Class 8 Mathematics Exam 2025: Questions & Detailed Solutions
Class 8 Mathematics Part-II Exam 2025: Questions & Detailed Solutions
- Question 1 – Choose the correct answer
- Question 2 – Short answers
- Question 3 – Algebra, arithmetic and geometry
- Question 4 – Applications of proportion and percentage
- Question 5 – Factorisation and algebraic simplification
- Question 6 – Geometry proofs
- Question 7 – Geometry proofs
- Question 8 – Constructions
1. Choose the correct answer
Q 1(i)
The percentage of 85 g out of 17 kg is — (a) 0.25 (b) 0.5 (c) 0.75 (d) 1.25.
Convert 17 kg into grams:
\(17\text{ kg}=17,000\text{ g}\).
Percentage \(=\dfrac{85}{17000}\times100=0.5\%\).
Q 1(ii)
The ratio of hydrogen and oxygen in water is 2:1. The percentage of hydrogen in water is — (a) 20% (b) \(33\frac13\%\) (c) \(66\frac23\%\) (d) 80%.
Total parts \(=2+1=3\). Hydrogen occupies 2 parts.
Hydrogen percentage \(=\dfrac{2}{3}\times100=66\frac23\%\).
Q 1(iii)
Find the least form of \(\dfrac{18a^4b^5c^2}{21a^7b^2}\).
Cancel common factors:
\[\frac{18a^4b^5c^2}{21a^7b^2}=\frac{18}{21}\frac{b^3c^2}{a^3}=\frac{6b^3c^2}{7a^3}.\]
Q 1(iv)
The G.C.D. of \(3a^2b^2c,\ 12a^2b^4c^2,\ 9a^5b^4\) is —
G.C.D. of numerical coefficients: \(\gcd(3,12,9)=3\).
Take the smallest powers common to all three expressions: \(a^2\) and \(b^2\). There is no common factor \(c\), because the third term has no \(c\).
Therefore, G.C.D. \(=3a^2b^2\).
Q 1(v)
In \(\triangle ABC\), \(AB=AC\). If \(\angle BAC=80^\circ\), find the measure of \(\angle ABC\).
Since \(AB=AC\), the base angles are equal:
\(\angle ABC=\angle ACB\).
Using the angle-sum property:
\[80^\circ+\angle ABC+\angle ACB=180^\circ.\]
Thus \(2\angle ABC=100^\circ\), so \(\angle ABC=50^\circ\).
Q 1(vi)
In \(\triangle ABC\), if \(AC>AB\), then — (a) \(x=2y\) (b) \(x=y\) (c) \(x=\frac35y\) (d) none of these.
2. Answer each of the following questions
Q 2(i)
\(12\frac12\%\) of what amount is Rs. 320?
Let the amount be \(x\).
\[12\frac12\%=\frac{25}{2}\%=\frac18.\]
Therefore, \(\frac18x=320\).
So \(x=320\times8=2560\).
Q 2(ii)
Express \(0.1\overline{6}\) in percentage.
Here \(0.1\overline6=0.1666\ldots=\frac16\).
Hence percentage \(=\frac16\times100=16\frac23\%\).
Q 2(iii)
Simplify: \(\dfrac{a-b-c}{a}+\dfrac{a+b+c}{a}\).
Take the common denominator \(a\):
\[\frac{a-b-c+a+b+c}{a}=\frac{2a}{a}=2.\]
Q 2(iv)
Find the G.C.D. of \((x^3-3x^2y)\) and \((x^2-9y^2)\).
Factor both expressions:
\[x^3-3x^2y=x^2(x-3y),\]
\[x^2-9y^2=(x-3y)(x+3y).\]
The common factor is \(x-3y\).
Q 2(v)
In \(\triangle ABC\), \(AB=AC\). If \(\angle BAC=70^\circ\), which side is greatest in length?
Since \(AB=AC\), the base angles are equal:
\[\angle ABC=\angle ACB=\frac{180^\circ-70^\circ}{2}=55^\circ.\]
The largest angle is \(70^\circ\), opposite side \(BC\). Therefore \(BC\) is the greatest side.
