Quadratic Equation

aX²+bX +c =0

a = coefficient of X²

b = coefficient of X

c = constant term

There will be five problems from this topic. It is the easiest one, so you can solve all problems within 5 to 7 minutes.

In this kind of problems two equations will be given as question, you have to solve the equation and find the relation between them

1.P > Q

2.P < Q

3.P ≥ Q

4.p ≤ Q

5.P = Q or relationship cannot be established

Example 1:P²+13P+40=0

                  Q²+7Q+12=0

Step1: let us take equ 1. P²+13P+40=0 in this equation coefficient of p, 13 should be split into two numbers in such a way that multiplication of both numbers should be equal to constant term 40 and addition of numbers should be equal to 13

13 can be split into (1,12) (2,11) (3,10) (4,9) (5,8) (6,7)

In these combination 5 and 8 only can give 40 while multiplying, so this is the number we are searching for,and since the coefficient of P² is 1 and there is no negative sign in the equation, we can directly write value of P by simply changing the sign

 P= -5,-8

( just for reference Actual procedure is  P²+5P+8P+40=0

P(P+5)+8(P+5)=0

(P+5)(P+8)=0

P=-5,-8 )

Step 2: Now equ2. Q²+7Q+12=0 similar process applicable for this equation to find Q, here coefficient of Q should be split into two numbers and multiplication of the numbers should give 12

7 can be split up into (1,6) (2,5) (3,4)

Combination of 3 and 4 alone satisfy our need i.e. giving 7 and 12 while adding and multiplying the numbers respectively, since there is no negative sign in the equation,we can directly write value of Q by changing sign .

Q=-3, -4

P= -5,-8

Obviously P < Q

Example 2: 2p²+12p+16=0

                   2q²+14q+24=0

Step 1: let us take equ1 since coefficient of p² is 2,we have to multiply 2 with constant number 16 , now as usual 12 should be split up into two numbers and multiplication of the numbers should give ( 2*16=) 32 .

4+8 =12

4*8=32

 12 can be split up into 4 and 8,

 now change the sign of numbers and divide it by 2 since the coefficient of p² is 2 ,

Thus the value of  p = -2,-4

Step 2: now take equ2. And follow the same procedure multiply 2 with constant number 24 . So addition of two numbers should be 14 and multiplication of numbers should be 48 (24*2)

6+8 =14  ;  6*8=48

Numbers are 6 and 8, now divide by 2, since the coefficient of q² is 2 so the value of q is -3,-4

here, one of the numbers of p and q are same but other number of p is greater than other number of q 

so ans is p ≥ q

If you have doubt in finding which one is greater, use this technique.

Number which is 1^st from right hand side is greater one.

Example 3:  x²-x-6=0

                    2y²+13y+21

Step1: Here constant number and coefficient of x are negative so the combination of numbers will be positive and negative.

Combination is 2,-3

Value of x=-2 , 3

Step2: proceed with equation 2

Combination is (6,7) ,

since the coefficient of y² is 2 divide the value of y by 2  

Y=  -(6/2) , -(7/2)

Y= -3 , -3.5

Relation is X > Y

Example 4: 12x²+11x+12=10x²+22x

                  13y²-18y+3=9y²-10y

Step1: convert this into normal quadratic equation form

2x²-11x+12=0

4y²-8y+3=0

Step2: now as usual normal procedure , here coefficient of x is negative and constant term is positive so both numbers will be negative

 Combination is -8,-3

And value of x= 4, 3/2

 Step3: proceed with equ 2

Combination is (-2,-6)

since coefficient of y² is 4 divide the value of y by 4 ,

And value of y = ½ , 3/2

 Relation is x ≥ y

 Example 5: 18/x² + 6/x -12/x² = 8/x²

                    Y³+9.68+5.64 =16.95

Multiply x² in equ1 it becomes 18+6X -12 =8

Solving this equations we get x = 1/3 = 0.33

Solving equ2 we get y³ = 1.63

          Y = 1.17

Relation is x < y

Example 6: (x+18)^1/2 = (144)^1/2 – (49)^1/2

                   Y² +409 = 473

By solving equ1, we get x = 7

Solving equ2 we get y = ± 8

Relation cannot be formed between x and y since x is both highest and lowest one.

