Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2017 is the best way to boost your preparation for the board exams.
ICSE Class 10 Maths Question Paper 2017 Solved
Section – A (40 Marks)
(Attempt all questions from this Section)
Question 1.
(a) If b is the mean proportion between a and c, show that: [3]
\(\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}=\frac{a^2}{c^2}\)
Solution:
Here, b is the mean proportion between a and c.
∴ b2 = ac
(b) Solve the equation 4x2 – 5x – 3 = 0 and give your answer correct to two decimal places. [4]
Solution:
Given equation is : 4x2 – 5x – 3 = 0
By using quadratic formula, we obtain
= 1.693 or -0.443
= 1.69 or -0.44
(c) AB and CD are two parallel chords of a circle such that AB = 24 cm and CD = 10 cm. If the radius of the circle is 13 cm, find the distance between the two chords. [3]
Solution:
Here, O is the centre of the given circle of radius 13 cm. AB and CD are two parallel chords, such that AB = 24 cm and CD = 10 cm.
Join OA and OC.
Since ON ⊥ AB and OM ⊥ CD.
∴ M, O and N are collinear and M, N are mid-points of CD and AB.
Now, in rt. ∠ed ∆ANO, we have
ON2 = OA2 – AN2
= 132 – 122
= 169 – 144 = 25
∴ ON = \(\sqrt{25}\) = 5 cm
Similarly, in rt. ∠ed ∆CMO, we have
OM2 = OC2 – CM2
= 132 – 52
= 169 – 25
= 144
OM = \(\sqrt{144}\) = 12cm
Hence, distance between the two chords
NM = NO + OM
= 5 + 12
= 17 cm
Question 2.
(a) Evaluate without using trigonometric tables, sin228° + sin262° + tan238° – cot2 52° + \(\frac{1}{4}\) sec2 30° [3]
Solution:
sin228° + sin262° + tan238° – cot2 52° + \(\frac{1}{4}\) sec2 30°
= sin2 28° + sin2 (90° – 28°) + tan2 38° – cot2 (90° – 38°) + \(\frac{1}{4}\left(\frac{2}{\sqrt{3}}\right)^2\)
= sin2 28° + cos2 28° + tan2 38° – tan2 38° + \(\frac{1}{4}\left(\frac{4}{3}\right)\)
[∵ sin (90° – θ) = cos θ, cot (90° – θ) = tan θ]
= 1 + \(\frac{1}{3}\) [sin2 θ + cos2 θ = 1]
= \(\frac{3+1}{3}\)
= \(\frac{4}{3}\)
= 1 \(\frac{1}{3}\)
(b) If A = \(\left[\begin{array}{ll}
1 & 3 \\
3 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
-2 & 1 \\
-3 & 2
\end{array}\right]\) and A2 – 5B2 = 5C.
Find matrix C where C is a 2 by 2 matrix. [4]
Solution:
(c) Jay a borrowed ₹ 50000for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹ 33000 at the end of the first year. Find the amount she must pay at the end of the second year to clear her debt. [3]
Solution:
Principal = ₹ 50000
Time = 1 year
Rate = 12%
Amount after one year = P(1 + \(\frac{\mathrm{R}}{\mathrm{100}}\))
= 50000 (1 + \(\frac{12}{100}\))
= 50000 × \(\frac{112}{100}\) = ₹ 56000
Amount paid at the end of first year = ₹ 33000
∴ Principal for second year = ₹ (56000 – 33000)
= ₹ 23000
Time = 1 year
Rate= 15%
Amount due after second year = P(1 + \(\frac{\mathrm{R}}{\mathrm{100}}\))
= ₹ 23000 [1 + \(\frac{15}{100}\))
= ₹ 23000 × \(\frac{115}{100}\)
= ₹ 26450.
Question 3.
(a) The catalogue price of a computer set is ₹ 42000. The shopkeeper gives a discount of 10% on the listed price. He further gives an off-season discount of 5% on the discounted price. However, sales tax at 8% is charged on the remaining price after the two successive discounts.
Find: [3]
(i) the amount of sales tax a customer has to pay.
