# Linear Equation Questions for SNAP PDF

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## Linear Equation Questions for SNAP PDF

Download important SNAP Linear Equation Questions PDF based on previously asked questions in SNAP and other MBA exams. Practice Linear Equation Questions and answers for SNAP and other exams.

Question 1: A leather factory produces two kinds of bags, standard and deluxe. The profit margin is Rs. 20 on a standard bag and Rs. 30 on a deluxe bag. Every bag must be processed on machine A and on Machine B. The processing times per bag on the two machines are as follows:

The total time available on machine A is 700 hours and on machine B is 1250 hours. Among the following production plans, which one meets the machine availability constraints and maximizes the profit?

a) Standard 75 bags, Deluxe 80 bags

b) Standard 100 bags, Deluxe 60 bags

c) Standard 50 bags, Deluxe 100 bags

d) Standard 60 bags, Deluxe 90 bags

Question 2: Every 10 years the Indian Government counts all the people living in the country. Suppose that the director of the census has reported the following data on two neighbouring villages Chota Hazri and Mota Hazri.
Chota Hazri has 4,522 fewer males than Mota Hazri.
Mota Hazri has 4,020 more females than males.
Chota Hazri has twice as many females as males.
Chota Hazri has 2,910 fewer females than Mota Hazri.
What is the total number of males in Chota Hazri?

a) 11,264

b) 14,174

c) 5,632

d) 10,154

Question 3: Consider the following steps :
1. Put x = 1, y = 2
2. Replace x by xy
3. Replace y by y +1
4. If y = 5 then go to step 6 otherwise go to step 5.
5. Go to step 2
6. Stop Then the final value of x equals

a) 1

b) 24

c) 120

d) 720

Question 4: The sum of the possible values of X in the equation |X + 7| + |X – 8| = 16 is:

a) 0

b) 1

c) 2

d) 3

e) None of the above

Question 5: In a school, students were called for the Flag Hoisting ceremony on August 15. After the ceremony, small boxes of sweets were distributed among the students. In each class, the student with roll no. 1 got one box of sweets, student with roll number 2 got 2 boxes of sweets, student with roll no. 3 got 3 boxes of sweets and so on. In class III, a total of 1200 boxes of sweets were distributed. By mistake one of the students of class III got double the sweets he was entitled to get. Identify the roll number of the student who got twice as many boxes of sweets as compared to his entitlement.

a) 22

b) 24

c) 28

d) 30

Question 6: Among Anil, Bibek, Charu, Debu, and Eswar, Eswar is taller than Debu but not as fat as Debu. Charu is taller than Anil but shorter than Bibek. Anil is fatter than Debu but not as fat as Bibek. Eswar is thinner than Charu, who is thinner than Debu. Eswar is shorter than Anil. Who is the thinnest person?

a) Bibek

b) Charu

c) Debu

d) Eswar

Question 7: The number of solutions $(x, y, z)$ to the equation $x – y – z = 25$, where x, y, and z are positive integers such that $x\leq40,y\leq12$, and $z\leq12$ is

a) 101

b) 99

c) 87

d) 105

Question 8: If $5^x – 3^y = 13438$ and $5^{x – 1} + 3^{y + 1} = 9686$, then x + y equals

Question 9: Let a, b, x, y be real numbers such that $a^2 + b^2 = 25, x^2 + y^2 = 169$, and $ax + by = 65$. If $k = ay – bx$, then

a) $0 < k \leq \frac{5}{13}$

b) $k > \frac{5}{13}$

c) $k = \frac{5}{13}$

d) k = 0

Question 10: The sum of the possible values of X in the equation |X + 7| + |X – 8| = 16 is:

a) 0

b) 1

c) 2

d) 3

e) None of the above

.Let x be no. of standard bags and y be no. of deluxe bags. According to given conditions we have 2 equations 4x+5y<=700 and 6x+10y<=1250. Here option A satisfies both the equations.

Let the number of males in Mota Hazri = x
No. of males in Chota Hazri = x – 4522
Let the number of females in Mota Hazri = y
No. of females in Chota Hazri = y – 2910
(y – 2910) = 2(x – 4522) => y = 2x – 9044 + 2910 = 2x – 6134
Also y = x + 4020
So, x + 4020 = 2x – 6134 => x = 10154
So, number of males in Chota Hazri = 10154 – 4522 = 5632

1. x=1 ; y=2

2. x=2 ; y=3

3. x=6 ; y=4

4. x=24 ; y=5

Hence when y=5 , x will be 24

Expression : $|x + 7| + |x – 8| = 16$

Case 1 : $x < -7$

=> $-(x + 7) – (x – 8) = 16$

=> $-2x + 1 = 16$

=> $x = \frac{-15}{2} = – 7.5$

Case 2 : $-7 \leq x < 8$

=> $(x + 7) – (x – 8) = 16$

=> $15 = 16$, which is not possible.

Case 3 : $x \geq 8$

=> $(x + 7) + (x – 8) = 16$

=> $2x – 1 = 16$

=> $x = \frac{17}{2} = 8.5$

$\therefore$ Sum of all possible values of $x = 8.5 – 7.5 = 1$

1+2+3+4+………………+n < 1200

and

1+2+3+4+………………+n+ (n+1) > 1200

We need to calculate value of n.

Using first and second inequality,

$\frac{n (n+1)}{2}$ < 1200

N = 48

Total number of boxes of sweets distributed is: 48*49/2 = 1176

So, roll no 24 got doubles number of boxes.

The question asks for the thinnest person. Therefore, let us arrange the persons in the decreasing order of their weight.

Eswar is not as fat as Debu.
Therefore, Debu > Eswar.

Anil is fatter than Debu but not as fat as Bibek.
Bibek > Anil > Debu > Eswar.

Eswar is thinner than Charu, who is thinner than Debu

Bibek > Anil > Debu > Charu > Eswar.

Therefore, Eswar is the thinnest among the five persons and hence, option D is the right answer.

x – y – z = 25 and $x\leq40,y\leq12$, $z\leq12$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99

$5^x – 3^y = 13438$ and $5^{x – 1} + 3^{y + 1} = 9686$

$5^{x} + 3^{y}*15 = 9686*5$

$5^{x} + 3^{y}*15 = 48430$

16*$3^y$=34992

$3^y$ = 2187

y = 7

$5^x$=13438+2187=15625

x=6

x+y = 13

Expression : $|x + 7| + |x – 8| = 16$

Case 1 : $x < -7$

=> $-(x + 7) – (x – 8) = 16$

=> $-2x + 1 = 16$

=> $x = \frac{-15}{2} = – 7.5$

Case 2 : $-7 \leq x < 8$

=> $(x + 7) – (x – 8) = 16$

=> $15 = 16$, which is not possible.

Case 3 : $x \geq 8$

=> $(x + 7) + (x – 8) = 16$

=> $2x – 1 = 16$

=> $x = \frac{17}{2} = 8.5$

$\therefore$ Sum of all possible values of $x = 8.5 – 7.5 = 1$

We hope this Linear Equation questions and answers for SNAP PDF will be helpful to you.