Quantitative Aptitude and Reasoning

161. A number is 124 more than its one-third. What is that number?

A number is 124 more than its one-third. What is that number?

[amp_mcq option1=”194″ option2=”180″ option3=”189″ option4=”186″ correct=”option4″]

This question was previously asked in
UPSC CAPF – 2022
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Let the number be $x$.
The problem states that “a number is 124 more than its one-third”.
This can be written as an equation:
$x = 124 + \frac{1}{3}x$

To solve for $x$, first move the term with $x$ to one side:
$x – \frac{1}{3}x = 124$

Combine the $x$ terms:
$\left(1 – \frac{1}{3}\right)x = 124$
$\left(\frac{3}{3} – \frac{1}{3}\right)x = 124$
$\frac{2}{3}x = 124$

Multiply both sides by $\frac{3}{2}$ to isolate $x$:
$x = 124 \times \frac{3}{2}$
$x = (124/2) \times 3$
$x = 62 \times 3$
$x = 186$

The number is 186.

– Translating a word problem into a mathematical equation.
– Solving a linear equation involving fractions.
To verify the answer, check if 186 is 124 more than its one-third:
One-third of 186 is $186 / 3 = 62$.
124 more than 62 is $124 + 62 = 186$.
The condition is satisfied, so the answer is correct.

162. The average age of father and elder son is 35 years, the average age o

The average age of father and elder son is 35 years, the average age of father and younger son is 32 years and the average age of the two sons is 17 years. What is the average age of the father and his two sons?

[amp_mcq option1=”30 years” option2=”27 years” option3=”28 years” option4=”29 years” correct=”option3″]

This question was previously asked in
UPSC CAPF – 2022
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Let the age of the father be $F$, the elder son be $E$, and the younger son be $Y$.
Given information:
1. Average age of father and elder son is 35 years: $(F + E)/2 = 35 \implies F + E = 70$.
2. Average age of father and younger son is 32 years: $(F + Y)/2 = 32 \implies F + Y = 64$.
3. Average age of the two sons is 17 years: $(E + Y)/2 = 17 \implies E + Y = 34$.

We need to find the average age of the father and his two sons, which is $(F + E + Y)/3$.

Adding the three equations:
$(F + E) + (F + Y) + (E + Y) = 70 + 64 + 34$
$2F + 2E + 2Y = 168$
$2(F + E + Y) = 168$
$F + E + Y = 168 / 2 = 84$.

The average age of the father and his two sons is $(F + E + Y)/3 = 84/3 = 28$ years.

– Understanding and setting up equations from average definitions.
– Solving a system of three linear equations with three variables.
– Calculating the final average.
We can also find individual ages:
$(F+E+Y) = 84$
Substitute $(E+Y)=34$: $F + 34 = 84 \implies F = 50$.
Substitute $F=50$ into $F+E=70$: $50 + E = 70 \implies E = 20$.
Substitute $F=50$ into $F+Y=64$: $50 + Y = 64 \implies Y = 14$.
Check with $E+Y=34$: $20 + 14 = 34$. The ages are 50, 20, and 14 years.
Average age of father and sons = $(50 + 20 + 14)/3 = 84/3 = 28$.

163. A runner’s average speed reduces by 25% every hour. If he runs 16 km i

A runner’s average speed reduces by 25% every hour. If he runs 16 km in the first hour and he runs for 3 hours, then what is his overall average speed?

[amp_mcq option1=”12 km/hr” option2=”12.33 km/hr” option3=”10.33 km/hr” option4=”13 km/hr” correct=”option2″]

This question was previously asked in
UPSC CAPF – 2022
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The runner runs for 3 hours. In the first hour, the speed is 16 km/hr, so the distance covered is $16 \times 1 = 16$ km. The speed reduces by 25% every hour. Speed in the second hour = $16 \times (1 – 0.25) = 16 \times 0.75 = 12$ km/hr. Distance covered in the second hour = $12 \times 1 = 12$ km. Speed in the third hour = $12 \times (1 – 0.25) = 12 \times 0.75 = 9$ km/hr. Distance covered in the third hour = $9 \times 1 = 9$ km. The total distance covered in 3 hours is $16 + 12 + 9 = 37$ km. The total time taken is 3 hours. The overall average speed is Total Distance / Total Time = $\frac{37}{3}$ km/hr. $\frac{37}{3} \approx 12.333$ km/hr.
Average speed is calculated as total distance divided by total time. The speed changes each hour, so we must calculate the distance covered in each hour and sum them up.
Note that average speed is not simply the average of the speeds in each hour (which would be $(16+12+9)/3 = 37/3 \approx 12.33$ km/hr in this specific case because each speed was maintained for one hour). The formula for average speed is always Total Distance / Total Time, which is the correct approach here.

