eLitmus Quantitative Aptitude Dashboard

Given that Arun\'s present age in years is 40% of Barun\'s. And after few years, Arun’s age will be half of Barun’s. We have to find by what percentage will Barun’s age increase during this period.
So Arun’s present age is 2x and Barun’s present age will be equal to 5x.
2(2x + y) = 5x + y
4x + 2y = 5x + y
4x + 2y = 5x + y
y = x
So 2x and 5x becomes 3x and 6x respectively.
By what percentage will Barun\'s age increase during this period is that x increases of 5x so it increases by 20%.
Hence 20% of Barun’s age will increase during this period.

et us first rewrite the numbers a touch – 40% of students in Class A are boys, and 37.5% of boys in class B are boys. The overall percentage of boys should lie between these two numbers.

Now, class A has at least twice as many students as class B. So, the overall weighted average should definitely be closer to the percentage of boys in class A, or closer to 40%.

Now, the number of students in class A can be much higher than the number in class B, in which case the overall percentage would practically be 40%, This is the maximum percentage that can be there.

For the minimum percentage, we need to consider the other extreme-where class A has exactly twice as many students as class B.
The weighted average would be 2∗40+1∗37.5/3 = 39.17%

Distance = (Diagonalofcube–Diameterofsphere)/2
= (5√3−5)/2 (Since diagonal of the cube is given as √3 x l)
= 5(√3−1)2

=2.5(√3 – 1)

In-radius of equilateral triangle of side a = a/2√3
Diameter of larger circle = a/2√3

Let us say common tangent PQ touches the two circle at R, center of smaller circle is I.
Now, PQ is parallel to BC. AR is perpendicular to PQ. Triangle PQR is also an equilateral triangle and AORID is a straight line. (Try to establish each of these observations. Just to maintain the rigour.)

AD = √3/2a

RD = a√3
AR = √3/2a - a√3 = 3a−2a/2√3 = a/2√3
AR = 13 AD
Radius of smaller circle = 13 radius of larger circle
Radius of smaller circle = 13 * a/2√3 = a/6√3
Area of smaller circle = πr²
π (a/6√3)² = πa²/108
Area of triangle = √3/4a²
Ratio = πa2108 : √3/4a² = π:27/√3

Number of rearrangements of word DEMAND = 6!/2! = 360
Number of rearrangements of word DEMAND where 2 D’s appear together = 5! = 120
Number of rearrangements of word DEMAND where 2D’s do not appear together =
360 – 120 = 240

Words starting with ‘A’; without two D’s adjacent to each other
Words starting with A: 5!/2! = 60
Words starting with A where 2 D’s are together = 4! = 24
Words starting with ‘A’, without two D’s adjacent to each other = 36

Next we have words starting with D.
Within this, we have words starting with DA: 4! words = 24 words
Then words starting with DE
Within this, words starting with DEA => 3! = 6 words
Then starting with DED – 3! = 6 words
Then starting with DEM
=> First word is DEMADN
=> Second is DEMAND

Rank of DEMAND = 36 + 24 + 6 + 6 + 2 = 74

Let us say A and B split their share of the task and started doing their respective shares simultaneously.

Let’s say A takes A days to finish the task. Therefore, B takes A + 12 days to finish the entire task.

A has to finish 40% of the task, since B is doing the rest. So A will only take 2A/5 number of days.

B only has to finish 60% of the task, so B will take (3(A+12)/5) number of days.

But as we know, B starts working along with A and finishes 12 days after A stops working.

So, (3(A+12)/5) = ((2A)/5+12)
3A + 36 = 2A + 60
A = 24; B = 36 days.

To begin with, let us ignore car A. Car B and car C travel in opposite directions.
Their relative speed = Sum of the two speeds = 45 + 54 kmph = 99kmph.
= 99 * 5/18 m/s = 55/2

m/s = 27.5m/s

The relative distance = 220m. So, time they will take to cross each other = 5/18

= 8 seconds

Now, car A has to overtake car B within 8 seconds. The relative distance = 50m
=> Relative speed should be at least 50/8

m/s = 6.25 m/s.
=>6.25 * 18/5kmph = 22.5 kmph

Car B travels at 45kmph, so car A should travel at at least 45 + 22.5 = 67.5kmph.

Let us x = h * a; y = h * b
a and b are co-prime. So, LCM of (x, y) = h * a * b

So, in essence h + h * a * b = 91. Or h(ab + 1) = 91
Now, 91 can be written as 1 * 91 or 7 * 13
Or, we can have HCF as 1, LCM as 90 -
There are 4 pairs of numbers like this (2, 45), (9, 10), (1, 90) and (5, 18)

We can have HCF as 7, ab + 1 = 13 => ab = 12 => 1 * 12 or 4 * 3

Or, the pairs of numbers are (7, 84) or (21, 28)

The third option is when HCF = 13, ab + 1 = 7 => ab = 6
Or (a, b) can be either (1, 6) or (2, 3)
The pairs possible are (13, 78) and (26, 39)
There are totally 8 options possible - (2, 45), (9, 10), (1, 90), (5, 18), (7, 84), (21, 28), (13, 78) and (26, 39).
8 Pairs.

We know, 935 = 5*11*17

Let p = hx
q = hy
Where h is the hcf of p and q.
Therefore, lcm of p and q= hxy

We have lcm = 935.
hxy = 935

If h = 1,
Then p = 935
q = 1
Sum of digits of q = 1

If h = 5,
p = 5*17 = 85
q = 5*11 = 55
Sum of digits of q = 10

If h = 11,
p = 11*17 = 187
q = 11*5 = 55
Sum of digits of q = 10

If h = 17,
p = 17*11 = 187
q = 17*5 = 85
Sum of digits of q = 13

Maximum possible sum of digits of q = 13.

a + b + c = 10. a, b, c are whole numbers. Now this is similar to the previous question that we solved by placing 10 sticks and simplifying.

We cannot follow an exactly similar approach, as in this case a, b and c can be zero. Let us modify the approach a little bit. Let us see if we can remove the constraint that a, b, c can be zero.

If we give a minimum of 1 to a, b, c then the original approach can be used. And then we can finally remove 1 from each of a, b, c. So, let us distribute 13 sticks across a, b and c and finally remove one from each.
a + b + c = 13. Now, let us place ten sticks in a row

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This question now becomes the equivalent of placing two \'+\' symbols somewhere between these sticks. For instance,
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This would be the equivalent of 4 + 5 + 4. or, a = 4, b = 5, c = 4.
There are 12 slots between the sticks, out of which one has to select 2 for placing the \'+\'s.
The number of ways of doing this is ¹²C₂.

Hence the answer is \"66\"