ADRE Higher Secondary (HSSLC) Level Paper III 2024 Complete Question Paper Solution with Explanations of General Mathematics Section

ADRE Higher Secondary (HSSLC) Level Paper III 2024 Complete Solved Paper with Detailed Explanations of General Mathematics Section

The General Mathematics section is a vital part of competitive examinations, designed to assess a candidate’s numerical ability, mathematical concepts, and problem-solving skills. This section tests how quickly and accurately candidates can perform calculations, interpret numerical data, and apply mathematical formulas to solve practical problems. Strong mathematical aptitude not only helps in scoring higher but also improves overall exam performance through better speed and accuracy.

The General Mathematics section typically includes topics such as arithmetic, percentage, profit and loss, ratio and proportion, averages, simple and compound interest, time and work, time, speed and distance, algebra, geometry, mensuration, number system, simplification, data interpretation, and basic statistics. Consistent practice, conceptual clarity, and effective time management are the keys to achieving a high score in this section.

General Mathematics Section

• Total Questions : 35
• Total Marks : 35
• Nagative Marking : 0.25%

Q1. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is

(A) 16% increase

(B) 16% decrease

(C) 32% decrease

(D) No change

Answer: (B) 16% decrease

Explanation

Method 1

• Assume the original number is 100.
• Increase by 40%
• So, 100 → 140

Now decrease 140 by 40%:

• 40% of 140 = 56
• 140 − 56 = 84

Compare with the original number:

• Original = 100
• Final = 84
• Loss = 16

So, the net change is 16% decrease.

Method 2

Quick Exam Trick

Whenever a number is increased by x% and then decreased by the same x%,

Net Change = −x²/100 percentage

Here,  x = 40

Therefore,  -(40) 2 / 100 = -1600 / 100 = -16 

So, the result is 16% decrease.

• [Minus means Decrease & Plus means Increase]
• Method 2 is easy, shortcut and simple

Q2. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.

(A) 25 years

(B) 35 years

(C) 50 years

(D) 45 years

Answer: (A) 25 years

Explanation

Method 1

Step 1: Find the total age of A, B and C.

• Average = 22
• Number of persons = 3

So, Total age = 22 × 3 = 66 years

Step 2: Find the total age of B and C.

• Average = 25
• Number of persons = 2

So, Total age = 25 × 2 = 50 years

Step 3: Find A's present age.

• A's present age = 66 − 50 = 16 years

Step 4: Find A's age after 9 years.

• 16 + 9 = 25 years

Method 2

Shortcut Method

Given:

• Average of A, B & C = 22
• Average of B & C = 25

Shortcut Formula:

  A's present age  = (Average of 3 × 3) − (Average of 2 × 2)

                                              = 22 × 3 − 25 × 2

                                            = 66 − 50

                                         = 16 years

After 9 years: = 16 + 9 = 25 years

One-Line Trick : 

(22×3)−(25×2)+9=25

Q3. The smallest among √(1/2), √(1/3), √(1/4) and √(2/3) is:

(A) √(1/2)

(B) √(1/3)

(C) √(1/4)

(D) √(2/3)

Correct Answer: (C) √(1/4)

Explanation

The square root (√) is an increasing function. This means:

If a < b, then √a < √b.

So, first compare the numbers inside the square roots:

• 1/2 = 0.50
• 1/3 ≈ 0.33
• 1/4 = 0.25 ( Smallest ) [ After Point it is 2 but others are more than 2 ]
• 2/3 ≈ 0.67

Since

1/4​ < 1/3​ < 1/2​ < 2/3​,

taking square roots gives:

√(1/4) < √(1/3)  < √(1/2) < √(2/3)

Therefore, the smallest is: (C) √(1/4)

[Note: 0.1 < 0.2 < 0.3 < 0.4 < 0.5 ans so on ]

Shortcut Trick

When all numbers are under the square root (√), compare the numbers inside the root.

The smallest number inside is 1/4, so the smallest value is √(1/4).

