Official Correct Answer: C. ### Core Concept
Use the cyclic pattern of remainders of powers modulo 7.
### Step-by-Step Solution
1. Since $10 \equiv 3 \pmod 7$, we have $10^{100} \equiv 3^{100} \pmod 7$.
2. By Fermat's little theorem, $3^6 \equiv 1 \pmod 7$.
3. Write $100 = 6 \times 16 + 4$. Then $3^{100} = (3^6)^{16} \cdot 3^4 \equiv 3^4 \pmod 7$.
4. $3^4 = 81 = 77 + 4$, so the remainder is 4.
### Why Not the Others
- 6 is the remainder of $3^3$, and 1 is the remainder of $3^6$, so both come from wrong exponent reductions.
- 3 is the remainder of $3^1$, which would apply if the exponent were taken as 1 more than a multiple of 6.
### Quick Check
The remainders of $3^n$ cycle as 3, 2, 6, 4, 5, 1. The exponent 100 leaves remainder 4 when divided by 6, so the 4th term is 4.
**Correct Answer: C**
Two places A and B are 45 kms apart and connected by a straight road. Anil goes from A to B while Sunil goes from B to A. Starting at the same time, they cross each other in exactly 1 hour 30 minutes. If Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A, the speed of Anil, in km per hour, is
Official Correct Answer: B. ### Core Concept
The meeting time gives the sum of the speeds. The difference in travel times then gives each speed.
### Step-by-Step Solution
1. They cover 45 km together in 1.5 hours, so the sum of their speeds is $\frac{45}{1.5} = 30$ km/h.
2. Let Anil's speed be $x$, so Sunil's speed is $30 - x$.
3. Anil takes $\frac{45}{x}$ hours to reach B, and Sunil takes $\frac{45}{30-x}$ hours to reach A. Anil is 1.25 hours later, so $\frac{45}{x} - \frac{45}{30-x} = \frac{5}{4}$.
4. This simplifies to $36(30 - 2x) = x(30 - x)$, i.e. $x^2 - 102x + 1080 = 0$.
5. The roots are $x = 12$ or $x = 90$. Since $x < 30$, $x = 12$.
### Quick Check
Anil at 12 km/h takes 3.75 h. Sunil at 18 km/h takes 2.5 h. The difference is 1.25 h = 1 h 15 min, as given.
**Correct Answer: B**
There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is
Official Correct Answer: 15. ### Core Concept
Convert each average condition into a relation between the running sums, then express every number in terms of the first number $a$.
### Step-by-Step Solution
1. Let the numbers be $a, b, c, d$. The first condition gives $\frac{a+b}{2} = a+1$, so $b = a+2$.
2. The average of the first two is $a+1$. The second condition makes the average of the first three $a+3$, so $a+b+c = 3a+9$.
3. Since $a+b = 2a+2$, we get $c = (3a+9)-(2a+2) = a+7$.
4. The third condition makes the average of the first four $a+3+3 = a+6$, so $a+b+c+d = 4a+24$.
5. Hence $d = (4a+24)-(3a+9) = a+15$.
6. The numbers are $a,\ a+2,\ a+7,\ a+15$. The largest is $a+15$ and the smallest is $a$, so the difference is $15$, whatever $a$ is.
### Quick Check
Take $a=1$: the numbers are 1, 3, 8, 16. The averages are 2, 4, 7, which increase by 1, 2, 3 in turn, as required. Also $16-1=15$.
**Correct Answer: 15**
The sum of all four-digit numbers that can be formed with the distinct non-zero digits a,b,c, and d , with each digit appearing exactly once in every number, is 153310+n , where n is a single digit natural number. Then, the value of (a+b+c+d+n) is
Official Correct Answer: 31. ### Core Concept
When four distinct digits are arranged in all $4! = 24$ ways, each digit appears in each place value exactly $3! = 6$ times. This gives a closed form for the total.
### Step-by-Step Solution
1. Each digit appears 6 times in each of the units, tens, hundreds and thousands places.
2. The sum of all 24 numbers is $6(a+b+c+d)(1000+100+10+1) = 6666\,(a+b+c+d)$.
3. Let $S = a+b+c+d$. Then $6666S = 153310 + n$, so $6666S$ is a multiple of 6666 lying between $153310$ and $153319$, because $n$ is a single digit from 1 to 9.
