Quantitative Aptitude PYQs

Given:

Divisors = 60, 135, 235

Same remainder in each case.

Formula Used:

If a number N gives the same remainder R when divided by a, b, and c, then (a - b), (b - c), and (c - a) will exactly divide (N - R).

The greatest such number is the HCF of (a - b), (b - c), and (c - a).

Calculations:

a = 60, b = 135, c = 235

a - b = 60 - 135 = -75

|a - b| = 75

b - c = 135 - 235 = -100

|b - c| = 100

c - a = 235 - 60 = 175

We need to find the HCF of 75, 100, and 175.

Prime factorization of 75 = 3 × 5 × 5 = 3 × 5²

Prime factorization of 100 = 2 × 2 × 5 × 5 = 2² × 5²

Prime factorization of 175 = 5 × 5 × 7 = 5² × 7

The common prime factor with the lowest power is 5².

HCF(75, 100, 175) = 5² = 25

∴ The greatest number is 25.

Given:

Isosceles triangle ABC with AB = AC

Altitude AD = 9 cm

Area of triangle ABC = 36 cm²

Formula used:

Area of triangle = 1/2 × base × height

Pythagorean theorem: a² + b² = c²

Calculations:

Area = 1/2 × base × height

⇒ 36 = 1/2 × BC × 9

⇒ BC = 8 cm

In right triangle ABD, AD is the altitude:

Let BD = x cm

Then BC = 2x = 8 cm ⇒ x = 4 cm

Using Pythagorean theorem in triangle ABD:

AB² = AD² + BD²

⇒ AB² = 9² + 4²

⇒ AB² = 81 + 16

⇒ AB² = 97

⇒ AB = √97 cm

Perimeter of triangle ABC = AB + AC + BC

⇒ Perimeter = 2AB + BC

⇒ Perimeter = 2×√97 + 8

⇒ Perimeter = 8 + 2√97 cm

∴ The correct answer is option (3).

Shortcut Trick

Factorize the numbers: 148 = 4 × 37 and 205 = 5 × 41.

Since there are no common prime factors, the numbers 148 and 205 are co-prime.

The Least Common Multiple (LCM) of co-prime numbers is their product.

LCM = 148 × 205 = 30340.

∴ The correct answer is 30340.

Alternate Method

Given: The two numbers are 148 and 205.

Formula Used: LCM = Product of the highest powers of all prime factors present in the numbers.

Prime factorization of 148:

⇒ 148 = 2 × 74

⇒ 148 = 2 × 2 × 37 = 2² × 37¹

Prime factorization of 205:

⇒ 205 = 5 × 41 = 5¹ × 41¹

Finding the LCM:

⇒ LCM = 2² × 5¹ × 37¹ × 41¹

⇒ LCM = 4 × 5 × 37 × 41

⇒ LCM = 20 × 1517

⇒ LCM = 30340

∴ The correct answer is 30340.

Additional Information

Relationship between HCF and LCM

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.

LCM of Co-prime Numbers

Two numbers are co-prime if their HCF is 1. For such numbers, the LCM is always the product of the numbers.

Division Method for LCM

LCM can also be found by dividing the numbers by common prime factors until all quotients are 1, then multiplying the divisors.

Given:

Winning candidate votes percentage = 60%

Winning candidate votes difference = 18,600

Formula used:

Let total votes be V

Winning candidate votes = 0.60V

Losing candidate votes = 0.40V

Difference = Winning candidate votes - Losing candidate votes

Calculation:

Difference = 0.60V - 0.40V = 18,600

⇒ 0.20V = 18,600

⇒ V = 93,000

Losing candidate votes = 0.40V

⇒ Losing candidate votes = 93,000 × 0.40

⇒ Losing candidate votes = 37,200

∴ The correct answer is option (3).

Given:

Total bananas = 12

Bananas sold at 20% profit = x

Bananas sold at 20% loss = 12 - x

Overall profit % = 10%

Formula Used:

Profit = Selling Price (SP) - Cost Price (CP)

Profit % = (Profit / CP) × 100

Overall Profit = Total SP - Total CP

Overall Profit % = (Overall Profit / Total CP) × 100

Calculations:

Let CP per banana be ₹1.

