Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

Other Exercises

Question 1.
Find the seventh term from the end of the series :
√2, 2, 2√2,……32
Solution:
√2, 2, 2√2,……32
Here a = √2
r = \(\frac { 2 }{ \surd 2 } =\surd 2\)
and l =32
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q1.1

Question 2.
Find the third term from the end of the GP.
\(\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,….162\)
Solution:
G.P is \(\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,….162\)
a = \(\\ \frac { 2 }{ 27 } \)
r = \(\frac { 2 }{ 9 } \div \frac { 2 }{ 27 } \)
= \(\frac { 2 }{ 9 } \times \frac { 27 }{ 2 } \)
= 3
l = 162
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q2.1

Question 3.
For the G.P. \(\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } …..81\)
find the product of fourth term from the beginning and the fourth term from the end.
Solution:
\(\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } …..81\)
a = \(\\ \frac { 2 }{ 27 } \)
r = \(\frac { 1 }{ 9 } \div \frac { 1 }{ 27 } \)
= \(\frac { 1 }{ 9 } \times \frac { 27 }{ 1 } \)
= 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q3.1

Question 4.
If for a G.P., pth, qth and rth terms are a, b and c respectively ;
prove that :
{q – r) log a + (r – p) log b + (p – q) log c = 0
Solution:
In a G.P
Tp = a,
Tq = b,
Tr = c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q4.2

Question 5.
If a, b and c in G.P., prove that : log an, log bn and log cn are in A.P.
Solution:
a, b, c are in G.P.
Let A and R be the first term and common ratio respectively.
Therefore,
a = A
b = AR
c = AR2
log a = log A
log b = log AR = log A + log R
log c = log AR2 = log A + 2log R
log a, log b and log c are in A.P.
If 2log b = log a + log c
If 2[logA + logR] = log A + log A + 2log R
If 2log A + 2log R = 2log A + 2log R
which is true.
Hence log a, log b and log c are in A.P.

Question 6.
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
Solution:
Let a, b, c are in G.P.
Then b2 = ac …(i)
Now ax, bx + cx will be in G.P. if (bx)2 = ax.cx
=> (bx)2 = ax.cx
=>(b2)= (ac)x
Hence ax, bx, cx are in G.P. (∴ b2 = ac)
Hence proved.

Question 7.
If a, b and c are in A.P. a, x, b are in G.P. whereas b, y and c are also in G.P. Show that : x2, b2, y2 are in A.P.
Solution:
2 b = a + c _(i)
a, x, b are in G.P.
x2 = ab _(ii)
and b, y, c in G.P.
y2 = bc _(iii)
Now x2 + y2 = ab + bc
= b(a + c)
= b x 2b [from(i)]
= 2 b2
Hence x2, b2, y2 are in G.P.

Question 8.
If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that :
(i)\(\frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 2 }{ b } \)
(ii)\(\frac { a }{ x } +\frac { c }{ y } =2\)
Solution:
a, b, c are in G.P.
b2 = ac
a, x, b, y, c are in A.P.
2x = a + b and 2y = b + c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q8.2

Question 9.
If a, b and c are in A.P. and also in G.P., show that: a = b = c.
Solution:
a, b, c are in A.R
2 b = a + c ….(i)
Again, a, b, c are in G.P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q9.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E.

Other Exercises

Question 1.
The distance by road between two towns A and B is 216 km., and by rail it is 208 km. A car travels at a speed of x km/hr. and the train travels at a speed which is 16 km/hr faster than the car. Calculate:
(i) the time taken by the car to reach town B from A, in terms of x ;
(ii) the time taken by the train, to reach town B from A, in terms of x.
(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x, and solve it.
(iv) Hence, find the speed of the train. [1998]
Solution:
Distance between two stations by road = 216 km and by rail = 208 km.
Speed of car = x km/hr.
and speed of train = (x + 16) km/ hr.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q1.1
⇒ 8x + 3456 = 2×2 + 32x
⇒ 2x² + 32x – 8x – 3456 = 0
⇒ 2x² + 24x – 3456 = 0
⇒ x² + 12x – 1728 = 0 (Dividing by 2)
⇒ x² + 48x – 36x – 1728 = 0
⇒ x (x + 48) – 36 (x + 48) = 0
⇒ (x + 48 ) (x – 36) = 0
Either x + 48 =0, then x = – 48 which is not possible.
or x – 36 = 0, then x = 36
(iv) Speed of train = x + 16 = 36 + 16 = 52 km/hr.

Question 2.
A trader buys x articles for a total cost of Rs. 600.
(i) Write down the cost of one article in terms of x. If the cost per article were Rs. 5 more, the number of articles that can be bought for Rs. 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it for x. [1999]
Solution:
C.P. of x articles = Rs. 600
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q2.1
⇒ 600x = (x – 4) (600 + 5x) (By cross multiplication)
⇒ 600x = 600x + 5x² – 2400 – 20x
⇒ 5x² – 20x – 2400 = 0
(ii) x² – 4x – 480 = 0
⇒ x² – 24x + 20x – 480 = 0
⇒ x (x – 24) + 20 (x – 24) = 0
⇒ (x – 24) (x + 20) = 0 (Zero Product Rule)
Either x – 24 = 0, then x = 24
or x + 20 = 0, then x = -20 Which is not possible.
Hence no. of articles = 24

Question 3.
A hotel bill for a number of people for overnight stay is Rs. 4,800. If there were 4 people more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight. [2000]
Solution:
Amount of the bill = Rs. 4800
Let the number of persons staying overnight = x
Then amount to be paid by each person
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q3.2
⇒ x (x + 12) – 8 (x + 12) = 0
⇒ (x + 12) (x – 8) = 0 (Zero Product Rule)
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 8 = 0, then x = 8
Hence no. of persons staying overnight = 8

Question 4.
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression lor the time taken for :
(i) the omvard journey;
(ii) the return journey.
If the return journey took 30 minutes less than the on ward journey, write down an equation in x and find its value. |2002]
Solution:
Distance between A and B = 400 km.
Speed of aeroplane onward journey = x km/hr.
and Speed of aeroplane on return journey = (x + 40) km/hr.
Now time taken for onward journey = \(\frac { 400 }{ x }\) hrs.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q4.2
Which is not possible being negative
or x – 160 = 0, then x = 160
x = 160

Question 5.
Rs. 6,500 were divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs. 30 less. Find the original number of persons.
Solution:
Let original number of persons = x
Amount = Rs. 6,500
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q5.1
⇒ 30x² + 450x – 97500 = 0 (Dividing by 30)
⇒ x² + 15x – 3250 = 0
⇒ x² + 65x – 50x – 3250 = 0
⇒ x (x + 65) – 50 (x + 65) = 0
⇒ (x + 65) (x – 50) = 0
Either x + 65 = 0, then x = -65 which is not possible.
or x – 50 = 0, then x = 50
Original number of persons = 50

Question 6.
A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km. away in time, it has to increase its speed by 250 km./hr. from its usual speed. Find its usual speed.
Solution:
Let the usual speed of plane = x km/hr.
Distance = 1500 km.
Increased speed = (x + 250) km./hr.
Now, according to the condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q6.1
⇒ x² + 250x – 750000 = 0
⇒ x² + 1000x – 750x – 750000 = 0
⇒ x (x + 1000) – 750 (x + 1000) = 0
⇒ (x + 1000) (x – 750) = 0
Either x + 1000 = 0, then x = – 1000 But it is not possible.
or x – 750 = 0, then x = 750
Usual speed of plane = 750 km/hr

Question 7.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the average speed of each train.
Solution:
Let the speed of first train = x km/hr.
Then speed of second train = (x – 5) km/ hr.
In 2 hours, first train will travel = 2x km.
and second train will travel = 2 (x – 5) km.
According to the condition,
2x + 2 (x – 5) = 50
⇒ 2x + 2x – 10 = 50
⇒ 4x = 50 + 10 = 60
x = 15 km/hr.
and speed of second train = 15 – 5 = 10 km/hr.