4. Since $6666 \times 23 = 153318$, this is the multiple that fits. The neighbouring multiples, $6666 \times 22 = 146652$ and $6666 \times 24 = 159984$, are outside the range.
5. So $S = 23$ and $n = 153318 - 153310 = 8$.
6. Therefore $a+b+c+d+n = 23 + 8 = 31$.
### Quick Check
$6666 \times 23 = 153318 = 153310 + 8$, and $n = 8$ is a single-digit natural number.
**Correct Answer: 31**
In September, the incomes of Kamal, Amal and Vimal are in the ratio 8:6:5. They rent a house together, and Kamal pays 15% , Amal pays 12% and Vimal pays 18% of their respective incomes to cover the total house rent in that month. In October, the house rent remains unchanged while their incomes increase by 10%,12% and 15% , respectively. In October, the percentage of their total income that will be paid as house rent, is nearest to
Official Correct Answer: A. ### Core Concept
Find the total rent from the September percentages. The rent stays fixed, so compare it with the total October income.
### Step-by-Step Solution
1. Take the September incomes as 800, 600 and 500, which is the ratio $8:6:5$.
2. Rent paid: Kamal $0.15\times800=120$, Amal $0.12\times600=72$, Vimal $0.18\times500=90$.
3. Total rent $=120+72+90=282$.
4. October incomes: Kamal $800\times1.10=880$, Amal $600\times1.12=672$, Vimal $500\times1.15=575$.
5. Total October income $=880+672+575=2127$.
6. Rent as a percentage of income $=\dfrac{282}{2127}\times100\approx13.26\%$.
### Why Not the Others
- 14.84 is too high, because the rent share falls when incomes rise.
- 12.75 is too low, because $282/2127$ is clearly above 13%.
- 15.18 is higher than the September share (about 14.8%), so it cannot be right after incomes rise.
### Quick Check
$2127\times0.1326\approx282$.
**Correct Answer: A**
An amount of Rs 10000 is deposited in bank A for a certain number of years at a simple interest of 5% per annum. On maturity, the total amount received is deposited in bank B for another 5 years at a simple interest of 6% per annum. If the interests received from bank A and bank B are in the ratio 10 : 13, then the investment period, in years, in bank A is
Official Correct Answer: B. ### Core Concept
Set up both simple interests in terms of the unknown period $t$ in bank A, then use the given ratio of interests.
### Step-by-Step Solution
1. Interest from bank A for $t$ years at 5%: $10000 \times 0.05 \times t = 500t$.
2. The amount deposited in bank B is $10000 + 500t$.
3. Interest from bank B for 5 years at 6%: $(10000+500t) \times 0.06 \times 5 = 0.3(10000+500t) = 3000 + 150t$.
4. The ratio condition gives $\frac{500t}{3000+150t} = \frac{10}{13}$.
5. Cross-multiplying: $6500t = 30000 + 1500t$, so $5000t = 30000$ and $t = 6$.
### Why Not the Others
- A) 4: A's interest would be 2000 and B's 3600, a ratio of 5 : 9, not 10 : 13.
- C) 3: A's interest would be 1500 and B's 3450, a ratio of about 10 : 23.
- D) 5: A's interest would be 2500 and B's 3750, a ratio of 2 : 3, not 10 : 13.
### Quick Check
For $t=6$, A's interest is 3000 and B's is $3000+900 = 3900$. The ratio $3000 : 3900 = 10 : 13$.
**Correct Answer: B**
Official Correct Answer: 38. ### Core Concept
A common root $\alpha$ satisfies all three equations. Subtracting pairs of equations eliminates $\alpha^2$ and leaves linear relations in $\alpha$.
### Step-by-Step Solution
1. Let the common negative root be $\alpha$. Then:
$$\alpha^2 + m\alpha + 9 = 0,\quad \alpha^2 + n\alpha + 17 = 0,\quad \alpha^2 + (m+n)\alpha + 35 = 0$$
2. Subtract the second equation from the third: $m\alpha + 18 = 0$, so $m\alpha = -18$.
3. Subtract the first equation from the third: $n\alpha + 26 = 0$, so $n\alpha = -26$.
4. Substitute $m\alpha = -18$ into the first equation: $\alpha^2 - 18 + 9 = 0$, so $\alpha^2 = 9$. Since the root is negative, $\alpha = -3$.
