Official Correct Answer: B. ### Core Concept
The problem is an application of the **Pigeonhole Principle**, which states that if \( n \) items are put into \( m \) containers, then at least one container must contain more than \( \left\lfloor \frac{n}{m} \right\rfloor \) items.
### Step-by-Step Solution
1. **Total Students and Months**: We have 100 students and 12 months.
2. **Even Distribution**: Divide 100 by 12 to find the base number of students per month.
\[
\frac{100}{12} \approx 8.333
\]
3. **Integer Division**: Since we can't have a fraction of a student, we take the integer part, which is 8.
4. **Remaining Students**: Calculate the remainder when 100 is divided by 12.
\[
100 = 12 \times 8 + 4
\]
So, 4 students remain.
5. **Distribute Remaining Students**: These 4 students must be added to 4 different months, increasing those months' count by 1.
6. **Maximum Students in a Month**: The months with the extra students will have \( 8 + 1 = 9 \) students.
7. **Conclusion**: The smallest possible value of \( x_0 \) is 9.
### 30-Second Shortcut
- Divide total students by months: \( 100 \div 12 \approx 8.33 \).
- Since you can't have a fraction, round up the remainder: \( 100 - (12 \times 8) = 4 \).
- These 4 students must go into separate months, making the maximum \( 8 + 1 = 9 \).
### Final Answer
Correct Answer: Option B
A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?
Official Correct Answer: 62. ### Core Concept
Work backwards from the empty stock. Each child takes half of the current stock plus 1, so the stock before that child can be recovered from the stock after.
### Step-by-Step Solution
1. If the stock before a child is $x$, the child gets $\frac{x}{2}+1$ and the stock left is $\frac{x}{2}-1$.
2. Call the stock left $r$. Then $r=\frac{x}{2}-1$, so $x=2(r+1)$.
3. After the 5th child the stock is 0. Before the 5th child: $2(0+1)=2$.
4. Before the 4th child: $2(2+1)=6$.
5. Before the 3rd child: $2(6+1)=14$.
6. Before the 2nd child: $2(14+1)=30$.
7. Before the 1st child: $2(30+1)=62$.
### Quick Check
Start with 62. The children get 32, 16, 8, 4, 2 toffees (half plus one each time), and $32+16+8+4+2=62$, so the stock is exhausted after 5 children. ✓
**Correct Answer: 62**
Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru's investment, will their balances, i.e., principal plus accumulated interest, be equal?
Official Correct Answer: 12. ### Core Concept
The problem involves calculating the time when two investments with different principals, interest rates, and start times will have equal balances. We use the **Simple Interest Formula**:
\[ \text{Simple Interest} = \frac{P \times R \times T}{100} \]
where \( P \) is the principal, \( R \) is the annual interest rate, and \( T \) is the time in years.
### Step-by-Step Solution
1. **Calculate Veeru's Investment Growth:**
- Principal (\( P_V \)) = Rs 10,000
- Rate (\( R_V \)) = 5% per annum
- Time = \( T \) years
- Simple Interest for Veeru after \( T \) years:
\[ I_V = \frac{10000 \times 5 \times T}{100} = 500T \]
- Total Amount (\( A_V \)) after \( T \) years:
\[ A_V = 10000 + 500T \]
2. **Calculate Joy's Investment Growth:**
- Joy invested 2 years after Veeru, so her investment time is \( T - 2 \) years.
- Principal (\( P_J \)) = Rs 8,000
- Rate (\( R_J \)) = 10% per annum
- Simple Interest for Joy after \( T - 2 \) years:
\[ I_J = \frac{8000 \times 10 \times (T - 2)}{100} = 800(T - 2) \]
- Total Amount (\( A_J \)) after \( T - 2 \) years:
\[ A_J = 8000 + 800(T - 2) = 8000 + 800T - 1600 = 6400 + 800T \]
3. **Set the Total Amounts Equal and Solve for \( T \):**
\[ A_V = A_J \]
\[ 10000 + 500T = 6400 + 800T \]
- Subtract \( 500T \) from both sides:
\[ 10000 = 6400 + 300T \]
- Subtract \( 6400 \) from both sides:
\[ 3600 = 300T \]
- Divide both sides by \( 300 \):
\[ T = \frac{3600}{300} = 12 \]
### 30-Second Shortcut
- Recognize that Joy's investment grows faster due to a higher interest rate but starts later.
