CAIIB ABM case studies

When chiken prices rise 40%, the quantity of KFC fried chicken

supplied rises by 20%. Calculate the price

elasticity of supply.

a. 0.25

b. 0.50

c. 0.75

d. 0.85

Ans - a

Solution :

Price Elasticity of Supply = (% change in quantity supplied) / (%

change in price)

= 20/40 = 0.5

When the price of a commodity falls from Rs. 60 per unit to

Rs. 48 per unit, the quantity supplied falls by

20%. Calculate the price elasticity of supply.

a. 1

b. 1.5

c. 2

d. 2.5

Ans – a

495

Solution :

Price Elasticity of Supply = (% change in quantity supplied) / (%

change in price)

= 20/((60-48)*100/60)

= 20/(12*100/60)

= 20/20

= 1

21 bricks have a mean mass of 24.2 kg, and 29 similar bricks

have a mass of 23.6 kg. Determine the

mean mass of the 50 bricks.

a. 18.35 kg

b. 20.35 kg

c. 23.85 kg

d. 32.85 kg

Ans - c

Solution :

Mean value = ((21 x 24.2) + 29 x 23.6 )) / (21+29)

= 1192.6 / 50

= 23.85 kg

.............................................

Net income available to stockholders is ₹150 and total assets

are ₹2,100 then return on total assets

would be ......

a. 0.07%

b. 7.14%

c. 0.05 times

d. 7.15 times

Ans - b

Solution

Return on Assets Ratio=Net Income/Average Total Assets*100

=150/2100*100

=0.0714

=7.14

.............................................

Given,

Recoveries of loan and advance - Rs. 3000 Crores

Misc capital receipt - Rs. 600 Crores

Market loAns - Rs. 600 Crores

Short term borrowings - Rs. 1200 Crores

External assistance (Net) - Rs. 500 Crores

State provident fund - Rs. 600 Crores

Other receipts (Net) - Rs. 1200 Crores

Securities issued against small savings - Rs. 600 Crores

Recoveries of short term loans and advances from states and

loans to govt servants - Rs. 1000 Crores

Total Non Tax Revenue - Rs. 5000 Crores

Net Tax Revenue - Rs. 2000 Crores

Draw down cash balance - Rs. 4000 Crores

Calculate Debt Receipt ...

a. Rs 2500 Crores

b. Rs 3700 Crores

c. Rs 4700 Crores

d. Rs 5400 Crores

Ans - c

.............................................

Calculate Non Debt Receipt ...

a. Rs 2500 Crores

b. Rs 3700 Crores

c. Rs 4700 Crores

d. Rs 5400 Crores

Ans - a

.............................................

Calculate Capital Receipt ...

a. Rs 4700 Crores

b. Rs 5400 Crores

c. Rs 6200 Crores

d. Rs 7200 Crores

Ans - c

.............................................

Solution :

1. Debt Receipt = Market Loans + Short Term Borrowings +

External assistance(NET) + Securities issued

against Small savings + State provident fund + other

Receipts(Net)

= 600 + 1200 + 500 + 600 + 600 + 1200

= 3700 Crores

2. Non Debt Receipt = Recoveries of loan & advances (deduct

recoveries of short term loans & advance

from state and loans to govt sarvants) + MISC Capital receipts

= (3000-1000)+500

= 2500 Crores

3. Capital Receipt = Non Debt Receipt + Debt Receipt

= 3700 + 2500

= 6200 Crores

Given,

Currency with public - Rs. 120000 Crores

Demand deposit with banking system - Rs. 200000 Crores

Time deposits with banking system - Rs. 250000 Crores

Other deposit with RBI - Rs. 300000 Crores

Savings deposit of post office savings banks - Rs. 100000

Crores

All deposit with post office savings bank excluding NSCs - Rs.

50000 Crores

1. Calculate M1.

a. Rs. 570000 Crores

b. Rs. 620000 Crores

c. Rs. 670000 Crores

d. Rs. 720000 Crores

Ans - b

.............................................

2. Calculate M2.

a. Rs. 570000 Crores

b. Rs. 620000 Crores

c. Rs. 670000 Crores

d. Rs. 720000 Crores

Ans - d

.............................................

3. Calculate broad money M3.

a. Rs. 570000 Crores

b. Rs. 620000 Crores

c. Rs. 670000 Crores

d. Rs. 870000 Crores

Ans - d

.............................................

