1). 2x + 3y = 29 and y = x + 3, what is the value of x?
Correct Answer: 4
2x+3y = 29 ...(i)
and y = x+3 ...(ii)
Putting the value of y from Eq. (i) to Eq. (ii), we get
2x+3y = 29
\( \Large 2x+3 \left(x+3\right)=29 \)
2x+3x+9 = 29 = 5x = 20
Therefore, x = 4 
2). Deepak has some hens and some goats. If the total number of animal heads is 90 and the total number of animal feet is 248, what is the total number of goats Deepak has?
A). 32 
B). 36 
C). 34 
D). Cannot be determined 
Correct Answer: 34 Let hens = H, goats = G
According to the question,
H + G = 90 ...(i)
2H + 40 = 248 ...(ii)
On multiplying Eq. i) by 2 and subtracting from Eq. (ii), we get
by solving 1 and 2 we get
2G = 68
G = 34

3). The sum of the two digits is 15 and the difference between them is 3. What is the product of the digits?
A). 56 
B). 63 
C). 42 
D). 54 
Correct Answer: 54 Let the number be x and y.
Then, according to the question,
x + y = 15 ...(i)
x  y = 3 ...(ii)
On adding Eqs. (i) and (ii), we get
2x =18 = x = 9
On putting the value of x in Eq. (i), we get
y = 6
Therefore, Product = xy = 9 x 6 = 54

4). The cost of 21 pencils and 9 clippers is Rs.819. What is the total cost of 7 pencils and 3 clippers together?
A). Rs.204 
B). Rs.409 
C). Rs.273 
D). Rs.208 
Correct Answer: Rs.273
Let cost of 1 pencil and 1 clipper be p and c, respectively
Now. according to the question,
\( \Large 21p + 9c = Rs.819 \)
\( \Large 3 \left(7p+3c\right)= Rs.819 \)
\( \Large 7p + 3c = Rs.273 \)
Cost of 7 pencils and 3 clippers = Rs.273. 
5). The value of k for which kx+ 3yk+ 3=0 and 12x+ky=k, have infinite solutions, is
Correct Answer: 6
For infinite solution
\( \Large \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \)
= \( \Large \frac{K}{12}=\frac{3}{K}=\frac{K+3}{K} \)
= \( \Large \frac{K}{12}=\frac{3}{K}=K^{2}=36 \)
Therefore, \( \Large K = \sqrt{36} = 6 \) 
6). In a rare coin collection, there is one gold coin for every three nongold coins. 10 more gold coins are added to the collection and the ratio of gold coins to nongold coins would be 1 : 2. Based on the information; the total number of coins in the collection now becomes.
A). 90 
B). 80 
C). 60 
D). 50 
Correct Answer: 90
Let the number of gold coins initially be x
and the number of nongold coins be y.
According to the question,
3x = y
When 10 more gold coins, total number gold coins become x + 10
and the number nongold coins remain the same at y.
Now, we have \( \Large 2 \left(10+x\right)=y \)
Solving these two equations, we get
x = 20 and y = 60.
Total number of coins in the collection at the end is equal to
x+10+y = 20+10+60 = 90. 
7). If \( \Large \frac{\sqrt{3+x}+\sqrt{3x}}{\sqrt{3+x}\sqrt{3x}}=2 \), then x is equal to
A). \( \Large \frac{5}{12} \) 
B). \( \Large \frac{12}{5} \) 
C). \( \Large \frac{5}{7} \) 
D). \( \Large \frac{7}{5} \) 
Correct Answer: \( \Large \frac{12}{5} \) Given, \( \Large \frac{\sqrt{3+x}+\sqrt{3x}}{\sqrt{3+x}\sqrt{3x}}=2 \)
Let \( \Large \sqrt{3+x}=a \ and \ \sqrt{3x}=b \)
Then, \( \Large \frac{a+b}{ab}=\frac{2}{1} \)
\( \Large a+b=2a2b=a=3b \)
On squaring both sides, we get
\( \Large \sqrt{3+x}= \left(3\sqrt{3x}\right)^{2} \)
= \( \Large 3+x = 9 \left(3x\right) \)
= \( \Large 3+x=279x \)
= 10x = 24
\( \Large x = \frac{12}{5} \)

8). In an examination, a student scores 4 marks for every correct answer and losses 1 mark for every wrong answer. A student attempted all the 200 questions stud and scored 200 marks. Find the number of questions he answered correctly.
A). 82 
B). 80 
C). 68 
D). 60 
Correct Answer: 80
Let the number of correct answers be x
and number of wrong answers be y
Then, 4x  y = 200 ...(i)
and x + y = 200 ...(ii)
On adding Eqs. (i) and (ii). we get
4x  y = 200
x + y = 200
5x = 400
Therefore, x = 80 
9). The graphs of ax + by = c, dx + ey = f will be
I. parallel, if the system has no solution.
II. coincident, if the system has finite numbers of solutions.
III. intersecting, if the system has only one solution.
Which of the above statements are correct?
A). Only I and II 
B). Only ll and Ill 
C). Only I and III 
D). I, II and Ill 
Correct Answer: Only I and III ax + by = c and dx + ey = f
\( \Large \frac{a_{1}}{a_{2}}=\frac{a}{d}, \frac{b_{1}}{b_{2}}=\frac{b}{e}, \frac{c_{1}}{c_{2}}=\frac{c}{f} \)
Because, \( \Large \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)
Therefore, \( \Large \frac{b}{e} = \frac{c}{f} \)
It represents a pair of parallel lines.
Because, \( \Large \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \)
Therefore, \( \Large \frac{a}{d} = \frac{b}{e}\)
Therefore, system has unique solution and represents a pair of intersecting lines.

10). If \( \Large 3^{x+y}=81 \) and \( \Large 81^{xy}=3 \), then what is the value of x?
A). \( \Large \frac{17}{16} \) 
B). \( \Large \frac{17}{8} \) 
C). \( \Large \frac{17}{4} \) 
D). \( \Large \frac{15}{4} \) 
Correct Answer: \( \Large \frac{17}{8} \) Given,
\( \Large 3^{x+y}=51 \)
= \( \Large 3^{x+y}=3^{4} \)
= x + y = 4 ...(i)
and \( \Large 81^{xy}=3 \ or \ \left(3^{4}\right)^{xy}=3^{1} \)
= \( \Large xy = \frac{1}{4} \) ...(ii)
On solving the Eqs. (i) and (ii), we get
\( \Large 2x = \frac{17}{4} = x = \frac{17}{8} \)
