Numbers Questions in English

(C) Let the $5$ successive odd integers be $x, x+2, x+4, x+6,$ and $x+8.$
According to the problem,the sum of these marks is $185.$
$x + (x+2) + (x+4) + (x+6) + (x+8) = 185$
$5x + 20 = 185$
$5x = 185 - 20$
$5x = 165$
$x = 33$
The highest score is $x+8 = 33 + 8 = 41.$

(A) To compare the given numbers,we express them with a common index (root).
The numbers are $4^{1/3}, 2^{1/2}, 3^{1/6}, 5^{1/4}$.
The denominators of the exponents are $3, 2, 6, 4$. The $LCM$ of these denominators is $12$.
Now,rewrite each number with the exponent $1/12$:
$4^{1/3} = 4^{4/12} = (4^4)^{1/12} = 256^{1/12}$
$2^{1/2} = 2^{6/12} = (2^6)^{1/12} = 64^{1/12}$
$3^{1/6} = 3^{2/12} = (3^2)^{1/12} = 9^{1/12}$
$5^{1/4} = 5^{3/12} = (5^3)^{1/12} = 125^{1/12}$
Comparing the bases: $256 > 125 > 64 > 9$.
Therefore,the descending order is $\sqrt[3]{4} > \sqrt[4]{5} > \sqrt{2} > \sqrt[6]{3}$.

(B) Let the expression be $f(n) = 3^{2n} + 9n + 5$.
We can rewrite the expression as $f(n) = 3^{2n} + 9n + 3 + 2$.
Now,we can factor out $3$ from the first three terms:
$f(n) = 3(3^{2n-1} + 3n + 1) + 2$.
Since $3(3^{2n-1} + 3n + 1)$ is clearly divisible by $3$,the remainder when $f(n)$ is divided by $3$ is $2$.

(B) prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
The sequence of prime numbers begins as $2, 3, 5, 7, 11, \dots$
Since $2$ is the first number in this sequence,it is the smallest prime number.
Therefore,the least prime number is $2$.

(C) First,observe the difference between the divisor and the corresponding remainder:
$12 - 8 = 4$
$18 - 14 = 4$
$36 - 32 = 4$
$45 - 41 = 4$
Since the difference is constant $(4)$,the required number is $\operatorname{LCM}(12, 18, 36, 45) - 4$.
Calculate the $\operatorname{LCM}$ of $12, 18, 36, 45$:
$12 = 2^2 \times 3$
$18 = 2 \times 3^2$
$36 = 2^2 \times 3^2$
$45 = 3^2 \times 5$
$\operatorname{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$.
Therefore,the required number $= 180 - 4 = 176$.

(C) To calculate $204 \times 197$,we can use the algebraic identity $(x + a)(x - b) = x^2 + x(a - b) - ab$.
Here,let $x = 200$,$a = 4$,and $b = 3$.
$204 \times 197 = (200 + 4)(200 - 3)$
$= 200^2 + 200(4 - 3) - (4 \times 3)$
$= 40000 + 200(1) - 12$
$= 40000 + 200 - 12$
$= 40200 - 12$
$= 40188$

(D) The length of the metre stick is $1 \ m$,but it is $1 \ cm$ short. Therefore,the actual length of the stick is $100 \ cm - 1 \ cm = 99 \ cm$.
When the cloth is measured,the reading on the stick shows $100 \ m$. Since each 'metre' on this faulty stick is actually $99 \ cm$,the total length of the cloth is $100 \times 99 \ cm = 9900 \ cm$.

(C) We can express $1001$ as $(1000 + 1)$.
Using the algebraic identity $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,where $a = 1000$ and $b = 1$:
$(1000 + 1)^3 = (1000)^3 + 3(1000)^2(1) + 3(1000)(1)^2 + (1)^3$
$= 1,000,000,000 + 3(1,000,000) + 3(1000) + 1$
$= 1,000,000,000 + 3,000,000 + 3,000 + 1$
$= 1,003,003,001$.