TIPS to find combination:

1. If the coefficient of x or y and constant term is negative, then one number will be positive and other will be negative.
2. If the coefficient of x or y and constant term is positive, combination will be positive.
3. If the coefficient of x or y and constant term are positive and negative then the combination of numbers will be both positive and negative.

Easy way to solve Quadratic Equation for SBI PO

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Tricks to solve Train Problems

Introduction

A train is said to have crossed an object (stationary or moving) only when the last coach (end) of the train crosses the said object completely. It implies that the total length of the train has crossed the total length of the object.

In case of Train, The Distance covered by the Train = Length of Train + Length of Object

You have to keep in mind some Formulae which are given below:

Conversion of  km/hr into meter/sec

Following formula is used in this case

Conversion of meter/sec into km/h 

The formula is:

We can find the basic formula for the time required for a train to cross different type of objects.

Different types of Objects

On the basis of various types of objects that a train has to cross, we find the following different cases:

Question
A train 110 meter long travels at 60 km/h. How long does it take to cross,
a) a telegraph post
b) a man running at 6 km/h in the same direction
c) a man running at 6 km/h in the opposite direction
d) a platform 240 meter long
e) another train 170 meter long standing on another parallel track
f) another train 170 meter long, running at 54 km/h in same direction
g) another train 170 meter long, running at 80 km/h in opposite direction

Solution
a) We have to convert the speed of train from km per hour to meter per second by applying following formula:
Speed of Train = 60 km/h × 5/18 m/sec 

The telegraph post is a stationary object with negligible length  so following formula will be applied:
t = Length of Train/Speed of Train
Crossing Time =
        

b) The man is running in the same direction that means object is moving but of negligible length. Hence, formula will be:
Time = Length of Train / Speed of Train - Speed of Man
Crossing Time = 

c) The man is running in the opposite direction so following formula will be applied:
Crossing Time = Length of Train / Speed of Train + Speed of Man
Crossing Time = 

d) The platform is stationary but of some length so, following formula will be applied:
Crossing Time = Length of Train + Length of Platform / Speed of Train

e) Another Train is standing (stationary), so following formula will be applied:
Crossing Time = Length of First Train + Length of Second (Stationary) Train / Speed of Train

f) Another train is running in the same direction, then following formula will be applied:
Crossing Time = (Length of First Train + Length of Second Train) / (Speed of First Train - Speed of Second Train)
Crossing Time = 

168 Seconds = 2 minutes 48 seconds

g) Another train is running in opposite direction so following formula will be applied

Crossing Time = Length of First Train + Length of Second Train / Speed of 

Tricky Question 

You have to learn the basic formulae only then you can easily solve any question from this chapter.
Question
A train passes a platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/h, find the length of the platform.
Solution
Speed = Distance /Time
Because Time is given in seconds so we have to convert the speed into meter/second.
Speed (Velocity) = 54 × 5/18 = 15 meter per second

Tricks to solve Train Problems

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Formula used in Mensuration

Mensuration is one of the topics in Quants. Now a days questions coming from this chapter also. Here we are providing you Some important formulas which is used in mensuration which are useful in SSC, Railways and other exams.

Rectangle :

• Area = lb
• Perimeter  = 2(l+b)

Square :

• Area = a×a
• Perimeter = 4a

Parallelogram

• Area = l × h
• Perimeter = 2(l+b)

Triangle :

• Area =b×h/2 or √s(s-a)(s-b)(s-c)…………….where s=a+b+c/2

Right angle Triangle :

• Area =1/2(bh)
• Perimeter = b+h+d

Isosceles right angle triangle :

• Area = ½. a²
• Perimeter = 2a+d……………………….where d=a√2

Equilateral Triangle :

• Area = √3. a²/4 or  ½(ah)….where h = √3/2
• Perimeter = 3a

Trapezium :

• Area = 1/2h(a+b)
• Perimeter = Sum of all sides

Rhombus :

• Area = d1 × d2/2
• Perimeter = 4l

Quadrilateral

• Area =1/2 × Diagonal × (Sum of offsets)

Kite :

• Area  = d1×d2/2
• Perimeter  = 2 × Sum on non-adjacent sides

Circle :

