16
Q:

The Students in three classes are in the ratio 2 : 3 : 5. If 20 students are increased in each class, the ratio of changes to 4 : 5 : 7. The total number of students before the increase was

Q:

Two solutions have milk & water in the ratio 7:5 and 6:11. Find the proportion in which these two solutions should Be mixed so that the resulting solution has 1 part milk and 2 parts water ?

 A) 7 : 12 B) 8 : 13 C) 9 : 4 D) 2 : 5

Explanation:

To get the solution that contains 1 part of milk and two parts of water,
they must be mixed in the ratio as
7x+6x/5y+11y = 1/2
26x = 16y
x/y = 16/26
x/y = 8/13

1 46
Q:

A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions made by the larger wheel is ?

 A) 11 B) 7 C) 9 D) 13

Explanation:

Here the number of cogs is inveresly proportional to number of revolutions

=> More cogs less revolution
=> 6 : 14 :: x : 21
=> (21x6)/14 = 9

1 61
Q:

At a game of billiards, A can give B 20 points in 60 and A can give C  15 points in 60. How many points can C give B in a game of 90 ?

 A) 9 points B) 10 points C) 11 points D) 8 points

Explanation:

a:b = 60:40
a:c = 60:45
c/a x a/b = 45/60 x 60/40 = 45/40 = 90/80
So C gives B => 10 points.

2 93
Q:

Total number of students in 3 classes of a school is 333 . The number of students in class 1 and 2 are in 3:5 ratio and 2 and 3 class are 7:11 ratio . What is the strength of class that has highest number of students ?

 A) 125 B) 155 C) 135 D) 165

Explanation:

Ratio C1:C2 = 3:5 and C2:C3 = 7:11
So C1:C2:C3 = 21 : 35 : 55
Let the strength of three classes are 21x, 35x and 55x respectively, then
Given that 21x + 35x + 55x = 333
=> 111x = 333 or x=3
So strength of the class with highest number of students =55x = 55x3 = 165.

2 95
Q:

A man spend Rs. 810 in buying trouser at Rs. 70 each and shirt at Rs. 30 each. What will be the ratio of trouser and shirt when the maximum number of trouser is purchased ?

 A) 1:3 B) 2:1 C) 3:2 D) 2:3

Explanation:

Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7

We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7

Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9

Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.