CSC - Aptitude
1.
One
year ago, the ratio of Roonie’s and Ronaldo’s age was 6: 7 respectively. Four
years hence, this ratio would become 7: 8. How old is Ronaldo ?
Ans-36
years
Let the ages
of Roonie and Ronaldo one year ago be X and Y.
One year ago
the ratio of their ages was X : Y
:: 6 : 7
This will give you the value of Y as 7X/6 ----------- (a)
After four
years from now the ratio of the ages will become (X+5) :
(Y+5) :: 7 : 8
è 8X + 40 = 7Y
+ 35 -> 7Y = 8X + 5 Substituting the
value of Y from (a)
We have 7 x
7X/6 = 8X + 5 -> 49X = 48X + 30 -> X = 30 and Y = 35.
Ronaldo’s
present age is Y+1 -> 35 + 1 = 36 years.
2.
In
what ratio must rice at Rs. 9.30 per kg be mixed with rice at Rs. 10.80 per kg
so that the mixture be worth Rs. 10 per kg ?
Ans-8:7
9.30 10.80
10.00
0.80 0.70
Using the Rule of Alligation we find the
ratio as 8 : 7.
3.
One
side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the
area of the field.
Ans-120m2
Using
Pythagoras theorem we find the length of the other side as 8 m and hence the
area is
15 x 8 = 120
sq.m
4.
A
rectangular grassy plot 110 m. by 65 m has a gravel path 2.5 m wide all round
it on the inside. Find the cost of gravelling the path at 80 paise per sq. meter
Ans- Rs 680
The overall
area of the grassy plot is 110 x 65
-> 7150 sq. meters.
The gravel
path is 2.5 m wide all around. This would reduce the dimensions of the
remaining grassy plot by 5 m on all sides.
Hence the
area of the grassy plot after applying the gravel path is 105 x 60 -> 6300 sq. meters.
Hence, the
area of the gravel path is 7150 – 6300 =
850 sq.mtrs. The cost of gravelling the
path at the rate of 0.80 paise per sq.mtr
will be 850 x 0.80 -> Rs
680.00
5.
There
are two sections A and B of a class consisting of 36 and 44 students respectively.
If the average weight of section A is 40kg and that of section B is 35kg, find
the average weight of the whole class?
Ans-37.25 kg
The total
weight of Section A -> 40 x 36 -> 1440 kg
The total
weight of Section B -> 35 x 44 -> 1540 kg.
Total weight
of both sections (80 students)
-> 2980 kg
Hence the
average weight of whole class ->
2980 / 80 = 37.25 kgs
6.
The
milk and water in two vessels A and B are in the ratio 4 : 3 and 2: 3
respectively. In what ratio, the liquids in both the vessels be mixed to obtain
a new mixture in vessel C containing half milk and half water?
Ans-7 : 5
This is a
tricky question. In vessel A milk is 4 parts out of total 7 parts.
In vessel B
milk is 2 parts out of total 5 parts. Hence, the temptation to answer the
question will be 1 : 1
The question
is in what ratio the liquids in both the vessels are to be mixed. If the
liquids are mixed in the ratio of the total parts in the two vessels then the
vessel C into which these are poured will have half milk and half water.
Hence the
answer is -> 7 : 5.
7.
Of the
three numbers, second is twice the first and is also thrice the third. If the
average of the three numbers is 44.Find the largest number.
Ans-72
Let the
first number be ‘x’. Then the second number is ‘2x’ and the third number be ‘y’.
Since the
average of the three numbers is 44 the total of the three numbers is 132.
Hence we
have the equation x + 2x + y = 132. We
are informed the second number is thrice the third number. Hence, we have 2x = 3y or y = 2x/3.
Substituting this value of Y
We have a
new equation x + 2x + 2x/3 = 132.Solving
we get the value of ‘x’ as 36.
Applying
this value, we get the three numbers as
36, 72 and 24.
Hence the
largest number is 72.
8.
Two
cards are drawn at random from a pack of 52 cards. What is the probability that
either both are black or both are queen?
Ans-55/221
Probability
(Black or Queen) = P(Black ) + P(Queen) –
P(Black, Queen)
è 26C2 + 4C2 –
2C2 -> 660/2652 -> 55/221
9.
A bag
contains 6 white and 4 black balls. 2 balls are drawn at random. Find the
probability that they are of same color.
Ans-7/15
(6C2 + 4C2)/10C2
10.
A,
Band C are three contestants in a km race. If A can give B a start of 40
m and A can give C a start of 64m how many meters’ start
can B give C ?
Ans: 25 meters.
When A covers 1000 meters, B covers 960
meters and C covers 936 meters.
Thus when B runs 960 meters C can run only
936 meters.
In other words in a race of 960 meters B can
give C a start of 24 meters.
Thus in a race of 1000 meters ( One Km) B can
give C a start of
(24 * 1000)/ 960 -> 25 meters.
No comments:
Post a Comment