Q4. If x : y : z = 3 : 4 : 7, then (x+y+z) / z is equal to:

(A) 2

(B) 3

(C) 4

(D) 5

Answer: 2

Solution

Given:   x:y:z=3:4:7

Let, x=3k, y=4k, z=7k

Then,   (x+y+z) / z = (3k + 4k + 7k ) / 7k = 14k / 7k = 2

Shortcut Trick

When given a ratio, substitute the ratio values directly:

( 3+ 4 + 7 ) / 7 = 14 / 7 = 2

Q5. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is

(A) 18

(B) 20

(C) 22

(D) 24

Correct Answer: (B) 20

Solution

Let the numbers be:

2X and 3X

Since 2 and 3 are co-prime,

So, 

• HCF = X
• LCM = 6X

Given:

HCF × LCM = 96

=> X × 6X=96

=> 6X2=96

=> X2=16

=> X = 4

Therefore, the numbers are:

• 2X = 2 x 4 = 8
• 3X = 3x 4 = 12

Therefore, Sum = 8 + 12 = 20

[ Remember : HCF × LCM = Product of the two numbers ]

Q6. The value of ∛(8²) is:

(A) 1

(B) 3

(C) 2

(D) 4

Correct Answer: (D) 4

Solution

∛(8²) = ∛64 =  ∛(4³) = (4³)^(1/3) = 4^(3×1/3) = 4¹ = 4

Rule to remember:

ⁿ√(aᵐ) = (aᵐ)^(1/n) = a^(m/n)

Q7. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹ 120 on every doubling. A person invests an amount of ₹ 105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is

(A) ₹ 55

(B) ₹ 0

(C) ₹ 270

(D) ₹ 45

Correct Answer: (B) ₹ 0

Solution

Shortcut Trick

Given amount ₹ 105.

• Month 1: 105×2−120=90
• Month 2: 90×2−120=60
• Month 3: 60×2−120=0

[ 2 is multiplied due to double and do minus 120 as question says 120 deduction ]

Q8. The count of prime numbers between 80 and 100 is

(A) 2

(B) 5

(C) 3

(D) 4

Correct Answer: (C) 3

Solution

Prime numbers between 80 and 100 are:

• 83 , 89 & 97

There are 3 prime numbers.

Q9. Find the value of [(10⁷ ÷ 10²) + (10³ ÷ 10⁻²)] ÷ 10⁵

(A) 100

(B) 200

(C) 2

(D) 1

Correct Answer: (C) 2

Solution:

[(10⁷ ÷ 10²) + (10³ ÷ 10⁻²)] ÷ 10⁵

= [10^(7−2) + 10^(3−(−2))] ÷ 10⁵

= [10⁵ + 10⁵] ÷ 10⁵

= (100000 + 100000) ÷ 100000

= 200000 ÷ 100000

= 2

Shortcut:

10⁷ ÷ 10² = 10⁵

10³ ÷ 10⁻² = 10⁵

So,

(10⁵ + 10⁵) ÷ 10⁵

= 2 × 10⁵ ÷ 10⁵

= 2

Q10. If A : B = 3 : 1, then 144A : 36B is

(A) 4

(B) 12

(C) 3

(D) 16

Correct Answer: (B) 12

Solution:

Given , A : B = 3 : 1

Therefore, A = 3 , B = 1

So,  

144A : 36B

= (144 × 3) : (36 × 1)

= 432 : 36

= 12 : 1

= 12

Shortcut:

144A : 36B

= 144/36 : B /A

= 144 / 36 x A /B

= 144 / 36 x A : B

= 4 x 3

= 12 

[ Now , If you understood, we can write more simply

144A : 36B

= (144 ÷ 36) × (A : B)

= 4 × 3

= 12   ]

Q11. The fraction equivalent of the recurring number 2 .̅19̅   is

(A) 271/99

(B) 217/99

(C) 219/90

(D) 291/90

Correct Answer: (B) 217/99

Solution:

Let x = 2.191919...

Then,

100x = 219.191919...

Subtract:

100X − X = 219.191919... − 2.191919...

99X = 217

X = 217/99

Therefore, the fraction equivalent of  2.̅19̅    is    217/99.

Shortcut:

For 2.̅19̅,

Fraction = (219 − 2) / 99

= 217 / 99

Q12. The largest of 2⁶, 3⁵, 4⁴ and 5³ is

(A) 2⁶

(B) 3⁵

(C) 4⁴

(D) 5³

Correct Answer: (C) 4⁴

Solution:

2⁶ = 64

3⁵ = 243

4⁴ = 256

5³ = 125

Since 256 is the largest,

4⁴ > 3⁵ > 5³ > 2⁶

Therefore, the largest number is 4⁴.

Answer: (C) 4⁴

Q13. For a bicycle rider, it is seen that for every two complete pedaling, the front wheel of the bicycle makes 3 complete turns. If the radius of the bicycle wheel is 70 cm, the distance (in metres) covered by the bicycle in 10 complete pedaling is

(A) 7π

(B) 14π

(C) 21π

(D) 28π

Correct Answer: (C) 21π

Solution:

For every 2 complete pedalings, the wheel makes 3 complete turns.

