Linear Equations and Polynomials

M. Velrajan

1. Two railway tickets from city A to B and three tickets from city A to C cost Rs.177. Three

tickets from city A to B and two tickets from city A to C cost Rs.173. The fare for city B from

city A will be Rs.       (1)25  (2)27  (3)30  (4)33

Let x, y be the fare for A to B and A to C respectively. Then 2x + 3y = 177 and 3x + 2y = 173.

To get x from the above equations we have to eliminate y by multiplying the first equation by 2

and the second by 3. 

4x + 6y = 177 × 2

9x + 6y = 173 × 3

Subtracting the first from the second equation 5x = 173 × 3 - 177 × 2 = 165, hence x = 33.

Aliter :  Instead of writing the two equations and multiplying the first equation by 2

and the second by 3 and then subtracting the first from second, by working out it in mind we

have to write directly 5AB = 173 × 3 - 177 × 2 = 165, AB = 33

2. In a class if 5 students are seated in a bench then 2 benches will be vacant. If 4 students are seated in

a bench then 10 students will not have benches to sit. Then the number of students and benches in the

class are    (a)20 and 8             (b)40 and 10           (c)90 and 20            (d) 20 and 90

If x and y are the number of students and benches then x = 5(y - 2) and x - 10 = 4y

i.e. x = 5y - 10 and x = 4y + 10. Hence 5y - 10 = x = 4y + 10

Hence 5y - 4y = 10 + 10 i.e. y = 20 and x = 100 - 10 = 90  Ans. (c)

Aliter : (Benches - 2) × 5 = students (vacant benches is subtracted from the total benches) 

and benches × 4 = students -10 

                          (number of students without benches is subtracted from the total students).

Now check these conditions among answer choices.

(8 - 2) × 5 = 30, so (a) is not the answer.

(10 - 2) × 5 = 40, so (a) is not the answer.

(20 - 2) × 5 = 90 and 20 × 4 = 80 = (90 - 10), so (c) satisfies the conditions.  Ans. (c)

3. In an examination for every correct answer 2 marks are awarded for every wrong answer 1/ 2 mark is

reduced. If a student answers all 120 questions and secures 45 marks then the no. of correct answered

question is  (a)45                    (b)42                 (c)36                            (d)38

If x is the number of correct answered questions then

2x - 12(120 - x) = 45 i.e. 4x - 120 + x = 90

5x = 210, hence x = 42

Aliter: If all the questions are answered wrongly, marks secured = - ½ × 120 = - 60.

Marks increase due to correct answers = 45 - (-60) = 105

Marks increase for one correct answer = 2 + ½ = 5/2

Number of correct answered questions = 105/(5/2) = (2/5) × 105 = 42

4. Divide Rs. 700 among A, B, C so that A gets 2 times more than B and B gets 2 times to C. 

How much of the amount they are getting? 

(a)100, 200, 400   (b) 200, 300, 200  (c)300, 200, 200    (d)400, 200,100

A gets > B gets > C gets. In the answer choices only (d) satisfies this condition

Aliter : B = 2C, A = 2B = 4C

A + B + C = 700 i.e. 7C = 700, hence C = 100 and A = 400, B = 200 Ans. (d)

5. Divide Rs. 680 among A, B, C so that A gets 3 times more to B and B gets 4 times to C.

(a) Rs.160,  40,  480  (b) Rs.480,  160, 40 (c) Rs.480, 40,  160 (d) Rs.160,  480,  .40

B = 4C, A = 3B. Among the answer choices, only b satisfies this.  Ans. (b)

6.   A sum of Rs. 53 is divided among A,B,C in such a way that A gets Rs.7 more than what B gets and

B gets Rs. 8 more than what C gets. The ratio of their share is.   

(a)16 : 9 : 18        (b)25 : 18 : 10     (c)18 : 25 : 10  (d)15 : 8 : 30 

A > B > C.   Among the answer choices, only b satisfies this.  Ans. (b)

7. 5% income of X is equal to 15% income of Y and 10% income of Y is equal to 20% income of Z. 

If income of Z is Rs. 3,000 then total income of X, Y and Z in Rupees is

(a)18,000                  (b)12,000                     (c)27,000                         (d0. 16,000

10Y = 20Z, Y = 2Z = 2 × 3000 = 6000.

5X = 15Y, X = 3Y = 18000. 

X + Y + Z = 18000 + 6000 + 3000 = 27000

8. A voluntary organisation planted a total of 106 trees along the roadside. Some of the trees were fruit 

bearing trees. If the number of non – fruit bearing trees was two more than thrice the number of fruit 

bearing trees, what was the number of fruit bearing trees planted?   (a)20    (b)22      (c)24       (d)26

If x is the number of fruit bearing trees then number of non-fruit trees = 3x + 2, hence 

x + 3x + 2 = 106  i.e. 4x = 104, hence x = 26

Aliter : Number of non - fruit bearing trees = 3 × Number of fruit bearing trees + 2

4 × Number of fruit bearing trees + 2 = total trees = 106.

Number of fruit bearing trees =  (106-2)/4 = 104/4 = 26

9. A person is having some amount with him. He gave one-third of the amount as loan to A and took

back 54 more. He gave half of the amount with him as loan to B and took back 25 more. 

When he met C, the amount left was Rs.262. What was the amount with the person when he met A?

(a)620                          (b) 630 (c) 640 (d) 650

Let x be the amount. After giving to A and took back 54 more, the amount with him is ⅔ x + 54.

