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Sept Math JJ 50-100 (with answer and explain)

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Here are the answers & explanations for 51-100 Math JJ. Please verify & confirm them.


51) How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?
A) 15
B) 16
C) 17
D) 18
E) 19


Solution

since the problem tests the y=xq + r concept, we have:

y=3q +1, where 3 is the divisor and 1 is the remainder. now for this to become true, 3 cannot be a divisor of y. y will be 1 greater than it would be if 3 would be a divisor of y. so i started with 4, next one 7 etc. to save calculation time, i set up an equality:

0 <= 3q + 1 <= 50

solve for q and you get 16.3. but that's not in the answer choice. one might tend to pick 16, however, the y=xq+r concept says that when you divide a smaller number by a greater number, the quotient q is zero and what's left is the remainder. so the only other number we didn't account for yet is the number 1. so we have:

1=3q+1, which becomes 1=3(0)+1. so what's left is the remainder 1. thus the answer is 17, not 16. ; C


52) A straight line in the xy-plane has a slope of 2 and a y-intercept of 2. On this line, what is the x-coordinate of the point whose y-coordinate is 500 ?



A. 249

B. 498

C. 676

D. 823

E. 1,002

Solution



Equation can be written in the form y=mx+c, where M=slope and c is y intercept.
substituting the given values 500 = 2x+2 => 448 = 2x => x = 249; A


53. A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the bushes at random, one at a time, and plant them in a row, what is the probability that the 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

Solution

If you want to apply combination formula total number of ways in which you can arrange 4 things with 2 of each type= 4!/2!*2! = 6

General formula is: n!/(a!*b!*c!..) where a, b and c.. are the number of objects which are of the same kind.

the number of combination when 2 Rosebushes are together in the middle = 1 , there for P=1/6 ; B

Its very easy to visualize if u draw four boxes.


54.If (10^50) -74 is written as an integer in base 10 notation , what is the sum of the digits in that integer?

Solution

10^n-74 = 9999...9926 (the number of 9 is n-2)

there fore n=50 : 9 * 48 = 432

432 +2+6 = 440



55.If A and C are integers, is A>0?
(1) A+C>50
(2) C>48

Solution

Stmt 1 : A can hold a negative/positive value,
e.g A = -1; C= 52; => A+C = 51
e.g A = 4 C= 47 => A+C = 51

Stmt 2 : Using C value alone without any relation between A/C, A value cannot be determined

Stmt 1& 2 => A can hold a negative/positive value,
e.g A = -1; C= 52; => A+C = 51
e.g A = 4 C= 49 => A+C = 53

=> E


56.

A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?



A. 20

B. 36

C. 48

D. 60

E. 84

Solution

Assume y = ax + b (y=s scale, x = r scale)

30 = 6a + b

60 = 24a + b

Solving these 2 eqns =>b=20 and a = 10/6

100 = (10/6)* x + 20

x=48; C




57. A certain characteristic in a large population has a distribution that is symmetric about the mean M. If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distribution is less than M+D?

A. 16%
B. 32%
C. 48%
D. 84%
E. 92%

Solution

Basically - they tell you that between M-D and M+D there is 68% of the population, therefore, 2*D=68%, or D=34%.

that been said, less then M - 50 %

----> Less then M+D = 50% + 34% = 84% - answer is D.

|

| | |

----------------|------------------------

D M D & ;nbs p;

34% 34%

----------> = 50%-34%=16%

----------------> = 50%

------------------------> = 84%

58. Is the integer N odd?
(1) N is divisible by 3
(2)2N is divisible by twice as many positive integer as N.

Solution

Stmt 1 :Mutiples of 3 : 3, 6, 9…..; Both even/ odd numbers are divided by 3; => A is not enough
Stmt 2 : Only odd numbers multipled by 2 will have twice the factors as that of the odd number; e.g 3 = 3,1; 6= 6,3,2,1; 9=9,3,1;18=18,9,6,3,2,1

=>B


59. In a circle, QS and RT all are diameters, also mutually perpendicular, the QR arc length is 8√2, What is the area of the circle

A. 256∏
B. 256/∏
C. 512/∏
D. 512/∏
E. 512∏



Solution


The length of an arc is given by (x/360) * 2 * pi * r where x is the measure of the angle. We know x is 90.

90/360 * 2 * pi * r = 8√2 => Pi/2*r = 8√2

r = 16√2/Pi
therefore area = 16√2/Pi * 16√2/Pi * Pi =>512/Pi; D


60. Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?

A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3

Solution

Possible ways to get sum 8 are (3,5),(5,3),(4,4),(2,6),(6,2) i.e 5

Out of these ways in which 1 card is 5 are 2 --> (3,5),(5,3)

probability = 2/5 ; D



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61. m, n are positive integers, r is remainder, what is r of (5*10^n + m) /3?
1) m =1;
2) n = 10

Solution

Stmt 1: 10 to any power always starts with 1 and will end with 0’s. When that number is multipled by 5, 1 will be replaced by 5. the sum of the digits in that number will be 5. When m = 1 is added the sum of the digits will be 6; there for the reminder when devided by 3 will be 0;
e.g : assuming n = 4 ; 10 = 10000; 5*10000= 50000; 50000+1 =50001; 50001 is devisible by 3;

Stmt 2 : as mentioned above value of n doesn’t matter;

There fore A


62. What is remainder if the number N is divided by 12?
1. Remainder is 1 if the number is divided by 6
2. Remainder is >6 if the number is divided by 12

Solution

Stmt 1: N = 6k+1; N can be 7,13,19; When N is devided by 12, the reminders are 7,1,7… stmt alone is not enough
Stmt 2 : N = 12K+R, where R > 6; R cannot be determined

Stmt 1 & 2 : 6k+1; N can be 7,13,19; When N is devided by 12, the reminders are 7,1,7, when Stmt 2 is applied reminder cannot be 1; thr fore reminder is 7;

C


63. Is the tallest man of A group is taller than tallest man of B group?
(1): The shortest man of A group is shorter than tallest man of B group
(2) A group people heights RANGE is bigger than B group

Solution

Statement 1 of course is not enough since it tells us nothing about the height of the tallest man.

Statement 2 although seeming right at first sight, doesn't actually tells us if the groups overlap, contain in each etc.

Even when taking both statements we still cannot determine. We don't know if group be is contained in the range of group A. Let's take for example the following situation: the tallest men of groups A and B are of the same height, but the shortest of group A is shorter than the shortest of group B. It abides with both statements. Therefore, E.

It's E.


64. If n is an integer greater than 6, which of the following must be divisible by 3?

A) n(n+1) (n-4)

B) n(n+2) (n-1)

C) n(n+3) (n-5)

D) n(n+4) (n-2)

E) n(n+5) (n-6)

Solution

B fails for n = 8 => 8*10*7
C fails for n = 7 => 7*10*2
D fails for n = 7 => 7*11*5
E fails for n = 8 => 8*13*2

65. The ratio of the number of women to the number of men to the number of children in a room is 5:2:7, respectively. What is the total number of people in the room?
(1) The total number of women and children in the room is 12.
(2) There are fewer than 4 men in the room.

Solution`

W:M:C :: 5:2:7 => W:M:C :: 5K:2K:7K;

Stmt 1 : 5K+7K = 12 => K = 1; total strength => 5K+2K+7K = 14K = 14;
Stmt 2 : 2K = 3 or 2 or 1 =>
K cannot be 3/2 as 5K will be 5*3/2 => women should be a +ve interger;
K cannot be 1/2 as 5K will be 5*1/2 => women should be a +ve interger;
=> 2K = 2; K = 1 ; total strength => 5K+2K+7K = 14K = 14;


D


66. If S is an infinite set of real numbers, is there a number in S that is less than every other number in S?
1) Every number in S is an integer.
2) Every number in S is positive.

Solution`

Stmt 1 : every integer range : -∞ to ∞; There for -∞ is the smallest; Pls CONFIRM
Stmt 2 : Positive number range : 1 to ∞; there for there should be a smallest number;

D


67. All the employees donate for the charity concert held in the city. 1/4 of the employees contributes below 200, 2/3 of the employees contribute 200 to 1,000, The rest contribute above 1000. Mean value of contributions greater than 200 is 360 dollar . What is the mean value of contributions greater than 1,000?
(1) Mean value of contributions less than 200 is 180
(2) Mean value of contributions between 200~1000 is 540

Solution`

X employees ; 1- 1/4 - 2/3 = 12/12 - 3/12 - 8/12 = 1/12

X/12 of people donated above 1000

1/4 * 180 + 2/3 * 540 + 1/12 * X = 360

so we can get X, and we can compare X with 1000.

