ICSE Class 10 Maths Mid-Term 2026: Important Questions, Formulas & Must-Practise Topics

You may have finished the chapters for your ICSE Class 10 Maths Mid-Term Exam, but revising them effectively requires more than simply going through the textbook again. Which formulas should you revise first? Which question types need more practice? And which topics should you prioritise before the exam? Focusing on the right areas can make your revision more useful when time is limited.

To make this preparation easier, PW has brought its Class 10 Mid-Term Marathon, a dedicated revision session covering important concepts, formulas, previous-year questions, and specimen-paper questions. The session also features 3-marker, 5-marker, and 7-marker questions with solutions, helping you practise different question formats and apply the concepts while solving problems. 

Important Formulas for ICSE Class 10 Maths Mid-Term

Before solving the practice questions, revise the important formulas and concepts from the chapters covered in the marathon. Keeping these formulas fresh can help you approach numerical, algebraic, and application-based questions with greater confidence.

GST

• GST Amount: GST = Taxable Value × Rate of GST / 100

• CGST and SGST: For an intra-state transaction, GST is divided equally between CGST and SGST.

• IGST: For an inter-state transaction, the applicable GST is charged as IGST.

• Amount Paid by Consumer: Taxable Value + GST

Banking

For a Recurring Deposit account:

• Interest: I = P × n(n + 1)/2 × r/1200

• Maturity Value: Maturity Value = Total Deposits + Interest

• Total Deposits: Monthly Deposit × Number of Months

Here, P is the monthly deposit, n is the number of months, and r is the annual rate of interest.

Shares and Dividend

• Market Value at Premium: MV = NV + Premium

• Market Value at Discount: MV = NV − Discount

• Sum Invested: Number of Shares × Market Value

• Dividend per Share: NV × Dividend Rate / 100

• Annual Dividend: Number of Shares × Dividend per Share

• Rate of Return: Annual Income / Sum Invested × 100

Here, NV represents the nominal value and MV represents the market value.

Linear Inequations

While solving linear inequations:

• Adding or subtracting the same quantity on both sides does not change the direction of the inequality.

• Multiplying or dividing both sides by a positive number does not change the direction of the inequality.

• Multiplying or dividing both sides by a negative number reverses the inequality sign.

• When taking reciprocals of positive quantities, the direction of the inequality is reversed.

• Represent solutions on a number line using an appropriate point or circle.

• Distinguish between the replacement set and the solution set when required.

Quadratic Equations

For ax² + bx + c = 0:

• Quadratic Formula: x = [−b ± √(b² − 4ac)] / 2a

• Discriminant: D = b² − 4ac

• Real and Distinct Roots: D > 0

• Real and Equal Roots: D = 0

• No Real Roots: D < 0

Quadratic equations can also be solved by factorisation or by reducing an equation to quadratic form.

Ratio and Proportion

For a proportion:

a : b = c : d

Important properties include:

• Invertendo: b : a = d : c

• Alternendo: a : c = b : d

• Componendo: (a + b) : b = (c + d) : d

• Dividendo: (a − b) : b = (c − d) : d

• Componendo-Dividendo: (a + b) : (a − b) = (c + d) : (c − d)

Also revise fourth proportional, third proportional, mean proportional, and continued proportion.

Remainder and Factor Theorem

• Remainder Theorem: When f(x) is divided by (x − a), the remainder is f(a).

• Factor Theorem: If f(a) = 0, then (x − a) is a factor of f(x).

• To completely factorise a cubic polynomial, first identify a possible linear factor using the Factor Theorem and then divide the polynomial by that factor.

Matrices

• Order of a Matrix: Number of rows × Number of columns

• Row Matrix: A matrix having only one row.

• Column Matrix: A matrix having only one column.

• Square Matrix: Number of rows = Number of columns.

• Rectangular Matrix: Number of rows ≠ Number of columns.

• Null Matrix: All elements are zero.

• Diagonal Matrix: All non-diagonal elements are zero.

• Identity Matrix: A square matrix with 1s on the principal diagonal and 0s elsewhere.

• Two matrices are equal when they have the same order and corresponding elements are equal.

• Matrix multiplication is possible only when the number of columns of the first matrix equals the number of rows of the second matrix.

Arithmetic Progression

• nth Term: aₙ = a + (n − 1)d

• Sum of n Terms: Sₙ = n/2 [2a + (n − 1)d]

• Sum Using Last Term: Sₙ = n/2 (a + l)

• nth Term from the End: l − (n − 1)d

Here, a is the first term, d is the common difference, and l is the last term.

Geometric Progression

• Common Ratio: r = Second Term / First Term

• nth Term: aₙ = arⁿ⁻¹

• Sum of n Terms: Sₙ = a(rⁿ − 1)/(r − 1), when r > 1

• Alternatively, Sₙ = a(1 − rⁿ)/(1 − r), when r < 1

Here, a is the first term and r is the common ratio.

