The external radius of a hollow sphere made of 1 c m thick lead sheet is 6 c m . By melting the sphere, a solid cylindrical rod of 2 c m radius is formed. What is the length of the rod ?

Given that, The external radius of a hollow sphere made of 1 c m thick lead sheet is 6 c m . By melting the sphere, a solid cylindrical rod of 2 c m radius is formed. Volume of hollow sphere = 4 3 π R 3 - r 3 , where R is the radius of bigger sphere and r is the radius of small sphere. = 4 3 π 6 3 - 5 3 And volume of cylindrical rod = π R 1 2 H , where R 1 is the radius of cylindrical rod and H is length of the rod, Then according to the question, 4 3 π 6 3 - 5 3 = π 2 2 H 4 3 216 - 125 = 4 H 91 3 = H H = 30 1 3 c m Therefore, the length of the rod is 30 1 3 c m .

In an art competition the marks obtained are given below. From the marks, find the mean. Marks 0 or greater than 0 10 or greater than 10 20 or greater than 20 30 or greater than 30 40 or greater than 40 50 or greater than 50 Number of competitors 40 36 22 11 2 0

Given that in an art competition the marks obtained are given below. Marks 0 or greater than 0 10 or greater than 10 20 or greater than 20 30 or greater than 30 40 or greater than 40 50 or greater than 50 Number of competitors 40 36 22 11 2 0 Class interval Mid-point x i Frequency f i cumulative frequency c . f x i f i 0 - 10 5 4 40 20 10 - 20 15 14 36 210 20 - 30 25 11 22 275 30 - 40 35 9 11 315 40 - 50 45 2 2 90 ∑ f i = 40 ∑ x i f i = 910 We know that Arithmetic mean = ∑ x i f i ∑ f i Thus, 910 40 = 22 . 75 Hence, the mean from the given frequency distribution table is 22 . 75 .

Solve from the statistical data given below: Find median Class interval 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 40 - 45 Frequency 5 6 15 10 5 4 3 2

Given that, Class interval 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 40 - 45 Frequency 5 6 15 10 5 4 3 2 Now, let prepare the cumulative frequency distribution table for the given data, Class interval Frequency f Cumulative frequency 5 - 10 5 5 10 - 15 6 5 + 6 = 11 15 - 20 15 11 + 15 = 26 20 - 25 10 26 + 10 = 36 25 - 30 5 36 + 5 = 41 30 - 35 4 41 + 4 = 45 35 - 40 3 45 + 3 = 48 40 - 45 2 48 + 2 = 50 N = ∑ f = 50 N 2 = 50 2 = 25 ∴ Median = l + N 2 - F f m × i , where l is the boundary of the median class, F is the cumulative frequency of the class preceding the median class, f m is the frequency of the median class, i is the length of the median class, N = Total frequency Here, l = 15 , F = 11 , f m = 15 , i = 20 - 15 = 5 , N = 50 On substituting the values, we get, Median = 15 + 25 - 11 15 × 5 = 15 + 4 . 67 = 19 . 67 (approx) Hence, the median for the given data is 19 . 67 (approx).

Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 9: THIRD SUMMATIVE TEST

Author: Bidyut Guha Thakurta & Amitava Mitra

Bidyut Guha Thakurta Mathematics Solutions for Exercise - Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 9: THIRD SUMMATIVE TEST

Attempt the practice questions from Exercise 9: THIRD SUMMATIVE TEST with hints and solutions to strengthen your understanding. SCHOOL MATHEMATICS Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 9: THIRD SUMMATIVE TEST with Hints & Solutions

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 13

The external radius of a hollow sphere made of $1 c m$ thick lead sheet is $H = 30 \frac{1}{3} c m$ . By melting the sphere, a solid cylindrical rod of $2 c m$ radius is formed. What is the length of the rod ?

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 13

The upper surface of a metallic sphere is cut out in such a way that the curved surface of the new sphere is half the previous one. What will be the ratio of the volume of the cut-out portion and the remaining portion of the sphere ?

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 14

The angle of depression of the top and foot of a lamp post as seen from the roof of a house $11$ metre high are $30 °$ and $60 °$ . What is the height of the lamp post ?

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 14

From a point on a horizontal line lying in the same plane with the chimney if $50$ metres proceeded towards the chimney, the angle of elevation of the top becomes $60 °$ from $30 °$ . What is the height of the chimney ?

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 15

In an art competition the marks obtained are given below. From the marks, find the mean.

Marks	$0$ or greater than $0$	$10$ or greater than $10$	$20$ or greater than $20$	$30$ or greater than $30$	$40$ or greater than $40$	$50$ or greater than $50$
Number of competitors	$40$	$36$	$22$	$11$	$2$	$0$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 15

Solve from the statistical data given below:

Find median

Class interval	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$	$35 - 40$	$40 - 45$
Frequency	$5$	$6$	$15$	$10$	$5$	$4$	$3$	$2$

MEDIUM

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 15

In an Entrance Examination the frequency distribution of the marks obtained is given below, and from this find the mode.

Age (in years)	$16 - 18$	$18 - 20$	$20 - 22$	$22 - 24$	$24 - 26$
Number of candidates	$45$	$75$	$38$	$22$	$20$

MEDIUM

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 35 - Third Summative Test>THIRD SUMMATIVE TEST>Q 4

$A B$ is the diameter of a circle with centre $O$ . $P$ is any point on the circumference of the circle. If $∠ P O A = 120 °$ , then find $∠ P B O$ .

Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 9: THIRD SUMMATIVE TEST

Bidyut Guha Thakurta Mathematics Solutions for Exercise - Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 9: THIRD SUMMATIVE TEST

Questions from Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 9: THIRD SUMMATIVE TEST with Hints & Solutions

Learn and Prepare for any exam you want !