Therefore, for 10 complete pedalings,

Distance Covered by The Wheel = (3/2) × 10 = 15 turns   [ The Concept is, Wheel turns = 3 turns / 2 pedalings x 10 pedalings = 15 turns,  ]

Given, Radius of the wheel = 70 cm = 0.7 m

Circumference of the wheel = 2πr  [ This is Formulae ]

                                                                             = 2π × 0.7

                                                                             = 1.4π m

Therefore, Distance covered = 1.4π x 15= 21π m

Q14. The base and the height of a right-angled triangle are equal in magnitude and the length of the third side is 2√2 cm. The area of the triangle is

(A) 2 sq. cm

(B) 2√2 sq. cm

(C) 4 sq. cm

(D) 4√2 sq. cm

Correct Answer: (A) 2 sq. cm

Solution:

Method 1

Let the base = height = x cm.

Using Pythagoras' theorem,

x² + x² = (2√2)²

2x² = 8

x² = 4

x = 2 cm

Area of the triangle

= ½ × base × height

= ½ × 2 × 2

= 2 sq. cm

Therefore, the area of the triangle is 2 sq. cm.

Method 2

For an isosceles right triangle,

Hypotenuse = Side × √2

So,

Side = Hypotenuse / √2

               = (2√2)/√2 = 2 cm

Area = ½ × 2 × 2 = 2 sq. cm

Q15. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is

(A) 35

(C) 49

(B) 42

(D) 56

Correct Answer: (C) 49

Solution:

After removing all even numbers, the remaining numbers are:

1, 3, 5, 7, 9, 11, 13, ...

The first 7 numbers are:

1, 3, 5, 7, 9, 11, 13

Sum

= 1 + 3 + 5 + 7 + 9 + 11 + 13

= 49

Therefore, the required sum is 49.

Shortcut:

The sum of the first n odd numbers is n².  [ This is a Formulae ]

Here, n = 7

Sum = 7² = 49

Q16. Two numbers A and B are first added and squared to get C. In the next step, B is subtracted from A and the result is squared to get D. The result C − D expressed in terms of A and B is

(A) A(A + B)

(B) 4(A − B)

(C) AB/4

(D) 4AB

 Correct Answer: (D) 4AB

Solution:

C = (A + B)²

D = (A − B)²

Therefore,

C − D = (A + B)² − (A − B)²

                = (A² + 2AB + B²) − (A² − 2AB + B²)

                 = A² + 2AB + B² − A² + 2AB − B²

                 = 4AB

Therefore, the value of C − D is 4AB.

Shortcut:

Using the formulae: (a + b)² − (a − b)² = 4ab

C − D = (A + B)² − (A − B)²

                 = 4AB

Q17. Find the least number by which 1250 must be multiplied to make it a perfect square.

(A) 4

(C) 5

(B) 3

(D) 2

Correct Answer: (D) 2

Solution:

Prime factorization of 1250:

1250 = 2 × 5 × 5 × 5 × 5

= 2 × 5⁴

For a perfect square, the power of every prime must be even.

Here,

2¹ × 5⁴

The power of 5 is even, but the power of 2 is odd.

So, multiply by 2.

1250 × 2 = 2500 = 50²

Therefore, the least number required is 2.

Shortcut:

1250 = 2 × 5⁴

Only the power of 2 is odd.

Multiply by 2 to make it even , 2² × 5⁴  , which gives a perfect aquare (50²) 

Q18. When the numerator of a fraction is multiplied by 4 and the denominator by 9, the fraction reverses. The fraction is

(A) 4/5

(B) 2/3

(C) 5/4

(D) 3/2

Correct Answer: (D) 3/2

Solution:

Let the fraction be A/B.

According to the question,

(4A)/(9B) = B/A

Cross-multiplying,

4A² = 9B²

Taking square roots,

2A = 3B

A/B = 3/2

Therefore, the required fraction is 3/2.

Q19. At what percentage of simple interest does an amount of money double in 12 years?

(A) 9½%

(B) 8½%

(C) 8⅓%

(D) 9⅓%

Correct Answer: (C) 8⅓%

Solution:

Given,   Simple Interest = Principal    (Since the amount becomes double.)

Therefore, Using the formula,

      SI = (P × R × T) / 100

=> P = (P × R × 12) / 100  [  Given, Time = 12 years ]

=> R = 100 / 12

             = 8⅓%

Therefore, the required rate of simple interest is 8⅓%.

