Question:
In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are:
Detailed Solution:
Let the number of girls be \(G\) and the number of boys be \(B\). Since the total number of students is 250: \[ G + B = 250 \] – The percentage of girls is between 44% and 60%. Thus: \[ G \in [0.44 \times 250, 0.60 \times 250] \implies G \in [110, 150] \] – The number of boys: \[ B = 250 – G \implies B \in [100, 140] \]
Let \(S\) represent the students opting for swimming, \(R\) represent the students opting for running, and \(S \cap R\) represent the students opting for both activities. Using the given percentages: – Students opting for swimming: \[ S = 0.50B + 0.80G \] – Students opting for running: \[ R = 0.70B + 0.60G \] – Using the principle of inclusion-exclusion: \[ |S \cap R| = S + R – 250 \]
To find the minimum and maximum \( |S \cap R| \), compute \(S\) and \(R\) for the boundary values of \(G\) and \(B\): – For \(G = 110\) and \(B = 140\): \[ S = 0.50(140) + 0.80(110) = 70 + 88 = 158 \] \[ R = 0.70(140) + 0.60(110) = 98 + 66 = 164 \] \[ |S \cap R| = 158 + 164 – 250 = 72 \] – For \(G = 150\) and \(B = 100\): \[ S = 0.50(100) + 0.80(150) = 50 + 120 = 170 \] \[ R = 0.70(100) + 0.60(150) = 70 + 90 = 160 \] \[ |S \cap R| = 170 + 160 – 250 = 80 \]
Thus, the minimum and maximum possible values of \( |S \cap R| \) are \(72\) and \(80\), respectively.
Correct Answer: Option 4 (72 and 80, respectively).
Question:
Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is:
Detailed Solution:
Let the cost price (C.P.) of the product be \( x \), and the marked price (M.P.) be \( 1.20x \) (as the marked price is set to include a 20% profit).
Ravi gets a 10% discount on the marked price, so the selling price (S.P.) becomes: \[ \text{S.P.} = \text{M.P.} \times (1 – 0.10) = 1.20x \times 0.90 = 1.08x \]
Ravi saves Rs 15 due to the 10% discount on the marked price. Thus, the discount is: \[ \text{Discount} = \text{M.P.} \times 0.10 = 15 \] Substituting \( \text{M.P.} = 1.20x \): \[ 1.20x \times 0.10 = 15 \implies 0.12x = 15 \] Solving for \( x \): \[ x = \frac{15}{0.12} = 125 \]
Therefore, the cost price is \( x = 125 \), and the selling price is: \[ \text{S.P.} = 1.08x = 1.08 \times 125 = 135 \]
The profit made by Gopi is: \[ \text{Profit} = \text{S.P.} – \text{C.P.} = 135 – 125 = 10 \]
Correct Answer: Option 3 (10).
Question: Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of 5 : 3. If the total cultivation area of wheat and mustard are in the ratio 11 : 9, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is:
Solution:
Let the cultivation areas of wheat and mustard on Rajesh’s land be \( x \) and \( y \), respectively.
Let the cultivation areas of wheat and mustard on Vimal’s land be \( 5k \) and \( 3k \), respectively, since the ratio for Vimal is \( 5 : 3 \).
Total cultivation areas:
– Wheat: \( x + 5k \)
– Mustard: \( y + 3k \)
Given: \( \frac{x + 5k}{y + 3k} = \frac{11}{9} \) and \( x + y = 20 \).
For Vimal: \( 5k + 3k = 30 \implies k = \frac{15}{4} \).
Substituting \( k = \frac{15}{4} \):
\( \frac{x + \frac{75}{4}}{y + \frac{45}{4}} = \frac{11}{9} \).
Cross-multiplying and solving:
\( 9(4x + 75) = 11(4y + 45) \implies 36x – 44y = -180 \).
With \( x + y = 20 \):
\( x = 20 – y \).
Substituting: \( 9(20 – y) – 11y = -45 \implies y = 11.25 \), \( x = 8.75 \).
Ratio of wheat to mustard for Rajesh: \( \frac{8.75}{11.25} = \frac{7}{9} \).
Final Answer: 7 : 9
Question:
After two successive increments, Gopal’s salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was:
Detailed Solution:
Let the initial salary be \( x \), and let the percentage increase in the first increment be \( r\% \). Then, the percentage increase in the second increment is \( 2r\% \).
After the first increment, the salary becomes: \[ x \times \left(1 + \frac{r}{100}\right) \] After the second increment, the salary becomes: \[ x \times \left(1 + \frac{r}{100}\right) \times \left(1 + \frac{2r}{100}\right) \]
This final salary is given to be 187.5% of the initial salary: \[ x \times \left(1 + \frac{r}{100}\right) \times \left(1 + \frac{2r}{100}\right) = 1.875x \] Dividing through by \( x \): \[ \left(1 + \frac{r}{100}\right) \times \left(1 + \frac{2r}{100}\right) = 1.875 \] Expanding the left-hand side: \[ 1 + \frac{r}{100} + \frac{2r}{100} + \frac{2r^2}{100^2} = 1.875 \] Simplify: \[ 1 + \frac{3r}{100} + \frac{2r^2}{10000} = 1.875 \] Subtract 1 from both sides: \[ \frac{3r}{100} + \frac{2r^2}{10000} = 0.875 \] Multiply through by 10000 to eliminate fractions: \[ 300r + 2r^2 = 8750 \] Simplify: \[ 2r^2 + 300r – 8750 = 0 \] Divide through by 2: \[ r^2 + 150r – 4375 = 0 \]
Solve this quadratic equation using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \] Here, \( a = 1 \), \( b = 150 \), and \( c = -4375 \). Substituting: \[ r = \frac{-150 \pm \sqrt{150^2 – 4(1)(-4375)}}{2(1)} \] \[ r = \frac{-150 \pm \sqrt{22500 + 17500}}{2} \] \[ r = \frac{-150 \pm \sqrt{40000}}{2} \] \[ r = \frac{-150 \pm 200}{2} \] Taking the positive root (since percentages cannot be negative): \[ r = \frac{50}{2} = 25 \]
Correct Answer: Option 4 (25).
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. What is the largest number in the original set of three numbers?
Question:
Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is \( 25 : 36 \), then the minimum number of years required for the value of the investment to exceed Rs 20000 is:
Interactive Quiz
A certain amount of water was poured into a 300-liter container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in liters, that was initially poured into the container was?
Solution:
Let the amount of water initially poured into the container be \(x\) liters.
Step 1: The container initially contains \(x\) liters of water and \(300 – x\) liters of milk.
Step 2: An amount \(2x\) liters of the solution is removed, containing water and milk in the same ratio.
Water removed: \( \frac{2x^2}{300} \), Milk removed: \( \frac{600x – 2x^2}{300} \).
Step 3: The container is refilled with \(2x\) liters of water.
Step 4: The resulting solution has 72% milk. Solving the quadratic equation:
\[ x^2 – 450x + 12600 = 0 \]We find \(x = 30\).
Final Answer: The initial amount of water was \(30\) liters.