Solution :

1. M1 = currency with public + demand deposit with the

banking system + other deposits with RBI

M1 = 120000+200000+300000

M1 = 620000

485

2. M2 = M1+Savings deposit of post office savings banks

So,

M2 = 620000+100000

M2 = 720000 Crores

3. M3 = M1+Time deposit with banking system

So,

M3 = 620000+250000

M3 = 870000 Crores

Given

1. Consumptions - Rs. 50000

2. Gross investment - Rs. 40000

3. Govt spending - Rs. 10000

4. Export - Rs. 90000

5. Import - Rs. 60000

6. Indirect Taxes - Rs. 10000

7. Subsidies(on production and import) - RS. 5000

8. Compensation of employee - Rs. 500

9. Property Income - Rs. 500

7,8,9 - Net receivable from aboard

10.Total capital gains from overseas investment - Rs. 15000

11.Income earned by foreign national domestically - Rs. 5000

1. Calculate GDP

a. Rs. 125000

b. Rs. 130000

c. Rs. 135000

d. Rs. 140000

Ans - b

.............................................

2. Calculate GDP at cost factor

a. Rs. 125000

b. Rs. 130000

c. Rs. 135000

d. Rs. 140000

Ans - c

.............................................

3. Calculate GNP

a. Rs. 110000

b. Rs. 120000

c. Rs. 130000

d. Rs. 140000

Ans - d

.............................................

479

Solution :

1. GDP = Consumption + Gross investment + Government

spending + (Exports - Imports)

GDP = C+I+G+(X-M)

= 50000+40000+10000+(90000-60000)

= 130000

2. GDP at factor rate

= GDP-(Indirect taxes-subsidies)

= 130000-(10000-5000)

= 135000

3. GNP=GDP+NR(total capital gains from Overseas

investment-income earned by foreign national

domestically)

= 130000 + (15000-5000)

= 140000

.............................................

Savings deposit of post office savings banks - Rs. 60000 Crores

All deposit with post office savings bank excluding NSCs - Rs.

50000 Crores

Calculate M4.

a. Rs. 750000 Crores

b. Rs. 800000 Crores

c. Rs. 810000 Crores

d. Rs. 870000 Crores

Ans - b

Solution :

M4 = M3+All deposit with post office savings bank excluding

NSCs

M3 = M1+Time deposit with banking system

M1 = currency with public + demand deposit with the banking

system + other deposits with RBI

M1 = 90000+180000+260000

M1 = 530000

So,

M3 = M1+Time deposit with banking system

M3 = 530000+220000

M3 = 750000 Crores

So,

M4 = M3+All deposit with post office savings bank excluding

NSCs

M4 = 750000+50000

M4 = 800000 Crores

Calculate the present value of 6 year bond with 9 per cent

coupon rate with FV Rs. 1000/-. Current

interest rate is 12 per cent.

a. Rs.843.83

b. Rs.1025.57

c. Rs.876.66

d. Rs.768.68

Ans - c

Solution

FV = 1000

Coupon Rate (CR) = 0.9%

t = 6 year

R (YTM) = 0.12

Annual interest rate payable=1000*9%=90

Principal repayment at the end of 6 year = Rs. 1000

=90 (PVIFA, 12%,6 years)+1000(PVIF,12%, 6 Years)

PVIFA= ((1+r)^t -1)/ r+ PVIF=1/(1+R)^t

=90(1.12^6-1/0.12*(1.12)^6+1000(1/1.12^6)

=90*1.97382-1/0.12*1.1.97382+1000(1/1.97382)

=90*0.97382/0.12*1.97382+1000*0.0.50663

=90*0.97382/0.23685+506.63

=90*4.11154+506.63

=370.03+506.63

=876.66

Mr x is to receive Rs. 10000, as interest on bonds by end of

each year for 5 years @ 5% roi. Calculate the

present value of the amount he is to receive.

a. 43925

b. 43295

c. 49325

d. 49235

Ans - b

Explanation :

Here,

P = 10000

R = 5% p.a.

T = 5 Y

PV = P / R * [(1+R)^T - 1]/(1+R)^T

PV to be received, if the amount invested at end of each year:

So,

FV = (100000÷0.05) * {(1+0.05)^5 – 1} ÷ (1+0.05)^5

= 43295

A company has net worth of Rs. 15 lac, term liabilities are Rs.

