Important Formulae
Consider the following two system of equations:
a₁x + b₁y = c₁
a₂x + b₂y = c₂
System of Equations are consistent but there can be number of solutions.
( a₁ / a₂) = (b₁ / b₂) ≠ (c₁ / c₂)
( a₁ / a₂) = (b₁ / b₂) ≠ (c₁ / c₂)
System of Equations are inconsistent and NO solution.
(a₁ / a₂) ≠ (b₁ / b₂)
System of Equations are consistent but has unique solution.
Examples
Example 1. A fraction becomes 2/3 if its numerator is increased by 1 and the denominator by 2. It becomes 3/4 if its numerator is increased by 4 and the denominator by 5. Find the fractions.
(A) 3/5 (B) 2/7 (C) 4/5 (D) 5/7
Solution:
Let the numerator be x and the denominator be y
(x + 1) / (y + 1) = 2/3 ⇒ 3x - 2y = 1 ----------- (1)
(x + 4) / (y + 5) = 3/4 ⇒ 4x - 3y = -1 ----------- (2)
3 * (1) - 2 * (2)
⇒ 9x - 6y = 3
8x - 6y = -2
-----------------
x = 5
-----------------
From (1), 3*(5) - 2* (y) = 1
⇒ y = 7 ∴ The fraction is 5/7. Answer: D
Example 2. Five years ago, the age of a man was three years more than four time the age of his son. Three years hence, the age of the man will be six years less than thrice the age of his son. After how many years from now will their combined age be 80 years?
(A) 16 (B) 18 (C) 20 (D) 22
Solution:
Let 'f' be the present age of the father and that of the son be 's'. Then,
f - 5 = 4 (s-5) + 3
⇒ f - 4s = -12 -------- (1)
f + 3 = 3 (s + 3) - 6
f - 3s = 0 -------- (2)
(1) - (2) ⇒ f - 4s = -12
f - 3s = 0
---------------
-s = -12
---------------
∴ s = 12
From equation (2) f = 3s ⇒ f = 3 (12) = 36
Let, after x years, the sum of their ages be 80 years.
f + s + 2x = 80
36 + 12 + 2x = 80 ∴ 2x = 80 - 48 ∴ x = 16
∴ After 16 years their combined age will be 80 years. Answer: A
Example 3. Praveen takes 2 hours more than Prakash to cover a distance of 600 km. If instead Praveen doubles his speed he would reach the destination 4 hours before Prakash. Find Prakash's speed.
(A) 50 km/hr (B) 120 km/hr (C) 100 km/hr (D) 60 km/hr
Solution:
Let, Praveen's speed be x km/hr and that of Prakash be y km/hr.
Given, (600/x) - (600/y) = 2
⇒ (1/x) - (1/y) = 1/300 -------- (1)
and ⇒ (600/y) - (600/2x) = 4
⇒ (1/y) - (1/2x) = 2/300 -------- (2)
(1) + (2) ⇒ (1/x) - (1/2x) = (1/300) + (2/300)
⇒ x = 50
From (1/50) - (1/y) = 1/300 ⇒ (1/y) = 5/300 = 1/60
∴ y = 12
∴ Prakash's speed is 60 km/hr. Answer: D
Example 4. Vinay has only Rs.2 and Rs.5 coins with him. If he has in all 57 coins worth Rs.150 with him. How many Rs.2 coins does he have?
(A) 20 (B) 45 (C) 12 (D) 40
Solution:
Let the number of Rs.2 coins be x and Rs.5 coins be y.
x + y = 57 ---- (1)
2x + 5y = 150 ---- (2)
5 * (1) - (2)
⇒ 5x + 5y = 285
2x + 5y = 150
-------------------------
3x = 135
-------------------------
x = 45
∴ The number of Rs.2 coins with Vinay is 45. Answer: B
Example 5. The difference between a three-digit number and the number formed by reversing its digits is 396. The difference of the hundreds and the units digit, is one less than the sum of the units and tens digits. Also, the hundreds digit is twice the units digit. Find the number.
(A) 412 (B) 814 (C) 418 (D) 612
Solution:
Let the three digit number be 100x + 10y + z
Now (100x + 10y + z) - (100z + 10y + z) = 396
⇒ 99 (x - z) = 396 ⇒ x - z = 4 -------- (1)
Also x - z = y + z - 1
As the hundred's digit is twice the unit's digit, x > z.
