Quantitative Aptitude and Reasoning

171. A shopkeeper gives two consecutive discounts of 10% and 5% respectivel

A shopkeeper gives two consecutive discounts of 10% and 5% respectively on his items. He then adds 20% GST on his items. If an item has marked price ₹2,000, how much more or less of the actual price of the item a customer has to pay?

[amp_mcq option1=”2·6% less” option2=”2·6% more” option3=”Same price” option4=”5·2% more” correct=”option2″]

This question was previously asked in
UPSC CAPF – 2021
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The customer has to pay 2.6% more than the marked price.
To calculate the final price, apply consecutive discounts first to the marked price, and then add the GST to the discounted price. Compare this final price to the original marked price.
Marked Price = ₹2,000.
Price after 1st discount (10%): ₹2000 * (1 – 0.10) = ₹2000 * 0.90 = ₹1800.
Price after 2nd discount (5%): ₹1800 * (1 – 0.05) = ₹1800 * 0.95 = ₹1710. This is the price before GST.
Price after adding 20% GST: ₹1710 * (1 + 0.20) = ₹1710 * 1.20 = ₹2052.
The final price paid by the customer is ₹2052.
The original marked price was ₹2000.
Difference = Final Price – Marked Price = ₹2052 – ₹2000 = ₹52.
The customer pays ₹52 more.
Percentage difference = (Difference / Marked Price) * 100 = (52 / 2000) * 100 = (52/20) = 2.6%.
The customer pays 2.6% more than the actual (marked) price.

172. How many different words, with or without meaning, can be formed by us

How many different words, with or without meaning, can be formed by using the letters of the word COVID?

[amp_mcq option1=”60″ option2=”150″ option3=”100″ option4=”120″ correct=”option4″]

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UPSC CAPF – 2021
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The correct answer is D) 120.
The word COVID has 5 letters: C, O, V, I, D. All five letters are distinct. To find the number of different words that can be formed by using all the letters, we need to find the number of permutations of these 5 distinct letters. The number of permutations of ‘n’ distinct objects is given by n!.
In this case, n = 5.
Number of words = 5! = 5 × 4 × 3 × 2 × 1.
5! = 20 × 6 × 1 = 120.
This is a basic problem involving permutations. If there were repeated letters in the word, the formula would be adjusted by dividing by the factorial of the counts of repeated letters.

173. Suppose the nth term of a series is $1+\frac{n}{2}+\frac{n^2}{2}$. If

Suppose the nth term of a series is $1+\frac{n}{2}+\frac{n^2}{2}$. If there are 20 terms in the series, then the sum of the series is equal to

[amp_mcq option1=”1360″ option2=”1450″ option3=”1500″ option4=”1560″ correct=”option4″]

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The correct answer is D) 1560.
The nth term of the series is $a_n = 1+\frac{n}{2}+\frac{n^2}{2}$. We need to find the sum of the first 20 terms, $S_{20} = \sum_{n=1}^{20} a_n$.
$S_{20} = \sum_{n=1}^{20} (1+\frac{n}{2}+\frac{n^2}{2}) = \sum_{n=1}^{20} 1 + \frac{1}{2}\sum_{n=1}^{20} n + \frac{1}{2}\sum_{n=1}^{20} n^2$
Using standard summation formulas:
$\sum_{n=1}^{N} c = Nc$
$\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$
$\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$
Here, N = 20.
$\sum_{n=1}^{20} 1 = 20 \times 1 = 20$
$\sum_{n=1}^{20} n = \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = 10 \times 21 = 210$
$\sum_{n=1}^{20} n^2 = \frac{20(20+1)(2 \times 20 + 1)}{6} = \frac{20 \times 21 \times 41}{6} = \frac{10 \times 7 \times 41}{1} = 2870$
Now substitute these values back into the sum expression:
$S_{20} = 20 + \frac{1}{2}(210) + \frac{1}{2}(2870)$
$S_{20} = 20 + 105 + 1435$
$S_{20} = 125 + 1435 = 1560$.
The problem requires separating the given nth term into terms based on powers of n and applying standard summation formulas for constants, linear terms, and quadratic terms. Careful calculation is needed for each part and then summing them up.

174. If first March of a year is Sunday, which day will be the first Februa

If first March of a year is Sunday, which day will be the first February of the next year?