Therefore, answer to this question would be C


68. M is a positive integer, How many figures does M have?

1) M is 3 figures

2) M^2 is 5 figures

Solution

Stmt 1 : e.g M = 100; M = 1000000= >7 figures; M =200; M = 27000000=> 8 figures Thr fore Stmt 1 is not enough
Stmt 2 : e.g M = 100; M = 10000= >5 figures; M will be 5 figures till apprx M = 300; again the M of M=100 and M=300 will not have same number of digits;

E;




69. Last year in a group of 30 businesses, 21 reported a net profit and 15 had investments in foreign markets. How many of the businesses did not report a net profit nor invest in foreign markets last year?

(1) Last year 12 of the 30 businesses reported a net profit and had investments in foreign markets.

(2) Last year 24 of the 30 businesses reported a net profit or invested in foreign markets, or both.

Solution

Sets; Given A=21; B=15; Required : 30-(A+B-A[Intersection]B)

Stmt 1 : Given A[Intersection]B = 12; There fore required value can be calculated
Stmt 2 : Given A+B-A[Intersection]B = 24; There fore required value can be calculated


70. Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits?

(1) P>Q

(2) Q=8

Solution


Stmt 1 : P>Q; 10/3 is infinite; 4/2 is finite; thr fore Stmt 1 alone is not enough
Stmt 2 : Any number devided by 2,4,8,10 will always result in finite number of nonzero digits; There fore stmt 2 is enough

B;


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71. From May 1 to May 30 in the same year, the balance in a checking account increased. What was the balance in the checking account on May 30?

1) If, during this period of time, the increase in the balance in the checking account had been 12 percent, then the balance in the account on May 30 would have been $504

2) During this period of time, the increase in the balance in the checking account was 8 percent.

Solution


Stmt 1 : Given amount on may 30 and the increase from may 1st; Using this statement balance on May 1 can be calculated; i.e X+X*12/100 = 504; But the given percent is assumption and therefore the balance on 30 cannot be calculated; Stmt 1 alone is not enough
Stmt 2 : The actual percent increase is given; But the initial value on May 1 st is not known; Therefore Stmt 2 alone is not enough



Stmt 1 & 2 : initial amount and the actual increase given; Therefore the balance can be calculated; C




72. During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob read per week?

(1) Twice the average number of books that Carolyn read per week was greater than 5 less than twice the average number of books that Jacob read per week.

(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more than Jacob.

Solution

Books read by Carolyn : C; Books read by Jacob : J

Stmt 1 : given 2* C/10 > 2* J/10 -5; When J = 10 & M = 30 the equation satisfies; again J = 50 & M = 30 the equation satisfies; The greatest of J & M cannot be determined
Stmt 2 : The last 5 weeks comparision is given; But any number of books could have been read in the first 5 weeks; Therefore Stmt 2 is not enough

E;

73. In a certain game played with red chips and blue chips, each red chip has a point value of X and each blue chip has a point value of Y, where X> Y and X and Y are positive integers. If a player has 5 red chips and 3 blue chips, what is the average (arithmetic mean ) point value of the 8 chips that the player has?

1) The average point value of one red chip and one blue chip is 5.

2) The average point value of the 8 chips that the player has is an integer.

Solution

To find : (5R +3B)/8; X > Y; again X+Y = 10; So X is always greater than 5. its range : 6-9;

Stmt 1 : R + B= 10; we take 3 such combinations the sum will be 30; Left out 2 R’s; CANNOT BE DETERMINED
Stmt 2 : CANNOT DETERMINE;

Stmt 1 & 2 : 30+2R = 8 * Integer; 2R can b3 12;14;16;18 of which only 30+18 = 8*integer

C

74. At a Wall Street company, 70 percent of this year new employees are graduates of business schools and the remainders are graduates of liberal arts colleges. If 550 new employees were hired this year, what is the difference between the number of new business school employees and the number of new liberal arts employees?

