A. 32
B. 48
C. 64
D. 96
E. 55
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box , if at least one black ball is to be included in the draw?
A. 32
B. 48
C. 64
D. 96
E. 55
A. 32
B. 48
C. 64
D. 96
E. 55
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box , if at least one black ball is to be included in the draw?
A. 32
B. 48
C. 64
D. 96
E. 55
My Take: 'C'
Black Balls - 3
Non Black Balls - 2 +4 = 6
Combinations for at least 1 black from a total of 3 balls
1-black + 2-Other = 3C1 * 6C2 = 3 * 15 = 45
2-black + 1-Other = 3C2 * 6C1 = 3 * 6 = 18
3-black + 0-Other = 3C3 * 6C0 = 1 * 1 = 1
Total = 64.
YEP, That's the OA.
Quote:
Originally Posted by jishnu.nandy View Post
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box , if at least one black ball is to be included in the draw?
A. 32
B. 48
C. 64
D. 96
E. 55
My Take: 'C'
Black Balls - 3
Non Black Balls - 2 +4 = 6
Combinations for at least 1 black from a total of 3 balls
1-black + 2-Other = 3C1 * 6C2 = 3 * 15 = 45
2-black + 1-Other = 3C2 * 6C1 = 3 * 6 = 18
3-black + 0-Other = 3C3 * 6C0 = 1 * 1 = 1
Total = 64.
i believe the answer shoud be 64...........
though your approach is correct ..........
just one more way of solving the probs
total ball =9
total number of ways in which 3 balls can be selected = 9C3=84
ways in which none of the black is selected = 6C3=20
ways in which at least 1 black is selected = Total ways - no balck ball selected = 84-20= 64
Pls explain...
9.The table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both chess and drama, 5 students are in both chess and math, and 6 students are in both drama and math. How many different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84
Club
Number of Students
Chess
40
Drama
30
Math
25
11. In a nationwide poll, N people were interviewed. If 41
of them answered "yes" to question 1, and of those, 31
answered "yes" to question 2, which of the following expressions represents the number of people interviewed who did not answer "yes" to both questions? (A) 7N
(B) 76N
(C) 125N
(D) 127N
(E) 1211N
13. The ratio of two quantities is 3 to 4. If each of the quantities is increased by 5, what is the ratio of these two new quantities?
(A) 43
(B) 98
(C) 1918
(D) 2423
(E) It cannot be determined from the information given.
9.The table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both chess and drama, 5 students are in both chess and math, and 6 students are in both drama and math. How many different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84
Club
Number of Students
Chess
40
Drama
30
Math
25
11. In a nationwide poll, N people were interviewed. If 41
of them answered "yes" to question 1, and of those, 31
answered "yes" to question 2, which of the following expressions represents the number of people interviewed who did not answer "yes" to both questions? (A) 7N
(B) 76N
(C) 125N
(D) 127N
(E) 1211N
13. The ratio of two quantities is 3 to 4. If each of the quantities is increased by 5, what is the ratio of these two new quantities?
(A) 43
(B) 98
(C) 1918
(D) 2423
(E) It cannot be determined from the information given.
Pls explain...
9.The table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both chess and drama, 5 students are in both chess and math, and 6 students are in both drama and math. How many different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84
Club
Number of Students
Chess
40
Drama
30
Math
25
11. In a nationwide poll, N people were interviewed. If 41
of them answered yes to question 1, and of those, 31
answered yes to question 2, which of the following expressions represents the number of people interviewed who did not answer yes to both questions? (A) 7N
(B) 76N
(C) 125N
(D) 127N
(E) 1211N
13. The ratio of two quantities is 3 to 4. If each of the quantities is increased by 5, what is the ratio of these two new quantities?
(A) 43
(B) 98
(C) 1918
(D) 2423
(E) It cannot be determined from the information given.
My Take:
1. Ans:C
n(AUBUC) = n(A) + n(B) + n(C) - n(A&B;) - n(B&C;) - n (C&A;) + n(A&B;&C;)
= 40 + 30 + 25 - (10+6+5) + 0 = 95 - 21 = 74
2. it is in 'N" and 41 31 - i cudnt understand the question.
