## SOLVED CAIIB COMBINED PAPER 8:

**1. A bond has been issued with a face value of Rs. 1000 at 10% Coupon for 3 years. The required rate of return is 8%. What is the value of the bond if the Coupon amount is payable on half-yearly basis?**

a. 1520

b. 1052

c. 1205

d. 1025

Ans - b

Explanation :

Here,

FV = 1000

CR = 10% half-yearly = 5% p.a.

Coupon = FV × CR = 50

R = 8% yearly = 4% p.a.

t = 3 years

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

= 1052

2. In Random Sampling, we pick up one, we note it down, put it back with the remaining and pick the next one. This is called...

a. Sampling with replacement

b. Simple random sampling

c. Sampling without replacement

d. None of the above

Ans - a

3. 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 nF ) 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 is the 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

4. Asha wants to receive a fixed amount for 15 years by investing Rs. 9 lacs @ 9% roi. How much she will receive annually?

a. 116153

b. 111563

c. 115163

d. 111653

Ans - d

Explanation :

Here,

P = 9 lac

R = 9% p.a.

T = 15 yrs

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

EMI = 900000 * 0.09 * 1.0915 ÷ (1.0915 – 1)

= 111653

5. 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%

6. A cash flow that is expected to grow at a constant rate forever, is called.

a. Annuity

b. Perpetuity

c. Growing annuity

d. Growing perpetuity

Ans : d

7. A firm needs Rs. 170000 to replace its machinery at the end of 5 years. At 12% roi, how much it should contribute every month?

a. 2802

b. 2082

c. 2820

d. 2028

Ans - b

Explanation :

Here,

FV = 170000

R = 12% p.a. = 0.01% monthly

T = 5 Y = 60 months

(Here, the firm has to contribute monthly, so we have converted rate and time to monthly equivalent values)

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

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

170000 = P * (1.0160 -1) ÷ 0.01

170000 = P * 81.66967

P = 170000 / 81.66967

= 2082

8. We have six students say A, B, C, D, E, F participating in a quiz contest. Out of six students only two can reach to the final. What is the probability of reaching to the final of each student ?

a. 1/2

b. 2/3

c. 1/3

d. 1/6

Ans – c

Solution :

Since out of 6, 2 can reach the final.

Hence sample space is

n(S) = 6 c2 = 6!/(6-b.!×2! = 15

Here event of occurrence of probability of each student out of six (A B C D E F) = (AB AC AD AE AF) = n(E) = 5

Now P(E) = 5/15 = 1/3

9. Ram purchased two bonds bond-1 & bond-2 with face value of Rs. 1000 each and Coupon of 8% and maturity of 4 years & 6 years respectively. If YTM is increased by 1%, the % change in prices of bond-1 & bond-2 would be ......

a. 2.39 & 4.84

b. 3.29 & 4.84

c. 3.29 & 4.48

d. 2.39 & 4.48

Ans - c

Explanation :

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

Bond 1:

If YTM is 9%, then bond’s price

= [80 × (1.09^4 – 1) ÷ 0.09 + 1000] ÷ 1.09^4

= 967.64

Bond 2:

If YTM is 9%, then bond’s price

= [80 × (1.09^6 – 1) ÷ 0.09 + 1000] ÷ 1.09^6

= 955.14

So, % change in price of bond 1

= (1000 – 967.04) ÷ 1000

= 0.03296

= 3.29%

& % change in price of bond 2

= (1000 – 955.14) ÷ 1000

= 0.04486

= 4.48%

10. 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 :

Since here total sample space is ( 4+5 ) = 9. Out of 9 , 3 ( 1 black & 2 red are expected to be drawn )

Hence sample space n( S ) = 9c3 or it can be written c( 9, 3 ).

c(9, c. = 9!/(9-c.!×3! i.e 84

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

n( B ) = 4 c1 i.e. 4

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

n( R) = 5 c2 = 5!/3!×2! = 10

Then P( B U R ) = n( B) × n( R ) / n ( S)

i.e 4×10/84 = 10/21

11. If inflation rate is higher in an economy, the discount rate should generally,

a. Be lower

b. Be higher

c. Be Stable

d. Be fluctuating

Ans - B

12. For carrying out his studies, a student borrows Rs. 3 lac from a bank at concessional rate of 5% p.a. for 4 years of his professional course. What is the total amount payable by him at the end of the 4th year?

a. 1298038

b. 1280838

c. 1293038

d. 1283038

Ans - c

Explanation :

Here,

P = 3 lac

R = 5% p.a.

