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Exercise 5.4 (Revised) - Chapter 5 Arithmetic Equations class 10 ncert solutions Maths - SaraNextGen [2024-2025]


Exercise 5.4 (Revised) : Chapter 5 - Arithmetic Equations - Ncert Solutions class 10 - Maths

Question 1:

Which term of the A.P. 121, 117, 113 … is its first negative term?

[Hint: Find n for an < 0]

Answer:

Given A.P. is 121, 117, 113 …

a = 121

= 117 − 121 = −4

an = a + (− 1) d

= 121 + (n − 1) (−4)

= 121 − 4n + 4

= 125 − 4n

We have to find the first negative term of this A.P.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2583/Chapter%205_html_m10433be9.gif

Therefore, 32nd term will be the first negative term of this A.P.

 

Question 2:

The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Answer:

We know that,

an = a + (− 1) d

a3 = a + (3 − 1) d

a3 = a + 2d

Similarly, a7 = a + 6d

Given that, a3 + a7 = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 − 4d (i)

Also, it is given that (a3) × (a7) = 8

(a + 2d) × (+ 6d) = 8

From equation (i),

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_42ec0cc.gif

From equation (i),

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_62337806.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_m4f834044.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_m32a05eb7.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_3b2fa20c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_4170f91f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_m2694699a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_79133e8b.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_m9988ffe.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_m6bdb6af0.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2586/Chapter%205_html_m20752670.gif

 

Question 3:

A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2591/Chapter%205_html_m22149725.gif m apart, what is the length of the wood required for the rungs?

[Hint: number of rungshttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2591/Chapter%205_html_74cabb49.gif ]

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2591/Chapter%205_html_m193818b0.jpg

Answer:

It is given that the rungs are 25 cm apart and the top and bottom rungs arehttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2591/Chapter%205_html_m22149725.gif m apart.

∴ Total number of rungshttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2591/Chapter%205_html_m55048702.gif https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2591/Chapter%205_html_m53d5f239.gif

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, = 45

Last term, l = 25

n = 11

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2591/Chapter%205_html_62aab945.gif

Therefore, the length of the wood required for the rungs is 385 cm.

 

Question 4:

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

Find this value of x.

[Hint Sx − 1 S49 − Sx]

Answer:

The number of houses was

1, 2, 3 … 49

It can be observed that the number of houses are in an A.P. having a as 1 and d also as 1.

Let us assume that the number of xth house was like this.

We know that,

Sum of n terms in an A.P.https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2597/Chapter%205_html_m409edaf2.gif

Sum of number of houses preceding xth house = Sx − 1

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2597/Chapter%205_html_60b2b5.gif

Sum of number of houses following xth house = S49 − Sx

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2597/Chapter%205_html_m50c96183.gif

It is given that these sums are equal to each other.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2597/Chapter%205_html_4f611e83.gif

However, the house numbers are positive integers.

The value of x will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

 

Question 5:

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m3c88992d.gif m and a tread of https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m5a4d85ce.gif m (See figure) calculate the total volume of concrete required to build the terrace.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_4cd512da.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m7469e79b.jpg

From the figure, it can be observed that

1st step is https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m5a4d85ce.gif m wide,

2nd step is 1 m wide,

3rd step is https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m47d40970.gif m wide.

Therefore, the width of each step is increasing by https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m5a4d85ce.gif m each time whereas their height https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m3c88992d.gif m and length 50 m remains the same.

Therefore, the widths of these steps are

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m41b881.gif

Volume of concrete in 1st step https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_1316f6a4.gif

Volume of concrete in 2nd step https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m34103f75.gif

Volume of concrete in 3rd step https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_78edcd5d.gif

It can be observed that the volumes of concrete in these steps are in an A.P.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_m7435c6e2.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_485bcd21.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/132/2603/Chapter%205_html_7ff05627.gif

Volume of concrete required to build the terrace is 750 m3.

Also Read : Exercise-6.1-(Revised)-Chapter-6-Triangles-class-10-ncert-solutions-Maths

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