**1.**If $2x + 3y = 12$, $3x - 4y = 1$, then the values of $x$ and $y$ are

a. 2, 3

b. 6, 0

c. 0, 4

d. 9, -2

Answer: B

Explanation:

$2x + 3y = 12$ - - - (1)

$3x - 4y = 1$ - - - (2)

Multiplying the first equation by 4, and the second by 3, we get

$8x + 12y = 48$

$9x - 12y = 3$

Adding above two equations,

$17x = 51$

$\Rightarrow x = 3$

Substituting in the first equation,

$2 \times 3 + 3y = 12$

$3y = 12$

$\Rightarrow y = 4$

$\therefore (x,y) = (3, 4)$

Explanation:

$2x + 3y = 12$ - - - (1)

$3x - 4y = 1$ - - - (2)

Multiplying the first equation by 4, and the second by 3, we get

$8x + 12y = 48$

$9x - 12y = 3$

Adding above two equations,

$17x = 51$

$\Rightarrow x = 3$

Substituting in the first equation,

$2 \times 3 + 3y = 12$

$3y = 12$

$\Rightarrow y = 4$

$\therefore (x,y) = (3, 4)$

**2.**If 1 is added to the denominator of fraction, the fraction becomes $\displaystyle\frac{1}{2}$. If 1 is added to the numerator, the fraction becomes 1. The fraction is

a. $\displaystyle\frac{4}{7}$

b. $\displaystyle\frac{5}{9}{\rm{ }}$

c. $\displaystyle\frac{2}{3}$

d. $\displaystyle\frac{{10}}{{11}}$

Answer: C

Explanation:

Let the required fraction be $\displaystyle\frac{a}{b}$.

$\Rightarrow \displaystyle\frac{a}{{b + 1}} = \displaystyle\frac{1}{2} \Rightarrow 2a-b = 1$ - - - - (1)

$ \Rightarrow \displaystyle\frac{{a + 1}}{b} = 1$ $ \Rightarrow a-b =-1$ - - - - (2)

Solving (1) & (2) we get $a = 2, b=3$

Fraction = $\displaystyle\frac{a}{b}{\rm{ = }}\displaystyle\frac{2}{3}$

Explanation:

Let the required fraction be $\displaystyle\frac{a}{b}$.

$\Rightarrow \displaystyle\frac{a}{{b + 1}} = \displaystyle\frac{1}{2} \Rightarrow 2a-b = 1$ - - - - (1)

$ \Rightarrow \displaystyle\frac{{a + 1}}{b} = 1$ $ \Rightarrow a-b =-1$ - - - - (2)

Solving (1) & (2) we get $a = 2, b=3$

Fraction = $\displaystyle\frac{a}{b}{\rm{ = }}\displaystyle\frac{2}{3}$

**3.**The sum of two numbers is twice their difference. If one of the numbers is 10, the other number is

a. ${\rm{3}}\displaystyle\frac{1}{3}$

b. 30

c. ${\rm{ - 3}}\displaystyle\frac{1}{3}$

d. a or b

Answer: D

Explanation:

Let the other number be $a$ and $a > 10$

Given that, ${{10 + a = 2(a-10)}} \Rightarrow {{a = 30}}$

If $a < 10$,

${{10 + a = 2}}({{10-a)}} \Rightarrow $${{a = }}\dfrac{{10}}{3} = 3\dfrac{1}{3}$

Explanation:

Let the other number be $a$ and $a > 10$

Given that, ${{10 + a = 2(a-10)}} \Rightarrow {{a = 30}}$

If $a < 10$,

${{10 + a = 2}}({{10-a)}} \Rightarrow $${{a = }}\dfrac{{10}}{3} = 3\dfrac{1}{3}$

**4.**The sum of squares of two numbers is 80 and the square of their difference is 36. The product of the two numbers is.

a. 22

b. 44

c. 58

d. 116

Answer: A

Explanation:

Let the numbers be x and y . Then

${{{x}}^2} + {y^2} = 80$ and ${({{x - y}})^2} = 36$

${(x - y)^2} = 36 \Rightarrow {x^2} + {y^2} - 2xy = 36$

$ \Rightarrow 2xy = ({{{x}}^2} + {y^2}) - 36 = 80 - 36 = 44$

$ \Rightarrow xy = 22$

Explanation:

Let the numbers be x and y . Then

${{{x}}^2} + {y^2} = 80$ and ${({{x - y}})^2} = 36$

${(x - y)^2} = 36 \Rightarrow {x^2} + {y^2} - 2xy = 36$

$ \Rightarrow 2xy = ({{{x}}^2} + {y^2}) - 36 = 80 - 36 = 44$

$ \Rightarrow xy = 22$

**5.**A man has some hens and some cows. If the number of heads is 50 and number of feet is 142. The number of cows is:

a. 26

b. 24

c. 21

d. 20

Answer: C

Explanation:

Let the mans has $x$ hens and $y$ cows.

Then total heads = $ x + y = 50 $ - - - (1)

total feet = $2x + 4y = 142$ - - - (2)

Solving above two equations, we get $x = 29$, $y = 21$

So cows are 21.

Let, the man has hens only.

Then total heads = 50 $\times$ 1 = 50.

And, legs = 50 $\times$ 2 = 100 which is short by 42 from the actual legs i.e. = 142.

Now, replacement of one cow with one hen means the same number of heads and two more legs.

Therefore, Hens replaced with cows = $\displaystyle\frac{{{{42}}}}{{{2}}}$ = 21

Therefore, Cows = 21

Explanation:

Let the mans has $x$ hens and $y$ cows.

Then total heads = $ x + y = 50 $ - - - (1)

total feet = $2x + 4y = 142$ - - - (2)

Solving above two equations, we get $x = 29$, $y = 21$

So cows are 21.

*Alternative method:*Let, the man has hens only.

Then total heads = 50 $\times$ 1 = 50.

And, legs = 50 $\times$ 2 = 100 which is short by 42 from the actual legs i.e. = 142.

Now, replacement of one cow with one hen means the same number of heads and two more legs.

Therefore, Hens replaced with cows = $\displaystyle\frac{{{{42}}}}{{{2}}}$ = 21

Therefore, Cows = 21

**6.**A sum of Rs. 350 made up of 110 coins, which are of either Rs. 1 or Rs. 5 denomination. How many coins are of Rs. 5?

a. 52

b. 60

c. 62

d. 72

Answer: B

Explanation:

Let, the number of Rs.1 coins are $x$ and Rs.5 coins are $y$.

Then $x + y = 110$

$x + 5y = 350$

Solving above two equations, we get $y = 60$. So number of Rs.5 coins are 60.

Let, all the coins are of Rs.1 denomination.

Then, total value of 110 coins = 110 $\times$ 1 = Rs. 110 which is short from Rs. 350 by Rs.350 - Rs.110 = Rs. 240.

Now, replacing 1 one-rupee coin with five-rupee coin means Rs. 4 extra.

Therefore, Five rupee coins = $\displaystyle\frac{{{{240}}}}{{{4}}}$ = 60 coins

Explanation:

Let, the number of Rs.1 coins are $x$ and Rs.5 coins are $y$.

Then $x + y = 110$

$x + 5y = 350$

Solving above two equations, we get $y = 60$. So number of Rs.5 coins are 60.

*Alternative method:*Let, all the coins are of Rs.1 denomination.

Then, total value of 110 coins = 110 $\times$ 1 = Rs. 110 which is short from Rs. 350 by Rs.350 - Rs.110 = Rs. 240.

Now, replacing 1 one-rupee coin with five-rupee coin means Rs. 4 extra.

Therefore, Five rupee coins = $\displaystyle\frac{{{{240}}}}{{{4}}}$ = 60 coins