ICSE Class 10 Maths: 3, 5 & 7-Mark Important Questions

Most Important 3-Marker Questions

1. Rohit buys an article for ₹600 and pays GST at 10% on the amount. Find the total amount paid by Rohit.

Solution:

GST = 10% of ₹600

= 10/100 × 600

= ₹60

Total amount paid = ₹600 + ₹60

Answer: ₹660

2. Mr. Suresh has a Recurring Deposit account and deposits ₹2,000 per month. The bank pays interest at 8% p.a. If he receives ₹1,040 as interest at maturity, find the time period of the account.

Solution:

Using the RD interest formula:

I = P × n(n + 1)/2 × r/1200

Substituting the values:

1040 = 2000 × n(n + 1)/2 × 8/1200

1040 = 20n(n + 1)/3

n(n + 1) = 156

n² + n − 156 = 0

(n − 12)(n + 13) = 0

Therefore:

n = 12 or n = −13

Since the number of months cannot be negative:

n = 12 months

Answer: The time period is 12 months or 1 year.

3. Find the sum of money required to buy 50 shares of ₹40 each when the market value of each share is ₹38.50.

Solution:

Sum invested = Number of shares × Market Value

= 50 × ₹38.50

Answer: ₹1,925

4. Find the value of k for which 3x² − 6x + k = 0 has real and distinct roots.

Solution:

For real and distinct roots:

D > 0

Here,

a = 3, b = −6 and c = k

D = b² − 4ac

= (−6)² − 4(3)(k)

= 36 − 12k

For distinct real roots:

36 − 12k > 0

36 > 12k

k < 3

Answer: k < 3

5. Using the Factor Theorem, show that (x + 3) is a factor of 3x³ + 2x² − 19x + 6.

Solution:

Let:

f(x) = 3x³ + 2x² − 19x + 6

For (x + 3) to be a factor, put x = −3.

f(−3) = 3(−3)³ + 2(−3)² − 19(−3) + 6

= −81 + 18 + 57 + 6

= 0

Since f(−3) = 0, by the Factor Theorem:

(x + 3) is a factor of f(x).

Most Important 5-Marker Questions

1. A shopkeeper buys goods worth ₹4,000 and sells them at a profit of 20% within the same state. If the rate of GST is 5%, find (i) the selling price, (ii) CGST and SGST, and (iii) the total amount paid by the customer.

Solution:

Profit = 20% of ₹4,000

= ₹800

Selling Price = ₹4,000 + ₹800

= ₹4,800

Since the transaction is within the same state:

GST = 5%

CGST = 2.5%

SGST = 2.5%

CGST = 2.5% of ₹4,800

= ₹120

SGST = ₹120

Total GST = ₹120 + ₹120 = ₹240

Total amount paid by the customer:

= ₹4,800 + ₹240

Answer:

• Selling Price = ₹4,800

• CGST = ₹120

• SGST = ₹120

• Total Amount = ₹5,040

2. Anwesha opened a Recurring Deposit account of ₹1,000 per month for 1 year at 5% p.a. If the bank reduces the rate to 4% p.a., find the monthly deposit needed to earn the same interest over the same period.

Solution:

For 12 months:

n = 12

Original monthly deposit = ₹1,000

Original rate = 5%

Using:

I = P × n(n + 1)/2 × r/1200

Original interest:

= 1000 × 12 × 13/2 × 5/1200

= ₹325

Let the new monthly deposit be ₹x.

At 4% interest:

325 = x × 12 × 13/2 × 4/1200

325 = 0.26x

x = 325/0.26

x = ₹1,250

Answer: The new monthly deposit should be ₹1,250.

3. Mr. Gupta invested ₹33,000 in buying shares of a company of nominal value ₹100 each at a premium of 10%. If the company declared a dividend of 12%, find the number of shares purchased and his annual dividend.

Solution:

Nominal Value = ₹100

Premium = 10% of ₹100 = ₹10

Market Value = ₹100 + ₹10

= ₹110

Number of shares:

= ₹33,000/₹110

= 300 shares

Dividend per share:

= 12% of ₹100

= ₹12

Annual dividend:

= 300 × ₹12

= ₹3,600

Answer:

• Number of shares = 300

• Annual dividend = ₹3,600

4. The difference of two natural numbers is 5, and the sum of their reciprocals is 3/10. Find the two numbers.

Solution:

Let the smaller number be x.

Then the larger number = x + 5.

According to the question:

1/x + 1/(x + 5) = 3/10

[(x + 5) + x]/[x(x + 5)] = 3/10

(2x + 5)/(x² + 5x) = 3/10

10(2x + 5) = 3(x² + 5x)

20x + 50 = 3x² + 15x

3x² − 5x − 50 = 0

(3x + 10)(x − 5) = 0

Therefore:

x = −10/3 or x = 5

Since the numbers are natural numbers:

x = 5

The other number:

= 5 + 5

= 10

Answer: The two numbers are 5 and 10.