(A)55

(B)220

(D)240

(D)385

(E)440

Solution



Let X be total number of new employees; given .7X = Business schools; .3X = liberal arts; X = 550; difference between Business schools joiners and liberal arts joiners
=> .7X-.3X = 0.4X ; => 0.4*550=>220

B


75. Group of numbers A and group of numbers B: there a numbers in A, The median is 85, the average value is 82. There are b numbers in B, the median is 78, the average value is 75. Is the median number greater than the average value, after A and B are mixed.

(1) a+b=97

(2) a=42,b=37

Solution



Stmt 1 : Average of the combined cannot be found unless each groups strength is known
Stmt 2 : Average of the combined can be found. But Median of the combined cannot be found.

Stmt1 & 2 : Using stmt2 we already have a+b; Stmt1 is not required and by using that along with stmt2, it doesn’t give any further information.

E.



76. The telephone number, first three is fixed, for example 921-XXXX, other four numbers are random arrangement, and the 921 is an area code. If we need 282, 000 telephone numbers, how many area codes do we need? We can choose among 28, 29, 2,800, etc.

A. 28
B. 29
C. 30
D. 290
E. 300



Solution



Lets say there is a pool of 3 digit area codes available. Now, lets say I pick a 3 digit area code from the pool which is 921. Now for 921 I can assign 4 different numbers as suffix. ie., I can assign, 921-0000 to 921-9999. For every area code I can assign 10,000 numbers.

Now, I need to assign numbers for 282,000 residences. How many area codes do I need? 282,000/10,000 = 28.2. Now I can't get a fraction of the area codes.

so, answer is 29. B


77. The length, width and height of cuboid are: 8, 10, and 12. The Wall of the cuboid is thick measuring 0.5, What is the radius of biggest circular cylinder that can be placed inside?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5



Solution

Internal length = 8-0.5-0.5 = 7; 0.5 is thickness; width = 9; height = 11;
The biggest radius of cylinder placed on base with L/W = 7/2 = 3.5; height of the cylinder will be 11; volume = ∏rh = 3.5*3.5*11*∏
The biggest radius of cylinder placed on base with W/H = 9/2 = 4.5; height of the cylinder will be 7; volume = ∏rh = 4.5*4.5*7*∏
The biggest radius of cylinder placed on base with L/H = 7/2 = 3.5; height of the cylinder will be 9; volume = ∏rh = 3.5*3.5*9*∏

the second cylinder will have max volume;

Ans : D


78. What are units number and tens number of 123456789 ?

A. 1 & 2
B. 2 & 3
C. 3 & 2
D. 2 & 1
E. 9 & 7



Solution

Calculate 89*89=> 7921 => 1 & 2 => A

79. P = sum of positive odd numbers which are < 50

Q = sum of positive even numbers which are < 50

The result of P-Q =?

A. 23
B. 24
C. 25
D. 26
E. 27



Solution

Odd numbers …1,3,5…49
Even Numbers ..2,4,6…50

In 50 there are 25 even and 25 odd numbers and every even number is 1 greater than odd number. There fore the 25*1 = 25;

C;

80. m and n are integers, is m^n an integer?

1) n^m is positive

2) n^m is an integer

Solution

Stmt1 : n^m is positive
m can b +ive or –ive
m can be even or odd; if m is even n can b -ive or +ive; if m is odd n is +ive; m^n can not b deduced; if n is -ive it not a integer if n is +ive it is a integer;

Stmt 2 n^m is an integer ; m has to be positive it can be even or odd;

both eqn are also insufficient to answer the Q?


E


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81. If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisible by exactly 4 different prime numbers.

Solution

Need to find if k is a square i.e 1,4,9,16,25….

Stmt 1 : There are many non-squares/ squares divisible by 4;
Stmt 2 : When 4 prime numbers are the only factors the numbers can never be square as the prime numbers are not divisible by any number;

Stmt1 & Stmt2 cannot be used together to determine;


82. An object thrown directly upward is at a height of h feet after t seconds, where h = -16 (t - 3)^2 + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height? When does it reaches its maximum height?

