ICSE Class 10 Maths Long Answer Questions 2 – ICSE Board
ICSE Class 10 Maths long-answer solutions
ICSE Class 10 Maths competency focused long-answer questions test whether you can read a situation, choose the correct formula, and show the working in a clear order. This page gives step-by-step solutions for the Long Answer Questions 2 set, covering commercial mathematics, polynomials, arithmetic progression, coordinate geometry, similarity, locus, constructions, mensuration, heights and distances, and statistics.
The solutions below are written in a board-answer style: formula first, substitution next, calculation after that, and a final answer at the end. For extra practice across the subject, use the ICSE solutions hub, the Class 10 ICSE solutions section, and the ICSE Class 10 Maths solutions page.
Formula and method reference
Use this table before attempting the worked solutions. It helps you decide which syllabus idea applies to each part of the question.
| Area of question | Method or formula | Where it is used |
|---|---|---|
| GST and discount | Selling price after discount =\text{marked price}-\text{discount}; GST =\frac{\text{rate}}{100}\times\text{taxable value} | Question 110 |
| Factor theorem | If f(a)=0, then x-a is a factor of f(x). For px+q, use x=-\frac{q}{p}. | Question 111 |
| Arithmetic progression | a_n=a+(n-1)d, and S_n=\frac{n}{2}\{2a+(n-1)d\} | Question 112 |
| Straight lines | For y=mx+c, m is slope and c is the y-intercept. | Question 113 |
| Similarity | Corresponding sides of similar triangles are proportional; area ratio is the square of the corresponding side ratio. | Question 114 |
| Mensuration | Sphere volume =\frac{4}{3}\pi r^3, cone volume =\frac{1}{3}\pi r^2h, cylinder volume =\pi r^2h | Question 117 |
| Trigonometry | \tan\theta=\frac{\text{opposite side}}{\text{adjacent side}} | Question 118 |
| Statistics | For grouped data, class mark =\frac{\text{lower limit}+\text{upper limit}}{2}; median class contains the \frac{N}{2}-th item. | Question 119 |
Concept snapshot for mixed long answers
Think of a long-answer problem as a train with several stations. You do not jump to the last station. First identify the topic, then write the formula, then substitute the data, then simplify, and only then state the final answer. In this set, Question 117 is a good example: the clay changes shape, but the volume stays unchanged. That one idea connects sphere, cone and cylinder in the same answer.
Question 110: GST, discount and premium
Question: On seeing the display board outside Pearl Stationary Shop, Chetan buys 1 dozen pens, 20 pencils and 5 rainbow cover notebooks. Pens cost ₹5 each with 5\% discount on a dozen and 18\% GST. Pencils cost ₹7 each with 10\% discount on 20 pencils and 12\% GST. Each notebook costs ₹200, has a premium of ₹50, and has 12\% GST. The shopkeeper gives a further 2\% discount on the total bill but wrongly charges 18\% GST on all items. Find the correct amounts and decide whether Chetan was overcharged.
Step 1: Calculate the taxable value of pens after the displayed discount.
\text{Price of }12\text{ pens}=12\times ₹5=₹60
\text{Discount}=5\%\text{ of }₹60=\frac{5}{100}\times 60=₹3
\text{Taxable value of pens}=₹60-₹3=₹57
Step 2: Add correct GST on pens.
\text{GST on pens}=18\%\text{ of }₹57=\frac{18}{100}\times 57=₹10.26
\text{Pens total}=₹57+₹10.26=₹67.26
Step 3: Calculate the taxable value of pencils after discount.
\text{Price of }20\text{ pencils}=20\times ₹7=₹140
\text{Discount}=10\%\text{ of }₹140=\frac{10}{100}\times 140=₹14
\text{Taxable value of pencils}=₹140-₹14=₹126
\text{GST on pencils}=12\%\text{ of }₹126=\frac{12}{100}\times 126=₹15.12
\text{Pencils total}=₹126+₹15.12=₹141.12
Step 4: Add the notebook premium before GST.