5. Then $m = \frac{-18}{-3} = 6$ and $n = \frac{-26}{-3} = \frac{26}{3}$.
6. So $2m + 3n = 12 + 26 = 38$.
### Quick Check
The second equation gives $9 + n(-3) + 17 = 0$, so $n = \frac{26}{3}$, which matches step 5.
**Correct Answer: 38**
Official Correct Answer: 31. ### Core Concept
Convert every logarithm to base 10 with the change-of-base rule, then solve for $\log_{10} x$.
### Step-by-Step Solution
1. Let $L = \log_{10} x$.
2. $\log_{100} x = \dfrac{\log_{10} x}{\log_{10} 100} = \dfrac{L}{2}$ and $\log_{1000} x = \dfrac{L}{3}$.
3. Substitute: $4L + 4\cdot\dfrac{L}{2} + 8\cdot\dfrac{L}{3} = 13$.
4. This gives $4L + 2L + \dfrac{8L}{3} = \dfrac{26L}{3} = 13$, so $L = \dfrac{3}{2}$.
5. Then $x = 10^{3/2} = 10\sqrt{10} \approx 31.62$.
6. The greatest integer not exceeding $x$ is 31.
### Quick Check
$\sqrt{10} \approx 3.162$, so $x \approx 31.62$. This lies between 31 and 32.
**Correct Answer: 31**
Official Correct Answer: B. ### Core Concept
Combine the two equations so that a sum of squares equals zero. Each square must then be zero.
### Step-by-Step Solution
1. The equations are $4(x^2+y^2+z^2) = a$ and $4(x-y-z) = 3 + a$.
2. Subtract the second from the first: $4(x^2+y^2+z^2) - 4(x-y-z) = -3$.
3. Divide by 4 and group the terms: $(x^2 - x) + (y^2 + y) + (z^2 + z) = -\dfrac{3}{4}$.
4. Complete each square: $\left(x-\tfrac12\right)^2 + \left(y+\tfrac12\right)^2 + \left(z+\tfrac12\right)^2 - \dfrac34 = -\dfrac34$.
5. So $\left(x-\tfrac12\right)^2 + \left(y+\tfrac12\right)^2 + \left(z+\tfrac12\right)^2 = 0$. Real squares sum to zero only if each is zero, so $x = \tfrac12$, $y = -\tfrac12$, $z = -\tfrac12$.
6. Then $a = 4\left(\tfrac14+\tfrac14+\tfrac14\right) = 3$.
### Quick Check
$4(x-y-z) = 4\left(\tfrac12+\tfrac12+\tfrac12\right) = 6 = 3 + a$ with $a = 3$.
**Correct Answer: B**
Official Correct Answer: B. ### Core Concept
Complete the square to see the region as a ring (annulus) around a centre, then use the line to cut it.
### Step-by-Step Solution
1. Rewrite the middle expression: $x^2+y^2+4x-4y = (x+2)^2 + (y-2)^2 - 8$.
2. The condition $-4 \le (x+2)^2 + (y-2)^2 - 8 \le 0$ becomes $4 \le (x+2)^2 + (y-2)^2 \le 8$.
3. This is a ring centred at $(-2, 2)$ with inner radius 2 and outer radius $2\sqrt2$.
4. The ring's area is $\pi(8 - 4) = 4\pi$.
5. The boundary line $y = x + 4$ passes through the centre, since $2 = -2 + 4$.
6. A line through the centre splits the ring symmetrically, so $y \ge x+4$ keeps exactly half of it.
7. The required area is $\dfrac{4\pi}{2} = 2\pi$.
### Why Not the Others
- $4\pi$ is the full ring, ignoring the line.
- $\pi$ and $3\pi$ do not match half of $4\pi$.
**Correct Answer: B**
A glass is filled with milk. Two-thirds of its content is poured out and replaced with water. If this process of pouring out two-thirds the content and replacing with water is repeated three more times, then the final ratio of milk to water in the glass, is
Official Correct Answer: B. ### Core Concept
The problem involves repeated dilution where a fraction of the mixture is replaced with water. The formula to determine the remaining concentration of the original substance after each replacement is:
$$ \text{Remaining concentration} = \left(1 - \frac{\text{Fraction replaced}}{}\right)^{\text{Number of replacements}} $$
### Step-by-Step Solution
1. **Initial Setup**: The glass is initially filled with milk. Let the initial volume be \( V \).
2. **First Replacement**: Two-thirds of the milk is poured out, leaving one-third. The remaining milk is \( \frac{1}{3}V \).