(ii) the total price to be paid by the customer for the computer set.
Solution:
List price of the computer = ₹ 42000
Rate of discount = 10%
Total discount = ₹ \(\frac{10}{100}\) × 42000
= ₹ 4200
Discounted price = ₹ (42000 – 4200)
= ₹ 37800
Another off-season discount = 5%
Total off-season discount = ₹ \(\frac{5}{100}\) × 37800
= ₹ 1890
Remaining price after two successive discounts = ₹ 35910
Total sales tax at 8% = ₹ \(\frac{8}{100}\) × 35910
= ₹ 2872.80
Total price of the computer set
= ₹ (35910 + 2872.80)
= ₹ 38782.80
Hence, amount of sales tax is ₹ 2872.80 and total price to be paid by the customer for the computer set is ₹ 38782.80.
(b) P(1, -2) is a point on the line segment A (3, -6) and B(x, y) such that AP : PB is equal to 2 : 3. Find the coordinates of B. [4]
Solution:
Here, AP : PB = 2 :3, therefore, P divides AB in the ratio 2 : 3
Thus, coordinates of P are : P \(\left(\frac{2 x+9}{2+3}, \frac{2 y-18}{2+3}\right)\)
Also, coordinates of P are P(1, -2)
∴ \(\frac{2 x+9}{5}\) = 1 and \(\frac{2 y-18}{5}\) = -2
2x = 5 – 9 and 2y = 18 – 10
x = \(\frac{-4}{2}\) and y = \(\frac{8}{2}\)
x = -2 and y = 4
Hence, coordinates of B are B (-2, 4).
(c) The marks of 10 students of a class in an examination arranged in ascending order is as follows : [3]
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Solution:
Here, number of students are 10 i.e., even number of observations.
∴ Median = \(\frac{1}{2}\)
(\(\frac{10^{\text {th }}}{2}\) observation + \(\left(\frac{10}{2}+1\right)^{\text {th }}\) observation)
= \(\frac{1}{2}\) (5th observation + 6th observation)
∴ 48 = \(\frac{1}{2}\) (x + x + 4)
96 = 2x + 4
92 = 2x
⇒ x = 46
Thus, the observations are 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.
Hence, the mode of the given data is 46.
Question 4.
(a) What must be subtracted from 16x3 – 8x2 + 4x + 7 so that the resulting expression has 2x + 1 as a factor ? [3]
Solution:
Let p(x) = 16x3 – 8x2 + 4x + 7
and g(x) = 2x + 1
Put 2x + 1 = 0 ⇒ x = –\(\frac{1}{2}\)
∴ Remainder = p\(\left(-\frac{1}{2}\right)\)
= 16\(\left(-\frac{1}{2}\right)^3\) – 8\(\left(-\frac{1}{2}\right)^2\) + 4\(\left(-\frac{1}{2}\right)\) + 7
= 16 × \(\left(-\frac{1}{8}\right)\) – 8 × \(\frac{1}{4}\) – 4 × \(\frac{1}{2}\) + 8
= -2 – 2 – 2 + 7
= -6 + 7 = 1
Hence, 1 is subtracted from p(x), so that g(x) is a factor of p(x).
(b) In the given figure ABCD is a rectangle. It consists of a circle and two semi-circles each of which are of radius 5 cm. Find the area of the shaded region. Give your answer correct to three significant figures. [4]
Solution:
Here, radius of a circle and two semi-circles = 5 cm
∴ Length of the rectangle = 5 + 10 + 5
= 20 cm
Breadth of the rectangle = 10 cm
Now, area of the shaded part = Area of rectangle -2 × Area of circle
= 20 × 10 – 2 × \(\frac{22}{7}\) × 5 × 5
= 200 – \(\frac{1100}{7}\) = 200 – 157.14
= 42.86 cm2 or 42.9 cm2.
