Raju has 9000 with him and he wants to buy a mobile handset; but he finds that he has only 75% of the amount required to buy the handset. Therefore, he borrows 2000 from a friend. Then
UPSC-2019
Raju still does not have enough amount to buy the handset.
Raju has exactly the same amount as required to buy the handset.
Raju has enough amount to buy the handset and he will have 500 with him after buying the handset.
Raju has enough amount to buy the handset and he will have 1000 with him after buying the handset.
To solve this problem, we need to calculate the actual cost of the mobile handset and then determine if Raju has enough money after borrowing from his friend. We will use the given percentage to find the total cost.
A) Raju still does not have enough amount to buy the handset. This is correct because he needs ₹12,000 but only has ₹11,000.
In an examination, A has scored 20 marks more than B. If B has scored 5% less marks than A, how much has B scored ?
UPSC-2019
We are given two pieces of information: the difference in marks between A and B, and the percentage difference in marks for B compared to A. We need to find the marks scored by B. We can set up equations based on these statements and solve for the unknown marks.
Let A's score be \(A\) and B's score be \(B\).
From the first statement, "A has scored 20 marks more than B", we can write the equation:
\[ A = B + 20 \quad \text{(Equation 1)} \]From the second statement, "B has scored 5% less marks than A", we can express B's score in terms of A's score:
\[ B = A - (5\% \text{ of } A) \]\[ B = A - \left(\frac{5}{100} \times A\right) \]\[ B = A - \frac{A}{20} \]\[ B = \frac{20A - A}{20} \]\[ B = \frac{19A}{20} \quad \text{(Equation 2)} \]Now we have a system of two linear equations with two variables. Substitute Equation 1 into Equation 2:
\[ B = \frac{19(B + 20)}{20} \]Multiply both sides by 20 to eliminate the denominator:
\[ 20B = 19(B + 20) \]Distribute 19 on the right side:
\[ 20B = 19B + (19 \times 20) \]\[ 20B = 19B + 380 \]Subtract \(19B\) from both sides to solve for \(B\):
\[ 20B - 19B = 380 \]\[ B = 380 \]So, B has scored 380 marks.
To verify, if B scored 380, then A scored \(380 + 20 = 400\). Let's check if B's score is 5% less than A's score:
\[ 5\% \text{ of } 400 = \frac{5}{100} \times 400 = 5 \times 4 = 20 \]\[ A - 20 = 400 - 20 = 380 \]This matches B's score, so our calculation is correct.
Correct Option: B) 380
Rakesh had money to buy 8 mobile handsets [of a specific company. But the retailer offered very good discount on that particular handset. Rakesh could buy 10 mobile handsets with the amount he had. What was the discount the retailer offered ?
UPSC-2019
The problem asks us to calculate the percentage discount offered by a retailer. Rakesh initially had enough money to buy 8 mobile handsets. Due to a discount, he was able to buy 10 handsets with the same amount of money. We need to find the percentage discount.
Let the original price of one mobile handset be \(P\).
Let the amount of money Rakesh had be \(M\).
Rakesh had enough money to buy 8 mobile handsets at the original price.
\[ M = 8 \times P \quad \text{(Equation 1)} \]With the same amount of money \(M\), Rakesh could buy 10 mobile handsets after the discount.
Let the discounted price of one mobile handset be \(P'\).
\[ M = 10 \times P' \quad \text{(Equation 2)} \]From Equation 1 and Equation 2, we can equate the expressions for \(M\):
\[ 8 \times P = 10 \times P' \]Divide both sides by 10 to find \(P'\):
\[ P' = \frac{8}{10} \times P \]\[ P' = 0.8 \times P \]The discount amount per handset is the original price minus the discounted price:
\[ \text{Discount Amount} = P - P' \]\[ \text{Discount Amount} = P - 0.8P \]\[ \text{Discount Amount} = 0.2P \]Percentage discount is calculated as the discount amount divided by the original price, multiplied by 100.
\[ \text{Percentage Discount} = \frac{\text{Discount Amount}}{\text{Original Price}} \times 100 \% \]\[ \text{Percentage Discount} = \frac{0.2P}{P} \times 100 \% \]\[ \text{Percentage Discount} = 0.2 \times 100 \% \]\[ \text{Percentage Discount} = 20 \% \]B) The discount offered by the retailer was 20%.
The total emoluments of two persons are the same, but one gets allowances to the extent of 65% of his basic pay and the other gets allowances to the extent of 80% of his basic pay. The ratio of the basic pay of the former to the basic pay of the latter is:
UPSC-2016
The problem states that the total emoluments of two persons are the same. Emoluments consist of basic pay plus allowances. The allowances are given as a percentage of the basic pay. We need to find the ratio of their basic pays. We will set up equations for the total emoluments of both persons and then equate them to find the required ratio.
Let the basic pay of the first person be \(B_1\) and their allowances be \(A_1\).
Let the basic pay of the second person be \(B_2\) and their allowances be \(A_2\).