3. **Subsequent Replacements**: This process is repeated three more times. Each time, the remaining milk is multiplied by \( \frac{1}{3} \).
4. **Final Concentration**: After four replacements, the remaining milk is:
\[
\left(\frac{1}{3}\right)^4 V = \frac{1}{81}V
\]
5. **Ratio Calculation**: The ratio of milk to water is:
\[
\frac{\frac{1}{81}V}{\frac{80}{81}V} = 1 : 80
\]
### 30-Second Shortcut
Each replacement retains \( \frac{1}{3} \) of the milk. After four replacements, the milk fraction is \( \left(\frac{1}{3}\right)^4 = \frac{1}{81} \). Thus, the ratio is \( 1 : 80 \).
### Final Answer Confirmation
Correct Answer: B
Consider two sets A={2,3,5,7,11,13} and B={1,8,27}. Let f be a function from A to B such that for every element b in B , there is at least one element a in A such that f(a)=b. Then, the total number of such functions f is
Official Correct Answer: B. ### Core Concept
The problem requires counting the number of onto (surjective) functions from set \( A \) to set \( B \). The formula for the number of onto functions from a set with \( m \) elements to a set with \( n \) elements is given by the inclusion-exclusion principle:
$$
\sum_{k=0}^{n} (-1)^k \binom{n}{k} (n - k)^m
$$
### Step-by-Step Solution
1. **Identify the sizes of sets**: \( |A| = 6 \) and \( |B| = 3 \).
2. **Apply the inclusion-exclusion formula**:
- For \( k = 0 \): \( (-1)^0 \binom{3}{0} (3 - 0)^6 = 1 \times 1 \times 729 = 729 \)
- For \( k = 1 \): \( (-1)^1 \binom{3}{1} (3 - 1)^6 = -1 \times 3 \times 64 = -192 \)
- For \( k = 2 \): \( (-1)^2 \binom{3}{2} (3 - 2)^6 = 1 \times 3 \times 1 = 3 \)
- For \( k = 3 \): \( (-1)^3 \binom{3}{3} (3 - 3)^6 = -1 \times 1 \times 0 = 0 \)
3. **Sum the results**: \( 729 - 192 + 3 + 0 = 540 \).
### 30-Second Shortcut
Alternatively, using Stirling numbers of the second kind \( S(6, 3) = 90 \), the number of onto functions is \( 3! \times 90 = 6 \times 90 = 540 \).
### Final Answer
Correct Answer: Option B
The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is
Official Correct Answer: A. ### Core Concept
A box inscribed in a sphere has its space diagonal equal to the sphere's diameter. Use $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$.
### Step-by-Step Solution
1. Let the edges be $a, b, c$. A box has 4 edges of each length, so $4(a+b+c) = 144$ and $a+b+c = 36$.
2. The surface area gives $2(ab+bc+ca) = 846$.
3. Then $a^2+b^2+c^2 = 36^2 - 846 = 1296 - 846 = 450$.
4. The space diagonal is $d = \sqrt{450} = 15\sqrt2$, and this equals the sphere's diameter.
5. The radius is $r = \dfrac{15\sqrt2}{2}$, so $r^3 = \dfrac{3375 \cdot 2\sqrt2}{8} = \dfrac{3375\sqrt2}{4}$.
6. The volume is $\dfrac43 \pi r^3 = \dfrac43 \pi \cdot \dfrac{3375\sqrt2}{4} = 1125\pi\sqrt2$.
### Quick Check
The diagonal squared is $450$, so $r^2 = \dfrac{450}{4} = 112.5$. Then $r \approx 10.6$ and $\dfrac43\pi r^3 \approx 5000$, which matches $1125\pi\sqrt2 \approx 4998$.
**Correct Answer: A**
The selling price of a product is fixed to ensure 40% profit. If the product had cost 40% less and had been sold for 5 rupees less, then the resulting profit would have been 50%. The original selling price, in rupees, of the product is
Official Correct Answer: D. ### Core Concept
Express everything in terms of the original cost price $C$, then use the second profit condition to form one equation.
