Ques 1.
X, Y and Z are situated at the bank of river which is flowing at a constant rate. Y is at an equal distance with X and Z. A swimmer Aviral takes 10 h to swim from X to Y and Y to X. Also, he takes 4 h to swim from X to Z. What is the ratio of speed of Aviral in still water and speed of stream?
Solution :
Speed with stream " u " =a+b
Speed against stream "v"=a-b
Distance between X to Z = 2D
Speed against stream "v"=a-b
Distance between X to Z = 2D
==>2D/u = 4 hours
==> D=2u
==> D=2u
D/u + D/v = 10 hours
==>2u/u + 2u/v = 10
==>u/v=4
put u nd v value.
==>a+b/a-b = 4
==>a/b = 5/3
==>2u/u + 2u/v = 10
==>u/v=4
put u nd v value.
==>a+b/a-b = 4
==>a/b = 5/3
Ques 2.
If s1 nd s2 be the surface area of a sphere and the curved surface area of the circumscribed cylinder respectively, then s1 is equal to?
Solution :
Radius of cylinder= r
Height=2r
S1= 4πr²
S2=2πr×2r=4πr²
Height=2r
S1= 4πr²
S2=2πr×2r=4πr²
S1=S2
Ques 3.
8,12,24,60,180,630 ?
Solution :
12= 8*1.5
24= 12*2
60= 24*2.5
180= 60*3
630= 180*3.5
630*4 = 2520
24= 12*2
60= 24*2.5
180= 60*3
630= 180*3.5
630*4 = 2520
Ques 4.
A train starts from Jammu for Srinagar at 13:30 and reaches at 17:30. Another train starts from Srinagar at 15:30 and reaches Jammu at 19:00. At what time both train meet?
Solution :
Speed of first train D/4
Speed of second train D/7/2=2D/7
Second train starts 2 hrs after first train so,distance travelled by first train by that time is (D/4)*2=D/2
Speed of second train D/7/2=2D/7
Second train starts 2 hrs after first train so,distance travelled by first train by that time is (D/4)*2=D/2
So,for remaining distance time taken by both trains is equal to meet each other so, (2D/7)*T=D/2-(D/4)*T
T=14/15=56 mins
13:30+56mins=16:26
Ques 5.
A and B can do a piece of work in 30 days .A having worked for 16 days , B finished the remaining work alone in 44 days .In how many days shall B finish the whole work alone?
Solution :
A and B did work in 30 days ==> 30A+30B=1 --- i
A works for 16 days and B works for 44 day ==>16B+44B=1 --- ii
solving A,B from i and ii
We get 1/60
so B done 1/60 days
Ques 6.
A contractor employed 25 labourers on a job. He was paid 275 for the work. After retaining 20% of this sum, he distributed the remaining amount amongst the labourers. If the number of men to women labourers was in the ratio 2 : 3 and their wages in the ratio 5 : 4, what wages did a woman labourer get ?
Solution :
Men 25×2/5= 10
Women 25×3/5=15
Wage each man 5x
Woman 4x
Women 25×3/5=15
Wage each man 5x
Woman 4x
20% of amount retaining = 275* 20% = 55
Remaining amount = 275-55=220
==>10×5x + 15×4x= 220
==>110x= 220
==>x= 2
Wages of women = 4X = 4×2=8
==>110x= 220
==>x= 2
Wages of women = 4X = 4×2=8
Ques 7.
A certain amount was to be distributed among A,B,C in the ratio 3:4:5 respectively,but was erroneously distributed in the ratio 9:4:7 respectivly. As a result of this B got 260 less. What is the amount?
Solution :
3+4+5 = 12
9+4+7 = 20
( 3:4:5 ) *20 = 60:80:100 = 240 parts
( 9:4:7 ) *12 = 108:48:84
Error in B = 80-48 = 32 parts
B less than 260 with error.
32 parts. ---- 260
240 parts. ---- x
==> Total amount (x) = (260*240)/32 = 1950
Ques 8.
If the coin is to be tossed three times, what is the probability that on at least one of the tosses of the coin will turn up Tails ?
Solution :
1 - 1/8 = 7/8
Ques 9.
If the income tax is increased by 19% the net income is reduced by 1%. The rate of income tax is?
Solution :
1/(19+1) x 100 =1/20 x100=5%
Ques 10.
Asha got married 9yrs ago.today her age is 4/3 times her age at the time of marriage.at present her daughter's age is 1/6th of her age.wat was her daughter's age two yrs ago?
Solution :
x=4/3(x-9)
==>x=4/3x-12
==>12=1/3x
==>X=36
==>x=4/3x-12
==>12=1/3x
==>X=36
Daughter age = 36/6=6
Daughter age two years ago = 6-2=4
Daughter age two years ago = 6-2=4