DXC Placements Papers and Previous Years Papers 2021

Solution: (SP(A))/(CP(A)) = 120/100____________ (1)

(SP(B))/(CP(B)) = 160/100____________ (2)

(SP(B))/(sp(A)) = 250/100____________

Given (1)/ (2) =(CP(B))/(CP(A)) x100/250 = 120/100 x 100/160 =(CP(B))/(CP(A)) = 15/8 = (CP(B)-CP(A))/(CP(A)) x100 =(15-8)/8 x 100 =87.5%

We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080.
Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.
So h³(ab) = 1080 = (2³)(3³)(5).
We need to write 1080 as a product of a perfect cube and another number.

Four cases:
1. h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding co-prime a,b such that a*b = 1080).

2. h = 2, We need to find number of ways of writing (3³) * (5) as a product of two co-prime numbers. This can be done in two ways - 1 and (3³) * (5) , (3³) and (5)
number of pairs = 2, number of ordered pairs = 4

3. h = 3, number of pairs = 2, number of ordered pairs = 4

4. h = 6, number of pairs = 1, number of ordered pairs = 2

Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.
The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30).

q can be 1.
If q =2, number would be of the form 28n + 2 which is a multiple of 2.

Similarly, when q = 4, number would be of the form 28n + 4 which is again a multiple of 2. Any number of the form 28n + an even number will be a multiple of 2.

When q = 7, number would be of the form 28n + 7 which is a multiple of 7.

So, the only remainders possible are remainders that share no factors with 28. Or numbers that are co-prime to 28.

There is a formula for this and a shorter way of finding the number of numbers co-prime to a given natural number. A more detailed discussion on this is provided here and here.

1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25 and 27. q can take 12 different values

Solution:

The difference in the simple interest and compound interest for two years is on account of the interest paid on the first year\'s interest, when interest is reckoned using compound interest, interest being compounded annually.
Hence 12% of simple interest = 90 => simple interest = =750.

As the simple interest for a year = 750 @ 12% p.a., the principal = = Rs.6250.

If the principal is 6250, then the amount outstanding at the end of 3 years = 6250 + 3(simple interest on 6250) + 3 (interest on simple interest) + 1 (interest on interest on interest) = 6250 + 3(750) + 3(90) + 1(10.80) = 8780.80.

Total distance travelled by both of them for 22nd meeting = 1400 + (21 * 2 * 1400) = 43 * 1400
Distance travelled by each will be in proportion of their speed :
Therefore, distance travelled by A = 50/(50 + 20) * 43 * 1400 = 43000 (Note - Always do complicated calculations at last because things cancel out generally)
Now, for every odd multiple of 1400, A will be at Q and for every even multiple of 1400 A will be at P. So, at 42000 Km (1400 x 30, even multiple) A will beat P. So at their 22 meeting, A will be 1000 Km from P, therefore, 400 Km from Q.

A can win in the following scenarios:
A strikes in first shot.
A misses in the first shot, B misses in the second, A strikes in the third.
A misses in the 1st shot, B misses in the 2nd, A misses the 3rd, B misses the 4th and A strikes the 5th.
And so on…
P(A in 1^st shot) = 0.6
P(A in 3^rd shot) = 0.4 * 0.2 * 0.6 {A misses, then B misses and then A strikes}
P(A in 5^st shot) = 0.4 * 0.2 * 0.4 * 0.2 * 0.6 {A misses, then B misses and then A misses, then B misses, then A strikes}
Overall probability = Sum of all these

= 0.6 + 0.4 * 0.2 * 0.6 + 0.4 * 0.2 * 0.4 * 0.2 * 0.6 + …
Which is nothing but
0.6 + (0.4 * 0.2) * 0.6 + (0.4 * 0.2)² * 0.6 + (0.4 * 0.2)³ * 0.6…

This is an infinite geometric progression with first term 0.6 and common ratio 0.4 * 0.2
Required Probability = a1−r = 0.61−0.08 = 0.60.92 = 3046 = 1523

log₂x + log₄x = log_0.25√6
We can rewrite the equation as:
log₂x + log2xlog24 = log_0.25√6
log₂x * 32 = log_0.25√6
=> log₂x * 3 = 2log_0.25√6
=> log₂x³ = -log₄6
=> log₂x³ = −log26log24
=> log₂x³ = −log266

=> 2log₂x³ = -log₂6
2log₂x³ + log₂6 = 0
log₂6X⁶ = 0
6x⁶ = 1
x⁶ = 16
x = 16−−√6

SP of trader A = 1000 (1 + x).
Profit of trader A = 1000 (1 + x) – 1000.
MP of trader B = 2000 (1 + 2x).
SP of trader B = 2000 (1 + 2x) (1 – x).

Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000
1000x = 2000 – 4000x² + 4000x – 2000x – 2000
4000xx² -1000x = 0
1000x (4x – 1) = 0
=> x = 25%

t₈|=|t₁₆|. This can happen under two scenarios t₈ = t₁₆ or t₈ = – t₁₆.
If t₈ = t₁₆, the common difference would be 0 suggesting that t₃ would be equal to t₇.
However, we know t₃ is not equal to t₇, so the common difference cannot be zero.
This tells us that t₈ = – t₁₆ Or, t₈ + t₁₆ = 0.
If t₈ + t₁₆ = 0, then t₁₂ = 0. t₁₂ = t₈ + 4d, and t₁₆ – 4d So, t₁₂ = t8+t162
For any two terms in an AP, the mean is the term right in between them.
So, t₁₆ is the arithmetic mean of t₈ and t₁₆.
So, t₁₂ = 0.
Now, S₂₃ = 23 * t₁₂. We know that average of n terms in an A.P. is the middle term.
This implies that sum of n terms in an A.P., is n times the middle term. So, S₂₃ = 0.

In-radius of equilateral triangle of side a = a2√3
Diameter of larger circle = a2√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 = √32

a
RD = a√3
AR = √32a - a√3 = 3a−2a2√3 = a2√3
AR = 13 AD
Radius of smaller circle = 13 radius of larger circle
Radius of smaller circle = 13 * a2√3 = a6√3
Area of smaller circle = πr²
π (a6√3)² = πa2108
Area of triangle = √34a²
Ratio = πa2108 : √34a² = π:27√3