CTS - Reasoning - 4

             CTS – Reasoning - 4

Read the following information and then answer the questions that follow:

In a certain society, there are two marriage groups, red and brown. No marriage is permitted within a group. On marriage, males become part of their wives groups; women remain in their own group. Children belong to the same group as their parents. Widowers and divorced males revert to the group of their birth. Marriage to more than one person at the same time and marriage to a direct descendant are forbidden

1.       A brown female could have had
I. A grandfather born Red
II. A grandmother born Red
III. Two grandfathers born Brown

(A) I only      (B) III only           (C) I, II and III     (D) I and II only
Ans: (D)

2.       A male born into the brown group may have

(A)   An uncle in either group

(B)   A brown daughter

(C)   A brown son

(D)   A son-in-law born into red group

Ans: (A)

3.       Which of the following is not permitted under the rules as stated.
(A) A brown male marrying his father's sister
(B) A red female marrying her mother's brother
(C) A widower marrying his wife's sister
(D) A widow marrying her divorced daughter's ex-husband
Ans: (B)

4.       If widowers and divorced males retained their group they had upon marrying which of the following would be permissible ( Assume that no previous marriage occurred)

(A)   A woman marrying her dead sister's husband

(B)   A woman marrying her divorced daughter's ex-husband

(C)   A widower marrying his brother's daughter

(D)   A woman marrying her mother's brother who is a widower.

Ans: (D)

There are six steps that lead from the first to the second floor. No two people can be on the same step
Mr. A is two steps below Mr. C
Mr. B is a step next to Mr. D
Only one step is vacant ( No one standing on that step )
Denote the first step by step 1 and second step by step 2 etc.

5.        If Mr. A is on the first step, which of the following is true?
(a) Mr. B is on the second step.
(b) Mr. C is on the fourth step.
(c) A person Mr. E, could be on the third step
(d) Mr. D is on a higher step than Mr. C.
Ans: (d)

6.       If Mr. E was on the third step & Mr. B was on a higher step than Mr. E which step must be vacant
(a) step 1
(b) step 2
(c) step 4
(d) step 5
(e) step 6
Ans: (a)

7.       If Mr. B was on step 1, which step could A be on?
(a) 2&5 only
(b) 3&5 only
(c) 3&4 only
(d) 4&5 only
(e) 2&4 only
Ans: (c)

8.       If there were two steps between the step that A was standing and the step that B was standing on, and A was on a higher step than D , A must be on step
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
Ans: (c)

9.       Which of the following is false
i. B&D can be both on odd-numbered steps in one configuration
ii. In a particular configuration A and C must either both an odd numbered steps or both an even-numbered steps
iii. A person E can be on a step next to the vacant step.

(a) i only
(b) ii only
(c) iii only
(d) both i and iii
Ans: (c)

Six swimmers A, B, C, D, E, F compete in a race. The outcome is as follows.
i. B does not win.
ii. Only two swimmers separate E & D
iii. A is behind D & E
iv. B is ahead of E , with one swimmer intervening
v. F is a head of D

10.   Who stood fifth in the race ?
(a) A
(b) B
(c) C
(d) D
(e) E
Ans: (e)

11.   How many swimmers separate A and F ?
(a) 1
(b) 2
(c) 3
(d) 4
(e) cannot be determined
Ans: (d)

12.   The swimmer between C & E is
(a) none
(b) F
(c) D
(d) B
(e) A
Ans: (a)

13.   If the end of the race, swimmer D is disqualified by the Judges then swimmer B finishes in which place
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Ans: (b)

14.   A farmer built a fence around his 17 cows, in a square shaped region. He used 27 fence poles on each side of the square. How many poles did he need altogether?
Ans.104 poles
As he used 27 poles on each side, excluding the two corner poles, he had used 25 poles on each side.  As the region is square shaped, he had used 25 poles on each side totalling in all 25 * 4 = 100 poles. In addition he used four poles at the four corners.

Thus the total number o poles used is 100 + 4 = 104 poles.

15.   On the first test of the semester, Kiran scored 60%. On the last test of the semester, Kiran scored 75%. By what percent did Kiran's score improve?
Ans: 25%
The percentage increase between the first and the last tests is 15%.

On a base of 60% the increase is 15%. Hence the percentage increase is

15 * 100/60 -> 25%.

16.   A group consists of equal number of men and women. Of them 10% of men and 45% of women are unemployed. If a person is randomly selected from the group. Find the probability of the selected person to be an employee.
Ans:29/40

Let us assume the number of Men and Women at 100 each.

Men:  Number of unemployed 10% = 10

Women: Number of unemployed 45% = 45.

Men:       Total employed 100 – 10 = 90

Women: Total employed 100 – 45 = 55
Total men & women employed = 90 + 55 = 145

The probability of the person picked being an employed = 145/200

è 29/40

17.   Randy's chain of used car dealership sold 16,400 cars in 1998. If the chain sold 15,744 cars in 1999, by what percent did the number of cars sold decrease?
Ans: 4%

The decrease in number of cars sold between 1998 and 1999 is 16400 – 15744 = 656.

On a base of 16400 the decrease is 656.

Hence the percentage of decrease is 656 * 100/16400 -> 4%

18.   A radio when sold at a certain price gives a gain of 20%. What will be the gain percent, if sold for thrice the price?
A) 260%
B) 150%
C) 100%
D) 50%
E) None of these
Ans: 260%
Let the cost price be 100. Since the gain was 20% it was sold for 120.

If it is sold at thrice the price, then the sale price would be 120 * 3 = 360

So the percentage gain will be 360 – 100 = 260.

Since the base taken is 100 the gain percentage is 260%.

19.   If the Arithmetic mean is 34 and geometric mean is 16 then what is greatest number in that series of two numbers?
Ans: 64

Let the numbers be x, y
Then arithmetic mean = (x+y)/2 =34  →  x+y =68
Also geometric mean =√(xy)=16
or xy=16²=256
Hence, substituting the value of ‘y’ as (68-x) we have
x(68−x)=256
or x²−68x+256=0
(x−64)(x−4)=0
Hence x=64 or x=4
and y=4 or 64
Larger number is 64

20.   The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?
Ans: 250

Distance covered in one hour is = 66 * 1000 meters.

Distance to be covered in one minute = (66*1000)/60 -> 1100 m

The diameter of the wheel is 140 cm and the radius is 70 cm.

The distance covered in one revolution is ( 2 * 22/7 * 70) = 440 cm or 4.4 m.

So to cover 1100 m the wheel should make  1100 / 4.4 = 250 revolutions.