Class 10

[CBSE 2025] In order to organise, Annual Sports Day, a school prepared an eight lane. . .

In order to organise, Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below :

The length of the innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane.

Based on given information, answer the following questions, using the concept of Arithmetic Progression.

(i) What is the length of the 6th lane? [1 Marks]

(ii) How long is the 8th lane than that of the 4th lane? [1 Marks]

(iii) (a) While practicing for a race, a student took one round each in the first six lanes. Find the total distance covered by the student. [2 Marks]

(b) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student. [2 Marks]

Solution :

Here a=400 m

d=7.6 m

(i) What is the length of the 6th lane? Ans : an=a+(n-1)d a6=400+(6-1)7.6 a₆=400+5×(7.6)
a₆=400+38
a₆=438

(ii) How long is the 8th lane than that of the 4th lane? Ans : a₈-a₄ = a+7d-(a+3d) = a+7d-a-3d = 4d = 4(7.6)

= 30.4 m

(iii) (a) While practicing for a race, a student took one round each in the first six lanes. Find the total distance covered by the student.

Ans : S_n=n/2[2a+(n-1)d]
S₆=6/2[2(400)+(6-1)7.6]
S₆=3[800+5×7.6]
S₆=3[800+38]
S₆=3[838]
S₆=2514 m

(iv) (b) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student.
Ans :
S₈=8/2[2(400)+(8-1)7.6]
S₈=4[800+7×7.6]
S₈=4[800+53.2]
S₈=4[853.2]
S₈=3412.8 m

S₃=3/2[2(400)+(3-1)7.6]
S₃=1.5[800+2×7.6]
S₃=1.5[800+15.2]
S₃=1.5[815.2]
S₃=1222.8 m

S₈-S₃ =3412.8-1222.8 =2190 m

[CBSE 2025] In an equilateral triangle of side 10 cm, equilateral triangles of side 1 . . .

In an equilateral triangle of side 10 cm, equilateral triangles of side 1 cm are formed as shown in the figure below, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row and so on.

Based on the given information, answer the following questions using Arithmetic Progression.

(i) How many triangles will be there in bottom most row?

(ii) How many triangles will be there in fourth row from the bottom?

(iii) (a) Find the total number of triangles of side 1 cm each till 8th row.

(b) How many more number of triangles are there from the 5th row to the 10th row than in the first 4 rows? Show working.

[CBSE 2025] A school is organising a charity run to raise funds for a local hospital. The run . . .

A school is organising a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organisers decided to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.

Based on the information given above, answer the following questions :

(i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. [1 Mark]

(ii) Determine the distance of the 8th round. [1 Mark]

(iii) (a) Find the total distance run after completing all 10 rounds. [2 Marks]

(iii) (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner ? [2 Marks]

The minimum age of children eligible to participate in a painting competition . . .

Question : The minimum age of children eligible to participate in a painting competition is 8 years. It is observed that the age of the youngest boy was 8 years and the ages of the participants, when seated in order of age, have a common difference of 4 months. If the sum of the ages of all the participants is 168 years, find the age of the eldest participant in the painting competition.

CBSE Board Exam 2025

Solution :

Minimum age of children
= 8 years
= 8×12 = 96 months

So,

First Term of the AP (a) = 96 months
Common difference (d) = 4 months

Sum of the ages of all the participants (S_n)
= 168 years
= 168×12
= 2016 months

Let total number of participants be 'n'.

S_n=n/2[2a+(n-1)d]

2016=n/2[2(96)+(n-1)(4)]
2016=n/2[192+4n-4]
2016=n/2[188+4n]
2016=[n/2]×2[94+2n]
2016=n[94+2n]
2016=94n+2n²
2n²+94n-2016=0
2[n²+47n-1008]=0
n²+47n-1008=0
n²+(63-16)n-1008=0
n²+63n-16n-1008=0
n(n+63)-16(n+63)=0
(n+63)(n-16)=0
n+63=0
n=-63
∵n≠-63
So n=-63 is rejected.

n-16=0
n=16

So, there are 16 students.

Now,
a_n=a+(n-1)d
a16=96+(16-1)4
a16=96+15×4
a16=96+60
a16=156
=156 months
=156/12
= 13 years

So, age of eldest child is 156 months or 13 years.

If a=2^2×3^x, b=2^2×3×5, c=2^2×3×7 and . . .

Question : If a=2²×3^x, b=2²×3×5, c=2²×3×7 and LCM (a, b, c) = 3780, then x is equal to

(A) 1
(B) 2
(C) 3
(D) 0

CBSE Sample Question Paper 2025-2026

Solution :

a=2²×3^x

b=2²×3×5

c=2²×3×7

LCM (a,b,c) = 2²×3^x×5×7 — (1)

LCM (a,b,c) = 3780 (Given)
LCM (a,b,c) = 2²×3³×5×7 — (2)

From (1) and (2)
2²×3^x×5×7=2²×3³×5×7
3^x=3³

Clearly x=3

So, (C) 3, would be the correct option.

If a=5^13×7, b=3×5^25×7^6, c=5^n×7^4 . . .

Question : If a=5¹³×7, b=3×5²⁵×7⁶, c=5ⁿ×7⁴ and LCM (a,b,c) = 3×5³¹×7⁶, then n=?

