =2\sqrt3
Diagonal of square= side x\sqrt2
=2\sqrt6
Radius of Circle = Diagonal of Square
2\sqrt6
Area of circle =\pi r^2
=\pi(2\sqrt6)^2
=24\pi

Q.14: The sum of three sides of an isosceles triangle is 20 cm, and the ratio of an equal side to the base is 3 : 4, The altitude of the triangle is:
(A) 3\sqrt3cm
(B) 4\sqrt5cm
(C) 3\sqrt5cm
(D) 2\sqrt5cm
(SSC CHSL August 2021)

Ans : (D) 2\sqrt5cm
Ration of side and base = 3 : 4
Sides 3x, 3x, 4x
Sum of sides = 20
3x + 3x + 4x = 20
10x = 20
x = 2
Sides = 6, 6, 8
Height =\sqrt{6^2 - 4^2}
= 36 – 16
=\sqrt{20}
=2\sqrt5

Q.15: If the volume of a sphere is 697\frac{4}{21}cm3 , then its radius is:
(\text{Take} \pi=\frac{22}{7})
(A) 5 cm
(B) 6 cm
(C) 4.5 cm
(D) 5.5 cm
(SSC CHSL August 2021)

Ans : (D) 5.5 cm
Volume of Sphere =\frac43\pi r^3
\frac43\pi r^3 = 697\frac{4}{21}
r3 =\frac{14641\times3\times7}{21\times4\times22}
r3 =\frac{1331}{8}
r =\frac{11}{2}= 5.5 cm

Q.16: The area of the largest that can be inscribed in a semi-circle of radius 6 cm is:
(A) 38cm2
(B) 35cm2
(C) 36cm2
(D) 34cm2
(SSC CHSL August 2021)

Ans : (C) 36cm2
Base of Tringle = Diameter of semi-circle
Base of Triangle = 12
Height of Triangle = radius of semi-circle
Height= 6
Area of Triangle=\frac12 x base x height
=\frac12 x 12 x 6 = 36 cm2

Q.18: The square of the diagonal of a cube is 2175 cm2 . What is the total surface area (in cm2 ) of the cube ?
(A) 4272
(B) 4305
(C) 4350
(D) 4530
(SSC CHSL August 2021)

Ans : (C) 4350
Diagonal of square =a\sqrt3
(a\sqrt3)^2 = 2175
a2 =\frac{2175}{3}
a2 = 725
total surface area = 6a2
= 6 x 725 = 4350

Q.19: A rectangle with perimeter 50 cm has its sides in the ratio 1 : 4. What is the perimeter of a square whose area is the same as that of the rectangle ?
(A) 45 cm
(B) 36 cm
(C) 50 cm
(D) 40 cm
(SSC CHSL August 2021)

Ans : (D) 40 cm
Let sides are x , 4x
Perimeter = 2(l + b)
2(x + 4x) = 50
5x = 25, x = 5
Length= 20 cm
Breath= 5 cm
Area= l x b = 20 x 5 = 100
As per question, Area of Rectangle = Area of Square
a2 = 100, a = 10
Perimeter of square= 4 x side = 4 x 10 = 40

Q.20: The inner and outer radius of two concentric are 6.7 cm and 9.5 cm, respectively. What is the difference between their circumferences (in cm ) ?\text {Take} \pi=\frac{22}{7}
(A) 10.4
(B) 17.6
(C) 6.5
(D) 20.5
(SSC CHSL August 2021)

Ans : (B) 17.6
Circumference of circle =2\pir
Difference of Circumference
=2\piR – 2\pir
=2\times\frac{22}{7} [ 9.5 – 6.7]
=2\times\frac{22}{7}\times 2.8
= 17.6

 

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