Simultaneous Linear Equations Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions. In this article you will learn how to solve MCQs on Simultaneous Linear Equations easily. Visit official Website CISCE for detail information about ICSE Board Class-9.
Simultaneous Linear Equations Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-5 | Simultaneous Linear Equations |
| Writer | RS Aggrawal |
| Topics | Solving MCQs on Simultaneous Linear Equations |
| Academic Session | 2024-2025 |
How to Solve MCQs on Simultaneous Linear Equations
Solving MCQs problems on Simultaneous Linear Equations is very easy if you know how to create formula using variable. Suppose unknown as variable and create equation using variable according to given condition. After creating both equations solve it by any suitable method of solving Simultaneous Linear Equations. At last focus on option given, Choose the most suitable option among them.
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Multiple Choice Questions:
(Simultaneous Linear Equations Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions)
Que-1: The solution of the simultaneous equations 3x-2y=5 and x+2y=-1 is :
(a) x=1, y=1 (b) x=1, y=-1 (c) x=-1, y=1 (d) x=-1, y=-1
Sol: (b) x=1, y=-1
Reason: 3x-2y = 5 …… (i)
x+2y = -1 …. (ii)
On adding both equations we get,
4x = 4
x = 1
Put the value of x in eqn (ii) we get,
1 + 2y = -1
2y = -2
y = -1.
Que-2: The solution of the simultaneous equation (x/2)-(y/3)=0 and (3x/2)+(2y/3)+10=0 is :
(a) x=4, y=6 (b) x=4, y=-6 (c) x=-4, y=6 (d) x=-4, y=-6
Sol: (d) x=-4, y=-6
Reason: (x/2) – (y/3) = 0 ……. (i)
x/2 = y/3
x = 2y/3
(3x/2) + (2y/3) + 10 = 0 ……. (ii)
Put the value of x in second equation we get,
2y/2 + 2y/3 + 10 = 0
(3y+2y)/3 + 10 = 0
5y/3 + 10 = 0
5y/3 = -10
y = -10 x 3/5
y = -6.
Put the value of y in x = 2y/3 we get,
x = {2(-6)}/3
x = -4.
Que-3: The solution of the simultaneous equations 2x+(1/y)=0 and 3x+(1/2y)=-2 is :
(a) x=1, y=1/2 (b) x=1, y=-1/2 (c) x=-1, y=1/2 (d) x=-1, y=-1/2
Sol: (c) x=-1, y=1/2
Reason: 2x+(1/y) = 0 ….. (i)
1/y = -2x
3x+(1/2y) = -2 ……. (ii)
Put the value of 1/y in eqn (ii) we get,
3x + {1/2 x 1/y} = -2
3x + {1/2 x -2} = -2
3x – x = -2
2x = -2
x = -1
Put the value of x in 1/y = -2x we get,
1/y = -2(-1)
1/y = 2
y = 1/2.
Que-4: If (x/6)+6=y, 3x+4=1+y, then;
(a) x=12, y=8 (b) x=10, y=8 (c) x=8, y=12 (d) x=12, y=6
Sol: (a) x=12, y=8
Reason: (x/6)+6 = y
y-6 = x/6
y-x/6 = 6
6y-x = 36—->(1)
3x+4 = 1+y
3x/4 = 1+y
3x/4-y = 1
3x-4y = 4—–>(2)
equating 1 and 2
14y = 112
y = 8
substitute in eq 1 we get
6×8-x = 36
x = 48-36 = 12
x = 12 and y = 8
Que-5: If x and y are real numbers and (2x-1)^2 + (3y-1)^2 = 0, then [(1/x^2)+(1/y^2)]
(a) 25 (b) 13 (c) 1/13 (d) -1/13
Sol: (b) 13
Reason: (2x-1)^2 = 0
2x-1 = 0
x = 1/2
(3y-1)^2 = 0
3y-1 = 0
y = 1/3
Therefore,
(1/(x ^ 2) + 1/(y ^ 2)) = (2)^2 + (3)^2
(1/(x ^ 2) + 1/(y ^ 2)) = 4+9
(1/(x ^ 2) + 1/(y ^ 2)) = 13
Que-6: The solution of √5x – √7y = 0 and √3y + √13x = 0
(a) x=0, y=0 (b) x=0, y=1 (c) x=1, y=0 (d) x=1, y=-1
Sol: (a) x = 0, y = 0
Reason: √5x – √7y = 0 ….. (i)
√3y + √13x = 0 ……. (ii)
Multiply by √3 in eqn (i) and by √7 in eqn (ii) we get,
√15x – √21y = 0
√21y + √91x = 0
subtract both equation we get,
√15x – √91x = 0
x (√15-√91) = 0
x = 0
y = 0.
