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Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: C) 3

Explanation:

Area of the park = (60 x 40) = 2400 $m^{2}$

Area of the lawn = 2109 $m^{2}$

Area of the crossroads = (2400 - 2109) = 291 $m^{2}$

Let the width of the road be x metres. Then,

$60 x + 40 x - X^{2} = 291$

$x^{2} - 100 x + 291 = 0$

(x - 97)(x - 3) = 0
x = 3.

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

A sum was put at simple interest at a certain rate for 10 years . Had it been put at 5% higher rate , it would have fetched Rs.600 more. What was the Sum?

A) Rs.1200	B) Rs.1300
C) Rs.1400	D) Rs.1500

Answer & Explanation Answer: A) Rs.1200

Explanation:

At 5% more rate, the increase in S.I for 10 years = Rs.600 (given)

So, at 5% more rate, the increase in SI for 1 year = 600/10 = Rs.60/-

i.e. Rs.60 is 5% of the invested sum

So, 1% of the invested sum = 60/5

Therefore, the invested sum = 60 × 100/5 = Rs.1200

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Filed Under: Simple Interest
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

The population of a town was 3600 three years back. It is 4800 right now. What will be the population three years down the line, if the rate of growth of population has been constant over the years and has been compounding annually?

A) Rs.600	B) Rs,6400
C) Rs.6500	D) Rs.6600

Answer & Explanation Answer: B) Rs,6400

Explanation:

The population grew from 3600 to 4800 in 3 years. That is a growth of 1200 on 3600 during three year span.

Therefore, the rate of growth for three years has been constant.

The rate of growth during the next three years will also be the same.

Therefore, the population will grow from 4800 by $4800 * \frac{1}{3}$ = 1600

Hence, the population three years from now will be 4800 + 1600 = 6400

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Filed Under: Compound Interest
Exam Prep: Bank Exams
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Q:

A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs 360 more. Find the sum.

A) Rs.4000	B) Rs.9000
C) Rs.5000	D) Rs.6000

Answer & Explanation Answer: D) Rs.6000

Explanation:

Let sum = P and original rate = R. Then

[(P * (R+2) * 3)/100] - [ (P * R * 3)/100] = 360

3P*(R+2) - 3PR = 36000

3PR + 6P - 3PR = 36000

6P = 36000

P = 6000

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Filed Under: Simple Interest
Exam Prep: GRE
Job Role: Bank PO

Q:

A father left a will of Rs.35 lakhs between his two daughters aged 8.5 and 16 such that they may get equal amounts when each of them reach the age of 21 years. The original amount of Rs.35 lakhs has been instructed to be invested at 10% p.a. simple interest. How much did the elder daughter get at the time of the will?

A) 17.5 lakhs	B) 21 lakhs
C) 15 lakhs	D) 20 lakhs

Answer & Explanation Answer: B) 21 lakhs

Explanation:

Let Rs.x be the amount that the elder daughter got at the time of the will. Therefore, the younger daughter got (3,500,000 - x).

The elder daughter’s money earns interest for (21 - 16) = 5 years @ 10% p.a simple interest.

The younger daughter’s money earns interest for (21 - 8.5) = 12.5 years @ 10% p.a simple interest.

As the sum of money that each of the daughters get when they are 21 is the same,

$x + \frac{5 * 10 * x}{100} = (3, 500, 000 - x) + \frac{12.5 * 10 * (3, 500, 000 - x)}{100}$

$x + \frac{50}{x} = 3, 500, 000 - x + \frac{125}{100} * 3, 500, 000 - \frac{125 x}{100}$

$2 x + \frac{50 x}{100} + \frac{125 x}{100} = 3, 500, 000 * (1 + \frac{5}{4})$

$\frac{200 x + 50 x + 125 x}{100} = \frac{9}{4} * (3, 500, 000)$

=> $x = 2, 100, 000 = 21 l a k h s$

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Filed Under: Simple Interest
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Q:

A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?

A) Rs.35	B) Rs.245
C) Rs.350	D) cannot be determined

Answer & Explanation Answer: D) cannot be determined

Explanation:

We need to know the S.I, principal and time to find the rate. Since the principal is not given, so data is inadequate.

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Filed Under: Simple Interest
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Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A) 2.91 m	B) 3 m
C) 5.82 m	D) None of these

Answer & Explanation Answer: B) 3 m

Explanation:

Area of the park = (60 x 40) = 2400 sq.m

Area of the lawn = 2109 sq.m

Area of the crossroads = (2400 - 2109) = 291 sq.m

Let the width of the road be x metres. Then,

60x + 40x - (x * x) = 291

=>(x * x) - 100x + 291 = 0

=>(x - 97)(x - 3) = 0

=>x=3

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Filed Under: Volume and Surface Area

Q:

A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The number of points, the team scored was :

A) 665	B) 657
C) 658	D) 678

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