Question 8.
The sum S of first n even natural numbers is given by the relation S = n (n + 1). Find n if the sum is 420.
Solution:
S = n (n + 1) and x = 420
⇒ n (n + 1) = 420
⇒ n² + n – 420 = 0
⇒ n² + 21n – 20n – 420 = 0
⇒ n (n + 21) – 20 (n + 21) = 0
⇒ (n + 21) (n – 20) = 0
Either n + 21 =0, then n = -21 which is not possible as it is negative
or n – 20 = 0 then n = 20

Question 9.
The Sum of the ages of a father and his son is 45 years. Five year ago, the product of their ages (in years) was 124. Determine their present ages.
Solution:
Let age of father = x years
Then age of his son = (45 – x) years (sum = 45 years)
5 years ago,
The age of father = (x – 5) years
and age of son = 45 – x – 5 = (40 – x) years
According to the condition,
(x – 5) (40 – x) = 124
⇒ 40x – x² – 200 + 5x = 124
⇒ -x² + 45x – 200 – 124 = 0
⇒ -x² + 45x – 324 = 0
⇒ x² – 45x + 324 = 0
⇒ x² – 36x – 9x + 324 = 0
⇒ x (x – 36) – 9 (x – 36) = 0
⇒ (x – 36) (x – 9) = 0
Either x – 36 = 0, then x = 36
or x – 9 = 0, then x = 9, but it is not possible as age of father cannot be less than his son.
Age of father = 36 years
and age of son = 45 – 36 = 9 years

Question 10.
In an auditorium, seats were arranged in rows and columns. The number of rows w as equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement. [2003]
Solution:
Let the number of rows in the auditorium = x
No. of seats in each row = x
and no. of total seats in the auditorium =
x x x = x²
In second case,
No. of rows = 2x
and no. of seats in each row = x – 10
Then the total seats will = x² + 300
Now, according to the condition,
2x (x – 10) = x² + 300
⇒ 2x² – 20x = x² + 300
⇒ 2x² – x² – 20x – 300 = 0
⇒ x² – 30x + 10x – 300 = 0
⇒ x (x – 30) + 10 (x – 30) = 0
⇒ (x – 30) (x + 10) = 0 (Zero Product Rule)
Either x – 30 = 0, then x = 30
or x + 10 = 0, then x = -10 Which not possible.
(i) No. of rows in original arrangement = 30
(ii) and no. of seats after re-arrangements = x² + 300 = (30)2 + 300 = 900 + 300 = 1200

Question 11.
Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, how many days will Mohan alone complete the work?
Solution:
Let time taken by Mohan = x days
Time taken by Manoj = (x + 16) days
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q11.1
⇒ 30x + 240 = x² + 16x
⇒ x² + 16x – 30x – 240 = 0
⇒ x² – 14x – 240 = 0
⇒ x² – 24x + 10x – 240 = 0
⇒ x (x – 24) + 10 (x – 24) = 0
⇒ (x + 10) (x – 24) = 0
Either x + 10 = 0, then .x = -10 But it is not possible
or x – 24 = 0, then x = 24
Mohan can do the work in = 24 days

Question 12.
Two years ago. a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.
Solution:
2 years ago,
let son’s age = x
Man’s age = 3x
Son’s present age = x + 2
and man’s age = 3x² + 2
and 3 years after,
Son’s age = x + 2 + 3 = x + 5
Man’s age 3x² + 2 + 3 = 3x² + 5
According to condition,
3x² + 5 = 4 (x + 5)
⇒ 3x² + 5 = 4x + 20
⇒ 3x² – 4x + 5 – 20 = 0
⇒ 3x² – 4x – 15 = 0
⇒ 3x² – 9x + 5x – 15 = 0
⇒ 3x (x – 3) + 5 (x – 3) = 0
⇒ (x – 3) (3x + 5) = 0
Either x – 3 = 0, then x = 3
or 3x + 5 = 0, then 3x = -5 ⇒ x = \(\frac { -5 }{ 3 }\)
But it is not possible.
x = 3
Son’s present age = x + 2 = 3 + 2 = 5 years
and man’s present age = 3x² + 2 = 3(3)² + 2 = 27 + 2 = 29 years

Question 13.
In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both. the fraction reduces by \(\frac { 1 }{ 14 }\) Find the fraction.
Solution:
In first case.
Let numerator of a fraction = x
then, its denominator = x + 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q13.2

Question 14.
In a two digit number, the ten’s digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Solution:
Difference of digits = 6
Let one’s digit = x
Then ten’s digit = x + 6
and number = x + 10 (x + 6) = x + 10x + 60 = 11x + 60
But product of digits = 27
x (x + 6) = 27
⇒ x² + 6x – 27 = 0
⇒ x² + 9x – 3x – 27 = 0
⇒ x (x + 9) – 3 (x + 9) = 0
⇒ (x + 9) (x – 3) = 0
Either x + 9 = 0, then x = – 9 But it is not possible
or x – 3 = 0, then x = 3
Number = 11x + 60 = 11 x 3 + 60 = 33 + 60 = 93

Question 15.
Some school children on an excursion by a bus to a picnic spot at a distance of 300 km. While returning, it was raining and the bus had to reduce its speed by 5 km/ hr and it took two hours longer for returning. Find the time taken to return.
Solution:
Distance = 300 km.
Let speed of the bus = x km/hr.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q15.1
⇒ x² – 5x = 750
⇒ x² – 5x – 750 = 0
⇒ x² – 30x + 25x – 750 = 0
⇒ x (x – 30) + 25 (x – 30) = 0
⇒ (x – 30) (x + 25) = 0
Either x – 30 = 0, then x = 30
or x + 25 = 0, then x = -25, but it is not possible as it is negative
Speed of the bus = 30 km/hr
and time taken while returning = \(\frac { 300 }{ x }\) = \(\frac { 300 }{ 25 }\) = 12 hours

Question 16.
Rs. 480 is divided equally among ‘x’ children. If the number of children were 20 more then each would have got Rs. 12 less. Find ‘x’.
Solution:
Total amount = Rs. 480
Number of children = x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q16.1
⇒ x² + 20x – 800 = 0
⇒ x² + 40x – 20x – 800 – 0
⇒ x (x + 40) – 20 (x + 40) = 0
⇒ (x + 40) (x – 20) = 0
Either x + 40 = 0, then x = – 40 which is not possible being negative
or x – 20 = 0 then x = 20
Number of children = 20

Question 17.
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’ (2016)
Solution:
Let the original speed be x km/hr
Time taken by the bus with moving at speed x km/h = \(\frac { 240 }{ x }\)
Time taken by the bus with moving at speed (x – 10) km/h = \(\frac { 240 }{ x – 10 }\)
According to the given condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q17.1
⇒ x (x – 10) = 10 x 120
⇒ x² – 10x = 1200
⇒ x² – 10x – 1200 = 0
⇒ x² – 40x + 30x – 1200 = 0
⇒ x (x – 40) + 30 (x – 40) = 0
⇒ (x – 40) (x + 30) = 0
⇒ x – 40 = 0 or x + 30 = 0
⇒ x = 40 or x = -30
Since, the speed cannot be negative, the uniform speed is 40 km/h

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24E.