\text{Basic price of }5\text{ notebooks}=5\times ₹200=₹1000
\text{Premium}=5\times ₹50=₹250
\text{Taxable value of notebooks}=₹1000+₹250=₹1250
\text{GST on notebooks}=12\%\text{ of }₹1250=\frac{12}{100}\times 1250=₹150
\text{Notebooks total}=₹1250+₹150=₹1400
Step 5: Add the selling prices before GST.
\text{Total selling price}=₹57+₹126+₹1250=₹1433
Final answer for (a)(i): Total selling price of all items as per the displayed offers is ₹1433.
Step 6: Add the correct GST totals.
\text{Correct total before further }2\%\text{ discount}=₹67.26+₹141.12+₹1400=₹1608.38
Final answer for (a)(ii): Correct amount including GST, before the extra 2\% discount, is ₹1608.38.
Step 7: Calculate the actual bill when 18\% GST is wrongly charged on every item.
\text{Pens with wrong GST}=₹57+\frac{18}{100}\times 57=₹67.26
\text{Pencils with wrong GST}=₹126+\frac{18}{100}\times 126=₹148.68
\text{Notebooks with wrong GST}=₹1250+\frac{18}{100}\times 1250=₹1475
\text{Wrong GST total before }2\%\text{ discount}=₹67.26+₹148.68+₹1475=₹1690.94
\text{Further discount}=2\%\text{ of }₹1690.94=\frac{2}{100}\times 1690.94=₹33.82
\text{Actual amount charged}=₹1690.94-₹33.82=₹1657.12
Final answer for (a)(iii): The shopkeeper charged ₹1657.12.
Step 8: Compare the charged amount with the correct amount. Using the amount in part (a)(ii),
\text{Overcharge}=₹1657.12-₹1608.38=₹48.74
Step 9: If the same promised 2\% discount is also applied to the correct bill, the fair amount would be:
₹1608.38-\frac{2}{100}\times 1608.38=₹1608.38-₹32.17=₹1576.21
\text{Overcharge on this stricter bill audit}=₹1657.12-₹1576.21=₹80.91
Final answer for (b): Yes, Chetan was overcharged. Compared with part (a)(ii), the overcharge is ₹48.74. If the promised 2\% discount is applied to the correct GST bill, the overcharge is ₹80.91.
Question 111: Remainder and factor theorem
Question: Using the remainder and factor theorem, show that 2x+3 is a factor of 2x^2+11x+12. Hence, factorise it completely. What must be multiplied to the given polynomial so that x^2+3x-4 is a factor of the resulting polynomial? Also write the resulting polynomial.
Step 1: Put the linear factor equal to zero.
2x+3=0 \Rightarrow x=-\frac{3}{2}
Step 2: Substitute x=-\frac{3}{2} in f(x)=2x^2+11x+12.
f\left(-\frac{3}{2}\right)=2\left(-\frac{3}{2}\right)^2+11\left(-\frac{3}{2}\right)+12
=2\cdot\frac{9}{4}-\frac{33}{2}+12=\frac{9}{2}-\frac{33}{2}+\frac{24}{2}=0
Step 3: Since the remainder is 0, 2x+3 is a factor.
Step 4: Factorise the quadratic completely.
2x^2+11x+12=2x^2+8x+3x+12
=2x(x+4)+3(x+4)=(2x+3)(x+4)
Final answer: 2x^2+11x+12=(2x+3)(x+4).
Step 5: Factorise the required divisor polynomial.
x^2+3x-4=x^2+4x-x-4=x(x+4)-1(x+4)=(x-1)(x+4)
Step 6: The given polynomial already contains the factor x+4. To make it divisible by (x-1)(x+4), multiply it by x-1.
(2x^2+11x+12)(x-1)=2x^3-2x^2+11x^2-11x+12x-12
=2x^3+9x^2+x-12
Final answer: Multiply the given polynomial by x-1. The resulting polynomial is 2x^3+9x^2+x-12.
Question 112: Arithmetic progression
Question: The sequence 2,9,16,\ldots is given. Identify whether it is an AP or GP, find the 20-th term, find the difference between the sums of the first 22 and 25 terms, test whether 102 belongs to the sequence, and state what happens if k is added to each term.
Step 1: Check the common difference.
9-2=7,\qquad 16-9=7
Final answer for (a): The sequence is an arithmetic progression because the common difference is 7.