3. Ans: E
let the intial quantities be 3a : 4a
new quantities = 3a+5 : 4a+5 --- until we know 'a' we cannot get this values..
Hi,
Can someone please help me out with these questions, the first one totally stumped me.
1. The height of women in Dewaria follows a normal distribution with mean 160 cm and standard deviation of 6 cm. In a normal distribution , only 0.0063% of the population is not within 4 standard deviations of the mean. if 5 women are more than 184 cm tall, approximately how many women live in Dewaria?
A.16,000
B.40,000
C.80,000
D.100,000
E.160,000
2. How many positive integers less than 100 have exactly 4 odd factors but no even factors:
A.13
B.14
C.15
D.16
E.17
Thanks.
Hi,
Can someone please help me out with these questions, the first one totally stumped me.
1. The height of women in Dewaria follows a normal distribution with mean 160 cm and standard deviation of 6 cm. In a normal distribution , only 0.0063% of the population is not within 4 standard deviations of the mean. if 5 women are more than 184 cm tall, approximately how many women live in Dewaria?
A.16,000
B.40,000
C.80,000
D.100,000
E.160,000
2. How many positive integers less than 100 have exactly 4 odd factors but no even factors:
A.13
B.14
C.15
D.16
E.17
Thanks.
Are the OA s C) and C)
Hi,
Can someone please help me out with these questions, the first one totally stumped me.
1. The height of women in Dewaria follows a normal distribution with mean 160 cm and standard deviation of 6 cm. In a normal distribution , only 0.0063% of the population is not within 4 standard deviations of the mean. if 5 women are more than 184 cm tall, approximately how many women live in Dewaria?
A.16,000
B.40,000
C.80,000
D.100,000
E.160,000
2. How many positive integers less than 100 have exactly 4 odd factors but no even factors:
A.13
B.14
C.15
D.16
E.17
Thanks.
Hi both these questions look pretty tough..
1. only 0.0063% of the population is not within 4 standard deviations of the mean.
so, 0.0063% population is outside 4 S.D
since it is given as normal distribution, both the sides it must be same. i.e it is similar on either sides of the bell curve.
Given: 5 women are more than 184 cm tall --- since 184 = 160 + 4 * 6 (4 standard deviations of the mean. ). it shows that 5 women should be less than 160-24.
Hence, .0063% of population = 10
so.. population approximately = 160,000 ... answer 'E'.
2. Trying the second one.. all i can do is count.. there must be some mathematical theorm to do this in a easier way....
italian7745 SaysAre the OA s C) and C)
The OA's are E and C.
For the first one alchemist-mba has explained it pretty well. For the second one can you explain how did you arrive at C? Thanks.
Hi Guys, need ur help....
1. The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
2. If d-9 = 2d, then d=
(A) -9
(B) -3
(C) 1
(D) 3
(E) 9
won't there be 2 solns...as in...1) d-9=2d; d=3 & 2) -d+9=2d; d=3
3. Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24
Thnx,
Sumit
The OA's are E and C.
For the first one alchemist-mba has explained it pretty well. For the second one can you explain how did you arrive at C? Thanks.
I couldn t find any particular theorem to solve this question:-
1) We cannot take any even No. from 1-100,no even no. can have all odd factors.
2) We cannot take any Prime No. from 1-100.as prime No. is divisible by 1 and itself.
If you evaluate rest of the Nos - you will come up with the answer...
Hi Guys, need ur help....
1. The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
2. If d-9 = 2d, then d=
(A) -9
(B) -3
(C) 1
(D) 3
(E) 9
won't there be 2 solns...as in...1) d-9=2d; d=3 & 2) -d+9=2d; d=3
3. Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24
Thnx,
Sumit
1. f(n) = 2^x 3^y 5^z
f(m) = 9 * f(v) = 9 * (2^x 3^y 5^z) = 2^x 3^(y+2) 5^z
so m = x y+2 z and v = x y z - -the only diff is in tenth place.. hence it must be 20 and D is the answer.