### Step-by-Step Solution
1. Original selling price: $SP = 1.4C$, since the profit is 40%.
2. The new cost is 40% less, so it is $0.6C$.
3. The new selling price is $1.4C - 5$, and it gives a 50% profit on the new cost: $1.4C - 5 = 1.5 \times 0.6C = 0.9C$.
4. So $0.5C = 5$, which gives $C = 10$.
5. The original selling price is $1.4 \times 10 = 14$.
### Quick Check
The new cost is 6 and the new selling price is $14 - 5 = 9$. The profit is $3/6 = 50\%$, as required.
**Correct Answer: D**
A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional 3 kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is
Official Correct Answer: A. ### Core Concept
Work backwards from the empty stock. Each customer takes half of what is present plus 3 kg, so the stock before each sale can be recovered step by step.
### Step-by-Step Solution
1. Let the stock before a sale be $x$. After the sale, $x - \left(\frac{x}{2} + 3\right) = \frac{x}{2} - 3$ remains.
2. Third customer: nothing is left, so $\frac{x}{2} - 3 = 0$ and $x = 6$ kg before the third sale.
3. Second customer: $\frac{x}{2} - 3 = 6$, so $x = 18$ kg before the second sale.
4. First customer: $\frac{x}{2} - 3 = 18$, so $x = 42$ kg initially.
### Quick Check
Start with 42. The first customer takes $21 + 3 = 24$, leaving 18. The second takes $9 + 3 = 12$, leaving 6. The third takes $3 + 3 = 6$, leaving 0.
**Correct Answer: A**
If (a+bn) is the positive square root of (29−125) , where a and b are integers, and n is a natural number, then the maximum possible value of (a+b+n) is
Official Correct Answer: B. ### Core Concept
Square $a + b\sqrt{n}$ and match the rational and irrational parts with $29 - 12\sqrt{5}$. Then find the choice of $a$, $b$, $n$ that gives the largest sum.
### Step-by-Step Solution
1. Expanding gives $(a + b\sqrt{n})^2 = a^2 + b^2 n + 2ab\sqrt{n}$.
2. The surd part must be $-12\sqrt{5}$, so $\sqrt{n}$ must be a multiple of $\sqrt{5}$. Write $\sqrt{n} = m\sqrt{5}$, so $n = 5m^2$.
3. Matching parts gives $a^2 + 5(bm)^2 = 29$ and $2a(bm) = -12$. With $c = bm$, this means $a^2 + 5c^2 = 29$ and $ac = -6$.
4. The solutions are $(a, c) = (3, -2)$ or $(-3, 2)$. The root must be positive, and $3 - 2\sqrt{5} < 0$, so $a = -3$ and $c = 2$. The root is $-3 + 2\sqrt{5} \approx 1.47$.
5. Since $bm = 2$ with integer $b$, the options are:
- $m = 1$, $b = 2$, $n = 5$, giving $a + b + n = -3 + 2 + 5 = 4$.
- $m = 2$, $b = 1$, $n = 20$, giving $a + b + n = -3 + 1 + 20 = 18$.
6. Here $-3 + \sqrt{20}$ is the same positive root written differently. Larger $m$ would make $b$ non-integer, so the maximum is 18.
**Correct Answer: B**
Official Correct Answer: 10. Consider. ABCD means AB and CD are the lengths, each 56 \text{ cm}, and BC and AD are the breadths, each 45 \text{ cm}. So, if I imagine the rectangle, A is the bottom-left corner, B is the bottom-right, C is the top-right, and D is the top-left. E is the midpoint of CD, so that would be halfway along the top side.
Now, triangle ADE is formed by points A, D, and E. Place point A at (0, 0). Then, since AB is 56 \text{ cm}, point B would be at (56, 0). BC is 45 \text{ cm}, so point C is at (56, 45), and point D is at (0, 45). E is the midpoint of CD, so CD goes from (56, 45) to (0, 45). The midpoint would be at ((56+0)/2, (45+45)/2) = (28, 45).
So, the coordinates are:
- A: (0, 0)
- D: (0, 45)
- E: (28, 45)
Now, We find the lengths of the sides of triangle ADE. Computing each side one by one.
First, side AD is from A(0,0) to D(0,45). That's straightforward; it's just the vertical distance, which is 45 \text{ cm}.
Next, side AE is from A(0,0) to E(28,45). To find this length, Using the distance formula: \sqrt{(x2 - x1)^2 + (y2 - y1)^2}. Plugging in the values, it's \sqrt{(28 - 0)^2 + (45 - 0)^2} = \sqrt{784 + 2025} = \sqrt{2809}. Hmm, 2809 is 53 squared because 53*53 is 2809. So, AE is 53 \text{ cm}.
