MATHS : QUIZ for SSC

1. What day of the week does May 28 2006 fall on
A. Saturday
B. Monday
C. Sunday
D. Thursday

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?
A. 30
B. 45
C. 60
D. 80

3. 01-Jan-2007 was Monday. What day of the week lies on 01-Jan-2008?
A. Wednesday
B. Sunday
C. Friday
D. Tuesday
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?
A. 2.8 min
B. 4.2 min
C. 4.8 min
D. 5.6 min

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:
A. 15360
B. 153600
C. 30720
D. 307200
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).
A. 45 years
B. 60 years
C. 40 years
D. 50 Years
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?
A. 2.91 m
B. 3 m
C. 5.82 m
D. None of these
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:
A.Rs. 456
B. Rs. 458
C. Rs. 558
D. Rs. 568
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?
A. 154°
B. 180°
C. 170°
D. 160°
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:
A. 10%
B. 10.08%
C. 20%
D. 28%

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%

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