[amp_mcq option1=”Friday” option2=”Tuesday” option3=”Saturday” option4=”Monday” correct=”option2″]

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UPSC CAPF – 2021
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If the first of March of a year is Sunday, the first of February of the next year will be Tuesday.
– We need to find the number of days from March 1st of year Y to February 1st of year Y+1.
– The period covers the rest of year Y starting from March 1st, plus January and the first day of February in year Y+1.
– Number of days from March 1st (Y) to Feb 1st (Y+1):
– March (Y): 31 days
– April (Y): 30 days
– May (Y): 31 days
– June (Y): 30 days
– July (Y): 31 days
– August (Y): 31 days
– September (Y): 30 days
– October (Y): 31 days
– November (Y): 30 days
– December (Y): 31 days
– January (Y+1): 31 days
– February (Y+1): 1 day (up to the 1st)
– Total number of days = 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 31 + 1 = 338 days.
– The number of “odd days” is the remainder when the total number of days is divided by 7.
– 338 ÷ 7 = 48 with a remainder of 2.
– The number of odd days is 2.
– Since March 1st of the starting year is Sunday, February 1st of the next year will be Sunday + 2 days.
– Sunday + 2 days = Tuesday.
– This calculation does not depend on whether year Y or year Y+1 is a leap year, because the period starts after February 29th in year Y (if it’s a leap year) and ends before February 29th in year Y+1 (if it’s a leap year).
A standard non-leap year has 365 days (52 weeks and 1 odd day). A leap year has 366 days (52 weeks and 2 odd days). The day of the week advances by one day for a non-leap year and by two days for a leap year when crossing a full year. Here we are crossing almost a full year (March 1 to Feb 1 of next year).

175. A tree is at present 9 feet tall. If every year it grows 1/9 th of its

A tree is at present 9 feet tall. If every year it grows 1/9 th of its height, what will be the height of the tree after three years?

[amp_mcq option1=”12 feet” option2=”12.34 feet” option3=”13 feet” option4=”13.10 feet” correct=”option2″]

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If a tree is 9 feet tall and grows 1/9th of its height every year, its height after three years will be approximately 12.34 feet.
– The current height of the tree is H₀ = 9 feet.
– The growth each year is 1/9th of the height at the beginning of that year. This is a compound growth pattern.
– After 1 year, height H₁ = H₀ + (1/9)H₀ = H₀(1 + 1/9) = H₀(10/9).
– After 2 years, height H₂ = H₁ + (1/9)H₁ = H₁(1 + 1/9) = H₁(10/9) = H₀(10/9)(10/9) = H₀(10/9)².
– After 3 years, height H₃ = H₂ + (1/9)H₂ = H₂(1 + 1/9) = H₂(10/9) = H₀(10/9)³.
– Substitute the initial height H₀ = 9 feet:
– H₃ = 9 * (10/9)³ = 9 * (1000 / 729).
– H₃ = 9000 / 729.
– Simplify the fraction by dividing numerator and denominator by 9:
– H₃ = 1000 / 81.
– Calculate the decimal value: 1000 ÷ 81 ≈ 12.34567…
– Rounding to two decimal places, the height is approximately 12.35 feet. Among the given options, 12.34 feet is the closest value.
The alternative interpretation, where the tree grows 1/9th of the *original* height (1 foot) each year, would result in a height of 9 + 3*1 = 12 feet after 3 years. Since 12.34 is an option and 12 is also an option, the wording “1/9 th of its height” generally implies growth relative to the current height unless otherwise specified, leading to the compounding calculation.

176. A and B together can finish a job in 20 days. B and C together can fin

A and B together can finish a job in 20 days. B and C together can finish the same job in 30 days. If A and C together can finish it in 24 days, in how many days can A alone finish the job?

