Simplification & Approximation Questions and Answers

   

Preparation Video and Questions on Simplification & Approximation for Competitive Exams.

What is Campus Recruitment Process?

Generally, a Campus Placement Process has four components:

 * Pre Placement Talk 

 * Aptitude test 

 * Group Discussion 

 * Interview which can be various combinations of Technical and HR rounds 

Aptitude Test for Placements:

The Aptitude test consists of basic problems in quantitative ability, logical reasoning, and verbal ability. The primary objective of this test is to check the problem-solving skills of the test-taker. Any job out there is the problem-solving skills. It is essential to understand what the problem is and how the problem should be solved. So, the application of the learning is the Aptitude test.

Take a Free Campus Placement Mock Test

Please read the below blogs to know the detailed campus recruitment process of respective companies: 

Amazon Recruitment Process TCS Recruitment Process
IBM Recruitment Process Capgemini Recruitment Process
Wipro Recruitment Process Deloitte Recruitment Process
Tech Mahindra Recruitment Process Infosys Recruitment Process
Cognizant Recruitment Process Accenture Recruitment Process

Tips and Tricks to solve the Simplification and Approximation Questions for Campus Recruitment Test

  • If you want to stand out & beat the competition in your Campus Placements, the first step is to ace the Written Aptitude Test. Quantitative Aptitude contributes a lot to increase your scores. No aptitude test is complete without questions based on Simplification & Approximations.
  • Methods of learning in one chapter can be used in other chapters.
  • Whether the Question is simple or difficult you can know the options.
  • Focus on all areas in quantitative will help you crack quantitative section

Watch the below video to learn Tips & Tricks to crack Questions on Simplification and Approximation in Competitive Exams.

The Following questions are based on Simplification and Approximation:

Simplification involving Fraction:

1.4 1/7-8 2/3+? 4/7-6 11/33=15 5/7

A. 46

B. 36

C. 26

D. 33

Answer: C

Explanation:

According to the given question

Isolate whole and given fraction

Ler 'x' be the number to find

 4                    1/7   

-8                    -2/3

 x                     4/7

-6                    -1/3

-10+x-1=15        5/7

From the above  point,  '-1' will be combing the fraction part of (-2/3,-1/3) it will get  '-1'

x-11=15

x=26

2. 4/3-11/9+22/27-?+19/6 =35/27

A. 44/227

B. 23/27

C. 76/27

D. 53/27

Answer: C

Explanation:

Let 'x' be a number to find

4/3-11/9+22/7+19/6-35/27=x

4/3→1+1/3

-11/9→(-1-2/9)

19/6→3+1/6

35/27→(-1-8/27)

so according to the question

1+4/3-1-2/9+22/7+3+1/6-1-8/27=x

x=2+(18-12+44+9-16)/54

x=2+43/54

x=76/27

Simplification involving percentages

3. 8% of 720÷9% of 360=? *10% of 36

A. 20/81

B. 30/81

C. 50/81

D. 40/81

Answer: D

Explanation:

Let 'x' be a number to find

8% of 720 is same as 160% of 36

9% of 360 is same as 90% of 36

10% of 36=3.6

(160%  of 36)/(90% 0f 36)→16/9

16/9=x*3.6

x=16/32.4=160/324=40/81

4.12 1/2% of 800+3 3/4% of 1600=?% of 1000

A.40

B.15

C.25

D.16

Answer: D

Explanation:

Let 'x' be a number to find

25%=1/4th part    →12 1/2 %=1/8th part

37.5%=3/8thpart  →(3 3/4%)or(3.75%)=3/80th part

1/8th of 800 is 100

3/80 of 1600 is 60

100+60=x% of 1000

x=16%

5. 8 1/3 of 2400+6 1/4% of  3200=?

A.300

B.400

C.500

D.600

Answer: B

Explanation:

Let 'x' be a number to find

25/3%of 2400+25/4% of 3200=x

25%=1/4thpart

so the given equation

25*(1/3)% of 2400+25*(1/4)%of 3200

(1/4)*(1/3)*2400+(1/4)(1/4)*3200=200+200=400

Simplification using Algebraic

6. (175^4-25^4)/(175²-25²)=?

