Remaining money = 7044 - 2000 = Rs. 5044
If each installment is of Rs. P
When the amount is Rs. P at the end of first, second and third year at the rate of 5% then principal will be - P/(1 + 5/100), P/(1 +5/100)2 and P/(1 +5/100)3
? P/(1 +5/100) + P/(1 + 5/100)2 + P/(1 +5/100)3 = 5044
? P(20/21) + P(20/21)2 + P(20/21)3 = 5044
? P(20/21) { 1 +(20/21) + (20/21)2} = 5044
? P(20/21) {1 +20/21 +400/441} = 5044
? P(20/21) {441 + 420 + 400/441} =5044
? P(20/21) (1261/441) =5044
? P = (5044 x 21 x 441)/ (20 x 1261) = Rs. 1852.20
Let MP = ? N
? SP = N x (100 - 10)/100 = ? 9N/10
? CP = (9N/10) x [100/(100 + 20)] = ? 3N/4
Thus, CP has to be increased by = [(N - 3N/4) / (3N/4)] x 100%
= [(N/4) / (3N/4)] x 100 %
= 331/3 %
Let the original price of pant and shirt be = ? N.
? Cost price of point and shirt = [N x (100 - 25)]/100 = ? 3N/4
And selling price of shirt and pant = (3N/4) x (100 + 40)/100 = (3N/4) x (140/100)
= ? 21N/20
Hence, required percentage
= [(21N/20 - N)/ N] x 100 % = 100 / 20 % = 5%
? Marked price of an article = ? 50
? SP of an article = 50 x (100 - 20)/100
= (50 x 80)/100
= ? 40
Hence, cost price of an article = (40 x 100)/(100 + 25)
= (40 x 100)/125
= ? 32
Series pattern
+3, +6, +9, +12, +15, +18
? Missing term = 57 + 18 = 75
Given that : - log10 ( x2 - 6x + 45 ) = 2
? x2 - 6x + 45 = 102 = 100
? x2 - 6x - 55 = 0
? ( x - 11 ) ( x + 5 ) = 0
? x = 11 or x = - 5.
Hence , the values of x are 11 and - 5.
A : B : C = 90 : 75 : 60
B:C = 75 / 60 = ( 75 x 100?75 ) / ( 60 x 100?75) = 100 / 80
Hence, in a game of 100 points, B can give C(100 - 80)
= 20 points.
The man takes 3600 s for 14.4 km
The man will take 88 s for
14.4 x (88/3600) = 352/1000 km = 352 m
Now, circumference of circular field = 352 m
? 2?r = 352 m
2 x (22/7) x r = 352
? r = 56 m
Therefore, area of the field = ?r2
= (22/7) x 56 x 56
= 8 x 22 x 56 m2
= 9856 sq m.
Let the money at 10% be Rs. P, Then the money at 9% is Rs. (2600 - P )
? (P x 10 x 5)/100 = [(2600 - P ) x 9 x 6]/100
or 104 x P = 2600 x 54
or P = (2600 x 54) / 104 = Rs.1350
b = (36 -16)m = 20 m,
h = 8 m.
∴ Volume of the box = (32 x 20 x 8) m3 = 5120 m3.
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