Other Exercises

Question 1.
The following distribution represents the height of 160 students of a school
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q1.1
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :
(i) The median height.
(ii) The inter quartile range.
(iii) The number of students whose height is above 172 cm
Solution:
The cumulative frequency table may be prepared as follows :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q1.3
Now, we take height along x-axis and number of students along the y-axis. Now, plot the point (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152) and (180, 160). On the graph paper and join them with free hand.
(i) Here N = 160 ⇒ \(\frac { N }{ 2 }\) = 80
Which is even now take a point A on the y-axis representing 80. Through A draw horizontal line meeting the ogive at B. From B, draw BC ⊥ x-axis, meeting the x-axis at C.
The abscissa of C is 157.5 So, median = 157.5 cm
(ii) Proceeding in the same way as we have done in above, we have, Q1 = 152 and Q3 = 164
So, inter quartile range = Q3 – Q1 = 164 – 152 = 12 cm
(iii) From the ogive, we see that the number of students whose height is less than 172 is 145.
No. of students whose height is above 172 cm = 160 – 145 = 15

Question 2.
The following table gives the weekly wages of workers in a factory.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q2.1
Calculate : (i) the mean, (ii) the modal class, (iii) the number of workers getting weekly w ages below Rs. 80 and (iv) the number of workers get¬ting Rs 65 or more but less than Rs. 85 as weekly wages. [2002]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q2.2
(ii) Modal class 55-60 (It has maximum frequency)
(iii) No. of workers getting wages below Rs. 80 = 60
(iv) No. of worker getting Rs. 65 is more but less than 85 as weekly wages = 37

Question 3.
Draw an ogive for the data given below and from the graph determine :
(i) the median marks,
(ii) the number of students who obtained more than 75% marks ?
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q3.3
Through 60.5 th marks, draw a line segment parallel to x-axis which meets the curve at A.
From A, draw a line segment perpendicular to, x-axis meeting at B.
∴ B is the median = 43 (approx.)
No. of students who obtained upto 75% marks in the test =111
∴ No. of students who obtained more than 75% = 120- 111 =9

Question 4.
The mean of 1, 7, 5, 3, 4, and 4 is m. The numbers 3,2,4,2,3,3 and p have mean m-1 and median q. Find p and q.
Solution:
Mean of 1,7, 5, 3,4, and 4 = \(\frac { 24 }{ 6 }\) =4
∴ m = 4.
Now mean of 3,2,4,2,3,3 and p = m- 1= 4- 1 = 3
i.e. 17+p = 3xn when n =7
17 + p = 3×7 = 21
⇒ p = 21 – 17 = 4
Median of 3, 2,4,2, 3, 3 and 3 is q
Arranging in ascending order, 2,2, 3,3,3,3,4,4
Mean = 4th terms is 3.
∴ q = 3

P.Q.
The marks of 200 students in a test were recorded as follows :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp1.1
Construct the cumulative frequency table. Draw an ogive and use it to find :
(i) the median and
(ii) the number of students who score more than 35 % marks.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp1.2
Through 100 th scores, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line segment perpendicular to it which meets at B.
∴ Median = 52.5
No. of students who score more than 35% marks.
= 200 – 28 = 172
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp1.3

Question 5.
In a malaria epidemic, the number of cases diagnosed were as follows :
Date (July) 1 2 3 4 5 6 7 8 9 10 11 12 Number 5 12 20 27 46 30 31 18 11 5 0 1 On what days does the mode, the upper and lower quartiles occur ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q5.1
(i) Mode = 5th July (because it has the maximum frequencies i,e. 46)

Question 6.
The incomes of the parents of 100 students in a class in a certain university are tabulated below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.1
(i) Draw a cumulative frequency curve to estimate the median income.
(ii) If 15 % of the students ae given freeships on the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.
(iii) Calculate the Arithmetic mean.
Solution:
(Cummulative Frequancy table)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.2
(i) No. of terms= 100
∴ Mean = \(\frac { 100 }{ 2 }\) = 50th term
Through 50 mark, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B, B is the median.
∴ B = 17.6 thousands
(ii) Upper quartile = 100 x \(\frac { 3 }{ 4 }\) = 75th term
From the Curve Q3 = 23.2
(iii) Lower-quartile = 100 x \(\frac { 1 }{ 4 }\)=25th term
From the curve Q1 = 12.8
∴ Inter-quartile range = Q3– Q1 = 23.2 – 12.8
= 10.4 thousands
(iv) 15% of 100 students = \(\frac { 100 x 15 }{ 100 }\) = 15
From C.F. 15, draw a horizontal line which intersects the curve at P. From P, draw a perpendicular to x-axis meeting it at 11.2
∴ Freeship to parents = Rs. 11.2 thousands upto
the income of Rs. 11.2 thousands
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.4

Question 7.
The marks of 20 students in a test were as follows : 2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20. Calculate-
(i) the mean
(ii) the median
(iii) the mode. [2002]
Solution:
Arranging in ascending order,
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 15, 16, 16, 18, 19, 20
No. of terms = 20
Ix = 2 + 6 + 8 + 9+ 10+ 11 + 11 + 12+ 13 + 13 + 14 + 14 + 14 + 15 + 15 + 15 + 16 + 16 + 18 +
19 + 20 = 257
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q7.1
(iii) Mode = 15 (as it has maximum frequency i.e. it has 3)

Question 8.
The marks obtained by 120 students in a Mathematics test are given below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q8.1
Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate :
(i) the median
(ii) the number of students who obtained more than 75% marks in a test ?
(iii) the number of students who did not pass in the test if the pass percentage was 40. [2002]
(iv) the lower quartile
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q8.3
Lower Quartile (Q1)
∴ \(\frac { N }{ 4 }\)  = \(\frac { 120 }{ 4 }\)  = 30
From a point B (30) on v-axis, draw a line parallel to x- axis meeting the curve at Q and from Q Draw a line parallel to .Y-axis meeting it at 30.
∴ Lower quartile = 30
Through 60.5th marks, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line segment perpendicular to, x-axis meeting at B.
∴ B is the median = 43 (approx.)
No. of students who obtained upto 75% marks in the test = 110
∴ No. of students who obtained more than 75% = 120- 110 = 10
No. of students who obtained less than 40% marks in the test = 52 (∴ in the graph x = 40, y = 52)

P.Q.
Find the mean for the following frequency distribution: [2003]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp2.2

P.Q.
Draw a histogram and hence estimate the mode for the following frequency distribution: [2003]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp3.1
Solution:
(i) Draw the histogram.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp3.2
(ii) In the highest rectangle which represents the modal class draw two lines points AC and BD intersecting at P.
(iii) From P, draw a perpendicular to x-axis meeting at Q.
(iv) value of Q is the mode which is = 23

P.Q.
For die following set of data find the median :
10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9 and 15.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q4.1

P.Q.
For the following frequency distribution draw a histogram. Hence, calculate the mode.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp5.1
Solution:
Histogram :
(i) Draw a histogram and make the upper corner of the rectangle
(ii) With maximum frequency A and B. Also upper corners of the two other rectangles as C and D which are the next and to maximum rectangle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp5.2
(iii) Join AD and BC which intersect at P.
(iv) From P, draw PM ⊥ X – axis
OM = 13
Hence mode = 13

Question 9.
Using a graph paper, draw an Ogive for the following distribution which shows a record of the weight in kilograms of 200 students.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.1
Use your Ogive to estimate the following :
(i) The percentage of students weighing 55 kg or more,
(ii) The weight above which the heaviest 30% of the students fall.
(iii) The number of students who are (a) under-weight and (b) over-weight, if 55.70 kg is considered as standard weight. (2005)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.2
Plot the points (45. 5), (50, 22), (55, 44). (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200) on the graph and join them in free hand to get an ogive as shown.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.3
(i) From the graph, number of students weighing 55 kg or more = 200 – 44 = 156
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.4
∴ Heaviest 60 students in weight = 9 + 20 + 31 = 60 (From the graph, the required weight is 65 kg or more but less than 80 kg)
(iii) Total number of students who are (i) under weight = 47 and (ii) over weight = 152 (∴ Standard weight is 55.70 kg)

P.Q.
Using step deviation method, find the mean of the following distribution.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp6.2

