# Basic Numeracy - Online Test

Q1. It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
Answer : Option C
Explaination / Solution:

On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday.

Q2. What was the day of the week on 28th May, 2006?
Answer : Option D
Explaination / Solution:

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2)  6 odd days

Jan.  Feb.   March    April    May
(31 +  28  +  31   +   30   +   28 ) = 148 days
148 days = (21 weeks + 1 day)  1 odd day.

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

Given day is Sunday.

Q3. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
Answer : Option A
Explaination / Solution:

 Required run rate = 282 - (3.2 x 10) = 250 = 6.25 40 40

Q4. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
Answer : Option A
Explaination / Solution:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009. Required sale = Rs. [ (6500 x 6) - 34009 ]

= Rs. (39000 - 34009)

= Rs. 4991.

Q5. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
Answer : Option B
Explaination / Solution:

1 hectare = 10,000 m2

So, Area = (1.5 x 10000) m2 = 15000 m2.

 Depth = 5 m = 1 m. 100 20 Volume = (Area x Depth) = 15000 x 1 m3 = 750 m3. 20

Q6. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
Answer : Option C
Explaination / Solution:

2(15 + 12) x h = 2(15 x 12) h = 180 m = 20 m. 27 3 Volume = 15 x 12 x 20 m3 = 1200 m3. 3

Q7. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Answer : Option A
Explaination / Solution:

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

Q8. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
Answer : Option C
Explaination / Solution:

Clearly, the numbers are (23 x 13) and (23 x 14). Larger number = (23 x 14) = 322.

Q9. (17)3.5 x (17)? = 178
Answer : Option D
Explaination / Solution:

Let (17)3.5 x (17)x = 178.

Then, (17)3.5 + x = 178. 3.5 + x = 8 x = (8 - 3.5) x = 4.5

Q10. Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
Answer : Option C
Explaination / Solution:

xz = y2 10(0.48z) = 10(2 x 0.70) = 101.40 0.48z = 1.40 z = 140 = 35 = 2.9 (approx.) 48 12