- Set up the equation for total amounts and solve for \( T \) by equating the two expressions.
- The difference in their interest accumulations will give the time when they equalize.
### Final Answer Confirmation
After solving, we find that \( T = 12 \) years. Verifying with both amounts confirms they are equal at this time.
**Correct Answer: 12**
Official Correct Answer: D. ### Core Concept
Take logarithms to a common base (base 2) on both sides, so the exponent equation becomes an algebraic equation in $y^2$.
### Step-by-Step Solution
1. Take $\log_2$ of both sides of $2^{y^2\log_3 5}=5^{\log_2 3}$: $$y^2\log_3 5=\log_2 3\cdot\log_2 5.$$
2. Change the base: $\log_3 5=\dfrac{\log_2 5}{\log_2 3}$.
3. Substitute: $$y^2\cdot\frac{\log_2 5}{\log_2 3}=\log_2 3\cdot\log_2 5.$$
4. Cancel $\log_2 5$ (non-zero): $y^2=(\log_2 3)^2$.
5. So $y=\pm\log_2 3$. Since $y$ is negative, $y=-\log_2 3=\log_2\left(\frac{1}{3}\right)$.
### Why Not the Others
- A: $-\log_2(1/5)=\log_2 5$ is positive.
- B: $\log_2(1/5)=-\log_2 5$ is negative but its square is $(\log_2 5)^2$, not $(\log_2 3)^2$.
- C: $-\log_2(1/3)=\log_2 3$ is positive.
**Correct Answer: D**
Leaving home at the same time, Amal reaches the office at 10:15am if he travels at 8km/hr , and at 9:40am if he travels at 15km/hr. Leaving home at 9:10, at what speed, in km/hr, must be travel so as to reach office exactly at 10am?
Official Correct Answer: D. ### Core Concept
The distance to the office is fixed, so the difference in travel times at the two speeds gives the distance. That fixes the usual leaving time and the speed needed for the new schedule.
### Step-by-Step Solution
1. Arrival times differ by 10:15 − 9:40 = 35 minutes $= \frac{7}{12}$ hr, because the departure time is the same in both cases.
2. Let the distance be $d$ km. Then $\frac{d}{8} - \frac{d}{15} = \frac{7}{12}$, so $\frac{7d}{120} = \frac{7}{12}$ and $d = 10$ km.
3. At 8 km/hr the trip takes $\frac{10}{8} = 1.25$ hr, so Amal normally leaves at 10:15 − 1:15 = 9:00 am. This is only a consistency check.
4. Leaving at 9:10 and arriving at 10:00 leaves 50 minutes $= \frac{5}{6}$ hr.
5. Required speed $= \frac{10}{5/6} = 12$ km/hr.
### Why Not the Others
- 14 km/hr covers 10 km in about 43 minutes, so he arrives too early.
- 13 km/hr takes about 46 minutes, so he arrives too early.
- 11 km/hr takes about 54.5 minutes, so he arrives late.
### Quick Check
$10 \div 12 = \frac{5}{6}$ hr $= 50$ min, and 9:10 + 50 min = 10:00 am.
**Correct Answer: D**
A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to
Official Correct Answer: D. ### Core Concept
For a fixed distance, time is inversely proportional to speed. The scheduled time follows from the delay at one-third speed, and the speed increase follows from the time left after the stop.
### Step-by-Step Solution
1. Let the usual time for the journey be $t$ and the usual speed be $v$. At $\frac{v}{3}$ the time is $3t$.
2. The delay is $3t - t = 2t = 30$ min, so $t = 15$ min.
3. On the return trip the train runs 5 minutes at usual speed and then stops for 4 minutes, so 9 minutes of the 15-minute schedule are gone.