After giving B and took back, amount with him is ½ (⅔ x + 54) + 25 = 262

⅓ x + 27 + 25 = 262, hence x = 3(262 - 27 - 25) = 630

10. A sum of Rs.312 was divided among 100 boys and girls in such a way that each boy gets Rs. 3.60

and each girl gets Rs. 2.40 the number of girls is       (a)35            (b)40            (c)60                 (d)65

Amount for 100 boys = 360

Difference due to less amount for all girls = 360 – 312 = 48

Difference for one girl = 3.6 - 2.4 = 1.2

Number of girls = 48/1.2 = 40

Aliter : Let x be the number of girls then 3.6(100 - x) + 2.4x = 312

- 1.2x = 312 - 360 = - 48, hence x = 40

11. If a particular amount distributed to each of 14 students is Rs. 80 more than the amount distributed

to each of 18 students, find the amount   (a) 5040     (b) 3150      (c) 2520       (d) 4200

If the particular amount is x then given that x/14 = x/18 + 80

Hence 18x = 14x + 80  × 14  × 18

4x = 80  × 14  × 18 and hence x = 80  × 7  × 9 = 80  × 63 = 5040

Aliter : Amount distributed to the additional 4 students among the 18 students =  80  × 14

Amount = (80  × 14/4) × 18 = 80 × 63 = 5040

12. A scored 30% marks and failed by 15 marks. B scored 40% marks and obtained 35 marks more

than that required to pass. The pass percentage is       (a)45%         (b)40%            (c)35%        (d)33%

Clearly the pass mark lies between 30% and 40% and it is 15 marks away from 30% and 

35 marks less to 40%. So pass is not in the middle of 30% and 40%. Hence it is not 35%.

In the answer choices only 35% and 33% lie between 30% and 40%. Hence it is 33%

Aliter :

30% marks = Passmark - 15

40% marks = Passmark + 35

Subtracting first from second equation, 10% marks = 35 - (- 15) = 35 + 15 = 50

Since 50 marks = 10% marks, 15 marks = (10/50)× 15 = 3%

Pass percentage =   30% + 3% = 33% 

Aliter : If x and y are the pass mark and the total mark then 

30/100 y = x - 15 and  40/100 y = x + 35

30y = 100x - 1500 and 40y = 100x + 3500

Subtracting the first from the second equation, 10y = 5000, y = 500

Hence 150 = x - 15 i.e. x = 165

Pass percentage = (165/500) × 100 = 33%

13. There are 6 persons: A, B, C, D, E and F. A has 3 items more than C. D has 4 items less than B. E has 6 items

less than F. C has 2 items more than F. F has 3 items more than D. Which one of the following figures cannot be

equal to the total no. of items possessed by all the 6 persons?   (a) 41 (b) 47    (c) 53   (d) 58

A = C + 3, D = B - 4, E = F - 6, C = F + 2, F = D + 3

Since D = B - 4, 

          F = B - 4 + 3 = B - 1, 

          C = B - 1 + 2 = B + 1, 

          E = B - 1 - 6 = B - 7, 

          A = B + 1 + 3 = B + 4

Hence A + B + C + D + E + F = 6B - 7

Hence total number + 7 = 6B, a multiple of 6.

41 + 7, 47 + 7, 53 + 7 are multiples of 6, but 58 + 7 is not a multiple of 6. Ans. (d)

14. The degree of the expression 2+3x+4xy+5y2+8x2y3 is a)8    b) 6    c) 5    d) 3

Degree is 2 + 3 = 5

15. The roots of the equation x3 + 2x2 - x - 2 are  a) 1, 1, 2    b) 1, – 1, 2    c) 1, – 1, –2    d) 1, 1, – 2

If x = 1, then x3 + 2x2 - x - 2 = 0, hence 1 is a root.

If x = 2, then x3 + 2x2 - x - 2 = 12, hence 2 is not a root. Hence (a) and (b) are not the answer.

If x = - 1, then x3 + 2x2 - x - 2 = 0, hence - 1 is a root. Hence (d) is not the answer and hence 

(c) is the answer. Of course  we can verify that - 2 is also a root, but it is not necessary.

Aliter : Since the given polynomial is of degree 3, 

sum of the roots = - (coefficient of x2) / (coefficient of x3) = -2 

In the answer choices only for 1, -1, - 2 the sum is -2. So, (c) is the answer.

16. If x - 2 is a factor of x3 - 3x2 - 4x + 12 then the other factors are 

a) x + 2, x - 3    b) x + 2, x + 3     c) x - 2, x - 3    d) x - 2, x + 3

Sum of the roots = - (coefficient of x2) / (coefficient of x3) = - (-3)/1 = 3.

Since x - 2 is a factor, 2 is one of the roots.

From the answer choices the other possible roots are - 2, 3 or - 2, - 3 or 2, 3 or 2, - 3.

Along with the root 2 only - 2, 3 have sum = 3.

Hence the other factors are x + 2 and x - 3.

Of course product of roots = - (constant term) / (coefficient of x3) = - 12/1 = -12 is

also satisfied by 2, - 2, 3.

Aliter : x3 - 3x2 - 4x + 12 = (x - 2)(x2 - x- 6) = (x - 2)(x + 2)(x - 3)

17. Find the point of intersection of the straight line 9x – y - 2 = 0 and 2x + y - 9 = 0

(a)(-1, 7)         (b)(7 , 1)                 (c)(1, 7)                    (d)( -1, -7)

The point of intersection satisfies the two equations .

(-1, 7) and (7, 1) do not satisfy the first equation, but (1, 7) satisfies both equations. 

Ans. (c). Of course (-1, -7) do not satisfy the first equation.