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


23 Questions MCQ Test Olympiad Preparation for Class 10 | Linear Equation In Two Variables - Olympiad Level MCQ, Class 10 Mathematics


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This mock test of Linear Equation In Two Variables - Olympiad Level MCQ, Class 10 Mathematics for Class 10 helps you for every Class 10 entrance exam. This contains 23 Multiple Choice Questions for Class 10 Linear Equation In Two Variables - Olympiad Level MCQ, Class 10 Mathematics (mcq) to study with solutions a complete question bank. The solved questions answers in this Linear Equation In Two Variables - Olympiad Level MCQ, Class 10 Mathematics quiz give you a good mix of easy questions and tough questions. Class 10 students definitely take this Linear Equation In Two Variables - Olympiad Level MCQ, Class 10 Mathematics exercise for a better result in the exam. You can find other Linear Equation In Two Variables - Olympiad Level MCQ, Class 10 Mathematics extra questions, long questions & short questions for Class 10 on EduRev as well by searching above.
QUESTION: 1

The number of solutions of the equation 2x + y = 40, where both x and y are positive integers and x ≤ y is :

Solution:

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.

QUESTION: 2

A confused bank teller transposed the rupees and paise when he cashed a cheque for Mansi, giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise, Mansi noticed that she was left with exactly three times as much as the amount on the cheque. Which of the following is a valid statement about the cheque amount?

Solution:

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

QUESTION: 3

John inherited $25000 and invested part of it in a money market account, part in municipal bonds, and part in a mutual fund. After one year, he received a total of $ 1620 in simple interest from the three investments. The money market paid 6% annually, the bonds paid 7% annually, and the mutual funds paid 8% annually. There was $ 6000 more invested in the bonds than the mutual funds. The amount John invested in each category are in the ratio:

Solution:

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. 

QUESTION: 4

Which one of the following conditions must p,q and r satisfy so that the following system of linear simultaneous
equations has atleast one solution, such that p + q + r = 0?
x + 2y – 3z = p ; 2x + 6y – 11z = q ; x – 2y + 7z = r

Solution:

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

QUESTION: 5

If x and y are integers, then the equation 5x + 19y = 64 has :

Solution:

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

QUESTION: 6

Rozly can row downstream 20km in 2 hours, and the upstream 4km in 2 hours. What will be the speed of rowing in still water?       

Solution:

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.  

QUESTION: 7

Study the question and statements given below : Decide whether any information provided in the statement(s)
is rebundant and/or can be dispensed with, to answer it.If 7 is added to numerator and denominator each of fraction a/b, will the new fraction be less than the original one? (Assume both a and b to be positive)
Statement-I : a = 73, b = 103
Statement-II : The average of a and b is less than b.
Statement-III : a – 5 is greater than b – 5.

Solution:
QUESTION: 8

A cyclist drove 1 km, with the wind in his back, in 3 min and drove the same way back, against the wind in 4 min. If we assume that the cyclist always puts constant force on the pedals, how much time would it take to drive 1 km without wind?

Solution:
QUESTION: 9

A person buys 18 local tickets for Rs. 110. Each first class ticket costs Rs. 10 and each second class ticket costs Rs. 3. What will another lot of 18 tickets in which the number of first class and second class tickets are interchanged cost?

Solution:

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

QUESTION: 10

Rajesh walks to and fro to a shopping mall. He spends 30 min. shopping. If he walks at a speed of 10 km/h, he returns to home at 19:00h. If he walks at 15 km/h, he returns at 18:30 h. How fast must he walk in order to return home at 18:15 h?

Solution:

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:

QUESTION: 11

A single reservoir supplies the petrol to the whole city, while the reservoir is fed by a single pipeline filling the reservoir with the stream of uniform volume. When the reservoir is full and if 40000 litres of petrol is used daily, the supply fails in 90 days. If 32000 litres of petrol is used daily, the supply fails in 60 days. How much petrol can be used daily without the supply ever failing?

Solution:

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

QUESTION: 12

Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other
after one hour and the first horse reaches B, 5/6 hours before the second horse reaches A. If the distance
between A and B is 50 km. What is the speed of the slower horse?

Solution:

QUESTION: 13

A man row downstream at 12 km/h and upstream at 8 km/h. What is the speed of man in still water?

Solution:

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

QUESTION: 14

A motor boat takes 12 hours to go downstream and it takes 24 hours to return the same distance. What is the time taken by boat in still water?

Solution:

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


QUESTION: 15

Equation yx2 + xy2 = 2xy : x, y = 0 is

Solution:

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.

QUESTION: 16

Sum of two integers is 88. If the greater is divided by the smaller, the quotient is 5 and the remainder is 10. The greater integer is :

Solution:

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

QUESTION: 17

The length of the sides of a triangle are 3x + 2y, 4x +  y and 3(x + 1) +  (y – 1). If the triangle is equilateral, then its side is

Solution:

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

QUESTION: 18

The largest angle of a triangle is twice the sum of the other two. The smaller angle is one fourth of the largest.
The largest angle is :

Solution:

↪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°

QUESTION: 19

In town, 2/3 of men are married to 3/7 of the women. In the town total population is more than 1000. If all
marriages happen within the town. The smallest possible number of total population is (assume there are only
adults in the town) :

Solution:

QUESTION: 20

The solution of the equations :  , 7x + 8y + 5z = 62 is :

Solution:

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

QUESTION: 21

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. Find the number of students in the class.       

Solution:

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

QUESTION: 22

The solution of the equations :  = 110,  = 132,  =  is :

 

Solution:
QUESTION: 23

Four men earn as much in a day as 7 women. 1 woman earns as much as 2 boys. If 6 men, 10 women and 14 boys work together for 8 days to earn Rs. 2200, then what will be the earning of 8 men and 6 women working together for 10 days?

Solution:

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