164. Suppose a bank gives an interest of 10% per annum compounded annually

Suppose a bank gives an interest of 10% per annum compounded annually for a fixed deposit for a period of two years. What should be the simple interest rate per annum if the maturity amount after two years is to remain the same?

[amp_mcq option1=”10%” option2=”10.5%” option3=”11%” option4=”12%” correct=”option2″]

This question was previously asked in
UPSC CAPF – 2022
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Let the principal amount be P. For compound interest at 10% per annum compounded annually for 2 years, the maturity amount is $A_{CI} = P(1 + \frac{10}{100})^2 = P(1.1)^2 = 1.21P$. For simple interest over 2 years with an annual rate $R_{SI}$, the maturity amount is $A_{SI} = P + \text{Interest} = P + \frac{P \times R_{SI} \times 2}{100} = P(1 + \frac{2R_{SI}}{100})$. For the maturity amounts to be the same, $1.21P = P(1 + \frac{2R_{SI}}{100})$. Dividing by P (assuming P > 0), we get $1.21 = 1 + \frac{2R_{SI}}{100}$. Subtracting 1 from both sides, $0.21 = \frac{2R_{SI}}{100}$. Multiplying by 100, $21 = 2R_{SI}$. Therefore, $R_{SI} = \frac{21}{2} = 10.5$. The simple interest rate should be 10.5% per annum.
The problem requires comparing the maturity amounts obtained from compound interest and simple interest over the same period and finding the equivalent simple interest rate that yields the same amount.
Over a period of more than one year, compound interest will always yield a higher maturity amount than simple interest for the same principal and nominal rate, because interest earned in previous periods also earns interest. To get the same maturity amount, the simple interest rate must be higher than the compound interest rate (except for the first year, where they are equal).

165. A solid spherical ball made of iron is melted and two new balls are ma

A solid spherical ball made of iron is melted and two new balls are made whose diameters are in the ratio of 1:2. The ratio of the volume of the smaller new ball to the original ball is

[amp_mcq option1=”1:3″ option2=”1:5″ option3=”2:9″ option4=”1:9″ correct=”option4″]

This question was previously asked in
UPSC CAPF – 2022
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Let the radius of the original ball be $R$. Its volume is $V_{orig} = \frac{4}{3}\pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = \frac{4}{3}\pi r_1^3$ and $V_2 = \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (2r_1)^3 = \frac{4}{3}\pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi (8r_1^3) = \frac{4}{3}\pi (r_1^3 + 8r_1^3) = \frac{4}{3}\pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $\frac{V_1}{V_{orig}} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi (9r_1^3)} = \frac{r_1^3}{9r_1^3} = \frac{1}{9}$. The ratio is 1:9.
When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V \propto r^3$).
If two spheres have radii in the ratio $r_a:r_b = k:l$, their volumes are in the ratio $V_a:V_b = r_a^3:r_b^3 = k^3:l^3$. In this problem, the ratio of radii of the two new spheres is 1:2, so their volumes are in the ratio $1^3:2^3 = 1:8$. The total volume of the two new spheres is proportional to $1+8=9$ units. The smaller new sphere has a volume proportional to 1 unit. The original sphere’s volume is equal to the sum of the volumes of the new spheres, which is proportional to 9 units. Thus, the ratio of the smaller new sphere’s volume to the original sphere’s volume is 1:9.

166. A person buys an item from a shop for which the shopkeeper offers a di

A person buys an item from a shop for which the shopkeeper offers a discount of 10% on the marked price. The person pays using an e-wallet which gives 10% cash back. Which one of the following is the value of effective discount?

[amp_mcq option1=”20%” option2=”18%” option3=”19%” option4=”21%” correct=”option3″]

This question was previously asked in
UPSC CAPF – 2022
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Let the marked price be Rs. 100. The shopkeeper offers a 10% discount, so the price after the shop discount is $100 \times (1 – 0.10) = 100 \times 0.90 = Rs. 90$. The person pays Rs. 90 using an e-wallet which gives 10% cashback on the amount paid. The cashback amount is $90 \times 0.10 = Rs. 9$. The total reduction from the marked price is the shop discount (Rs. 10) plus the cashback (Rs. 9), which is $10 + 9 = Rs. 19$. The effective discount is the total reduction as a percentage of the marked price: $(\frac{19}{100}) \times 100\% = 19\%$.
The e-wallet cashback is applied to the price *after* the initial discount, not the original marked price. Effective discount is the total benefit (discount + cashback) expressed as a percentage of the original price.
A common mistake is to simply add the percentages (10% + 10% = 20%), which is incorrect because the second percentage is applied to a reduced amount. This problem involves successive discounts/benefits. If the marked price is M, the price paid is $M(1-d_1)$. The cashback is $C = M(1-d_1) \times c$, where $d_1$ is the shop discount rate and $c$ is the cashback rate. The effective price paid is $M(1-d_1) – C = M(1-d_1) – M(1-d_1)c = M(1-d_1)(1-c)$. The total reduction is $M – M(1-d_1)(1-c)$. For this problem, it is $M – M(0.9)(0.9) = M – 0.81M = 0.19M$. The effective discount rate is $0.19 \times 100\% = 19\%$.