Total CP = ₹12.

SP of x bananas at 20% profit = x × (1 + 20/100) = x × 1.2 = 1.2x

SP of (12 - x) bananas at 20% loss = (12 - x) × (1 - 20/100) = (12 - x) × 0.8 = 9.6 - 0.8x

Total SP = 1.2x + 9.6 - 0.8x = 0.4x + 9.6

Overall Profit % = 10%

10 = ((0.4x + 9.6) - 12) / 12 × 100

⇒ 10 × 12 = (0.4x - 2.4) × 100

⇒ 120 = 40x - 240

⇒ 120 + 240 = 40x

⇒ x = 360 / 40

⇒ x = 9

∴ The value of x is 9.

Given:

A pipe is cut into two pieces. The smaller piece is 40% of the length of the pipe.

Formula used:

Percentage by which the smaller piece is shorter than the larger piece = (Difference in lengths / Length of larger piece) × 100

Calculations:

Let the total length of the pipe be 100 units.

Smaller piece = 40 units

Larger piece = 100 - 40 = 60 units

Difference in lengths = 60 - 40 = 20 units

⇒ Percentage = (20 / 60) × 100

⇒ Percentage = \(33\frac{1}{3}%\)

∴ The correct answer is option 1.

Given:

There are 36 words in 9 lines printed in a column of a book.

We need to find the number of words in a column of 51 lines.

Formula used:

Number of words per line = Total words / Number of lines

Total words in 51 lines = Number of words per line × Number of lines

Calculation:

Number of words per line = 36 / 9

⇒ Number of words per line = 4

Total words in 51 lines = 4 × 51

⇒ Total words in 51 lines = 204

∴ The correct answer is option (3).

Given:

Side a = 3 cm

Side b = 6 cm

Side c = 7 cm

Formula used:

Heron's formula: Area = \(\sqrt{s(s-a)(s-b)(s-c)} \)

Where s = semi-perimeter = \(\dfrac{a+b+c}{2}\)

Calculations:

s = \(\dfrac{3+6+7}{2}\)

⇒ s = 8 cm

Area = \(\sqrt{8(8-3)(8-6)(8-7)} \)

⇒ Area = \(\sqrt{8 \times 5 \times 2 \times 1} \)

⇒ Area = \(\sqrt{80} \)

⇒ Area = \(4\sqrt{5} \) cm²

∴ The correct answer is option (4).

Given:

Principal (P) = ₹1,300

Simple Interest (SI) = ₹650

Time (t) = 10 years

Formula used:

SI = \( \frac{P \times r \times t}{100} \)

Where, r = rate

Calculation:

650 = \( \frac{1300 \times r \times 10}{100} \)

⇒ 650 = \( 130 \times r \)

⇒ r = \( \frac{650}{130} \)

⇒ r = 5%

∴ The correct answer is option (4).

Given:

A sum of ₹1,900 is divided among A, B, and C in such a way that B gets ₹100 more than A and C gets ₹200 more than B.

Formula used:

Let A's share be x.

Then, B's share = x + 100

And C's share = x + 100 + 200 = x + 300

Calculation:

Total sum = A's share + B's share + C's share

1900 = x + (x + 100) + (x + 300)

⇒ 1900 = 3x + 400

⇒ 3x = 1500

⇒ x = 500

A's share = x = 500

B's share = x + 100 = 500 + 100 = 600

C's share = x + 300 = 500 + 300 = 800

Ratio of their shares = 500 ∶ 600 ∶ 800

⇒ 5 ∶ 6 ∶ 8

∴ The correct answer is option (1).

Given:

6 boys can complete a work in 18 hours.

Formula used:

Work = Number of Boys x Time

Work = Constant

Calculation:

6 boys x 18 hours = 16 boys x x hours

⇒ 6 x 18 = 16 x x

⇒ 108 = 16x

⇒ x = 108 ÷ 16

⇒ x = 6.75 hours or 6 (3/4) hours

∴ The correct answer is option 3.

Given:

Prime number greater than 7

Formula used:

If a prime number is greater than 7, it can be expressed in the form of 6k ± 1 where k is a positive integer.