Step 2: Use a_n=a+(n-1)d, where a=2 and d=7.
a_{20}=2+(20-1)7=2+133=135
Final answer for (b): The 20-th term is 135.
Step 3: Find S_{22}.
S_{22}=\frac{22}{2}\{2(2)+(22-1)7\}=11(4+147)=11\times 151=1661
Step 4: Find S_{25}.
S_{25}=\frac{25}{2}\{2(2)+(25-1)7\}=\frac{25}{2}(4+168)=\frac{25}{2}\times 172=2150
S_{25}-S_{22}=2150-1661=489
Final answer for (c): The difference is 489.
Step 5: Test whether 102 is a term.
102=2+(n-1)7
102=2+7n-7=7n-5 \Rightarrow 107=7n \Rightarrow n=\frac{107}{7}=15\frac{2}{7}
Final answer for (d): 102 is not a term of the sequence because n is not a positive integer.
Step 6: Add k to every term.
(9+k)-(2+k)=7,\qquad (16+k)-(9+k)=7
Final answer for (e): The new sequence is still an AP, with common difference 7.
Question 113: Straight lines and ratio
Question: Lines L_1 and L_2 are x-y=1 and x+y=5. They intersect at Q(3,2). Find the equation of L_3, parallel to L_1 with y-intercept 3; find k if L_3 meets L_2 at P(k,4); and find R and the ratio PQ:QR if L_2 meets the x-axis at R.
Step 1: Write L_1 in slope-intercept form.
x-y=1 \Rightarrow y=x-1
Step 2: The slope of L_1 is 1. Since L_3\parallel L_1, the slope of L_3 is also 1.
L_3:y=x+3
Final answer for (a): L_3 is y=x+3.
Step 3: Find the intersection of L_2 and L_3.
x+y=5,\qquad y=x+3
x+(x+3)=5 \Rightarrow 2x=2 \Rightarrow x=1
y=1+3=4
Final answer for (b): P=(1,4), so k=1.
Step 4: Find the point where L_2 cuts the x-axis. On the x-axis, y=0.
x+0=5 \Rightarrow x=5
Final answer for the coordinate: R=(5,0).
Step 5: Compare P(1,4), Q(3,2) and R(5,0).
\frac{1+5}{2}=3,\qquad \frac{4+0}{2}=2
Step 6: Since Q is the midpoint of PR, the two segments are equal.
Final answer for (c): R=(5,0) and PQ:QR=1:1.
Question 114: Similar triangles and areas
Question: In the figure, AD\parallel GE\parallel BC, DE=18\text{ cm}, EC=3\text{ cm}, and AD=35\text{ cm}. Find AF:FC, EF, \text{area}(\text{trapezium }ADEF):\text{area}(\triangle EFC), and BC:GF.
Step 1: Add the two parts on DC.
DC=DE+EC=18+3=21\text{ cm}
Step 2: Since EF\parallel AD, triangles \triangle FCE and \triangle ACD are similar.
\frac{FC}{AC}=\frac{EC}{DC}=\frac{3}{21}=\frac{1}{7}
Step 3: Let FC=x. Then AC=7x, so AF=AC-FC=7x-x=6x.
Final answer for (a): AF:FC=6:1.
Step 4: Use the corresponding side ratio to find EF.
\frac{EF}{AD}=\frac{EC}{DC}=\frac{1}{7}
\frac{EF}{35}=\frac{1}{7}\Rightarrow EF=5\text{ cm}
Final answer for (b): EF=5\text{ cm}.
Step 5: Use the area ratio of similar triangles.
\frac{\text{area}(\triangle ACD)}{\text{area}(\triangle FCE)}=\left(\frac{DC}{EC}\right)^2=\left(\frac{21}{3}\right)^2=49
Step 6: Let \text{area}(\triangle FCE)=a. Then \text{area}(\triangle ACD)=49a.
\text{area}(\text{trapezium }ADEF)=49a-a=48a
Final answer for (c): \text{area}(\text{trapezium }ADEF):\text{area}(\triangle EFC)=48:1.
Step 7: Since GF\parallel BC, triangles \triangle AGF and \triangle ABC are similar.