(c) Solve the following inequation and represent the solution set on a number line. [3]
-8\(\frac{1}{2}\) < – \(\frac{1}{2}\) – 4x ≤ 7\(\frac{1}{2}\), x ∈ I
Solution:
Given inequation is :
– 8\(\frac{1}{2}\) < – \(\frac{1}{2}\) – 4x ≤ 7\(\frac{1}{2}\), x ∈ I
⇒ –\(\frac{17}{2}\) < –\(\frac{1}{2}\) – 4x ≤ \(\frac{15}{2}\)
⇒ \(\frac{1}{2}\) – \(\frac{17}{2}\) < -4x ≤ \(\frac{1}{2}\) + \(\frac{15}{2}\)
⇒ – 8 < – 4x ≤ 8 ⇒ \(\frac{-8}{-4}\) > \(\frac{-4x}{-4}\) ≥ \(\frac{8}{-4}\)
2 > x ≥ -2
Solution set on number line
Section – B (40 Marks)
Attempt any four questions from this section.
Question 5.
(a) Given matrix B = \(\). Find the matrix X if, X = B2 – 4B. [4]
Hence, solve for a and b given X\(\left[\begin{array}{l}
a \\
b
\end{array}\right]=\left[\begin{array}{l}
5 \\
50
\end{array}\right]\)
Solution:
(b) How much should a man invest in ₹ 50 shares selling at ₹ 60 to obtain an income of ₹ 450, if the rate of dividend declared is 10%. Also, find his yield percent, to the nearest whole number. [3]
Solution:
Market value of a share = ₹ 60
Face value of a share = ₹ 50
Rate of dividend = 10%
Total income = ₹ 450
If income is ₹ 5, then investment = ₹ 60
If income is ₹ 1, then investment = ₹ \(\frac{60}{5}\)
= ₹ 12
If income is 1450, then investment = ₹ 12 × 450
= ₹ 5400
Thus, total investment is ₹ 5400
Yield percent = ₹ \(\frac{450}{5400}\) × 100
= 8.33
= 8 (to the nearest whole number)
(c) Sixteen cards are labelled as a, b, c … m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is: [3]
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median ?
Solution:
Total number of cards = 16
(i) Number of vowels = 4 (a, e, i, o)
Probability = \(\frac{4}{16}\) = \(\frac{1}{4}\)
(ii) Number of consonant = 16 – 4 = 12
Probability = \(\frac{12}{16}\) = \(\frac{3}{4}\)
(iii) Probability (none of the letters of the word median)
= \(\frac{10}{16}\) = \(\frac{5}{8}\)
Question 6.
(a) Using a ruler and a compass construct a triangle ABC in which AB = 7 cm, ∠CAB = 60° and AC = 5 cm. Construct the locus of: [4]
(i) points equidistant from AB and AC.
(ii) points equidistant from BA and BC.
Hence, construct a circle touching the three sides of the triangle internally.
Solution:
Steps of Construction :
- Draw a line segment AB = 7 cm.
- At A construct an angle of 60° such that AC = 5 cm.
- Join BC to get ∆ABC.
- Draw angle bisector of ∠BAC, which is the required locus of the points equidistant from AB and AC.
- Draw angle bisector of ∠ABC, which is the required locus of the points equidistant from BA and BC.
- Let the two angle bisectors intersect each other in I.
- Through I, draw ID ⊥ AB. With 1 as centre and radius = ID draw a circle which touches all the sides of the ∆ABC internally.
(b) A conical tent is to accommodate 77 persons. Each person must have 16 m3 of air to breathe. Given the radius of the tent as 7m, find the height of the tent and also its curved surface area. [3]
Solution:
Total number of persons accommodated = 77
Volume of air required for each person = 16 m3
Volume of the conical tent = 77 × 16
= 1232 m3
Radius of the tent = 7 m
Let h be the height of the conical tent
∴ \(\frac{1}{3}\)πr2h = 1232
∴ \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 × h = 1232
⇒ h = \(\frac{1232 \times 7 \times 3}{22 \times 7 \times 7}\)
= 24 m
Now, Slant height (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{7^2+24^2}\)
= \(\sqrt{49+576}\)
= \(\sqrt{625}\) = 25 m
∴ Curved surface area = πrl
= \(\frac{22}{7}\) × 7 × 25
= 550 m2
(c) If = \(\frac{7 m+2 n}{7 m-2 n}=\frac{5}{3}\), use properties of proportion to find : [3]
(i) m : n
(ii) \(\frac{m^2+n^2}{m^2-n^2}\)
Solution:
\(\frac{7 m+2 n}{7 m-2 n}=\frac{5}{3}\)
Using componendo and dividendo, we have
[Using componendo and dividendo]
Question 7.
