Question: 36 -
If V = 12R / (r + R) , then R =
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Vr / (12 - V)
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V
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V / (r - 12 )
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Vr + (V /12 )
Answer:
Vr / (12 - V)
Solution:
We have to rearrange the equation to make R the subject.
Start by cross multiplying by (r + R); V (r + R) = 12R
Multiply out the bracket Vr + VR = 12R
We have to rearrange the equation to make R the subject.
Start by cross multiplying by (r + R); V (r + R) = 12R
Multiply out the bracket Vr + VR = 12R
Question: 37 -
A piece of ribbon 4 yards long is used to make bows requiring 15 inches of ribbon for each. What is the maximum number of bows that can be made?
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10
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8
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9
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11
Answer:
9
Solution:
The maximum number of bows will be 4 yards
=> ( 4 x 36 inches) divided by 15 inches.
This gives 9.6. But as a fraction of a bow is no use,
we can only make 9 bows.
The maximum number of bows will be 4 yards
=> ( 4 x 36 inches) divided by 15 inches.
This gives 9.6. But as a fraction of a bow is no use,
we can only make 9 bows.
Question: 38 -
n is a whole number which when divided by 4 gives 3 as remainder. What will be the remainder when 2*n is divided by 4 ?
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1
-
2
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0
-
3
Answer:
2
Solution:
Let n=4*q + 3. Then, 2*n = 8*q + 6 = 4(2*q + 1) + 2.
Thus when 2*n is divided by 4, the reminder is 2.
Let n=4*q + 3. Then, 2*n = 8*q + 6 = 4(2*q + 1) + 2.
Thus when 2*n is divided by 4, the reminder is 2.
Question: 39 -
Which of the following can be used to illustrate that not all prime numbers are odd?
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1
-
3
-
2
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4
Answer:
2
Solution:
The only even numbers in the list are 2 and 4,
but 4 is not a prime. So 2 can be used to illustrate
the statement that all primes are not odd.
The only even numbers in the list are 2 and 4,
but 4 is not a prime. So 2 can be used to illustrate
the statement that all primes are not odd.
Question: 40 -
A perfect cube is an integer whose cube root is an integer. For example, 27, 64 and 125 are perfect cubes. If p and q are perfect cubes, which of the following will not necessarily be a perfect cube?
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pq+27
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(-p)
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8p
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pq
Answer:
pq+27
Solution:
pq is a cube and so is 27, but their sum
need not be a cube. Consider the case
where p =1 and q = 8, the sum of pq and 27
will be 35 which has factors 5 x 7 and is not a cube.
pq is a cube and so is 27, but their sum
need not be a cube. Consider the case
where p =1 and q = 8, the sum of pq and 27
will be 35 which has factors 5 x 7 and is not a cube.