2. True.. it will have two solutions.. d=-9 or 3. isn't there any prior information with regard to d?
3. I cudnt find any that matches the answer..
The length of each red stick is 19 inches less that the average length of the sticks in Box W
=> r = (average length of box W) - 19
=> average length of box W = r + 19
and 6 inches greater than the average length of the sticks in Box V.
=> r = (average length of box V) + 6
=> average length of box V = r - 6
What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
average length of box W - average length of box V = (r+19) - (r-6) = 25..
1. f(n) = 2^x 3^y 5^z
f(m) = 9 * f(v) = 9 * (2^x 3^y 5^z) = 2^x 3^(y+2) 5^z
so m = x y+2 z and v = x y z - -the only diff is in tenth place.. hence it must be 20 and D is the answer.
2. True.. it will have two solutions.. d=-9 or 3. isn't there any prior information with regard to d?
3. I cudnt find any that matches the answer..
The length of each red stick is 19 inches less that the average length of the sticks in Box W
=> r = (average length of box W) - 19
=> average length of box W = r + 19
and 6 inches greater than the average length of the sticks in Box V.
=> r = (average length of box V) + 6
=> average length of box V = r - 6
What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
average length of box W - average length of box V = (r+19) - (r-6) = 25..
I think the answer for the 2nd question will be d=3 that means option d)...
It is true that we are getting 2 Solns d=3 and d=-9 but when we put back the same values in the question d=-9 is not satisfying it
mod=2d now, mod=2*-9 ,which is not true so D is the answer
I think the answer for the 2nd question will be d=3 that means option d)...
It is true that we are getting 2 Solns d=3 and d=-9 but when we put back the same values in the question d=-9 is not satisfying it
mod=2d now, mod=2*-9 ,which is not true so D is the answer
Absolutely... how did i miss this :-(....
1. f(n) = 2^x 3^y 5^z
f(m) = 9 * f(v) = 9 * (2^x 3^y 5^z) = 2^x 3^(y+2) 5^z
so m = x y+2 z and v = x y z - -the only diff is in tenth place.. hence it must be 20 and D is the answer.
I understand the first part where you say
"f(m) = 9 * f(v) = 9 * (2^x 3^y 5^z) = 2^x 3^(y+2) 5^z
so m = x y+2 z and v = x y z"
Can you please further ellaborate on the conclusion that you are drawing from the above solution i.e.
"-the only diff is in tenth place.. hence it must be 20 and D is the answer."
Thanks.
I understand the first part where you say
"f(m) = 9 * f(v) = 9 * (2^x 3^y 5^z) = 2^x 3^(y+2) 5^z
so m = x y+2 z and v = x y z"
Can you please further ellaborate on the conclusion that you are drawing from the above solution i.e.
"-the only diff is in tenth place.. hence it must be 20 and D is the answer."
Thanks.
Lets take an example ....
if the first number is 234.. second number must be 254
difference.. 254 - 234 = 20...
Lets take an example ....
if the first number is 234.. second number must be 254
difference.. 254 - 234 = 20...
let m = x1y1z1 and v be x2y2z2
given that 2^x1.3^y1.5^z1 = 2^x2.3^2+y2.5^z2
put m = x1y1z1 = 121 f(m) = 235
so for f(v) = 235, v = x2y2z2 = 101
hence m-v = 20
i cud nt solve the ques at first attempt. i guess the seeing the ans before first dilutes our approach towards more difficult PS
thanks
Guess. My first question here...
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
A. 10(
1)
B. 5
C. 10(
1)
D. 5( 1)
E. 5( 1)
I answered the question but got a different answer as OA. Can someone pls help?
Guess. My first question here...
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
A. 10(
B. 5
C. 10(
D. 5(
E. 5(
I answered the question but got a different answer as OA. Can someone pls help?
I think the answer should be E.
As the sphere will be touching all the 6 faces, the shortest distance will be the distance between any edge of cube and the point where sphere touched the cube minus the radius of sphere.
This distance will be half of a diagonal of cube i.e sqrt (10^2 + 10^2) = 5.
Shortest distance = 5.
- 5 = 5( - 1)Whats the OA ?