Numbers Questions in English

(C) Let the two natural numbers be $x$ and $y$.
Given $xy = 17$.
Since $17$ is a prime number,the only natural numbers whose product is $17$ are $1$ and $17$.
Thus,$x = 1$ and $y = 17$.
We need to find the sum of the reciprocals of their squares,which is $\frac{1}{x^2} + \frac{1}{y^2}$.
Substituting the values,we get $\frac{1}{1^2} + \frac{1}{17^2} = 1 + \frac{1}{289}$.
Calculating the sum: $1 + \frac{1}{289} = \frac{289 + 1}{289} = \frac{290}{289}$.

(C) Let the number be $x$.
According to the problem,doubling the number and adding $20$ is $2x + 20$.
Multiplying the number by $8$ and taking away $4$ is $8x - 4$.
Equating the two expressions: $2x + 20 = 8x - 4$.
Rearranging the terms: $8x - 2x = 20 + 4$.
$6x = 24$.
Dividing by $6$: $x = 4$.

(B) Let the natural number be $x$.
According to the problem,the equation is:
$3x^2 - 4x = x + 50$
Rearranging the terms to form a quadratic equation:
$3x^2 - 5x - 50 = 0$
Factoring the quadratic equation:
$3x^2 - 15x + 10x - 50 = 0$
$3x(x - 5) + 10(x - 5) = 0$
$(3x + 10)(x - 5) = 0$
This gives $x = -\frac{10}{3}$ or $x = 5$.
Since the number must be a natural number,we discard the negative fraction.
Therefore,the natural number is $5$.

(C) Let the four numbers be $a, b, c,$ and $d.$
According to the problem,$a+3 = b-3 = 3c = d/3 = k.$
From this,we can express each variable in terms of $k$:
$a = k-3$
$b = k+3$
$c = k/3$
$d = 3k$
Given that the sum of the numbers is $64$:
$(k-3) + (k+3) + (k/3) + 3k = 64$
$5k + k/3 = 64$
$(15k + k)/3 = 64$
$16k/3 = 64$
$k = (64 \times 3) / 16 = 12.$
Now,substitute $k=12$ to find the numbers:
$a = 12-3 = 9$
$b = 12+3 = 15$
$c = 12/3 = 4$
$d = 3 \times 12 = 36.$
The numbers are $9, 15, 4,$ and $36.$
The largest number is $36$ and the smallest is $4.$
The difference is $36 - 4 = 32.$

(B) Let the digit in the ten's place be $x$ and the digit in the unit's place be $y$. The original number is $10x + y$.
According to the problem,$y = 2x + 1$ (Equation $1$).
When the digits are interchanged,the new number is $10y + x$.
The difference between the new number and the original number is $(10y + x) - (10x + y) = 9y - 9x$.
According to the problem,this difference is $1$ less than the original number:
$9y - 9x = (10x + y) - 1$
$8y - 19x = -1$ (Equation $2$).
Substituting $y = 2x + 1$ into Equation $2$:
$8(2x + 1) - 19x = -1$
$16x + 8 - 19x = -1$
$-3x = -9$
$x = 3$.
Now,find $y$ using Equation $1$:
$y = 2(3) + 1 = 7$.
The original number is $10(3) + 7 = 37$.

(B) Let the two-digit number be $10x + y$,where $x$ is the digit in the ten's place and $y$ is the digit in the unit's place.
According to the problem,if the unit's digit is halved $(y/2)$ and the ten's digit is doubled $(2x)$,the new number is $10(2x) + y/2$.
This new number is equal to the number obtained by interchanging the digits,which is $10y + x$.
So,$10(2x) + y/2 = 10y + x$.
$20x + y/2 = 10y + x$.
Multiply the entire equation by $2$ to eliminate the fraction: $40x + y = 20y + 2x$.
Rearranging the terms: $40x - 2x = 20y - y$.
$38x = 19y$.
Dividing both sides by $19$,we get $2x = y$.
This means the digit in the unit's place $(y)$ is twice the digit in the ten's place $(x)$.

(B) Let the three consecutive odd numbers be $(x-2)$,$x$,and $(x+2)$.
According to the problem,the sum of these three numbers is $20$ more than the first number:
$(x-2) + x + (x+2) = (x-2) + 20$
Simplifying the left side: $3x = x + 18$
Subtract $x$ from both sides: $2x = 18$
Divide by $2$: $x = 9$
Thus,the middle number is $9$.

