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DE Shaw Previous Placement Paper - Aptitude - Interview Preparation MCQ


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10 Questions MCQ Test Company Wise Preparation - DE Shaw Previous Placement Paper - Aptitude

DE Shaw Previous Placement Paper - Aptitude for Interview Preparation 2024 is part of Company Wise Preparation preparation. The DE Shaw Previous Placement Paper - Aptitude questions and answers have been prepared according to the Interview Preparation exam syllabus.The DE Shaw Previous Placement Paper - Aptitude MCQs are made for Interview Preparation 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for DE Shaw Previous Placement Paper - Aptitude below.
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DE Shaw Previous Placement Paper - Aptitude - Question 1

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 1

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.

DE Shaw Previous Placement Paper - Aptitude - Question 2

What is the difference between the compound interests on Rs. 5000 for 1 years at 4% per annum compounded yearly and half-yearly?

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 2

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.

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DE Shaw Previous Placement Paper - Aptitude - Question 3

A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is?

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 3

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.

DE Shaw Previous Placement Paper - Aptitude - Question 4

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 4

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.

DE Shaw Previous Placement Paper - Aptitude - Question 5

A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 5

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.

DE Shaw Previous Placement Paper - Aptitude - Question 6

The ratio between the perimeter and the breadth of a rectangle is 5: 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 6

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.

DE Shaw Previous Placement Paper - Aptitude - Question 7

A shopkeeper allows a discount of 10% on the marked price and still gains 17% on the whole. Find at what percent above the cost price did he mark his goods.

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 7

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.

DE Shaw Previous Placement Paper - Aptitude - Question 8

A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

Detailed Solution for DE Shaw Previous Placement Paper - Aptitude - Question 8

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.

*Answer can only contain numeric values
DE Shaw Previous Placement Paper - Aptitude - Question 9

One rectangular plate with length 8 inches, breadth 11 inches and 2 inches thickness is there. what is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate?


*Answer can only contain numeric values
DE Shaw Previous Placement Paper - Aptitude - Question 10

In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?


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