\frac{GF}{BC}=\frac{AF}{AC}=\frac{6x}{7x}=\frac{6}{7}
Final answer for (d): BC:GF=7:6.
Question 115: Locus construction
Question: Construct the locus of a point at a fixed distance 4.5\text{ cm} from a fixed point O. Draw AB=6\text{ cm}, where A and B are points on this locus. Construct the locus of points equidistant from A and B, mark its intersections with the first locus as P and Q, join PA, construct the locus of points equidistant from AP and AB, and mark its intersection with the circle as R. Measure AR.
Step 1: Draw a circle with centre O and radius 4.5\text{ cm}. This is the locus of all points 4.5\text{ cm} from O.
Step 2: Mark a chord AB=6\text{ cm} on the circle.
Step 3: Draw the perpendicular bisector of AB. This is the locus of all points equidistant from A and B.
Step 4: Let this perpendicular bisector meet the circle at P and Q.
Step 5: Join PA. The locus of points equidistant from the two lines AP and AB is the angle bisector of \angle PAB.
Step 6: Draw the angle bisector of \angle PAB. Let it meet the original circle again at R.
Final answer: On measurement, AR\approx 4.8\text{ cm}. A small variation is possible because this is a ruler-and-compass construction.
Question 116: Hexagon and tangents
Question: Construct a regular hexagon ABCDEF of side 4.3\text{ cm}, construct its circumscribed circle, and draw tangents to the circle at B and C. If the tangents meet at P, measure \angle BPC.
Step 1: Draw AB=4.3\text{ cm}.
Step 2: Construct the regular hexagon ABCDEF, keeping each side equal to 4.3\text{ cm}. Each interior angle of a regular hexagon is 120^\circ.
Step 3: Draw perpendicular bisectors of two sides, such as AB and BC, to locate the centre O of the circumscribed circle.
Step 4: With centre O and radius OB, draw the circumscribed circle through all six vertices.
Step 5: Draw OB and OC. At B, draw a line perpendicular to OB. At C, draw a line perpendicular to OC. These are the tangents at B and C.
Step 6: Let the two tangents meet at P. In a regular hexagon, the central angle \angle BOC=60^\circ.
\angle BPC=180^\circ-\angle BOC=180^\circ-60^\circ=120^\circ
Final answer: \angle BPC=120^\circ.
Question 117: Mensuration with the same volume
Question: A teacher uses the same terracotta clay to form different solids. First, she forms a sphere of radius 7\text{ cm}. Then she makes a right circular cone with base radius 14\text{ cm}. Find the height of the cone. If the same clay is turned into a right circular cylinder of height \frac{7}{3}\text{ cm}, find the radius of the cylinder. Also compare the total surface areas of the sphere and cylinder.
Step 1: Since the same clay is used, volume remains unchanged.
\text{Volume of sphere}=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(7)^3=\frac{1372\pi}{3}\text{ cm}^3
Step 2: Equate the cone volume to the sphere volume. Let the height of the cone be h\text{ cm}.
\frac{1}{3}\pi(14)^2h=\frac{1372\pi}{3}
196h=1372 \Rightarrow h=7\text{ cm}
Final answer for cone: The height of the cone is 7\text{ cm}.
Step 3: Let the cylinder radius be R\text{ cm}. Its height is \frac{7}{3}\text{ cm}.
\pi R^2\cdot\frac{7}{3}=\frac{1372\pi}{3}
7R^2=1372 \Rightarrow R^2=196 \Rightarrow R=14\text{ cm}
Final answer for cylinder radius: The radius of the cylinder is 14\text{ cm}.
Step 4: Compare total surface areas.
\text{TSA of sphere}=4\pi(7)^2=196\pi
\text{TSA of cylinder}=2\pi R(R+h)=2\pi(14)\left(14+\frac{7}{3}\right)
=28\pi\cdot\frac{49}{3}=\frac{1372\pi}{3}
\text{Sphere TSA}:\text{Cylinder TSA}=196\pi:\frac{1372\pi}{3}=588:1372=3:7
Final answer: Height of cone =7\text{ cm}, cylinder radius =14\text{ cm}, and \text{TSA of sphere}:\text{TSA of cylinder}=3:7.