5. If (8x + 13y) : (8x − 13y) = 9 : 7, use the properties of proportion to find x : y.

Solution:

Given:

(8x + 13y)/(8x − 13y) = 9/7

Using componendo-dividendo:

[(8x + 13y) + (8x − 13y)] / [(8x + 13y) − (8x − 13y)] = (9 + 7)/(9 − 7)

16x/26y = 16/2

16x/26y = 8

16x = 208y

x/y = 13

Answer: x : y = 13 : 1

Most Important 7-Marker Questions

1. The marked price of a TV set is ₹20,000. A shopkeeper marks it up to ₹24,000 and sells it to a consumer after a discount of 10% on the marked price, within the same state. If GST is 12%, find (i) the price after discount, (ii) the GST charged, and (iii) the total amount paid by the consumer.

Solution:

Marked-up price = ₹24,000

Discount = 10% of ₹24,000

= ₹2,400

Price after discount:

= ₹24,000 − ₹2,400

= ₹21,600

GST = 12% of ₹21,600

= ₹2,592

Total amount paid:

= ₹21,600 + ₹2,592

= ₹24,192

Answer:

• Price after discount = ₹21,600

• GST charged = ₹2,592

• Total amount paid = ₹24,192

2. Using the Remainder and Factor Theorem, completely factorise 3x³ + 2x² − 19x + 6.

Solution:

Let:

f(x) = 3x³ + 2x² − 19x + 6

Try x = 2:

f(2) = 3(2)³ + 2(2)² − 19(2) + 6

= 24 + 8 − 38 + 6

= 0

Therefore, by the Factor Theorem:

(x − 2) is a factor.

Dividing the polynomial by (x − 2):

3x³ + 2x² − 19x + 6 ÷ (x − 2)

gives:

3x² + 8x − 3

Now factorise:

3x² + 8x − 3

= 3x² + 9x − x − 3

= 3x(x + 3) − 1(x + 3)

= (x + 3)(3x − 1)

Therefore:

3x³ + 2x² − 19x + 6 = (x − 2)(x + 3)(3x − 1)

3. Aman has 500 shares of a company of nominal value ₹100 each, quoted at ₹120, paying a 10% dividend. When the market price rises to ₹200, he sells all his shares. He invests half the sale proceeds in shares of nominal value ₹10 at market value ₹25 paying a 12% dividend, and the remaining half in shares of nominal value ₹400 at market value ₹500 paying a 9% dividend. Find the change in his income.

Solution:

Original number of shares = 500

Original nominal value = ₹100

Original dividend = 10%

Original annual income:

= 500 × ₹100 × 10/100

= ₹5,000

When the market price rises to ₹200:

Sale proceeds = 500 × ₹200

= ₹1,00,000

Half of the sale proceeds:

= ₹1,00,000/2

= ₹50,000

First Investment

Market value = ₹25

Number of shares purchased:

= ₹50,000/₹25

= 2,000 shares

Dividend per share:

= 12% of ₹10

= ₹1.20

Annual income:

= 2,000 × ₹1.20

= ₹2,400

Second Investment

Market value = ₹500

Number of shares purchased:

= ₹50,000/₹500

= 100 shares

Dividend per share:

= 9% of ₹400

= ₹36

Annual income:

= 100 × ₹36

= ₹3,600

New total annual income:

= ₹2,400 + ₹3,600

= ₹6,000

Change in income:

= ₹6,000 − ₹5,000

= ₹1,000 increase

4. Find the sum of all multiples of 7 lying between 300 and 700.

Solution:

The first multiple of 7 greater than 300 is:

7 × 43 = 301

The last multiple of 7 less than 700 is:

7 × 99 = 693

Therefore:

a = 301

l = 693

d = 7

Using:

aₙ = a + (n − 1)d

693 = 301 + (n − 1)7

392 = 7(n − 1)

n − 1 = 56

n = 57

Now use:

Sₙ = n/2(a + l)

S₅₇ = 57/2(301 + 693)

= 57/2 × 994

= 57 × 497

Answer: The sum is 28,329.

Note: For more important questions and a deeper understanding of the concepts covered here, watch the PW Class 10 Mid-Term Marathon video and revise along with the explanations.

How Does the PW Marathon Bring Your Mid-Term Revision Together?

The PW Mid-Term Marathon brings important ICSE Class 10 Maths concepts, formulas, and question types together for focused revision. Instead of revising every chapter from the beginning, you can use the marathon to:

• Revise Important Formulas and Concepts: Refresh formulas from GST, Banking, Shares and Dividend, Quadratic Equations, AP, GP, and other chapters covered in the marathon.

• Understand How Formulas Are Applied: See how formulas and properties are used in numerical and application-based questions.

• Practise Different Question Types: Work through commercial mathematics problems, algebraic questions, inequations, matrix operations, factorisation, and progression-based questions.

• Strengthen Problem-Solving Steps: Follow the working used to solve questions involving quadratic equations, proportions, GST calculations, shares, and recurring deposits.

• Identify Topics That Need More Revision: Note the questions, formulas, or concepts where you make mistakes and return to those topics before your mid-term exam.

The PW Mid-Term Marathon can help you combine formula revision with question practice across the chapters covered in the session. Attempt the questions yourself first, check the solutions afterwards, and revisit the topics where you need more practice before your ICSE Class 10 Maths Mid-Term Exam.