(B) Let the two numbers be $x$ and $y$.
According to the problem,we have the following equations:
$(1) \quad \frac{1}{5} x = \frac{5}{8} y$
$(2) \quad x + 35 = 4 y$
From equation $(1)$,we can express $x$ in terms of $y$:
$x = \frac{5}{8} y \times 5 = \frac{25}{8} y$
Now,substitute this value of $x$ into equation $(2)$:
$\frac{25}{8} y + 35 = 4 y$
Rearranging the terms to solve for $y$:
$35 = 4 y - \frac{25}{8} y$
$35 = \left( \frac{32 - 25}{8} \right) y$
$35 = \frac{7}{8} y$
$y = \frac{35 \times 8}{7}$
$y = 5 \times 8 = 40$
Therefore,the second number is $40$.

(D) Let the number be $x$.
According to the problem,the sum of the number and its square is $182$.
So,$x^2 + x = 182$.
Rearranging the equation,we get $x^2 + x - 182 = 0$.
To solve this quadratic equation,we look for two numbers whose product is $-182$ and whose sum is $1$.
These numbers are $14$ and $-13$.
So,$x^2 + 14x - 13x - 182 = 0$.
$x(x + 14) - 13(x + 14) = 0$.
$(x - 13)(x + 14) = 0$.
This gives $x = 13$ or $x = -14$.
Since $13$ is one of the options,the correct number is $13$.

(B) Let the number be $x$.
According to the problem,the equation is:
$\frac{x-4}{6} = 8$
Multiply both sides by $6$:
$x - 4 = 8 \times 6$
$x - 4 = 48$
Add $4$ to both sides to find $x$:
$x = 48 + 4 = 52$
Now,we need to find the result if $2$ is subtracted from the number and then divided by $5$:
$\frac{x - 2}{5} = \frac{52 - 2}{5}$
$= \frac{50}{5} = 10$

(B) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the product of the digits is $8$,so $xy = 8$.
When $18$ is added to the number,the digits are reversed,which means the new number becomes $10y + x$.
Therefore,$(10x + y) + 18 = 10y + x$.
Rearranging the terms: $10y - y + x - 10x = 18$,which simplifies to $9y - 9x = 18$.
Dividing by $9$,we get $y - x = 2$,or $y = x + 2$.
Substitute $y = x + 2$ into the product equation $xy = 8$:
$x(x + 2) = 8$
$x^2 + 2x - 8 = 0$
$(x + 4)(x - 2) = 0$.
Since $x$ must be a positive digit,$x = 2$.
Then $y = 2 + 2 = 4$.
The number is $10(2) + 4 = 24$.

(B) Let the two-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the difference between the number and the number obtained by interchanging its digits is $36$:
$| (10x + y) - (10y + x) | = 36$
$| 9x - 9y | = 36$
$| x - y | = 4$
Given the ratio of the digits is $1:2$,we have $y/x = 1/2$ or $x/y = 1/2$. Thus,$x = 2y$ or $y = 2x$.
Substituting $x = 2y$ into $| x - y | = 4$,we get $| 2y - y | = 4$,which means $y = 4$. Then $x = 2(4) = 8$.
The digits are $8$ and $4$.
The sum of the digits is $x + y = 8 + 4 = 12$.
The difference of the digits is $x - y = 8 - 4 = 4$.
The difference between the sum and the difference of the digits is $12 - 4 = 8$.

(D) Let the three consecutive odd integers be $x$,$(x+2)$,and $(x+4)$.
According to the problem,three times the first integer is $3$ more than twice the third integer.
$3x = 2(x+4) + 3$
$3x = 2x + 8 + 3$
$3x - 2x = 11$
$x = 11$
The three consecutive odd integers are $11$,$13$,and $15$.
The third integer is $x + 4 = 11 + 4 = 15$.