A.30,000

B.40,000

C.50,000

D.60,000

Answer: A

Explanation:

By using the formula (a^4-b^4)/(a²+b²)=(a²-b²)(a²+b²)/(a²+b²)=(a²-b²)

Then  (175²-25²)=(175+25)(175-25)=200*150=30,000

Inequalities based on Quadratic equations

7. 2x²-5x+2=0, 3y²-10y+3=0

A. x<y

B. x>y

C. No relationship

D. x=y

Answer: C

Explanation:

2x²–5x+2=0→eq1

3y²-10y+3=0→eq2

 The Product of the roots in eq1=2/2=1

 The Product of the roots in eq2=3/3=1

Both x and y has one positive root and one negative root

   x     y     

>0    >0

<0    <0

 So from the above box, we can determine

In the first case, x has the positive value and y has the negative value 

In the second case, x has the positive value and y has the positive value 

so the relationship cannot be determined

8. x²-2x+15=0,  y²+4y-12=0?

A.x<y

B.x>y

C.No relationship

D.x=y

Answer: C

Explanation:

x²-2x+15=0→eq1

y²+4y-12=0→eq2

The Product of the roots in eq1=-15

 The Product of the roots in eq2=-12

Both x and y has one positive root and one negative root

    x    y

>0     >0

<0     <0

 In the first case, x has the positive value and y has the negative value 

 In the second case, x has the positive value and y has the positive value 

Hence there is no relationship

9.16x²-10x+1=0, 4y²+17y-4=0

A.x<y

B.x>y

C.No relationship

D.x=y

Answer: B

Explanation:

16x²–10x+1=0ceq1

4y²-17y+4=0→eq2

The Product of the roots in eq1=1/16

The Product of the roots in eq2=1

Both x and y has one positive root and one negative root

Sum of the roots=10/16>0 from eq1

Sum of the roots=-17/4<0 from eq2

Both  x1 and  x2 greater than zero

Both  y1  and  y2 less than zero

Then x>y

10. x²=1024 y=√1024

A.x<y

B.x>y

C.No relationship

D.x=y

Answer: C

Explanation:

x²=1024

x=±32

y=√1024

y=32>0

Root of any positive number is always positive  

In one case x and y are equal

In other case y>x

so y>=x then there is no relationship

11. ==729  y³=27   

A. x<y

B. x>y

C. No relationship

D. x=y

Answer: C

Explanation:

x³=±√729=±27

x=3√±27=±3

y=3

In one case x and y are equal

In other case y>=x

so y>=x then there is no relationship

12.15x²-x-6=0; 30y²-y-20=0

A.x<y

B.x>y

C.No relationship

D.x=y

Answer: C

Explanation:

15x²-x-6=0→eq1

30y²-y-20=0→eq2

 The Product of the roots in eq1=-6/15<0

 The Product of the roots in eq2=-20/30<0

     x     y

x1>0   y1>0

x2<0  y2<0

 In the first case, x has the positive value and y has the negative value 

 In the second case, x has the positive value and y has the positive value 

Hence there is no relationship

13. x²+x-6=0; y²+3y-15=0

A. x<y

B. x>y

C. No relationship

D. x=y

Answer: C

Explanation:

x²+x-6=0→eq1

y²+3y-15=0→eq2

 The Product of the roots in eq1=-6<0

 The Product of the roots in eq2=-15<0

     x     y

x1>0   y1>0

x2<0   y2<0

In the first case, x has the positive value and y has the negative value 

In the second case, x has the positive value and y has the positive value 

Hence there is no relationship

14. x²-7x-198=0; y²+26y+165=0

A.x<y

B.x>y

C.No relationship

D.x=y

Answer: C

Explanation:

x²-7x-198=0→eq1

y²+26y-165=0→eq2

 The Product of the roots in eq1=-198<0

 The Product of the roots in eq2=-165<0

    x     y

x1>0   y1<0

x2<0   y2<0

Here both the values of  y are negative

x1>0(so, x1 greater than both y1 and y2

y²+26y-169=4

(y+13)²=2²

y+13=±2

y=-13-2 (or) y=-13+2

y=-15,-11

x²-7x-198=0

x²-18x+11x-198=0

x(x-18)+11(x-18)=0

x=-11,18

Hence x>=y

Hence there is no relation

15. x²-2x-80=0; y²+2y-80=0

A.x<y

B.x>y

C.No relationship

D.x=y

Answer: C

Explanation:

x²-2x-80=0→eq1

y²+2y-80=0→eq2

 The Product of the roots in eq1=-80<0

 The Product of the roots in eq2=-80<0

    x           y

x1>0     y1>0

x2<0     y2<0

In the first case, x has the positive value and y has the negative value 

In the second case, x has the positive value and y has the positive value 

Hence there is no relationship

Simplification using BODMAS

16.18*14÷7*4-40*1 3/4–4*5=?

A. 100

B. -80

C. 60

D. 150

Answer: C

Explanation:

We have to follow the order to solve the problem

Brackets: simplify the values inside the brackets before you discard them

Order: complete the calculations involving  powers or roots

Division and multiplication: Both are of equal priority; go left to right

Addition and Multiplication: Both are of equal priority; go left to right

18*14÷7*4-40*1 3/4–4*5=?

252÷7*4-40*8/5-20

36*4-64-20

144-84=60

17. 4+21*5=?

A.107

B.108

C.109

D.106

Answer: C

Explanation:

4+105=109

18. 21-16÷(5+3)=?

A.12

B.19

C.15

D.16

Answer: B

Explanation:

21-16÷8=21-2=19

19. 11+7*(2+14)=?

A.124

B.123

C.125

D.128

Answer: B

Explanation:

11+112=123

20. (108+207)-550÷52+ 10=?

A.301

B.303

C.300

D.302

Answer: B

Explanation:

315-550÷25+10=315-22+10=303

21. 6+6÷6+5*5-5=?

A.20

B.27

C.24

D.23

Answer: B

Explanation:

6+1+25-5=7+20=27

22. 4*3÷2*3*5÷6*8=?

A.120

B.180

C.150

D.90

Answer: A

Explanation:

12÷2*3*5÷6*8=6*3*5÷6*8=90÷6*8=15*8=120

23. 360÷2÷3÷4=?

A.15

B.16

C.19

D. 20

Answer: A

Explanation:

360÷2÷3÷4=180÷3÷4=60÷4=15

24.16+4*5÷2.5-860÷86*5+40÷2

A.-5

B.-6

C.-4

D.-3

Answer: B

Explanation:

According to the question 16+20÷2.5-860÷86*5+40÷2

16+20÷2.5-10*5+40÷2

16+8-50+20=24-30=-6

25. 240÷15*4÷8-12*3÷4++15-16=?

A.-2

B.-3

C.-4

D.2

Answer: A

Explanation:

26. 420*5÷6+25*2÷4*8-300÷75÷5

A. 449.2

B. 550

C. 448

D. 447.2

Answer: A

Explanation:

2100÷6+25*2÷4*8-300÷75÷5

350+50÷4*8-300÷75÷5

350÷12.5*8-4÷5

350÷100-0.8=449.2

Simplification involving parenthesis

27. 320-{160*5÷(460÷23*10)}*10*(4+8÷2)

A.1

B.0

C.2

D.3

Answer: B

Explanation:

320-{800÷(20*10)}*10*(4+4)

320-{800÷200}*10*8

320-4*10*8

320-320=0

28.175+[360-{180*4-(32-(72÷5*25))}]

A.146

B.145

C.143

D.148

Answer: C

Explanation:

175+[360-{180*4-(32-360)}]

175+[360-{180*4-328}]

175+[360-{720-328}]

175+360-392=143

29.1000*[4000÷{720*(160÷(320÷4)}]

A.25000/9

B.32000/9

C.24000/9

D.52000/9

Answer: A

Explanation:

1000*[4000÷{720*(160÷80)

1000*[4000÷{720*2}]

1000*[4000÷1440]

1000*4000/1440=25000/9

Practice Top 30 Company's Aptitude Questions with 135 Test Papers
Read 126 times Last modified on Friday, 31 August 2018 14:57
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