P.Q.
The daily wages of 80 workers in a building project are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.1
Using graph paper, draw an Ogive for the above distribution.
Use your Ogive to estimate : (i)the median wages of the workers.
(ii) the percentage of workers who earn more than Rs. 75 day.
(iii) the upper quartile wage of the workers.
(iv) the lower quartile wage of the workers.
(v) Inter quartile range.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.3
Now plot the points (40, 6), (50, 16), (60, 31), (70, 50), (80, 62), (90, 70) (100, 76). and (110, 80) on the graph and join them with free hand to get an ogive as shown.
(i) Median : \(\frac { N }{ 2 }\) = \(\frac { 80 }{ 2 }\) = 40
From 40 on y-axis, draw a line parallel to x- axis meeting the curve at P. From P, draw PL ⊥ x-axis
Then L is the median which is 65
∴ Median = Rs. 65
(ii) No. of workers earning more than Rs. 75 per day
From 75 on v-axis, draw a perpendicular meeting the curved at Q and from Q( draw a line parallel to x-axis which meet y-axis at B which is 57
∴No. of workers getting more than Rs. 75 per day = 80 – 57 = 23
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.4
From 60 on y-axis, draw- a line parallel to x- axis which meets the curve at R. From R, draw a perpendicular on x-axis meeting it at N.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.5
From 20 on y-axis, draw a line parallel to x- axis meeting the curve at S. From S, draw a perpendicular on x-axis meeting it at T.
T is the lower quartile (Q1) which is 53.5
∴ Q1 = Rs. 53.50
(v) Inter quartile range = Q3 – Q1 = Rs. 78-53.50 = Rs. 24.50

Question 10.
The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q10.2
Mode = Marks with maximum frequency is 6 ∴ Mode = 6

Question 11.
The mean of the following distribution is 52 and the frequency of class interval 30-40 is Find ‘f’ .Find ‘ f ‘.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q11.2

Question 12.
The monthly income of a group of 320 employees in a company is given below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q12.1
Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm= 50 employees on the other axis.
From the graph determine
(i)the median wage
(ii)the number of employees whose income is below Rs. 8500.
(iii)If the salary of a senior employee is above Rs. 11,500, find the number of senior employees in the company.
(iv) the upper quartile.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q12.3
(i)For median wage, Take OP =\(\frac { 320 }{ 2 }\) =160 on y-axis, Draw a line PQ || x-axis and from Q,
draw QM ⊥ x-axis, abcissa of M point is 9400 ⇒ Median = Rs. 9400
(ii) Take OM’ = 8500 on.t-axis. Draw Q’M’|| toy-axis and P’Q’ || X-axis
Where ordinate of P’ is 92.5
There are approximately 93 employees whose monthly wage is below Rs. 8500
(iii) There are approximately 18 employees whose salary is above Rs. 11500.
(iv) Upper quartile
Mark a point A ony-axis on \(\frac { 3N }{ 4 }\) = \(\frac { 3 x 320 }{ 4 }\)= 240 and draw a line AB || X-axis, then draw BB’
⊥ x-axis abscissa of B’ is upper quartile i.e., Rs. 10250.

Question 13.
A Mathematics aptitude test of 50 students was recorded as follows :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q13.1
Draw a histogram for the above data using a graph paper and locate the mode.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q13.2
1. Draw the histogram with given data.
2. Inside the highest rectangle which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper comer C and D or adjacent rectangle which intersect at K.
3. Draw KL ⊥ X-axis.
Value ofL is the mode which is 82.5 (approx).

Question 14.
Marks obtained by 200 students in an examination are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q14.1
Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph, determine
(i) The median marks.
(ii) The number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q14.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q14.3
(i) Median is 57.
(ii)44 students failed.
(iii) No. of students who secured grade one = 200 – 188 = 12

Question 15.
Marks obtained by 40 students in a short assessment are given below, where a and ft are two missing data.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q15.1
If the mean of the distribution is 7.2, find a and b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q15.3

Question 16.
Find the mode and median of the following frequency distribution :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q16.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q16.2

Question 17.
The median of the following observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q17.2

Question 18.
The numbers 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x. (2014)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q18.1

Question 19.
(Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q19.1
Draw a histogram representing the above distribution and estimate the mode from the graph. (2014)
Solution:
Steps of construction :
(i) Draw a line BC = 6.5 cm.
(ii) Centre B and C draw arcs AB = 5.5 cm and AC = 5 cm
(iii) Join AB and AC, ABC is the required triangle,
(iv) Draw the angle bisetors of B and C. Let these bisectors meet at O.
(v) Taking O as centre. Draw a incircle which touches all the sides of the ∆ ABC.
(vi) From O draw a perpendicular to side BC which cut at N.
(vii) Measure ON which is required radius of the incircle.
ON = 1.5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q19.2

Question 20.
The marks obtained by 100 students in a Mathematics test are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q20.1
Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm 10 units on both axes).
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students failed, If the pass percentage was 35. (2014)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q20.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q20.3
N= 100
Median = \(\frac { 100 }{ 2 }\) = 50th term Median = 45
(ii) Lower quartile : (Q1)
N = 100
⇒ \(\frac { N }{ 4 }\) = \(\frac { 100 }{ 4 }\) = 25
∴ Q1 = 32
(iii) Mo. of students with 85% less = 94
∴ More than 85% marks = 100 94 6
(iv) Number of students who did not pass = 30

Question 21.
The marks obtained by 30 students in a eiass assessment of 5 marks is given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q21.1
Calculate the mean, median and mode of the above distribution. (2015)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q21.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q21.3
Which is between 10 and 20
∴ Median = 3
(ii) Mode frequency of 3 is the greatest
∴ Mode = 3

Question 22.
The weight of 50 workers is given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q22.1
Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use the ogive drawn to estimate the following:
(i) the upper and lower quartiles.
(ii) if weighing 95 Kg and above is considered overweight, find the number of workers who are overweight. (2015)
Solution:
Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42) (110, 47) and (120, 50) on the graph and join them in order with free hand.
This is the required ogive
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q22.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q22.3
(i) Upper Quartile = \(\frac { 50 x 3 }{ 4 }\) th term = \(\frac { 150 }{ 4 }\) th = -37.5th term
Lower Quartile = \(\frac { 50 }{ 4 }\) th = 12.5th term
Upper quartile is 42 kg and lower quartile is 72 kg.
(ii) 95 kg and above are over weight
∴ No. of over weight students are 50 – 39 = 11 students.

Question 23.
The mean of following number is 68. Find the value of ‘x’. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence estimate the median. (2016)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q23.1

Question 24.
The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis). (2016)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.1
Use your graph to estimate the following :
(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.3
Through mark 80 on y-axis, draw a horizontal line which meets the ogive drawn at point Q.
Through Q, draw a vertical line which meets the x-axis at the mark of 43(app.).
∴ Median = 43
(ii) Since the number of terms = 160
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.5
(iii) Since 85% scores = 85% of 100 = 85
Through mark for 85 on x-axis, draw a vertical line which meets the ogive drawn at point B.
Through the point B, draw a horizontal line which meets thej-axis at the mark of 148 = 160- 148= 12
So, the number of shooters who obtained more than 85% score is 12.

Question 25.
The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to :
(i) Frame a frequency distribution table
(ii) To calculate mean
(iii) To determine the Modal class
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q25.2
(iii) Here the maximum class frequency is 8, and the class corresponding to this frequency is 20-30. So, the modal class is 20-30.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B

Other Exercises

Question 1.
Which term of the G.P. :
– 10, \(\frac { 5 }{ \surd 3 } ,-\frac { 5 }{ 6 } ,….-\frac { 5 }{ 72 } ? \)
Solution:
– 10, \(\frac { 5 }{ \surd 3 } ,-\frac { 5 }{ 6 } ,….\)
Here a = – 10
r = \(\frac { 5 }{ \surd 3 } \div \left( -10 \right) \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q1.1
n – 1 = 4
=> n = 4 + 1 = 5
It is 5th term

Question 2.
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
Solution:
In a G.P.
T5 = ar4 = 81
T2 = ar = 24
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q2.2

Question 3.
Fourth and seventh terms of a G.P. \(\\ \frac { 1 }{ 18 } \) are \(– \frac { 1 }{ 486 } \) respectively. Find the GP.
Solution:
In a G.P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q3.1

Question 4.
If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term
Solution:
In a G.P.
T1 = 2, and T3 = 8
=>a = 2 and ar² = 8
Dividing, we get
r² = \(\\ \frac { 8 }{ 2 } \) = 4 = (2)²
r = 2
Second term = ar = 2 x 2 = 4

