ICSE Class 10

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

Question 1.
In the given figure, O is the centre of the circle. If ∠ABC = 20°, then ∠AOC is equal to
(a) 20°
(b) 40°
(c) 60°
(d) 10°

Solution:
In the given figure,
Arc AC subtends ∠AOC at the centre
and ∠ABC at the remaining part of the circle
∠AOC = 2∠ABC = 2 × 20° = 40° (b)

Question 2.
In the given figure, AB is a diameter of the circle. If AC = BC, then ∠CAB is. equal to
(a) 30°
(b) 60°
(c) 90°
(d) 45°

Solution:
In the given figure,
AB is the diameter of the circle and AC = BC
∠ACB = 90° (angle in a semi-circle)
AC = BC

Question 3.
In the given figure, if ∠DAB = 60° and ∠ABD = 50° then ∠ACB is equal to
(a) 60°
(b) 50°
(c) 70°
(d) 80°

Solution:
In the given figure,
∠DAB = 60°, ∠ABD = 50°
In ∆ADB, ∆ADB = 180° – (60° + 50°)
= 180° – 110° = 70°
∠ACB = ∠ADB
(angles in the same segment) = 70° (c)

Question 4.
In the given figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to
(a) 50°
(b) 40°
(c) 60°
(d) 70°

Solution:
In the given figure, O is the centre of the circle.
In ∆OAB,
∠OAB = 40°
But ∠OBA = ∠OAB = 40°

Question 5.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to
(a) 80°
(b) 50°
(c) 40°
(d) 30°
Solution:
ABCD is a cyclic quadrilateral,
AB is the diameter of the circle circumscribing it
∠ADC = 140°, ∠BAC = Join AC

Question 6.
In the given figure, O is the centre of the circle. If ∠BAO = 60°, then ∠ADC is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 120°

Solution:
In the given figure, O is the centre of the circle ∠BAO = 60°

Question 7.
In the given figure, O is the centre of the circle. If ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to
(a) 30°
(b) 45°
(c) 90°
(d) 60°

Solution:
In the given figure, O is the centre of the circle


∠CAO = 105° – 45° = 60° (d)

Question 8.
In the given figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, then ∠PRQ is
(a) 60°
(b) 45°
(c) 30°
(d) 15°

Solution:
In the given figure, O is the centre of the circle
Chord PQ = radius of the circle
∆OPQ is an equilateral triangle
∴∠POQ = 60°
Arc PQ subtends ∠POQ at the centre and
∴∠PRQ at the remaining part of the circle
∴∠PRQ = \(\\ \frac { 1 }{ 2 } \) ∠POQ = \(\\ \frac { 1 }{ 2 } \) x 60° = 30° (c)

Question 9.
In the given figure, if O is the centre of the circle then the value of x is
(a) 18°
(b) 20°
(c) 24°
(d) 36°

Solution:
In the given figure, O is the centre of the circle.
Join OA.

Question 10.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
From Q, length of tangent PQ to the circle = 24 cm
and QO = 25 cm

Question 11.
From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
Let point P is 13 cm from O, the centre of the circle
Radius of the circle (OQ) = 5 cm
PQ and PR are tangents from P to the circle
Join OQ and OR

Question 12.
If angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
Angles between two radii OA and OB = 130°
From A and B, tangents are drawn which meet at P

Question 13.
In the given figure, PQ and PR are tangents from P to a circle with centre O. If ∠POR = 55°, then ∠QPR is
(a) 35°
(b) 55°
(c) 70°
(d) 80°

Solution:
In the given figure,
PQ and PR are the tangents to the circle from a point P outside it

Question 14.
If tangents PA and PB from an exterior point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 100°
Solution:
Length of tangents PA and PB to the circle from a point P
outside the circle with centre O, and inclined an angle of 80°

Question 15.
In the given figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA ⊥ PB, then the length OP is equal to
(a) 5 cm
(b) 10 cm
(c) 7.5 cm
(d) 5√2 cm

Solution:
In the given figure,
PA and PB are tangents to the circle with centre O.
Radius of the circle is 5 cm, PA ⊥ PB.

