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Q:

What will be the day of the week 15th August, 2010?

A) Sunday	B) Saturday
C) Wednesday	D) Monday

Answer & Explanation Answer: A) Sunday

Explanation:

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 = 4 odd days.

Jan. Feb. Mar. Apr. May. Jun. Jul. Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 = 0 odd days.

Given day is Sunday

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Filed Under: Calendar

Q:

The calendar of the year 2024 can be used again in the year?

A) 2030	B) 2052
C) 2048	D) 2036

Answer & Explanation Answer: B) 2052

Explanation:

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2: Add +6 to the Given Year.

Solution :

Given Year is 2024, Which is a leap year.

So, Add +28 to the given year (i.e 2024 + 28) = 2052

Therfore, The calendar of the year 2024 can be used again in the year 2052.

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Filed Under: Calendar

Q:

The calendar of the year 2006 was used before in the year?

A) 1990	B) 1995
C) 2002	D) 2000

Answer & Explanation Answer: B) 1995

Explanation:

Given year 2006, when divided by 4 leaves a remainder of 2.

NOTE: When remainder is 2, 11 is subtracted to the given year to get the result.

So, 2006 - 11 = 1995

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Filed Under: Calendar

Q:

The calendar of the year 1897 can be used again in the year?

A) 1908	B) 1901
C) 1903	D) 1926

Answer & Explanation Answer: C) 1903

Explanation:

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2: Add +6 to the Given Year.

Solution :

Given Year is 1897, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year.

Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903

Therfore, The calendar of the year 1897 can be used again in the year 1903

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Filed Under: Calendar

Q:

If log 2 = 0.3010 and log 3 = 0.4771, the values of log₅ 512 is

A) 2.875	B) 3.875
C) 4.875	D) 5.875

Answer & Explanation Answer: B) 3.875

Explanation:

ANS: log₅512 = log512/log5 = $\frac{\log 2^{9}}{\log (10 / 2)}$ = $\frac{9 \log 2}{\log 10 - \log 2}$ = $\frac{9 * 0.3010}{1 - 0.3010}$ =2.709/0.699 =2709/699 =3.876

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

Q:

At what time, in minutes, between 3 o'clock and 4 o'clock, both the needles will coincide each other?

A) 11 4/11	B) 13 4/11
C) 15 4/11	D) 16 4/11

Answer & Explanation Answer: D) 16 4/11

Explanation:

At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand.
To be coincident, it must gain 15 min. spaces.55 min. are gained in 60 min.
15 min. are gained in
= (60/55 x 15)min
=16+4/11
The hands are coincident at 16 + 4/11 min past 3

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

Q:

The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq.cm. Find its height?

A) 3 cm	B) 4 cm
C) 6 cm	D) 8 cm

Answer & Explanation Answer: C) 6 cm

Explanation:

Let height =x Then, base =2x

2x * x = 72

=>x=6

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

Q:

A room 5m 55cm long and 3m 74cm broad is to be paved with square tiles.Find the least number of square tiles required to cover the floor?

A) 156	B) 166
C) 176	D) 186

Answer & Explanation Answer: C) 176

Explanation:

Area of the room=(544 * 374)

size of largest square tile= H.C.F of 544 & 374 = 34 cm

Area of 1 tile = (34 x 34) $c m^{2}$

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

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

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