Question 5.
The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.
Solution:
Let a be first term and r be common ratio, then
T3 = ar2
T8 = ar7
T3 x T8 = ar2 x ar7
= 243
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q5.1

Question 6.
Find the geometric progression with 4th term = 54 and 7th term = 1458.
Solution:
In a G.P.
T4 = 54 and T7 = 1458
Let a be the first term and r be the common
ratio, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q6.1

Question 7.
Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
Solution:
In a G.P.
T2 = 6,
T5 = 9 x T3
Let a be the first term and r be the common ratio
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q7.1

Question 8.
The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.
Solution:
In a G.P.
T4= 10,
T7 = 80 and l = 2560
Let a be the first term and r be the common ratio. Therefore
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q8.2

Question 9.
If the 4th and 9th terms of a G.P. are 54 and 13122 respectively, find the GP. Also, find its general term.
Solution:
In a G.P.
T4 = 54 and T9 = 13122
Let a be the first term and r be the common ratio
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q9.2

Question 10.
The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : q² = pr.
Solution:
In a G.P.
T5 = p,
T8 = q and T11 = r
To show that q² = pr
Let a be the first term and r be the common ratio, therefore
ar4 = p, ar7 = q and ar10 = r
Squaring the ar7 = q
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B Q10.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A

Other Exercises

Question 1.
Find, which of the following sequence form a G.P. :
(i) 8, 24, 72, 216, ……
(ii) \(\\ \frac { 1 }{ 8 } \),\(\\ \frac { 1 }{ 24 } \),\(\\ \frac { 1 }{ 72 } \),\(\\ \frac { 1 }{ 216 } \)
(iii) 9, 12, 16, 24,…..
Solution:
(i) 8, 24, 72, 216,……
Here, a = 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q1.2

Question 2.
Find the 9th term of the series :
1, 4, 16, 64,…….
Solution:
In G.P. 1, 4, 16, 64,….
Here first term (a) = 1
and common ratio (r) = \(\\ \frac { 4 }{ 1 } \) = 4,
T9 = arn – 1 = 1 x 49 – 1 = 1 x 48 = 48
= 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4
= 65536

Question 3.
Find the seventh term of the G.P. :
1 , √3, 3, 3√3…….
Solution:
G.P. is 1 , √3, 3, 3√3
Here first term (a) = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q3.2

Question 4.
Find the 8th term of the sequence :
\(\\ \frac { 3 }{ 4 } \),\(1 \frac { 1 }{ 2 } \),3……
Solution:
G.P. = \(\\ \frac { 3 }{ 4 } \),\(1 \frac { 1 }{ 2 } \),3…….
= \(\\ \frac { 3 }{ 4 } \),\(\\ \frac { 3 }{ 2 } \),3…….
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q4.1

Question 5.
Find the 10th term of the G.P. :
12, 4, \(1 \frac { 1 }{ 3 } \),……
Solution:
G.P. = 12, 4, \(1 \frac { 1 }{ 3 } \),……..
= 12, 4, \(\\ \frac { 4 }{ 3 } \),…..
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q5.2

Question 6.
Find the nth term of the series :
1, 2, 4, 8 …….
Solution:
1, 2, 4, 8,……
Here, a = 1,r = \(\\ \frac { 2 }{ 1 } \) = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q6.1

Question 7.
Find the next three terms of the sequence :
√5, 5, 5√5…..
Solution:
√5, 5, 5√5……
Here a = √5 and r = \(\frac { 5 }{ \surd 5 }\) = √5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q7.1

Question 8.
Find the sixth term of the series :
22, 23, 24,….
Solution:
22, 23, 24,……
Here, a = 22, r = 23 ÷ 22 = 23 – 2 = 21 = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q8.1

Question 9.
Find the seventh term of the G.P. :
√3 + 1, 1, \(\frac { \surd 3-1 }{ 2 } \),…….
Solution:
√3 + 1, 1, \(\frac { \surd 3-1 }{ 2 } \),…….
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q9.1

Question 10.
Find the G.P. whose first term is 64 and next term is 32.
Solution:
First term of a G.P. (a) = 64
and second term (ar) = 32
G.P. will be 64, 32, 16, 8, 4, 2, 1,…….

Question 11.
Find the next three terms of the series:
\(\\ \frac { 2 }{ 27 } \),\(\\ \frac { 2 }{ 9 } \),\(\\ \frac { 2 }{ 3 } \),…..
Solution:
G.P. is \(\\ \frac { 2 }{ 27 } \),\(\\ \frac { 2 }{ 9 } \),\(\\ \frac { 2 }{ 3 } \),…..
a = \(\\ \frac { 2 }{ 27 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A Q11.2

Question 12.
Find the next two terms of the series
2 – 6 + 18 – 54…..
Solution:
G.P. is 2 – 6 + 18 – 54 +………
Here a = 2 and r = \(\\ \frac { -6 }{ 2 } \) = – 3
Next two terms will be
– 54 x ( – 3) = + 162
162 x ( – 3) = – 486
Next two terms are 162 – 486

 

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D.

Other Exercises

Question 1.
The sum S of n successive odd numbers starting from 3 is given by the relation :
S = n (n + 2). Determine n, if the sum is 168.
Solution:
S = n (n + 2) and S = 168
⇒ n (n + 2) = 168
⇒ n² + 2n – 168 = 0
⇒ n² + 14n – 12n – 168 = 0
⇒ n (n + 14) – 12 (n + 14) = 0
⇒ (n + 14) (n – 12) = 0
Either n + 14 = 0, then n = -14 which is not possible as n is positive.
or n – 12 = 0, then n = 12
Hence n = 12

Question 2.
A stone is thrown vertically downwards and the formula d = 16t² + 4t gives the distance, d metres, that it falls in t seconds. How long does it take to fall 420 metres ?
Solution:
d = 16t² + 4t, d = 420 m
Distance = 420 m.
6t² + 4t = 420
⇒ 16t² + 4t – 420 = 0
⇒ 4t² + t – 105= 0 (Dividing by 4)
⇒ 4t² + 21t – 20t – 105 = 0
⇒ t (4t + 21) – 5 (4t + 21) = 0
⇒ (4t + 21) (t – 5) = 0
Either 4t + 21 = 0, then 4t = -21 ⇒ t = \(\frac { -21 }{ 4 }\)
But it is not possible as time can not be negative.
or t – 5 = 0 , then t = 5
t = 5 seconds

Question 3.
The product of the digits of two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.
Solution:
Let ten’s digit = x
then unit’s digit = 2x + 2
According to the condition,
x (2x + 2) = 24
⇒ 2x² + 2x – 24 = 0
⇒ x² + x – 12 = 0 (Dividing by 2)
⇒ x² + 4x – 3x – 12 = 0
⇒ x (x + 4) – 3 (x + 4) = 0
⇒ (x + 4) (x – 3) = 0
Either x + 4 = 0, then x = – 4, which is not possible.
x – 3 = 0, then x = 3.
Ten’s digit = 3
and unit’s digit = 3 x 2 + 2 = 6 + 2 = 8
Number = 8 + 10 x 3 = 8 + 30 = 38

Question 4.
The ages of two sisters are 11 years and 14 years. In how many years time will the product of their ages be 304 ?
Solution:
Let the number of years = x
Age of first sister = 11 + x
and of second sister = 14 + x
Now according to the condition,
(11 + x) ( 14 + x) = 304
⇒ 154 + 11x + 14x + x² = 304
⇒ x² + 25x – 150 = 0
⇒ x² + 30x – 5x – 150 = 0
⇒ x (x + 30) – 5 (x + 30 ) = 0
⇒ (x + 30) (x – 5) = 0
Either x + 30 = 0 , then x = -30 But it is not possible as can’t be in negative
or x – 5 = 0 , then x = 5
Number of years = 5