10 lac. Fixed Assets worth Rs. 16 lac and

current assets are Rs. 25 lac. There is no intangible assets or

the non current assets. Calculate it's net

working capital.

a. 6 lac

b. 7 lac

c. 8 lac

d. 9 lac

Ans - d

Total Assets = Total liabilities

Total Assets = Fixed Assets + current assets

= 16 + 25

= 41

So total liabilites must be 41 lac

Now out of 41 lac, the long term liability is 25 lac (15 + 10)

Hence CL = 41 - 25 = 16 lac

Now we have CA = 25 lac and CL = 16 lac

NWC = 25 - 16

= 9 lac

.............................................

Calculate Standard Error from the given data : X = 10,

20,30,40,50

a. 6.1071

b. 6.0711

c. 7.1071

d. 7.0711

Ans - d

Explanation :

Total Inputs (N) = (10,20,30,40,50)

Total Inputs (N) =5

First find Mean:

Mean (xm) = (x1+x2+x3...xn)/N

Mean (xm) = 150/5

Mean (xm) = 30

Then find SD:

SD = √(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))

= √(1/(5-1)((10-30)2+(20-30)2+(30-30)2+(40-30)2+(50-30)2))

= √(1/4((-20)2+(-10)2+(0)2+(10)2+(20)2))

= √(1/4((400)+(100)+(0)+(100)+(400)))

= √(250)

= 15.811

Then Find Standard Error:

Standard Error=SD / √(N)

= 15.8114/√(5)

= 15.8114/2.2361

= 7.0711

A sack contains 4 black balls 5 red balls. What is probability to

draw 1 black ball and 2 red balls in one

draw ?

a. 12/21

b. 9/20

c. 10/21

d. 11/20

Ans – c

Solution :

Out of 9, 3 (1 black & 2 red. are expected to be drawn)

Hence sample space

n(S) = 9c3

= 9!/(6!×3!)

= 362880/4320

= 84

Now out of 4 black ball 1 is expected to be drawn hence

nb. = 4c1

= 4

Same way out of 5 red balls 2 are expected be drawn hence

n(R) = 5c2

= 5!/(3!×2!)

= 120/12

= 10

Then P(B U R) = n(B)×n(R)/n(S)

i.e 4×10/84 = 10/21

........................................................

Quantity supplied of a product at Rs. 8 per unit is 200 Units. If

the price elasticity of supply is 1.5, what

will be the quantity supplied at Rs. 10 per unit?

a. 150

b. 175

c. 250

d. 275

Ans - d

Solution :

Price Elasticity of Supply = (% change in quantity supplied. / (%

change in price)

1.5 = ((x-200)*100/200)/((10-8)*100/8)

1.5 = ((x-200)/2)/(200/8)

1.5 = ((x-200)/2)/25

1.5 = (x-200)/50

75 = x-200

x = 75+200

x = 275

X opened a recurring account with a bank to deposit Rs. 16000

by the end of each year @ 10% roi. How

much he would get at the end of 3rd year?

a. 52960

b. 52690

c. 52069

d. 52096

Ans - a

Explanation :

Here,

P = 16000

R = 10% p.a.

T = 3 yrs

FV = P / R * [(1+R)^T - 1]

FV = 16000 * (1.13 – 1) ÷ 0.1

= 52960

.............................................

Find Correlation coefficient for X and Y values given below :

X= (1,2,3,4,5)

Y= {11,22,34,43,56}

a. 0.8899

b. 0.9989

c. 1.0899

d. 1.0989

Ans - b

Explanation :

Step 1: Find Mean for X and Y

X=15/5=3

Y=166/5=33.2

Step 2: Calculate Standard Deviation for Y inputs:

σx=

359

√(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))

=√(1/(5-1)((11-33.2)2+(22-33.2)2+(34-33.2)2+(43-33.2)2+(56-

33.2)2))

=√(1/4((-22.2)2+(-11.2)2+(0.8)2+(9.8)2+(22.8)2))

=√(1/4((492.84)+(125.44)+(0.64)+(96.04)+(519.84)))

=√(308.7)

=17.5699

Step 3: Standard Deviation for X Inputs:

σx=

√(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))

=√(1/(5-1)((1-3)2+(2-3)2+(3-3)2+(4-3)2+(5-3)2))