(a) A page from a savings bank account passbook is given below: [5]
(i) Calculate the interest for the 6 months from January to June 2016, at 6% per annum.
(ii) If the account is closed on 1st July 2016, find the amount received by the account holder.
Solution:
Minimum balance for the month Jan., 2016 = ₹ 5600
Minimum balance for the month Feb., 2016 = ₹ 4100
Minimum balance for the month Mar., 2016 = ₹ 4100
Minimum balance for the month Apr., 2016 = ₹ 2000
Minimum balance for the month May, 2016 = ₹ 8500
Minimum balance for the month June, 2016 = ₹ 10000
Total = ₹ 34300
Principal = ₹ 34300
Rate = 6% p.a.
Time = \(\frac{1}{12}\) year
∴ I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{34300 \times 6 \times 1}{100 \times 12}\)
= ₹ 171.50
Amount on 1st July 2016 = ₹ (34300 + 171.50)
= ₹ 34471.50
(b) Use a graph paper for this question (Take 2 cm =1 unit on both x and y axis) [5]
(i) Plot the following points :
A(0, 4), B(2, 3), C(1, 1) andD(2, 0).
(ii) Reflect points B, C, D on the y-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.
(iii) Join the points A, B, C, D, D’, C’,B’ and A in order, so as to form a closed figure. Write down the equation of the line of symmetry of the figure formed.
Solution:
(i) On graph
(ii) B'(- 2, 3), C'(-1, 1), D'(-2, 0)
(iii) Equation of the line of symmetry is x = 0
Question 8.
(a) Calculate the mean of the following distribution using step deviation method. [4]
| Marks | Number of Students |
| 0-10 | 10 |
| 10-20 | 9 |
| 20-30 | 25 |
| 30-40 | 30 |
| 40-50 | 16 |
| 50-60 | 16 |
Solution:
Let assumed (a) = 35
∴ Mean = a + \(\frac{\sum f_i u_i}{\sum f_i}\) × h
= 35 – \(\frac{37}{100}\) × 10 = 35 – 3.7 = 31.3
(b) In the given figure PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that: [3]
(i) BD is a diameter of the circle.
(ii) ABC is an isosceles triangle.
Solution:
Here, PQ is a tangent to the circle at A.
∠BAQ = 30°. Since AB is angle bisector of ∠CAQ.
∴ ∠CAB = ∠BAQ = 30°
Again, ∠PAC = 180° – ∠CAQ
= 180° – 30° – 30°
= 120°
Also, AD is angle bisector of ∠PAC
∴ ∠PAD = ∠CAD = 60°
Since angles in the corresponding alternate segment are equal
∴ ∠ADB = ∠BAQ = 30°
and ∠DBA = ∠PAD = 60°
Also, angles in same segment are equal
∴ ∠DCA = ∠DBA = 60°
and ∠ACB = ∠ADB = 30°
Now, ∠DCB = ∠DCA + ∠ACB
= 60° + 30° = 90°
We know that angle in a semi-circle is right angle.
Thus, BD is a diameter of the circle.
In AACB, ∠ACB = ∠CAB = 30°
Hence, ∆ABC is an isosceles triangle.
(c) The printed price of an air conditioner is ₹ 45000/-. The wholesaler allows a discount of 10% to the shopkeeper. The shopkeeper sells the article to the customer at a discount of 5% of the marked price. Sales tax (under VAT) is charged at the rate of 12% at every stage. Find: [3]
(i) VAT paid by the shopkeeper to the government.