Question 5.
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
Solution:
One year ago, let the age of son = x years
and age of his father = 8x.
But present age of father is = (8x + 1) years
8x + 1 = (x + 1)²
⇒ x² + 2x + 1 = 8x + 1
⇒ x² + 2x + 1 – 8x – 1 = 0
⇒ x² – 6x = 0
⇒ x (x – 6) = 0
Either x = 0, which is not possible.
or x – 6 = 0, then x = 6
Present age of father = 8x + 1 = 8 x 6 + 1 = 48 + 1 = 49 years.
and age of son = x + 1 = 6 + 1 = 7 years

Question 6.
The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Solution:
Let age of son = x
Then age of father will be = 2x²
8 years hence,
age of son = x + 8
and age of father = 2x² + 8
According to the condition,
2x² + 8 = 3 (x + 8) + 4
⇒ 2x² + 8 = 3x + 24 + 4
⇒ 2x² + 8 – 3x – 28 = 0
⇒ 2x² – 3x – 20 = 0
⇒ 2x² – 8x + 5x – 20 = 0
⇒ 2x (x – 4) + 5 (x – 4) = 0
⇒ (x – 4) (2x + 5) = 0
Either x – 4 = 0, then x = 4
or 2x + 5 = 0, then 2x – 5 ⇒ x = \(\frac { -5 }{ 2 }\)
Which is not possible being negative
x = 4
Present age of son = 4 years
and age of father = 2x² = 2 (4)² = 2 x 16 = 32 years

Question 7.
The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return down-stream to the original point in 4 hours 30 minutes, find the speed of the stream.
Solution:
Let the speed of stream = x km/hr.
Distance = 30 km.
Speed of boat in still water = 15 km/hr.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D Q7.1
⇒ 9x² – 2025 + 1800
⇒ 9x² – 225 = 0
⇒ x² – 25 = 0
⇒ (x)² – (5)² = 0
⇒ (x + 5) (x – 5) = 0
Either x + 5 = 0, then x = -5 which is not possible.
or x – 5 = 0, then x = 5
Speed of stream = 5 km/hr.

Question 8.
Mr. Mehra sends his servant to the market to buy oranges worth Rs. 15. The servant having eaten three oranges on the way, Mr. Mehra pays 25 paise per orange more than the market price. Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.
Solution:
No. of oranges received by Mr. Mehra = x
No. of oranges eaten by the servant = 3
Total no. of oranges bought = x + 3
Total cost = Rs. 15
Price of one orange = Rs. \(\frac { 15 }{ x + 3 }\)
Now according to the sum,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D Q8.1
⇒ x² + 3x = 180
⇒ x² + 3x – 180 = 0
⇒ x² + 15x – 12x – 180 = 0
⇒ x (x + 15) – 12 (x + 15) = 0
⇒ (x + 15) (x – 12) = 0
Either x + 15 = 0, then x = – 15 which is not possible
or x – 12 = 0, then x = 12
x = 12

Question 9.
Rs. 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Solution:
Let the number of children = x
Amount to be divided = Rs. 250
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D Q9.1
⇒ 6250 x 2 = x² + 25x
⇒ x² + 25x – 12500 = 0
⇒ x² + 125x – 100x – 12500 = 0
⇒ x (x + 125) – 100 (x + 125) = 0
⇒ (x + 125) (x – 100) = 0
Either x + 125 = 0 then x = -125 which is not possible.
or x – 100 = 0, then x = 100
No. of children = 100

Question 10.
An employer finds that if he increases the weekly wages of each worker by Rs. 5 and employs five workers less, he increases his weekly wage bill from Rs. 3,150 to Rs. 3,250. Taking the original weekly wage of each worker as Rs. x; obtain an equation in* and then solve it to find the weekly wages of each worker.
Solution:
In first case,
Let weekly wages of each employee = Rs. x
and number of employees = y
and weekly wages = 3150
xy = 3150 ⇒ y = \(\frac { 3150 }{ x }\) ….(i)
In second case,
Weekly wages = x + 5
and number of employees = y – 5
and weekly wages = 3250
(x + 5) (y – 5) = 3250
⇒ xy + 5y – 5x – 25 = 3200
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D Q10.1
⇒x (x + 70) – 45 (x + 70) = 0
⇒ (x + 70) (x – 45) = 0
Either x + 70 = 0, then x = -70 which is not possible being negative
or x – 45 = 0, then x = 45
Weekly wages per worker = Rs. 45

Question 11.
A trader bought a number of articles for Rs. 1,200. Ten were damaged and he sold each of the remaining articles at Rs. 2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it.
Solution:
Let number of articles = x
C.P. = Rs. 1200
Profit = Rs. 60
S.P. = Rs. 1200 + 60 = Rs. 1260
No. of articles damaged = 10
Remaining articles = x – 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D Q11.1
⇒ 2x² – 80x – 12000 = 0
⇒ x² – 40x – 6000 = 0 (Dividing by 2)
⇒ x² – 100x + 60x – 6000 = 0
⇒ x (x – 100) + 60 (x – 100) = 0
⇒ (x – 100) (x + 60) = 0
Either x – 100 = 0, then x = 100
or x + 60 = 0, then x = – 60 which is not possible.
Number of articles = 100

Question 12.
The total cost price of a certain number of identical articles is Rs. 4,800. By selling the article at Rs. 100 each, a profit equal to the cost price of 15 articles is made. Find the number of articles bought.
Solution:
Total cost of some articles = Rs. 4800
Let number of articles = x
S.P. of one article = Rs. 100
S.P. of x articles = Rs. 100x
Profit = Cost price of 15 articles
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D Q12.1
⇒ x² = 48x + 720 (Dividing by 100)
⇒ x² – 48x – 720 = 0
⇒ x² – 60x + 12x – 720 = 0
⇒ x (x – 60) + 12 (x – 60) = 0
⇒ (x – 60) (x + 12) = 0
Either x – 60 = 0, then x = 60
or x + 12 = 0, then x = -12 Which is not possible.
x = 60
Number of articles = 60

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

Other Exercises

Question 1.
The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Solution:
Let the first term and common difference of an A.P. be a and d
As, we know that,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.1

Question 2.
If the third and the 9th terms of an A.P. term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Let the first term and common difference of an A.P. be a and d.
As, we know that,
an = a + (n – 1 )d
a3 = a + (3 – 1 )d = a + 2d
Similarly,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.3

Question 3.
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Number of terms in an A.P. = 50
T3= 12, l = 106
To find T29
Let a be the first term and d be the common difference
=> a + 2d = 12 …(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q3.1

Question 4.
Find the arithmetic mean of :
(i) – 5 and 41
(ii) 3x – 2y and 3x + 2y
(iii) (m + n)² and (m – n)²
Solution:
(i) Arithmetic mean between – 5 and 41
= \(\\ \frac { -5+41 }{ 2 } \)
= \(\\ \frac { 36 }{ 2 } \)
= 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q4.1

Question 5.
Find the sum of first 10 terms of the A.P. 4 + 6 + 8 +…..
Solution:
A.P. = 4 + 6 + 8 +…….
Here, a = 4, d = 6 – 4 = 2, n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q5.1

Question 6.
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 60.
Solution:
Sum of first 20 terms of an A.P. in which
a = 3 and a20 = 60
a20 = a + (20 – 1) x d
60 = 3 + 19 x d
19d = 60 – 3
d = \(\\ \frac { 57 }{ 19 } \)
= 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q6.1

Question 7.
How many terms of the series 18 + 15 + 12 +…..when added together will give 45 ?
Solution:
A.P. is 18 + 15 + 12 +…..
Here, a = 18, d = 15 – 18 = – 3
Given : Sn = 45
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q7.1

Question 8.
The nth term of a sequence is 8 – 5n. Show that the sequence is an A.P.
Solution:
Given, an = 8 – Sn
a1 = 8 – 5 x (1) = 8 – 5 = 3
a2 = 8 – 5 x (2) = 8 – 10 = – 2
a3 = 8 – 5 x (3) = 8 – 15 = – 7
We see that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q8.1

Question 9.
The the general term (nth term) and 23rd term of the sequence 3, 1, – 1, – 3,……
Solution:
The progression 3, 1, – 1, – 3,…..is A.P.
with first term (a) = 3 and common difference (d) = 1 – 3 = – 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q9.1

