ML Aggarwal Solution Class 10 Chapter 2 Banking Test

Test

Question 1

Mr. Dhruv deposits Rs 600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.

Sol :

Deposit per month = Rs 600

Rate of interest = 10% p.a.

Period (n) = 5 years 60 months.

Total principal for one month

$=₹ 600 \times \frac{n(n+1)}{2}=₹ 600 \times \frac{60(60+1)}{2}$

$=₹ \frac{600 \times 60 \times 61}{2}=₹ 1098000$

Interest $=\frac{PRT}{100}$ $=\frac{1098000 \times 10 \times 1}{100 \times 12}=₹ 9150$

∴Amount of maturity $=₹ 600 \times 60+₹ 9+50$

=₹ 36000+₹ 9150=₹ 45150


Question 2

Ankita started paying Rs 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs 500 per month in a $2 \frac{1}{2}$ years recurring deposit. The bank paid 10% p.a. simple interest for both. At maturity who will get more money and by how much?

Sol :

In case of Ankita,

Deposit per month = Rs 400

Period (n) = 3 years = 36 months

Rate of interest = 10%

Total principal for one month

$=400 \times \frac{n(n+1)}{2}=400 \times \frac{36(36+1)}{2}$

$=₹ \frac{400 \times 36 \times 37}{2}=₹ 266400$

Interest $=\frac{PRT}{100}$ $=\frac{266400 \times 10 \times 1}{100 \times 12}=₹ 2220$

∴Amount of maturity =₹ 400 \times 36+₹ 2220

=₹ 14400+₹ 2220=₹ 16620

In case of Anshul, 

Deposit p.m.=₹ 500

Rate of interest =10 %

Period $(n)=2 \frac{1}{2}$ years =30 months

∴Total principal for one month

$=₹ 500 \times \frac{n(n+1)}{2}=500 \times \frac{30(30+1)}{2}$

$=₹ \frac{500 \times 30 \times 31}{2}=₹ 232500$

Interest $=\frac{232500 \times 10 \times 1}{100 \times 12}=₹ 1937.50$

Amount of maturity =₹ 500 × 30+₹ 1937.50

=₹ 15000+₹ 1937.50=₹ 16937.50

At maturity Anshul will get more amount

Difference =₹ 16937.50-₹ 16620.00

=₹ 317.50


Question 3

Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and deposits Rs 800 per month. If she gets Rs 48200 at the time of maturity, find

(i) the rate of simple interest,

(ii) the total interest earned by Shilpa

Sol :

Deposit per month (P) = Rs 800

Amount of maturity = Rs 48200

Period (n)=4 years =48 months

Let rate of interest be R % p.a.

Total principal for one month

$=\frac{P(n)(n+1)}{2}=\frac{800 \times 48 \times(48+1)}{2}$ $=₹ \frac{800 \times 48 \times 49}{2}=₹ 940800$

Total deposit =₹ 800 \times 48=₹ 38400

∴Interest earned =₹ 48200-₹ 38400=₹ 9800

(i) Rate of interest $=\frac{\text { S.I. } \times 100}{P \times T}$

$=\frac{9800 \times 100 \times 12}{940800 \times 1}=12.5 \%$

(ii) Total interest earned by Shilpa $=₹ 9800$


Question 4

Mr. Chaturvedi has a recurring deposit account in Grindlay’s Bank for $4 \frac{1}{2}$ years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly instalment.

Sol :

Let each monthly instalment = Rs x

Rate of interest = 11 %

Period $(n)=4 \frac{1}{2}$ years or 54 months,

Total principal for one month

$=₹ x \times \frac{n(n+1)}{2}=₹ x \times \frac{54(54+1)}{2}$

$=x \times \frac{54 \times 55}{2}=1485 x$

Interest $=\frac{1485 x \times 11 \times 1}{100 \times 12}=13 \cdot 6125 x$

∴Total amount of maturity $=54 x+13 \cdot 6125 x$

=67.6125x

∴67.6125 x=101418.75

$x=\frac{101418 \cdot 75}{67 \cdot 6125}=₹ 1500$

∴Deposit per month =₹ 1500


Question 5

Rajiv Bhardwaj has a recurring deposit account in a bank of Rs 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs 15450 as maturity amount, find the total time for which the account was held.

Sol :

Deposit during the month (P) = Rs 600

Rate of interest = 7% p.a.

Amount of maturity = Rs 15450

Let time = n months

∴Total principal $=\frac{\mathrm{P}(n)(n+1)}{2}$

$=\frac{600 \times n(n+1)}{2}=\frac{600\left(n^{2}+n\right)}{2}=300\left(n^{2}+n\right)$

∴Interest $=\frac{\text { PRT }}{100}=\frac{300\left(n^{2}+n\right) \times 7 \times 1}{100 \times 12}$

$=\frac{7}{4}\left(n^{2}+n\right)$

∴$600 n+\frac{7}{4}\left(n^{2}+n\right)=15450$

⇒2400n+7n2+7n=61800

⇒7n2+2407 n-61800=0

⇒7n2-168 n+2575 n-61800=0

⇒7 n(n-24)+2575(n-24)=0

⇒(n-24)(7 n+2575)=0

Either n-24=0, then n=24

or 7 n+2575=0, then

$7 n=-2575 \Rightarrow n=\frac{-2575}{7}$

Which is not possible being negative.

∴n=24

∴Period=24 months or 2 years

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more