The total emoluments for the first person, \(E_1\), are the sum of their basic pay and allowances:
\[ E_1 = B_1 + A_1 \]The first person gets allowances to the extent of 65% of their basic pay. So,
\[ A_1 = 0.65 \times B_1 \]Substitute \(A_1\) into the equation for \(E_1\):
\[ E_1 = B_1 + 0.65 B_1 = 1.65 B_1 \]Similarly, for the second person, the total emoluments \(E_2\) are:
\[ E_2 = B_2 + A_2 \]The second person gets allowances to the extent of 80% of their basic pay. So,
\[ A_2 = 0.80 \times B_2 \]Substitute \(A_2\) into the equation for \(E_2\):
\[ E_2 = B_2 + 0.80 B_2 = 1.80 B_2 \]The problem states that the total emoluments of the two persons are the same, i.e., \(E_1 = E_2\).
Equate the expressions for \(E_1\) and \(E_2\):
\[ 1.65 B_1 = 1.80 B_2 \]We need to find the ratio of the basic pay of the former to the basic pay of the latter, which is \(B_1 : B_2\) or \(\frac{B_1}{B_2}\).
\[ \frac{B_1}{B_2} = \frac{1.80}{1.65} \]To simplify the ratio, multiply the numerator and denominator by 100 to remove decimals:
\[ \frac{B_1}{B_2} = \frac{180}{165} \]Divide both the numerator and the denominator by their greatest common divisor. Both are divisible by 5:
\[ \frac{180 \div 5}{165 \div 5} = \frac{36}{33} \]Both 36 and 33 are divisible by 3:
\[ \frac{36 \div 3}{33 \div 3} = \frac{12}{11} \]So, the ratio \(B_1 : B_2\) is \(12 : 11\).
Correct Option: D) 12:11
Two numbers X and Y are respectively 20% and 28% less than a third number Z. By what percentage is the number Y less than the number X?
UPSC-2016
To solve this problem, we will assume a convenient value for the third number Z. Then, we will calculate X and Y based on the given percentage reductions. Finally, we will determine the percentage by which Y is less than X using the formula for percentage decrease.
Assume a value for Z: Let the third number \(Z = 100\).
Calculate X: Number X is 20% less than Z. \[ X = Z - (20\% \text{ of } Z) = 100 - \left(\frac{20}{100} \times 100\right) = 100 - 20 = 80 \]
Calculate Y: Number Y is 28% less than Z. \[ Y = Z - (28\% \text{ of } Z) = 100 - \left(\frac{28}{100} \times 100\right) = 100 - 28 = 72 \]
Find the difference between X and Y: Difference \( = X - Y = 80 - 72 = 8 \)
Calculate the percentage by which Y is less than X: To find by what percentage Y is less than X, we compare the difference to X. \[ \text{Percentage less} = \frac{\text{Difference}}{X} \times 100\% \] \[ \text{Percentage less} = \frac{8}{80} \times 100\% = \frac{1}{10} \times 100\% = 10\% \]
Correct Option: B) 10%
The monthly average salary paid to all the employees of a company was Rs. 5000. The monthly average salary paid to male and female employees was Rs. 5200 and Rs. 4200 respectively. Then the percentage of males employed in the company is:
UPSC-2016
This problem can be solved using the concept of weighted averages or by applying the alligation rule. We will use the alligation rule as it provides a quick and efficient way to find the ratio of two quantities when their individual averages and the combined average are known.
Let the average salary of all employees be \(A = \text{Rs. } 5000\).
Let the average salary of male employees be \(A_M = \text{Rs. } 5200\).
Let the average salary of female employees be \(A_F = \text{Rs. } 4200\).
Let the number of male employees be \(N_M\) and the number of female employees be \(N_F\).
According to the alligation rule, we can set up the following diagram:
\[ \begin{array}{ccc} \text{Male Average} & & \text{Female Average} \\ 5200 & & 4200 \\ & \diagdown & \diagup \\ & 5000 & \\ & \diagup & \diagdown \\ (5000 - 4200) & : & (5200 - 5000) \\ 800 & : & 200 \end{array} \]
The ratio of the number of male employees to female employees is \(N_M : N_F\). From the alligation rule, this ratio is:
\[ N_M : N_F = (A - A_F) : (A_M - A) \]
\[ N_M : N_F = (5000 - 4200) : (5200 - 5000) \]
\[ N_M : N_F = 800 : 200 \]
Simplifying the ratio:
\[ N_M : N_F = 4 : 1 \]
This means for every 4 male employees, there is 1 female employee. The total number of parts in the ratio is \(4 + 1 = 5\).
The percentage of male employees is the number of male parts divided by the total parts, multiplied by 100:
\[ \text{Percentage of Males} = \frac{N_M}{N_M + N_F} \times 100\% \]
\[ \text{Percentage of Males} = \frac{4}{4 + 1} \times 100\% \]
\[ \text{Percentage of Males} = \frac{4}{5} \times 100\% \]
\[ \text{Percentage of Males} = 0.8 \times 100\% = 80\% \]
Correct Option: B) 80%