(D) Let the digits of the two-digit number be $x$ and $y$.
Given that the sum of the digits is $x + y = 15$ $(1)$.
Given that the difference between the digits is $x - y = 3$ $(2)$ or $y - x = 3$ $(3)$.
Case $1$: Solving $(1)$ and $(2)$:
Adding the equations,$2x = 18 \Rightarrow x = 9$.
Substituting $x = 9$ into $(1)$,$9 + y = 15 \Rightarrow y = 6$.
Thus,the number is $96$.
Case $2$: Solving $(1)$ and $(3)$:
Adding the equations,$2y = 18 \Rightarrow y = 9$.
Substituting $y = 9$ into $(1)$,$x + 9 = 15 \Rightarrow x = 6$.
Thus,the number is $69$.
Since both $96$ and $69$ satisfy the given conditions,the specific two-digit number cannot be uniquely determined.
Therefore,the correct option is $(d)$.

(B) Let the two numbers be $x$ and $y$.
Given that the product of the numbers is $xy = 45$ and the sum of their squares is $x^2 + y^2 = 106$.
We know the identity $(x + y)^2 = x^2 + y^2 + 2xy$.
Substituting the given values: $(x + y)^2 = 106 + 2(45) = 106 + 90 = 196$.
Thus,$x + y = \sqrt{196} = 14$.
We have $x + y = 14$ and $xy = 45$.
These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - 14t + 45 = 0$.
Factoring the equation: $(t - 9)(t - 5) = 0$.
Therefore,the numbers are $5$ and $9$.

(B) Let the two numbers be $x$ and $y$.
Given that the product of the numbers is $x y = 120$.
Given that the sum of their squares is $x^2 + y^2 = 289$.
We know the algebraic identity $(x + y)^2 = x^2 + y^2 + 2xy$.
Substituting the given values into the identity:
$(x + y)^2 = 289 + 2(120)$
$(x + y)^2 = 289 + 240$
$(x + y)^2 = 529$
Taking the square root on both sides,we get $x + y = \sqrt{529} = 23$.
Therefore,the sum of the numbers is $23$.

(C) Let the second number be $x$.
According to the problem:
First number $= 2x$
Third number $= \frac{1}{3} \times (2x) = \frac{2}{3}x$
The sum of the three numbers is $264$,so:
$2x + x + \frac{2}{3}x = 264$
Combine the terms:
$3x + \frac{2}{3}x = 264$
$\frac{9x + 2x}{3} = 264$
$\frac{11x}{3} = 264$
Solve for $x$:
$11x = 264 \times 3$
$11x = 792$
$x = \frac{792}{11} = 72$
Therefore,the second number is $72$.

(B) Let the number be $x$.
According to the problem,the equation is:
$\frac{2}{3}x - 50 = \frac{x}{4} + 40$
Rearrange the terms to bring $x$ variables on one side:
$\frac{2}{3}x - \frac{x}{4} = 50 + 40$
$\frac{2}{3}x - \frac{x}{4} = 90$
Find a common denominator for the fractions on the left side,which is $12$:
$\frac{8x - 3x}{12} = 90$
$\frac{5x}{12} = 90$
Multiply both sides by $12$:
$5x = 90 \times 12$
$5x = 1080$
Divide by $5$:
$x = \frac{1080}{5}$
$x = 216$
Thus,the number is $216$.

(B) Let the number be $x$.
According to the problem,the equation is:
$\left[\left(x + 2 \frac{1}{2}\right) \times 4 \frac{1}{2} + 3\right] \div 1 \frac{1}{5} = 25$
Convert mixed fractions to improper fractions:
$\left[\left(x + \frac{5}{2}\right) \times \frac{9}{2} + 3\right] \div \frac{6}{5} = 25$
Multiply both sides by $\frac{6}{5}$:
$\left(x + \frac{5}{2}\right) \times \frac{9}{2} + 3 = 25 \times \frac{6}{5}$
$\left(x + \frac{5}{2}\right) \times \frac{9}{2} + 3 = 30$
Subtract $3$ from both sides:
$\left(x + \frac{5}{2}\right) \times \frac{9}{2} = 27$
Multiply both sides by $\frac{2}{9}$:
$x + \frac{5}{2} = 27 \times \frac{2}{9}$
$x + \frac{5}{2} = 6$
Subtract $\frac{5}{2}$ from both sides:
$x = 6 - \frac{5}{2} = \frac{12 - 5}{2} = \frac{7}{2} = 3 \frac{1}{2}$

