Equations mcq

Question 1.
A man went to the Reserve Bank of India with ₹ 1,000. He asked the cashier to give him ₹ 5 and ₹ 10 notes only in return. The man got 175 notes in all. Find how many notes of ₹ 5 and ₹ 10 did he receive? 
(a) (25,150)
(b) (40,110)
(c) (150,25)
(d) None
Answer:
Tricks :
Go by choices
For (a) 25 × 5 + 150 × 10 ≠ Rs. 1000
(b) 40 × 5 + 110 × 10 ≠ Rs 1000
(c) 150 × 5 + 25 × 10 = Rs 1000
(c) is correct

Question 2.
A man rowing at the rate of 5 km in an hour in still water takes thrice as much time in going 40 km up the river as in going 40 km down. Find the rate at which the river flows: 
(a) 9 km/hr
(b) 10 km/hr
(c) 12 km/hr
(d) None
Answer:
Tricks : Go by choices
Given Distance = 40 km
(a) t = 409−5 (upstream) = 10hrs
Down Stream t = 409+5=4014=207 hrs
∴ (a) is not correct.

(b) Down stream t = 4010+5=4015=83
Upstream, t = 4010−5=405hrs
Clearly; 8 = 3 × 83
∴ (b) is correct

Question 3.
The point of intersection of the lines 2x – 5y = 6 and x + y = 3 is . 
(a) (0, 3)
(b) (3, 0)
(c) (3, 3)
(d) (0, 0)
Answer:
(b) Tricks : Go by choices.
Intersecting Point lies on both straight lines. It will satisfy both eqns.
∴ For (a) (0,3) Point 2 × 0 – 5 × 3 ≠ 6
∴ It is incorrect
Option (b) (3 ; 0) satisfies both eqns.
∴ (b) is correct.

Question 4.
If the length of a rectangle is 5 cm more than the breadth and if the perimeter of the rectangle is 40 cm, then the length & breadth of the rectangle will be: 
(a) 7.5 cm, 2.5 cm
(b) 10cm, 5cm
(c) 2.5 cm, 7.5cm
(d) 15.5cm, 10.5cm
Answer:
(c) Tricks : Go by choices.
1st condition = length is 5cm more than breadth
All option satisfy this condition
II Perimeter = 2(l+b) = 40 of rectangle
Only option (c) satisfies it
∴ (c) is correct.

Detail Method
Let breadth = x ;Length = x + 5
Perimeter = 40
2 (x+ 5 + x) = 40
or 2x + 5 = 20
or x = 152 = 7.5cm
∴ Length = x + 5 = 7.5 + 5 = 12.5
Breadth = x = 7.5 cm
∴ (c) is correct

Question 5.
For all λ ∈ R , the line (2 + λ)x + (3 – λ) y + 5 = 0 passing through a fixed point, then the fixed point is _________.
(a) (1, 1)
(b) (-1, -1)
(c) (1, -1)
(d) (-1, 1)
Answer:
(b)
Tricks : Go by choices
For option (b) Point (-1 ; -1) satisfies the eqn.
LHS = =(2 + λ)x + (3 – λ)y + 5
or (2 + 2)(-1) + (3 – 2)(-1) + 5
= -2 – λ – 3 + λ + 5 = 0 = RHS.
∴ (b) is correct

Question 6.
If kx – 4 = (k – 1).x which of the following is true. 
(a) x = -5
(b) x = -4
(c) x = -3
(d) x = 4
Answer:
(d) is correct kx – 4 = (k – 1) x
or kx – 4 = kx – x
or -4 = -x
∴ x = 4

Question 7.
If the equations kx + 2y = 5, 3x + y = 1 has no solution then the value of k is: 
(a) 5
(b) 2/3
(c) 6
(d) 3/2
Answer:
(c) is correct kx + 2y = 5
3x + y = 1
They have no soln. (given) k3=21≠51 ⇒ -k = 6

Question 8.
The equation x + 5y = 33; x+yx−y=133 has the solution (x,y) as:
(a) (4, 8)
(b) (8, 5)
(c) (4, 16)
(d) (16, 4)
Answer:
(b) is correct
Tricks: Go by choices
For LHS = x + 5y = 8 + 5 × 5 = 33
and x+yx−y+8+58−5=133
Clearly (b) satisfies both eqns.

Question 9.
The age of a person is 8 years more than thrice the age of the sum of his two grandsons who were twins. After 8 years his age will be 10 years more than twice the sum ofthe ages of his grandsons. Then the age of the person when the twins were bom is: 
(a) 86 yrs
(b) 73 yrs
(c) 68 yrs
(d) 63 yrs
Answer:
(b) Let age of 1st grandson = x
∴ Person’s Age = P = 3(x + x) + 8
P = 6x + 8
After 8 years P + 8 = 2 [x + 8 + x + 8] + 10 = 2(2x+ 16)+ 10
or 6x + 8 + 8 = 4x + 32 + 10
or 2x = 42 – 16 =26
∴ x = 13
∴ Age of person when grandsons were born = 6x + 8 – x = 6 × 13 + 8 – 13 = 73
(b) is correct

Question 10.
In a school number of students in each section is 36. If 12 new students are added, then the number of sections are increased by 4 and the number of students in each section becomes 30. The original number of section at first is: 
(a) 6
(b) 10
(c) 14
(d) 18
Answer:
(d); Let original No. of sections = x
Total students = 36x
Again Qts.
36x + 12 = (x+4).30
or 36x + 12 = 30 x + 120
or 6x= 108 ⇒ x = 18
Tricks: Go by choices

