ANSWERS AND SOLUTION:
1(D) Explanation :
When a 5 cc cube is sliced into 1 cc cubes, we will get 5*5*5 = 125 1 cc cubes.
In each side of the larger cube, the smaller cubes on the edges will have more than one of their sides painted.
Therefore, the cubes which are not on the edge of the larger cube and that lie on the facing sides of the larger cube will have exactly one side painted.
In each face of the larger cube, there will be 5*5 = 25 cubes. Of these, there will be 16 cubes on the edge and 3*3 = 9 cubes which are not on the edge.
Therefore, there will be 9 1-cc cubes per face that will have exactly one of their sides painted.
In total, there will be 9*6 = 54 such cubes.
Speed of A : Speed of B = 7/3 : 1 = 7 : 3
It means, in a race of 7 m, A gains (7-3)=4 metre
If A needs to gain 80 metre, race should be of 7/4×80 = 140 metre
3 (A) Explanation :
The shape of the area used for growing cabbages has remained a square in both the years.
Let the side of the square area used for growing cabbages this year be X ft.
Therefore, the area of the ground this year = X^2 sq.ft.
And let the side of the square area used for growing cabbages last year be Y ft.
Therefore, the area of the ground used last year = Y^2 sq.ft.
As the number of cabbages grown has increased by 211, the area would have increased by 211 sq ft as each cabbage takes 1 sq ft space.
Hence X^2 – Y^2 = 211 => (X + Y)(X – Y) = 211.
211 is a prime number and hence it will have only two factors. i.e., 211 = 211*1.
This can be represented as (106 + 105)*(106-105).
(X + Y)(X – Y) = (106 + 105)(106 – 105).
From this we can deduce that X = 106 and Y = 105.
Therefore, number of cabbages produced this year = X^2 = 106^2 = 11236.
Clearly, A beats B by 4 seconds
Now find out how much B will run in these 4 seconds
Speed of B = Distance/Time taken by B=224/32=28/4=7 m/s
Distance covered by B in 4 seconds = Speed × time =7×4=28 metre
i.e., A beat B by 28 metre
5 (C) Explanation :
Let us assume that Jose will take x days to complete the task if he works alone and that Jane will take y days to complete the task if she worked alone.
From the information provided in the first statement of the question, we know that they will complete the task in 20 days, if they worked together on the task.
Therefore,1/x + 1/y = 1/20
From the second statement, we can conclude that x/2 + y/2 = 45
Or, x + y = 90 => x = 90 – y.
Substituting the value of x as 90 – y in the first equation
we get 1/(90-x) + 1/y = 1/20
Or y^2 – 90 + 1800 = 0.
Factorizing and solving for y, we get y = 60 or y = 30.
If y = 60, then x = 90 – y = 90 – 60 = 30 and
If y = 30, then x = 90 – y = 90 – 30 = 60.
As the question clearly states that Jane is more efficient than Jose, the second answer is the only possible alternative.
Hence, Jose will take 60 days to complete the task if he worked alone and Jane will take only 30 days to complete the same task.
6 (C)Explanation :
Ram takes 24 days to complete the work, if he works alone.
As Krish is twice as efficient as Ram is, Krish will take half the time to complete the work when Krish works alone, i.e., in 12 days.
Ram completes 1/24 th of the work in a day.
Krish completes 1/12 th of the work in a day.
When they work together, they will complete 1/24 + 1/12 = 1/8 th work in a day.
Therefore, when they work together they will complete the work in 8 days.
7( A)Explanation :
Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b
Work done by 6 men and 8 women in 1 day = 1/10
=> 6m + 8b = 1/10
=> 60m + 80b = 1 — (1)
Work done by 26 men and 48 women in 1 day = 1/2
=> 26m + 48b = ½
=> 52m + 96b = 1— (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day
= 15/100 + 20/200 =1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days
8 (A) Explanation :
We know that (A U B) = A + B – (A n B), where (A U B) represents the set of people who have enrolled for at least one of the two subjects Math or Economics and (A n B) represents the set of people who have enrolled for both the subjects Math and Economics.
(A U B) = A + B – (A n B) => (A U B) = 40 + 70 – 15 = 95%
That is 95% of the students have enrolled for at least one of the two subjects Math or Economics.
Therefore, the balance (100 – 95)% = 5% of the students have not enrolled for either of the two subjects.
Let six years ago the age of Kunal and Sagar are 6x and 5x resp.
So Sagar age is (5x+6) = 16
Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm