Quantitative Aptitude PYQs

Concept:

HCF: It is the product of the smallest power of the common factors involved in

the prime factorization.

HCF of (am - 1) & (an - 1) is aHCF (m, n) -1

Calculation:

We have 235 - 1 &  291 - 1

By using the above concept

HCF = [2HCF (35, 91) - 1]

Let's find the HCF of (35, 91) first

35 = 5 × 7

91 = 13 × 7

HCF of (35, 91) = 7

Now, the required value

HCF = (2⁷ - 1) = 128 - 1 = 127.

Calculation:

We need to count the number of trailing zeros in that product.

It is obtained by multiplying a factor of 10, which is equivalent to multiplying

by 2 and 5.

Since every even number contributes a factor of 2 and every multiple of 5

contributes a factor of 5, we need to determine the number of multiples of 5

among the numbers from 1 to 29.

The numbers 5, 10, 15, 20, and 25 have one factor of 5 each, while 25 contributes

an additional factor of 5. Hence, there are 6 factors of 5 in total.

Therefore, the largest power of 10 that divides the product of the numbers

from 29 to 1 is 10⁶.

∴ The largest power of 10 that divides the given product is 6.

Shortcut Trick

When dividing 65 by 11, the nearest multiple of 11 is 66, which gives a negative remainder of −1.

Substitute this negative remainder into the given expression: (−1)⁹⁹.

Since 99 is an odd power, (−1)⁹⁹ = −1.

To convert the negative remainder into a positive one, add the divisor: 11 + (−1) = 10.

∴ The correct answer is 10.

Alternate Method

Given:

Expression = 65⁹⁹ ÷ 11

Concept Used:

If an expression is of the form (ax − 1)ⁿ ÷ a, the remainder is (−1)ⁿ.

If n is an odd number, remainder is −1, which means the actual positive remainder is (a − 1).

Calculation:

⇒ 65⁹⁹ ÷ 11

⇒ Expressing 65 in terms of a multiple of 11: 65 = 66 − 1 = 11 × 6 − 1

⇒ (11 × 6 − 1)⁹⁹ ÷ 11

⇒ Applying the concept, the remainder depends only on the (−1) term.

⇒ Remainder = (−1)⁹⁹

⇒ Remainder = −1

⇒ As the remainder cannot be negative, we add the divisor (11) to it.

⇒ Final positive remainder = 11 − 1 = 10

∴ The correct answer is 10.

Additional Information

Fermat's Little Theorem

If p is a prime number and a is not divisible by p, then a^p−1 leaves a remainder of 1 when divided by p.

Wilson's Theorem

For any prime number p, the expression (p − 1)! leaves a remainder of −1 when divided by p.

Negative Remainder Concept

A negative remainder −R indicates that the number is R less than a multiple of the divisor D. The true positive remainder is calculated as D − R.

Shortcut Trick

For a quadratic equation ax² + bx + c = 0, the difference of roots is given by |α − β| = √D / |a|.

Rewrite the equation: x² − bx + (c − 5) = 0, where a = 1, b = −b, and constant term = (c − 5).

Difference = 5 ⇒ 5 = √((−b)² − 4 × 1 × (c − 5)) / 1

Squaring both sides: 25 = b² − 4c + 20 ⇒ b² = 4c + 5.

∴ The correct answer is b² = 4c + 5.

Alternate Method

Given:

Equation: x² − bx + c = 5

Difference of roots (|α − β|) = 5

Formula Used:

1. Sum of roots (α + β) = − (coefficient of x) / (coefficient of x²)
2. Product of roots (αβ) = (constant term) / (coefficient of x²)
3. (α − β)² = (α + β)² − 4αβ

Calculations:

The standard form of the equation is x² − bx + (c − 5) = 0.

⇒ Sum of roots (α + β) = −(−b) / 1 = b

⇒ Product of roots (αβ) = (c − 5) / 1 = c − 5

⇒ Given |α − β| = 5

⇒ (α − β)² = 5² = 25

Substituting sum and product values:

⇒ b² − 4(c − 5) = 25

⇒ b² − 4c + 20 = 25

⇒ b² − 4c = 25 − 20

⇒ b² − 4c = 5

⇒ b² = 4c + 5

∴ The correct answer is b² = 4c + 5.

Additional Information

Discriminant and Difference of Roots

The discriminant (D) is b² − 4ac. The difference of roots is directly related to D by the formula |α − β| = √D / |a|.

Nature of Roots

If D > 0, roots are real and distinct. If D = 0, roots are real and equal. If D < 0, roots are imaginary/complex conjugates.

Vieta's Formulas

For any polynomial equation, the relations between coefficients and roots are established by Vieta's formulas (Sum = −b/a, Product = c/a).

Calculation:

Let, x number of persons taking milk only.

According to the question

60 + 15 + (x - 5) + (50 - x) + x = 150

120 + x = 150

x = 150 - 120 = 30

∴ The required value is 30.