Que-7: The solution of 0.4x+3y=1.2 and 7x-2y=17/6 is :
(a) x=1/3, y=1/2 (b) x=1/2, y=1/3 (c) x=1/3, y=1 (d) x=0, y=1/2
Sol: (b) x=1/2, y=1/3
Reason: 0.4x + 3y = 1.2 ………. (i)
7x-2y = 17/6 ……….. (ii)
Multiply by 10 in eqn (i) and 6 in eqn (ii) we get,
4x+30 = 12 …….. (iii)
42x-12y = 17 …….. (iv)
Multiply by 42 in eqn (iii) and by 4 in eqn (iv) we get,
168x + 1260y = 504 ………. (v)
168x – 48y = 68 ………… (vi)
Subtract the eqn (vi) from eqn (v) we get,
1308y = 436
y = 436/1308
y = 1/3
Put the value of y in eqn (i) we get,
0.4x + 3(1/3) = 1.2
0.4x + 1 =1.2
0.4x = 1.2 – 1
0.4x = 0.2
x = 0.2/0.4
x = 1/2.
Que-8: 2 tables and 3 chairs together costs Rs1075 and 3 tables and 8 chairs together costs Rs1875. The cost of 4 tables and 5 chairs together will be :
(a) Rs2750 (b) Rs2705 (c) Rs2075 (d) Rs2057
Sol: (c) Rs2075
Reason: let price of 1 table = x
price of 1 chair = y
price of 2 tables, 3 chairs = ₹1075
2x + 3y = 1075 … (1)
price of 3 tables, 8 chairs = ₹1875
3x + 8y = 1875 … (2)
(1) × 3 : 6x + 9y = 3225
(2) × 2 : 6x + 16y = 3750
subtracting two equations,
-7y = -525
divide by 7 on both sides,
y = 75
sub y in (1),
2x + 3(75) = 1075
2x + 225 = 1075
2x = 1075 – 225
2x = 850
divide by 2 on both sides,
x = 425
cost of 1 table : x = ₹425
cost of 1 chair : y = ₹75
cost of 4 table + cost of 5 chair
= 4x + 5y
= 4(425) + 5(75)
= 1700 + 375
= ₹2075.
Que-9: The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Then the original number is :
(a) 90 (b) 18 (c) 81 (d) 54
Sol: (b) 18
Reason: Let unit digit = x
Tens digit = y
Number will 10 times the tens digit + unit times the unit digit
Hence number will 10 y + x
Sum of digits are 9
So that
X + y = 9 ………….(1)
nine times this number is twice the number obtained by reversing the order of the digits
9 (10 y + x ) = 2 (10 x + y )
90 y + 9 x = 20 x + 2y
88 y – 11 x = 0
Divide by 11 we get
8 y – x = 0 …………..(2)
X + y = 9 ………….(1)
Adding both equations we get
9 y = 9
Y = 9/9 = 1
Plug this value in equation first we get
X+ y = 9
X + 1 = 9
X = 8
So our original number is 10 y + x = 10*1 + 8 = 18
Que-10: Five year ago, Bharat was thrice as old as Rajat. Ten year later, Bharat will be twice as old as Rajat. The difference between their present ages is :
(a) 10 year (b) 20 year (c) 30 year (d) 35 year
Sol: (c) 30 years
Reason: Let the age of A be a, and age of B be b
Now, a−5 = 3(b−5)
a = 3b−10 (1)
After 10 years,
a+10 = 2(b+10)
a = 2b+10 (2)
From 1 and 2,
3b−10 = 2b+10
b = 20
And, a = 2b+10 = 2×20+10 = 50 years
Thus, age of A = 50 years and age of B = 20 years.
