Q 1-Find the average of first 10 multiples of 7?

A - 35.5

B - 37.5

C - 38.5

D - 40.5

Answer - B

Explanation

"{7(1+2+3+...+10)}/10
={7(10(10+1))}10×2
={7(10(10+1))}/10×2
={7(110)}/10×2=38.5"

Q 2-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?

A - 6.25

B - 6.5

C - 6.75

D - 7

Answer - A

Explanation

"Required run rate =		[282 - (3.2 x 10)]/40		=	250/40	   = 6.25"

Q 3-A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?

A - 200/7 years

B - 222/7 years

C - 229/7 years

D - None of these

Answer - B

Explanation

"Required Average= [(67 X 2 + 35 X 2 + 6 X 3)/(2+2+3)] 
hence,=[(134+70+18)/7) =222/7 years
"

Q 4-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 ?

A - 4991

B - 5467

C - 5987

D - 6453

Answer - 

Explanation

" 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."

Q 5-A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning

A - 39

B - 40

C - 41

D - None of these

Answer - A

Explanation

"Let the average after 7th inning = x

Then average after 16th inning = x - 3

therefore 16(x-3)+87 = 17x

therefore  x = 87 - 48 = 39"

Q 6-After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member ?

A - 12

B - 13

C - 14

D - 15

Answer - D

Explanation

"i) Let the ages of the five members at present be a, b, c, d & e years. 
And the age of the new member be f years. 
ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1) 
iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years 
So their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2) 
==> a + b + c + d + e = 5x + 15 
==> a + b + c + d = 5x + 15 - e ------ (3) 
iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above, 
The new average is: (5x + 15 - e + f)/5 
Equating this to the average age of x years, 3 yrs, ago as in (2) above, 
(5x + 15 - e + f)/5 = x 
==> (5x + 15 - e + f) = 5x 
Solving e - f = 15 years. 
Thus the difference of ages between replaced and new member = 15 years."

Q 7-The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers :

A - 4

B - 8

C - 12

D - 16

Answer - B

Explanation

"Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.
Then  [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.
or 5x + 20 = 305 or x = 57.
So, required difference = (57 + 8) - 57  = 8"

Q 8-The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team ?

A - 20 years

B - 21 years

C - 23 years

D - 25 years

Answer - C

Explanation

"Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years."

Q 9-The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4 ?

A - 76

B - 79

C - 85

D - 87

Answer - A

Explanation

"Average = total runs / no.of innings = 32 
So, total = Average x no.of innings = 32 x 10 = 320.
 
Now increase in avg = 4runs. So, new avg = 32+4 = 36runs 
Total runs = new avg x new no. of innings = 36 x 11 = 396 
Runs made in the 11th inning = 396 - 320 = 76"

Q 10-A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is :

A - 45

B - 40

C - 39

D - 37

Answer - B

Explanation

"Let there be x pupils in the class.
Total increase in marks = (x * 1/2) = x/2.
x/2 = (83 - 63) => x/2 = 20 => x = 40."

Q 11-The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is :

A - 45 kg

B - 46 kg

C - 47 kg

D - 48 kg

Answer - A

Explanation


Weight of the teacher = (35.4 x 25 - 35 x 24) kg = 45 kg.

Q 12-The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person ?

A - 70 kg

B - 75 kg

C - 80 kg

D - 85 kg

Answer - D

Explanation

"Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg."

Q 13-The average of 7 consecutive numbers is 20. The largest of these numbers is :

A - 21

B - 22

C - 23

D - 24

Answer - C

Explanation

"Let the numbers be x, x + 1, x + 2,  x + 3, x + 4, x + 5 and x + 6,
Then   (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.
or  7x + 21 = 140 or 7x = 119 or x =17.
Latest number = x + 6 = 23."

Q 14-Out of 9 persons, 8 persons spent Rs. 30 each for their meals. The ninth one spent Rs. 20 more than the average expenditure of all the nine. The total money spent by all of them was :

A - Rs. 292.50

B - Rs. 297.50

C - Rs. 298

D - Rs. 298.50

Answer - A

Explanation

"Let the average expenditure be Rs. x Then,

Total money spent = 9x =  Rs. (9 x 32.5O) = Rs 292. 50"

Q 15-The mean temperature of Monday to Wednesday was 37C and of Tuesday to Thursday was 34C . If the temperature on Thursday was (4/5) th that of Monday, the temperature on Thursday was?

A - 36

B - 38

C - 39

D - None of these

Answer - A

Explanation

"M+T+W =(37 X 3)=111C -----------(1) 
T+W+Th=(34 X 3)C=102C ->T+W+(4/5)M = 102}C --------------(2) 
(1)- (2) 
gives 1/5 th of temperature on Monday =111C-102C=9C 
->Temperature on Monday = 45C 
-> Temperature on Thursday = (4/5) * 45=36C
"

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