Q1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

A. [latex]\frac {1}{2}[/latex] B. [latex]\frac {2}{5}[/latex] C. [latex]\frac {8}{15}[/latex] D. [latex]\frac {9}{20}[/latex]

Answer: D
Explanation: Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P (E) = [latex]\frac {n (E)}{n (S)} [/latex]= [latex]\frac {9}{20} [/latex]
Q2. A bag contains 2 red, 3 green, and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A. [latex]\frac {10}{21}[/latex] B. [latex]\frac {11}{21}[/latex] C. [latex]\frac {2}{7}[/latex] D. [latex]\frac {5}{7}[/latex]

Answer: A
Explanation: Total number of balls = (2 + 3 + 2) = 7
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7
= [latex]{7}_{{C}_{2}}[/latex]
= [latex]\frac {(7 \times 6)}{(2 \times 1)}[/latex]
= 21
E = Event of drawing 2 balls, none of which is blue.
therefore, n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls.
= [latex]{5}_{{C}_{2}}[/latex]
= [latex]\frac {(5 \times 4)}{(2 \times 1)}[/latex]
= 10
P(E) = [latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {10}{21}[/latex]
Q3. In a box, there are 8 red, 7 blue, and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A. [latex]\frac {1}{3}[/latex] B. [latex]\frac {3}{4}[/latex] C. [latex]\frac {7}{19}[/latex] D. [latex]\frac {8}{21}[/latex]

Answer: A
Explanation: The number of balls = (8 + 7 + 6) = 21
Let = Event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
therefore, n(E) = 7
P(E) = [latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {7}{21} = \frac {1}{3} [/latex]
Q4. What is the probability of getting a sum 9 from two throws of a dice?

A. [latex]\frac {1}{6}[/latex] B. [latex]\frac {1}{8}[/latex] C. [latex]\frac {1}{9}[/latex] D. [latex]\frac {1}{12}[/latex]

Answer: C
Explanation: In two throws, n(S) = [latex] (6 \times 6) = 36[/latex]
E = event of getting a sum = {(3, 6), (4, 5), (5, 4), (6, 3)}.
therefor, P(E) = [latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {4}{36} = \frac {1}{9} [/latex]
Q5. Three unbiased coins are tossed. What is the probability of getting at most two heads?

A. [latex]\frac {3}{4}[/latex] B. [latex]\frac {1}{4}[/latex] C. [latex]\frac {3}{8}[/latex] D. [latex]\frac {7}{}[/latex]

Answer: D
Explanation: Here S = {TTT, TTH, THT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHt}
therefor, P(E) = [latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {7}{8} [/latex]

Compound Interest Formulas:
Annually compounded interest formula:
Amount = P (1 + [latex]\frac {R}{100})^{n}[/latex]
Half-yearly compounded interest formula:
Amount = P [1 + [latex]\frac {(\frac {R}{2})}{100}]^{2n}[/latex]
Quarterly compounded interest formula:
Amount = P [1 + [latex]\frac {(\frac {R}{4})}{100}]^{4n}[/latex]
Compounded interest formula with varying interest rates:
Amount = P (1 + [latex]\frac {{R}_{1}}{100}) (1 + \frac {{R}_{2}}{100}) (1 + \frac {{R}_{3}}{100})[/latex]
Here, Principal sum = P, Interest = R % per annum, Time = n years
Q1. You invest Yen 1000 in a 100-year bond but only receive 1% annually. How much would you have received interest in total after 100 years?

A. 1705 B. 999 C. 562 D. 100

Answer: A
Explanation: Use the annual formula above to come to the answer.
[latex]{1 . 01}^{100} \times 10000 - 10000 = 1704 . 81 = 1705 [/latex]
Q2. Igor the loanshark is harassing you because you haven't paid back that loan of yours. He is not very clever so he says "pay me 10 % on 1000 that you borrowed". You have had the loan for 1 year so you ask if it is calculated annually? He thinks you are stupid and says "of course it is" and kicks you out. If he would have said quarterly, then how much more in interest would you have had to pay?