T = 4 yrs

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

FV = 300000*(1.054 – 1) ÷ 0.05

= 1293038

a. 1520

b. 1052

c. 1205

d. 1025

Ans - b

Explanation :

Here,

FV = 1000

CR = 10% half-yearly = 5% p.a.

Coupon = FV × CR = 50

R = 8% yearly = 4% p.a.

t = 3 years

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

= 1052

2. In Random Sampling, we pick up one, we note it down, put it back with the remaining and pick the next one. This is called...

a. Sampling with replacement

b. Simple random sampling

c. Sampling without replacement

d. None of the above

Ans - a

3. 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 nF ) 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 is the 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

4. Asha wants to receive a fixed amount for 15 years by investing Rs. 9 lacs @ 9% roi. How much she will receive annually?

a. 116153

b. 111563

c. 115163

d. 111653

Ans - d

Explanation :

Here,

P = 9 lac

R = 9% p.a.

T = 15 yrs

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

EMI = 900000 * 0.09 * 1.0915 ÷ (1.0915 – 1)

= 111653

5. 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%

6. A cash flow that is expected to grow at a constant rate forever, is called.

a. Annuity

b. Perpetuity

c. Growing annuity

d. Growing perpetuity

Ans : d

7. A firm needs Rs. 170000 to replace its machinery at the end of 5 years. At 12% roi, how much it should contribute every month?

a. 2802

b. 2082

c. 2820

d. 2028

Ans - b

Explanation :

Here,

FV = 170000

R = 12% p.a. = 0.01% monthly

T = 5 Y = 60 months

(Here, the firm has to contribute monthly, so we have converted rate and time to monthly equivalent values)

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

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

170000 = P * (1.0160 -1) ÷ 0.01

170000 = P * 81.66967

P = 170000 / 81.66967

= 2082

8. We have six students say A, B, C, D, E, F participating in a quiz contest. Out of six students only two can reach to the final. What is the probability of reaching to the final of each student ?

a. 1/2

b. 2/3

c. 1/3

d. 1/6

Ans – c

Solution :

Since out of 6, 2 can reach the final.

Hence sample space is

n(S) = 6 c2 = 6!/(6-b.!×2! = 15

Here event of occurrence of probability of each student out of six (A B C D E F) = (AB AC AD AE AF) = n(E) = 5

Now P(E) = 5/15 = 1/3

9. Ram purchased two bonds bond-1 & bond-2 with face value of Rs. 1000 each and Coupon of 8% and maturity of 4 years & 6 years respectively. If YTM is increased by 1%, the % change in prices of bond-1 & bond-2 would be ......

a. 2.39 & 4.84

b. 3.29 & 4.84

c. 3.29 & 4.48

d. 2.39 & 4.48

Ans - c

Explanation :

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

Bond 1:

If YTM is 9%, then bond’s price

= [80 × (1.09^4 – 1) ÷ 0.09 + 1000] ÷ 1.09^4

= 967.64

Bond 2:

If YTM is 9%, then bond’s price

= [80 × (1.09^6 – 1) ÷ 0.09 + 1000] ÷ 1.09^6

= 955.14

So, % change in price of bond 1

= (1000 – 967.04) ÷ 1000

= 0.03296

= 3.29%

& % change in price of bond 2

= (1000 – 955.14) ÷ 1000

= 0.04486

= 4.48%

10. 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 :

Since here total sample space is ( 4+5 ) = 9. Out of 9 , 3 ( 1 black & 2 red are expected to be drawn )

Hence sample space n( S ) = 9c3 or it can be written c( 9, 3 ).

c(9, c. = 9!/(9-c.!×3! i.e 84

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

n( B ) = 4 c1 i.e. 4

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

n( R) = 5 c2 = 5!/3!×2! = 10

Then P( B U R ) = n( B) × n( R ) / n ( S)

i.e 4×10/84 = 10/21

11. If inflation rate is higher in an economy, the discount rate should generally,

a. Be lower

b. Be higher

c. Be Stable

d. Be fluctuating

Ans - B

12. For carrying out his studies, a student borrows Rs. 3 lac from a bank at concessional rate of 5% p.a. for 4 years of his professional course. What is the total amount payable by him at the end of the 4th year?

a. 1298038

b. 1280838

c. 1293038

d. 1283038

Ans - c

Explanation :

Here,

P = 3 lac

R = 5% p.a.

T = 4 yrs

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

FV = 300000*(1.054 – 1) ÷ 0.05

= 1293038