Question 10.
Which term of the sequence 3, 8, 13,…..is 78 ?
Solution:
Let 78 be the nth term
a = 3, d = 8 – 3 = 5, an = 78, n = ?
a + (n – 1)d = an
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q10.1

Question 11.
Is – 150 a term of 11, 8, 5, 2,….. ?
Solution:
11, 8, 5, 2,….1st term, a = 11
Common difference, d = 8 – 11 = – 3
an = – 150
=> a + (n – 1 )d = – 150
=> 11 + (n – 1) ( – 3) = – 150
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q11.1

Question 12.
How many two digit numbers are divisible by 3 ?
Solution:
Numbers divisible by 3 are 3, 6, 9, 12,….
Hence, lowest two digit number divisible by 3 = 12
and highest two digit number divisible by 3 = 99
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q12.1

Question 13.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Multiples of 4 between 10 and 250 are
12, 16, 20, 24,……, 248
Here, a = 12, d = 4, l = 248
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q13.1

Question 14.
The sum of the 4th term and the 8th term of an A.P. is 24 and the sum of 6th term and the 10th term is 44. Find the first three terms of the A.P.
Solution:
In an A.P.
T4 + T8 = 24
T6 + T10 = 44
Let a be the first term and d be the common difference
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q14.1

Question 15.
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Solution:
Given a14 = 140
we know, an = a + (n – 1) x d
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q15.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C.

Other Exercises

Question 1.
The speed of an ordinary train is x km/hr and that of an express train is (x + 25) km per hour.
(i) Find the time taken by each train to cover 300 km.
(ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.
Solution:
Speed of an ordinary train = x km/ hr
and speed of an express train = (x + 25) km/hr.
(i) Time taken by ordinary train to cover 300 km = \(\frac { 300 }{ x }\) hr
(ii) Time taken by express train to cover 300 km = \(\frac { 300 }{ x + 25 }\) hr
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q1.1
⇒ 7500 = 2x² + 50x
⇒ 2x² + 50x – 7500 = 0
⇒ x² + 25x – 3750 = 0 (Dividing by 2)
⇒ x² + 75x – 50x – 3750 = 0
⇒ x (x + 75) -50 (x + 75) = 0
⇒ (x + 75) (x – 50) = 0
Either x + 75 = 0 then x = -75 But it is not possible,
or x – 50 = 0 then x = 50
Speed of an ordinary train = 50 km/ hr
and speed of express train = 50 + 25 = 75 km/ hr

Question 2.
If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.
Solution:
Let the speed of ear = x km/hr
time taken to cover 36 km = \(\frac { 36 }{ x }\) hr
In second case,
Speed of car = (x + 10) km/hr
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q2.1
⇒ 3x² + 30 x = 3600
⇒ 3x² + 30x – 3600 = 0 .
⇒ x² + 10x – 1200 = 0 (Dividing by 3)
⇒ x² + 40x – 30x – 1200 = 0
⇒ x (x + 40) – 30 (x + 40) = 0
⇒ (x + 40) (x – 30) = 0
Either x + 40 = 0 then x = – 40 But it is not possible,
or x – 30 = 0, then x = 30
Speed of car = 30 km/hr.

Question 3.
If the speed of an aeroplane is reduced by 40 km/per hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.
Solution:
Let the speed of aeroplane = x km/hr.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q3.1
⇒ x² – 40x = 144000
⇒ x² – 40x – 144000 = 0
⇒ x² – 400x + 360x – 144000 = 0
⇒ x (x – 400) + 360 (x – 400) = 0
Either x – 400 = 0, then x = 400
or x + 360 = 0, then x = – 360, But it is not possible.
x = 400
Hence speed of aeroplane = 400 km/hr.

Question 4.
A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Solution:
Let the original speed of the car be x km/h.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q4.1
⇒ 1200x + 14400 – 1200x = 5x² + 60x
⇒ 14400 = 5x² + 60x
⇒ x² + 12x – 2880 = 0 (Dividing by 5)
⇒ x² + 60x – 48x – 2880 = 0
⇒ x (x + 60) – 48 (x + 60) = 0
⇒ (x + 60) (x – 48) = 0
⇒ x = 48 or x = -60
⇒ x = 48 (Rejecting x = -60, being speed)
Hence, original speed of the car = 48 km/h.

Question 5.
A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of ‘x’ km/hr. and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs. 30 minutes to cover the whole distance; find ‘x’
Solution:
Distance = 12 km.
Speed for the first half distance = x km/hr.
and for the second half distance = (x + 2) km/hr.
Total time taken = 2 hrs. 30 minutes.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q5.1
But it is not possible.
Speed for first half distance (x) = 4 km/hr

Question 6.
A car made a run of 390 km in ‘x’ hours. If the speed had been 4 km per hour more, it would have taken 2 hours less for the journey. Find ‘x’.
Solution:
Distance = 390 km
time = x hours
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q6.1
⇒ (x – 15) (x + 13) = 0
Either x – 15 = 0, then x = 15
or (x + 13) = 0 then x = -13 Which is not possible.
Value of x = 15

Question 7.
A goods train leaves a station at 6 p.m., followed by an express train which leaves at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the trains remain constant between the two stations; calculate their speeds.
Solution:
Departure of goods train = 6 p.m.
and departure of express train = 8 p.m.
Speed of express train is more than goods trains by 20 km/hr
Total distance = 1040 km
Let speed of goods train = x km/hr
Then speed of express train = (x + 20) km/hr
Difference of time taken = 8 p.m. – 6 p.m. + 36 minutes = 2 hours 36 minutes
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q7.1
⇒ x – (x + 100) – 80 (x + 100) = 0
⇒ (x + 100) (x – 80) = 0
Either x + 100 = 0, then x = -100 but it is not possible being negative
or x – 80 = 0, then x = 80
Speed of goods train = 80 km/hr
and speed of express train = 80 + 20 = 100 km/hr
⇒ 13x² + 507x – 35100 = 0
⇒ x² + 39x – 2700 = 0 (Dividing by 13)
⇒ x² + 75x – 36x – 2700 = 0
⇒ x (x + 75) – 36(x + 75) = 0
⇒ (x + 75) (x – 36) = 0
Either x + 75 = 0, then x = – 75 Which is not possible,
or x – 36 = 0 then x = 36
Speed of goods train = 36 km/hr
and speed of express train = 36 + 39 = 75 km/hr.

Question 8.
A man bought an article for Rs. x and sold it for Rs. 16. If his loss was x percent, find the cost price of the article.
Solution:
C.P. of article = Rs. x
S.R = Rs. 16
Loss = C.P. – S.P. = Rs. (x – 16)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q8.1
⇒ x² – 100x + 1600 = 0
⇒ x² – 20x – 80x + 1600 = 0
⇒ x (x – 20) – 80 (x – 20) = 0
⇒ (x – 20) (x – 80) = 0
Either x – 20 = 0, then x = 20
or x – 80 = 0, then x = 80
Cost Price = Rs. 20 or Rs. 80

Question 9.
A trader bought an article for Rs. x and sold it for Rs. 52, thereby making a profit of (x – 10) per cent on his outlay. Calculate the cost price.
Solution:
Let C.P. = Rs. x
S.R = Rs. 52
Profit = S.P – C.P. = Rs. 52 – x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q9.1
x² – 10x = 5200 – 100x
⇒ x² – 10x + 100x – 5200 = 0
⇒ x² + 90x – 5200 = 0
⇒ x2 + 130x – 40x – 5200 = 0
⇒ x (x + 130) – 40(x + 130) = 0
⇒ (x + 130) (x – 40 ) = 0
Either x + 130 = 0 , then x = – 130 which is not possible.
or x – 40 = 0 then x = 40
Cost price = Rs. 40

