**Q1. A path of uniform width surrounds a**

circular park. The difference of internal and external circumferences of this

circular path is 132 metres. Its width is: (Take π=22/7)

circular park. The difference of internal and external circumferences of this

circular path is 132 metres. Its width is: (Take π=22/7)

(a) 22m

(b) 20 m

(c) 21m

(d) 24m

**Sol. Let the internal radius of the park be r and the external**

radius (with the path) be R.

radius (with the path) be R.

**The difference between the internal and external circumferences is**

132m.

132m.

**so 2πR-2πr = 132**

**=>2π(R-r) = 132**

**=R – r = 132/2π**

**= (132 * 7)/ (2 * 22) = 21**

**Hence, the width of path**

**= 21 meters**

**Q2. A is thrice as good as workman as B**

and therefore, able to finish a job in 60 days less than B. Working together

they will do it in:

and therefore, able to finish a job in 60 days less than B. Working together

they will do it in:

(a) 120/7 days

(b) 90/4days

(c) 25 days

(d) 30 days

**Sol. Let A can finish the whole work in x days**

so B can finish the same work in 3x days.

According to question.

3x – x = 60

so x = 30 days

so A can finish the work in 30 days and B can finish the same work in 90 days.

so (A + B)’s one day’s work = 1/30 + 1/90 = 4/90

so B can finish the same work in 3x days.

According to question.

3x – x = 60

so x = 30 days

so A can finish the work in 30 days and B can finish the same work in 90 days.

so (A + B)’s one day’s work = 1/30 + 1/90 = 4/90

**so (A + B)’s can finish the complete work in = 90/4 days.**

**Q3. A man, a woman and a boy can**

complete a work in 20 days. 30 days and 60 days respectively. How many boys

must assist 2 men and 8 women so as to complete the work in 2 days?

complete a work in 20 days. 30 days and 60 days respectively. How many boys

must assist 2 men and 8 women so as to complete the work in 2 days?

(a) 8

(b) 12

(c) 4

(d) 6

**Sol. Part of work done by 2 men and 8 women in 2 days.**

= 2(2/20 + 8/30)

= 2(1/10 + 8/20) = 2((3+8)/30)

= 22/30 = 11/15

Remaining work = 1 – 11/15 = 4/15

work done by 1 boy in 2 days

= 2/60 = 1/30

= 2(2/20 + 8/30)

= 2(1/10 + 8/20) = 2((3+8)/30)

= 22/30 = 11/15

Remaining work = 1 – 11/15 = 4/15

work done by 1 boy in 2 days

= 2/60 = 1/30

**Number of boys done 4/15 work = 4*30/15**

**= 8 boys**

**Q4. 8% of the voters in an election did**

not cast their votes. In this election, there were only two candidates. The

winner by obtaining 48% of the total votes defeated his contestant by 1100

votes. The total number of voters in the election was:

not cast their votes. In this election, there were only two candidates. The

winner by obtaining 48% of the total votes defeated his contestant by 1100

votes. The total number of voters in the election was:

(a) 21000

(b) 23500

(c) 22000

(d) 27500

**Sol. Let the total number of votes be 100.**

**Number of cast votes = 100-8 = 92**

**so Number of votes obtained by the winner = 48**

**Number of votes obtained by the loser = 92 – 48 = 44**

**If the difference of win be 4 votes, total voters = 100**

**so When the difference be 1100 votes, total voters**

**= 100/4 * 1100 = 27500**

**Q5. A, B and C start at the same time in**

the same direction to run around a circular stadium. A completes a round in 252

seconds, B in 308 seconds and C in 198 seconds, all starting at the same point.

After what time will they next meet at the starting point again?

the same direction to run around a circular stadium. A completes a round in 252

seconds, B in 308 seconds and C in 198 seconds, all starting at the same point.

After what time will they next meet at the starting point again?