Then, side DE is from D(0,45) to E(28,45). Since both points are on the same horizontal line, the distance is just the difference in the x-coordinates. So, DE = 28 - 0 = 28 \text{ cm}.
Triangle ADE has sides AD = 45 \text{ cm}, AE = 53 \text{ cm}, and DE = 28 \text{ cm}.
AD is vertical, so yes, 45 \text{ cm}. AE is from (0,0) to (28,45), which we calculated as 53 \text{ cm}. DE is horizontal from (0,45) to (28,45), which is 28 \text{ cm}. Now, We find the radius of the incircle of triangle ADE. the radius of the incircle (r) of a triangle is given by the formula:
r = A / s
where A is the area of the triangle, and s is the semi-perimeter.
So, first, let's find the semi-perimeter (s). The semi-perimeter is half the sum of all sides.
s = (AD + AE + DE) / 2 = (45 + 53 + 28) / 2
Compute that: 45 + 53 is 98, plus 28 is 126. So, s = 126 / 2 = 63 \text{ cm}.
Next, The area is given by (A) of triangle ADE. Since I have the coordinates of all three points, Using the shoelace formula to find the area.
The shoelace formula is given by:
A = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2|
Plugging in the coordinates:
x1 = 0, y1 = 0 (point A)
x2 = 0, y2 = 45 (point D)
x3 = 28, y3 = 45 (point E)
So,
A = |(0*(45 - 45) + 0*(45 - 0) + 28*(0 - 45)) / 2|
= |(0 + 0 + 28*(-45)) / 2|
= |( -1260 ) / 2|
= | -630 | = 630 \text{ cm}²
Alternatively, since triangle ADE is a right triangle? Is angle at D a right angle? AD is vertical, DE is horizontal, so yes, angle at D is 90 degrees. So, triangle ADE is a right-angled triangle at D.
Therefore, the area can also be calculated as (base * height) / 2. Here, base is DE = 28 \text{ cm}, height is AD = 45 \text{ cm}.
So, area A = (28 * 45) / 2 = (1260) / 2 = 630 \text{ cm}². Now, using the formula for the inradius:
r = A / s = 630 / 63 = 10 \text{ cm}.
So, the radius of the incircle is 10 \text{ cm}.
Official Correct Answer: 66. ### Core Concept
Set up the ratio with a common multiplier and use the unsold ratio to form a linear equation.
### Step-by-Step Solution
1. Apples : mangoes $= 5 : 2$, so let apples $= 5k$, mangoes $= 2k$ and oranges $= o$. Then $7k + o = 187$.
2. After selling, unsold apples are $5k - 75$ and unsold oranges are $\frac{o}{2}$.
3. The ratio is $3 : 2$, so $\frac{5k - 75}{o/2} = \frac{3}{2}$. This gives $2(5k - 75) = \frac{3o}{2}$, so $20k - 300 = 3o$.
4. Substitute $o = 187 - 7k$: $20k - 300 = 561 - 21k$. So $41k = 861$ and $k = 21$.
5. Apples $= 105$, mangoes $= 42$, oranges $= 187 - 147 = 40$.
6. Unsold fruits are $105 - 75 = 30$ apples, $42 - 26 = 16$ mangoes and $40/2 = 20$ oranges.
7. Total unsold $= 30 + 16 + 20 = 66$.
### Quick Check
Unsold apples to unsold oranges is $30 : 20 = 3 : 2$, as given.
**Correct Answer: 66**
Official Correct Answer: B. ### Core Concept
Write every term as a power of the same base, $\frac{1}{8}$, then equate exponents.
### Step-by-Step Solution
1. Since $32768 = 2^{15} = 8^5$, we get $\left(\frac{1}{32768}\right)^{1/3} = \left(\frac{1}{8}\right)^{5/3}$.
2. In the same way, $\left(\frac{1}{32768}\right)^{1/k} = \left(\frac{1}{8}\right)^{5/k}$.
3. The equation becomes $\left(\frac{1}{8}\right)^{k+\frac{5}{3}} = \left(\frac{1}{8}\right)^{1+\frac{5}{k}}$.
4. Equate the exponents: $k + \frac{5}{3} = 1 + \frac{5}{k}$.
5. Multiply by $3k$ (with $k \neq 0$): $3k^2 + 5k = 3k + 15$, so $3k^2 + 2k - 15 = 0$.
6. The discriminant is $4 + 180 = 184 > 0$, so both roots are real. The sum of the roots is $-\frac{2}{3}$.
### Quick Check
Sum of roots $= -\frac{b}{a} = -\frac{2}{3}$.