4. In those 5 minutes of running, one-third of the distance is covered, so $10$ minutes' worth of distance at usual speed remains.
5. Only $15 - 9 = 6$ minutes remain on the schedule.
6. The speed ratio is $\frac{10}{6} = \frac{5}{3}$, so the speed must rise by $\frac{2}{3} \approx 66.7\%$.
7. This is nearest to 67%.
### Why Not the Others
- 58% and 61% would leave the train late.
- 50% would need 6.67 minutes for the remaining distance, which is more than the 6 minutes available.
### Quick Check
At $\frac{5}{3}v$ the remaining 10 minutes' worth of distance takes $10 \times \frac{3}{5} = 6$ minutes, and $9 + 6 = 15$.
**Correct Answer: D**
Official Correct Answer: B. ### Core Concept & Formula
- For any real $x$, by the AM-GM inequality on positive terms:
$$2^x + 2^{-x} \geq 2 \sqrt{2^x \cdot 2^{-x}} = 2$$
Equality holds if and only if $2^x = 2^{-x} \implies x = 0$.
- For the RHS:
$$2 - (x - 2)^2 \leq 2$$
Equality holds if and only if $(x - 2)^2 = 0 \implies x = 2$.
### Step-by-Step Solution
1. **Analyze Left-Hand Side (LHS)**:
$$\text{LHS} = 2^x + 2^{-x} \geq 2$$
The minimum value of LHS is $2$, occurring at $x = 0$.
2. **Analyze Right-Hand Side (RHS)**:
$$\text{RHS} = 2 - (x - 2)^2 \leq 2$$
The maximum value of RHS is $2$, occurring at $x = 2$.
3. **Determine Intersection**:
For $\text{LHS} = \text{RHS}$, both must simultaneously equal $2$.
However, LHS equals $2$ only at $x = 0$, while RHS equals $2$ only at $x = 2$.
Since $0 \neq 2$, no real number $x$ satisfies the equation.
$$\text{Number of real solutions} = 0$$
### 30-Second Shortcut
- $\min(\text{LHS}) = 2$ at $x=0$. $\max(\text{RHS}) = 2$ at $x=2$. No common solution $\implies 0$.
### Final Answer
**Option B**
A solution, of volume 40 litres, has dye and water in the proportion 2:3. Water is added to the solution to change this proportion to 2:5. If one-fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2:3?
Official Correct Answer: 8. ### Core Concept
When water is added, the quantity of dye stays the same. Taking out a fraction of the solution removes dye and water in the same ratio. Then dye is added until the ratio is back to $2:3$.
### Step-by-Step Solution
1. In the original 40 litres, dye $= 40 \times \frac{2}{5} = 16$ L and water $= 24$ L.
2. For a ratio of $2:5$ with 16 L of dye, water must be $16 \times \frac{5}{2} = 40$ L. So 16 L of water is added, and the total becomes 56 L.
3. Removing one-fourth leaves three-fourths: 42 L in total, with dye $= 12$ L and water $= 30$ L.
4. Let $x$ litres of dye be added. Then $\frac{12 + x}{30} = \frac{2}{3}$, so $12 + x = 20$ and $x = 8$.
### Quick Check
The new mixture has 20 L of dye and 30 L of water, a ratio of $2:3$.
**Correct Answer: 8**
Let A,B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is
Official Correct Answer: D. ### Core Concept
The two conditions give two linear equations in three unknowns. The requirement that $A$, $B$, $C$ are positive integers then leaves only one solution.
### Step-by-Step Solution
1. The first condition is $A + \frac{B + C}{2} = 5$, which gives $2A + B + C = 10$.
2. The second condition is $B + \frac{A + C}{2} = 7$, which gives $A + 2B + C = 14$.
3. Subtracting the first from the second gives $B - A = 4$, so $B = A + 4$.
4. Substituting into the first equation gives $2A + A + 4 + C = 10$, so $C = 6 - 3A$.
5. $C$ must be a positive integer, so $A = 1$ (since $A = 2$ gives $C = 0$). Then $C = 3$ and $B = 5$.