Question 118: Heights and distances
Question: A tree TS of height 30\text{ m} stands in front of a building AB. Rohit stands at R, 150\text{ m} from the foot of the building, and observes the angle of elevation of the top of the building as 30^\circ. Neha stands at N on the same straight line and observes that the top of the building and the top of the tree have the same elevation of 60^\circ. Find the height of the building and the required distances.
Step 1: In \triangle ARB, use \tan 30^\circ=\frac{AB}{AR}.
\tan 30^\circ=\frac{1}{\sqrt{3}}=\frac{AB}{150}
AB=\frac{150}{\sqrt{3}}=50\sqrt{3}\text{ m}\approx 86.60\text{ m}
Final answer for (a): Height of the building =50\sqrt{3}\text{ m}\approx 86.60\text{ m}.
Step 2: From Neha’s position, the angle of elevation of the building is 60^\circ. Use \tan 60^\circ=\frac{AB}{AN}.
\sqrt{3}=\frac{50\sqrt{3}}{AN} \Rightarrow AN=50\text{ m}
Final answer for (b)(i): Distance between Neha and the foot of the building =50\text{ m}.
Step 3: Find the distance between Rohit and Neha.
RN=AR-AN=150-50=100\text{ m}
Final answer for (b)(ii): RN=100\text{ m}.
Step 4: Neha also sees the top of the tree at 60^\circ. Use \tan 60^\circ=\frac{ST}{NT}.
\sqrt{3}=\frac{30}{NT} \Rightarrow NT=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{ m}\approx 17.32\text{ m}
Final answer for (b)(iii): Distance between Neha and the tree =10\sqrt{3}\text{ m}\approx 17.32\text{ m}.
Step 5: Assuming the tree lies between the building and Neha on the same straight line, find the distance between the building and the tree.
AT=AN-NT=50-10\sqrt{3}\text{ m}
AT\approx 50-17.32=32.68\text{ m}
Final answer for (b)(iv): Distance between the building and the tree =50-10\sqrt{3}\text{ m}\approx 32.68\text{ m}.
Question 119: Statistics and median class
Question: A life insurance agent records the age distribution of 100 policy holders, where f is an unknown frequency.
| Age in years | No. of policy holders |
|---|---|
| 15-20 | 7 |
| 20-25 | 12 |
| 25-30 | 15 |
| 30-35 | 22 |
| 35-40 | f |
| 40-45 | 14 |
| 45-50 | 8 |
| 50-55 | 4 |
The question statement also mentions a mean of 35.65 years. The frequency table and that mean do not agree mathematically. The solution below shows the correct check instead of hiding the inconsistency.
Step 1: Use the given total number of policy holders.
7+12+15+22+f+14+8+4=100
82+f=100 \Rightarrow f=18
Final answer for (a), using total frequency: f=18.
Step 2: Check the mean statement using class marks 17.5,22.5,27.5,32.5,37.5,42.5,47.5,52.5.
\sum fx=7(17.5)+12(22.5)+15(27.5)+22(32.5)+18(37.5)+14(42.5)+8(47.5)+4(52.5)
\sum fx=3380
\bar{x}=\frac{3380}{100}=33.8
Step 3: Therefore, with the printed frequencies and 100 policy holders, the mean is 33.8, not 35.65. The mean value appears to be a misprint in the source question.
Step 4: Make the cumulative frequency table using f=18.
| Age in years | Frequency | Cumulative frequency |
|---|---|---|
| 15-20 | 7 | 7 |
| 20-25 | 12 | 19 |
| 25-30 | 15 | 34 |
| 30-35 | 22 | 56 |
| 35-40 | 18 | 74 |
| 40-45 | 14 | 88 |
| 45-50 | 8 | 96 |
| 50-55 | 4 | 100 |
Step 5: Since N=100, the median class contains the \frac{N}{2}=50-th item.
Step 6: The cumulative frequency just before 30-35 is 34, and the cumulative frequency at 30-35 is 56. So the 50-th item lies in 30-35.
Final answer for (b): The median class is 30-35.
