c-> structured oriented lang
Pascal-> Procedure oriented lang
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
Options:
A) EDRIRL
B) DCQHQK
C) ESJFME
D) DEQJQM
Ans: A)
The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
@ Aprameya Shyam
length=x does not imply breadth=36-x
2x+4y=36.
3 possible answers are
(2,8) -> 16sqcm.
(4,7) -> 28sqcm.
(6,6) -> 36sqcm
3, 7, 15, 27, 63, 127, 255
B. 27
( Number × 2 ) + 1
3 × 2 + 1 = 7
7 × 2 + 1 = 15
15 × 2 + 1 = 31 ( not 27 )
31 × 2 + 1 = 63
63 × 2 + 1 = 127
127 × 2 + 1 = 255
16
D
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
these are question tht are seem to be like it consumes
time. so dont see jus the question and run to the next .
ans. a) 12
if 9 balls are added then the ratio to the combination
becomes 2:4:3.
9 balls make the ratio 3 for grey balls in tht mixture.
so the factor is 9/3 = 3 (in tht mixture)
so the same factor has to be maintained through out the
ratio. so black balls is 4 * 3 = 12 and white balls is 2 *
3 = 6.
the answer is 34
35000