Q 2(vi)
In \(\triangle ABC\), \(BC\) is produced to \(D\). If \(\angle ACD=112^\circ\) and \(\angle ABC=60^\circ\), find \(\angle BAC\).
By the exterior-angle theorem,
\[\angle ACD=\angle BAC+\angle ABC.\]
Therefore \(112^\circ=\angle BAC+60^\circ\).
So \(\angle BAC=52^\circ\).
3. Answer any ten questions
Q 3(i)
The length of each arm of a square was increased by 10%. Find the percentage increase in area.
Let the original side be \(s\). Original area \(=s^2\).
New side \(=110\%\text{ of }s=1.1s\).
New area \(=(1.1s)^2=1.21s^2\).
Increase \(=1.21s^2-s^2=0.21s^2\), i.e. 21% of the original area.
Q 3(ii)
When water freezes into ice, it increases in volume by 10%. Find in percentage how much it will decrease in volume if the ice melts into water.
Let the original water volume be 100 units. Ice volume becomes 110 units.
On melting, it returns to 100 units. Decrease from the ice volume:
\[\frac{110-100}{110}\times100=\frac{10}{110}\times100=9\frac1{11}\%.\]
Q 3(iii)
In a certain type of brass, the ratio of copper and zinc is 5:2. What will be the ratio of copper and zinc in 28 kg of brass if 4 kg of copper is added to it?
Total parts \(=7\). In 28 kg, copper \(=28\times\frac57=20\) kg and zinc \(=28\times\frac27=8\) kg.
After adding 4 kg copper: copper \(=24\) kg, zinc \(=8\) kg.
Thus ratio \(=24:8=3:1\).
Q 3(iv)
Resolve into factors: \(6x^2-x-15\).
We need two numbers whose product is \(6(-15)=-90\) and sum is \(-1\): they are \(-10\) and \(9\).
\[6x^2-10x+9x-15=2x(3x-5)+3(3x-5).\]
Therefore \((2x+3)(3x-5)\).
Q 3(v)
Find the G.C.D. of \(2ax(a-x)^2\) and \(4a^2x(a-x)^3\).
Numerical G.C.D. is \(2\). The common variable factors are \(a\), \(x\), and \((a-x)^2\).
Hence G.C.D. \(=2ax(a-x)^2\).
Q 3(vi)
Express in reduced form: \(\dfrac{a+1}{a-2}\times\dfrac{a^2-a-2}{a^2+a}\).
Factor:
\[a^2-a-2=(a-2)(a+1),\qquad a^2+a=a(a+1).\]
Thus
\[\frac{a+1}{a-2}\times\frac{(a-2)(a+1)}{a(a+1)}=\frac{a+1}{a}.\]
Q 3(vii)
Simplify: \(\dfrac{a}{a^2+ab}-\dfrac{b}{(a+b)^2}\).
Since \(a^2+ab=a(a+b)\),
\[\frac{a}{a(a+b)}-\frac{b}{(a+b)^2}=\frac1{a+b}-\frac{b}{(a+b)^2}.\]
Taking the common denominator:
\[\frac{a+b-b}{(a+b)^2}=\frac{a}{(a+b)^2}.\]
Q 3(viii)
Resolve into factors by expressing as the difference of two squares: \(x^2-2x-3\).
Complete the square:
\[x^2-2x-3=(x-1)^2-4=(x-1)^2-2^2.\]
Using \(A^2-B^2=(A-B)(A+B)\):
\[(x-3)(x+1).\]
Q 3(ix)
Find the L.C.M. of \((x^2y^2-x^2)\) and \((xy^2-2xy+x)\).
Factor:
\[x^2y^2-x^2=x^2(y-1)(y+1),\]
\[xy^2-2xy+x=x(y-1)^2.\]
Take the highest powers of all factors:
\[\mathrm{LCM}=x^2(y-1)^2(y+1).\]
Q 3(x)
Resolve into factors: \(a^2+1-\dfrac6{a^2}\).
Put over the common denominator \(a^2\):
\[a^2+1-\frac6{a^2}=\frac{a^4+a^2-6}{a^2}.\]
Let \(t=a^2\). Then \(t^2+t-6=(t+3)(t-2)\).