A. 38 B. 3. 81 C. 103 D. 100

Answer: B
Explanation: Use the quarterly yearly formula above to come to the answer.
[latex] \frac {10}{4} = 2. 5 % [/latex]
[latex]{1. 025}^{4} \times 1000 [/latex] = 1103. 81 - 1000 = 103. 81 With 10 % annual you would have paid [latex] 1000 \times 0.1 [/latex]= 100
Q3. You stumble upon a great investment proposal where you are offered 25% annually for ever. How long would it take to increase 1000 to 10000 i.e. 10 times in years?

A. 15 years B. under 9 years C. just over 10 years D. 7 years

Answer: C
Explanation: [latex]{1.25}^{9} = 9. 313 and {1. 25}^{10} [/latex] = 11. 641 i.e. it would take just over 10 years to do it. [latex]{1.0125}^{40} \times 25000 - 25000 = 16090 [/latex]
Q4. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

A. 17 kg B. 20 kg C. 26 kg D. 31 kg

Answer: D
Explanation: Let A, B, C represent their respective weights.
Then, we have: A + B + C = [latex](45 \times 3)[/latex] = 135 .... (i)
A + B = [latex](40 \times 2) [/latex]= 80 .... (ii)
B + C = [latex](43 \times 2) [/latex]= 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
B's weight = 31 kg.
Q5. If you can choose how frequently the interest rate on a loan is compounded that you have to pay back to someone else. In order to pay the least amount of interest would you want it rather to:

A. Compound Monthly B. Annually C. Half - Yearly D. It Makes No Difference

Answer: B
Explanation: By compounding annually you would pay less interest than monthly compounding for example so B is the right answer.

Q1. A sum of money at simple interest amounts to 815 in 3 years and to 854 in 4 years. The sum is:

A. 650 B. 690 C. 698 D. 700

Answer: C
Explanation: S.I. for 1 year = (854 - 815) = 39.
S.I. for 3 years = [latex] (39 \times 3) [/latex]= 117.
Principal = (815 - 117) = 698.
Q2. Mr. Thomas invested an amount of 13,900 divided into two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be 3508, what was the amount invested in Scheme B?

A. 6400 B. 6500 C. 7200 D. 7500

Answer: A
Explanation: Let the sum invested in Scheme A be x and that in Scheme B be (13900 - x).
Then, [latex] \frac {(x \times 14 \times 2)}{100} + \frac {((13900 - x) \times 11 \times 2)}{100} [/latex] = 3508
28 x - 22 x = 350800 - [latex](13900 \times 22) [/latex]
6 x = 45000
x = 7500.
So, sum invested in Scheme B = (13900 - 7500) = 6400.
Q3. A sum fetched a total simple interest of 4016.25 at the rate of 9 %.p.a. in 5 years. What is the sum?

A. 4462.50 B. 8032.50 C. 8925 D. None of these

Answer: C
Explanation: Principal = [latex] \frac {(100 \times 4016.25)}{(9 \times 5)} [/latex]
= [latex] \frac {401625}{45} [/latex]
= 8925.
Q4. How much time will it take for an amount of 450 to yield 81 as interest at 4.5% per annum of simple interest?

A. 3.5 years B. 4 years C. 4.5 years D. 5 years

Answer: B
Explanation: Time = [latex]\frac {(100 \times 81)}{(450 \times 4.5)} years [/latex] = 4 years.
Q5. Reena took a loan of 1200 with simple interest for as many years as the rate of interest. If she paid 432 as interest at the end of the loan period, what was the rate of interest?

A. 3.6 B. 6 C. 18 D. Cannot be determined

Answer: B
Explanation: Let rate = R % and time = R years.
Then, [latex] \frac {(1200 \times R \times R)}{100} [/latex]= 432
[latex]12 {R}^{2} [/latex] = 432
[latex] {R}^{2} [/latex] = 36
R = 6.