Given:
In a group of 15 students,
7 have studied Latin,
8 have studied Greek,
3 have not studied either.
To find:
The number of students who studied both Latin and Greek.
Solution:
In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.
Therefore,
n(A∪B) = 15 – 3
n(A∪B) = 12
7 have studied Latin,
n(A) = 7
8 have studied Greek,
n(B) = 8
n(A∩B) is the number of students who studied both Latin and Greek.
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 7 + 8 – 12
n(A∩B) = 15 – 12
n(A∩B) = 3
The number of students who studied both Latin and Greek is 3
Final answer:
3 of them studied both Latin and Greek.
Thus, the correct answer .3
R
T,P,R,Q,S
12 %
6084
7431 is largest number and 1347 is the least number from digits 1,3,4,7
so 7431-1347=6084
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
32
Solution:
life as a boy = 1/4
life as a youth = 1/8
life as an active man = 1/2
sum of life as boy, youth and active man = 1/4 + 1/8 + 1/2 = 7/8
life as an old man = 1 − 7/8 = 1/8
1/8 Wrinkle’s life (as an old man) is 8 years.
and 1/2 = 1/8 *4
So, 1/2 Wrinkle’s Age (as active man) = 8*4 = 32years.
Lawyer
The batsman on 98 is on strike. He hits the ball and they run 3. UNFORTUNATELY one of the batsmen doesn`t turn correctly for one of the runs and the umpire calls ONE SHORT and awards only two runs. Therefore the first batsman has his century. There is now 1 ball remaining and one run is required to win. The batsman on strike, however is now the one on 97 runs. He now either hits a 4 or a 6. They win the game and both batsmen scored centuries.
Read more: 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries … – 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries as well win the match ?
A
let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer
Let the principal be P and rate of interest be R%.
Therefore Required ratio
=P×R×6/100/P×R×9/100
=6PR/9PR
=6/9
=2:3
126
cos if one of the factor is given for hcf and lcm and multiply hcf and lcm and them divide it by the given factor
so 18*3780/540=18*7=126
Answer: 600m
Explanation:
Distance covered by Amar = 18/4.8 (1.6km) = 3/8(1600) = 600 m
8 days