please solve these two questions and give detailed solution
1) 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`^2 + QQ`^2
c) RS^2 = PP`^2 + AB^2
d) RS^2 = BS^2 + PP`^2
2) if TA and TB are tangents to a circle C(O,r) from a external point T and OT intersects the circle at P , then AP bisects the angle TAB.
a) always true
b) always false
c) cant be determined
d) some fact missing.
PLEASE GIVE DETAILED SOLUTION !!!!!
can u explain this one ---
in the ans they have given--
( (4c4 x 4c1 x 5!) / 3! ) + ( (4c2 x 4c3 x 5!)/3! ) -(4c2 x 4c3 x 2c2 x 2!)
main problem is there with the last part(red one)
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
can u explain this one ---
in the ans they have given--
( (4c4 x 4c1 x 5!) / 3! ) + ( (4c2 x 4c3 x 5!)/3! ) -(4c2 x 4c3 x 2c2 x 2!)
main problem is there with the last part(red one)
can u explain this one ---
in the ans they have given--
( (4c4 x 4c1 x 5!) / 3! ) + ( (4c2 x 4c3 x 5!)/3! ) -(4c2 x 4c3 x 2c2 x 2!)
main problem is there with the last part(red one)
yaar thats not my approach, sorry I dont know how the guy who did it approached the problem
please solve these two questions and give detailed solution
1) 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`^2 + QQ`^2
c) RS^2 = PP`^2 + AB^2
d) RS^2 = BS^2 + PP`^2
first draw circle on given premise.
RS^2-AB^2=(RS+AB)(Rs-AB)
from fig RS=RA+AB+BS
=>(RA+AB+BS+AB)(RA+SB)
=>(2RA+2AB)(2RA)
bcoz RA=SB
=>4RA(RB)
Now RP^2=RAxRB
Why?=> property of a tangent
=>4RP^2
=>4(PP`/2)^2
=>PP`^2
=>RS^2-AB^2=PP`^2
=>RS^2=PP`^2+AB^2
the highest power on 990 that will exactly divide 1090! is
a.101
b.100
c.108
d.109
e.110
11. a set S is formed by including some of the first One thousand natural numbers. S contains the maximum numbers such that they satisfy the following conditions
1. no number of set S is prime
2. when the numbers of the set S are selected two at a time, we always see a co prime numbers
what is the number of elements in the set S
a. 11
b 12
c. 13
d. 7
***************
2.1090=9*11*10
11 is highest so we have 1090/11 + 1090/121=99+9=108(take quotitent)
11.
*I am getting 10 as ans*
see in s u can have only nos tht are coprime nd no prime which means tht if in s we put 25 so no other multiple of5
then we put 36 so no multiple of 2,3,6
similarly 49 then we can put squares of all prime no s less than 1000 as they would be coprime nd not prime so in that way we will see that we have 10 nos
Hi i have the right solution for problem no 11.
The 2 conditions stated are
a) No prime numbers in the set.
b) Always coprimes are to be selected from the set --> implying the set shoud have co-primes only.
A little bit of introspection suggests that only the squares of the prime numbers would satisfy both the conditions.
All squares if prime numbers are co-prime to each other...
Prime No Square of the Prime No
ie- 1 ---- 1
2 ---- 4
3 ---- 9
5 ---- 25
7 ---- 49
11 ---- 121
13 ---- 169
17 ---- 289
19 ---- 361
23 ---- 529
29 ---- 841
31 ---- 961
======
12 numbers ---- > (Observe that all are coprime)
======
Cheers....
@EagleMenace thanks for the solution
what abt the other one ???
2) if TA and TB are tangents to a circle C(O,r) from a external point T and OT intersects the circle at P , then AP bisects the angle TAB.
a) always true
b) always false
c) cant be determined
d) some fact missing.
PLEASE GIVE DETAILED SOLUTION !!!!!
@EagleMenace thanks for the solution
what abt the other one ???
2) if TA and TB are tangents to a circle C(O,r) from a external point T and OT intersects the circle at P , then AP bisects the angle TAB.
a) always true
b) always false
c) cant be determined
d) some fact missing.
PLEASE GIVE DETAILED SOLUTION !!!!!
thats always true bhai..
here TA = TB; TAP is congruent to TBP; angle TAP = angleTBP; angleTBP = angle PAB
Plz help me solve
1. one fashion house has to make 810 dresses and another one 900 dresses during the same period of time. In the first house the order was ready 3 days ahead of time and in second house ,6 days ahead of time. How many dresses did each fashion house make a day if the second house made 21 dresses more a day than the first?
1. 54 and 75
2. 24 and 48
3. 44 and 68
4. 4 and 25
5. none of these
2. An enterprise got a bonus and decided to share it in equal parts between the exemplary workers.It turned out however that there were 3 more exemplary workers than it had been assumed. In that case each worker would have got Rs 4 less. The administration had found the possibility to increase the total sum of bonus by 90 rs and as a result each exemplary worker got rs 25. how many people got the bonus?
1. 9
2. 18
3. 8
4. 16
5. 20
Plz help me solve
1. one fashion house has to make 810 dresses and another one 900 dresses during the same period of time. In the first house the order was ready 3 days ahead of time and in second house ,6 days ahead of time. How many dresses did each fashion house make a day if the second house made 21 dresses more a day than the first?