• Area =  πr^2 or πd^2/4
• Circumference = 2πr  or πd
• Area of sector of a circle = (θπr^2 )/360

Frustum :

• Curved surface area = πh(r₁+r₂)
• Surface area = π( r₁²+ h(r₁+r₂) + r₂²)

Cube :

• Volume: V = l³
• Lateral surface area = 4a²
• Surface Area: S = 6s²
• Diagonal (d) = √3l

Cuboid :

• Volume of cuboid: lbh
• Total surface area = 2 (lb + bh + hl) or 6l²
• Length of diagonal =√(l^2+b^2+h^2)

Right Circular Cylinder :

• Volume of Cylinder = π r² h
• Lateral Surface Area (LSA or CSA) = 2π r h
• Total Surface Area = TSA = 2 π r (r + h)
• Volume of hollow cylinder = π r h(R² – r²)

Right Circular cone :

• Volume = 1/3 π r²h
• Curved surface area: CSA=  π r l
• Total surface area = TSA = πr(r + l )

Sphere

• Volume: V = 4/3 πr³
• Surface Area: S = 4πr²

Hemisphere :

• Volume = 2/3 π r³
• Curved surface area(CSA) = 2 π r²
• Total surface area = TSA = 3 π r²

Prism :

• Volume = Base area x h
• Lateral Surface area = perimeter of the base x h

Pyramid:

• Volume of a right pyramid = (1/3) × area of the base × height.
• Area of the lateral faces of a right pyramid = (1/2) × perimeter of the base x slant height.
• Area of whole surface of a right pyramid = area of the lateral faces + area of the base.

Tetrahedron :

• Area of its slant sides = 3a²√3/4
• Area of its whole surface = √3a² 
• Volume of the tetrahedron = (√2/12) a ³

Regular Hexagon :

• Area =  3√3 a² / 2
• Perimeter  = 6a

Some other Formula :

• Area of Pathway running across the middle of  a rectangle = w(l+b-w)
• Perimeter of Pathway around a rectangle field = 2(l+b+4w)
• Area of Pathway around a rectangle field =2w(l+b+2w)
• Perimeter of Pathway inside a rectangle field =2(l+b-4w)
• Area of Pathway inside a rectangle field =2w(l+b-2w)
• Area of four walls = 2h(l+b)

Formula used in Mensuration

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Partnership

Partnership Problems:

Profit is directly proportional to Time and Investments.
So, Profit ∝ Time  Profit ∝ Investments
So, Profit ∝ (Time × Investments )
Example 1: Three partners A, B and C invest Rs.1500, Rs.1200 and Rs.1800 respectively in a company. How should they divide a profit of Rs.900?
Solution: Given, there is no time given, we can say profit is proportional to investment.
Ratio of profit = ratio of investment
Profit ratio  of A:B:C = 1500:1200:1800 =5:4:6
so, total profit is 5+4+6 = 15 i.e. equal to 900
profit of A = (5/15)× 900 = 300
profit of B = (4/15)× 900 = 240
profit of C = (6/15)× 900 = 360
Example 2: In a company A invested Rs.1500 for 4 months and B invested Rs.1200 for 6 months and C invested Rs.3600 for 2 months. If company has a profit of Rs.680. What will be the share of A,B and C?
Solution:
Ratio of profit A:B:C = (1500 × 4):(1200 × 6):(3600 × 2)
                                     = 60:72:72
                                     = 5:6:6
total profit is 5+6+6 = 17 i.e. equal to 680.
we can say, 17 = 680
                   1 = 40
profit of A is 5, so 5× 40 = 200
profit of B is 6, so 6× 40 = 240
profit of C is 6, so 6 × 40 = 240
Note: Read questions carefully. If we can calculate capital invested and time for which capital invested. We can easily calculate share in profit.
Example 3: A and B enter into a partnership with Rs.50000 and Rs.75000 respectively in a company for a year. After 7 months, C get into partnership with them with Rs.30000 and A withdraws his contribution after 9 months. How would they share their profit of Rs.2600 at the end of the year?
Solution: A, B and C do business for 1 year but, A contributed Rs.50000 for 9 months, B contributed 75000 for 12 months and C invested Rs.30000 for 5 months not for 7 months.So ratio of profit  A:B:C = 50×9 : 75×12 : 30×5
                                          = 15 : 30 : 5
Hence total profit is (15+30+5) = 50 which is equal to 2600
So share of A = (15/50)× 2600 = 780
 share of B = (30/50)× 2600 = 1560
 share of C = (5/50) × 2600 = 260
Example 4: A, B and C started a company in which A invested (1/3)^rd of the capital for (1/4)^th of the time, B invested (1/2)^nd of the capital for (1/6)th of the time and C invested the remaining capital for whole of the time. If the profit at the end of the year is Rs.1200. How would they share it?
Solution:  A invested (1/3)^rd of the capital and B invested (1/2)^nd of the capital
So, remaining capital invested by C = 1-((1/3)+(1/2)) = 1/6
Ratio of profit A: B:C = (1/3)× (1/4) : (1/2)× (1/6) : (1/6)× 1
                                      =  (1/12):(1/12):(1/6)
                                      = 1 : 1 : 2
A’s share = (1/4)× 1200 = 300
B’s share = (1/4)× 1200 = 300
C’s share = (1/2)× 1200 = 600
Example 5: A and B rent a field for 11 months. A puts 100 bags for 9 months. How many bags can be put by B for 3 months if the ratio of their rent is 2:3?
Solution: Let B puts X bags.
the ratio of rent of A : B is 2: 3
so, (100×9) : (X × 3 ) = 2 : 3
        X = 450 bags
Example 6: If A and B entered into partnership and invested their capital in ratio of 19:15. At the end of 19 months B withdraws his capital. If they share profit in ratio of 3:2, then for how many months A invested his ratio?
Solution: Let A invested for X months.
Ratio of profit A : B = X × 19 : 19 × 15
So, 19X : 19×15 = 3:2
 X = 22(1/2) months
Example 7: Sandeep, Vineet and Shekhar are three partners. Sandeep receives 1/5 of the profit and Vineet and Shekhar share the remaining profit equally. If Vineet’s income is increased by Rs.650 when the profit rises from 10% to 15%. Find the capitals invested by Sandeep, Vineet and Shekhar and total capital invested.
Solution: As given, profit share of Sandeep is 1/5, remaining profit (1-1/5) = 4/5 is shared between Vineet and Shekar equally.
So, profit share of Vineet = 2/5 and profit share of Shekhar = 2/5
when profit % increases, Vineet’s income increase by Rs.650
 (15%-10%) = 5% = 650
100% = 13000
So, Vineet’s capital = 13000
i.e (2/5) of total capital = 13000
total capital = 32500
and Shekhar’s captal = 13000
Sandeep’s capital i.e (1/5) of total capital or ½ of (Vineet or Shekhar’s Capital) = 6500
Example 8: A, B and C start a business. Twice the capital of A is equal to thrice the capital of B and Capital of B is four times of the capital of C. What will be A’s share if the profit earned is Rs. 2,75,00
Solution: Let capital of C is C.
Given, 2A=3B and B = 4C
So, 2A = 3× 4C = 12 C
      A = 6C
Hence ratio of capital A : B : C = 6 : 4 : 1
so, Share of A = (6/11)×2,75,000 = 1,50,000
Example 9: A and B are partners in a business. They invest in the ratio 5 : 6, at the end of 8 months B withdraws. If they receive profits at the end of year in the ratio of 5 : 9, find how long A’s investment was used? (SBI PO Pre 2016 Memory based)
Solution:  Let A’s investment used for X months.
Given, ratio of invest (A : B) = 5 : 6
 ratio of time =  X : 8
ratio of profit = 5X : 6×8 and given ratio of profit = 5 : 9
so 5X/48 = 5/9
X = 48/9
X = 16/3 months
Example 10: A, B and C started a business with their investments in the ratio 1 : 2 : 4. After 6 months A invested the half amount more as before and B invested same the amount as before while C withdrew (1/4)^th of his investment after the 9 months. Find the ratio of their profits at the end of the year. (SBI Clerk Mains)
Solution: Ratio of investments A:B:C = 1:2:4, there is no changes in the investment of A and B up to 6 months and in investment of C up to 9 months.
At the end of 6 months, A invested half the amount more as before so A’s investment = 1 +(1/2)
Similarly B invest the same amount more as before = 2 + 2 = 4
But, C withdraw the (1/4)^th of the amount after 9 months = 4 – 1 = 3
ratio of profit = (1×6 + (3/2)× 6) : (2× 6 + 4× 6) : (4× 9+3× 3)  
                         = 15 : 36 : 45
                         = 5 : 12 : 15
Example 11: A sum of money is divided amongst P, Q and R in the ratio of 3 : 4 : 5. Another amount is divided amongst A and B in the respective ratio of 2 : 1. If B got Rs. 1050 less than Q, what is the amount received by R?
Solution: Let sum of money divided amongst P,Q and R is 3x, 4x and 5x respectively and sum of money divided amongst A and B is 2y and y respectively.
4x – y = 1050
another relation between x and y cannot be established. So,it cannot be determined.
Directions (12-15): In the following table, the investments and profit of three persons is given for different years in a joint business.