Concept:

 Total investment = Investment × Time  of Investment 

Calculation:

A's total investment = 2 × 6 = 12 unit

B's total investment = 3 × 5 = 15 unit

C's total investment = 4 × 4 = 16 unit

D's total investment = 5 × 3 = 15 unit

E's total investment = 6 × 2 = 12 unit

C has the highest total investment, which is 16 unit.

Since, the profit is directly proportional to the total investment.

Therefore, C would receive the highest amount of profit.

Concept:

Prime Number:

• A prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself.
• In other words, a prime number is only divisible by 1 and itself.
• To determine whether a number is prime or not: Test its divisibility by all the integers from 2 to the square root of the number.
• If none of these integers divide the number, then the number is prime.

 

Calculation:

437 = 19 × 23    

It is not a prime number 

1073 = 29 × 37 

It is not a prime number 

Check for 797:   √797 ≈ 28.2.

The prime numbers below 28 are

2, 3, 5, 7, 11, 13, 17, 19, 23

797 is not divisible by any of the given numbers 

Therefore, we can conclude that 797 is a prime number.

∴ Out of the three given numbers, only 1 i.e. 797 number is prime.

Calculation:

Let's assume that C can finish the work in x days.

Since B can do the work 1.5 times as fast as C,

B can finish the work in (\(\frac{x}{1.5}\)) i.e. \(\frac{2x}{3}\) days.

Similarly,

A can do the work twice as fast as B, 

A can finish the work in \( \frac{\frac{2x}{3}}{2}=\frac{x}{3}\) days.

Now, let's consider the combined rate of work of all three persons:

One day work of A = 3/x

One day work of B = 3/2x

One day work of C = 1/x

According to the question

3/x + 3/(2x) + 1/x = 1/12

Multiplying both sides by 12x gives:

36 + 18 + 12 = x

Therefore, x = 66.

Hence, C can finish the work alone in 66 days.

Shortcut TrickRatio of Speed is also known as the ratio of efficiency.

Ratio of efficiency of A : B : C = 6 : 3 : 2

Total work = efficiency × days = 11 × 12

Time taken by C = Total work/Efficiency of C

⇒ (11 × 12)/2 = 66 days

∴ The correct answer is 66 days.

Calculation:

Let's assume that the 2-digit number is represented as 10x + y,

where x and y are the tens and units digits respectively.

We are given that the sum of the digits is 12, so we can write:

x + y = 12

Digits are reversed, the new number is represented as 10y + x.

According to the question

10y + x = 10x + y + 18

Simplifying this equation, we get:

9y - 9x = 18

Dividing both sides by 9, we get:

y - x = 2

∴ The difference between the digits in the number is 2.

Concept:

Relative speed:

If two trains are traveling with speeds of magnitude u & v respectively, then the relative speed will be

u + v,  if both trains are traveling in the opposite direction.

u - v,  if both trains are traveling in the same direction.

Formula used:

Speed = distance/ times

Calculation:

Let the speed of train 1 (of length 200 m) is 'u' m/s & speed of train 2

(of length 150 m) be 'v' m/s.

Given that, the man is sitting in the second train & a train of length 200 meters across a man (here, 150 m is irrelevant data).

By using the above concept,

When both trains are traveling in the same direction,

u - v = 200/20

u - v = 10           -----(1)

When both trains are traveling in the opposite direction,

u + v = 200/10 

u + v =  20         -----(2)

Adding both equations, we will get

2u = 30 or u =15 m/s

Since, 1 m/s = 3.6 km/hr

u = 15 × 3.6 = 54 km/hr

∴ The correct answer is 54 km/hr.

Calculation:

2a = 3b = 6c = 9d = 12e = 18f = k (let)

a = k/2, b = k/3, c = k/6, d = k/9, e = k/12, f =k/18

LCM of (2, 3, 6, 9, 12, 18) is 36.

Multiply the a, b, c, d, e & f by 36, 

a = 18k, b = 12k, c = 6k, 

d = 4k, e = 3k, f = 2k

Hence, the required value

(a + b)/(c + d + e + f)

(18k + 12k)/(6k + 4k + 3k + 2k)

30k/15k = 2

∴ (a + b)/(c + d + e + f) = 2

Formula used:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

Solution of quadratic equation ax² + bx + c = 0 is given by

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

Given that 

a + b + c = 0

b = - (a + c)              ------(1)

Root of quadratic equation ax2 + bx + c = 0  are

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {(a+c) \pm \sqrt{[-(a+c)]^2-4ac} \over 2a}\)

\(x = {(a+c) \pm \sqrt{a^2+2ac+c^2-4ac} \over 2a}\)

\(x = {(a+c) \pm \sqrt{a^2-2ac+c^2} \over 2a}\)

\(x = {(a+c) \pm \sqrt{(a-c)^2} \over 2a}\)

\(x = {(a+c) \pm {(a-c)} \over 2a}\)

Considering +ve & -ve signs alternatively,

\(x = {(a+c) +{(a-c)} \over 2a}\) & \(x = {(a+c) -{(a-c)} \over 2a}\) 

\(x = \frac{{2a}}{ 2a}\) & \(x = \frac{{2c}}{ 2a}\)

 \(x = 1, \ \ x = \frac{c}{a}\)

Formula used:

Profit % = [(SP - CP)/CP] × 100

Calculation:

The vendor initially bought 100 bananas at a cost of 1 Rs each, so the total cost is 100 Rs.

