DE Shaw Previous Placement Paper - Aptitude - Interview Preparation MCQ

Let S be the sample space.
Then, n(S) = 52C2 = (52 x 51)/(2x1) = 1326.
Let E = event of getting 2 kings out of 4.
⇒ n(E) = 4C2 = (4 x 3)/ (2 x 1) = 6.
⇒ P(E) =n(E)/n(S) =6/1326 = 1/221.

C.I. when interest compounded yearly
= Rs. [ 5000 x (1 + 4/100) x {1 + (1/2 x 4)/ 100}]
= Rs. 5000 x 26/25 x 51/50
= Rs. 5304.
and, C.I. when interest is compounded half-yearly
= Rs.5000 x (1 + 2 /100) ^3.
= Rs. 5000 x 51/50 x 51/50 x 51/50.
= Rs. 5306.04
The required Difference = Rs. (5306.04 - 5304) = Rs. 2.04.

A can do a work in 15 days and B in 20 days.
A’s 1 day’s work = 1/15.
B′s 1 day′s work = 1/20
(A+B)’s 1 day’s work = (1/15 + 1/20) = 7/60.
(A+B)’s 4 day’s work= (7/60 × 4) = 7/15.
Remaining work= (1-7/15) = 8/15.

Let the marks secured by them be x and (x + 9) respectively.
Then, sum of their marks = x + (x + 9) = 2x + 9
Given, (x + 9) was 56% of the sum of their marks.
⇒ (x+9) = 56/100(2x+9).
⇒ (x+9) = 14/25(2x+9).
⇒ 25x + 225 = 28x + 126.
⇒ 3x = 99.
⇒ x = 33.
Then, (x + 9) = 33 + 9 = 42
Hence, their marks are 33 and 42.

By the rule of allegation, we have:
Profit on 1st part Profit on 2nd part
8% 18%
\ /
Mean Profit
14%
/ \
4 6
Ratio of 1st and 2nd parts = 4 : 6 = 2 : 3.
Quantity of 2nd kind = (3/5 ) x 1000 kg = 600 kg.

According to the question,
⇒ 2(l + b)/b = 5/1
⇒ 2l + 2b = 5b
⇒ 3b = 2l
⇒ b = (2/3) l
Given, Area = 216 cm2
So, l x b = 216
⇒ l x (2/3) l = 216.
⇒ l^2 = 324
⇒ l = 18 cm.

Let the cost price be 100. Then SP = 117.
Let the marked price be x.
So, 90% of x = 117
=> x = 130.
Therefore, he marked his goods 30% above the cost price.

Length of the platform = speed of train * extra time taken to cross the platform.
Length of platform = 72 kmph * 12 seconds.
.'. 72 kmph = 5/18 x 72 = 20 m/sec.
Therefore, length of the platform = 20 m/s * 12 sec = 240 meters.