Question 16.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Solution:
AB is the diameter of a circle with radius 5 cm
At A, XAY is a tangent to the circle
CD || XAY at a distance of 8 cm from A
Join OC

Question 17.
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other is
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
Radii of two concentric circles are 4 cm and 5 cm
AB is a chord of the bigger circle
which is tangent to the smaller circle at C.
Join OA, OC

Question 18.
In the given figure, AB is a chord of the circle such that ∠ACB = 50°. If AT is tangent to the circle at the point A, then ∠BAT is equal to
(a) 65°
(b) 60°
(c) 50°
(d) 40°

Solution:
In the given figure, AB is a chord of the circle
such that ∠ACB = 50°
AT is tangent to the circle at A
AT is tangent and AB is a chord
∠ACB = ∠BAT = 50°
(Angles in the alternate segments) (c)

Question 19.
In the given figure, O is the centre of a circle and PQ is a chord. If the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is
(a) 100°
(b) 80°
(c) 90°
(d) 75°

Solution:
In the given figure, O is the centre of the circle.
PR is tangent and PQ is chord ∠RPQ = 50°
OP is radius and PR is tangent to the circle

Question 20.
In the given figure, PA and PB are tangents to a circle with centre O. If ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°

Solution:
In the given figure,
PA and PB are tangents to the circle with centre O.
∠APB = 50°
But ∠AOB + ∠APB = 180°
∠AOB + 50° = 180°
⇒ ∠AOB = 180° – 50° = 130°
In ∆OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA
But ∠OAB + ∠OBA = 180° – ∠AOB
= 180° – 130° = 50°
∠OAB = \(\frac { { 50 }^{ 0 } }{ 2 } \) = 25° (a)

Question 21.
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at point D, E and F respectively. If BD = 4 cm, DC = 3 cm and CA = 8 cm, then the length of side AB is
(a) 12 cm
(b) 11 cm
(c) 10 cm
(d) 9 cm

Solution:
In the given figure,
sides BC, CA and AB of ∆ABC touch a circle at D, E and F respectively.
BD = 4 cm, DC = 3 cm and CA = 8 cm

Question 22.
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at the points P, Q and R respectively. If PC = 5 cm, AR = 4 cm and RB = 6 cm, then the perimeter of ∆ABC is
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm

Solution:
In the given figure, sides BC, CA and AB of ∆ABC
touch a circle at P, Q and R respectively
PC = 5 cm, AR = 4 cm, RB = 6 cm

Question 23.
PQ is a tangent to a circle at point P. Centre of circle is O. If ∆OPQ is an isosceles triangle, then ∠QOP is equal to
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Solution:
PQ is tangent to the circle at point P centre of the circle is O.

Question 24.
In the given figure, PT is a tangent at T to the circle with centre O. If ∠TPO = 25°, then the value of x is
(a) 25°
(b) 65°
(c) 115°
(d) 90°

Solution:
In the given figure, PT is the tangent at T to the circle with centre O.
∠TPO = 25°
OT is the radius and TP is the tangent

Question 25.
In the given figure, PA and PB are tangents at ponits A and B respectively to a circle with centre O. If C is a point on the circle and ∠APB = 40°, then ∠ACB is equal to
(a) 80°
(b) 70°
(c) 90°
(d) 140°

Solution:
In the given figure,
PA and PB are tangents to the circle at A and B respectively
C is a point on the circle and ∠APB = 40°
But ∠APB + ∠AOB = 180°

Question 26.
In the given figure, two circles touch each other at A. BC and AP are common tangents to these circles. If BP = 3.8 cm, then the length of BC is equal to
(a) 7.6 cm
(b) 1.9 cm
(c) 11.4 cm
(d) 5.7 cm