Shortcut:

For simple interest, when the amount doubles

Rate = 100 / Time

= 100 / 12

= 8⅓%

Q20. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is

(A) 1 : 6

(B) 2 : 9

(C) 1 : 9

(D) 6 : 9

Correct Answer: (C) 1 : 9

Solution:

Area of a circle = πr²

Ratio of radii = 1 : 3

Therefore,

Ratio of areas

= 1² : 3²

= 1 : 9

Hence, the ratio of their areas is 1 : 9.

Shortcut:

For a circle,  Area ∝ (Radius)²

So, simply square the ratio of the radii.

       1 : 3

⇒ 1² : 3²

⇒ 1 : 9

Q21. 100% of 100 when added to 200% of 200 would result

(A) 300

(B) 400

(C) 500

(D) 600

Correct Answer: (C) 500

Solution:

100% of 100

= (100/100) × 100

= 100

200% of 200

= (200/100) × 200

= 2 × 200

= 400

Required sum

= 100 + 400

=500

Shortcut:

100% of any number = the number itself.

200% of any number = 2 × the number.

So,

100 + (2 × 200)

= 100 + 400

= 500

Q22. If a + b = 8, b + c = 20 and a + c = 18, then the value of a² + b² is

(A) 34

(B) 43

(C) 46

(D) 64

Correct Answer: (A) 34

Solution:

Given,

a + b = 8

b + c = 20

a + c = 18

Add the first and third equations and subtract the second:

(a + b) + (a + c) − (b + c) = 8 + 18 − 20

= > 2a = 6

=> a = 3

Now,

         a + b = 8

=> 3 + b = 8

=> b = 5

Therefore,

a² + b²  = 3² + 5²

                    = 9 + 25

                    = 34

Therefore, the correct answer is 34.

Q23. If 2ˣ = 32, then the value of x is

(A) 4

(B) 5

(C) 6

(D) 7

 Correct Answer: (B) 5

Solution:

Given, 2ˣ = 32

Since, 32 = 2⁵

Therefore,  2ˣ = 2⁵

Hence, x = 5

Therefore, the correct answer is 5.

Q24. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is

(A) 17

(B) 21

(C) 23

(D) 19

Correct Answer: (C) 23

Solution:

Let the five consecutive odd numbers be:

x − 4, x − 2, x, x + 2, x + 4

Their sum is 95.

So, (x − 4) + (x − 2) + x + (x + 2) + (x + 4) = 95

 => 5x = 95

=> x = 19

Therefore, the numbers are:

15, 17, 19, 21, 23

The largest number is 23.

Answer: (C) 23

Shortcut:

The middle number (X) = Total Sum ÷ Number of Terms

                                                             = 95 ÷ 5

                                                            = 19

So , X = 19

Therefore, Largest number = 19 + 4 = 23

Q25. If the decimal number 34p5 is divisible by 9, then the value of p is

(A) 8

(B) 7

(C) 4

(D) 6

Correct Answer: (D) 6

Solution:

A number is divisible by 9 if the sum of its digits is divisible by 9.

Sum of the digits = 3 + 4 + p + 5 = 12 + p

To be divisible by 9,

12 + p = 18 , [ Since 18 is the next number divisible by 9 ]

=> p = 6

Therefore, the required value of p is 6.

Q26. The maximum area of the circle that can be drawn within a square of side 4 cm is

(A) 4π² cm²

(B) 4π cm²

(C) π/4 cm²

(D) π²/4 cm²

Correct Answer: (B) 4π cm²

Solution:

The largest circle that can be drawn inside a square is the inscribed circle.

Diameter of the circle = Side of the square = 4 cm

Therefore, Radius = 4cm / 2 = 2 cm

Area of the circle = πr²

                                             = π × 2²

                                              = 4π cm²

Therefore, the maximum area of the circle is 4π cm².

Answer: (A) 4π cm²

Q27. If today is Sunday, then the day after 92 days will be

(A) Saturday

(B) Friday

(C) Sunday

(D) Monday

Correct Answer: (D) Monday

Solution:

We know that, there are 7 days in a week and Same Day comes after a week.

So, 92 ÷ 7 = 13 weeks and 1 day

Remainder = 1

So, after 92 days, the day will be 1 day after Sunday.

Sunday + 1 day = Monday

Therefore, the required day is Monday.

Q28. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is

(A) 4

(B) 2

(C) –2

(D) –4

Correct Answer: (A) 4

Solution:

Let the number be x.

According to the question,

x² − 3x = x

x² − 4x = 0

x(x − 4) = 0

x = 0 or x = 4

Since 0 is not among the options, the required number is 4.

Q29. In Roman numeral, XCI represents the decimal number

(A) 41

(B) 21

(C) 91

(D) 111

Correct Answer: (C) 91

Solution:

In Roman number X = 10 & C = 100

Therefore , XC = 100 − 10 = 90

I = 1

Therefore,

XCI = 90 + 1

= 91

Hence, the decimal number is 91.