(A) Let the ten's digit be $x$ and the unit's digit be $y$. The number is $10x + y$.
Given that the unit's digit exceeds the ten's digit by $2$, so $y = x + 2$.
The sum of the digits is $x + y = x + (x + 2) = 2x + 2$.
The product of the number and the sum of its digits is $144$, so $(10x + y)(x + y) = 144$.
Substituting $y = x + 2$ into the equation:
$(10x + x + 2)(x + x + 2) = 144$
$(11x + 2)(2x + 2) = 144$
$(11x + 2) \cdot 2(x + 1) = 144$
$(11x + 2)(x + 1) = 72$
$11x^2 + 11x + 2x + 2 = 72$
$11x^2 + 13x - 70 = 0$
Factoring the quadratic equation:
$(11x + 35)(x - 2) = 0$
Since $x$ must be a positive integer, $x = 2$.
Then $y = x + 2 = 2 + 2 = 4$.
The number is $10(2) + 4 = 24$.

(B) Let the two numbers be $x$ and $y$.
According to the problem,the sum of the numbers is $x + y = 22$ (Equation $1$).
Also,five times one number is equal to $6$ times the other,so $5x = 6y$ (Equation $2$).
From Equation $2$,we can express $x$ as $x = \frac{6}{5}y$.
Substitute this into Equation $1$:
$\frac{6}{5}y + y = 22$
$\frac{6y + 5y}{5} = 22$
$\frac{11y}{5} = 22$
$11y = 110$
$y = 10$.
Now,substitute $y = 10$ into Equation $1$:
$x + 10 = 22$
$x = 12$.
The two numbers are $12$ and $10$. The bigger number is $12$.

(A) Let the two numbers be $x$ and $y$,where $x > y$.
According to the problem:
$x + y = 33$ --- $(1)$
$x - y = 15$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(x + y) - (x - y) = 33 - 15$
$2y = 18$
$y = 9$
Substituting $y = 9$ into equation $(1)$:
$x + 9 = 33$
$x = 33 - 9 = 24$
The two numbers are $24$ and $9$. The smaller number is $9$.

(C) Let the number be $x$.
According to the problem,the sum of one-half and one-fifth of the number exceeds one-third of the number by $7 \frac{1}{3} = \frac{22}{3}$.
So,the equation is: $\frac{x}{2} + \frac{x}{5} = \frac{x}{3} + \frac{22}{3}$.
Rearranging the terms to solve for $x$: $\frac{x}{2} + \frac{x}{5} - \frac{x}{3} = \frac{22}{3}$.
The least common multiple of $2, 5,$ and $3$ is $30$.
$\frac{15x + 6x - 10x}{30} = \frac{22}{3}$.
$\frac{11x}{30} = \frac{22}{3}$.
$x = \frac{22}{3} \times \frac{30}{11} = 2 \times 10 = 20$.
Therefore,the number is $20$.

(A) Let the positive number be $x$.
According to the problem,two-third of the number is equal to $\frac{25}{216}$ of its reciprocal.
So,$\frac{2}{3} x = \frac{25}{216} \times \frac{1}{x}$.
Multiplying both sides by $x$,we get $\frac{2}{3} x^2 = \frac{25}{216}$.
Now,$x^2 = \frac{25}{216} \times \frac{3}{2}$.
$x^2 = \frac{25}{72 \times 2} = \frac{25}{144}$.
Taking the square root of both sides,$x = \sqrt{\frac{25}{144}} = \frac{5}{12}$ (since the number is positive).

(A) The given expression is $7372 \times 7372 + 7372 \times 628$.
Using the distributive property of multiplication over addition,$a \times b + a \times c = a \times (b + c)$.
Here,$a = 7372$,$b = 7372$,and $c = 628$.
So,the expression becomes $7372 \times (7372 + 628)$.
Calculating the sum inside the parentheses: $7372 + 628 = 8000$.
Now,multiply: $7372 \times 8000 = 58976000$.
Thus,the correct option is $A$.

(A) Let the missing number be $x$.
Then,the equation is: $9999 + 8888 + 777 + x = 19700$.
First,calculate the sum of the known numbers: $9999 + 8888 + 777 = 19664$.
Now,substitute this sum into the equation: $19664 + x = 19700$.
To find $x$,subtract $19664$ from $19700$: $x = 19700 - 19664$.
Therefore,$x = 36$.