(D) Given the addition:
$5A7 + 335 = 8B2$
Looking at the units place: $7 + 5 = 12$, so the last digit is $2$ and there is a carry of $1$ to the tens place.
Looking at the tens place: $A + 3 + 1 = B$ (where $1$ is the carry from the units place).
So, $A + 4 = B$.
Looking at the hundreds place: $5 + 3 = 8$, which matches the result.
Since $8B2$ is divisible by $3$, the sum of its digits must be a multiple of $3$:
$8 + B + 2 = 10 + B = \text{multiple of } 3$.
Possible values for $B$ are $2, 5, 8$.
Case $1$: If $B = 2$, then $A + 4 = 2 \Rightarrow A = -2$ (Not possible as $A$ is a digit).
Case $2$: If $B = 5$, then $A + 4 = 5 \Rightarrow A = 1$.
Case $3$: If $B = 8$, then $A + 4 = 8 \Rightarrow A = 4$.
The largest possible value of $A$ is $4$.

(C) number is divisible by $25$ if its last two digits are $00, 25, 50,$ or $75$.
Checking the options:
$A) 303310$: Last two digits are $10$ (not divisible by $25$).
$B) 373355$: Last two digits are $55$ (not divisible by $25$).
$C) 303375$: Last two digits are $75$ (divisible by $25$).
$D) 22040$: Last two digits are $40$ (not divisible by $25$).
Therefore,$303375$ is divisible by $25$.

(D) To find the units digit of the product $3 \times 38 \times 537 \times 1256$,we only need to multiply the units digits of each number.
The units digits are $3, 8, 7,$ and $6$.
Multiplying these: $3 \times 8 = 24$ (units digit is $4$)
Next,multiply this result by the next units digit: $4 \times 7 = 28$ (units digit is $8$)
Finally,multiply this result by the last units digit: $8 \times 6 = 48$ (units digit is $8$)
Therefore,the units digit of the product is $8$.

(D) Let the two-digit number be represented as $(10x + y)$,where $x$ is the tens digit and $y$ is the units digit.
When the positions of the digits are interchanged,the new number becomes $(10y + x)$.
Adding the original number and the new number,we get:
$(10x + y) + (10y + x) = 11x + 11y$
$= 11(x + y)$
Since the sum is $11(x + y)$,the resultant number will always be divisible by $11$.

(A) To find the unit's digit of $(57)^{25}$,we look at the unit's digit of powers of $7$.
$7^{1} = 7$
$7^{2} = 49$ (unit's digit is $9$)
$7^{3} = 343$ (unit's digit is $3$)
$7^{4} = 2401$ (unit's digit is $1$)
The cycle of unit's digits is $(7, 9, 3, 1)$ with a period of $4$.
For $(57)^{25}$,we divide the exponent $25$ by the period $4$: $25 = 4 \times 6 + 1$.
Since the remainder is $1$,the unit's digit of $(57)^{25}$ is the same as the unit's digit of $7^{1}$,which is $7$.
Therefore,the unit's digit of $[(57)^{25} - 1]$ is $7 - 1 = 6$.

(C) Let the three-digit number $N$ be represented as $100x + 10z + y$,where $x$ is the digit in the hundred's place,$z$ is the digit in the ten's place,and $y$ is the digit in the unit's place.
Given the expression $N - 100x - y$,we substitute the value of $N$:
$N - 100x - y = (100x + 10z + y) - 100x - y$
Simplifying the expression:
$= 100x - 100x + 10z + y - y$
$= 10z$
Since the result is $10z$,the expression is always divisible by $10$.

(B) Given that $n$ is a whole number such that $n = 4k + 3$ for some integer $k ge 0$.
We need to find the remainder when $2n$ is divided by $4$.
Substitute the expression for $n$ into $2n$:
$2n = 2(4k + 3) = 8k + 6$.
Now,divide $8k + 6$ by $4$:
$8k + 6 = 4(2k) + 4 + 2 = 4(2k + 1) + 2$.
Since $4(2k + 1)$ is perfectly divisible by $4$,the remainder is $2$.
Illustration: Let $n = 7$ (which leaves a remainder of $3$ when divided by $4$).
Then $2n = 14$.
Dividing $14$ by $4$ gives $14 = 4 \times 3 + 2$,so the remainder is $2$.

(A) We know that $m^{2}-n^{2} = (m-n)(m+n)$.
Given that $(m-n)$ is an even number,let $(m-n) = 2k$ for some integer $k$.
Since $(m+n) = (m-n) + 2n$,and $(m-n)$ is even,$(m+n)$ must also be an even number because the sum of two even numbers is even.
Let $(m+n) = 2j$ for some integer $j$.
Therefore,$(m^{2}-n^{2}) = (2k)(2j) = 4kj$.
This shows that $(m^{2}-n^{2})$ is always divisible by $4$.