Solution:
In the given figure, two circles touch each other at A.
BC and AP are common tangents to these circles
BP = 3.8 cm

Question 27.
In the given figure, if sides PQ, QR, RS and SP of a quadrilateral PQRS touch a circle at points A, B, C and D respectively, then PD + BQ is equal to
(a) PQ
(b) QR
(c) PS
(d) SR

Solution:
In the given figure,
sides PQ, QR, RS and SP of a quadrilateral PQRS
touch a circle at the points A, B, C and D respectively
PD and PA are the tangents to the circle
∴ PA = PD …(i)
Similarly, QA and QB are the tangents
∴ QA = QB …(ii)
Now PD + BQ = PA + QA = PQ (a)
[From (i) and (ii)]

Question 28.
In the given figure, PQR is a tangent at Q to a circle. If AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(b) 35°
(d) 45°

Solution:
In the given figure, PQR is a tangent at Q to a circle.
Chord AB || PR and ∠BQR = 70°
BQ is chord and PQR is a tangent
∠BQR = ∠A

Question 29.
Two chords AB and CD of a circle intersect externally at a point P. If PC = 15 cm, CD = 7 cm and AP = 12 cm, then AB is
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) none of these

Solution:
In the given figure,
two chords AB and CD of a circle intersect externally at P.
PC = 15 cm, CD = 7 cm, AP = 12 cm
Join AC and BD

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS 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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

Question 1.
(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD

Solution:
(a) ∆ABC is an equilateral triangle.

Question 2.
(a) In the figure given below, AB is a diameter of the circle. If AE = BE and ∠ADC = 118°, find
(i) ∠BDC (ii) ∠CAE.

(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.

Solution:
(a) Join DB, CA and CB.
∠ADC = 118° (given)
and ∠ADB = 90°

Question 3.
(a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC).
(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.

Solution:
(a) Given: O is the centre of the circle.
To Prove : ∠AOC = 2 (∠ACB + ∠BAC).
Proof: In ∆ABC,
∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle)

Question 4.
(a) In the figure (i) given below, AB is the diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find :
(i)∠ADC (ii) ∠DAC.
(b) In the figure (ii) given below, the centre O of the smaller circle lies on the circumference of the bigger circle. If ∠APB = 70° and ∠BCD = 60°, find :
(i) ∠AOB (ii) ∠ACB
(iii) ∠ABD (iv) ∠ADB.

Solution:
(a) AB is diameter and DC || AB,
∠CAB = 25°, join AD,BD

Question 5.
(a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.
(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.

Solution:
(a) Given : ABDC is a cyclic quadrilateral AB = CD.
To Prove: AD = BC.

Question 6.
A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.
Solution:
Join OT, OP = 13 cm and TP = 12 cm

Question 7.
Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:
Given: Two circles with centre O and O’
touch each other internally at P.

Question 8.
From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP.
(ii) OP is the perpendicular bisector of the chord AB.
Solution:
Given: From a point P, outside the circle with centre O.
PA and PB are the tangents to the circle,
OA, OB and OP are joined.

Question 9.
(a) The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that:
AP : BQ = PC : CQ.

(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.

Solution:
(a) Given: Two circles with centres A and B
and a transverse common tangent to these circles meets AB at C.

Question 10.
In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.

Solution:
In the given figure, two chords with centre A and B touch externally.
PM is a tangent to the circle with centre A
and QN is tangent to the circle with centre B.
PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm.
We have to find AB.

Question 11.
Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.
Solution:
∵ AB and CD are two chords of a circle
which intersect each other at P, outside the circle.

Question 12.
(a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle

(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also, find the length of the tangent drawn from P to the circle. .

Solution:
Given : (a) AB is a chord of a circle with centre O
and PT is tangent and CD is the diameter of the circle
which meet at P.
AP = 16 cm, AB = 12 cm, OP = 2 cm
∴PB = PA – AB = 16 – 12 = 4 cm
∵ABP is a secant and PT is tangent.
∴PT² = PA × PB.