Hints:

In Roman numerals:

• When a smaller numeral comes before a larger numeral, it is subtracted.
• When a smaller numeral comes after a larger numeral, it is added.

Examples:

Q30. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is

(A) 9 m

(B) 5 m

(C) 11 m

(D) 12 m

 Correct Answer: (D) 12 m

Solution:

Wall

|

|\

|      \

|           \ 13 m (Ladder)

|5 m       \

|                      \

+---------------- Ground

<---12 m--->

          Base

The ladder, wall, and ground form a right-angled triangle.

Hypotenuse = 13 m

Height = 5 m

Using Pythagoras' theorem,

(Base)² + (Height)² = (Hypotenuse)²

Base² + 5² = 13²

Base² + 25 = 169

Base² = 144

Base = √144

= 12 m

Therefore, the horizontal distance is 12 m.

Shortcut:

Remember the Pythagorean triplet: 5, 12, 13

If two of them is given , the third will be from the triplet.

Since 5 & 12 is given the third will be 13. which takes deci second to solve. 

Note : Some More Pythagorean triplet

• 3,4,5
• 5, 12, 13
• 6,8, 10
• 7 ,24, 25

Q31. The median of the sequence of numbers   2, –1, 3, 1, –2, 5, 6   is

(A) 1

(B) –1

(C) 2

(D) 3

Correct Answer: (C) 2

Solution:

Original sequence:   2, –1, 3, 1, –2, 5, 6

Step 1: Arrange the numbers in ascending order.

Ascending order:  –2, –1, 1, 2, 3, 5, 6

Step 2: Count the numbers.

There are 7 numbers.

Median = Middle number

Therefore , Median = 2

Q32. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is

(A) 22 m

(B) 44 m

(C) 122 m

(D) 144 m

Correct Answer: (B) 44 m

Solution:

The person is always 7 m away from a fixed point.

Therefore, he is moving along the circumference of a circle of radius 7 m.

He returns to the starting point, so he completes one full circle.

Distance travelled = Circumference of the circle

                                                  = 2πr

                                                 = 2 × (22/7) × 7

                                                = 44 m

Therefore, the approximate distance travelled is 44 m.

Q33. A cube of volume 1 m³ is cut equally into smaller cubes of volume 1 cm³ each. The number of smaller cubes found is

(A) 10,000

(B) 1,00,000

(C) 1,000

(D) 10,00,000

Correct Answer: (D) 10,00,000

Solution:

1 m = 100 cm

Therefore,

1 m³ = (100 cm)³

= 100 cm × 100 cm × 100cm

= 10,00,000 cm³

Q34. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is

(A) 24

(B) 18

(C) 48

(D) 36

Correct Answer: (A) 24

Solution:

The smallest number divisible by 6, 8 and 12 is their LCM.

Since , LCM of 6, 8 & 12 is 24

Therefore, the smallest positive integer divisible by 6, 8 and 12 is 24.

Q35. The difference between the smallest 3-digit even natural number and the largest 2-digit even natural number is

(A) 1

(B) 2

(C) 3

(D) 4

Correct Answer: (B) 2

Solution:

Smallest 3-digit even natural number = 100

Largest 2-digit even natural number = 98

Difference = 100 − 98

= 2

Therefore, the required difference is 2.

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FAQs

Q1. Which subjects are covered in Questions 1–30 of the ADRE HSSLC Level 2024 Social Studies section?

Ans: Questions 1–30 mainly cover Geography, Indian Geography, General Science, Indian Awards, and important world facts, helping candidates test their basic social studies knowledge. 

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Q2. Are the answers to ADRE HSSLC 2024 Social Studies Questions  1–30   officially verified?

Ans: Yes. The correct answers are verified using the official ADRE HSSLC Level 2024 final answer key, while the explanations are written independently to help candidates understand the concepts.

Q3. Are detailed explanations provided for every question?

Ans: Yes. Every question includes a detailed explanation to help candidates understand why the correct option is right and improve their preparation for future ADRE examinations.

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Ans: Yes. Previous year question papers help candidates understand the exam pattern, important topics, and frequently asked questions, making them an essential resource for ADRE preparation.

Q5. Where can I find solutions for the remaining ADRE HSSLC Level 2024 questions?

Ans: You can explore the complete subject-wise solutions on RozgarAssam, including Social Studies, General Knowledge, Logical Reasoning & Mental Ability, General English, and General Mathematics, along with detailed explanations for all 150 questions.