(A) Let the missing digit be represented by $x$,so the expression becomes $(60x6) \times 111 = 666666$.
Divide both sides by $111$:
$60x6 = \frac{666666}{111}$
Performing the division:
$666666 \div 111 = 6006$
Comparing $60x6$ with $6006$,we find that the missing digit $x = 0$.

(A) Let the missing digit be $x$. The equation is $3149 \times (100 + 10x + 5) = 425115$.
Divide both sides by $3149$:
$100 + 10x + 5 = \frac{425115}{3149}$
$105 + 10x = 135$
$10x = 135 - 105$
$10x = 30$
$x = 3$
Therefore,the missing digit is $3$.

(D) Let the age of Mr. Manoj be $(10x + y)$ years.
Then,his wife's age is $(10y + x)$ years.
According to the problem,$\frac{1}{11}$ of the sum of their ages is equal to the difference between their ages:
$\frac{1}{11} (10x + y + 10y + x) = (10x + y) - (10y + x)$
$\frac{1}{11} (11x + 11y) = 9x - 9y$
$x + y = 9(x - y)$
$x + y = 9x - 9y$
$10y = 8x$
$\frac{x}{y} = \frac{10}{8} = \frac{5}{4}$
Since $x$ and $y$ are single digits,we take $x = 5$ and $y = 4$.
Therefore,Mr. Manoj's age is $54$ years and his wife's age is $45$ years.
The difference between their ages is $54 - 45 = 9$ years.

(B) Let $d$ be the divisor and $q$ be the quotient.
In a long division process,the dividend $D$ can be expressed as $D = d \times q + r_{last}$.
Given $D = 380606$ and $r_{last} = 413$,we have $d \times q = 380606 - 413 = 380193$.
Since $380193$ is odd,the divisor $d$ must be odd. This eliminates $d = 3372$.
Also,$380193$ does not end in $0$ or $5$,so $d$ cannot be a multiple of $5$. This eliminates $c = 4215$.
We are left with $451$ and $843$. We check the first step of the division:
If $d = 843$,then the first part of the dividend $3806$ divided by $843$ gives $3806 = 843 \times 4 + 434$. This matches the first remainder $434$.
If $d = 451$,then $3806 = 451 \times 8 + 198$. This does not match the first remainder $434$.
Thus,the divisor is $843$.

(C) Given that $\frac{x}{y} = \frac{3}{4}$,we can express $x$ as $x = \frac{3}{4}y$.
Substitute $x = \frac{3}{4}y$ into the expression $\frac{y-x}{y+x}$:
$\frac{y - \frac{3}{4}y}{y + \frac{3}{4}y} = \frac{\frac{1}{4}y}{\frac{7}{4}y} = \frac{1}{7}$.
Now,add this to the first term:
$\frac{6}{7} + \frac{1}{7} = \frac{7}{7} = 1$.

(C) Let the three consecutive even natural numbers be $2n, 2n+2$,and $2n+4$,where $n$ is a natural number.
The product $P = 2n(2n+2)(2n+4) = 8n(n+1)(n+2)$.
Since $n, (n+1)$,and $(n+2)$ are three consecutive integers,their product $n(n+1)(n+2)$ is always divisible by $3! = 6$.
Therefore,the product $P$ is divisible by $8 \times 6 = 48$.
Testing with the first three consecutive even numbers: $2 \times 4 \times 6 = 48$.
Testing with the next set: $4 \times 6 \times 8 = 192$,which is $48 \times 4$.
Thus,the largest natural number that always divides the product is $48$.

(B) Let the missing number be $x$.
Then the equation becomes $\frac{x}{21} \times \frac{x}{189} = 1$.
Multiplying the numerators and denominators,we get $\frac{x^2}{21 \times 189} = 1$.
Therefore,$x^2 = 21 \times 189$.
We can factorize $189$ as $21 \times 9$.
So,$x^2 = 21 \times (21 \times 9) = 21^2 \times 3^2$.
Taking the square root of both sides,$x = 21 \times 3 = 63$.

(D) Let the divisor be $D$,quotient be $Q$,and remainder be $R$.
Given that $R = 48$.
According to the problem,the divisor is $5$ times the remainder,so $D = 5 \times R = 5 \times 48 = 240$.
Also,the divisor is $12$ times the quotient,so $D = 12 \times Q$.
Substituting the value of $D$,we get $240 = 12 \times Q$,which implies $Q = 240 / 12 = 20$.
The formula for the dividend is: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$.
Substituting the values: $\text{Dividend} = (240 \times 20) + 48 = 4800 + 48 = 4848$.

