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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
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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
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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 non-officers is Rs. 250. If the number of officers is 15, then find the number of non-officers 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
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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
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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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