(a) 31

(b) 25
(c) 13
(d) 34

Doubt by Muskan

Solution :

a=5¹³×7
b=3×5²⁵×7⁶
c=5ⁿ×7⁴

^{LCM(a,b,c) =}3×5ⁿ×7⁶^{— (1)
LCM (a,b,c) =}3×5³¹×7⁶ ^{— (2) [Given]

From (1) and (2)}

3×5ⁿ×7⁶ = 3×5³¹×7⁶
5ⁿ = 5³¹
n=31

Hence, (a) 31, would be the correct option.

Prove that sinA(1+tanA)+cosA(1+cotA) . . .

Question : Prove that sinA(1+tanA)+cosA(1+cotA) = secA+cosecA

Doubt by Afifa

Solution : OR

sinA(1+tanA)+cosA(1+cotA) = secA+cosecA

LHS
sinA(1+tanA)+cosA(1+cotA)
=sinA(1+sinA/cosA) + cosA(1+cosA/sinA)

[∵tanA=sinA/cosA, cotA=cosA/sinA]
=sinA[(cosA+sinA)]/cosA+cosA[(sinA+cosA)]/sinA
=sinA[(sinA+cosA)]/cosA+cosA[(sinA+cosA)]/sinA

=sinA+cosA[sinA/cosA+cosA/sinA]
=sinA+cosA[(sin²A+cos²A)/sinAcosA]
=sinA+cosA[1/sinAcosA]
[∵sin²A+cos²A=1]

=(sinA+cosA)/sinAcosA
=sinA/sinAcosA + cosA/sinAcosA
= 1/cosA + 1/sinA

= secA+cosecA
[∵1/cosA=secA and 1/sinA=cosecA]
= RHS

LHS=RHS
Hence Proved.

In ΔABC, ∠ACB=90° and DCEF is a square . . .

Question : In ΔABC, ∠ACB=90° and DCEF is a square, as shown in figure. Prove that DE²=AD×BF

Doubt by Afifa

Solution :

DCEF is a square.
DE=EF=FC=CD (All sides of square are equal)

In ΔACB and ΔADE
∠A=∠A (Common)
∠ACB=∠ADE (Each 90°)
ΔACB~ΔADE (By AA Similarity Criteria) — (1)

In ΔACB and ΔEFB
∠B=∠B (Common)
∠ACB=∠EFB (Each 90°)
ΔACB~ΔEFB (By AA Similarity Criteria) — (2)

From equation (1) & (2)

ΔADE~ΔEFB

AD/EF=ED/BF (By CPCT)
AD×BF=EF×ED
AD×BF=EF×DE
AD×BF=DE×DE [∵EF=DE]
AD×BF=DE²
DE²=AD×BF

Hence Proved.

If the ratio of the sum of the first n terms of two A.P.s is (7n+1) . . .

Question : If the ratio of the sum of the first n terms of two A.P.s is (7n+1):(4n+27), then find the ratio of their 9th terms.

Doubt by Shaurya

Solution :

Let us consider two AP's having first term and common difference as a and a' and d and d' respectively.

S_n:S_n'=(7n+1):(4n+27)
n/2[2a+(n-1)d] : n/2[2a'+(n-1)d']=(7n+1):(4n+27)
[2a+(n-1)d]:[2a'+(n-1)d']=(7n+1):(4n+27)

2[a+{(n-1)/2}d]:2[a'+{(n-1)/2}d']=(7n+1):(4n+27)
[a+{(n-1)/2}d]:[a'+{(n-1)/2}d']=(7n+1):(4n+27) — (1)

Now in order to find the ratio of 9th term
{(n-1)/2} should be equal to 8
{(n-1)/2}=8
n-1=16
n=16+1
n=17

Putting n=17 in equation (1)

a+8d:a'+8d=[7(17)+1]:[4(17)+27]
a+(9-1)d : a'+(9-1)d'=[119+1] : [68+27]
a₉:a'₉=120:95
a₉:a'₉=24:19

Hence, the ratio of 9th term is 24:19

Divide 32 into four parts which are in A.P. such that the product

Question : Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7:15.

Doubt by Shaurya

Solution :

Let the four parts which are in AP are
a₁=a-3d
a₂=a-d
a₃=a+d
a₄=a+3d

ATQ

a₁+a₂+a₃+a₄=32

a-3d+a-d+a+d+a+3d=32

4a=32

a=32/4

a=8 — (1)

(a₁×a₄):(a₂×a₃) = 7:15

(a₁×a₄)/(a₂×a₃) = 7/15

(a-3d)(a+3d)/(a-d)(a+d)=7/15
(a²-9d²)/(a²-d²)=7/15
(a²-9d²)/(a²-d²)=7/15
15(a²-9d²)/=7×(a²-d²)
15a²-135d²=7a²-7d²
15a²-7a²=135d²-7d²
8a²=128d²
8(8)²/128=d² [From equation (1)]
8×64/128=d²
8/2=d²
4=d²
d²=4
d=±√4

d=±2

When d=+2, a=8
a₁=a-3d = 8-3(2) = 8-6 = 2

a₂=a-d = 8-2 = 6
a₃=a+d = 8+2 = 10
a₄=a+3d = 8+3(2) = 8+6 = 14

When d=-2, a=8

a₁=a-3d = 8-3(-2) = 8+6 = 14

a₂=a-d = 8-(-2) = 8+2 = 10
a₃=a+d = 8+(-2) = 8-2 = 6
a₄=a+3d = 8+3(-2) = 8-6 = 2

Hence, the required four parts are 2, 6, 10, 14 or 14, 10, 6, 2

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