(B) To find the number that leaves a remainder of $6$ when divided by $9, 11,$ and $13$,we first calculate the Least Common Multiple $(LCM)$ of $9, 11,$ and $13$.
Since $9, 11,$ and $13$ are coprime,their $LCM = 9 \times 11 \times 13 = 1287$.
$A$ number that leaves a remainder of $6$ when divided by $9, 11,$ and $13$ is of the form $(LCM + 6) = 1287 + 6 = 1293$.
We need to subtract a number from $1294$ to get $1293$.
Required number $= 1294 - 1293 = 1$.

(B) Let the first part be $x$ and the second part be $(24 - x)$.
According to the problem,$7$ times the first part added to $5$ times the second part equals $146$.
So,the equation is: $7x + 5(24 - x) = 146$.
Expanding the equation: $7x + 120 - 5x = 146$.
Simplifying the terms: $2x + 120 = 146$.
Subtracting $120$ from both sides: $2x = 146 - 120$.
$2x = 26$.
Dividing by $2$: $x = 13$.
Therefore,the first part is $13$.

(A) Let the number be $x$.
According to the problem,$\frac{1}{3}$ of the number minus $\frac{1}{4}$ of the number equals $12$.
So,the equation is: $\frac{1}{3}x - \frac{1}{4}x = 12$.
To solve for $x$,find a common denominator for the fractions,which is $12$.
$\frac{4x}{12} - \frac{3x}{12} = 12$.
$\frac{x}{12} = 12$.
Multiplying both sides by $12$,we get $x = 12 \times 12 = 144$.
Therefore,the number is $144$.

(B) Let the number be $x$.
According to the problem,$\frac{4}{3} \times x = 64$.
To find $x$,multiply both sides by $\frac{3}{4}$:
$x = 64 \times \frac{3}{4} = 16 \times 3 = 48$.
Now,we need to find half of that number:
$\frac{1}{2} \times x = \frac{1}{2} \times 48 = 24$.

(D) Let the required fraction be $\frac{x}{y}$.
According to the first condition,adding $1$ to both the numerator and denominator gives $4$:
$\frac{x+1}{y+1} = 4 \Rightarrow x+1 = 4y+4 \Rightarrow x-4y = 3$ (Equation $1$).
According to the second condition,subtracting $1$ from both the numerator and denominator gives $7$:
$\frac{x-1}{y-1} = 7 \Rightarrow x-1 = 7y-7 \Rightarrow x-7y = -6$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$:
$(x-4y) - (x-7y) = 3 - (-6)$
$3y = 9 \Rightarrow y = 3$.
Substituting $y=3$ into Equation $1$:
$x - 4(3) = 3 \Rightarrow x - 12 = 3 \Rightarrow x = 15$.
The fraction is $\frac{15}{3}$. The numerator is $15$.

(C) Let the three numbers be $3x$,$4x$,and $5x$.
According to the problem,the sum of the largest $(5x)$ and the smallest $(3x)$ is equal to the sum of the third $(4x)$ and $52$.
So,the equation is: $5x + 3x = 4x + 52$.
Simplifying the equation: $8x = 4x + 52$.
Subtracting $4x$ from both sides: $4x = 52$.
Dividing by $4$: $x = 13$.
The smallest number is $3x = 3 \times 13 = 39$.

(A) Let the three numbers be $a, b,$ and $c$.
Given ratios are $\frac{a}{b} = \frac{2}{3}$ and $\frac{b}{c} = \frac{5}{3}$.
To find the combined ratio $a:b:c$,we make the value of $b$ common in both ratios.
Multiply the first ratio by $5$ and the second ratio by $3$:
$\frac{a}{b} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$
$\frac{b}{c} = \frac{5 \times 3}{3 \times 3} = \frac{15}{9}$
Thus,$a:b:c = 10:15:9$.
Let the numbers be $10x, 15x,$ and $9x$.
The sum of the numbers is $68$,so:
$10x + 15x + 9x = 68$
$34x = 68$
$x = 2$
The second number is $15x = 15 \times 2 = 30$.