(C) Let $I$,$C$,and $P$ be the number of ice-creams,cookies,and pastries respectively.
We are given that $I \ge 9$,$C \ge 9$,and $P \ge 9$.
Also,$I < C < P$ and $I + C + P = 32$.
Since $I \ge 9$,the smallest possible value for $I$ is $9$.
If $I = 9$,then $9 < C < P$ and $9 + C + P = 32$,which implies $C + P = 23$.
Since $C < P$,we test values for $C$ starting from $10$:
If $C = 10$,then $P = 23 - 10 = 13$. Here $9 < 10 < 13$ (Satisfied).
If $C = 11$,then $P = 23 - 11 = 12$. Here $9 < 11 < 12$ (Satisfied).
If $C = 12$,then $P = 23 - 12 = 11$. This contradicts $C < P$.
If $I = 10$,then $10 < C < P$ and $10 + C + P = 32$,which implies $C + P = 22$.
Since $C < P$,the smallest $C$ can be is $11$,then $P = 11$,which contradicts $C < P$.
Thus,the only possible values for $C$ are $10$ or $11$.

(A) Let the three consecutive even numbers be $2x, 2x+2$,and $2x+4$.
Then,$(2x)(2x+2)(2x+4) = 4032$ $...(1)$
Given that the product of the first and the third number is $252$,we have:
$(2x)(2x+4) = 252$ $...(2)$
Substituting the value from equation $(2)$ into equation $(1)$:
$(2x+2) \times 252 = 4032$
$2x+2 = \frac{4032}{252} = 16$
$2x = 16 - 2 = 14$
Thus,the second number is $2x+2 = 16$.
Five times the second number is $5 \times 16 = 80$.

(D) To find the remainder when $2^{23}$ is divided by $10$,we need to find the unit digit of $2^{23}$.
Observe the pattern of the unit digits of powers of $2$:
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16$ (unit digit is $6$)
$2^5 = 32$ (unit digit is $2$)
The unit digits repeat in a cycle of $4$: $(2, 4, 8, 6)$.
Divide the exponent $23$ by the cycle length $4$: $23 = 4 \times 5 + 3$.
The remainder is $3$,which means the unit digit of $2^{23}$ is the same as the unit digit of $2^3$.
Since $2^3 = 8$,the unit digit of $2^{23}$ is $8$.
When any number is divided by $10$,the remainder is equal to its unit digit.
Therefore,the remainder when $2^{23}$ is divided by $10$ is $8$.

(D) To find the unit digit of $(4387)^{245} \times (621)^{72}$,we focus on the unit digits of the bases.
$1$. For $(4387)^{245}$,the unit digit is the same as $7^{245}$. The powers of $7$ follow a cycle of $4$: $7^1 = 7$,$7^2 = 9$,$7^3 = 3$,$7^4 = 1$.
$2$. Divide the exponent $245$ by $4$: $245 = 4 \times 61 + 1$. The remainder is $1$. Thus,the unit digit of $(4387)^{245}$ is $7^1 = 7$.
$3$. For $(621)^{72}$,any power of a number ending in $1$ will always have a unit digit of $1$. Thus,the unit digit of $(621)^{72}$ is $1$.
$4$. The unit digit of the product is the product of the unit digits: $7 \times 1 = 7$.

(D) Let $a$ and $b$ be two odd numbers.
We know that the sum of two odd numbers is always an even number.
Since $a$ and $b$ are odd,$a+b$ is even.
Also,the product of two odd numbers is always an odd number,so $ab$ is odd.
Now,consider option $D$: $a+b+2ab$.
Since $a+b$ is even and $2ab$ is even (as any number multiplied by $2$ is even),the sum of two even numbers is always even.
Therefore,$a+b+2ab$ is an even number.

(C) We can write $2^{16}-1$ as a difference of squares:
$2^{16}-1 = (2^8)^2 - 1^2$
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we get:
$(2^8 - 1)(2^8 + 1)$
Since $2^8 = 256$,we have:
$(256 - 1)(256 + 1) = 255 \times 257$
Now,we check for divisibility by the given options:
$255 = 15 \times 17$.
Since $255$ is a factor of $2^{16}-1$,the expression is divisible by $17$.

