need help:
Q:Three numbers are selected at random from the set of numbers{1,2,3,.....,n}.Find the conditional probability that the third number lies between the first two,if the first number is known to be smaller than the second.
a. 1/6.
b. 1/3.
c. 1/2.
d. 1/4.
In a test in which 120 students appeared,90 passed in History,65 passed in Sociology and 75 passed in Political Science. 30 students passed in only one subject and 55 students in only two. Five students passed in no subject.
1. How many students passed in all 3 subjects?
a.25
b.30
c.35
d.data insufficient
2.Find the no.of students who passed in at least 2 subjects.
a.85
b.95
c.90
d.data insufficient
I have drawn the venn diagram but the answers are not matching..can someone provide detail soln of this ..?
In a test in which 120 students appeared,90 passed in History,65 passed in Sociology and 75 passed in Political Science. 30 students passed in only one subject and 55 students in only two. Five students passed in no subject.
1. How many students passed in all 3 subjects?
a.25
b.30
c.35
d.data insufficient
2.Find the no.of students who passed in at least 2 subjects.
a.85
b.95
c.90
d.data insufficient
I have drawn the venn diagram but the answers are not matching..can someone provide detail soln of this ..?
i got these answrs
1-B
2-A
wot r the answrs givn?
i got these answrs
1-B
2-A
wot r the answrs givn?
the answers r correct..can u plz explain how u got them?
From 4 gentleman and 4 ladies ,a committee of 5 is to be formed.committee consists of a president , a vice president and 3 secretaries.
a) what will be the number of ways of selecting committee with at least 3 women such that at least one woman holds the post of either a president or a vice president ? ans 512
b) find the number of ways of selecting the committee with maximum of 2 women and having at the maximum one woman holding one of the two posts on the committee?? ans 512
In a test in which 120 students appeared,90 passed in History,65 passed in Sociology and 75 passed in Political Science. 30 students passed in only one subject and 55 students in only two. Five students passed in no subject.
1. How many students passed in all 3 subjects?
a.25
b.30
c.35
d.data insufficient
2.Find the no.of students who passed in at least 2 subjects.
a.85
b.95
c.90
d.data insufficient
I have drawn the venn diagram but the answers are not matching..can someone provide detail soln of this ..?
bro here no need of drawing venn diagrams..
1. no of students who passed in all 3 subjects=115-30-55=30
2.no of students who passed in atleast 2 subjects= means 2 or 3 subjects=55+30=85
From 4 gentleman and 4 ladies ,a committee of 5 is to be formed.committee consists of a president , a vice president and 3 secretaries.
a) what will be the number of ways of selecting committee with at least 3 women such that at least one woman holds the post of either a president or a vice president ? ans 512
b) find the number of ways of selecting the committee with maximum of 2 women and having at the maximum one woman holding one of the two posts on the committee?? ans 512
a)in each case we can have women president or vice president
case1-3women2men
p-vp-s1-s2-s3
w1w2w3m1m2
w1m1w2w3m2
m1w1w2w3m2
=>4x3x2x4C2
=>4x4x3c2x3
=>4x4x3c2x3
add all three to get 432
case2-4women1men
p-vp-s1-s2-s3
w1w2w3w4m1
w1m1w2w3w4
m1w1w2w3w4
=>4x3x2c2x4
=>4x4x1
=>4x4x1
=>80
so total combination=> 432+80=512
b)
case1-1women and 4 men
p-vp-s1-s2-s3
w1m1m2m3m4
m1w1m2m3m4
m1m2w1m3m4
=>4x3x3C3
=>4x4x3c3
=>4x3x4x1
=>16+16+48=80
case2-2women and 3men
p-vp-s1-s2-s3
w1m1w2m2m3
m1w2w3m2m3
m1m2w1w2m3
=>4x4x3x3c2
=>4x4x3x3C2
=>4x3x4c2x2
=>144+144+144=432
so total combination=>80+432=512
plz help me solve this:
1) find last two digits of following
65*29*37*63*71*87
a. 05
b. 95
c. 15
d. 25
2)what is remainder when (1!)^3 + (2!)^3+____+(1152!)^3 is divided by 1152?
a. 125
b. 225
c. 325
d. 205
Pls explain how to solve them....
Ch-1.
LOD2 76) what is the total number of divisors of the number 12^33*34^23*2^47 ?
a) 4658
b) 9316
c) 2744
d) none of these
LOD2 85) Find the last two digits: (201*202*203*204*246*247*248*249)^2 ?
LOD3 20) Remainder : 43^101 + 23^101 divided by 66 ?
Thanx.