1. 54 and 75
2. 24 and 48
3. 44 and 68
4. 4 and 25
5. none of these
Its option 1 by going through the options..:):)
Plz help me solve
1. one fashion house has to make 810 dresses and another one 900 dresses during the same period of time. In the first house the order was ready 3 days ahead of time and in second house ,6 days ahead of time. How many dresses did each fashion house make a day if the second house made 21 dresses more a day than the first?
1. 54 and 75
2. 24 and 48
3. 44 and 68
4. 4 and 25
5. none of these
let the time taken by fashion house 1 and 2 be x-3 and x-6 days...
so no of dresses done/day by fashion house 1 = 810/x-3
and no of dresses done/day by fashion house 2=900/x-6
also its given that fashion 2 makes 21 dresses more a day
=>810/x-3=900/x-6-21
now solve through options
consider 810/x-3=54
so x=18
810/183=900/18-6-21
=>54=54
also u can solve through options but there is a risk of answer being none of these
ps: will solve second one soon
let the time taken by fashion house 1 and 2 be x-3 and x-6 days...
so no of dresses done/day by fashion house 1 = 810/x-3
and no of dresses done/day by fashion house 2=900/x-6
also its given that fashion 2 makes 21 dresses more a day
=>810/x-3=900/x-6-21
now solve through options
consider 810/x-3=54
so x=18
810/183=900/18-6-21
=>54=54
also u can solve through options but there is a risk of answer being none of these
ps: will solve second one soon
There's no risk of being 'None of these' because:
1. Only options 1 and 4 satisfy the criteria "the second house made 21 dresses more a day than the first?"
2. Option 4 can't be the right answer because 810 can't be divided by 4
3. Using option 1, 900 dresses can be made in 12 days by making 75 dresses/day. Since it makes dresses 6 days ahead of time, the actual time taken is 18 days.
Thus, the company which makes 810 dresses should make it in 15 days, i.e. 810/15 = 54 dresses/day.
Thus, option 1 is the right answer.
There's no risk of being 'None of these' because:
Thus, option 1 is the right answer.
thx for lucid explanation bhai
Plz help me solve
2. An enterprise got a bonus and decided to share it in equal parts between the exemplary workers.It turned out however that there were 3 more exemplary workers than it had been assumed. In that case each worker would have got Rs 4 less. The administration had found the possibility to increase the total sum of bonus by 90 rs and as a result each exemplary worker got rs 25. how many people got the bonus?
lets assume the no of worker the enterprise thought first to be n
bonus x/n-3=z---------1
bcoz of three extra workers x/n=z-4------------2
after the 90Rs increase
x+90/n=25-----------3
now solve through options, try for n=9 in 3
x=210..
substitute n and x in 1 and 2 equality wont satisty
try for other values x=18 for which equality satisfies
Plz help me solve
2. An enterprise got a bonus and decided to share it in equal parts between the exemplary workers.It turned out however that there were 3 more exemplary workers than it had been assumed. In that case each worker would have got Rs 4 less. The administration had found the possibility to increase the total sum of bonus by 90 rs and as a result each exemplary worker got rs 25. how many people got the bonus?
1. 9
2. 18
3. 8
4. 16
5. 20
Solving by options again:
Taking option 4 = 16
Since the average of 16 people is Rs. 25, the total becomes 16*25 = 400
Now, this amount was increased by Rs. 90, which means the earlier amount was Rs. 310, which had to be divided among 3 lesser people, i.e. 13
Since 310 can't be divided by 13, option 4 can't be the one.
Next, taking the option 2 = 18
Total amount = 18*25 = Rs. 450
Less 90 = Rs. 360
Thus, 360/18 people = Rs. 20
This, 360 had to be divided among 15 people, i.e. 360/15 = Rs. 24, which is Rs. 4 more than the average with 18 people
Thus, this option satisfies all criteria and is the answer.
PS: Believe me, by using options and plugging in arbitrary values, you can solve a lot many questions of QA and some in DI/LR too 😃
1. a dishonest milkman purchased milk at rs 10 /litre and mixed 5 litre of water in it. By selling the mixture at the rate of Rs 10/l he earns a profit of 25%. The quantity of amount of mixture that he had was?
a. 15l
b 20 l
c. 25 l
d. 30 l
e. none
I want to know how to solve this by using equations as I am able to solve it through options
1. a dishonest milkman purchased milk at rs 10 /litre and mixed 5 litre of water in it. By selling the mixture at the rate of Rs 10/l he earns a profit of 25%. The quantity of amount of mixture that he had was?
a. 15l
b 20 l
c. 25 l
d. 30 l
e. none
I want to know how to solve this by using equations as I am able to solve it through options
is the ans 20l??
scan002 Saysis the ans 20l??
ans as per book is 25 l
divyaakanksha Saysans as per book is 25 l
oh yah actually its 20l(milk) +5l(water)..Dint read the question properly
1. a dishonest milkman purchased milk at rs 10 /litre and mixed 5 litre of water in it. By selling the mixture at the rate of Rs 10/l he earns a profit of 25%. The quantity of amount of mixture that he had was?
a. 15l
b 20 l
c. 25 l
d. 30 l
e. none
I want to know how to solve this by using equations as I am able to solve it through options
Solution: Let the pure milk bought be : x
C.P= 10x
Impure mixture= x+5
S.P=(x+5)*10
Now,
(x+5)*10-10x=5/2*x (25% of 10x)
On solving x=20
Mixture x+5 =25
how did u get 5/2 * (25 % 10x)?