Hence
\[\boxed{\frac{(a^2+3)(a^2-2)}{a^2}}.\]
Q 3(xi)
In an isosceles obtuse-angled triangle, the measurement of an acute angle is \(\frac13\) of the measurement of the obtuse angle. Write the measurements of all angles.
Let the obtuse angle be \(O\). Since the triangle is isosceles, the two acute angles are equal; each is \(O/3\).
Using the angle sum:
\[O+\frac O3+\frac O3=180^\circ\Rightarrow\frac{5O}{3}=180^\circ.\]
Thus \(O=108^\circ\), and each acute angle is \(36^\circ\).
Q 3(xii)
Show that the hypotenuse of a right-angled triangle is the greatest side.
Let \(\triangle ABC\) be right-angled at \(A\). Then \(\angle A=90^\circ\), while \(\angle B\) and \(\angle C\) are each less than \(90^\circ\).
The side opposite the greater angle is greater. Since \(90^\circ\) is the greatest angle, the side opposite it, \(BC\), is the greatest side. Therefore \(BC\), the hypotenuse, is the greatest side.
4. Answer any four questions
Q 4(i)
A company has got the word of unloading goods from a ship in 10 days. 280 heads of people have been employed. After 3 days it is seen that \(\frac14\) of the work has been completed. How many heads of people are to be engaged to complete the work in time? [Using the Rule of Three]
280 people working for 3 days complete \(\frac14\) of the work.
Therefore, total work \(=280\times3\times4=3360\) person-days.
After 3 days, remaining work \(=\frac34\), requiring \(3360\times\frac34=2520\) person-days.
Remaining time \(=10-3=7\) days.
People required for remaining work \(=2520/7=360\).
Thus 360 people must be working during the remaining 7 days. Since 280 are already employed, 80 additional people are needed.
Q 4(ii)
A power-loom is \(2\frac14\) times more powerful than a hand-loom. 12 hand-looms weave 1080 metres of cloth in 18 days. How many power-looms will be required to weave 2700 metres in 15 days?
One hand-loom's daily output is proportional to \(1080/(12\times18)=5\) metres per day.
A power-loom is \(2\frac14=\frac94\) times as productive, so one power-loom produces \(5\times\frac94=\frac{45}{4}=11.25\) m/day.
Output needed per day for 2700 m in 15 days:
\[2700/15=180\text{ m/day}.\]
Number of power-looms:
\[180\div11.25=16.\]
Q 4(iii)
Due to use of high-yielding seed Utpalbabu has got 55% production hike in paddy cultivation. But for this the cost of cultivation has increased by 40%. Previously a yield of Rs. 3000 was produced by investing Rs. 1200. Calculate whether his income will be increased or decreased after using high-yielding seeds.
Old production value = Rs. 3000; old cost = Rs. 1200.
Old net income = \(3000-1200=Rs.1800\).
New production value after 55% increase:
\[3000\times1.55=Rs.4650.\]
New cultivation cost after 40% increase:
\[1200\times1.40=Rs.1680.\]
New net income:
\[4650-1680=Rs.2970.\]
Increase in income = \(2970-1800=Rs.1170\).
Percentage increase:
\[\frac{1170}{1800}\times100=65\%.\]
Q 4(iv)
In a vessel of beverage the ratio of syrup and water is 5:2. What part of the drink should be removed and replaced by water so that the volume of syrup and water becomes equal?
Let total volume be 1 unit. Initially syrup \(=5/7\) and water \(=2/7\).
Let fraction \(x\) of the drink be removed. The removed mixture has syrup \(5x/7\), so remaining syrup is \(\frac57(1-x)\). After replacing the removed amount with water, water becomes \(\frac27+x\).
For equal amounts:
\[\frac57(1-x)=\frac27+x.\]
Multiplying by 7: \(5-5x=2+7x\), so \(3=12x\), hence \(x=\frac14\).
Q 4(v)
Two different types of brass contain copper and zinc in the ratios 8:3 and 15:7 respectively. What is the ratio of copper and zinc when these two types of brass are mixed in the ratio 5:2?
Take 5 units of the first brass and 2 units of the second.
Copper = \(5\times\frac8{11}+2\times\frac{15}{22}=\frac{40}{11}+\frac{15}{11}=5\).