Can anyone explain how to get Euler Number for a given number?
plz help me solve this:
1) find last two digits of following
65*29*37*63*71*87
a. 05
b. 95
c. 15
d. 25
2)what is remainder when (1!)^3 + (2!)^3+.......+(1152!)^3 is divided by 1152?
a. 125
b. 225
c. 325
d. 205
1)last digit is 5x9x7x3x7
=>5(last digit of 9x5)x9(last digit of 7x7)x3
=>5x3
=>5
2)1152=2^7x3^2
so 4!^3 is divsible by 1152, u need to find sum of 1!^3 + 2!^3 + 3!^3 = 1 + 8 + 216 = 225.
Pls explain how to solve them....
Ch-1.
LOD2 76) what is the total number of divisors of the number 12^33*34^23*2^47 ?
a) 4658
b) 9316
c) 2744
d) none of these
LOD2 85) Find the last two digits: (201*202*203*204*246*247*248*249)^2 ?
LOD3 20) Remainder : 43^101 + 23^101 divided by 66 ?
Thanx.
Can anyone explain how to get Euler Number for a given number?
1)12^33*34^23*2^47
=>(2^2x3)^33 x (17x2)^23 x 2^47
=>2^66 x 3^33 x 17^23 x 2^23 x 2^47
=>2^136x3^33x17^23
=>(136+1)x(33+1)x(23+1)
=>11792
=>d)none of these
2)(201*202*203*204*246*247*248*249)^2
=>(201*202*203*204*246*247*248*249/100)
=>1x2x3x4x46x47x48x49
=>24x62x52
=>76
now last two digits of (76)^2 is ur answer
=>76
3) Remainder : 43^101 + 23^101 divided by 66 ?
this is property of a^n+b^n is divisible by a+b if n is odd, so remainder is 0
1)last digit is 5x9x7x3x7
=>5(last digit of 9x5)x9(last digit of 7x7)x3
=>5x3
=>5
2)1152=2^7x3^2
so 4!^3 is divsible by 1152, u need to find sum of 1!^3 + 2!^3 + 3!^3 = 1 + 8 + 216 = 225.
for the first ques..I need to find the last two digits
oh my again missed the question
65*29*37*63*71*87/100
=>(65x29/100)(37*63/100)x(71*87/100)
=>85x31x77/100
=>(85x31/100)x77
=>35x77/100
=>95
ps:for multiplication of two digits refer to vedic maths shortcuts to cut down on time
@avinav missed the trick, will remember from now on 😁
plz help me solve this:
1) find last two digits of following
65*29*37*63*71*87
a. 05
b. 95
c. 15
d. 25
To find the last two digits, we need to find the remainder when divided by 100
65*29*37*63*71*87 mod 100
Dividing by 5,
13*29*37*63*71*87 mod 20
Now, taking the remainder of each individual number when divided by 20
-7*9*-3*3*-9*7 mod 20
Simplifying,
-49*81*9 mod 20
We continue to take remainders till we get a final number,
-81 mod 20 = -1 mod 20 = 19 mod 20 (20-1 = 19)
Since we had divided by 5 earlier, we need to multiply by 5 to get the final answer
Thus, 19*5 = 95 is the final answer
Edit: @ EagleMenace, you don't need to multiply that much to get the final answer. A series of small steps would do the job.
Geometry Pre-Assess Test Q.11
In the figure given below, a circle is inscribed inside a square. In the gap between the circle and the square (at the corner) a rectangle measuring 20cm X 10cm is drawn such that the corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?
a) 30cm b) 40cm c)50cm d)None of these
PP' and QQ' are two direct common tangents to two circles intersecting at points A and B .The common chord on produced intersects PP' in R and QQ' in S .Which of the following is true ?
a RA^2+ BS^2 =AB^2
b RS^2=PP'+AB^2
c RS^2=PP'^2 + QQ'
d RS^2=BS^2+PP'^2
PS:- how to attach fig as thumbnails ???
plz help me solve this
1. find the sum of all three digit numbers that give remaindeer of 4 when divided by 5
a. 98270
b. 99270
c. 1,02,090
d. 90270
e. none ofthese
b) 99270
Soln:
5
5
5
99450
Subtract 1 from each number(21 to 200) since the remainder is 4 when divided by 5
99450 - 180 = 99270
Thanks,
Easwar
b) 99270
Soln:
5
5
5
99450
Subtract 1 from each number(21 to 200) since the remainder is 4 when divided by 5
99450 - 180 = 99270
Thanks,
Easwar
m not able to understand the solution..how did u get 5
plz help me solve this
1. Find the sum of all three digit numbers that give remaindeer of 4 when divided by 5
a. 98270
b. 99270
c. 1,02,090
d. 90270
e. None ofthese
104+109+......+999=99270