[amp_mcq option1=”35 2/7 days” option2=”37 1/2 days” option3=”34 2/7 days” option4=”33 2/7 days” correct=”option3″]

This question was previously asked in
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If A and B together finish a job in 20 days, B and C in 30 days, and A and C in 24 days, A alone can finish the job in 34 2/7 days.
– Let the amount of work done by A, B, and C in one day be a, b, and c respectively.
– A and B together finish the job in 20 days: a + b = 1/20 (Work done in one day)
– B and C together finish the job in 30 days: b + c = 1/30
– A and C together finish the job in 24 days: a + c = 1/24
– Add the three equations: (a + b) + (b + c) + (a + c) = 1/20 + 1/30 + 1/24
– 2a + 2b + 2c = (6 + 4 + 5) / 120 (LCM of 20, 30, 24 is 120)
– 2(a + b + c) = 15 / 120 = 1/8
– a + b + c = 1/16 (Combined work rate of A, B, and C in one day)
– To find the work rate of A (a), subtract the work rate of B and C together (b + c) from the combined work rate:
– a = (a + b + c) – (b + c)
– a = 1/16 – 1/30
– Find a common denominator (LCM of 16 and 30 is 240):
– a = (15/240) – (8/240) = (15 – 8) / 240 = 7/240
– A’s work rate is 7/240 of the job per day.
– The time taken by A alone to finish the job is the reciprocal of A’s work rate:
– Time for A = 1 / (7/240) = 240 / 7 days.
– Convert the improper fraction to a mixed number: 240 ÷ 7 = 34 with a remainder of 2.
– So, 240/7 days = 34 and 2/7 days.
Similarly, the time taken by B or C alone can be calculated:
b = (a+b+c) – (a+c) = 1/16 – 1/24 = (3 – 2)/48 = 1/48. Time for B = 48 days.
c = (a+b+c) – (a+b) = 1/16 – 1/20 = (5 – 4)/80 = 1/80. Time for C = 80 days.

177. The ratio of monthly incomes of A and B is 7 : 10. The ratio of their

The ratio of monthly incomes of A and B is 7 : 10. The ratio of their expenditures is 2 : 3. If each of A and B saves ₹ 1,000 per month, then what will be the monthly income of B?

[amp_mcq option1=”₹ 9,000″ option2=”₹ 10,000″ option3=”₹ 15,000″ option4=”₹ 12,000″ correct=”option2″]

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UPSC CAPF – 2021
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If the ratio of monthly incomes of A and B is 7:10, the ratio of their expenditures is 2:3, and each saves ₹1,000 per month, the monthly income of B is ₹10,000.
– Let the monthly income of A be 7x and the monthly income of B be 10x.
– Let the monthly expenditure of A be 2y and the monthly expenditure of B be 3y.
– Savings = Income – Expenditure.
– For A: 7x – 2y = 1000 (Equation 1)
– For B: 10x – 3y = 1000 (Equation 2)
– We need to solve this system of linear equations for x to find the incomes. Multiply Equation 1 by 3 and Equation 2 by 2 to eliminate y:
3 * (7x – 2y) = 3 * 1000 => 21x – 6y = 3000
2 * (10x – 3y) = 2 * 1000 => 20x – 6y = 2000
– Subtract the second new equation from the first new equation:
(21x – 6y) – (20x – 6y) = 3000 – 2000
21x – 20x = 1000
x = 1000.
– The monthly income of B is 10x.
– Monthly income of B = 10 * 1000 = ₹10,000.
We can also find the incomes and expenditures:
Income of A = 7 * 1000 = ₹7,000
Income of B = 10 * 1000 = ₹10,000
Substitute x=1000 into Equation 1: 7(1000) – 2y = 1000 => 7000 – 2y = 1000 => 2y = 6000 => y = 3000.
Expenditure of A = 2y = 2 * 3000 = ₹6,000. Savings of A = 7000 – 6000 = ₹1,000.
Expenditure of B = 3y = 3 * 3000 = ₹9,000. Savings of B = 10000 – 9000 = ₹1,000.
The savings match the given information, confirming the value of x.