(C) Let the required fraction be $\frac{x}{y}$.
According to the first condition,adding $1$ to the denominator gives $\frac{x}{y+1} = \frac{1}{2}$.
This simplifies to $2x = y + 1$,or $2x - y = 1$ (Equation $1$).
According to the second condition,adding $1$ to the numerator gives $\frac{x+1}{y} = 1$.
This simplifies to $x + 1 = y$,or $x - y = -1$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$:
$(2x - y) - (x - y) = 1 - (-1)$
$2x - y - x + y = 1 + 1$
$x = 2$.
Substituting $x = 2$ into Equation $2$:
$2 - y = -1$
$y = 3$.
Therefore,the fraction is $\frac{2}{3}$.

(B) Let the number be $x$.
According to the problem,$\frac{4}{5}x - \frac{2}{3}x = 8$.
To solve this,find a common denominator for the fractions,which is $15$.
$\frac{12x - 10x}{15} = 8$.
$\frac{2x}{15} = 8$.
Multiply both sides by $15$:
$2x = 8 \times 15 = 120$.
Divide by $2$:
$x = 60$.

(C) To find the sum of all prime numbers between $60$ and $80$,we first identify the prime numbers in this range.
Prime numbers are numbers greater than $1$ that have only two factors: $1$ and themselves.
The prime numbers between $60$ and $80$ are $61, 67, 71, 73,$ and $79$.
Now,we calculate their sum:
$61 + 67 + 71 + 73 + 79 = 351$.
Therefore,the sum of all prime numbers from $60$ to $80$ is $351$.

(B) The relationship between dividend,divisor,quotient,and remainder is given by the formula:
Dividend = (Divisor $\times$ Quotient) + Remainder
Given:
Dividend = $24446$
Quotient = $79$
Remainder = $35$
Let the divisor be $x$.
$24446 = (x \times 79) + 35$
$24446 - 35 = 79x$
$24411 = 79x$
$x = 24411 \div 79$
$x = 309$
Therefore,the divisor is $309$.

(A) The relationship between dividend,divisor,quotient,and remainder is given by the formula:
$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$
Given:
$\text{Divisor} = 456$
$\text{Quotient} = 120$
$\text{Remainder} = 333$
Substituting the values into the formula:
$\text{Dividend} = (456 \times 120) + 333$
$\text{Dividend} = 54720 + 333$
$\text{Dividend} = 55053$

(B) The relationship between dividend, divisor, quotient, and remainder is given by the formula:
Dividend $= (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$
Given:
Divisor $= 62$
Quotient $= 463$
Remainder $= 60$
Substituting these values into the formula:
Dividend $= (62 \times 463) + 60$
Dividend $= 28706 + 60$
Dividend $= 28766$
Therefore, the number is $28766$.

(C) Let the number be $N$. According to the problem,$N = 221q + 43$,where $q$ is the quotient.
We want to find the remainder when $N$ is divided by $17$.
$N = (13 \times 17)q + 43$.
Since $(13 \times 17)q$ is perfectly divisible by $17$,the remainder depends only on dividing $43$ by $17$.
$43 = 17 \times 2 + 9$.
Thus,the remainder is $9$.

(A) prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
To find the largest three-digit prime number,we check the numbers starting from the largest three-digit number,which is $999$.
$1$. $999$ is divisible by $3$ $(999 = 3 \times 333)$,so it is not prime.
$2$. $998$ is an even number,so it is divisible by $2$,hence not prime.
$3$. $997$ is not divisible by any prime number up to $\sqrt{997} \approx 31.57$ (i.e.,$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31$).
Since $997$ is not divisible by any of these primes,it is a prime number.
Therefore,$997$ is the largest three-digit prime number.

(A) Let the number be $x$. According to the problem,$7 \times x = 555...5$ (a number consisting only of fives).
To find the least such number,we start testing numbers consisting of $n$ fives,where $n = 1, 2, 3, ...$
We need to find the smallest $n$ such that $55...5$ (n times) is divisible by $7$.
$5/7$ (Remainder $5$)
$55/7 = 7$ remainder $6$
$555/7 = 79$ remainder $2$
$5555/7 = 793$ remainder $4$
$55555/7 = 7936$ remainder $3$
$555555/7 = 79365$ remainder $0$.
Since $555555$ is the smallest number consisting only of fives that is divisible by $7$,the quotient $79365$ is the least such number.
Therefore,the least number is $79365$.