Linear Equation In Two Variables - Olympiad Level MCQ, Class 10 Mathematics - Class 10 MCQ

2x + y = 40

x <=y

y = x + k; where k >=0

3x + k = 40

x = ( 40 - k ) / 3

x is a positive integer so takes value from 1 to 40 all multiples of 3

13 values.

let us consider

ab = Money in rs.

cd = money in paise.

so here we have  

3(ab.cd) = cd.ab - 0.50

3ab + 3cd/100 = cd + ab/100 - 0.50

299ab + 50 = 97cd

cd = (299ab + 50)/97

cd = 3ab + (8ab + 50)/97

to get cd as two digit number we have, ab = 18

which gives cd = 56

So money = Rs 18.56

Check

56.18 - .50 = 3(18.56)

55.68 = 55.68

Let

x = The amount of money invested in the money market account.
y = The amount of money invested in municipal bonds.
z = The amount of money invested in a mutual fund.

x+y+z = 25000    (1)

0.06x+0.07+0.08z = 1620  (2)

y-z = 6000   (3)

solve for x, y and z using any method from substitution, elimination or cross multiplication

 Answer is  $15,000 is invested in the monkey market account, $8,000 is invested in the municipal bonds, and $2,000 is invested in mutual funds. 

It is given that p + q + r ≠ 0

Multiply the first equation by 5, second by -2 and third by -1, the coefficients of x, y and z all add up to zero.

5x + 10y - 15z = 5p

-4x - 12y + 22z = -2q

-x + 2y - 7z = -r

Adding all three equations:

5p - 2q - r = 0

The correct option is A

5x + 19y = 64

If y = 1, there is an integer solution for x = 9, now if y changes (increases or decreases) by 5, then x will change (decrease or increase) by 19.

Looking at the options, if x = 256, you get y = 64.

So, there is a solution for 250 < x < 300

The speed of the rowing on still water denoted by M kmph.

The speed of the stream denoted by N kmph.

The upstream speed of Ritu’s rowing is (M – N) kmph

The downstream speed of Ritu’s rowing is (M + N) kmph

2(M+N)=20   (1)  

2(M-N)=4      (2)

Solving the equation, we get  

2(M+N)=20;

2(M-N)=4;  

M-N=2;

M=2+N;

Substituting M=2+N in equation (1)

2(M+N)=20

2(2+N+N)=20 ;  

N=4  

Substituting N = 4 in equation (1),  

2(M+N)=20

2(M+4)=20

2M+8=20

2M=20-8

2M=12

M=6  

the value of M = 6.

Therefore, the speed at which Ritu rows is 6 kmph and the speed of the current is 4 kmph.  

Let there are x first class ticket
Total cost =10x + 3(18−x)
⇒ 10x + 54 − 3x= 110
⇒ 7x = 56
⇒ x= 8

If the first class and second class tickets are interchanged.
Then total cost = 10 × 10 + 3 × 8 = 124

As per the question, let D be the total distance and ‘t’ is the time taken. 
So we have:
D=10t =15(t−0.5)
⇒ t=1.5 hrs
⇒ D=15 km

Now, for the condition given we have:

Let quantity of petrol filled every day = x litre

Total volume = v litre

When the reservoir is full and if 40,000 litres of petrol is used daily, the supply fails in 90 days

v = (40000- x)*90

If 32,000 litres of petrol is used daily, supply fails in 60 days. 

v = (32000- x)*60

(40000- x)*90 = (32000- x)*60

(40000- x)*3= (32000- x)*2

120000 - 3x = 64000 - 2x

x = 56000

Quantity of petrol filled every day = 56000 litre

Therefore 56000 litre can be used daily without supply ever failing

Let speed of man be xkm/hr and speed of stream be y km/hr .

the speed of man during downstream (x+y)km/hr and during upstream (x-y)km/hr.

now solve (x+y)=12and (x-y)=8 .

yow will get man speed 10 km/hr

If t1 and t2 are the upstream and down stream times. Then time taken in still water is given by

At first we will bring RHS TO LHS. then take XY as common. We will find that equation has become xy(x+y-2)=0.Hence x+y-2 =0 so it is a linear equation.