	Investments (In Rs.)			Profit (In Rs.)
Year	A	B	C	A	B	C
2010	15000	-----	23000	-----	82500	115000
2011	-----	6000	----	----	15000	17500
2012	-----	------	18000	42000	27000	24000
2013	-----	17000	10000	----	-----	14000
2014	11000	20000	----	----	----	----

Note:
1. Except year 2012, they invested the amounts for same period.
2. Some values are missing. You have to calculate these values per given data.
Example 12: If the total profit in 2011 is 45000, then find the ratio of the investment of B in 2010 to the investment of A in 2011.
Solution: profit of A in 2011 is 45000-(15000+17500) = 12500
B makes profit of 15000 by investing 6000
So, investment of A in 2011 = (6000/15000)× 12500 = 5000
In 2010, 23000 investment of C makes profit of Rs.115000
So, investment of B = (23000/115000)× 82500 = 16500
required ratio of (B:A) is 16500:5000 = 33:10
Example 13: If the total investment in 2014 is 46000, then the ratio of profit in 2014 is?
Solution: investment of C is 46000 – (20000+11000) = 15000
Time period is the same, so ratio of profit will be also same as ratio of investment = 11:20:15
Example 14: In year 2012 total investment of A and B is 30000, A and B invested their amount for 4 months and 6 months respectively then find the number of months that C invested his amount ?
Solution:  ratio of profit (A:B) = 42000: 27000
              A× 4 : B× 6 = 42000 : 27000
                 A : B = 21 : 9 = 7 : 3
 So, investment of A is 21000 and investment of B is 9000.
let C invested 18000 for X months.
So, (18000× X) : (21000 × 4) = 24000 : 42000
 X = (8/3) months, Hence C invested for 8/3 months.
Example 15: If the total profit in year 2013 is 58800 then the investment of A is?
Solution : Rs.10000 investment of C gives profit of Rs.14000
then, Rs.17000 investment of B will give profit of Rs. (14000/10000)× 17000 = 23800
So, profit of A is 58800 – (14000+23800) = 21000
Investment of A is = (14000/10000)×21000 = 15000

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Simple Interest

Simple Interest (SI)

If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI). (fix percentage of principal)
 Principal (sum)
Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.
 What is Amount?
The Addition of Simple Interest and Principle is called the Amount.
A = S.I + P ( Principle )
 Interest
Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed calculated on the basis of Principle.
 Time
This is the duration for which money is lend / borrowed.
 Rate of Interest
It is the rate at which the interest is charge on principal.
What is Per annul means?
"Rate of interest R%  per annum" means that the interest for one year on a sum. If not stated explicitly, rate of interest is assumed to be for one year.
 Formulas Need to Remember
S.I =[( P X R X T )/( 100 )].
Where P = Principle, R = Rate of per annul, T = Number of years
 From the above formula , we can derive the followings
P=(100×SI)/ RT
R=(100×SI)/ PT
T=(100×SI)/ PR

Some Tricks to Solve easily
Trick 1:
If a sum of money becomes “n” times in “T years” at simple interest, then the rate of interest per annum can be given be



