Percentage: Quantitaive Aptitude Question Answers Short Tricks

Here Check Percentage Based questions with  Answers, Percentage Problems Short Tricks, Percentage Question with Solution, How to Solve Percentage Based questions easily,  How to Solve Percentage Problems in less time, Best Method to Solve Percentage Problems so please go through this blog and any other further information please stay connect our blogs.

Q 1. The Salary of Aman is 50% more than the salary of the Rahul then how much % of the salary of Rahul is less than the salary of Aman.

Solution  

If we want to calculate more % then direct calculate at formula =

More % - Less %  × 100 

Less Percentage

Less Percentage  = Less % - More %/More %*100

Then  

150 -100/150*100  = 50/15*10

= 100/3 = 331/3  Ans.

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Q. 2 If the Numerator of the fraction is increased by 50% and denominator is increase by 200 then fraction become 4/7 find the original fraction.

Solution

Let the Numerator is X and Denominator is Y then According to question

150X/300Y = 4/7

X/Y = 8/7ans.

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Q. 3 If the length of rectangle is increased by 40% and breadth decrease by 20% then the area of rectangle.

Solution

This type of question you can solve easily at

+ A  +B  +AB/100    

here  A is Length & B is Breadth

and + shows the Increment and – shows the decrement in length & breadth

Then + 40 – 20 -40*20/100

=        +20 – 8 = + 12 Ans.

Q. 4 Each Side of Square is increased by 13% then what will be effect on its area.

Solution

We solve it above given formula directly

+ 13 + 13 +  13*13/100

= 26 + 1.69 = 27.69 Ans.

Q. 5 in the exam 35 % students passed the exam and 455 students are failed the exam then find the number of students appeared in the exam.

Solution

Let total students sits in exam is 100 then 35% is passed so failed students is 65%

By short tricks

65% = 455

1% = 455/65

& 100 % is = 455/65* 100 = 700 Ans.

Q.6 The Price of Sugar reduced by 25% then a man can buy 4 Kg more sugar t 420 Rs then calculate

1. Reduced Price    2. Original Price

Solution

420 * 25/100 = 420 * ¼  = 105

Then 4 kg sugar’s price = 105

1kg sugar price is = 105/4 = 26.25 per/kg Ans

2. 75%  = 26.25

     1% = 26.25/75 * 100 = 35 per/kg Ans.

Q. 7  Rahul spends 20% of his income on food and its 50% spends on the rent of the remaining 30% spends on games. If the total saving is 6300 then find monthly income of Rahul.

Solution

Let total income is 100 then 100 – 20 = 80

Then 50 % of 80 is = 50 * 80/100 = 40

Then 80 – 40  = 40

Now 40 of 30% = 12

40 – 12 = 28%

If 28%  = 6300

1% = 6300/28

& 100% = 6300/28*100 = 22500

Q. 8 Rahul expenses 20% of his salary on food and 16% on rent and 12% on games if the savings is 52000 Rs find his monthly income.

Solution

Saving = Income – Exp

Total Expenditure is = 20 + 16 +12 = 48%

Remaining is = 100 – 48 = 52

Then 52% = 52000

          1%  = 52000/52

100% is = 52000/52*100 = 100000Ans.

Q. 9 In the election a candidates got 84% vote and won election by 476 votes then find total number of voters.

Solution

Candidates who won get votes = 84%

And loser get                              = 16%

Then winner won by 68% votes

68% = 476

1%   = 476/68

And 100% = 476/68 * 100 = 700

Q. 10 A students has to obtained 33% of the total marks to passed the exam he got 125 marks and failed by 40 marks then  find the max marks.

Solution

125 + 40 = 165

33% = 165

1 % = 165/33

And 100% = 165/33 * 100 = 500 Ans.

NCERT Solutions for Class 10th Maths: Chapter 12, CBSE Board 10th Math Exercise 12.1 Solution

Hell Fridens this blog we provide complete solution of CBSE Board 10th Math Solution of Chapter 12. Please goo through this blog and check NCERT Solutions for Class 10th Exercise 12, NCERT Solutions of Class 10th of Exercise 12.1. CBSE 10th Math Exercise 12.1, CBSE 10th Class Math Chapter 12.1 Complete Solutions is given here. For other chapter solutions please stay connect this blog. If you have any doubt of NCERT Book Questions Please write comment in comment section. we will provide you complete solution.  

Q.1 The radii of two circles are 19 cm &  9 cm respectively. Calculate the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Answer

Let the radius of third circle is R.