6. Therefore $A + B = 1 + 5 = 6$.
### Why Not the Others
- 4, 5 and 7 cannot arise, because $A = 1$, $B = 5$ is the only positive-integer solution.
### Quick Check
$1 + \frac{5 + 3}{2} = 5$ and $5 + \frac{1 + 3}{2} = 7$.
**Correct Answer: D**
Official Correct Answer: 21. ### Core Concept
To solve the problem, we need to count all 3-digit numbers where the product of their digits is between 3 and 6 (inclusive). This involves enumerating all valid combinations of digits (hundreds, tens, units) that satisfy the product condition and then calculating the number of permutations for each combination.
### Step-by-Step Solution
1. **Identify Target Products**: The product of digits must be 3, 4, 5, or 6.
2. **Analyze Each Product**:
- **Product = 3**:
- Digits: 1, 1, 3
- Permutations: \( \frac{3!}{2!} = 3 \)
- **Product = 4**:
- Digits: 1, 1, 4 and 1, 2, 2
- Permutations: \( 3 + 3 = 6 \)
- **Product = 5**:
- Digits: 1, 1, 5
- Permutations: \( \frac{3!}{2!} = 3 \)
- **Product = 6**:
- Digits: 1, 1, 6 and 1, 2, 3
- Permutations: \( 3 + 6 = 9 \)
3. **Sum All Permutations**:
- Total numbers = 3 (for 3) + 6 (for 4) + 3 (for 5) + 9 (for 6) = 21.
### 30-Second Shortcut
Instead of listing all possibilities, recognize that the valid products are limited (3, 4, 5, 6). For each product, determine the possible digit triplets and their permutations, ensuring no digit is 0. Sum the permutations for each product to get the total count.
### Final Answer
Correct Answer: 21
A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A , both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60km/hr , then the speed of Car 2, in km/hr, is
Official Correct Answer: C. ### Core Concept
When two bodies start together from opposite ends and meet, the ratio of their speeds equals the square root of the inverse ratio of the times each takes after the meeting.
### Step-by-Step Solution
1. Let the meeting point be $M$. After meeting, Car 1 covers $MB$ in 45 min and Car 2 covers $MA$ in 20 min.
2. Before meeting, both cars travel for the same time $t$. Car 1 covers $AM$ at speed $v_1$, and Car 2 covers $BM$ at speed $v_2$.
3. So $AM = v_1 t = 20\,v_2$ and $BM = v_2 t = 45\,v_1$, because $AM$ is the distance Car 2 covers after meeting and $BM$ is the distance Car 1 covers after meeting.
4. Multiply the two relations: $v_1 v_2 t^2 = 900\,v_1 v_2$, so $t = 30$ min.
5. Then $v_1 \cdot 30 = 20\,v_2$, which gives $\dfrac{v_1}{v_2} = \dfrac{2}{3}$.
6. With $v_1 = 60$ km/hr, $v_2 = 60 \times \dfrac{3}{2} = 90$ km/hr.
### Quick Check
$\sqrt{20/45} = 2/3$, which matches $v_1 : v_2$.
**Correct Answer: C**
A solid right circular cone of height 27 cm is cut into pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225cc, the volume, in cc, of the original cone is
Official Correct Answer: A. ### Core Concept
A plane parallel to the base cuts off a smaller cone similar to the original. Volumes of similar solids are in the ratio of the cubes of their heights.
### Step-by-Step Solution
1. The cut is 18 cm above the base, so the small cone on top has height $27 - 18 = 9$ cm.
2. The ratio of heights is $9 : 27 = 1 : 3$, so the ratio of volumes is $1 : 27$.
3. Let the original volume be $V$. The small cone has volume $\dfrac{V}{27}$, and the frustum (the other piece) has volume $\dfrac{26V}{27}$.
4. The difference between the pieces is $\dfrac{26V}{27} - \dfrac{V}{27} = \dfrac{25V}{27}$.
5. Set $\dfrac{25V}{27} = 225$, so $V = \dfrac{225 \times 27}{25} = 9 \times 27 = 243$ cc.
### Quick Check
Small cone $= 9$ cc and frustum $= 234$ cc, and the difference is $225$ cc.
**Correct Answer: A**
Two persons are walking beside a railway track at respective speeds of 2 and 4km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross a stationary object?
Official Correct Answer: D. ### Core Concept
The train crosses a walker at the relative speed (train speed minus walker speed), and in each case it covers its own length. Equating the two lengths gives the train's speed.
### Step-by-Step Solution
1. Let the train's speed be $v$ km/hr and its length $L$ metres. Convert km/hr to m/s by multiplying by $\dfrac{5}{18}$.
2. Crossing the first person (2 km/hr) in 90 s gives $L = (v-2)\cdot\dfrac{5}{18}\cdot 90$.
3. Crossing the second person (4 km/hr) in 100 s gives $L = (v-4)\cdot\dfrac{5}{18}\cdot 100$.
4. Equate the two: $90(v-2) = 100(v-4)$, so $90v - 180 = 100v - 400$ and $v = 22$ km/hr.
5. Then $L = 20 \cdot \dfrac{5}{18} \cdot 90 = 500$ m.
6. Crossing a stationary object takes $\dfrac{500}{22 \cdot \frac{5}{18}} = \dfrac{500 \times 18}{110} \approx 81.8$ s.
7. The nearest option is 82 s.
### Quick Check
Crossing the second walker: $18 \times \tfrac{5}{18} \times 100 = 500$ m, which matches $L$.
**Correct Answer: D**
Official Correct Answer: D. ### Core Concept
$f(5+x) = f(5-x)$ for every real $x$ means the graph of $f$ is symmetric about the line $x = 5$, so the roots of $f(x) = 0$ come in mirror-image pairs about 5.
### Step-by-Step Solution
1. **Roots pair up:** if $5 + a$ is a root, then $f(5 - a) = f(5 + a) = 0$, so $5 - a$ is a root as well.
2. **Each pair adds up to 10:** $(5 + a) + (5 - a) = 10$.
3. **No root sits at 5:** the roots other than 5 always come in pairs, so if 5 were a root the number of distinct roots would be odd. With four distinct roots, they form exactly two pairs.
4. **Sum:** $10 + 10 = 20$.
### Answer
The sum of the four roots is 20.
**Correct Answer: D**
Official Correct Answer: C. ### Core Concept
The diagonals of a rhombus bisect each other at right angles, so they give the side and the area. The inradius then follows from area = inradius × semi-perimeter.
### Step-by-Step Solution
1. Half-diagonals are 6 and 8, so each side is $\sqrt{6^2+8^2}=10$ cm.
2. Area of rhombus $=\tfrac12\times12\times16=96$ sq cm.
3. Perimeter is $40$, so the semi-perimeter is $20$. For a rhombus with an inscribed circle, area $=r\times$ semi-perimeter, so $96=20r$ and $r=\tfrac{24}{5}$.
4. Area of circle $=\pi r^2=\tfrac{576\pi}{25}$.
5. Ratio $=\dfrac{576\pi/25}{96}=\dfrac{6\pi}{25}$.
### Quick Check
The inradius is also the altitude from the right-angle vertex to the hypotenuse: $\tfrac{6\times8}{10}=4.8=\tfrac{24}{5}$.
**Correct Answer: C**
Official Correct Answer: C. ### Core Concept
Express $64$ as a power of $x$. This needs the reciprocal of $7+4\sqrt{3}$, which is its conjugate because $(7+4\sqrt3)(7-4\sqrt3)=1$.