Trick 2:
If an amount P1 is lent out at simple interest of R1% per annum and another amount P2 at simple interest rate of R2% per annum, then the rate of interest for the whole sum can be given by



Trick 3: A sum of money at simple interest n1 itself in t1 year. It will become n2 times of itself in (If Rate is constant)
 

Trick 4:
In what time will the simple interest be “n” of the principal at “r %” per annum:-
rt =n x 100
 Trick 5:
If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R1, R2, ... , Rn respectively and time periods are T1, T2, ... , Tn respectively, then the ratio in which the sum will be divided in n parts can be given by

Ex. 1. Find the simple interest on Rs.400 for 5 years at 6 per cent.
Solution:

Interest for a number of days
When the time is given in days or in years and days, 365 days are reckoned to a year. But when the time is given in months and days, 12 months are reckoned to a year and 30 days to the month. The day on which the money is paid back should be include be but not the day on which it is borrowed, ie, in counting, the first day is omitted.
Ex. 2. Find the simple interest on Rs.306. 25 from March 3^rd to July 27^th at  per annum.
Solution:
Interest = Rs. 
=  = Rs. 4.59
To find principal:-
Since I =                             
Ex.3. What sum of money will produce Rs.143 interest in   years at  p.c. simple interest?
Solution:
Let the required sum be Rs. P. Then
Rs P = 
To find rate %:-
Since I =                              

Ex. 4. A sum of Rs.468.75 was lent out at simple interest and at the end of 1 year 8 months the total amount was Rs500. Find the rate of interest per cent annum.
Solution:
Here, P =Rs468.75, t =  or 
I = Rs.(500-468.75) = Rs.31.25
rate p.c. = 
To find Time:-
Since, I = P t r / 100                        
Ex. 5. In what time will Rs.8500 amount to Rs.15767.50 at  per cent per annum?
Solution:
Here , interest = Rs.15767.50 – Rs.8500 =Rs.7267.50

Miscellaneous Examples on Simple Interest:-
Ex.6: The simple interest on a sum of money is 1/9^th of the principal, and the number of years is equal to the rate per cent per annum. Find the rate per cent.
Solution:
Let principal = P, time = t year, rate = t
Then,                         
Hence, rate =  %
Direct formula:
Rate = time =  %
Ex. 7: The rate of interest for the first 2 yrs is 3% per annum, for the next 3 years is 8% per annum and for the period beyond 5 years 10% per annum.To fetch an interest of 1520 in six years, money did he deposit?
Solution
Let his deposit be = Rs 100
Interest for first 2 yrs = Rs 6
Interest for next 3 yrs = Rs 24
Interest for the last year = Rs 10
Total interest = Rs 40
When interest is Rs 40, deposited amount is Rs 100
when interest is Rs 1520, deposited amount =  = Rs 3800
Direct formula:
Principle = Interest*100/r₁t₁+r₂t₂+r₃t₃… =  =Rs 3800
 Ex.8: A sum of money doubles itself in 10 years at simple interest. What is the rate of interest?
Solution:
Let the sum be Rs 100
After 10 years it becomes Rs 200
Interest = 200 - 100 = 100
Then, rate =  = 
Direct formula:
Time × Rate = 100 (Multiple number of principal – 1)
Or, Rate = 
Using the above formula rate = 
Ex.9: A sum was put at a certain rate for 2 yrs. Had it been put at 3% higher rate, it would have fetched Rs 300 more. Find the sum.
Solution:
Let the sum be Rs x and the original rate be y% per annum. Then, new rate = (y+3) % per annum

xy +3x – xy =15,000 or, x =5000
Thus, the sum =Rs 5000
Gradestack Method : Direct Formula
Sum =  = 
Ex.10. The simple interest on a certain sum of money at 4% per annum for 4 yrs is Rs 80 more than the interest on the same sum for 3 yrs at 5% per annum. Find the sum.
Solution:
Let the sum be Rs.x , then, at 4% rate for 4 yrs the simple interest =  = Rs.  
At 5% rate for 3 yrs the simple interest =  = Rs 
Now, we have,   or, 
Or,  x= Rs 8000
Gradestack Method : For this type of question
Sum =  = Rs 8000
We hope you all like the post. Don’t forget to share your feedback with us in the comment box.

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