Ratios + Simple Equations
A fruit seller has a total of 187 fruits consisting of apples, mangoes and oranges. The number of apples and mangoes are in the ratio 5 : 2. After she sells 75 apples, 26 mangoes and half of the oranges, the ratio of number of unsold apples to number of unsold oranges becomes 3 : 2. The total number of unsold fruits is: ______ TITA
Solution
Detailed Solution
Total fruits = 187
Apples : Mangoes = 5 : 2
Let the number of apples = \( 5x \), mangoes = \( 2x \)
Then, oranges = \( 187 – (5x + 2x) = 187 – 7x \)
After selling:
- Apples sold = 75 → Unsold = \( 5x – 75 \)
- Mangoes sold = 26 → Unsold = \( 2x – 26 \)
- Oranges sold = half → Unsold = \( \frac{187 – 7x}{2} \)
Given: Ratio of unsold apples to unsold oranges is \( 3:2 \)
So, \( \frac{5x – 75}{\frac{187 – 7x}{2}} = \frac{3}{2} \)
Solve:
\( 2(5x – 75) = 3 \cdot \frac{187 – 7x}{2} \)
\( \Rightarrow 4(5x – 75) = 3(187 – 7x) \)
\( \Rightarrow 20x – 300 = 561 – 21x \)
\( \Rightarrow 41x = 861 \Rightarrow x = 21 \)
Now compute values:
- Apples = \( 5x = 105 \)
- Mangoes = \( 2x = 42 \)
- Oranges = \( 187 – 147 = 40 \)
Unsold Fruits:
- Apples = \( 105 – 75 = 30 \)
- Mangoes = \( 42 – 26 = 16 \)
- Oranges = \( \frac{40}{2} = 20 \)
Total unsold fruits = 30 + 16 + 20 = 66
Ratios + Percentages
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?
a) 13.26
b) 15.18
c) 12.75
d) 14.84
Solution
House Rent as Percentage of Total Income in October
Incomes in September:
Kamal : Amal : Vimal = 8 : 6 : 5
Let their incomes be:
Kamal = \( 8x \), Amal = \( 6x \), Vimal = \( 5x \)
Rent Contributions in September:
Kamal pays 15% → \( 0.15 \times 8x = 1.2x \)
Amal pays 12% → \( 0.12 \times 6x = 0.72x \)
Vimal pays 18% → \( 0.18 \times 5x = 0.9x \)
Total Rent: \( 1.2x + 0.72x + 0.9x = 2.82x \)
Incomes in October:
Kamal = \( 8x \times 1.10 = 8.8x \)
Amal = \( 6x \times 1.12 = 6.72x \)
Vimal = \( 5x \times 1.15 = 5.75x \)
Total Income in October: \( 8.8x + 6.72x + 5.75x = 21.27x \)
Rent as % of October’s total income:
\( \frac{2.82x}{21.27x} \times 100 = \frac{2.82}{21.27} \times 100 \approx 13.25\% \)
Averages + Equations
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 ________ TITA
Solution
Solution
Let the four numbers be \( a, b, c, d \).
Step 1:
Average of first two numbers is 1 more than the first number:
\[ \frac{a + b}{2} = a + 1 \Rightarrow a + b = 2a + 2 \Rightarrow b = a + 2 \quad \text{(1)} \]
Step 2:
Average of first three numbers is 2 more than average of first two:
\[ \frac{a + b + c}{3} = \frac{a + b}{2} + 2 \]
From Step 1, \( \frac{a + b}{2} = a + 1 \), so:
\[ \frac{a + b + c}{3} = a + 3 \Rightarrow a + b + c = 3a + 9 \]
Substitute \( b = a + 2 \):
\[ a + (a + 2) + c = 3a + 9 \Rightarrow 2a + 2 + c = 3a + 9 \Rightarrow c = a + 7 \quad \text{(2)} \]
Step 3:
Average of all four numbers is 3 more than average of first three:
\[ \frac{a + b + c + d}{4} = \frac{a + b + c}{3} + 3 \]
From above, \( \frac{a + b + c}{3} = a + 3 \), so:
\[ \frac{a + b + c + d}{4} = a + 6 \Rightarrow a + b + c + d = 4a + 24 \]
Substitute \( b = a + 2 \), \( c = a + 7 \):
\[ a + (a + 2) + (a + 7) + d = 4a + 24 \Rightarrow 3a + 9 + d = 4a + 24 \Rightarrow d = a + 15 \quad \text{(3)} \]
Step 4:
We now have:
- \( a \)
- \( b = a + 2 \)
- \( c = a + 7 \)
- \( d = a + 15 \)
So the smallest number is \( a \), and the largest is \( d = a + 15 \).