### Step-by-Step Solution
1. Since $4096=64^2$, we get $x=64^{2(7+4\sqrt3)}$.
2. Therefore $64=x^{\frac{1}{2(7+4\sqrt3)}}$.
3. Because $(7+4\sqrt3)(7-4\sqrt3)=49-48=1$, we have $\dfrac{1}{7+4\sqrt3}=7-4\sqrt3$.
4. So $64=x^{\frac{7-4\sqrt3}{2}}=x^{\frac72-2\sqrt3}$.
5. Using the laws of indices, $x^{\frac72-2\sqrt3}=\dfrac{x^{7/2}}{x^{2\sqrt3}}$, which is option C.
### Why Not the Others
- A: equals $x^{7-2\sqrt3}$, with the wrong exponent.
- B: equals $x^{\frac72-\frac{4}{\sqrt3}}$, and $\tfrac{4}{\sqrt3}\ne2\sqrt3$.
- D: equals $x^{7-4\sqrt3}$, which is $64^2$, not $64$.
**Correct Answer: C**
On a rectangular metal sheet of area 135sq in, a circle is painted such that the circle touches two opposite sides. If the are the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is
Official Correct Answer: A. ### Core Concept
The circle touches two opposite sides, so its diameter equals the rectangle's width. The area split fixes the circle's radius, and the length then follows from the total area.
### Step-by-Step Solution
1. Unpainted $=\tfrac23$ of painted, so painted : total $=3:5$. Painted area $=\tfrac35\times135=81$ sq in.
2. $\pi r^2=81$, so $r=\dfrac{9}{\sqrt\pi}$.
3. Width $=2r=\dfrac{18}{\sqrt\pi}$.
4. Length $=\dfrac{135}{18/\sqrt\pi}=\dfrac{15\sqrt\pi}{2}$.
5. Perimeter $=2\left(\dfrac{18}{\sqrt\pi}+\dfrac{15\sqrt\pi}{2}\right)=\dfrac{36}{\sqrt\pi}+15\sqrt\pi$.
6. Factor out $3\sqrt\pi$: $\dfrac{36}{\sqrt\pi}=3\sqrt\pi\cdot\dfrac{12}{\pi}$ and $15\sqrt\pi=3\sqrt\pi\cdot5$. So the perimeter is $3\sqrt\pi\left(5+\dfrac{12}{\pi}\right)$.
### Quick Check
Width $\times$ length $=\dfrac{18}{\sqrt\pi}\times\dfrac{15\sqrt\pi}{2}=135$.
**Correct Answer: A**
Official Correct Answer: 1. ### Core Concept
We can simplify the given equation by substituting \( y = x + \frac{1}{x} \), transforming it into a quadratic equation in \( y \). Solving for \( y \) and then for \( x \) will help determine the number of distinct real roots.
### Step-by-Step Solution
1. **Substitute \( y = x + \frac{1}{x} \):**
The equation becomes:
\[
y^2 - 3y + 2 = 0
\]
2. **Solve the quadratic equation for \( y \):**
\[
(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2
\]
3. **Case 1: \( y = 1 \)**
\[
x + \frac{1}{x} = 1 \implies x^2 - x + 1 = 0
\]
- Discriminant \( D = (-1)^2 - 4(1)(1) = -3 \)
- No real roots since \( D < 0 \)
4. **Case 2: \( y = 2 \)**
\[
x + \frac{1}{x} = 2 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0
\]
- Double root at \( x = 1 \)
- One distinct real root
### 30-Second Shortcut
Substitute \( y = x + \frac{1}{x} \) to reduce the equation to a quadratic. Solve for \( y \), then check each case for real roots. Only one case yields a real root, which is a double root, so it counts as one distinct real root.
### Final Answer
Correct Answer: 1
Official Correct Answer: 3. ### Core Concept
Rewrite the inequalities as a region between two graphs and integrate (or use areas of simple shapes) over the $x$-range where the region exists.
### Step-by-Step Solution
1. $|x|-y\leq 1$ gives $y\geq |x|-1$. Together with $0\leq y\leq 1$, each $x$ allows $y$ from $\max(0,\,|x|-1)$ up to $1$.