 Then Circumference of the circle of Radius is  = 2πR

Circumference of the circle of radius 19 cm  will be = 2π × 19 = 38π cm

 And 9cm Radius circle Circumference is = 2π × 9 = 18π cm

Sum of the circumference of these two circles is =  18 π + 38π = 56π cm

 Then Circumference of the third circle is = 2πR = 56π

⇒ 2πR = 56π cm

⇒ R = 28 cm Ans.

2.  The radii of two circles are 8 cm and 6 cm respectively. Calculate the radius of the circle having area equal to the sum of the areas of the two circles.

Explanation

We assume that radius of third circle is R.

Area of circle is = πR²

Radius 8 cm circle’s Area  = π × 82 = 64π cm2

 Radius 6 cm Circle Area  is = π × 62 = 36π cm2

Sum of the area of two circles  is = 36π cm2 + 64π cm2 = 100π cm2

⇒ πR² = 100π cm2

⇒ R² = 100 cm2

⇒ R = 10 cm Ans.

Radius of the new circle is 10 cm.

Q. 3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

                                         

Explanation

Diameter of Gold circle or first circle is = 21 cm

Then Radius of first circle, r₁  is = 21/2 cm = 10.5 cm

Each of  other bands is 10.5 cm wide,

Radius of Red or second circle, r₂ = 10.5 cm + 10.5 cm = 21 cm

Radius of third or Blue circle, r₃ = 21 cm + 10.5 cm = 31.5 cm

Radius of fourth or Black circle, r₄ = 31.5 cm + 10.5 cm = 42 cm

Radius of fifth or white  circle, r₅ is  = 42 cm + 10.5 cm = 52.5 cm

Area of gold region = π r₁² = 22/7*(10.5)²= 346.5 cm2

Area of red region = Area of second circle – Area of first circle = π r₂² – 346.5 cm2

                              = π(21)² – 346.5 cm2 = 1386 – 346.5 cm2 = 1039.5 cm2

Area of blue region = Area of third circle – Area of second circle = π r32 – 1386 cm2

                                          =  π(31.5)² – 1386 cm2 = 3118.5 – 1386 cm2 = 1732.5 cm2

Area of black region = Area of fourth circle – Area of third circle = π r₃² – 3118.5 cm2

                                             = π(42)² – 1386 cm2 = 5544 – 3118.5 cm2 = 2425.5 cm2

Area of white region = Area of fifth circle – Area of fourth circle = π r₄²– 5544 cm2

                                             = π(52.5)² – 5544 cm2 = 8662.5 – 5544 cm2 = 3118.5 cm2

Q.4. The wheels of a car are of diameter is 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km /H?

Explanation

Diameter of the wheels  = 80 cm

And the Circumference of wheels is = 2πr = 2r × π = 80 π cm

 Then Distance travelled by car in 10 minutes = (66 × 1000 × 100 × 10)/60 = 1100000 cm/s

 Formula “No. of revolutions = Distance travelled by car/Circumference of wheels”

                              = 1100000/80 π = (1100000 × 7)/(80×22) = 4375 Ans.

Q.5. Tick the correct answer in given below &  justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

          (A) 2 units                     (B) π units                  (C) 4 units              (D) 7 units

Answer

Let the radius of  circle r.

Perimeter of the circle = 2πr

Area of the circle = π r²

A/q,

2πr = π r²

⇒ 2 = r

The radius of the circle is 2 units Then (A) is correct.

CBSE Board 12th Sample Paper 2017 - All Subjects

CBSE Board 12^th Class Sample Paper  2017

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CBSE 10th Math Sample Paper PDF with Solution

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Q.1  Find the root of quadratic equation that is 5x² -13x-6 = 0

a. -2, 3/2

b. 2, -3/2

c. 2, 3/2

d. 3, -5/2

Q. 2 The nth term of an A.P is (2n+1) then what is the sum of first three term is

a.  15

b. 21

c. 12

d. 6n+3

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Q. 4 The circumference of  a circle is 22 cm then the area of quadrant is

a. 77/4

b. 77/8

c. 77/2

d. 77/16

Q. 5 the Distance of Point (3, -4) from x-axis is

a. -3

b.  3

c. 5

d. 4

Q.6  Solve the below given quadratic equation for x

X² – 4ax – b² +4a² = 0

Q. 7 If the sum of two natural no. is 8 and sum of their product is 15 then find out the natural no.

Q.8 Find the area of quadrilateral ABCD that vertices is given below

A ( -3, -1), b (-2, -4), c (4, -1) & d ( 3, 4)

Q.9 Find the coordinate of point P which lies in the line segment join the points of A ( -2, 2)  and B (2, -4)  such that AP  = 3/7AB

Q. 10