Let the integers=x and 88-x,
so, As given in the question->
5x+10=88-x,
6x=78,
x=13,
and the other number is 88-13=75,
since 75>13,
the greater integer is75,
Hence option 'B 'is correct

Since all sides of equilateral triangle are equal
3x+2y = 4x + 4/3y = 3(x+1) +3/2(y-1)
then,
3x + 2y = 4x + 4/3y
or,. 3x - 4x = 4/3y - 2y
or. - x = (4y - 6y)/3
or. - x = -2y/3
or. x = 2y/3. ( I )

again ,

4x + 4/3y = 3( x+1 ) + 3/2( y-1 )
or. 4x + 4/3y = 3x + 3 + 3/2y - 3/2
or 4x - 3x = 3/2y - 4/3y + 3 - 3/2
or x = ( 9y - 8y )/6 + ( 6 - 3 )/2
or. x = y/6 + 3/2 ( ii )

from equations ( I ) & ( ii ) we get ,
2y/3. = y/6 + 3/2
or. 2y/3 = ( y + 9) / 6
or 6 × 2y = 3( y + 9 )
or. 12y = 3y + 27
or. 12y - 3y = 27
or. 9y = 27
or. y = 3


putting the value of y in equation ( I )
x = (2 × 3)/3
x = 6/3
x = 2

hence ,

3x + 2y = 3 × 2 + 2× 3
= 6 + 6
= 12
therefore C is your correct answer

↪Let the two smaller angles be x and y respectively 

=> Now , according to the given question ->

=> Larger angle = 2 ( x + y ) ....................( 1 )

▪Applying the Angle Sum Property of a ∆ ,

=> 2 ( x + y ) + x + y = 180

=> 2x + 2y + x + y = 180

=> 3x + 3y = 180

-> Taking 3 common ,

=> 3 ( x + y = 60 

=> Hence ( x + y ) = 60 . Put this value in Eq. (1)

=> Larger Angle = 2 ( x + y )

=> 2 ( 60 )

=> 120°

=> Smallest Angle = ¼ × 120 = 30°

=> Third Angle = 180 - ( 120 + 30 )

=> 180 - 150 = 30°

♦Hence the Angles are 30° , 30° and 120°

Let x/4=y/3=z/2=k
x=4k
y=3k
z=2k
7x+8y+5z=62
⇒7*4k+8*3k+5*2k=62
⇒28k+24k+10k=62
⇒62k=62
⇒k=62/62
∴k=1

so x/4 = y/3 = z/2 = 1

x=4 y=3 z=2

5 customers in a Line

Total Number of customers = 60

Step-by-step explanation:

Customers are asked to stand in the lines. If one customer is extra in a line, then there would be two less lines. If one customer is less in line, there would be three more lines.

Let say There are  C customers in a Line  and total L number of lines

Total number of customers = ( customers in a line) * (number of Lines)

=>Total number of customers =  CL

If one customer is extra in a line, then there would be two less lines

=> Total number of customers = (C + 1)(L -2)

(C + 1)(L -2)  = CL

=> CL + L - 2C - 2 = CL

=> L - 2C  = 2   - eq 1

If one customer is less in line, there would be three more lines.

=> Total number of customers = (C - 1)(L +3)

(C - 1)(L +3) = CL

=> CL - L + 3C - 3 = CL

=> - L + 3C = 3    - eq 2

Adding eq 1 & eq 2

=> C = 5

L - 2(5) = 2

=> L = 12

5 customers in a Line

Total Number of customers =  CL = 5*12  = 60

Let earning of 1 boy in 1 day =Rs.x

now,8×[{(14×6)/4}+(10×2)+14]x = Rs.2200..

55x = 275..

so,x=5

Earning of 8 men and 6 women in 1day
={8×(7/4×10)}+{6×10} = 140+60 = 200

so, In 10 days= Rs.200×10 =Rs.2000