2. The lower bound $|x|-1$ must not exceed the upper bound $1$, so $|x|\leq 2$ and $x\in[-2,2]$.
3. For $|x|\leq 1$ the lower bound is $0$, so the height is $1$. This strip has width $2$ and area $2$.
4. For $1\leq |x|\leq 2$ the height is $1-(|x|-1)=2-|x|$. On each side this is a triangle with base $1$ and height $1$, so its area is $\tfrac12$. The two sides give $1$.
5. Total area $=2+1=3$.
### Quick Check
The region is the trapezoid with parallel sides $4$ (at $y=0$) and $2$ (at $y=1$) and height $1$. Its area is $\tfrac{4+2}{2}\times 1=3$.
**Correct Answer: 3**
In a group of people, 28% of the members are young while the rest are old. If 65% of the members are literates, and 25% of the literates are young, then the percentage of old people among the illiterates is nearest to
Official Correct Answer: C. ### Core Concept
Take the total as 100 and fill in a two-way table of young/old against literate/illiterate.
### Step-by-Step Solution
1. Let the total be 100. Young $=28$ and old $=72$.
2. Literates $=65$, so illiterates $=35$.
3. Young literates $=25\%$ of $65=16.25$.
4. Young illiterates $=28-16.25=11.75$.
5. Old illiterates $=35-11.75=23.25$.
6. Required percentage $=\dfrac{23.25}{35}\times 100\approx 66.4\%$, which is nearest to 66.
### Why Not the Others
- A (59), B (55) and D (62) are all well below $66.4\%$.
- 59 would result from wrongly using $23.25$ against a different base, but the base must be the 35 illiterates.
**Correct Answer: C**
Official Correct Answer: C. First, We analyze what the form 'aabb' means. It's a four-digit number where the first two digits are the same (both 'a') and the last two digits are also the same (both 'b'). So, examples would be 1122, 3344, etc. Also, since it's a 4-digit number, 'a' can't be zero. So 'a' ranges from 1 to 9, and 'b' can be from 0 to 9, but since the number has to be even, 'b' must be even. That means 'b' can be 0, 2, 4, 6, or 8.
Alright, so the number is of the form 'aabb'. Let me express this number mathematically. The first digit 'a' is in the thousands place, the second 'a' is in the hundreds place, the first 'b' is in the tens place, and the second 'b' is in the ones place. So, the number can be written as:
1000*a + 100*a + 10*b + b = 1100*a + 11*b
So, the number is 1100a + 11b. Since we're dealing with all such numbers, we need to find the mean of all these numbers.
To find the mean, We find the sum of all such numbers and then divide by how many there are.
First, let's figure out how many such numbers there are. Since 'a' can be from 1 to 9 and 'b' can be 0, 2, 4, 6, 8 (so 5 options), the total number of such numbers is 9*5 = 45.
So, there are 45 numbers in total.
Now, let's find the sum of all these numbers. Each number is 1100a + 11b. So, the sum will be the sum over all a and b of (1100a + 11b).
I can separate this into two sums: 1100*sum(a) over all a and b, plus 11*sum(b) over all a and b.
But wait, actually, for each a, b varies, so for each a, we have 5 different b's. So, for each a, the sum contributed by that a is 1100a * 5 (since b is added 5 times for each a). Similarly, for each a, the sum contributed by b is 11*(sum of b's) for that a.
So, Computing sum(a) over all numbers. Since a ranges from 1 to 9, and for each a, there are 5 b's, the total sum contributed by a is 5*(1 + 2 + 3 + ... + 9). The sum from 1 to 9 is (9*10)/2 = 45. So, 5*45 = 225. Then, multiplied by 1100, that's 225*1100.
Similarly, for b, for each a, the sum of b's is 0 + 2 + 4 + 6 + 8 = 20. So, for each a, the sum contributed by b is 11*20. Since there are 9 a's, the total sum from b is 9*11*20.
Compute these:
First, sum from a: 225*1100. Computing 225*1000 = 225,000 and 225*100 = 22,500. So, 225,000 + 22,500 = 247,500.
Next, sum from b: 9*11*20. Computing 9*20 = 180, then 180*11 = 1,980.