Question 10.
By selling a chair for Rs. 75, Mohan gained as much per cent as its cost. Calculate the cost of the chair.
Solution:
Let. C.P of chair = Rs. x
Profit = x %
S.P. = Rs. 75
Total profit = Rs. (75 – x)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C Q10.1
⇒ x² = 7500 – 100x
⇒ x² + 100x – 7500 = 0
⇒ x²+ 150x – 50 – 7500 = 0
⇒ x (x + 150) – 50 (x + 150) = 0
⇒ (x + 150) (x – 50) = 0
Either x + 150 = 0, then x = -150 which is not possible
or x – 50 = 0, Then x = 50
Cost price of chair = Rs. 50

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E

Other Exercises

Question 1.
Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h-1. The second car goes at a speed of 8 km h-1 in the first hour and thereafter increasing the speed by 0.5 km h-1 each succeeding hour. After how many hours will the two cars meet ?
Solution:
Speed of first car = 10 km/hr
Speed of second car = 8 km/hr in first hour
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q1.2

Question 2.
A sum of Rs 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs 20 less than its preceding prize; find the value of each of the prizes.
Solution:
Total amount (Sn) = Rs 700
Cost of each prize is Rs 20 less than its preceding price or d = – 20
d = – 20 and n = 7
\({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q2.1

Question 3.
An article can be bought by paying Rs 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs 3,000 and every other installment is Rs 100 less than the previous one, find :
(i) amount of installment paid in the 9th month
(ii) total amount paid in the installment scheme.
Solution:
Total price of an article = Rs 28000
No. of installments (n) = 12
First installment (a) = RS 3000
d = Rs 100
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q3.1

Question 4.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the first year.
(ii) the production in the 10th year.
(iii) the total production in 7 years.
Solution:
A manufacture of TV sets, he produces
No. of units in 3rd year = 600
No. of units in 7th year = 700
Let a be the first term and d be the common difference, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q4.1

Question 5.
Mrs. Gupta repays her total loan of Rs 1.18,000 by paying installments every month. If the installment for the first month is Rs 1,000 and it increases by RS 100 every month, what amount will she pay as the 30th installment of loan? What amount of loan she still has to pay after the 30th installment?
Solution:
Total loan to be paid by Mrs. Gupta = Rs 118000
Installment for the first month (a) = Rs 1000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q5.1

Question 6.
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be five times of the class to which the respective section belongs. If there are 1 to 10 classes in the school and each class has three sections, find how many trees were planted by the students?
Solution:
Number of classes = 10
Number of sections of each class = 3
Total number of sections = 10 x 3 = 30
Each section plant tree = 5 times of the class
Each section of 1st class will plant = 1 x 15 = 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E Q6.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24D.

Other Exercises

Question 1.
Find the mode of the following data:
(i) 7,9,8,7,7,6,8,10,7 and 6
(ii) 9,11,8,11,16,9,11,5,3,11,17 and 8
Solution:
(i) Mode = 7
because it occurs 4 times
(ii) Mode =11
because it occurs 4 times

Question 2.
The following table shows the frequency distribution of heights of 50 boys:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q2.1
Find the mode of heights.
Solution:
Mode is 122 because it occurs maximum times i.e its., frequency is 18.

Question 3.
Find the mode of following data, using a histogram:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q3.1
Solution:
Mode class = 20 – 30
Mode = 24
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q3.2
We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal x-axis. Which is the value of the mode = 24

Question 4.
The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q4.1
Solution:
Model class is = 30 – 35
and Mode = 34
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q4.2
We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal axis. Which is the value of the mode.

Question 5.
Find the median and mode for the set of numbers 2,2,3,5,5,5,6,8 and 9.
Solution:
Median = \(\frac { 9 +1 }{ 2 }\) = 5th term which is 5
Mode = 5, because it occurs in maximum times.

Question 6.
A boy scored following marks in various class tests during a term, each test being marked out of 20.
15,17,16,7,10,12,14,16,19,12,16.
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his total marks ?
(iv) What are his mean marks ?
Solution:
Arranging the given data in ascending order : 7, 10,12, 12,14, 15,16,16, 16, 17,19.
(i) Mode = 16 as it occurs in maximum times.
(ii) Median= \(\frac { 11 +1 }{ 2 }\) = 6th term which is 15
(iii) Total marks = 7 + 10+ 12+ 12+ 14+ 15+ 16 + 16+ 16+ 17+ 19= 154
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q6.1

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0,0,2,2,3,3,3,4,5,5,5,5,6, 6,7,8
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q7.1
(ii) Median = Mean of 8th and 9th term
= \(\frac { 4 +5 }{ 2 }\) = \(\frac { 9 }{ 2 }\) = 4.5
(iii) Mode = 5 as it occurs in maximum times.

Question 8.
At a shooting competition the score of a com-petitor were as given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q8.1
(i) What was his modal score ?
(ii) What was his median score ?
(iii) What was his total score ?
(iv) What was his mean score ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q8.2
(i) Modal score =4 as its frequency is 7, the maximum.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D Q8.3

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

Other Exercises

Question 1.
Find three numbers in A.P. whose sum is 24 and whose product is 440.
Solution:
Let three numbers be a – d, a, a + d
a – d + a + a + d = 24
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q1.1

Question 2.
The sum of three consecutive terms of an A.P. is 21 and the slim of their squares is 165. Find these terms.
Solution:
Let three consecutive numbers in A.P. are
a – d, a, a + d
a – d + a + a + d = 21
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q2.1

Question 3.
The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Solution:
Let the angles of a quadrilateral are
a, a + d, a + 2d, a + 3d
d= 20°
a + a + d + a + 2d + a + 3d = 360°
(Sum of angles of a quadrilateral)
=> 4a + 6d = 360°
=> 4a + 6 x 20° = 360°
=> 4a + 120° = 360°
=> 4a = 360 – 120° = 240°
a = \(\\ \frac { 240 }{ 4 } \) = 60°
Angles are 60°, 80°, 100°, 120°

Question 4.
Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.
Solution:
Number = 96
Let its four parts be a, a + d, a + 2d, a + 3d
a + a + d + a + 2d + a + 3d = 96
=> 4a + 6d = 96
=> 2a + 3d = 48 …(i)
Product of means : Product of extremes = 15 : 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q4.2

Question 5.
Find five numbers in A.P. whose sum is \(12 \frac { 1 }{ 2 } \) and the ratio of the first to the last terms is 2 : 3.
Solution:
Let 5 numbers in A.P. be
a, a + d, a + 2d, a + 2d, a + 4d
a + a + d + a + 2d + a + 3d + a + 4d =
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q5.2

Question 6.
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.
Solution:
Number = 207
Let part be a – d, a, a + d
a – d + a + a + d = 207
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q6.1

Question 7.
The sum of three numbers in A.P. is 15 the sum of the squares of the extreme is 58. Find the numbers
Solution:
Let three numbers in A.P. be a – d, a, a + d
a – d + a + a + d = 15
=> 3a = 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q7.1

Question 8.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Solution:
Let four numbers in A.P. be
a – 3d, a – d, a + d, a + 3d
a – 3d + a – d + a + d + a + 3d = 20
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q8.1

Question 9.
Insert one arithmetic mean between 3 and 13.
Solution:
Let A be the arithmetic mean between 3 and 13
\(\left( A=\frac { a+b }{ 2 } \right) \)
A = \(\\ \frac { 3+13 }{ 2 } \)
= \(\\ \frac { 16 }{ 2 } \)
= 8

Question 10.
The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.
Solution:
Angles of a polygon are in A.P.
and common difference (d) = 5°
Smallest angle (a) = 120°
Let n be the number of sides of the polygon then sum of angles = (2n – 4) x 90°
a = 120° and d = 5°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q10.2

Question 11.
\(\\ \frac { 1 }{ a } \), \(\\ \frac { 1 }{ b } \) and \(\\ \frac { 1 }{ c } \) are in A.P
S.T : bc, ca and ab are also in A.P
Solution:
\(\\ \frac { 1 }{ a } \), \(\\ \frac { 1 }{ b } \) and \(\\ \frac { 1 }{ c } \) are in A.P
We have to show that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D Q11.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D are helpful to complete your math homework.

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