So, difference between their ages is
A-B = 50 – 20 = 30 years.
Que-11: A bag contains some one-rupee coins and some fifty-paisa coins. The total amount is Rs140. If half of the one-rupee coins are replaced by fifty-paisa coins, then the amount becomes Rs115. The coins of each type in the big initially, were :
(a) one-rupee coins = 100 and fifty-paisa coins = 80
(b) one-rupee coins = 80 and fifty-paisa coins = 100
(c) one-rupee coins = 110 and fifty-paisa coins = 80
(d) one-rupee coins = 70 and fifty-paisa coins = 90
Sol: (a) one-rupee coins = 100 and fifty-paisa coins = 80
Reason: If half of the one-rupee coins are replaced by fifty-paisa coins, the total amount becomes Rs 115.
First, we can set up the initial equation based on the total amount:
x + 0.5y = 140 ……… (i)
Next, if half of the one-rupee coins are replaced by fifty-paisa coins, we have x/2 one-rupee coins left and y+(x/2) fifty-paisa coins. The new total amount is Rs 115. Therefore, we can write the second equation as:
(x/2) + 0.5(y+(x/2)) = 115
Simplify the second equation:
(x/2) + 0.5y + 0.25x = 115
0.75x + 0.5y = 115 ………….. (ii)
Subtract the eqn (ii) from eqn (i) we get,
0.25x = 25
x = 25/0.25
x = 100.
Now substitute x = 100 into equation (1):
100 + 0.5y = 140
0.5y = 40
y = 80
Therefore, the bag initially contains:
100 one-rupee coins
80 fifty-paisa coins
Que-12: X takes 3 hours more than Y to walk a distance of 30km, but if X doubles his race, he is able to be ahead of Y by 1*(1/2) hours, then speed of their walking will be :
(a) X’s speed = 10/3 km/hr and Y’s speed = 5 km/hr
(b) X’s speed = 5 km/hr and Y’s speed = 10/3 km/hr
(c) X’s speed = 10 km/hr and Y’s speed = 5/3 km/hr
(d) X’s speed = 5/3 km/hr and Y’s speed = 10 km/hr
Sol: (a) X’s speed = 10/3 km/hr and Y’s speed = 5 km/hr
Reason: Let the speed of X and Y be x km/hr and y km/hr respectively. Then,
Time taken by X to cover 30km = 30/x hrs,
and, Time taken by Y to cover 30km = 30/y hrs
By the given conditions, we have
(30/x) − (30/y) = 3
⇒ (10/x) − (10/y) = 1 .(i)
If X doubles his pace, then speed of X is 2x km/hr
Times taken by X to cover 30 km = 30/2x hrs
Times taken by Y to cover 30km =30/y hrs
According to the given conditions, we have
(30/y) − (30/2x) = 3/2
⇒ (10/y) − (5/x) = 1/2
⇒ (−10/x) + (20/y) = 1 …..(ii)
Putting 1/x = u and 1/y = v, in equations (i) and (ii) we get
10u − 10v = 1
⇒ 10u − 10v − 1 = 0 (iii)
−10u + 20v = 1
⇒ −10u + 20v − 1 = 0 (iv)
Adding equations (iii) and (iv), we get
10v − 2 = 0
⇒ v = 15
Putting v = 15 in equation (iii), we get
10u − 3 = 0
⇒ u = 3/10
Now, u = 3/10
⇒ 1/x = 3/10
⇒ x = 10/3 and,
v = 1/5
⇒ 1/y = 1/5
⇒ y = 5.
Hence, X’s speed = 10/3 km/hr and, Y’s speed = 5 km/hr.
Que-13: A boat take 10 hours to go 44km downstream and 30km upstream. Again, the same boat takes 13 hours to go 55km downstream and 40km upstream. The speed of the boat and the current will be :
(a) speed of boat = 3kmph, speed of current = 2kmph
(b) speed of boat = 6kmph, speed of current = 4kmph
(c) speed of boat = 9kmph, speed of current = 2kmph
(d) speed of boat = 8kmph, speed of current = 3kmph
Sol: (d) speed of boat = 8kmph, speed of current = 3kmph
Reason: Let the speed of the boat in still water be x km/hr and the speed of the stream by y km/hr.
