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$