(ii) The total amount paid by the customer inclusive of tax. [3]
Solution:
List price of air conditioner = ₹ 45000
Discount = 10%
Total discount = ₹ \(\frac{10}{100}\) × 45000
= ₹ 4500
.’. Selling price = ₹ (45000 – 4500)
= ₹ 40500
VAT paid = ₹ \(\frac{12}{100}\) × 40500
= ₹ 4860
Marked price of air conditioner = ₹ 45000
Discount = 5%
Total discount = ₹ \(\frac{5}{100}\) × 45000
= ₹ 2250
Selling price = ₹ (45000 – 2250)
= ₹ 42750
VAT paid = ₹ \(\frac{12}{100}\) × 42750
= ₹ 5130
Thus, VAT paid by the shopkeeper to the government
= ₹ (5130 – 4860) = ₹ 270
Total amount paid by the customer = ₹ (42750 + 5130)
= ₹ 47880
Question 9.
(a) In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of: [4]
(i) ∠BCD
(ii) ∠BOD
(iii) ∠OBD
Solution:
Here, ∠DAE = 70°
∴ ∠BAD = 180° – ∠DAE [a linear pair]
= 180° – 70° = 110°
ABCD is a cyclic quadrilateral
∴ ∠BCD + ∠BAD = 180°
∠BCD + 110° = 180°
⇒ ∠BCD = 180° – 110°
= 70°
Since angle subtended by an arc at the centre of a circle is twice the angle subtended at the remaining part of the circle.
∴ ∠BOD = 2∠BCD = 2 × 70° = 140°
In ∆OBD, OB = OD = radii of same circle.
∴ ∠OBD = ∠ODB
Thus, ∠OBD = \(\frac{1}{2}\) (180° – ∠BOD)
= \(\frac{1}{2}\) (180° – 140°) = \(\frac{1}{2}\) × 40° = 20°
(b) A(-1, 3), B(4, 2) and C(3, -2) are the vertices of a triangle. [3]
(i) Find the coordinates of the centroid G of the triangle.
(ii) Find the equation of the line through G and parallel to AC.
Solution:
A(- 1, 3), B(4, 2) and C(3, -2) are the vertices of ∆ABC.
∴ Coordinates of the centroid G of the ∆ABC are :
G\(\left(\frac{-1+4+3}{3}, \frac{3+2-2}{3}\right)\)
G(2, 1)
Here, line 7’ is drawn through G(2, 1) and parallel to the line AC.
∴ Slope of the line l = Slope of the line AC
= \(\frac{-2-3}{3+1}\) = \(\frac{-5}{4}\)
Thus, the equation of the required line l is :
y – 1 = \(\left(\frac{-5}{4}\right)\) (x – 2)
4y – 4 = -5x + 10
5x + 4y = 14
(c) Prove that: \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ [3]
Solution:
L.H.S.
Question 10.
(a) The sum ofthe ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages. [4]
Solution:
Let Vivek’s age be x years
∴ Amit’s age = 47 – x
Also, product of their ages = 550
∴ x(47 – x) = 550
47x – x2 = 550
⇒ x2 – 47x + 550 = 0
⇒ x2 – 25x – 22x + 550 = 0
⇒ x(x – 25) – 22(x – 25) = 0
⇒ (x – 25) (x – 22) = 0
⇒ x = 25
or x = 22
Since Vivek is elder brother of Amit.
Hence, age of Vivek is 25 years and age of Amit is 22 years.
(b) The daily wages of 80 workers in a project are given below: [6]
| Wages (in ₹) | No. of Workers |
| 400 – 450 | 2 |
| 450 – 500 | 6 |
| 500 – 550 | 12 |
| 550 – 600 | 18 |
| 600 – 650 | 24 |
| 650 – 700 | 13 |
| 700 – 750 | 5 |
Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = ₹ 50 on x-axis and 2 cm = 10 workers on y-axis. Use your ogive to estimate:
(i) the median wage of the workers.
(ii) the Ipwer quartile wage of workers.
(iii) the number of workers who earn more than ₹ 625 daily.
