Linear Equations Questions for MAH-CET

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Question 1: Rubina could get equal number of Rs. 55, Rs. 85 and Rs. 105 tickets for a movie. She spends Rs. 2940 for all the tickets. How many of each did she buy?

a) 12

b) 14

c) 16

d) Cannot be determined

e) None of these

1) Answer (A)

Solution:

Let x be the number of tickets bought by tickets.
Total money spent = 55x + 85x + 105x
245x = 2940
x = $\frac{2940}{245}$ = 12

Question 2: In a test, a candidate secured 468 marks out of maximum marks ‘A”. Had the maximum marks ‘A’ converted to 700, he would have secured 336 marks. What was the maximum marks of the test?

a) 775

b) 875

c) 975

d) 1075

e) None of these

2) Answer (C)

Solution:

When the maximum marks were 700 the secured marks would be 336
% of marks out of 700 = $\frac{336}{700}$ x 100 = 48%
Now 468 is 48% of maximum marks ‘A’
468 = $\frac{48}{100}$ x A
A = 468 x $\frac{100}{48}$
= 975

Question 3: Six – elevenths of a number is equal to twentytwo percent of the second number. The second number is equal to one-fourth of the third number. The value of the third number is 2400. What is 45% of the first number?

a) 109.8

b) 111.7

c) 117.6

d) 123.4

e) None of these

3) Answer (E)

Solution:

Let the 1st number be x, 2nd number be y
Third number 2400
$\frac{6}{11}$ of x = 22% of y
y = $\frac{1}{4}$th of the third number
2nd No. = y = 600
22% of 2nd No. =  $\frac{22}{100}$ x 600 = 132
$\frac{6}{11}$ of x = 132
x = 242
45% of 1st number = $\frac{45}{100}$ x 242 = 108.9

Question 4: Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy at least 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy?

a) Either 12 or 13

b) Either 11 or 12

c) Either 10 or 11

d) Either 9 or 11

e) Either 9 or 10

4) Answer (C)

Solution:

Number of Ice-creams < Cookies < Pastries
Sum must be 32
If ice-creams = 9
Cookies > Ice-creams= 10 or 11
Pastries > Ice-creams = 11 or 12

 

Question 5: The fare of a bus is Rs.X for the first five kilometres and Rs.13 per kilometer thereafter. If a passenger pays Rs.2,402 for a journey of 187 kilometres, what is the value of X?

a) Rs. 29

b) Rs. 39

c) Rs. 36

d) Rs. 31

e) None of these

5) Answer (C)

Solution:

Total cost of the journey = 2402
X + 182 x 13 = 2402
X = 2402 -2366 = 36

Question 6: The sum of the ages of 4 members of a family 5 years ago was 94 years. Today, when the daughter has been married off and replaced by a daughter-in-law the sum of their ages is 92. Assuming that there has been no other change in the family structure and all the people are alive, what is the different in the age of the daughter and the daughter-in-law?

a) 22 years

b) 11 years

c) 25 years

d) 19 years

e) 15 years

6) Answer (A)

Solution:

Sum of the ages five years ago = 94
Sum of their ages now = 94 + 20 = 114
Sum of their ages after the daughter is left and replaced by daughter-in-law = 92
Difference = 114 – 92 = 22

Question 7: Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150. The overall percentage marks obtained by Akash in the three subjects together were 54%. How many marks did he scored in subject C

a) 84

b) 86

c) 79

d) 73

e) None of these

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7) Answer (B)

Solution:

Percentage of marks obtained in A = $\frac{73}{150}$ x 100 = 48.66
Overall percentage = 54%
$\frac{A + B + C}{3}$ = 54%
$\frac{48.66 + 56 + C}{3}$ = 54%
C = 57.34%
Marks = 57.34% of 150 = 86

Question 8: The present age of Romil is one fourth of that of her father. After 6 years the father’s age will be twice the age of Kapil. If Kapil celebrated fifth birth day 8 years ago. What is Romil’s percent age?

a) 7 years

b) 7.5 years

c) 8 years

d) 8.5 years

e) None of these

8) Answer (C)

Solution:

Kapil’s present age = 8 + 5 = 13
After 6 years Kapil’s age = 19
Father’s age after six years = Twice of Kapil’s age = 2 x 19 =38
Father’s present age = 32
Romil’s age = $\frac{1}{4}$th of 32 = 8

Question 9: In an examination, 30% of total students failed in Hindi, 45% failed in English and 20% failed in both subjects. Find the percentage of those who passed in both the subjects.

a) 35.7 %

b) 35%

c) 40%

d) 45%

e) 44%

9) Answer (D)

Solution:

45% of the total students failed in Hindi
30% failed in English
20% failed in both
Therefore 55% of the total students failed in atleat one subject.
Hence 45% have passed in both

Question 10: Rita’s present age is four times her daughter’s present age and two-thirds of her mother’s present age. The total of the present ages of all of them is 154 years. What is the difference between Rita’s and her mohter’s present age?

a) 28 years

b) 34 years

c) 32 years

d) Cannot be determined

e) None of these

10) Answer (A)

Solution:

Let Rita’s age be R.
Mother’s age = 3R/2
Daughter’s age = R/4
Total = R + 3R/2 + R/4 = 11R/4 = 154 => R = 56
So, difference of ages = R/2 = 28 years
InstructionsInstructions: In the following question, two equations are given. You have to solve those equations and give your answer

Question 11: In the following question, two equations are given.
I. 10x² – 7x + 1 = 0
II. 35y² – 12y + 1 = 0

a) x < y

b) x > y

c) x = y

d) x ≥ y

e) x ≤ y

11) Answer (D)

Solution:

Solving 10x² – 7x + 1 = 0
x = $\frac{7\pm\sqrt{49-40}}{20}$ = $\frac{7 \pm 3}{20}$ = 1/5 or 1/2

Solving 35y² – 12y + 1 = 0
y = $\frac{12\pm\sqrt{144-140}}{70}$ = $\frac{12 \pm 2}{70}$ = 1/5 or 1/7

x $\geq$ y

Question 12: In the following question, two equations are given.
I. x² – 6x – 7 = 0
II. 2y² + 13y + 15 = 0

a) x < y

b) x > y

c) x = y

d) x ≥ y

e) x ≤ y

12) Answer (B)

Solution:

x² – 6x – 7 = 0
x = 7 , -1
2y² + 13y + 15 = 0
y = -5 , -3/2
Thus, x > y

Question 13: In the following question, two equations are given.
I. 6x² + 77x + 121 = 0
II. y² + 9y – 22 = 0

a) x < y

b) x > y

c) cannot be determined

d) x ≥ y

e) x ≤ y

13) Answer (C)

Solution:

6x² + 77x + 121 = 0
x = -11 , -11/6
y² + 9y – 22 = 0
y = -11 , 2

Thus, the relationship between x and y cannot be determined.

Question 14: In the following question, two equations are given.
I. 2x² + 3x – 20 = 0
II. 2y² + 19y + 44 = 0

a) x < y

b) x > y

c) cannot be determined

d) x ≥ y

e) x ≤ y

14) Answer (D)

Solution:

2x² + 3x – 20 = 0
x = -8 , 5
2y² + 19y + 44 = 0
y = -11 , -8

x ≥ y

Question 15: In the following question, two equations are given.
I. x² – 24x + 95 = 0
II. y² – 26y + 169 = 0

a) x < y

b) x > y

c) cannot be determined

d) x ≥ y

e) x ≤ y

15) Answer (C)

Solution:

x² – 24x + 95 = 0
x = 19 , 5
y² – 26y + 169 = 0
y = 13, 13

x < y or x>y

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