ML Aggarwal Solution Class 10 Chapter 2 Banking MCQs

MCQs

Question 1

If Sharukh opened a recurring deposit account in a bank and deposited Rs 800 per month for $1 \frac{1}{2}$ years,  then the total money deposited in the account is

(a) Rs 11400

(b) Rs 14400

(c) Rs 13680

(d) none of these

Sol :

Monthly deposit = Rs 800

Period $(n)=1 \frac{1}{2}$ years $=18$ months

∴ Total money deposit = Rs 800 × 18

= Rs 14400

Ans : (b)


Question 2

Mrs. Asha Mehta deposit Rs 250 per month for one year in a bank’s recurring deposit account. If the rate of (simple) interest is 8% per annum, then the interest earned by her on this account is

(a) Rs 65

(b) Rs 120

(c) Rs 130

(d) Rs 260

Sol :

Deposit per month (P) = Rs 250

Period (n) = 1 year = 12 months

Rate (r) = 8% p.a.

∴Interest $=\frac{\mathrm{P} \times n \times(n+1)}{2 \times 12} \times \frac{r}{100}$

$=\frac{250 \times 12+13}{2 \times 12} \times \frac{8}{100}=₹ 130$

Ans : (c)


Question 3

Mr. Sharma deposited Rs 500 every month in a cumulative deposit account for 2 years. If the bank pays interest at the rate of 7% per annum, then the amount he gets on maturity is

(a) Rs 875

(b) Rs 6875

(c) Rs 10875

(d) Rs 12875

Sol :

Deposit (P) = Rs 500 per month

Period (n) = 2 years = 24 months

Rate (r) = 7% p.a.

∴Interest $=\frac{P \times n \times(n+1)}{2 \times 12} \times \frac{r}{100}$

$=\frac{500 \times 24 \times 25 \times 7}{2 \times 12 \times 100}=₹ 875$

∴Maturity value =P×24+Interest

=₹ 500 \times 24+875=₹ 12000 × 875

=₹ 12875

Ans : (d)


Question 4

John deposited Rs 400 every month in a bank’s recurring deposit account for $2 \frac{1}{2}$ years . If he gets Rs 1085 as interest at the time of maturity, then the rate of interest per annum is

(a) 6%

(b) 7%

(c) 8%

(d) 9%

Sol :

Deposit (P) = Rs 400 per month

Period (n) $=2 \frac{1}{2}$ years $=3$ months

∴Interest = Rs 1085

Let r% be the rate of interest

Interest $=\frac{P \times n \times(n+1)}{2 \times 12} \times \frac{r}{100}$

$1085=₹ \frac{400 \times 30 \times 31 \times r}{2 \times 12 \times 100}$

$1085=155 r \Rightarrow r=\frac{1085}{155}=7$

∴Rate 7 % p.a.

Ans : (b)

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