12% i.e. 12 bananas lost during transportation.

Remaing bananas are 100 - 12 = 88.

Since the remaining bananas are sold at a 4 % profit

Selling Price of  88 bananas = 100 (1 + 4/100) = 104

Selling Price of 1 banana = 104/88 = 13/11

Selling Price of 100 bananas = 100 × (13/11) = 118.18

As the cost price of 100 bananas was 100 Rs.

Profit % = [(118.18 - 100)/100] × 100

∴ Profit % = 18.18% or \(18 \frac{2}{11} %\)

Formula used:

(a - b) (a + b) = a² - b²

(a² + 2ab + b²) = (a + b)²

Calculation:

 Given that,

α  = √[(5 + 3√2) (5 - 3√2)]

Since, (a - b) (a + b) = a2 - b2

α = √[(5² - (3√2)²]

α = √(25 - 18) = √7

Let the required square root is β 

β = √(8 + 2√7) = √(7 + 1 + 2√7)

β = √[(√7)² + 2.1.√7 +  (1)²]

β = √(√7 + 1)² 

Since, (a2 + 2ab + b2) = (a + b)2

β = √7 + 1

Concept:

Even Number: Any number that can be exactly divided by 2 is called an even number. Even numbers always end up with the last digit as 0, 2, 4, 6, or 8. 

Calculation:

Let's assume the odd number = 7

Even power of 7 = 7² = 49

49 = 8 × 6 + 1

Similarly, we can also check for another value.

∴ The correct answer is 1.

Calculation:

Statement 1:

Let n be 1

So, \(\frac{n\left(n^2+2\right)}{3} = ​​\frac{1\left(1^2+2\right)}{3} \)

⇒ \( ​​\frac{1\left(1+2\right)}{3} \)

⇒ \(​​\frac{3}{3} \)

⇒ 1

So, statement 1 is correct.

Statement 2: 

Let m be 3

So, \(\frac{\mathrm{m}^4+4 \mathrm{~m}^2+11}{16} = \frac{\mathrm{3}^4+4 (3)^2+11}{16} \)

⇒ \(\frac{81+4 \times9+11}{16} \)

⇒ \(\frac{81+36+11}{16} \)

⇒ \(\frac{128}{16} \)

⇒ 8

So, statement 2 is correct.

∴ The required answer is Option 3.

Concept:

Divisibility rule of 5: A number is divisible by 5 only if the unit digit is 0 or 5.

Calculation:

Let n - 1, n & n + 1 be 6, 7 & 8.

⇒ n² + 1 = 7² + 1 = 50

⇒ n2 - 1 = 72 - 1 = 48

⇒ n2 + n = 72 + 7 =  56

⇒ n2 - 7 = 72 - 7 = 42

∴ 5 divides n² + 1

Concept:

Divisibility rule of 11: The given number can only be completely divided by 11 if the difference between the sum of digits at the odd position and the sum of digits at the even position in a number is 0 or 11.

Calculation:

Let the required number be n.

According to the question, 

n the remainder 7 when divided by 11. Therefore, n - 7 will be completely divisible by 11.

Now, let's check the options

Option:1   99981 - 7 = 99974 which is not divisible by 11.

Option:2   99988 - 7 = 99981 which is not divisible by 11.

Option:3   99997 - 7 = 99990 which is divisible by 11.

Option:4   99999 - 7 = 99992 which is not divisible by 11.

∴ The required value is 99997.

Formula used:

Present Value = Future Value /\([1+\frac{R× T}{100}]\) 

Calculation:

Given that 

Future Value = Rs. 20,000

Interest Rate = 5%, Time = 5 years

Present Value of A's debt = 20,000 /[1 + (0.05 ×  5)]

⇒ 20,000 / (1 + 0.25) = Rs. 16,000

Again, 

Future Value = Rs. 12,000, 

Interest Rate = 5% , Time = 4 years

Present Value of B's debt = 12,000 /[1 + (0.05 × 4)]

⇒ 12,000 / (1 + 0.20) = Rs. 10,000

Therefore, In the settlement,

A needs to give B (16,000 - 10,000) = Rs. 6,000.

Given: 

The average mark in Mathematics of Section A comprising 30 students is 65

The average mark in Section B comprising 35 students is 70.

The entry of 47 marks was wrongly made as 74 

Formula used:

Average = Sum of all observations / Number of observation

Calculation:

By using the Average formula,

Total marks of students in Section A = 30 × 65 = 1950

Total marks of students in section B = 35 × 70 = 2450

Total marks of students in section A & B = 1950 + 2450 = 4400

But, the entry of 47 marks was wrongly made as 74

Actual marks =  4400 + 47 - 74 = 4373

Again, using the above formula

Average = 4373 / (30 + 35) = 67.28

∴ The average mark for both sections is 67.28.