Quick answer index
| Question | Main answer |
|---|---|
| 110 | Selling price =₹1433; correct GST total before extra discount =₹1608.38; actual charged =₹1657.12; overcharge =₹48.74 by part (a)(ii), or ₹80.91 after applying the promised discount to the correct bill. |
| 111 | 2x^2+11x+12=(2x+3)(x+4); multiply by x-1; resulting polynomial =2x^3+9x^2+x-12. |
| 112 | AP; a_{20}=135; S_{25}-S_{22}=489; 102 is not a term; adding k keeps it an AP. |
| 113 | L_3:y=x+3; k=1; R=(5,0); PQ:QR=1:1. |
| 114 | AF:FC=6:1; EF=5\text{ cm}; area ratio =48:1; BC:GF=7:6. |
| 115 | Required locus is a circle of radius 4.5\text{ cm}, then perpendicular bisector and angle bisector constructions; AR\approx 4.8\text{ cm}. |
| 116 | \angle BPC=120^\circ. |
| 117 | Cone height =7\text{ cm}; cylinder radius =14\text{ cm}; sphere TSA : cylinder TSA =3:7. |
| 118 | Building height =50\sqrt{3}\text{ m}\approx 86.60\text{ m}; AN=50\text{ m}; RN=100\text{ m}; NT=10\sqrt{3}\text{ m}; AT=50-10\sqrt{3}\text{ m}\approx 32.68\text{ m}. |
| 119 | Using total frequency, f=18; the printed mean 35.65 is inconsistent with the table; median class =30-35. |
Examiner’s mindset
In ICSE Class 10 Maths long-answer work, the final answer alone is not enough. The examiner looks for the method. In commercial mathematics, show the taxable value before GST. In polynomials, write the zero of the factor before substitution. In geometry, name the two triangles and state the reason for similarity. In mensuration, write that the volume remains the same when the same clay is reshaped. In statistics, do not ignore a contradiction in the data; show the check clearly.
Common mistakes students make
- Applying GST before discount: In Question 110, first apply the discount or premium to get the taxable value, then calculate GST.
- Using x=-\frac{2}{3} for 2x+3: The correct zero is x=-\frac{3}{2}. Reversing the fraction gives a wrong remainder.
- Comparing AP terms by ratio: The sequence 2,9,16,\ldots has a common difference, so use AP formulas, not GP formulas.
- Forgetting the square in area ratio: If two similar triangles have side ratio 7:1, their area ratio is 49:1, not 7:1.
- Not checking printed data: In Question 119, f=18 follows from 100 policy holders, but the printed mean does not match. A careful solution should mention this inconsistency.
Sources and syllabus alignment
The methods used here follow the standard ICSE Class 10 Mathematics treatment of commercial mathematics, algebra, coordinate geometry, similarity, constructions, mensuration, trigonometry and statistics. The authoritative references used for alignment are the official CISCE syllabus resources, standard Class 10 Mathematics textbooks prescribed in CISCE schools, and overlapping school-level treatments available through NCERT.
Frequently Asked Questions
How should I present ICSE Class 10 Maths long-answer solutions?
Write the formula or theorem first, substitute values clearly, show each calculation line, and end with a labelled final answer. In geometry, name the triangles, state the reason for similarity or parallel-line results, and then use the proportion.
Why are competency focused Maths questions longer than normal exercise questions?
Competency focused Maths questions usually combine two or more ideas, such as GST with discount, similarity with area ratio, or volume with surface area. The method is still syllabus-based; the extra work is in reading the data and arranging the steps.
What should I do if a data question seems inconsistent?
Check the total frequency, given mean and class marks separately. If the total and mean cannot both be true, state the inconsistency and solve from the condition that is directly usable, as shown in Question 119.
Do construction answers need calculations in ICSE Class 10 Maths?
A construction answer needs clear construction steps and the measured result. If a theorem explains the measurement, mention it; for example, the angle between two tangents equals 180^\circ minus the angle at the centre.
Which formulas are most useful for this long-answer set?
The key formulas are AP term and sum formulas, factor theorem, GST and discount calculation, similar-triangle side and area ratios, trigonometric ratios, and volume and surface area formulas for sphere, cone and cylinder.