**13.**The sum of squares of two numbers is 68 and the square of their difference is 36. The product of the two numbers is

a. 16

b. 32

c. 58

d. 104

Answer: A

Explanation:

Let the numbers be $x$ and $y$. Then

${x^2} + {y^2} = 68$

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

$ \Rightarrow 68 − 2xy = 36 \Rightarrow 2xy = 32$

$ \Rightarrow xy = 16$

Explanation:

Let the numbers be $x$ and $y$. Then

${x^2} + {y^2} = 68$

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

$ \Rightarrow 68 − 2xy = 36 \Rightarrow 2xy = 32$

$ \Rightarrow xy = 16$

**14.**The sum of seven numbers is 235. The average of the first three is 23 and that of the last three is 42. The fourth number is

a. 40

b. 126

c. 69

d. 195

Answer: A

Explanation:

Average of the first three is 23. Therefore their sum = 23 x 3

Average of the last three is 42. Therefore their sum = 42 x 3

Sum of all number = Sum of first three + Fourth number + Sum of last three.

$(23 \times 3 + a + 42 \times 3) = 235 \Rightarrow a = 40$

Explanation:

Average of the first three is 23. Therefore their sum = 23 x 3

Average of the last three is 42. Therefore their sum = 42 x 3

Sum of all number = Sum of first three + Fourth number + Sum of last three.

$(23 \times 3 + a + 42 \times 3) = 235 \Rightarrow a = 40$

**15.**Two numbers are such that the ratio between them is 3:5 but if each is increased by 10, the ratio between them becomes 5 : 7, the numbers are

a. 3, 5

b. 7, 9

c. 13, 22

d. 15, 25

Answer: D

Explanation:

Let the numbers be $3a$ and $5a$

Then $\displaystyle\frac{{3a + 10}}{{5a + 10}} = \displaystyle\frac{5}{7}$

$ \Rightarrow 7(3a + 10) = 5(5a + 10) \Rightarrow a = 5$

The numbers are 15 & 25.

Explanation:

Let the numbers be $3a$ and $5a$

Then $\displaystyle\frac{{3a + 10}}{{5a + 10}} = \displaystyle\frac{5}{7}$

$ \Rightarrow 7(3a + 10) = 5(5a + 10) \Rightarrow a = 5$

The numbers are 15 & 25.

**16.**A fraction becomes 4 when 1 is added to both the numerator and denominator, and it becomes 7 when 1 is subtracted from both the numerator and denominator. The numerator of the given fraction is :

a. 2

b. 3

c. 7

d. 15

Answer: D

Explanation:

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

Then $\displaystyle\frac{{a + 1}}{{b + 1}} = 4 \Rightarrow a − 4b = 3$

and $\displaystyle\frac{{a − 1}}{{b − 1}} = 7 \Rightarrow a − 7b = - 6$

Solving these equations we get,

a = 15

b = 3

Explanation:

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

Then $\displaystyle\frac{{a + 1}}{{b + 1}} = 4 \Rightarrow a − 4b = 3$

and $\displaystyle\frac{{a − 1}}{{b − 1}} = 7 \Rightarrow a − 7b = - 6$

Solving these equations we get,

a = 15

b = 3

**17.**$a$, $b$, $c$, $d$, $e$ are five consecutive odd numbers in the same order. If the sum of the first and fourth numbers is 100, Find the square of the fifth number.

a. 2809

b. 3025

c. 3249

d. 3481

Answer: B

Explanation:

Let the numbers be $2x+1$, $2x+3$, $2x+5$, $2x+7$, $2x+9$

Given that, $a + d$ = $(2x+1)+(2x+7)$ = $4x + 8$

$4x + 8 = 100$

$ \Rightarrow x = \dfrac{{92}}{4} = 23$

Fifth number = $2 \times 23 + 9 = 55$

$ \Rightarrow {55^2} = 3025$

Explanation:

Let the numbers be $2x+1$, $2x+3$, $2x+5$, $2x+7$, $2x+9$

Given that, $a + d$ = $(2x+1)+(2x+7)$ = $4x + 8$

$4x + 8 = 100$

$ \Rightarrow x = \dfrac{{92}}{4} = 23$

Fifth number = $2 \times 23 + 9 = 55$

$ \Rightarrow {55^2} = 3025$

**18.**Sum of four consecutive even numbers is 60. Find the product of second and third numbers.

a. 192

b. 168

c. 224

d. 288

Answer: C

Explanation:

Let the numbers be $2x$, $2x+2$, $2x+4$, $2x+6$

Given that, $a + e$ = $2x+(2x+2)+(2x+4)+(2x+6)$ = $8x + 12$ = 60

$ \Rightarrow x = \dfrac{{48}}{8} = 6$

Product of second and third numbers = $14 \times 16 = 224$

Explanation:

Let the numbers be $2x$, $2x+2$, $2x+4$, $2x+6$

Given that, $a + e$ = $2x+(2x+2)+(2x+4)+(2x+6)$ = $8x + 12$ = 60

$ \Rightarrow x = \dfrac{{48}}{8} = 6$

Product of second and third numbers = $14 \times 16 = 224$