⇒ y + z = 5 ---- (2)
x = 2z ---- (3)
From (1) and (3) we get,
2z - z = 4 ⇒ z = 4
⇒ x = 2 (4) = 8
From (2), y + z = 5, y + 4 = 5 ⇒ y = 1
∴ The required number is 814. Answer: B
Example 6. A, B and C are successive even positive integers in the ascending order. Four times C is 4 more than five times A. What is the value of B?
(A) 14 (B) 18 (C) 12 (D) 16
Solution:
A, B and C are successive even positive integers in the ascending order.
⇒ C - B = 2 and B - A = 2
⇒ C - A = 4
4C = 5A + 4
⇒ 4C - 5A = 4
⇒ 4C - 4A - A = 4
But C - A = 4
∴ 4(C - A) - A = 4
⇒ 4 (4) - A = 4
⇒ A = 12
∴ B = A + 2 = 12 + 2 = 14 Answer: A
Example 7. A test has 175 questions. A candidate gets 4 marks for each correct answer and loses 2 marks for each wrong answer and loses 1 mark for leaving the question unattempted. A student scores 405. On analysing his performance he concludes that he left unattempted 35 questions. How many questions did he mark wrong?
(A) 25 (B) 35 (C) 30 (D) 20
Solution:
Let the number of questions got right be x. The number of questions got wrong be y
x + y + 35 = 175 ⇒ x + y = 140 -------- (1)
4x - 2y - 35 = 405
4x - 2y = 440 ⇒ 2x - y = 220 -------- (2)
(1) + (2) ⇒
x + y = 140
2x - y = 220
--------------------
3x = 360 ⇒ x = 120
∴ y = 140 - 120 = 20 Answer: D
Example 8. The cost of 7 pencils, 5 erasers and 2 rulers is Rs.39. The cost of 5 pencils, 3 erasers and 4 rulers is Rs.39. The cost of 3 pencils, 5 erasers and 7 rulers is Rs.56. Find the cost of each eraser.
(A) 3 (B) 7 (C) 2 (D) 5
Solution:
Let the cost of each pencil, eraser and ruler be x, y and z respectively. Then,
7x + 5y + 2z = 39 -------- (1)
5x + 3y + 4z = 39 -------- (2)
3x + 5y + 7z = 56 -------- (3)
(1) - (2) ⇒
7x + 5y + 2z = 39
5x + 3y + 4z = 39
--------------------------
2x + 2y - 2z = 0
--------------------------
⇒ x + y = z -------- (4)
(2) + (3) ⇒
5x + 3y + 4z = 39
3x + 5y + 7z = 56
------------------------
8x + 8y + 11z = 95
------------------------
substitute (x + y = z) in above equation, then
8 (x + y) + 11z = 95
8z + 11z = 95 ⇒ z = 5 -------- (5)
Substitute (4) in (3)
3x + 5y + 7z = 56
3 (x + y) + 2y + 7z = 56
3z + 2y + 7z = 56
2y = 56 - 10z
Substitute z value from equation (5) in the above equation.
2y = 6 ⇒ y = 3
∴ The cost of each eraser is Rs.3. Answer: A
Example 9. Find p if the given system of equations has infinite number of solutions.
4x + py = 2 + 10y and px + 24y = 8
(A) 16 (B) -10 (C) -16 (D) Cannot be determinedSolution:
The first equation can be written to group the y term as:
4x + py - 10y = 2 ⇒ 4x + (p - 10)y = 2
The equations have infinite solutions. Then,
4/p = (p-10)/24 = 2/8
4/p = 2/8 ⇒ p = 16. Answer: A
Example 10. In an examination, 3/5th of the students that appeared failed by 20 marks and 1/5th of the students got 20 marks above the pass mark. Each of the remaining students got 40 marks above the pass mark. The maximum marks in the examination is 100. A total of 120 students appeared for the exam and their average mark is 62 marks. The pass mark is
(A) 66 (B) 56 (C) 64 (D) 62
Solution:
Let the pass mark be p.
As the total number of students is 120, 3/5th i.e., 72 students scored (p - 20) marks, 24 students scored (p + 20) marks and the remaining 24 students scored (p + 40) marks.
∴ The average marks of the students
= {72 (p - 20) + 24 (p + 20) + 24 (p + 40) }/120 = 62
∴ 72p - 1440 + 24p + 480 + 960 + 24p = 62 * (120)
⇒ 120p = 62 * (120)
⇒ p = 62 Answer: D
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