**Correct Answer: B**
Official Correct Answer: B. ### Core Concept
Write each term of the AP as $a + (n-1)d$ and use the two conditions to find $a$ and $d$.
### Step-by-Step Solution
1. The condition $x_5 = -4$ gives $a + 4d = -4$.
2. The second condition $2x_6 + 2x_9 = x_{11} + x_{13}$ becomes $2(a+5d) + 2(a+8d) = (a+10d) + (a+12d)$.
3. The left side is $4a + 26d$ and the right side is $2a + 22d$. So $2a + 4d = 0$, which gives $a = -2d$.
4. Substitute into $a + 4d = -4$: $2d = -4$, so $d = -2$ and $a = 4$.
5. Then $x_{100} = a + 99d = 4 + 99(-2) = 4 - 198 = -194$.
### Quick Check
$x_5 = 4 + 4(-2) = -4$, which matches the given value.
**Correct Answer: B**
Renu would take 15 days working 4 hours per day to complete a certain task whereas Seema would take 8 days working 5 hours per day to complete the same task. They decide to work together to complete this task. Seema agrees to work for double the number of hours per day as Renu, while Renu agrees to work for double the number of days as Seema. If Renu works 2 hours per day, then the number of days Seema will work, is
Official Correct Answer: 6. ### Core Concept
Convert each person's rate into work per hour, then use the stated working pattern to count the total work done.
### Step-by-Step Solution
1. Renu takes $15 \times 4 = 60$ hours for the task, so her rate is $\frac{1}{60}$ of the task per hour.
2. Seema takes $8 \times 5 = 40$ hours, so her rate is $\frac{1}{40}$ of the task per hour.
3. Renu works 2 hours per day. Seema works double that, so 4 hours per day.
4. Let Seema work $d$ days. Renu works double the days, so $2d$ days.
5. Renu's total hours are $2 \times 2d = 4d$, so she completes $\frac{4d}{60} = \frac{d}{15}$ of the task.
6. Seema's total hours are $4d$, so she completes $\frac{4d}{40} = \frac{d}{10}$ of the task.
7. Together they finish the task, so $\frac{d}{15} + \frac{d}{10} = \frac{d}{6} = 1$, which gives $d = 6$.
### Quick Check
Renu: 12 days $\times$ 2 hours $= 24$ hours, which is $\frac{24}{60} = 0.4$ of the task. Seema: 6 days $\times$ 4 hours $= 24$ hours, which is $\frac{24}{40} = 0.6$ of the task. The total is 1.
**Correct Answer: 6**
Official Correct Answer: 217. ### Core Concept
$a_n = \lfloor \sqrt{n} \rfloor$ stays constant between consecutive perfect squares, so group the values of $n$ by block.
### Step-by-Step Solution
1. $a_n = k$ exactly when $k^2 \le n \le (k+1)^2 - 1$, so each value $k$ appears $2k+1$ times.
2. Add up each block:
- $n = 1$ to $3$: value 1, 3 times, sum $3$.
- $n = 4$ to $8$: value 2, 5 times, sum $10$.
- $n = 9$ to $15$: value 3, 7 times, sum $21$.
- $n = 16$ to $24$: value 4, 9 times, sum $36$.
- $n = 25$ to $35$: value 5, 11 times, sum $55$.
- $n = 36$ to $48$: value 6, 13 times, sum $78$.
3. The last block is cut off at $n = 50$. For $n = 49$ and $50$ the value is 7, so it appears 2 times, sum $14$.
4. Total: $3 + 10 + 21 + 36 + 55 + 78 + 14 = 217$.
### Quick Check
The counts are $3 + 5 + 7 + 9 + 11 + 13 + 2 = 50$, which covers every term.
**Correct Answer: 217**
A company has 40 employees whose names are listed in a certain order. In the year 2022, the average bonus of the first 30 employees was Rs. 40000, of the last 30 employees was Rs. 60000, and of the first 10 and last 10 employees together was Rs.