Solution:
The cumulative frequency distribution for the given data is:
| Wages (in ₹) | No. of Workers | c.f | Points |
| 400 – 450 | 2 | 2 | (450, 2) |
| 450 – 500 | 6 | 8 | (500, 8) |
| 500 – 550 | 12 | 20 | (550, 20) |
| 550 – 600 | 18 | 38 | (600, 38) |
| 600 – 650 | 24 | 62 | (650, 62) |
| 650 – 700 | 13 | 75 | (700, 75) |
| 700 – 750 | 5 | 80 | (750, 80) |
| Total | 80 |
Plot the points (450, 2), (500, 8), (550, 20), (600, 38), (650, 62), (700, 75), (750, 80).
Join them free hand to get the required ogive.
Now, from the graph, we obtain :
(i) median wage of the workers = ₹ 605
(ii) lower quartile wage of workers = ₹ 550
(iii) Number of workers who earn more than ₹ 625 daily
= 80 – 50
= 30.
Question 11.
(a) The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number. [4]
Solution:
Let PQ be the light house of height 60 m, A and B are the two ships on the opposite sides of the light house, such that: ∠PAQ = 60°, ∠PBQ = 45°
In rt.∠ed ∆PQB, we have
\(\frac{\mathrm{PQ}}{\mathrm{QB}}\) = tan 45° = 1
⇒ PQ = QB = 60 m
In rt. ∠ed ∆PQA, we have
\(\frac{\mathrm{PQ}}{\mathrm{AQ}}\) = tan 60° = \(\sqrt{3}\)
AQ = \(\frac{P Q}{\sqrt{3}}=\frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=20 \sqrt{3}\)
= 20(1.73) = 34.6 m
Now, AB = AQ + QB = 60+ 34.6 = 94.6 m
Hence, the distance between the two ships is 95 m (nearest to whole number).
(b) PQR is a triangle. S is a point on the side QR of ∆PQR such that ∠PSR = ∠QPR. [3]
Given QP = 8 cm, PR = 6 cm and SR = 3 cm.
(i) Prove ∆PQR ~ ∆SPR
(ii) Find the length of QR and PS
(iii) \(\frac{\text { area of } \triangle \mathrm{PQR}}{\text { area of } \triangle \mathrm{SPR}}\)
Solution:
Given : QP = 8 cm, PR = 6 cm and SR = 3 cm
In ∆PQR and ∆SPR
∠QPR = ∠PSR (given)
∠QRP = ∠SRP (common)
∴ ∆PQR ~ ∆SPR (by AA similarity rule)
Since APQR-ASPR
∴ \(\frac{\mathrm{QR}}{\mathrm{PR}}\) = \(\frac{\mathrm{PR}}{\mathrm{SR}}\)
⇒ QR = \(\frac{P R \times P R}{S R}=\frac{6 \times 6}{3}\) = 12 cm
And \(\frac{\mathrm{PS}}{\mathrm{QP}}\) = \(\frac{\mathrm{SR}}{\mathrm{PR}}\)
⇒ PS = \(\frac{\mathrm{SR} \times \mathrm{QP}}{\mathrm{PR}}=\frac{3 \times 8}{6}\) = 4 cm
\(\frac{\text { area of } \triangle \mathrm{PQR}}{\text { area of } \triangle \mathrm{SPR}}=\frac{\mathrm{QP}^2}{\mathrm{PS}^2}=\frac{8^2}{4^2}=\frac{64}{16}=\frac{4}{1}\)
(c) Mr. Richard has a recurring deposit account in a bank for 3 years at 7.5% p.a. simple interest. If he gets ₹ 8325 as interest at the time of maturity, find : [3]
(i) The monthly deposit
(ii) The maturity value.
Solution:
Let the monthly deposit be ₹ x
Time = 3 years or 36 months, R = 7.5%, Interest = ₹ 8325.
∴ x × \(\frac{36(36+1)}{2}\) × \(\frac{1}{12}\) × \(\frac{7.5}{100}\) = 8325
x = \(\frac{8325 \times 100 \times 2 \times 12}{36 \times 37 \times 7.5}\)
= ₹ 2000
Thus, the monthly deposit is ₹ 2000
The maturity value = ₹ 36 × 2000 + ₹ 8325
= ₹ 72000 + ₹ 8325
= ₹ 80325.