1. a rectangular box is inscribed in a sphere of radius r cm.Surface area of the box is 384cm sq and the sum of the lengths of its 12 edges is 112 cm.The valus of r is...
1. 8
2. 103. 12
4. 14
Solution:let a,b,c be the edges of the cuboidal box 4(a+b+c)=112;c is the height
a+b+c=28. the box has 6 faces 2(ab+bc+ca)=384
ab+bc+ca=192
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
784=a^2+b^2+c^2+2*192==> a^2+b^2+c^2=400
let r be the radius of the sphere hence 2r=sqrt(a^2+b^2+c^2)=20
hence r=10
2. The surface of the water in the swimming pool is a rectangle 26m long and 10 m wide and the depth of the water increases uniformly from 1.6m at one end to 4.4m at the other end.The volume (m cube) of the water is the pool is
1. 364
2. 390
3. 7804. 1560
Solution volume of a rectangular cuboid is length* breadth *height
so first lets calculate 1.6*26*10=416
secondly calculate=(4.4-1.6)*26*10/2=364
add these two=
7803.The compound interest on a certain sum of money for 2 years at 5% per annum is 102.5.
Compound interest on the same sum for the same period at 4% per annum is
1. 80.8
2. 80.6
3. 81.64. 84.4
solution:let the sum be x . so ((1.05)^2)*x will be total amount
1.1025*x
==>0.1025x=102.5
x=1000
(1.04)^2=1.0816
==>0.0816x is the required answer =0.0816*1000=
81.64. two persons working for 2 hours a day assemble 2 machines in 2 days.The number of machine assembled by 6 persons working 6 hours a day in 6 days is...
1. 6
2. 18
3. 36
4. 54Solution: use chain rule
2*3*3*3=
545. A and B worked on ajob for 4 hours and finished half of it.A worked thrice as fast as B did. If B then left and A was joined by C and they finished the job in 1 hour,how long would it have taken C to do the whole job alone...
1. 3 hours
2. 7/2 hours
3. 13/16 hours
4. 32/13 hours
Solution: use traditional method
6. A shopkeeper placed on dispaly some dresses each with a marked price.She then posted a sign "1/4 off on these dresses". The cost of dresses was 2/3 of the price at which she actually sold them.Then the ratio of the cost to the MP was
1. 1:22. 2:3
3. 4:9
4. 3:5
solution: mp=x
sp=0.75x
cp=0.5x
7. Two candles have different lengths and different thicknesses.The shorter one would last 11 hours, the longer one would last for 7 hours.Both of them are lit at the same time and burn evenly.After 3 hour both have the same length remaining.The ratio of their original lengths is
1. 5:8
2. 7:11
3. 10:13
4. 11:14Solution: let the lengths burnt and radii of the two candles be l,r and L,R
respectively
1st candle burns l/11 in the per hour second burns L/7 per hour
==>
l-3l/11=L-3L/7
==>8l/11=4L/7
l/L=
11:148. A and B start their new jobs on the same day.A's schedule is 3 working days followed by 1 rest day,while B's schedule is 7 working days followed by 3 rest days.On how many of their first 1000 days do both have rest-days on the same day?
1. 40
2. 50
3. 1004. 120
Solution:
the days when they rest together follows the series
8,20,28,40,48,60...
separate it to 2 series
8,28,48....988(50 terms) number of such days =50
and 20,40, 60,...1000(50 terms) number of such days=50
total=100 days