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Question 1 of 5
1. Question
A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs and there by decreases his average by 0.4. Find out the number of wickets taken by him till the last match.
Correct
Answer : c
Let x be the number of wickets taken till the last match.
According to the question ,
( 12.4 x + 26 ) = 12( x + 5 )
12.4x + 26 = 12 x + 60
0.4 x = 34
∴X = ( 34 / 0.4 ) = 85
Incorrect
Answer : c
Let x be the number of wickets taken till the last match.
According to the question ,
( 12.4 x + 26 ) = 12( x + 5 )
12.4x + 26 = 12 x + 60
0.4 x = 34
∴X = ( 34 / 0.4 ) = 85

Question 2 of 5
2. Question
Five years ago, the average age of P and Q was 15 yr. Now, average age of P, Q and R is 20 yr. What would be the age of R after 10 yr?
Correct
Answer : c
According to the question,
[(P + Q)/2] – 5 = 15
(P + Q ) / 2 = 20
P + Q = 40 …………………( 1 )
And ( P + Q + R )/3 = 20
P + Q + R = 60 …………………. ( 2 )
On solving Eqns. ( 1 ) and ( 2 ), we get R = 20 yr
∴Age of R after 10 yr from now = 20 + 10 = 30 yr
Incorrect
Answer : c
According to the question,
[(P + Q)/2] – 5 = 15
(P + Q ) / 2 = 20
P + Q = 40 …………………( 1 )
And ( P + Q + R )/3 = 20
P + Q + R = 60 …………………. ( 2 )
On solving Eqns. ( 1 ) and ( 2 ), we get R = 20 yr
∴Age of R after 10 yr from now = 20 + 10 = 30 yr

Question 3 of 5
3. Question
The average salary of the entire staff in an office is Rs. 500 per day. The average salary of officers is Rs. 750 and that of nonofficers is Rs. 250. If the number of officers is 15, then find the number of nonofficers in the office.
Correct
Answer :c
Let number of non – officers = x
Then, 250x + 750 ´ 15 = 500 ( 15 + x )
250x + 11250 = 7500 + 250x
11250 – 7500 = 500x – 250x
3750 = 250x
X = (3750/250)
∴x = 15
Incorrect
Answer :c
Let number of non – officers = x
Then, 250x + 750 ´ 15 = 500 ( 15 + x )
250x + 11250 = 7500 + 250x
11250 – 7500 = 500x – 250x
3750 = 250x
X = (3750/250)
∴x = 15

Question 4 of 5
4. Question
The age of Mitali and Pooja is in the ratio 2:6. After 5 years, the ratio of their ages will become 6:8. Find the average of their ages after 10 years
Correct
Answer : a
Let the present age of Mitali and Pooja be 2x
and 6x years respectively.
After 5 years,( 2x + 5)/(6x + 5 ) = 6/8
16x + 40 = 36x + 30
20x = 10
∴ x = ½
Hence Present ages 1 and 3 years.
After 10 years, their ages will be 11 and 13 years.
∴Average of their ages = (11+13)/2 = 12 years
Incorrect
Answer : a
Let the present age of Mitali and Pooja be 2x
and 6x years respectively.
After 5 years,( 2x + 5)/(6x + 5 ) = 6/8
16x + 40 = 36x + 30
20x = 10
∴ x = ½
Hence Present ages 1 and 3 years.
After 10 years, their ages will be 11 and 13 years.
∴Average of their ages = (11+13)/2 = 12 years

Question 5 of 5
5. Question
Out of two sections A and B, 10 students of section B shift to A, as a result strength of A becomes 3 times the strength of B. But, if 10 students shift over from A to B, both A and B become equal in strength. Ratio of the number of students in section A that of section B is
Correct
Answer : a
A + 10 = 3 (B 10)
A – 3B = – 20 ………………( 1 )
And (A – 10) = (B + 10)
A – B = 20 ……………. ( 2 )
On subtracting Eq. (2) from Eq. (1), we get
A – 3B – A + B = – 20 – 20
– 2B = – 40
B = 20
Now, by using Eq. (ii),
A = 20 + 20 = 40.
: Ratio of the numbers of students of A and
B = 40 : 20 = 2:1
Incorrect
Answer : a
A + 10 = 3 (B 10)
A – 3B = – 20 ………………( 1 )
And (A – 10) = (B + 10)
A – B = 20 ……………. ( 2 )
On subtracting Eq. (2) from Eq. (1), we get
A – 3B – A + B = – 20 – 20
– 2B = – 40
B = 20
Now, by using Eq. (ii),
A = 20 + 20 = 40.
: Ratio of the numbers of students of A and
B = 40 : 20 = 2:1
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