# Most Important Maths Questions of 3 and 4 marks for Class 10 CBSE-1

10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks Part-1 are created here, after examining the last year’s maths question papers of class 10 CBSE Board.The reason of selecting of 3 and 4 marks maths questions from last year’s question papers is based on the fact that these maths questions of 3 and 4 marks represent 65-70 % of the total marks of the maths question paper, therefore for achieving maximum percentage in maths every student need special attention to 3 and 4 marks question since the practice of these maths questions leads the complete preparation of maths.

Here, CBSE class 10 maths most important questions and answers for the 2019-20 board part I published by Future Study Point are presented for the students of X th class who are going to appear in the 2020-21 board exam of CBSE.

All the questions in CBSE class 10 maths most important questions and answers for 2020-21 board-I are designed by the expert of the subject and their answers are readily explained in an ordered way.

These CBSE class 10 maths most important questions and answers for 2020-21 board-I are designed as per the updated syllabus of CBSE, so you need not to have any kinds of doubt in the selections of the questions. Don’t forget to make a comment and to subscribe us.

Why do you need to study 3 and 4 marks questions of maths

- These maths questions represent 65-70% of total marks
- The preparation of these questions also handles the preparation of other questions
- Helpful in increasing the marks in maths
- Increases the maths skill of the students

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**Most Important Maths Questions of 3 and 4 marks for Class 10 CBSE-1**

**Q1. Prove that n³ – n is divisible by 6, where n is a positive integer.**

Ans. n is given to us a positive integer

n³–n = n(n² – 1)= (n –1)n(n +1)

A number divisible by 2 and 3 both then it will be divisible by 6

So let us show (n –1)n(n +1) is divisible by 3 as well as by 2

Let it is divisible by 3, according to Euclid’s theorem the given expression can be written in the form of 3q + r ,then possible remainders will be 0, 1,2 and corresponding integers will be 3q, 3q +1, 3q +2

When n = 3q

(n–1)n(n +1) =(3q–1)3q(3q +1)

When n= 3q +1

(n–1)n(n +1) = (3q +1 –1)(3q +1)(3q + 1 +1)= 3q(3q+1)(3q +2)

When n = 3q + 2

(n–1)n(n +1) = (3q + 2 –1)(3q +2)(3q +2 +1)= (3q +1)(3q +2)(3q +3)=3(q +1)(3q +1)(3q +2)

In every case the given expression is divisible by 3

Now let the given expression is divisible by 2

Then according to Euclid’s theorem the given expression can be written as 2q+r where possible value of r = 0,1. So corresponding value of the expression will be 2q, 2q +1

(n–1)n(n +1) = (2q–1) 2q (2q +1)

(n–1)n(n +1)=(2q +1 –1) (2q +1) (2q +1 +1) =2q(2q +1)(2q + 2)

In evry case the expression is divisible by 2

We have proved that given number is divisible by 2 as well by 3, so it will also be divisible by 2 ×3=6.

Hence proved

**Q2. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of the number of houses preceding the house numbered x is equal to the sum of the number of houses following x**.

Ans. The houses are numbered as follows,together we are also given a house number x within 1 to 49.

1,2,3,4…….x–1,x,x +1,x+2………..47,48,49

We are given the sum of house numbers preceding the house numbered x = Sum of the house numbers followed by x

The houses from 1 to x – 1 will be as given bellow

Let S is the sum of the numbers of houses preceding the house no.x and S’ is the sum of the numbers of houses followed by x

a =1, d =1, l =x –1

And

According to question

(x –1)x = (49 – x)(x + 50)

x² –x = 49x +2450 –x² –50x

2x² = 245o

x ² =1225

x = √(1225)

x = 35

Therefore,the house no x = 35 exist as described in the question.

**Q3.The first and the last terms of an AP are 10 and 361 respectively. If its common difference is 9 then find the number of terms and their total sum?**

Ans. Let the total terms of the given AP in the question =n

a = 10, a_{n }= 361

d = 9

a_{n }= a + (n -1)d

361 = 10 + (n – 1) 9

9n – 9 +10 = 361

9n = 360

n = 40

Placing the value of n = 40 in the following formula

S_{40 }= 20 × 371 = 7420

Therefore the number of terms in AP is 40 and sum of the terms of AP is 7420

**Most Important Maths Questions of 3 and 4 marks for Class 10 CBSE-I**

**Q4. A motorboat whose speed is 24 km/hr in still water takes 1 hr more to go 32km upstream than to return downstream to the same spot. Find the speed of the stream.**

Ans. Let the speed of stream = x km/hr

Speed of motorboat in still water= 24 km/hr

Distance covered by the motorboat to go upstream = 32 km

The speed of the boat to go upstream = 24–x

Speed of boat towards downstraem= 24+x

Let time taken by motorboat to go upstream = t_{1 }and down stream = t_{2}

Time = Distance/Time taken

Therefore

According to the question

t_{1 }= t_{2 }+ 1

64x = 576 –x²

x² + 64x –576 = 0

x² + 72x – 8x – 576 = 0

x(x + 72) – 8(x + 72) = 0

(x + 72)( x – 8) =0

x = –72, x = 8

The speed of stream can not be negative,so avoiding –72

Therefore the speed of stream will be 8 km/hr

**Q5.A train was to cover a distance of 2800km in prefixed time duration. But due to the foggy weather, the speed of the train was slowed down which reduced the average speed of the train for the overall journey by 100 km/hr. The time duration for the journey also increased by 30 min. Determine the original speed and travel time taken by the train.**

Ans. Let the original speed of the train = x km/hr

The modified speed of the train due to fog = (x – 100 ) km/hr

Distance covered by the train = 2800 km

Let time taken by original speed is t_{1 }and by new speed is t_{2}

According to question time taken by new speed = 30 min + time taken by original speed

30 min = 30/60 = 1/2 hr

x² – 100x – 560000 =0

x² – 800x + 700x –560000 =0

x(x – 800) + 700(x – 800) =0

(x – 800)(x + 700) =0

x = 800, x = –700

The original speed of the train is 800 km/hr.

And travel time is taken by the train= 2800/800= 28/8= 3.5 hr

**Q6.If p ^{th}**

**,q**

^{th}and r^{th}terms of an AP are x, y and z consecutively prove that** x(q –r) + y(r – p) + z(p – q) = 0**

Answer.

As we know the n term of an AP = a + (n –1)d

x = a + (p – 1)d…….(1)

y = a + (q – 1)d……..(2)

z = a + (r – 1)d………(3)

Substituting the value of x, y, and z in RHS of the equation

[ a + (p – 1)d](q – r) + [a + (q – 1)d](r –p) +[a + (r – 1)d](p –q)

=a(q – r) + (p – 1)(q – r) d + a(r –p) + (q – 1)(r –p)d +a(p –q)+ (r – 1)(p –q)d

=a(q – r + r –p + p –q ) + d(pq –pr –q +r + qr –pq –r +p +pr –qr –p +q)

= a.0 + d.0 = 0 = R.H.S

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