Question 13.
In the figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and.the radius of the circle is 6 cm, compute the length of AB. Also, find the length of tangent drawn from X to the circle.

Solution:
Chord AB and diameter PQ meet at X
on producing outside the circle

Question 14.
(a) In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.
(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find:
(i)∠BEC (ii) ∠ACB
(iii) ∠BCD (iv) ∠CED.

Solution:
(a) (i) Given : ABCD is a cyclic quadrilateral.
Ext. ∠PBC = ∠ADC
⇒ 40° = ∠ADC

Question 15.
(a) In the figure (i) given below, APC, AQB and BPD are straight lines.
(i) Prove that ∠ADB + ∠ACB = 180°.
(ii) If a circle can be drawn through A, B, C and D, Prove that it has AB as a diameter

(b) In the figure (ii) given below, AQB is a straight line. Sides AC and BC of ∆ABC cut the circles at E and D respectively. Prove that the points C, E, P and D are concyclic.

Solution:
(a) Given: In the figure, APC,
AQB and BPD are straight lines.

Question 16.
(a) In the figure (i) given below, chords AB, BC and CD of a circle with centre O are equal. If ∠BCD = 120°, find
(i) ∠BDC (ii) ∠BEC
(iii) ∠AEC (iv) ∠AOB.
Hence Prove that AOAB is equilateral.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. The chord BC of the circle is parallel to the radius OD and the lines OC and BD meet at E. Prove that
(i) ∠CED = 3 ∠CBD (ii) CD = DA.

Solution:
(a) In ∆BCD, BC = CD
∠CBD = ∠CDB
But ∠BCD + ∠CBD + ∠CDB = 180°
(∵ Angles of a triangle)

Question 17.
(a) In the adjoining figure, (i) given below AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find
(i) ∠AOX (ii) ∠APY (iii) ∠BPY (iv) ∠OAX.

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find :
(i) ∠ADB (ii) ∠AOB (iii) ∠ACB (iv) ∠APB.

Solution:
(a) AB and XY are diameters of a circle with centre O.
∠APX = 30°.
To find :
(i) ∠AOX (ii) ∠APY
(iii) ∠BPY (iv) ∠OAX

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test 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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking EX 2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test

Question 1.
Shweta deposits Rs. 350 per month in a recurring deposit account for one year at the rate of 8% p.a. Find the amount she will receive at the time of maturity.
Solution:
Deposit per month = Rs 350,
Rate of interest = 8% p.a.
Period (x) = 1 year
= 12 months

Question 2.
Salom deposited Rs 150 per month in a bank for 8 months under the Recurring Deposit Scheme. ‘What will be the maturity value of his deposit if the rate of interest is 8% per annum ?
Solution:
Deposit per month = Rs. 150
Rate of interest = 8% per
Period (x) = 8 month

Question 3.
Mrs. Goswami deposits Rs. 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value. (2009)
Solution:
Deposit per month (P) = Rs. 1000
Period = 3 years = 36 months
Rate = 8%

Question 4.
Kiran deposited Rs. 200 per month for 36 months in a bank’s recurring deposit account. If the banks pays interest at the rate of 11% per annum, find the amount she gets on maturity ?
Solution:
Amount deposited month (P) = Rs. 200
Period (n) = 36 months,
Rate (R) = 11% p.a.
Now amount deposited in 36 months = Rs. 200 x 36 = Rs 7200

Question 5.
Haneef has a cumulative bank account and deposits Rs. 600 per month for a period of 4 years. If he gets Rs. 5880 as interest at the time of maturity, find the rate of interest.
Solution:
Interest = Rs. 58800
Monthly deposit (P) = Rs. 600

Question 6.
David opened a Recurring Deposit Account in a bank and deposited Rs. 300 per month for two years. If he received Rs. 7725 at the time of maturity, find the rate of interest per annum. (2008)
Solution:
Deposit during one month (P) = Rs. 300
Period = 2 years = 24 months.
Maturity value = Rs. 7725