167. Which one of the following is the difference of the sum of cubes of fi

Which one of the following is the difference of the sum of cubes of first ten natural numbers and the sum of squares of first ten natural numbers?

[amp_mcq option1=”2400″ option2=”2640″ option3=”2880″ option4=”2000″ correct=”option2″]

This question was previously asked in
UPSC CAPF – 2022
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The sum of the first ten natural numbers’ cubes is calculated using the formula $[\frac{n(n+1)}{2}]^2$ with n=10, giving $[\frac{10(11)}{2}]^2 = 55^2 = 3025$. The sum of the first ten natural numbers’ squares is calculated using the formula $\frac{n(n+1)(2n+1)}{6}$ with n=10, giving $\frac{10(11)(21)}{6} = \frac{2310}{6} = 385$. The difference is $3025 – 385 = 2640$.
The problem requires knowing the formulas for the sum of cubes and the sum of squares of the first ‘n’ natural numbers and applying them for n=10.
The formula for the sum of the first n natural numbers is $\frac{n(n+1)}{2}$. The sum of cubes of the first n natural numbers is the square of the sum of the first n natural numbers: $(\sum_{k=1}^n k)^2 = (\frac{n(n+1)}{2})^2$. The sum of squares of the first n natural numbers is $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.

168. If 35% of a number is 416 more than 27% of the same number, then the n

If 35% of a number is 416 more than 27% of the same number, then the number is

[amp_mcq option1=”5200″ option2=”2600″ option3=”3900″ option4=”3328″ correct=”option1″]

This question was previously asked in
UPSC CAPF – 2021
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The number is 5200.
Translate the word problem into an algebraic equation involving percentages of an unknown number.
Let the number be N.
35% of the number = 0.35 * N
27% of the number = 0.27 * N
The problem states that 35% of the number is 416 more than 27% of the same number.
So, 0.35N = 0.27N + 416
Subtract 0.27N from both sides:
0.35N – 0.27N = 416
0.08N = 416
To find N, divide 416 by 0.08:
N = 416 / 0.08
N = 416 / (8/100)
N = 416 * (100/8)
N = (416/8) * 100
N = 52 * 100
N = 5200
The number is 5200.

169. How much water is to be added to 75 ml of alcohol so that the mixture

How much water is to be added to 75 ml of alcohol so that the mixture contains 25% of alcohol?

[amp_mcq option1=”100 ml” option2=”225 ml” option3=”250 ml” option4=”125 ml” correct=”option2″]

This question was previously asked in
UPSC CAPF – 2021
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225 ml of water is to be added to the alcohol.
To achieve a target percentage of a component in a mixture, set up an equation where the amount of the component divided by the total volume of the mixture equals the target percentage (as a decimal).
Initial amount of alcohol = 75 ml.
The initial mixture is pure alcohol, so it contains 75 ml of alcohol and 0 ml of water, with a total volume of 75 ml.
Let ‘x’ be the amount of water added in ml.
After adding ‘x’ ml of water, the amount of alcohol remains 75 ml, and the total volume of the mixture becomes (75 + x) ml.
The target is for the mixture to contain 25% alcohol.
Percentage of alcohol = (Amount of alcohol / Total volume of mixture) * 100
25 = (75 / (75 + x)) * 100
Dividing both sides by 100:
0.25 = 75 / (75 + x)
Multiply both sides by (75 + x):
0.25 * (75 + x) = 75
18.75 + 0.25x = 75
Subtract 18.75 from both sides:
0.25x = 75 – 18.75
0.25x = 56.25
Divide by 0.25:
x = 56.25 / 0.25
x = 5625 / 25
x = 225
Therefore, 225 ml of water must be added.

170. The difference of compound interest and simple interest of a sum of mo

The difference of compound interest and simple interest of a sum of money at the rate of 5% per year for 2 years is ₹250. The sum is

[amp_mcq option1=”₹1,00,000″ option2=”₹80,000″ option3=”₹40,000″ option4=”₹1,20,000″ correct=”option1″]

This question was previously asked in
UPSC CAPF – 2021
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The principal sum of money is ₹1,00,000.
The difference between compound interest (CI) and simple interest (SI) for a sum P at rate R% per annum for 2 years is given by the formula: CI – SI = P * (R/100)^2.
Given:
Rate (R) = 5% per year
Time (T) = 2 years
Difference (CI – SI) = ₹250
Let the sum be P.
Using the formula: CI – SI = P * (R/100)^2
250 = P * (5/100)^2
250 = P * (1/20)^2
250 = P * (1/400)
P = 250 * 400
P = 1,00,000
Thus, the sum is ₹1,00,000.
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