\[ \text{Difference} = (a + 15) – a = \boxed{15} \]
Final Answer:
\[ \boxed{15} \]
Rectangles + Mensuration + Pythagoras Theorm
ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of ∆ADE is
Solution
Finding the Inradius of Triangle ADE
Given:
- Rectangle \(ABCD\) with \(AB = 56\) cm and \(BC = 45\) cm
- \(E\) is the midpoint of \(CD\)
- We are to find the inradius of triangle \(ADE\)
Step 1: Coordinates and Lengths
Let us place the rectangle on the coordinate plane:
- \(A(0, 0)\), \(B(56, 0)\), \(C(56, 45)\), \(D(0, 45)\)
- \(E\), the midpoint of \(CD\), is: \[ E = \left( \frac{0 + 56}{2}, \frac{45 + 45}{2} \right) = (28, 45) \]
Now find the sides of triangle \(ADE\):
- \(AD = 45\)
- \(DE = 28\)
- \(AE = \sqrt{(28)^2 + (45)^2} = \sqrt{784 + 2025} = \sqrt{2809} = 53\)
Step 2: Use Inradius Formula
For a triangle with sides \(a, b, c\), the inradius is:
\[ r = \frac{A}{s}, \quad \text{where } s = \frac{a + b + c}{2} \] Here, \[ a = 45, \quad b = 28, \quad c = 53 \Rightarrow s = \frac{45 + 28 + 53}{2} = \frac{126}{2} = 63 \]
Use Heron’s Formula to find area \(A\):
\[ A = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{63 \cdot 18 \cdot 35 \cdot 10} \] \[ = \sqrt{396900} = 630 \]
Step 3: Compute Inradius
\[ r = \frac{A}{s} = \frac{630}{63} = \boxed{10} \]
✅ Final Answer: \( \boxed{10 \text{ cm}} \)
Interest + Ratios
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
Solution
Finding the Investment Period in Bank A
Given:
- Principal \( P = ₹10,000 \)
- Bank A: Simple Interest at 5% for \( x \) years
- Total amount from Bank A deposited in Bank B for 5 years at 6%
- Interests from Bank A and Bank B are in ratio 10 : 13
Step 1: Interest from Bank A
\[ \text{SI}_A = \frac{P \cdot R \cdot T}{100} = \frac{10000 \cdot 5 \cdot x}{100} = 500x \]
Step 2: Amount Deposited in Bank B
Total Amount after Bank A:
\[ A = 10000 + 500x \]
Interest from Bank B:
\[ \text{SI}_B = \frac{(10000 + 500x) \cdot 6 \cdot 5}{100} = \frac{(10000 + 500x) \cdot 30}{100} = \frac{300000 + 15000x}{100} = 3000 + 150x \]
Step 3: Using the Given Ratio
\[ \frac{\text{SI}_A}{\text{SI}_B} = \frac{10}{13} \Rightarrow \frac{500x}{3000 + 150x} = \frac{10}{13} \] Cross-multiplying: \[ 13 \cdot 500x = 10(3000 + 150x) \Rightarrow 6500x = 30000 + 1500x \Rightarrow 5000x = 30000 \Rightarrow x = \boxed{6} \]
✅ Final Answer: \( \boxed{6 \text{ years}} \)
Ratios + Allegations
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
a. 1:27
b. 1:81
c. 1: 26
d. 1:80
Solution
2/3 removed so left milk is 1/3.
Repeating this 4 times.. Milk is 1/3×1/3×1/3×1/3 = 1/81
Final ratio of Milk to water will be 1:80
Profit Loss + Percentages
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
a. 15
b. 10
c. 20
d. 14
Solution
Profit and Selling Price Problem
Let the original cost price be \( x \)
Then the original selling price is:
\( \text{SP} = x + 0.4x = 1.4x \)
New cost price is 40% less than the original cost price:
\( \text{New CP} = x – 0.4x = 0.6x \)
New selling price is ₹5 less than the original SP:
\( \text{New SP} = 1.4x – 5 \)
Given: New profit is 50% on the new cost price:
\( \text{New SP} = \text{New CP} + 50\% \text{ of New CP} = 0.6x + 0.3x = 0.9x \)
So we form the equation:
\( 1.4x – 5 = 0.9x \)
Simplify:
\( 1.4x – 0.9x = 5 \Rightarrow 0.5x = 5 \Rightarrow x = 10 \)
Original selling price = \( 1.4x = 1.4 \times 10 = \boxed{14} \)
Final Answer: \( \boxed{14} \)
Indices + Quadratic
The sum of all real values of \( k \) for which
\( \left( \frac{1}{8} \right)^k \times \left( \frac{1}{32768} \right)^{\frac{1}{3}} = \frac{1}{8} \times \left( \frac{1}{32768} \right)^{\frac{1}{k}} \), is:
Solution
Sum of All Real Values of \( k \)
We are given the equation:
\[ \left( \frac{1}{8} \right)^k \cdot \left( \frac{1}{32768} \right)^{1/3} = \left( \frac{1}{8} \right) \cdot \left( \frac{1}{32768} \right)^{1/k} \]
Step 1: Convert all terms to base 2
\[ \frac{1}{8} = 2^{-3}, \quad \frac{1}{32768} = 2^{-15} \] \[ \Rightarrow (2^{-3})^k \cdot (2^{-15})^{1/3} = 2^{-3} \cdot (2^{-15})^{1/k} \] \[ \Rightarrow 2^{-3k – 5} = 2^{-3 – \frac{15}{k}} \]
Step 2: Equating the exponents
\[ -3k – 5 = -3 – \frac{15}{k} \Rightarrow 3k + \frac{15}{k} = -2 \]
Step 3: Clear the fraction
Multiply both sides by \( k \): \[ 3k^2 + 15 = -2k \Rightarrow 3k^2 + 2k – 15 = 0 \]
Step 4: Use sum of roots formula
This is a quadratic of the form \( ax^2 + bx + c = 0 \).
The sum of real roots is:
\[
-\frac{b}{a} = -\frac{2}{3}
\]
✅ Final Answer: \( \boxed{-\frac{2}{3}} \)
Logs DPAC Shortcut
If \( x \) is a positive real number such that
\( 4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13 \),
then the greatest integer not exceeding \( x \) is:
Solution
Greatest Integer Not Exceeding \( x \)
We are given the equation:
\[ 4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13 \]
Step 1: Convert all logs to base 10
\[ \log_{100} x = \frac{\log_{10} x}{\log_{10} 100} = \frac{\log_{10} x}{2} \quad \text{and} \quad \log_{1000} x = \frac{\log_{10} x}{\log_{10} 1000} = \frac{\log_{10} x}{3} \]
Step 2: Substitute and simplify
\[ 4 \log_{10} x + 4 \cdot \frac{\log_{10} x}{2} + 8 \cdot \frac{\log_{10} x}{3} = 4 \log x + 2 \log x + \frac{8}{3} \log x \] \[ = \left(6 + \frac{8}{3}\right) \log x = \frac{26}{3} \log x \]
Step 3: Solve the equation
\[ \frac{26}{3} \log x = 13 \Rightarrow \log x = \frac{13 \cdot 3}{26} = \frac{39}{26} = \frac{3}{2} \Rightarrow x = 10^{3/2} = \sqrt{1000} \approx 31.62 \]