50000
Next year, the average bonus of the first 10 employees increased by 100%, of the last 10 employees increased by 200% and of the remaining employees was unchanged. Then, the average bonus, in rupees, of all the 40 employees together in the year 2023 was
Official Correct Answer: C. ### Core Concept
Split the 40 employees into four blocks of 10 (positions 1-10, 11-20, 21-30, 31-40) and work with block totals instead of averages.
### Step-by-Step Solution
1. Let the 2022 bonus totals of the blocks be $F, M_1, M_2, L$ (in Rs.).
2. First 30 employees: $F+M_1+M_2 = 30 \times 40000 = 12{,}00{,}000$.
3. Last 30 employees: $M_1+M_2+L = 30 \times 60000 = 18{,}00{,}000$.
4. First 10 and last 10 together: $F+L = 20 \times 50000 = 10{,}00{,}000$.
5. Add steps 2 and 3: $F+L+2(M_1+M_2) = 30{,}00{,}000$, so $M_1+M_2 = 10{,}00{,}000$.
6. Then $F = 12{,}00{,}000 - 10{,}00{,}000 = 2{,}00{,}000$ and $L = 8{,}00{,}000$.
7. In 2023, a 100% increase doubles the first block: $F' = 4{,}00{,}000$. A 200% increase triples the last block: $L' = 24{,}00{,}000$. The middle 20 stay at $10{,}00{,}000$.
8. Total for 2023 $= 4{,}00{,}000 + 10{,}00{,}000 + 24{,}00{,}000 = 38{,}00{,}000$.
9. Average $= \dfrac{38{,}00{,}000}{40} = 95000$.
### Quick Check
The 2022 total is $2+10+8 = 20$ lakh, an average of 50000. The first-30 total is $2+10 = 12$ lakh and the last-30 total is $10+8 = 18$ lakh, matching the given averages.
**Correct Answer: C**
Official Correct Answer: D. ### Core Concept
Fermat's Little Theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:
$$ a^{p-1} \equiv 1 \mod p $$
### Step-by-Step Solution
1. **Apply Fermat's Little Theorem**:
Since 11 is prime and 3 is not divisible by 11:
$$ 3^{10} \equiv 1 \mod 11 $$
2. **Express 333 in terms of 10**:
Divide 333 by 10:
$$ 333 = 10 \times 33 + 3 $$
So,
$$ 3^{333} = 3^{10 \times 33 + 3} = (3^{10})^{33} \times 3^3 $$
3. **Simplify using Fermat's Theorem**:
$$ (3^{10})^{33} \equiv 1^{33} \equiv 1 \mod 11 $$
Thus,
$$ 3^{333} \equiv 1 \times 3^3 \mod 11 $$
4. **Calculate \( 3^3 \)**:
$$ 3^3 = 27 $$
5. **Find the remainder**:
$$ 27 \div 11 = 2 \text{ with a remainder of } 5 $$
So,
$$ 27 \equiv 5 \mod 11 $$
### 30-Second Shortcut
- Use Fermat's Little Theorem to reduce the exponent modulo 10.
- Compute \( 333 \mod 10 = 3 \).
- Calculate \( 3^3 = 27 \).
- Find \( 27 \mod 11 = 5 \).
### Final Answer
Correct Answer: Option D
Official Correct Answer: B. ### Core Concept
For $m^n$ to equal a product of prime powers, every prime exponent in the product must be divisible by $n$. The largest such $n$ comes from the highest common factor of the exponents. Here $n>1$ and the exponents are 25 and 40, so the common factors decide $n$.
### Step-by-Step Solution
1. Write $m^n = 2^{25} \times 3^{40}$. Since $m$ is a natural number, each prime in $m$ appears with an integer exponent, so $n$ must divide both 25 and 40.
2. The common factors of 25 and 40 are 1 and 5. Since $n>1$, $n=5$.
3. Take the fifth root: $m = 2^{25/5} \times 3^{40/5} = 2^5 \times 3^8$.
4. Compute $2^5 = 32$ and $3^8 = 6561$, so $m = 32 \times 6561 = 209952$.
5. Then $m - n = 209952 - 5 = 209947$.
### Why Not the Others
- A) 209937 would need $n=15$, which does not divide 25.
- C) 209942 would need $n=10$, which does not divide 25.
- D) 209932 would need $n=20$, which does not divide 25.
### Quick Check
$(2^5 \times 3^8)^5 = 2^{25} \times 3^{40}$, which matches the given value.
**Correct Answer: B**
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