So, total sum is 247,500 + 1,980 = 249,480.
Now, the mean is total sum divided by number of numbers, which is 45.
So, mean = 249,480 / 45.
Compute that. 249,480 divided by 45.
First, how many times 45 goes into 249,480.
Alternatively, divide numerator and denominator by 5: 249,480 / 5 = 49,896; 45 / 5 = 9.
So, now it's 49,896 / 9.
Compute 49,896 ÷ 9:
9*5,544 = 49,896 because 5,544*9: 5,000*9=45,000; 500*9=4,500; 40*9=360; 4*9=36. So, 45,000 + 4,500 = 49,500; 49,500 + 360 = 49,860; 49,860 + 36 = 49,896.
So, 49,896 / 9 = 5,544.
Therefore, the mean is 5,544.
Looking at the options, that's option C.
Sum from a: 5*(1+2+...+9) = 5*45 = 225. 225*1100 = 247,500. That seems right.
Sum from b: For each a, sum of b is 0+2+4+6+8=20. So, 20*11=220 per a. 9 a's, so 9*220=1,980. That adds up to 247,500 + 1,980 = 249,480.
Divide by 45: 24
An alloy is prepared by mixing three metals A,B and C in the proportion 3:4:7 by volume. Weights of the same volume of the metals A,B and C are in the ratio 5:2:6. In 130 kg of the alloy, the weight, in kg, of the metal C is
Official Correct Answer: C. ### Core Concept
The weight share of each metal is its volume share multiplied by its weight per unit volume.
### Step-by-Step Solution
1. Volumes of $A:B:C$ are $3:4:7$. Weights per unit volume are $5:2:6$.
2. Weight contributions are $3\times5=15$, $4\times2=8$ and $7\times6=42$.
3. So the weight ratio is $15:8:42$, with total $65$ parts.
4. Weight of $C=\dfrac{42}{65}\times 130=42\times 2=84$ kg.
### Why Not the Others
- 70 comes from using the volume ratio $7/14$ directly, which ignores the different densities.
- 48 and 96 do not match the $42:65$ weight share of $C$.
**Correct Answer: C**
A person spent Rs 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is
Official Correct Answer: 20000. ### Core Concept
The problem involves calculating the cost price of a desktop and a laptop based on their selling prices and overall profit. We use the concepts of profit and loss percentages and set up equations to solve for the unknown cost prices.
### Step-by-Step Solution
1. **Define Variables:**
- Let \( D \) be the cost price of the desktop.
- Let \( L \) be the cost price of the laptop.
- Given \( D + L = 50000 \).
2. **Express Selling Prices:**
- Desktop sold at 20% profit: Selling Price \( = D \times 1.2 \).
- Laptop sold at 10% loss: Selling Price \( = L \times 0.9 \).
3. **Total Selling Price:**
- Total Selling Price \( = 1.2D + 0.9L \).
- Overall profit is 2%, so Total Selling Price \( = 50000 \times 1.02 = 51000 \).
4. **Set Up Equations:**
- Equation 1: \( D + L = 50000 \).
- Equation 2: \( 1.2D + 0.9L = 51000 \).
5. **Solve the System of Equations:**
- From Equation 1: \( L = 50000 - D \).
- Substitute \( L \) into Equation 2:
\[
1.2D + 0.9(50000 - D) = 51000
\]
- Simplify:
\[
1.2D + 45000 - 0.9D = 51000
\]
\[
0.3D + 45000 = 51000
\]
\[
0.3D = 6000
\]
\[
D = \frac{6000}{0.3} = 20000
\]
### 30-Second Shortcut
- Let the cost prices be \( D \) and \( L \).
- Total cost \( D + L = 50000 \).
- Total selling price \( 1.2D + 0.9L = 51000 \).
- Subtract \( 0.9(D + L) = 45000 \) from total selling price: \( 0.3D = 6000 \) → \( D = 20000 \).
### Final Answer
The purchase price of the desktop is Rs 20000.
**Correct Answer: 20000**
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