(C) Let the two numbers be $x$ and $y$.
Given that the sum of the numbers is $x + y = 24$.
Given that the product of the numbers is $xy = 143$.
We need to find the sum of their squares,which is $x^2 + y^2$.
Using the algebraic identity: $(x + y)^2 = x^2 + y^2 + 2xy$.
Rearranging the identity to solve for $x^2 + y^2$: $x^2 + y^2 = (x + y)^2 - 2xy$.
Substituting the given values: $x^2 + y^2 = (24)^2 - 2(143)$.
Calculating the squares and product: $x^2 + y^2 = 576 - 286$.
Therefore,$x^2 + y^2 = 290$.

(D) The unit digit of $124$ is $4$. We need to find the unit digit of $4^{372} + 4^{373}$.
Observe the pattern of powers of $4$:
$4^1 = 4$ (unit digit $4$)
$4^2 = 16$ (unit digit $6$)
$4^3 = 64$ (unit digit $4$)
$4^4 = 256$ (unit digit $6$)
If the exponent is odd,the unit digit is $4$. If the exponent is even,the unit digit is $6$.
For $(124)^{372}$,the exponent $372$ is even,so the unit digit is $6$.
For $(124)^{373}$,the exponent $373$ is odd,so the unit digit is $4$.
The sum of the unit digits is $6 + 4 = 10$.
Therefore,the unit digit of the sum is $0$.

(A) Let the two numbers be $x$ and $y$.
According to the problem,the sum of the numbers is $(x + y)$.
Multiplying the sum by each number separately,we get:
$x(x + y) = 247$ $...(1)$
$y(x + y) = 114$ $...(2)$
Adding equation $(1)$ and equation $(2)$:
$x(x + y) + y(x + y) = 247 + 114$
$(x + y)(x + y) = 361$
$(x + y)^2 = 361$
Taking the square root on both sides:
$x + y = \sqrt{361} = 19$
Thus,the sum of the numbers is $19$.

(A) Let the number be $x$.
According to the problem,one-seventh of the number exceeds its eleventh part by $100$.
$\Rightarrow \frac{1}{7}x - \frac{1}{11}x = 100$
Taking the least common multiple $(LCM)$ of $7$ and $11$,which is $77$:
$\Rightarrow \frac{11x - 7x}{77} = 100$
$\Rightarrow \frac{4x}{77} = 100$
$\Rightarrow 4x = 7700$
$\Rightarrow x = \frac{7700}{4}$
$\Rightarrow x = 1925$
Thus,the required number is $1925$.

(B) Let $x = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$
Since the expression is infinite,we can write:
$x = \sqrt{6+x}$
Squaring both sides:
$x^2 = 6+x$
Rearranging into a quadratic equation:
$x^2 - x - 6 = 0$
Factoring the quadratic equation:
$(x-3)(x+2) = 0$
This gives $x = 3$ or $x = -2$.
Since the square root of a positive number must be positive,we ignore $x = -2$.
Therefore,$x = 3$.

(D) Given that the divisor is $a$ and the dividend is $b$.
According to the problem,the divisor is $4$ times the quotient,so the quotient $= \frac{a}{4}$.
The divisor is also twice the remainder,so the remainder $= \frac{a}{2}$.
Using the division algorithm: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$.
Substituting the values: $b = (a \times \frac{a}{4}) + \frac{a}{2}$.
$b = \frac{a^2}{4} + \frac{a}{2} = \frac{a^2 + 2a}{4}$.
$4b = a^2 + 2a = a(a + 2)$.
Therefore,$\frac{a(a+2)}{b} = 4$.

(C) number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a multiple of $11$.
For the number $738A6A$:
Sum of digits at odd places (from right to left) = $A + 8 + 7 = A + 15$.
Sum of digits at even places (from right to left) = $6 + A + 3 = A + 9$.
Difference = $(A + 15) - (A + 9) = 6$.
Since the difference is $6$,which is not $0$ or a multiple of $11$,let's re-evaluate the alternating sum based on positions:
Sum of digits at odd positions (1st,3rd,5th) = $A + 8 + 7 = A + 15$.
Sum of digits at even positions (2nd,4th,6th) = $6 + A + 3 = A + 9$.
Wait,the standard rule is (Sum of digits at odd places) - (Sum of digits at even places) = $0$ or $11k$.
$(A + 8 + 7) - (6 + A + 3) = 6$. This is constant. Let's check the positions again: $7(1), 3(2), 8(3), A(4), 6(5), A(6)$.
Sum of odd positions = $A + 8 + 7 = A + 15$.
Sum of even positions = $A + 6 + 3 = A + 9$.
Difference = $(A + 15) - (A + 9) = 6$. This implies the number is never divisible by $11$ for any $A$.
Re-reading the sequence: $7, 3, 8, A, 6, A$.
Odd positions: $A, A, 3$. Even positions: $6, 8, 7$.
Sum of odd positions = $2A + 3$.
Sum of even positions = $6 + 8 + 7 = 21$.
Difference = $(2A + 3) - 21 = 2A - 18$.
For divisibility by $11$,$2A - 18 = 0 \Rightarrow 2A = 18 \Rightarrow A = 9$.