178. Eight metallic balls of one centimetre radius each are melted into one

Eight metallic balls of one centimetre radius each are melted into one ball. The diameter of the new ball is

[amp_mcq option1=”2 cm” option2=”6 cm” option3=”4 cm” option4=”1 cm” correct=”option3″]

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When eight metallic balls of 1 cm radius each are melted into one ball, the diameter of the new ball is 4 cm.
– The volume of a sphere with radius ‘r’ is given by the formula V = (4/3)πr³.
– The volume of each small ball (radius r=1 cm) is V_small = (4/3)π(1)³ = (4/3)π cubic cm.
– When 8 such balls are melted, the total volume of metal is the sum of their volumes: Total Volume = 8 * V_small = 8 * (4/3)π cubic cm.
– This total volume is melted into a single new ball. Let the radius of the new ball be R. Its volume is V_new = (4/3)πR³.
– Equating the volumes: (4/3)πR³ = 8 * (4/3)π.
– Cancelling (4/3)π from both sides: R³ = 8.
– Taking the cube root: R = ³√8 = 2 cm.
– The question asks for the diameter of the new ball, which is twice the radius: Diameter = 2 * R = 2 * 2 cm = 4 cm.
When a substance is melted and recast, its volume remains conserved (assuming no loss of material in the process). This principle is fundamental in solving such mensuration problems involving volume changes.

179. Two years ago, the age of A was three times the age of B. If B is curr

Two years ago, the age of A was three times the age of B. If B is currently 9 years old, then after how many years, the age of A will be double of the age of B ?

[amp_mcq option1=”2 years” option2=”3 years” option3=”4 years” option4=”5 years” correct=”option4″]

This question was previously asked in
UPSC CAPF – 2020
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Let A be A’s current age and B be B’s current age.
We are given that B’s current age is 9 years, so B = 9.
Two years ago:
A’s age was A – 2.
B’s age was B – 2 = 9 – 2 = 7.
According to the problem, two years ago, the age of A was three times the age of B:
A – 2 = 3 * (B – 2)
A – 2 = 3 * 7
A – 2 = 21
A = 21 + 2 = 23.
So, A’s current age is 23 years.
We want to find the number of years (let’s call it x) after which A’s age will be double B’s age.
After x years:
A’s age will be A + x = 23 + x.
B’s age will be B + x = 9 + x.
We are given that after x years, A’s age will be double B’s age:
23 + x = 2 * (9 + x)
23 + x = 18 + 2x
Subtract x from both sides:
23 = 18 + 2x – x
23 = 18 + x
Subtract 18 from both sides:
x = 23 – 18
x = 5.
So, after 5 years, the age of A will be double the age of B.
– Set up variables for the current ages.
– Translate the information about past ages into equations based on the current age variables.
– Solve for the unknown current age(s).
– Set up equations for future ages based on adding an unknown number of years (x).
– Solve for x based on the given relationship between future ages.
Always verify the answer by plugging the calculated number of years back into the future age condition. After 5 years, A will be 23 + 5 = 28, and B will be 9 + 5 = 14. Since 28 = 2 * 14, the condition is met.

180. In a group of 100 children, 64 children like to play cricket, 53 child

In a group of 100 children, 64 children like to play cricket, 53 children like to play football and 20 children like to play both cricket and football. How many children do NOT like to play cricket or football ?

[amp_mcq option1=”3″ option2=”5″ option3=”7″ option4=”9″ correct=”option1″]

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This problem can be solved using the principle of inclusion-exclusion or a Venn diagram.
Let C be the set of children who like cricket and F be the set of children who like football.
Total children = 100.
Number of children who like cricket, |C| = 64.
Number of children who like football, |F| = 53.
Number of children who like both cricket and football, |C ∩ F| = 20.
The number of children who like at least one of the games (either cricket or football or both) is given by the union of the two sets:
|C U F| = |C| + |F| – |C ∩ F|
|C U F| = 64 + 53 – 20
|C U F| = 117 – 20
|C U F| = 97.
This means 97 children like either cricket or football or both.
The number of children who do NOT like to play cricket or football is the total number of children minus the number of children who like at least one game:
Children who like neither = Total children – |C U F|
Children who like neither = 100 – 97
Children who like neither = 3.
– Use the formula for the union of two sets: |A U B| = |A| + |B| – |A ∩ B|.
– Understand that the number of elements in the union represents those who like at least one of the items.
– Subtract the number who like at least one from the total number to find those who like neither.
This type of problem is a classic application of basic set theory. A Venn diagram could also be used: Draw two overlapping circles for Cricket and Football. The overlap is 20. The part of the Cricket circle only is 64 – 20 = 44. The part of the Football circle only is 53 – 20 = 33. The total inside the circles is 44 + 20 + 33 = 97. Those outside the circles are 100 – 97 = 3.
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