Therefore, the speed of the boat downstream = (x + y) km/hr and the speed of the boat upstream
= (x – y) km/hr
Now, time = distance/speed
Therefore, time taken by the boat to cover 30 km upstream = 30/(𝑥- 𝑦) hours and the time taken by the boat to cover 24km downstream = 44/(𝑥+𝑦) hours.
But the total time is taken by the boat to cover 30 km upstream and 44 km downstream is 10 hours.
∴ 30/(𝑥-𝑦) + 44/(𝑥+𝑦) = 10 …(i)
similarly by second condition,
40/(𝑥-𝑦) + 55/(𝑥-𝑦) = 13 ….(ii)
substituting 1/(𝑥-𝑦) = 𝑎 and 1/(𝑥-𝑦) = 𝑏 in equation (i) and (ii)
∴ 30a + 44b = 10 ….(iii)
40a + 55b = 13 ….(iv)
Equation (iii) x (iv) and eqation (iv) x (iii), we get
120a + 176b = 40 ….(v)
120a + 165b = 39 …(vi)
equation (v) – equation (vi) , we get
11b = 1
𝑏 = 1/11
substituting 𝑏 = 1/11 in equation (v), we get
120𝑎+176(1/11) = 40
120a = 40 – 16
120a = 24
𝑎 = 24/120
𝑎 = 1/5
Now, 1/(𝑥-𝑦) = 1/5 and 1/(𝑥+𝑦) = 1/11
x – y = 5 and x + y = 11
x + y = 11 …..(vii)
x – y = 5 ……(viii)
Adding equation (vii) and equation (viii) , we get
2x = 16
x = 8
Subsidity x = 8 in equation (vii) we get y = 3
∴ speed of the boat in still water is 8 km/hr and speed of the current is 3 km/hr
Que-14: 42 mangoes are to be distributed among some boys and girls. If each boy is given 3 mangoes, then each girls get 6 mangoes and if each boy gets 5 mangoes, then each girl gets 3 mangoes. The number of boys and girls will be :
(a) boys = 4, girls = 6
(b) boys = 6, girls = 4
(c) boys = 7, girls = 3
(d) boys = 3, girls = 7
Sol: (b) boys = 6, girls = 4
Reason: Let the boy be x and girls be y
3x+6y = 42 ………. (i)
5x+3y = 42 ……… (ii)
Multiply by 2 in eqn (ii) we get,
10x+6y = 84 ……….. (iii)
Adding eqn (i) and eqn (iii) we get,
7x = 42
x = 6
Putting the value of x in eqn (i) we get,
3(6)+6y = 42
18+6y = 42
6y = 42-18
6y = 24
y = 4.
Que-15: If ∠A = 2x°, ∠B = (6y+10)°, ∠C = (2x+y)°, ∠D = (x+10)° are the angles of a quadrilateral , then the value of the x and y will be
(a) x = 40°, y = 20° (b) x = 20°, y = 40° (c) x = 45°, y = 15° (d) x = 15°, y = 45°
Sol: (a) x = 40°, y = 20°
Reason: We know that, by property of cyclic quadrilateral,
Sum of opposite angles = 180°
∠A + ∠C = 2x+2x+y = 180
4x+y = 180 ……… (i)
And ∠B + ∠D = 6y+10+x+10 = 180
x+6y = 180-20
x+6y = 160 ………. (ii)
Multiply by 4 in eqn (ii) we get,
4x+24y = 640 ………. (iii)
On adding eqn (i) and eqn (iii) we get,
23y = 460
y = 20°
Putting the value of y in eqn (i) we get,
4x+20 = 180
4x = 180-20
x = 160/4
x = 40°.
– : End of Simultaneous Linear Equations Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths solutions :–
Return to : – RS Aggarwal Solutions for ICSE Class-9 Mathematics
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