**Solution :-**

Q1. Option A

Runs scored in the first 10 overs = 10 × 3.2 = 32

Total runs = 282

Remaining runs to be scored = 282 – 32 = 250

Remaining overs = 40

Run Rate needed = 250/40=6.25

Q2. Option B

Let the sale in the month = x

Then (6435+6927+6855+7230+6562+x)/6=6500

6435+6927+6855+7230+6562+x=6 ×6500

34009+x=39000

x=39000-34009=4991

Q3. Option D

Average of 20 numbers = 0

(Sum of 20 numbers)/20=0

Sum of 20 numbers = 0

Hence at most there can be 19 positive numbers

Q4. Option A

Number of members in the team = 11

Let the average age of the team = x

(Sum of ages of all 11 members)/11=x

Sum of the ages of all 11 members=11x

Ages of the captain = 26

age of the wicket keeper = 26+3=29

Sum of the ages of 9 members od the team excluding captain and wicket keeper

= 11x-26-29=11x-55

Average age of 9 members of the team excluding captain and wicket keeper

=(11x-55)/9

Given that

(11x-55)/9=(x-1)

11x-55=9(x-1)

11x-55=9x-9

2x=46

x=46/2=23years

Q5. Option C

Let monthly income of A = a

monthly income of B = b

monthly income of C = c

a+b=2×5050 -x

b+c=2×6250 -y

a+c=2×5200 -z

x+y-z

a+b+a+c-(b+c)=(2×5050)+(2×5200)-(2×6250)

2a=2(5050+5200-6250)

a=4000

Q6. Option B

Total Cost = 4000×3

Total Diesel used = 4000/7.5+4000/8+4000/8.5

Averag cost per litre of diesel

=(4000×3)/(4000/7.5+4000/8+4000/8.5)

=3/(1/7.5+1/8+1/8.5)

It is important how you proceed from this stage. Remember time is very important and if we solve this in the normal method, it may take lot of time. Instead, we can find out the approximate value easily and select the right answer from the given choices.

In this case, answer

=3/(1/7.5+1/8+1/8.5)

≈3/(1/8+1/8+1/8)≈3/(3/8)≈8

We got that answer as approximately equal to 8. From the given choices, the answer can be 8 or 7.98 or 8.1 . But which one from these?

It is easy to figure out. We approximated the denominator, (1/7.5+1/8+1/8.5) to 3/8

However

1/7.5+1/8.5

=1/(8-0.5)+1/(8+0.5)

=(8+0.5+8-0.5)/(8-0.5)(8+0.5)

=16/(8^2-〖0.5〗^2 )

=64/(64-0.25)

i.e 1/7.5+1/8.5=16/(64-0.25)

We know that 1/8+1/8=1/4=16/64

i.e 1/7.5+1/8.5>1/8+1/8

Early we had approximated the denominator to 3/8 . However, from the above mentioned equations, now we know that actual denominator is slightly greater than 3/8. It means answer is slightly lesser than 8. Hence we can pick the choice 7.98 as the answer

Q7. Option D

Let Kiran’s weight = x. Then

According to Kiran, 65 < x < 72 —-(equation 1)

According to brother, 60 < x < 70 —-(equation 2)

According to mother, x ≤ 68 —-(equation 3)

Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,we can write as

65<x≤6865<x≤68

i.e., x = 66 or 67 or 68

Average of different probable weights of Kiran = (66+67+68)/3=67

Q8. Option A

Average weight of 16 boys = 50.25

Total weight of 16 boys = 50.25 × 16

Average weight of remaining 8 boys = 45.15

Total weight of remaining 8 boys = 45.15 × 8

Total weight of all boys in the class = (50.25 × 16)+ (45.15 × 8)

Total boys = 16 + 8 = 24

Average weight of all the boys = ((50.25×16)+(45.15×8))/24

=((50.25×2)+(45.15×1))/3

=16.75×2+15.05

=33.5+15.05

=48.55

Q9 Option C

In a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday.

Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday

Average visitors on Sundays = 510

Total visitors of 5 Sundays = 510 × 5

Average visitors on other days = 240

Total visitors of other 25 days = 240 × 25

Total visitors = (510 × 5) + (240 × 25)

Total days = 30

Average number of visitors per day

=((510×5)+(240×25))/30

=((51×5)+(24×25))/3

=(17×5)+(8×25)

=85+200

=285

Q10. Option B

Let the total number of students = x

The average marks increased by 1/2 due to an increase of 83 – 63 = 20 marks.

But total increase in the marks = 1/2×x = x/2

Hence we can write as

x/2=20⇒x=20×2=40