=√(1/4((-2)2+(-1)2+(0)2+(1)2+(2)2))

=√(1/4((4)+(1)+(0)+(1)+(4)))

=√(2.5)

=1.5811

Σ((X - μx) (Y - μy))

=(1-3)(11-33.2)+(2-3)(22-33.2)+(3-3)(34-33.2)+(4-3)(43-

33.2)+(5-3)(56-33.2)

=(-2*-22.2) + (-1*-11.2) + (0* 0.8) + (1 *9.8) + (2* 22.8)

=44.4 + 11.2 + 0 + 9.8 + 45.6

=111

Correlation Coefficient = 111/((5-1)*1.5811*17.5699)

Correlation Coefficient (r) = 0.9989

Hence the correlation coefficient between the two given data

set is 0.9989

Albert purchased 8%, 3 years bond of Rs. 10 lac, with annual

interest payment and face value payable on

maturity. The YTM is assumed@ 6%. Calculate the duration

and modified duration.

a. 2.36

b. 2.79

c. 2.63

333

d. 2.97

Ans - c

Explanation :

Bond’s Duration = ΣPV×T ÷ ΣP

ΣP = 1053421

Now, a = 0.943396 and a^t = 0.839619

So, ΣPV×T = 80000 × 16.666 × (0.160381÷0.056604 –

2.518857) + 2518857

= 419370.767 + 25188579

= 2938227.77

So, Duration of the Bond

= 2938227.77 / 1053421

= 2.79 years

& Modified Duration

= Mckauley Duration ÷ (1 + R)

= 2.79 ÷ 1.06

= 2.63

Mr. Raj is to invest Rs. 100000 by end of each year for 5 years

@ 5% roi. How much amount he will

receive?

a. 556253

b. 553562

c. 552563

d. 555263

Ans - c

Explanation :

Here,

P = 1000000

R = 5% p.a.

T = 5 Y

FV = P / R * [(1+R)^T - 1]

FV, if invested at end of each year, is:

So,

FV = (100000÷0.05) * {{1+0.05}^5 – 1}

= 552563

An urn contains 10 black balls and 5 white balls. 2 balls are

drawn from the urn one after other without

replacement. What is the probability that both drawn are

black ?

a. 2/7

b. 3/7

c. 4/7

d. 6/7

Ans - b

Solution :

Let E and F denote respective events that first and second ball

drawn are black.

We have to find here P(E n F ) and P(E/F)

Now P(E) = P(Black in first drawn) = 10/15

Also given that the first ball is drawn i.e events E has occurred.

Now there are 9 black balls and 5 white

balls left in the urn. Therefore the probability that the second

ball drawn is black, given that the ball first

drawn is black nothing but conditional probability of F given

that E has occurred already.

Hence P(E/F) = 9/14

Now by the multiplication rule of probability

P(E n F) = P(E) × P(E/F)

= 10/15 × 9/14

= 3/7

........................................................

A person invested Rs. 100000 in a bank FDR @ 6% p.a. for 1

year. If interest is compounded on halfyearly

basis, the amount payable shall be ......

a. 109060

b. 100960

c. 103090

d. 106090

Ans – d

265

Explanation :

Here,

P = 100000

R = 6% half-yearly = 3%@ p.a. = 0.03 p.a.

T = 1 yr = 2 half yrs

FV = P * (1 + R)^T

So,

FV = 100000 * (1+0.03)^2

= 106090

Ranjit borrowed an amount of Rs. 50000 for 8 years @ 18%

roi. What shall be monthly payment?

a. 986

b. 968

c. 896

d. 869

Ans - a

Explanation :

Here,

P = 50000

R = 18% = 18 % ÷ 12 = 0.015% monthly

T = 8 yrs = 96 months

EMI = P * R * [(1+R)^T/(1+R)^T-1)]

EMI = 50000 * 0.015 * 1.01596 ÷ (1.01596 – 1

= 986

Ajit wants to receive Rs. 40000 p.a. for 20 years by investing

@ 5%. How much he will have to invest

now?

a. 498489

263

b. 498849

c. 498948

d. 498984

Ans - a

Explanation :

Here,

P = 40000

R = 5% p.a.

T = 20 yrs

PV = P / R * [(1+R)^T - 1]/(1+R)^T

PV = (40000 ÷ 0.05) * {(1.0520 – 1) ÷ 1.0520}

= 498489

.................