Zinc = \(5\times\frac3{11}+2\times\frac7{22}=\frac{15}{11}+\frac7{11}=2\).
Therefore copper:zinc = \(5:2\).
Q 4(vi)
40% of the gross receipts of a Tramway company is taken up in meeting the working expenses, 40% of the remainder goes to reserve fund and the balance is paid away as dividends to shareholders at the rate of \(3\frac13\%\) on their shares, the total value of which is Rs. 864000. Find the amount of the gross receipts.
Let gross receipts be \(R\).
After 40% working expenses, 60% remains. Of this remainder, 40% goes to reserve, leaving 60% of 60%:
\[0.60\times0.60R=0.36R.\]
Dividend rate = \(3\frac13\%=\frac{10}{3}\%\).
Dividend paid:
\[864000\times\frac{10}{300}=Rs.28800.\]
Therefore \(0.36R=28800\), so
\[R=\frac{28800}{0.36}=Rs.80000.\]
5. Answer any seven questions
Q 5(i)
Resolve into factors: \((x+1)(x+9)(x+5)^2+63\).
Since \((x+1)(x+9)=x^2+10x+9=(x+5)^2-16\),
\[(x+1)(x+9)(x+5)^2+63=((x+5)^2-16)(x+5)^2+63.\]
Let \(t=(x+5)^2\). Then \(t^2-16t+63=(t-7)(t-9)\).
Thus
\[((x+5)^2-7)((x+5)^2-9)\]
\[=(x^2+10x+18)(x+2)(x+8).\]
Q 5(ii)
Resolve into factors: \(x^2+20xy-96y^2\). (At first eliminate the variable from the last term.)
We need two terms whose product is \(-96y^2\) and sum is \(20y\): \(24y\) and \(-4y\).
\[x^2+24xy-4xy-96y^2\]
\[=x(x+24y)-4y(x+24y).\]
Q 5(iii)
Resolve into factors by expressing as the difference of two squares: \(3a^2-2a-5\).
\[3a^2-2a-5=3a^2-2a+\frac13-\frac{16}{3}.\]
\[=\frac{(3a-1)^2-16}{3}=\frac{(3a-5)(3a+3)}{3}.\]
Therefore \((a+1)(3a-5)\).
Q 5(iv)
Find the G.C.D. of \(8(x^2-4),\ 12(x^3+8),\ 36(x^2-3x-10)\).
Factor:
\[8(x^2-4)=8(x-2)(x+2),\]
\[12(x^3+8)=12(x+2)(x^2-2x+4),\]
\[36(x^2-3x-10)=36(x-5)(x+2).\]
The common numerical factor is \(\gcd(8,12,36)=4\), and the common algebraic factor is \(x+2\).
Q 5(v)
Find the L.C.M. of \(x^4+x^2y^2+y^4,\ x^3y+y^4,\ (x^2-xy)^3\).
Factor each:
\[x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2),\]
\[x^3y+y^4=y(x+y)(x^2-xy+y^2),\]
\[(x^2-xy)^3=x^3(x-y)^3.\]
Taking every factor to its highest power:
\[\mathrm{LCM}=x^3y(x-y)^3(x+y)(x^2+xy+y^2)(x^2-xy+y^2).\]
Q 5(vi)
Simplify:
\[\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)}.\]
Rewrite denominators using consistent signs:
\[(a-b)(a-c),\quad (b-c)(b-a)=-(b-c)(a-b),\quad (c-a)(c-b)=(a-c)(b-c).\]
Putting the three terms over the common denominator \((a-b)(a-c)(b-c)\), the numerator becomes
\[(b+c-a)(b-c)-(c+a-b)(a-c)+(a+b-c)(a-b),\]
which expands and cancels completely to 0.
Q 5(vii)
Resolve into factors: \(a(a+1)x^2-x-a(a-1)\).
Try factors whose product gives the constant term \(-a(a-1)\):
\[a(a+1)x^2-x-a(a-1)=(ax-a+x)(ax+a-1).\]
Expanding verifies the result:
\[(ax-a+x)(ax+a-1)=a(a+1)x^2-x-a(a-1).\]
Q 5(viii)
Simplify:
\[\frac1{x^2-8x+15}+\frac1{x^2-4x+3}-\frac2{x^2-6x+5}.\]
Factor denominators:
\[x^2-8x+15=(x-3)(x-5),\]
\[x^2-4x+3=(x-1)(x-3),\]
\[x^2-6x+5=(x-1)(x-5).\]
With common denominator \((x-1)(x-3)(x-5)\), the numerator is
\[(x-1)+(x-5)-2(x-3)=0.\]
Q 5(ix)
Resolve into factors: \(x^6y^6-9x^3y^3+8\).
Let \(z=x^3y^3\). Then
\[z^2-9z+8=(z-1)(z-8).\]
Therefore
\[(x^3y^3-1)(x^3y^3-8).\]
Using difference of cubes:
\[x^3y^3-1=(xy-1)(x^2y^2+xy+1),\]
\[x^3y^3-8=(xy-2)(x^2y^2+2xy+4).\]
6. Answer any two questions – Proofs
Q 6(i)
Prove that the lengths of opposite sides of two angles equal in measurement of a triangle are equal.
Let \(\triangle ABC\) have \(\angle B=\angle C\). We have to prove \(AC=AB\).
Draw the internal bisector \(AD\) of \(\angle A\).
In triangles \(ABD\) and \(ACD\):
- \(\angle BAD=\angle DAC\) (angle-bisector construction),
- \(\angle ABD=\angle ACD\) because \(\angle B=\angle C\),
- \(AD=AD\) (common side).
Therefore the two triangles are congruent by AAS. Hence corresponding sides are equal:
\[AB=AC.\]
Q 6(ii)
Prove that the sum of the measurements of three angles is two right angles.
Let \(\triangle ABC\) be any triangle. Through \(A\), draw a line parallel to \(BC\).
The angle made by \(AB\) with this parallel line equals \(\angle B\), and the angle made by the parallel line with \(AC\) equals \(\angle C\), by alternate interior angles.
These two angles together with \(\angle A\) form a straight angle:
\[\angle B+\angle A+\angle C=180^\circ.\]
Since \(180^\circ\) is two right angles, the sum of the three angles is two right angles.
Q 6(iii)
Prove that if two sides of a triangle are unequal in length, the angle opposite to the greater side is greater than the angle opposite to the smaller side.
Let \(AB>AC\) in \(\triangle ABC\). We must prove \(\angle C>\angle B\).
Choose a point \(D\) on \(AB\) such that \(AD=AC\), and join \(CD\).
Then \(\triangle ACD\) is isosceles, so
\[\angle ACD=\angle CDA.\]
Since \(D\) lies on \(AB\), \(\angle CDA\) is an exterior angle of \(\triangle BCD\). Hence
\[\angle CDA=\angle CBD+\angle BCD=\angle B+\angle BCD.\]
Also
\[\angle C=\angle ACD+\angle BCD=\angle CDA+\angle BCD=\angle B+2\angle BCD.\]
Therefore \(\angle C>\angle B\).
7. Answer any two questions – Geometry
Q 7(i)
In right-angled triangle \(ABC\), \(\angle BAC=90^\circ\) and \(D\) is a point on the hypotenuse such that \(BD=AD\); prove that \(D\) is the midpoint of \(BC\).
Counterexample: take a right triangle with \(A=(0,0)\), \(B=(3,0)\), \(C=(0,4)\). Then \(BC=5\).
A point \(D\) on \(BC\) can satisfy \(AD=BD\) without being the midpoint. In fact, parameterising \(D=B+t(C-B)\), the condition \(AD=BD\) gives
\[9-18t+16t^2=0,\]
whose roots are \(t=\frac38\) and \(t=\frac34\). Neither is \(\frac12\), the midpoint value.
Thus the printed assertion is false. A likely intended condition may have been different, but it should not be silently changed.
Q 7(ii)
In \(\triangle ABC\), the internal bisector of \(\angle ABC\) and external bisector of \(\angle ACB\) meet at \(D\). Prove that \(\angle BDC=\frac12\angle BAC\).
Let \(\angle A=A\), \(\angle B=B\), and \(\angle C=C\).
