**1. What day of the week does May 28 2006 fall on**

**2. Bucket P has thrice the capacity as bucket Q. It takes 80 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?**

**3. 01-Jan-2007 was Monday. What day of the week lies on 01-Jan-2008?**

**4. A water tank is two-fifth full. Pipe A can fill a tank in 12 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?**

**5.The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:**

**6. A certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T).**

**7.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?**

**8. A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:**

**9. An accurate clock shows 8 o’clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o’clock in the afternoon?**

**10. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:**

**1( C)**Explanation :28th May 2006 = (2005 years + period from 1-Jan-2006 to 28-May-2006)

We know that number of odd days in 400 years = 0

Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Number of odd days in the period 2001-2005

= 4 normal years + 1 leap year

= 4 x 1 + 1 x 2 = 6

Days from 1-Jan-2006 to 28-May-2006 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 28(may)

= 148

148 days = 21 weeks + 1 day = 1 odd day

Total number of odd days = (0 + 6 + 1) = 7 odd days = 0 odd day

0 odd day = Sunday

Hence May 28 2006 is Sunday.

**2.( C)**Explanation :Let capacity of bucket P = x

Then capacity of bucket Q = x/3

Given that it takes 80 turns for bucket P to fill the empty drum

=> capacity of the drum = 80x

Number of turns required if both P and Q are used = (80x)/(x+x/3)

=240x/3x+x = 240/4=60

**3.( D**)Explanation :Given that January 1, 2007 was Monday.

Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out

the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)

Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday

**4.( C**)Explanation :Since pipe B is faster than pipe A, the tank will be emptied.

Part filled by pipe A in 1 minute = 1/12

Part emptied by pipe B in 1 minute = 1/6

Net part emptied by pipe A and pipe B in 1 minute = 1/6-1/12=1/12

Time taken to empty 2/5 of the tank = (2/5)/(1/12)=2×12/5=4.8 min

** 5. (B)**Explanation:Perimeter = Distance covered in 8 min

. = (12000/60x 8)m = 1600 m.

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) meter square = 153600 meter square

**6.(D**)Explanation :Let the principal = Rs.x

and time = y years

Principal,x amounts to Rs.400 at 10% per annum in y years

Simple Interest = (400-x)

Simple Interest = PRT/100

(400-x)=(x×10×y)100

(400-x)=xy/10— (equation 1)

Principal,x amounts to Rs.200 at 4% per annum in y years

Simple Interest = (200-x)

Simple Interest = PRT/100

(200-x)=x×4×y/100

(200-x)=xy/25 — (equation 2)

(equation 1)/(equation 2)?(400-x)/(200-x) =(xy/10)/(xy/25)

(400-x)/(200-x)=25/10

(400-x)/(200-x) =5/2

800-2x=1000-5x

200=3x

x=200/3

Substituting this value of x in Equation 1, we get,(400-200/3)=((200/3)y)/10

?(400-200/3)=20y/3

?1200-200=20y

?1000=20y

y=1000/20 =50 years

**7.(B**)Explanation:Area of the park = (60 x 40) m2 = 2400 m^2.

Area of the lawn = 2109 m^2.

Area of the crossroads = (2400 – 2109) m^2 = 291 m^2.

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

60x + 40x – x^2 = 291

= x^2 – 100x + 291 = 0

= (x – 97)(x – 3) = 0

= x = 3.

**8.(D**)Explanation:Area to be plastered= [2(l + b) x h] + (l x b)

= {[2(25 + 12) x 6] + (25 x 12)} m^2

= (444 + 300) m^2

= 744 m^2.

Cost of plastering = 744 x75/100 = Rs. 558.

**9.( B**)Explanation :We know that Angle traced by hour hand in 12 hrs = 360°

From 8 to 2, there are 6 hours

The angle traced by the hour hand in 6 hours = 6×360/12=180°

**10.( D**)Explanation:Let original length = a and original breadth = b

Decrease in area = ab – (80/100 a x 90/100 b)

= (ab-18/25ab)

=7/25ab.

So Decrease % = (7/25 abx1/abx100)%= 28%

10-D

9-B

1.C

2.C

3.D

4.C

5.B

6.D

7.B

8.C

9.B

10.D

C c d c b d b c b d

Can you explain 1 question

C,c,d,c,b,d,c,b,d

C C D C B

* B * B D

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