Ratio And Proportion Online Test 2020 - Aptitude Questions And Answers MCQ, Online Quiz

Solution: log 75/16-2 log 5/9+log 32/243
= log 75/16-log(5/9)2+log32/243
= log 75/16-log25/81+log 32/243
= log(75/16*32/243*81/25)=log 2

Solution: Clearly, A beats B by 10 sec.
Distance covered by B in 10 sec.
= (1000/200 x 10 )m = 50 m.
Therefore A beats B by 50 metres.

Solution: While A covers 1000 m, B covers (1000 - 40) m = 960 m and
C covers (1000 - 64) m or 936 m.
When B covers 960 m, C covers 936 m.
metre's start can B give C(960 -936 )m or 24m.

Solution: 16th April, 2000 = (1999 years + Period from 1st Jan., 2000 to 16thA'
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
99 years = (24 leap years + 75 ordinary years)
= [(24 x 2) + (75 x 1)] odd days = 123 odd days
= (17 weeks + 4 days) = 4 odd days.
Jan. Feb. March April
31 + 29 + 31 + 16 = 107 days = (15 weeks + 2 days) = 2 odd,
Total number of odd days = (0 + 1 + 4 + 2) odd days = 7 odd days = 0 odd day.
Hence, the required day was 'Sunday'.

Solution: Suppose he buys each share for Rs. x.
Then, [25*(9/100)] = [x*(10/100)] or x = Rs. 22.50.
Cost of each share = Rs. 22.50.

Solution: Cost of 1 share = Rs. [(10-(3/4)) + (1/4)] = Rs. (19/2).
Cost of 96 shares = Rs. [(19/2)*96] = Rs. 912

Solution: By investing Rs. 136,
income obtained = Rs. 10.
By investing Rs. 6800,
income obtained = Rs. [(10/136)*6800] = Rs. 500.

Solution: Income from Rs. 100 stock = Rs. 8.
Income from Rs. 2500 = Rs. [(8/1000*2500) =Rs. 200.

Solution: By selling Rs. 100 stock ,
cash released = Rs. [(100-4)-(1/4)] = Rs. (383/4).
By selling Rs. 2400 stock,
cash released = Rs. [(383/4)*(1/100)*2400] = Rs 2298

Solution: Cash required to purchase Rs. 100 stock = Rs (107+(1/2)) = Rs. (215/2).
Cash required to purchase Rs. 100 stock = Rs [(215/2)*(1/100)*3200] = Rs. 3440.

Solution: A: B = 80 : 75, A : C = 80 : 65.
B/C = ( B/ A * A/C) = (75 / 80 * 80 / 65)
= 15/13 = 60 /52 = 60: 5 Therefore ,In a game of 60, B can give C 8 points.
.

Solution: log10 3+log10 (4x+1)=log10 (x+1)+1
Log10 3+log10 (4x+1)=log10 (x+1)+log10 (x+1)+log10
10 Log10 (3(4x+1))=log10 (10(x+1))
=3(4x+1)=10(x+1)=12x+3
=10x+10
=2x=7=x=7/2

Solution: WPrincipal = Rs. 16000; Time = 9 months =3 quarters
Rate = 20% per annum = 5% per quarter.
Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.
CI = Rs. (18522 - 16000) = Rs. 2522.

Solution: Let the sum be Rs. x. Then,
C.I. = x ( 1 + ( 10 /100 ))2 - x = 21x / 100,
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100
( x / 100 ) = 632 x = 63100.
Hence, the sum is Rs.63,100.

Solution: P = Rs.68000,R = 50/3% p.a and T = 9/12 years = 3/4years.
S.I. = (P*R*T)/100 = Rs.(68,000*(50/3)*(3/4)*(1/100))= Rs.8500.

Solution: Let the sum be Rs. x. Then,
C.I. = [ x * (1 + (( 50/(3*100))3 - x ] = ((343x / 216) - x) = 127x / 216
127x /216 = 1270 or x = (1270 * 216) / 127 = 2160.
Thus, the sum is Rs. 2160
S.I. = Rs ( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080.

Solution: S.I. for 1 1/2 years = Rs.(1164-1008) = Rs.156.
S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208
Principal = Rs. (1008 - 208) = Rs. 800.
Now, P = 800, T = 2 and S.l. = 208.
Rate =(100* 208)/(800*2)% = 13%

Solution: Let principal = P. Then, S.l. = P and T = 16 yrs.
Rate = (100 x P)/(P*16)% = 6 1/4 % p.a.

Solution: S.I on rs.7350 for 1 year=rs.(8575-7350)=rs.1225.
Rate=(100*1225/7350*1)%=16 2/3%
Let the sum be rs.x.then,
X(1+50/3*100)2=7350
X*7/6*7/6=7350
X=(7350*36/49)=5400.
Sum=rs.5400.

Solution: Let sum = Rs. x. Then, S.l. = Rs. 4x/9
Let rate = R% and time = R years.
Then, (x*R*R)/100=4x/9 or R2 =400/9 or R = 20/3 = 6 2/3.
Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months.