(a) 46 minutes 12 seconds

(b) 45 minutes 12 seconds

(c) 42 minutes 36 seconds

(d) 26 minutes 18 seconds

**Sol. Required time = LCM of 252.**

**308 and 198 seconds.**

**Now, 252 = 2 * 2* 3 * 3 * 7**

**308 = 2 * 2 * 7 * 11**

**198 = 2 * 3 * 3 * 11**

**so LCM = 2 * 2* 3 * 3 * 7 * 11**

**= 36 * 77 seconds**

**= (36 * 77)/60 minutes**

**= 231/5 = 46 minutes 12 seconds**

**Q6. A**

merchant marks his goods up by 60% and then offers a discount on the marked

price. If the final selling price after the discount results in the merchant

making no profit or loss, what was the percentage discount offered by the

merchant?

merchant marks his goods up by 60% and then offers a discount on the marked

price. If the final selling price after the discount results in the merchant

making no profit or loss, what was the percentage discount offered by the

merchant?

(a) 60

(b) 40

(c) 37.5

(d) Depends on the cost price

Sol. Let C.P = 100

Sol. Let C.P = 100

**Markup price = 100 ×1.6**

**= 160**

**S.P. = C.P.**

**Discount % = (60/160) × 100**

**= 37.5%**

**Q7. A trader**

makes a profit equal to the selling price of 75 articles when he sold 100 of

the articles. What percentage of profit did he make in the transaction?

makes a profit equal to the selling price of 75 articles when he sold 100 of

the articles. What percentage of profit did he make in the transaction?

(a) 33.33%

(b) 75%

(c) 300%

(d) 150%

**Sol. Let S be the Selling price of 1 article,Therefore, the selling price of 100 articles=100 SProfit earned by selling these 100 articles=selling price of 75 articles = 75 SWe know that, Selling Price (SP)=Cost Price (CP)+ProfitSo, we get100S = CP+75S. Hence, CP=100S-75S=25SProfit percentage = Profit/(Cost Price)×100%**

**=75S/25S×100%=300%**

**Q8. A sum amounts to Rs. 98010 in 2 years and to**

Rs. 107811 in 3 years compounded annually. What is the sum?

Rs. 107811 in 3 years compounded annually. What is the sum?

(a) Rs. 81000

(b) Rs. 84000

(c) Rs. 87000

(d) Rs. 90000

**Sol. P**

(1 + R/100) ^3 = 107811

P (1 + r/100) ^2 = 98010

(1 + R/100) ^3 = 107811

P (1 + r/100) ^2 = 98010

1 + r/100 = 107811/98010

r/100 = 9801/98010

r = 10%

**P × (1 + 10/100) ^2 = 98010
P = 98010 * 100/121 = 81000**

**Q9. The simple interest on a certain sum**

for 8 months at 4% per annum is Rs. 129 less than the simple interest on the

same sum for 15 months at 5% per annum. The sum is:

for 8 months at 4% per annum is Rs. 129 less than the simple interest on the

same sum for 15 months at 5% per annum. The sum is:

(a) Rs. 2,580

(b) Rs. 2400

(c) Rs. 2,529

(d) Rs. 3600

**Sol. Let the sum be Rs. x**

**(x * 5 * 15)/(100 * 12) – (x * 4 * 8)/(100 * 12) = 129**

**= x/(100 * 12) (75 – 32) = 129**

**X = (129 * 1200)/ 43**

**Rs. 3600**

**Q10.Compound interest (compounded annually) on a certain sum of money for**

2 years at 4% per annum is Rs. 102. The simple interest on the same sum for the

same rate and for the same period will be:

2 years at 4% per annum is Rs. 102. The simple interest on the same sum for the

same rate and for the same period will be:

(a) Rs. 99

(b) Rs. 101

(c) Rs. 100

(d) Rs. 98

Sol. Let principal be Rs. P

Sol. Let principal be Rs. P

According to question.

102 = p {(1+r/100) ^t- 1}

Where r = 4% and t

= 2 years

P = 102/ ((26/25) ^2- 1) =Rs.1250

so Simple Interest at rate of 4% for 2 years

= (1250 ×4 ×2)/100=Rs.100

According to question.

102 = p {(1+r/100) ^t- 1}

Where r = 4% and t

= 2 years

P = 102/ ((26/25) ^2- 1) =Rs.1250

so Simple Interest at rate of 4% for 2 years

= (1250 ×4 ×2)/100=Rs.100

**Required answer is Rs. 100**