Priyanka made an investment of Rs. 18000 and he expects a

return of Rs. 3000 p.a. For 12 years. What is

the present value and net present value of the cash flow @

10% discount rate?

a. 2114

b. 2414

c. 2441

d. 2141

255

Ans - c

Explanation :

PV = 20441

NPV = PV – 18000

= Rs. 2441

The cash flow expected from a project is Rs. 700, Rs. 1000 and

Rs. 1200 in the 1st, 2nd, & 3rd year. The

discounting factor @ 10% roi is 1.10, 1.21 and 1.331. What is

the total present value of these cash

flows?

a. 3264

b. 3246

c. 2346

d. 2364

Ans - d

Explanation :

NPV = Σ {C÷ (1+r)T} – 1

Total Present Value

= Σ {C÷ (1+r)T}

= (700 ÷ 1.1) + (1000 ÷ 1.21) + (1200 ÷ 1.331)

= Rs. 2364

.............................................

A 10%, 6-years bond, with face value of Rs. 1000 has been

purchased by Mr. x for Rs. 900. What is his

yield till maturity?

a. 12.47

b. 14.27

c. 11.74

d. 11.27

Ans - a

Explanation :

Here,

FV = 1000

CR = 10%

R (YTM) =?

T = 6 years

Coupon = FV × CR = 100

Bond’s price = 900

Since FV > Bond’s Value, Coupon rate < YTM (based on above

three observations)

So, we have to use trial and error method. We have to start

with a value > 10 and find the price until we

get a value < 900.

Bond Price = (1/(1+R)^t)((coupon*((1+R)^t-1)/R)+Face Value)

So,

If YTM = 11%, price =957.69 (> 900, so keep guessing)

If YTM = 12%, price = 917.78 (> 900, so keep guessing)

If YTM = 13%, price = 880.06 (< 900, so stop)

So, YTM must lie between 12 and 13.

So, using interpolation technique,

YTM

= 12 + (917.78 – 900) ÷ (917.78 – 880.06)

= 12 + 17.78 ÷ 37.72

= 12.47%

.............................................

A bond with a par-value of Rs. 100 is purchased for 95.92 and

it paid a Coupon rate of 5%. Calculate its

current yield.

a. 5.12

b. 5.21

c. 5.34

d. 5.43

Ans - b

Explanation :

Coupon = Face value × Coupon Rate

And annual interest paid = Market Price × Current Yield

5 = 95.92 × CY

CY = 0.0521 = 5.21%

My grandfather, starts giving me gifts of Rupees 1 lakh for the

next 4 years. If the interest rate is 10 per

cent pa, how much will I get at the end of 4 years?

a. 414600

b. 416400

c. 461400

d. 464100

Ans - d

Solution

FV=P/R {(1+r)^n-1}

=100000/0.10{(1+.1)^4-1}

=1000000 (1.4641-1)

=100000

Find Coefficient of Variance for the values given :

{13,35,56,35,77}

a. 0.4156

b. 0.5164

161

c. 0.5614

d. 0.6514

Ans - c

Explanation :

Number of terms (N) = 5

Mean:

Xbar = (13+35+56+35+77)/5

= 216/5

= 43.2

Standard Deviation (SD):

Formula to find SD is

σx= √(1/(N - 1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))

=√(1/(5-1)((13-43.2)2+(35-43.2)2+(56-43.2)2+(35-43.2)2+(77-

43.2)2))

=√(1/4((-30.2)2+(-8.2)2+(12.9)2+(-8.2)2+(33.8)2))

=√(1/4((912.04)+(67.24)+(163.84)+(67.24)+(1142.44)))

=√(588.2)

=24.2528

Coefficient of variation (CV)

CV = Standard Deviation / Mean

= 24.2528/43.2

= 0.5614

Hence the required Coefficient of Variation is 0.5614

Calculate the relative variability (coefficient of variance) for

the samples 60.25, 62.38, 65.32, 61.41, and

63.23 of a population

a. 0.0301

b. 0.0307

c. 0.0103

d. 0.0107

Ans - b

Solution

Mean = (60.25 + 62.38 + 65.32 + 61.41 + 63.23)/5

= 312.59/5

= 62.51

calculate standard deviation

= √( (1/(5 - 1)) * (60.25 - 62.51799)2 + (62.38 - 62.51799)2 +

(65.32 - 62.51799)2 + (61.41 - 62.51799)2 +

(63.23 - 62.51799)2)

