**7.**The difference between the squares of two consecutive numbers is 35. The numbers are

a. 14,15

b. 15,16

c. 17,18

d. 18,19

Answer: C

Explanation:

Let the numbers be $a$ and $a+1$

${(a + 1)^2} - {a^2} = 35$

$ \Rightarrow {a^2} + 2a + 1 - {a^2} = 35$

$ \Rightarrow 2a = 34$ or a = 17

The numbers are 17 & 18.

Explanation:

Let the numbers be $a$ and $a+1$

${(a + 1)^2} - {a^2} = 35$

$ \Rightarrow {a^2} + 2a + 1 - {a^2} = 35$

$ \Rightarrow 2a = 34$ or a = 17

The numbers are 17 & 18.

**8.**$a$, $b$, $c$, $d$, $e$ are five consecutive numbers in the same order. If $a + b + d + e = 152$, Find $a \times e$

a. 1400

b. 1440

c. 1480

d. 1520

Answer: B

Explanation:

Let the numbers be $x$, $x+1$, $x+2$, $x+4$, $x+4$

Given that, $a + b + d + e$ = $x+(x+1)+(x+3)+(x+4)$ = $4x + 8$

$4x + 8 = 152$

$ \Rightarrow x = \dfrac{{144}}{4} = 36$

$a \times e$ = $36 \times 40 = 1440$

Explanation:

Let the numbers be $x$, $x+1$, $x+2$, $x+4$, $x+4$

Given that, $a + b + d + e$ = $x+(x+1)+(x+3)+(x+4)$ = $4x + 8$

$4x + 8 = 152$

$ \Rightarrow x = \dfrac{{144}}{4} = 36$

$a \times e$ = $36 \times 40 = 1440$

**9.**The sum of a two digit number and the number formed by reversing the two digits is always divisible by

a. 9

b. 10

c. 11

d. 3

Answer: C

Explanation:

To solve questions which involve "a number" and its "digits", we have to take the decimal format of the number.

Let the number be $ab$. Then, decimal format of the number = ${\left( {ab} \right)_{10}}$ = $10a + b$

Similarly, a three digit number $abc$ can be written in decimal format as ${\left( {abc} \right)_{10}}$ $= {10^2} \times a + 10b + c$ $ = 100 \times a + 10b + c$

Now, the number with reversed digits will be, $ba = 10b + a$

Sum = $(10a + b) + (10b + a)$ $=11a + 11b = 11(a+b)$

Explanation:

To solve questions which involve "a number" and its "digits", we have to take the decimal format of the number.

Let the number be $ab$. Then, decimal format of the number = ${\left( {ab} \right)_{10}}$ = $10a + b$

Similarly, a three digit number $abc$ can be written in decimal format as ${\left( {abc} \right)_{10}}$ $= {10^2} \times a + 10b + c$ $ = 100 \times a + 10b + c$

Now, the number with reversed digits will be, $ba = 10b + a$

Sum = $(10a + b) + (10b + a)$ $=11a + 11b = 11(a+b)$

**10.**A two digit number when 18 is subtracted becomes another two digit number with reversed digits. How many such two digit numbers are possible?

a. 2

b. 3

c. 7

d. 8

Answer: C

Explanation:

Let the two digit number be $ab$ and the number formed after subtracting 18 to it is $ba$

So $ab - 18 = ba$

$ \Rightarrow (10a + b) - 18 = 10b + a$

$ \Rightarrow 9a-9b = 18 $

$ \Rightarrow a-b = 2 $

So we know that $ab$, $ba$ both are two digit numbers so a, b \( \ne \) 0.

$\begin{array}{|cc|cc|}\hline

a & b & a & b \\[4px]\hline

9 & 7 & 5 & 3 \\[4px]

8 & 6 & 4 & 2 \\[4px]

7 & 5 & 3 & 1 \\[4px]

6 & 4 & - \\\hline

\end{array}$

Total 7 combinations are possible. So correct option is c.

Explanation:

Let the two digit number be $ab$ and the number formed after subtracting 18 to it is $ba$

So $ab - 18 = ba$

$ \Rightarrow (10a + b) - 18 = 10b + a$

$ \Rightarrow 9a-9b = 18 $

$ \Rightarrow a-b = 2 $

So we know that $ab$, $ba$ both are two digit numbers so a, b \( \ne \) 0.

$\begin{array}{|cc|cc|}\hline

a & b & a & b \\[4px]\hline

9 & 7 & 5 & 3 \\[4px]

8 & 6 & 4 & 2 \\[4px]

7 & 5 & 3 & 1 \\[4px]

6 & 4 & - \\\hline

\end{array}$

Total 7 combinations are possible. So correct option is c.

**11.**24 is divided into two parts such that 7 times the first part added to 5 times the second part gives 146. The first part is

a. 11

b. 13

c. 16

d. 17

Answer: B

Explanation:

Let the first and second parts be $a$ and $24-a$, then

${{7a + 5(24 − a) = 146}}$

$ \Rightarrow {{7a + 120 − 5a = 146}}$

$ \Rightarrow {{2a = 26}}$ or a = 13

Explanation:

Let the first and second parts be $a$ and $24-a$, then

${{7a + 5(24 − a) = 146}}$

$ \Rightarrow {{7a + 120 − 5a = 146}}$

$ \Rightarrow {{2a = 26}}$ or a = 13

**12.**The product of two numbers is 120. The sum of their squares is 289. The sum of the two numbers is :

a. 20

b. 23

c. 169

d. None

Answer: B

Explanation:

Let the number be $x$ and $y$ . We know that,

${(x + y)^2} = ({x^2} + {y^2}) + 2xy = 289 + 2\times120$

$ = 289 + 240 = 529 \Rightarrow x + y = \sqrt {529} = 23$

Explanation:

Let the number be $x$ and $y$ . We know that,

${(x + y)^2} = ({x^2} + {y^2}) + 2xy = 289 + 2\times120$

$ = 289 + 240 = 529 \Rightarrow x + y = \sqrt {529} = 23$