Because \(BD\) bisects the internal angle at \(B\),
\[\angle DBC=\frac B2.\]
Since \(CD\) is the external angle bisector at \(C\), in the standard configuration in which the two bisectors meet outside the triangle,
\[\angle BCD=90^\circ+\frac C2.\]
Therefore, in \(\triangle BDC\),
\[\angle BDC=180^\circ-\frac B2-\left(90^\circ+\frac C2\right).\]
Hence
\[\angle BDC=90^\circ-\frac{B+C}{2}.\]
But \(A+B+C=180^\circ\), so
\[90^\circ-\frac{B+C}{2}=90^\circ-\frac{180^\circ-A}{2}=\frac A2.\]
Thus
\[\boxed{\angle BDC=\frac12\angle BAC}.\]
Q 7(iii)
In \(\triangle ABC\), the bisectors of \(\angle ABC\) and \(\angle ACB\) meet at \(I\). If \(AB>AC\), prove that \(IB>IC\).
Since \(AB>AC\), the angle opposite \(AB\) is greater than the angle opposite \(AC\):
\[\angle C>\angle B.\]
Because \(BI\) and \(CI\) are angle bisectors,
\[\angle IBC=\frac B2,\qquad \angle BCI=\frac C2.\]
In \(\triangle BIC\), the side \(IB\) is opposite \(\angle BCI=C/2\), while \(IC\) is opposite \(\angle IBC=B/2\).
Since \(C/2>B/2\), the side opposite it is greater:
\[IB>IC.\]
8. Answer any one question – Construction
Q 8(i)
Divide a line segment of length 12.6 cm into 7 equal parts, by scale and compass. Using the above construction draw an equilateral triangle of length 7.2 cm.
Part A: Divide 12.6 cm into 7 equal parts.
- Draw a line segment \(AB=12.6\) cm.
- From \(A\), draw any ray \(AX\) making an acute angle with \(AB\).
- On \(AX\), mark seven equal intervals using the compass: \(A_1,A_2,\ldots,A_7\).
- Join \(A_7\) to \(B\).
- Through \(A_1,A_2,\ldots,A_6\), draw lines parallel to \(A_7B\). They meet \(AB\) at six points, dividing \(AB\) into seven equal parts.
- Each part is \(12.6/7=1.8\) cm.
Part B: Construct an equilateral triangle of side 7.2 cm.
- Since one constructed part is 1.8 cm, four such parts give \(4\times1.8=7.2\) cm.
- Construct \(PQ=7.2\) cm by laying off four equal 1.8 cm intervals.
- With centre \(P\) and radius \(PQ\), draw an arc.
- With centre \(Q\) and the same radius, draw another arc cutting the first at \(R\).
- Join \(PR\) and \(QR\). Then \(PQ=QR=RP=7.2\) cm.
Q 8(ii)
Draw the median \(AD\) of a scalene triangle by compass and trisect \(AD\) in \(AE,EF,FD\). Join \(B\) and \(F\) and extend it to \(P\) such that it intersects \(AC\) at \(P\). Find the relation between \(AP\) and \(CP\) after measuring the length by scale.
Construction:
- Draw a scalene triangle \(ABC\).
- Construct the midpoint \(D\) of \(BC\); then \(AD\) is the median.
- Trisect \(AD\) so that \(AE=EF=FD\).
- Join \(BF\) and extend it to meet \(AC\) at \(P\).
Relation: Since \(AD\) is a median, \(D\) is the midpoint of \(BC\). Take vectors with \(A\) as origin. If the position vectors of \(B,C\) are \(\mathbf b,\mathbf c\), then
\[\mathbf D=\frac{\mathbf b+\mathbf c}{2}.\]
Since \(F\) is two-thirds of the way from \(A\) to \(D\),
\[\mathbf F=\frac23\mathbf D=\frac{\mathbf b+\mathbf c}{3}.\]
A point on line \(BF\) has the form \(\mathbf b+t(\mathbf F-\mathbf b)\). For it to lie on \(AC\), its coefficient of \(\mathbf b\) must be zero. This gives \(t=\frac32\), and the point becomes
\[\mathbf P=\frac12\mathbf c.\]
Thus \(P\) is the midpoint of \(AC\). Therefore
\[\boxed{AP=CP}.\]
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