(C) Let the two numbers be $a$ and $b$.
Given that the product of the numbers is $ab = 1575$ and their quotient is $\frac{a}{b} = \frac{9}{7}$.
Multiplying the two equations: $(ab) \times (\frac{a}{b}) = 1575 \times \frac{9}{7}$.
This simplifies to $a^2 = 225 \times 9 = 2025$.
Taking the square root,$a = \sqrt{2025} = 45$.
Now,substitute $a = 45$ into the quotient equation: $\frac{45}{b} = \frac{9}{7}$.
Solving for $b$: $b = \frac{45 \times 7}{9} = 5 \times 7 = 35$.
The sum of the numbers is $a + b = 45 + 35 = 80$.

(C) To find the remainder when $(67^{67} + 67)$ is divided by $68$,we can use the property of remainders.
We know that $(a - 1)^n$ divided by $a$ leaves a remainder of $(-1)^n$ when $a$ is the divisor.
Here,$a = 68$ and $n = 67$.
So,$67^{67} = (68 - 1)^{67}$.
When $(68 - 1)^{67}$ is divided by $68$,the remainder is $(-1)^{67} = -1$.
Now,we add the remaining term $67$ to this remainder:
Remainder $= -1 + 67 = 66$.
Since the remainder must be positive and less than the divisor,the final remainder is $66$.

(C) Let the given expression be $E = 3011 \times 3012$.
We can write $3012$ as $(3011 + 1)$.
So,$E = 3011 \times (3011 + 1) = (3011)^2 + 3011$.
To make this expression a perfect square,we need to subtract a value $x$ such that $(3011)^2 + 3011 - x = k^2$ for some integer $k$.
If we subtract $3011$,we get $(3011)^2 + 3011 - 3011 = (3011)^2$,which is a perfect square.
Thus,the least positive integer to be subtracted is $3011$.

(D) The average of the nine consecutive odd numbers is $\frac{621}{9} = 69$.
Since there are nine numbers,the middle number (the $5$th number) is $69$.
For consecutive odd numbers,the difference between consecutive terms is $2$. Thus,the smallest number of Set $A$ is $69 - (4 \times 2) = 69 - 8 = 61$.
The lowest number of the new set of six consecutive even numbers is $61 + 15 = 76$.
The six consecutive even numbers are $76, 78, 80, 82, 84, 86$.
The sum of these numbers is $\frac{n}{2}[2a + (n-1)d]$,where $n=6, a=76, d=2$.
Sum $= \frac{6}{2}[2(76) + (6-1)2] = 3[152 + 10] = 3[162] = 486$.

(D) Let the two consecutive even numbers be $x$ and $x+2$.
According to the problem,the sum of their squares is $6500$.
$x^{2} + (x+2)^{2} = 6500$
$x^{2} + x^{2} + 4x + 4 = 6500$
$2x^{2} + 4x - 6496 = 0$
Dividing the entire equation by $2$,we get:
$x^{2} + 2x - 3248 = 0$
To solve this quadratic equation,we factorize it:
$x^{2} + 58x - 56x - 3248 = 0$
$x(x + 58) - 56(x + 58) = 0$
$(x + 58)(x - 56) = 0$
This gives $x = 56$ or $x = -58$.
Since we are looking for positive even numbers,we take $x = 56$.
The two consecutive even numbers are $56$ and $58$.
The smaller number is $56$.

(D) Let the five consecutive even numbers of set-$A$ be $x, x+2, x+4, x+6,$ and $x+8$.
According to the question,their sum is $220$:
$x + (x+2) + (x+4) + (x+6) + (x+8) = 220$
$5x + 20 = 220$
$5x = 200$
$x = 40$.
So,the lowest number of set-$A$ is $40$.
Now,consider a different set of five consecutive numbers. Let the lowest number be $y$. The second lowest number is $y+1$.
According to the problem,the second lowest number is $37$ less than double the lowest number of set-$A$:
$y+1 = (2 \times 40) - 37$
$y+1 = 80 - 37 = 43$
$y = 42$.
The five consecutive numbers are $42, 43, 44, 45,$ and $46$.
Their sum is $42 + 43 + 44 + 45 + 46 = 220$.