= √( (1/4) * (-2.267992 + -0.137989992 + 2.802012 + -

1.107992 + 0.712012))

= √( (1/4) * (5.14377 + 0.01904 + 7.85126 + 1.22764 +

0.50695))

= √ 3.68716 σ = 1.92

calculate coefficient of variance

Coefficient of Variance = (Standard Deviation (σ) / Mean (μ))

= 1.92 / 62.51

= 0.0307

alculate the Linear Regression whose input values are

X = [5,20,40,80,100] & Y = { 10,20,30,40,50}

a. 10.0914 + 0.3894 X

b. 10.9194 + 0.8934 X

c. 10.9194 + 0.3894 X

d. 10.9994 + 0.8934 X

Ans - c

Solution

Linear regression formula

Y = A + BX

To find X Mean

Total Inputs(N) =(5,20,40,80,100)

Total Inputs(N)=5

Mean(xm)= (x1+x2+x3...xn)/N

Mean(xm)= 245/5

Mean(xm)= 49

To find Y Mean

Total Inputs(N) =(10,20,30,40,50)

Total Inputs(N)=5

Mean(Ym)= (y1+y2+y3...yn)/N

Mean(Ym)= 150/5

Mean(Ym)= 30

To find Slope

Slope = (ΣXY - N x X x Y)/(ΣX2 - N x X)

= [(50+400+1200+3200+5000) - (5 x 30 x

49)]/[(25+400+1600+6400+10000) -( 5x492)]

= (9850 - 7350) /(18425 - 12005)

= 2500/6420

= 0.3894

To find Intercept Value:

A = Ybar - Slope x Xbar

= 30 - (0.3894 x 49)

= 10.9194.

To find Linear Regression:

Y = A + Bx

Answer = 10.9194 + 0.3894 X

.............................................

Calculate the relative variability (coefficient of variance) for

the samples 60.25, 62.38, 65.32, 61.41, and

63.23 of a population

a. 0.0301

b. 0.0307

c. 0.0103

d. 0.0107

Ans - b

Solution

Mean = (60.25 + 62.38 + 65.32 + 61.41 + 63.23)/5

= 312.59/5

= 62.51

calculate standard deviation

= √( (1/(5 - 1)) * (60.25 - 62.51799)2 + (62.38 - 62.51799)2 +

(65.32 - 62.51799)2 + (61.41 - 62.51799)2 +

(63.23 - 62.51799)2)

= √( (1/4) * (-2.267992 + -0.137989992 + 2.802012 + -

1.107992 + 0.712012))

= √( (1/4) * (5.14377 + 0.01904 + 7.85126 + 1.22764 +

0.50695))

= √ 3.68716 σ = 1.92

calculate coefficient of variance

Coefficient of Variance = (Standard Deviation (σ) / Mean (μ))

= 1.92 / 62.51

= 0.0307

.............................................

Given,

Currency with public - Rs. 230000 Crores

Demand deposit with banking system - Rs. 320000 Crores

Time deposits with banking system - Rs. 360000 Crores

Other deposit with RBI - Rs. 420000 Crores

Savings deposit of post office savings banks - Rs. 140000

Crores

All deposit with post office savings bank excluding NSCs - Rs.

80000 Crores

Calculate M1.

a. Rs. 670000 Crores

b. Rs. 830000 Crores

c. Rs. 970000 Crores

d. Rs. 1020000 Crores

Ans - c

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Calculate M2.

a. Rs. 830000 Crores

b. Rs. 970000 Crores

c. Rs. 1110000 Crores

d. Rs. 1330000 Crores

Ans - c

.............................................

Calculate broad money M3.

a. Rs. 830000 Crores

b. Rs. 970000 Crores

c. Rs. 1110000 Crores

d. Rs. 1330000 Crores

Ans - d

.............................................

Solution :

M1 = currency with public + demand deposit with the banking

system + other deposits with RBI

3

M1 = 230000+320000+420000

M1 = 970000

M2 = M1+Savings deposit of post office savings banks

So,

M2 = 970000+140000

M2 = 1110000 Crores

M3 = M1+Time deposit with banking system

So,

M3 = 970000+360000

M3 = 1330000 Crores