A. 6
B. 86
C. 134
D. 150
E. 214




Solution

h = -16 (t - 3)^2 + 150, the maximum height is when t-3 = 0 or t = 3, in other words, from this fomula, when t = 3, it reaches its maximum height 150.

2 seconds after that, would be t = 5,

Therefore, h = -16 (5 - 3) ^ 2 + 150 = -16 * 4 + 150 = 86 ; B


83. if X>1 and Y>1, is X<Y?

(1) X/(XY+X)<1

(2) XY/Y-Y<1

Solution

Stmt1 : X/(Y+1) < 1; with either X/Y –ve; e.g : X=4; Y=-5; Or X=-4; Y=3
Stmt2 : X/(Y-1) < 1; with either X/Y –ve; e.g : X=4; Y=-3; Or X=2; Y=4

Using Stmt1 & 2 : Values of X/Y cannot be determined;

84. In x-y plate, is the slope of a line m less than 0?
(1) The line passes quadrant 3
(2) The line passes quadrant 4

Solution

y=mx+c; x co-ordinate should be less than 0

Stmt1 : Line passing through Quadrant 3 can pass through any Quadrant and slope cannot be determined
Stmt2 : Line passing through Quadrant 4 can pass through any Quadrant and slope cannot be determined

Stmt 1 & 2 : Again line passing through Quadrant 3,4 can pass through any other quadrant. Slope cannot be determined;

E



85. Three equal pipes to fill a tank, take 36hr. A similar pipe starts filling the tank. If all 4 start filling simultaneously how many hours less will they take to fill the same tank?

A. 8
B. 9
C. 27
D. 108
E. 120

Solution

1/X+1/X+1/X = 1/36 => 3/X = 1/36 => X = 108; Now 4/108 => 1/27; 4 pipes take 27 hours; there fore 36-27 = 9;

86. Can Y be evenly divided by 5?
1), X divided by 5, remain is 2,
2) XY can be evenly divided by 5

Solution

Stmt 1 :X = 5K+2; e.g. : 7,12,17… No information about Y; Cannot be determined
Stmt 2 : XY = 5K; What is X? Cannot be determined

Stmt 1 & Stmt2 => X is not multiple of 5; for XY to be multiple of 5; Y has to be multiple of 5; e.g. X = 7, 12, 17; Y = 10; 5; Here 10 is even multiple of 5; but not 5;

There fore E;

87. There are 90 students apply for the audience of a special lecture, How many of them applied successfully?
(1) 2/3 male student and 1/3 female student applies successfully
(2) 26 male students applied successfully

Solution

Stmt 1 : Actual male/ female not given; so cannot be determined
Stmt 2 : No information about female; so cannot be determined

Stmt 1 & 2 : 2/3M = 26 => M = 39; => F = 51; 1/3*51 = 17; 17+26 = 43

C



88. n and m are integers. if m=n/1800, what is the min. value of m?

A. 1
B. 3
C. 5
D. 15
E. 450

Solution

n = 1800*m => n= 2*2*2*3*3*5*5*m; to make 2*2*2*3*3*5*5*m perfect cube, m = 3*5 =>15

Reference Key: 15

89. x,n are positive integers and x=(n+1)(n-1)/24. r is arithmetical compliment (remainder), what's the value of r?
(1) 2 is not a factor of n
(2) 3 is not a factor of n

Solution

The expression can be reduced to (n^2 - 1)/24

From Stmt1 : nothing can be said; as n can be 3,5,7; when substituted the r values vary

From Stmt2 : nothing can be said; as n can be 4,5,7; when substituted the r values vary

combining the two num can't be even and also can not be divisible by 3

putting the values till 20

1,5,7,11,13,17,19 ...the reminder comes to 0 hence C



90. A and B make 1,000 parts together take X/3 hours, B makes this 1,000 parts alone take X hours, asked how many hours does A alone need to make this 1,000 parts in terms of X.

A. 2/X
B. 2X
C. X/2
D. 3X/4
E. 4X/3


Solution



1/A + 1/B = 3/X; 1/A + 1/X = 3/X => 1/A = 3/X-1/X => 1/A = 2/X; A => X/2 ; C

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