Calculation:

Let the prime number be 6k ± 1

When divided by 6:

⇒ (6k ± 1) ÷ 6

The remainder is either 1 or 5.

∴ The correct answer is option (2).

Given:

Monthly Income of Amara (Initial) = ₹80,200

Decrease in Income = 11%

Formula used:

New Income = Initial Income - Decrease Amount

Where, Decrease Amount = Initial Income × (Percentage Decrease / 100)

Calculations:

Decrease Amount = ₹80,200 × (11 / 100)

⇒ Decrease Amount = ₹80,200 × 0.11

⇒ Decrease Amount = ₹8,822

New Income = ₹80,200 - ₹8,822

⇒ New Income = ₹71,378

∴ The correct answer is option (4).

Given:

The difference between compound interest for a year payable half-yearly and simple interest for a year is ₹729.

Rate of interest (r) = 6%

Time (t) = 1 year

Formula used:

Compound Interest (CI) for half-yearly = P\((1+\frac{r/2}{100})^2\) - P

Simple Interest (SI) = \(\frac{P \times r \times t}{100}\)

Calculation:

Let the principal amount be P.

CI for half-yearly = P\((1+\frac{6/2}{100})^2\) - P

⇒ CI = P\((1+\frac{3}{100})^2\) - P

⇒ CI = P\((1+0.03)^2\) - P

⇒ CI = P\((1.03)^2\) - P

⇒ CI = P(1.0609) - P

⇒ CI = 1.0609P - P

⇒ CI = 0.0609P

SI = \(\frac{P \times 6 \times 1}{100}\)

⇒ SI = 0.06P

Difference = CI - SI = ₹729

⇒ 0.0609P - 0.06P = ₹729

⇒ 0.0009P = ₹729

⇒ P = \(\frac{729}{0.0009}\)

⇒ P = ₹8,10,000

∴ The correct answer is option 3: ₹8,10,000.

Given:

The number 0.0623 is how much less than \( \frac{1}{16} \)?

Formula used:

Difference = \( \frac{1}{16} \) - 0.0623

Calculations:

\( \frac{1}{16} = 0.0625 \)

Difference = 0.0625 - 0.0623

⇒ Difference = 0.0002

⇒ Difference in fraction form is \( \frac{2}{10000} \)

⇒ \( \frac{2}{10000} = \frac{1}{5000} \)

∴ The correct answer is option (4).

Given:

The average of 5 consecutive even numbers = 70

Formula used:

Average of consecutive even numbers = Middle number

Calculation:

If the average is 70, then the middle number is 70.

The numbers are consecutive even numbers, so we have:

Middle number = 70

Two numbers before 70: 66, 68

Two numbers after 70: 72, 74

⇒ The largest even number = 74

∴ The correct answer is option (2).

Calculation:

0.7 × 0.7 + 0.7 ÷ 7 + 7 ÷ 700

⇒ 0.7 × 0.7 + 0.1 + 0.01

⇒ 0.49 + 0.1 + 0.01

⇒ 0.60

∴ The correct answer is 0.60

Given:

The area of a triangle is 182 cm² and the length of one of its sides is 28 cm.

Formula used:

Area of a triangle = 0.5 × base × height

Calculations:

182 = 0.5 × 28 × height

⇒ 182 = 14 × height

⇒ height = 182 / 14

⇒ height = 13

∴ The correct answer is option (2).

Calculation:

Let the added value be x 

⇒ (3 + x)/(5 + x) = 5/6

⇒ 18 + 6x = 25 + 5x 

⇒ x = 7

∴ The required answer is 7.

Given:

The price of an article is first reduced by 25% and then increased by 35%.

If the resulting price is Rs. 405, find the original price of the article.

Calculations:

Let the original price be x.

After 25% decrease: Price = x × (1 − 25/100)

After 35% increase: Price = new price × (1 + 35/100)

Resulting price = x × (1 − 25/100) × (1 + 35/100)

405 = x × 0.75 × 1.35

⇒ 405 = x × 1.0125

⇒ x = 405 ÷ 1.0125

⇒ x = 400

Therefore, the correct answer is option (4).