Question 11.
A person on a tour has ₹ 9600 for his expenses. But the tour was extended for another 16 days, so he has to cut down his daily expenses by ₹ 20.The original duration of the tour had been ? 
(a) 48 days
(b) 64 days
(c) 80 days
(d) 96 days
Answer:
(c); Let No. of tour days = x
∴ Expense per day = 9600x
Now Expense per day = 9600x+16
From 9600x−9600x+16 = 20
Tricks Go by choices
From here we get
For (c) LHS
960080−960080+16 = 20 RHS
(c) is correct
Do by Calculator

Question 12.
If 2^x+y = 2^2x-y = √8 then the respective values of x and y are _________. 
(a) 1, 12
(b) 12, 1
(c) 12, 12
(d) None of these
Answer:
(a) is Correct. 2^x+y= 2^2x-y = √8 = 23−−√2 = 2^3/2
∴ x + y = 32 (1)
2x – y = 32 (2)
Tricks: Then Go by Choices
(a) satisfies (1) & (2) both.

Question 13.
Let E₁, E₂ are two linear equations in two variables x and y .(0,1) is a solution for both the equations E₁ & E₂ (2, -1) is a solution of equation E, only and (-2, -1) is a solution of equation E₂ only, then E₁, E₂ are _________. 
(a) x = 0, y = 1;
(b) 2x – y = -1, 4x + y = 1
(c) x + y = 1, x – y = -1
(d) x + 2y = 2, x + y = 1
Answer:
(c) is correct.
Tricks: Go by Choices
(0 ; 1) satisfies E₁ & E₂ both
(2,-1) satisfies 1st Eqn. 2 – 1 = 1 (True)
But (-2 ; -1) also satisfies E₂
i.e. -2 -(-1) = -1 (True)

Question 14.
Particular company produces some articles on a day. The cost of production per article is ₹ 2 more than thrice the number of articles and the total cost of production is ₹ 800 on a day then the number of articles is : 
(a) 16
(b) 14
(c) 18
(d) 15
Answer:
(a) is correct.
TRICKS : Go by choices Let (A) is correct.
So, cost per unit = 800/16 = ₹ 50
It is 2 more than 3 times of 16.
(as given in Qts.)

Question 15.
The sides of equilateral triangle are shortened by 3 units, 4 units, 5 units respectively then a right angle triangle is formed. The side of the equilateral triangle was: 
(a) 5
(b) 6
(c) 8
(d) 10
Answer:
Tricks : Go by Choices
For option (c)
1 st side of right angled Δ = 8 – 3 = 5
2nd side = 8 – 4 = 4
and 3rd side = 8 – 5 = 3
Here ; 5 ; 4 and 3 are making a right angled triangle.
So, 5² = 4² +3²
Hence, option (c) is correct.

Question 16.
If 3x+y+2x−y = -1 and 1x+y−1x−y=43 then (x, y) is: 
(a) (2, 1)
(b)( 1, 2)
(c) (-1,2)
(d) (-2, 1)
Answer:
Tricks: Go by Choices
Option (B)
31+2+21−2 = -1 (True)
11+2−11−2=13+1=43 (True)
So; option (B) is correct.

Question 17.
The line 3x + 2y = 6 intersects the line 3 x – y = 12 in _________ quadrant: 
(a) 1st
(b) 2nd
(c) 3rd
(d) 4th
Answer:
(d), Eqn. (1) – Eqn. (2); we get

From (1); 3x = 6 – 2y = 6 – 2(-2) = 10
∴ x = 103
∴ Co-ordinate of the point of intersection
= (x, y) = (103, -2)
It is in 4th Quadrant.

Question 18.
If 2^x+y = 2^2x-y = √8, then the respective values of x and y ………..
(a) 1, 12
(b) 12, 1
(c) 12, 12
(d) None
Answer:
(a)
2^x+y = 2^2x-y = √8 = (2³)^1/2
= 2^3/2
Tricks:- Go by Choices (GBC)
(a) 2¹⁺¹² = 2³² = RHS and 2^2×1-12 = 232 = RHS
∴ (a) is correct

Question 19.
If 3x+y+2x−y = -1; 1x+y−1x−y=43; then (x, y) is: 
(a) 2, 1
(b) 1, 2
(c) -1, 2
(d) -2, 1
Answer:
(b) is correctQuestion 20.

If the sides of an equilateral triangle are shortened by 3 units , 4 units and 5 units respectively and a right triangle is formed then the sides of an equilateral triangle is: 
(a) 6 units
(b) 7 units
(c) 8 units
(d) 10 units
Answer:
(c) is correct.
Tricks:- Go by Choices.
Check for option (a).
1st Side = 6 – 3 = 3
2nd Side = 6 – 4 = 2
3rd Side = 6 – 5 = 1
But 1² + 2² ≠ 3²
[Note Try to do it mentally]
So (a) is False.
(c) 1st Side = 8 – 3 = 5
2nd Side = 8 – 4 = 4
3rd Side = 8 – 5 = 3
But 3² + 4² = 5² (True)
(Pythagoras Formula)
(c) is correct.

Question 21.
A number consists of two digits such that the digit in one’s place is thrice the digit at ten’s place. If 36 be added then the digits are reversed. Find the number: 
(a) 62
(b) 26
(c) 39
(d) None
Answer:
(b)
Tricks :- Go by choices
(a) 62 → 2 ≠ 3 × 6 (False)
and 62 + 36 = 98 ≠ 26 (False)

(b) 26 Clearly 6 = 3 × 2 (True)
and 26 + 36 = 62 (Orders of digits reversed)
So; (b) is correct.