Arithmetic Aptitude Questions & Answers (Compiled from UPSC, SSC ,PSC ,IBPS previous question papers)

I.  x²+x-20=0

   x²+5x-4x-20=0

  x(x+5)-4(x+5)=0

  (x+5)(x-4)=0

  x=4,-5

II.  y²-y-30=0

    y²-6y+5y-30=0

   y(y-6)+5(y-6)=0

   (y-6)(y+5)=0

   y = -5, 6

∴ Relationship cannot be decided

I. 6x²-29x+35=0

  6x²-15x-14x+35=0

 3x(2x-5)-7(2x-5)=0

 (2x-5)(3x-7)=0

 x = 2.3, 2.5

II.  3y²-11y+10=0

 3y²-6y-5y+10=0

 3y(y-2)-5(y-2)=0

 (y-2)(3y-5)=0

 y=1.6, 2

 ∴ x>y

I. 2x²+18x+40=0

  2x²+10x+8x+40=0

  2x(x+5)+8(x+5)=0

 (2x+8)(x+5)=0

 (x+5)(2x+8)=0

  x = -4, -5

II.  2y²+15y+27=0

     2y²+9y+6y+27=0

    y(2y+9)+3(2y+9)=0

    (2y+9)(y+3)=0

    (y+3)(2y+9)=0

     y = -4.5, -3

Relationship cannot be decided

I. 5x²+11x+6=0

5x²+6x+5x+6=0

x(5x+6)+1(5x+6)=0

(5x+6)(x+1)=0

x=-1, -6/5

II. y²-34y-336=0

y²-42y+8y-336=0

y(y-42)+8(y-42)=0

(y-42)(y+8)=0

y=42, -8

 Relationship cannot be decided

Let sia had the money initially = INR x

Her brother received the money = INR ^x/₄

According to question,

^x/₄-[^x/₄×¹/₈+^x/₄×¹/₄]=125

8x-x-2x÷32=125

5x= 125×32

x= INR 800/-

6(×1-1)5(×1+1)6(×2-2)10(×2+2)22

2(×1+2)4(×2+3)11(×3+4)37(×4+5)153(×5+6)771

120(÷5)24(÷4)6(÷3)2(÷2)1(÷1)

Total number of population of village C = 200

Number of males in village C = 200×3/5=120

Number of females in village C = 200-120=80

Let total number of population of village D = x

Number of illiterate population in village D = 120

x×40/100=120

x=300

Number of males in village D = 300×2/3=200

Number of females in village D = 300-200=100

∴ Required difference = (120+200)-(80+100)

=320-180

=140

Total population of village B = 500

Total population of village C = 200

∴Required percentage = 500-200/200×100

=300/2=150%

Let the total population of village B = x

Number of the illiterate population of village B = 350

∴ x×70/100=350

Number of literate population of village B

= 500-350 = 150

Number of literate males = 150×2/3=100

Number of literate females = 150-100=50

Total number of females = 500×2/5=200

∴ Required Percentage = 50/200×100 = 25%

Let the total population of village A =x

Number of illiterate population of village A = 360

∴ x×⁶⁰/₁₀₀₌360

x=600

Number of literate population of village A

= 600-360 = 240

Let the total population of village C =y

Number of illiterate population of village C = 50

∴ y×25/100 = 50

y=200

Number of literate population of village C

= 200-50 = 150

∴ Required ratio = 240/150

8:5

Required percentage = 120-50÷120×100

= ⁷⁰/₁₂₀×100

=58¹/₃%

Let the length of the rectangle = l cm

and breadth of the rectangle = b cm

Given, l = a

b = a/2

According to question,

a²+l×b=48

a²+a×^a/₂=48

3a² = 48×2

a² = 32 

a² = 36

∴ a = 6 cm

(3 × 9.2) ÷ ²/₉ of 62 = ?

3×10÷²/₉×63=?

³⁰/₁₄=?

?=2.4 ≈2

(24.99% of 900.911) ÷ (10.30% of 25.011) = ?⁴

[²⁵/₁₀₀×900]÷[¹⁰/₁₀₀×25] = ?⁴

?⁴ = 90 = 81

?⁴ = (3)⁴

∴ ?=3

√149-(113.79+65.89) ÷ ? = 6

√149-(114+66)÷?=6

12-¹⁸⁰/_?=6

6×?=180

?=30

Let the quantity of water in the mixture = x litre

and the quantity of milk in the mixture = 4x litre

According to question,

[3x-4(³/₄)+3]÷[x-4(¹/₄)]=⁴/₁

3x=4x-4

x=4

∴ Initial quantity of water in the mixture = 4 litre

According to question,

^4000×7×2/₁₀₀=1/2[x×7×5/100]

X=¹⁶⁰⁰⁰/₅

x=INR 3200/-

Let the cost price of the tablet = INR x

and cost price of the mobile = INR 1.5x

Total cost price of both the articles

= x+15x= 2.5x

Total selling price of both the articles

= x×⁹²/₁₀₀+^1.5x×110/₁₀₀

=INR 2.57 x

Gain = (2.57-2.50)x = 0.07x

∴ Required gain percentage = ^0.07/_2.5×100

=2.8%

Required ratio = ^300+240/_120+180

=⁵⁴⁰/₃₀₀

=9:5

Required Percentage = 60/270×100

=22.25

=22¹/₄

Required difference = (240+200)-(120+90)

= 440-210

=230

Required Percentage = ^180-150/₁₅₀×100

=³⁰/₁₅₀×100

=20%