Question 7.
Mr. Gupta-opened a recurring deposit account in a bank. He deposited Rs. 2500 per month for two years. At the time of maturity he got Rs. 67500. Find :
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.
Solution:
Deposit per month = Rs. 2500
Period = 2 years = 24 months
Maturity value = Rs. 67500

Question 8.
Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs 800 per month for \(1 \frac { 1 }{ 2 } \) years. If he received Rs 15084 at the time of maturity, find the rate of interest per annum.
Solution:
Money deposited by Shahrukh per month (P)= Rs 800
r = ?

Question 9.
Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs 1200 as interest at the time of maturity, find:
(i) the monthly instalment
(ii) the amount of maturity. (2016)
Solution:
Interest = Rs 1200
Period (n) = 2 years = 24 months
Rate (r) = 6% p.a.

Question 10.
Mr. R.K. Nair gets Rs 6,455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly instalment.
Solution:
Let monthly instalment is Rs P
here n = 1 year = 12 months
n = 12

Question 11.
Samita has a recurring deposit account in a bank of Rs 2000 per month at the rate of 10% p.a. If she gets Rs 83100 at the time of maturity. Find the total time for which the account was held.
Solution:
Deposit per month = Rs 2000,
Rate of interest = 10%, Let period = n months

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax EX 1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test

Question 1.
1. A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax is 8%, find:
(i) the VAT paid by the shopkeeper. .
(ii) the total amount that the consumer pays for the washing machine.
Solution:
(i) S.P. of washing machine
= \(\left( 1-\frac { 10 }{ 100 } \right) \) x ₹18000

Question 2.
A manufacturing company sold an article to its distributor for ₹22000 including VAT. The distributor sold the article to a dealer for ₹22000 excluding tax and the dealer sold it to a consumer for ₹25000 plus tax (under VAT). If the rate of sales tax (under VAT) at each stage is 10%, find :
(i) the sale price of the article for the manufacturing company.
(ii) the amount of VAT paid by the dealer.
Solution:
S.P. of an article for a manufacturer = ₹22000 including VAT
C.P. for the distributor = ₹22000
Rate of VAT = 10%
S.P. for the distributor of ₹22000 excluding VAT

Question 3.
The marked price of an article is ₹7500. A shopkeeper sells the article to a consumer at the marked prices and charges sales tax at . the rate of 7%. If the shopkeeper pays a VAT of ₹105, find the price inclusive of sales tax of the article which the shopkeeper paid to the wholesaler.
Solution:
Marked price of an article = ₹7500
Rate of S.T. = 7%

Question 4.
A shopkeeper buys an article at a discount of 30% and pays sales tax at the rate of 6%. The shopkeeper sells the article to a consumer at 10% discount on the list price and charges sales tax at the’ same rate. If the list price of the article is ₹3000, find the price inclusive of sales tax paid by the shopkeeper.
Solution:
List price of an article = ₹3000
Rate of discount = 30%
and rate of S.T. = 6%

Question 5.
Mukerjee purchased a movie camera for ₹27468. which includes 10% rebate on the list price and then 9% sales tax (under VAT) on the remaining price. Find the list price of the movie camera.
Solution:
Let list price of the movie camera = x
Rebate = 10%

Question 6.
A retailer buys an article at a discount of 15% on the printed price from a wholesaler. He marks up the price by 10%. Due to competition in the market, he allows a discount of 5% to a buyer. If the buyer pays ₹451.44 for the article inclusive of sales tax (under VAT) at 8%, find :
(i) the printed price of the article
(ii) the profit percentage of the retailer.
Solution:
(i) Let the printed price of the article = ₹100
Then, retailer’s cost price
= ₹100-₹15 = ₹85
Now, marked price for the retailer
= ₹100 + ₹10 = ₹110
Rate of discount allowed = 5%

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test are helpful to complete your math homework.

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