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Q:

In the match between India and Pakistan at Jaipur on 02-10-1983, which new rule was introduced

A) Limit of overs wes reduced to 50 overs	B) The rule of field restriction was taken
C) over throw runs were batsmans score	D) Noballs & wides were debited to bowlers analysis

Answer & Explanation Answer: D) Noballs & wides were debited to bowlers analysis

Explanation:

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Filed Under: Sports

Q:

What is the rate of simple interest?
I. The total interest earned was Rs. 4000.
II. The sum was invested for 4 years.

A) I alone sufficient while II alone not sufficient to answer	B) II alone sufficient while I alone not sufficient to answer
C) Either I or II alone sufficient to answer	D) Both I and II are not sufficient to answer

Answer & Explanation Answer: D) Both I and II are not sufficient to answer

Explanation:

We know that, R = (100 x S.I) / (P x T)

Now I gives, S.I = Rs. 4000.

II gives, T = 4 years.

But, P is unknown. So, we cannot find R.

So, given data is insufficient to get R.

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

Q:

A sum at simple interest at $13 \frac{1}{2}$ % per annum amounts to Rs. 2502.50 after 4 years. Find the sum

A) Rs.1625	B) Rs1525
C) Rs.1425	D) Rs.1325

Answer & Explanation Answer: A) Rs.1625

Explanation:

Let sum be Rs.x then, S.I. =Rs. [ x * (27/2) * 4 * (1/100) ] = Rs. 27x/50

Amount =Rs [ x + (27x/50)] = Rs.77x/50

=> 77X/50 = 2502.50

=>X= (2502.50 * 50)/77 = 1625

Hence Sum = Rs.1625

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

Q:

Find: S.l on Rs 68000 at $16 \frac{2}{3}$ 1 % per annum for 9 months

A) 6500	B) 7500
C) 8500	D) 9500

Answer & Explanation Answer: C) 8500

Explanation:

P = 68000, R = $\frac{50}{3}$ % & T = 9 months (3/4 years)

$S . I = \frac{p \times r \times t}{100} = [\frac{(6800 \times \frac{50}{3} \times \frac{3}{4})}{100}] = R s . 8500$

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

Q:

A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at $6 \frac{1}{4}$ p.a for 2 years. Find his gain in the transaction per year

A) Rs. 112.50	B) Rs.150.25
C) Rs.167.50	D) Rs.170

Answer & Explanation Answer: A) Rs. 112.50

Explanation:

$g a i n i n 2 y e a r s = R s . [(5000 * \frac{25}{4} * \frac{2}{100}) - (\frac{5000 * 4 * 2}{100})]$

= Rs. (625 - 400)

= Rs. 225

$g a i n i n 1 y e a r = R s . (\frac{225}{2}) = 112.50$

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

Q:

A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:

A) 5%	B) 8%
C) 12%	D) 15%

Answer & Explanation Answer: C) 12%

Explanation:

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.

S.I. for 5 years = Rs. $\frac{2205}{3} \times 5$ = Rs.3675

Principle = Rs.(9800-3675) = Rs.6125

Hence, Rate = $[\frac{100 \times 3675}{6125 \times 5}]$ =12%

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

Q:

The diagonal of the floor of a rectangular closet is $7 \frac{1}{2}$ feet. The shorter side of the closet is $4 \frac{1}{2}$ feet. What is the area of the closet in square feet?

A) 5 .5	B) 13.5
C) 27	D) 37

Answer & Explanation Answer: C) 27

Explanation:

Other side= $\sqrt{[{(\frac{15}{2})}^{2} - (\frac{9^{2}}{2})]} f t$ = $\sqrt{\frac{225}{4} - \frac{81}{4}} f t$ = $\sqrt{\frac{144}{4}}$ = 6ft

Area of closet =(6 x 4.5)sq.ft = 27 sq.ft

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Filed Under: Volume and Surface Area

Q:

The principal that amounts to Rs.4913 in 3 years at $4 \frac{1}{3}$ % per annum compound interest compounded annually is

A) Rs.3096	B) Rs.4076
C) Rs.4085	D) Rs.4096

Answer & Explanation Answer: D) Rs.4096

Explanation:

principal = $A = 4913, T = 3, R = 4 \frac{1}{3}$

$P = {\frac{A}{(\frac{1 + R}{100})}}^{T}$

= Rs.4096

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Filed Under: Compound Interest
Exam Prep: GRE

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