(A) Let the number be $x$. According to the division algorithm,we can write:
$x = 296k + 75$,where $k$ is an integer.
We want to find the remainder when $x$ is divided by $37$.
We can rewrite $296$ as $37 \times 8$ and $75$ as $(37 \times 2) + 1$.
Substituting these into the equation:
$x = (37 \times 8)k + (37 \times 2) + 1$
$x = 37(8k + 2) + 1$
This shows that $x$ is of the form $37m + 1$,where $m = 8k + 2$.
Therefore,when the number is divided by $37$,the remainder is $1$.

(C) Let the two numbers be $a$ and $b$.
According to the problem,the sum of the numbers is $a + b = 12$ and the product of the numbers is $ab = 35$.
We need to find the sum of their reciprocals,which is $\frac{1}{a} + \frac{1}{b}$.
By taking the common denominator,we get $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$.
Substituting the given values,we get $\frac{12}{35}$.

(C) To compare the numbers $\sqrt{5}, \sqrt[3]{4}, \sqrt[5]{2}, \sqrt[7]{3}$,we express them with a common index (root).
The indices are $2, 3, 5, 7$. The Least Common Multiple $(LCM)$ of these indices is $2 \times 3 \times 5 \times 7 = 210$.
Now,convert each number to the $210^{th}$ root:
$1$. $\sqrt{5} = 5^{1/2} = 5^{105/210} = \sqrt[210]{5^{105}}$
$2$. $\sqrt[3]{4} = 4^{1/3} = 4^{70/210} = \sqrt[210]{4^{70}}$
$3$. $\sqrt[5]{2} = 2^{1/5} = 2^{42/210} = \sqrt[210]{2^{42}}$
$4$. $\sqrt[7]{3} = 3^{1/7} = 3^{30/210} = \sqrt[210]{3^{30}}$
Comparing the radicands: $5^{105}$ is clearly the largest value among $5^{105}, 4^{70}, 2^{42}, 3^{30}$.
Therefore,$\sqrt{5}$ is the greatest number.

(C) Let the two numbers be $x$ and $y$.
Given the ratio of their difference,sum,and product is $1: 7: 24$.
Let the common multiplier be $a$. Then:
$x - y = 1a$
$x + y = 7a$
$xy = 24a$
Adding the first two equations: $(x - y) + (x + y) = 1a + 7a \Rightarrow 2x = 8a \Rightarrow x = 4a$.
Subtracting the first from the second: $(x + y) - (x - y) = 7a - 1a \Rightarrow 2y = 6a \Rightarrow y = 3a$.
Now,substitute $x$ and $y$ into the product equation:
$xy = (4a)(3a) = 12a^2$.
Since $xy = 24a$,we have $12a^2 = 24a$.
Dividing by $12a$ (assuming $a \neq 0$),we get $a = 2$.
The product of the numbers is $xy = 24a = 24 \times 2 = 48$.

(D) Let the five consecutive even numbers of set $A$ be $x, x+2, x+4, x+6, x+8$.
According to the question,the sum is $220$:
$x + (x+2) + (x+4) + (x+6) + (x+8) = 220$
$5x + 20 = 220$
$5x = 200$
$x = 40$.
The lowest number of set $A$ is $40$.
Let the second lowest number of the new set $B$ be $y$. According to the question,$y = (2 \times 40) - 37 = 80 - 37 = 43$.
Since the set $B$ consists of five consecutive numbers,if the second lowest number is $43$,the set is $42, 43, 44, 45, 46$.
The sum of these numbers is $42 + 43 + 44 + 45 + 46 = 220$.

(A) Let the first number be $f$ and the second number be $s$.
According to the problem,we have the following system of linear equations:
$2f + 3s = 100$ $...(1)$
$3f + 2s = 120$ $...(2)$
To solve this system,multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$4f + 6s = 200$ $...(3)$
$9f + 6s = 360$ $...(4)$
Subtract equation $(3)$ from equation $(4)$:
$(9f - 4f) + (6s - 6s) = 360 - 200$
$5f = 160$
$f = 32$
Now,substitute $f = 32$ into equation $(1)$:
$2(32) + 3s = 100$
$64 + 3s = 100$
$3s = 36$
$s = 12$
The two numbers are $32$ and $12$. Comparing these,the larger number is $32$.