A man buys ‘a’ kg of mustard, and their seeds weigh 40% of their total weight. The seed crushed for first time to yield oil, which is 20% of weight of seed. While the crushed waste material is treated once again & it yields 10% of oil by weight. If the amount of oil obtained at 2nd time is 25 kg. Find the value of a (in kg).
Weight of seeds = `(40a)/100` After crushing them for first time, oil collected = `20/100` `xx` `(4a)/10` = `(8a)/100` Remaining material is `(40a)/100` - `(8a)/100` = `(32a)/100` When it is again treated, oil obtain is `10/100` `xx` `(32a)/100` = `(320a)/10000` Now, `(320a)/10000` = 25 kg a = 781.25 kg
In a village, two contestants (C & D) are contesting in an election. 35% of the registered voters cast their votes in the election and C wins the election by 700 votes. If C had received 20% less votes, C’s votes would have been equal to D’s votes. How many registered voters are there in the village?
The marks scored by a student in three subjects are in the ratio 3 : 6 : 2. Student scored an overall aggregate of 55% in the exam. If the maximum marks in each subject are the same, if the passing marks is 33% in how many subjects did student failed ?
Monthly income of Amit and Bhola is Rs. (x + 120) and Rs. x, respectively. If monthly income of Amit is increased by 16% while that of Bhola is decreased by 32% then the ratio of their respective income becomes 9:5, respectively. Find the monthly income of Amit.
P gave 60% of the amount he had to Q. Q gave 2/7th of that amount to R. After paying Rs.250 to the shopkeeper out of the amount he gets from Q, R is now left with Rs.1250. How much amount did P have?
Let total amount P have Rs. x R have = Rs.1250 Now according to the question, => [x × (60/100) × (2/7)] – 250 = 1250 => x × (60/100) × (2/7) = 1500 => x = 8750 Therefore, P have Rs.8750.
A and B together have total of Rs.4000 out of which they donated 20% to the orphanage school. The remaining amount is to be then redistributed between them in such a manner that A gets 50% more amount than B. If the amount received by A is Rs.2X, then find the value of [(X/12) + 8].
Let the amount received by B be Rs.P. Therefore, amount received by A = 1.50 x P = Rs.1.5P According to the question, => (P + 1.5P) = 4000 x 0.8 => 2.5P = 3200 => P = 1280 Share of A = 3200 – 1280 = Rs.1920 So, 2X = 1920 => X = 960 Required value = [(X/12) + 8] = 80 + 8 = 88
The total population of villages A and B together is 75000. ‘y’% and (y+15)% of the population of village A and B are working professionals. The ratio between the population of village A and C is 9:10 respectively. If 70% of the population of village C are working professionals which is equal to 35000 and the total non working professionals population of village B and A together is 25500, then find out the value of ‘y’.
If 70% of the population of village C are working professionals which is equal to 35000.
70% of the population of village C = 35000
population of village C = 50000
The ratio between the population of village A and C is 9:10 respectively.
population of village A = (50000/10)x9 = 45000
The total population of villages A and B together is 75000.
population of village B = 75000-45000 = 30000
‘y’% and (y+15)% of the population of village A and B are working professionals. The total non working professionals population of village B and A together is 25500.
45000 of y% + 30000 of (y+15)% = 75000-25500
450 x y + 300 x (y+15) = 49500
3 x y + 2 x (y+15) = 330
3y+2y+30 = 330
3y+2y = 330-30
5y = 300
Value of ‘y’ = 60
Anil spends 20% of the monthly income on the reconstruction of his house, 15% on basic needs, 10% of the remaining on travelling. If he spends (100/9)% of the remaining income on food and he saves Rs.7280, then find the monthly income of Anil.
Let monthly income of Anil = Rs. 100x Amount spend on reconstruction and on basic needs = 100x × [(20 + 15)/100] = 35x Amount spend on travelling = (100x – 35x) × (10/100) = 6.5x Amount spend on food = (100x – 35x – 6.5x) × (1/9) = 6.5x According to the question, => 100x – 35x – 6.5x – 6.5x = 7280 => x = 140 Therefore, his monthly income = 100 × 140 = Rs.14000
In the given expression a2bc3, the value of ‘a’ and ‘b’ increases 20% and 30% individually and the value of ‘c’ decreases 16.67%, then what will be the % increase in the given expression.
ATQ, we can say that 20% = 1/5 30% = 3/10 16.67% = 1/6 ⇒ [{(6/5)2 × (13/10) × (5/6)3} – 1] × 100 ⇒ [{(36 × 13 × 125)/(25 × 10 × 216)} – 1] × 100 ⇒ (1/12) × 100 ⇒ (25/3)% ∴ The expression a2bc3, increases by (25/3)%
Rohit spent (R-4)% of his monthly income on food. Out of the remaining (R+7)% was spent on travelling. (1/6) of the remaining was spent on education. (2R–1)% of the remaining was spent on house rent and after that the remaining amount was saved by him. If the annual savings of Rohit is Rs. 402480, then find out the monthly expenditure on travelling. It is assumed that the monthly salary of Rohit is 1600 times of the difference between the both of the roots of ‘Y’.
(Y2 - 540Y + 72000 = 0)
(Y2 - 540Y + 72000 = 0) Y2 - 540Y + 72000 = 0 Y2 - (300+240)Y + 72000 = 0 Y2 - 300Y - 240Y + 72000 = 0 Y(Y- 300) - 240(Y- 300) = 0 (Y - 300) (Y - 240) = 0 Y = 300, 240 It is assumed that the monthly salary of Rohit is 1600 times of the difference between the both of the roots of ‘Y’. monthly salary of Rohit = 1600x(300-240) = 1600x60 = 96000 Rohit spent (R-4)% of his monthly income on food. Out of the remaining (R+7)% was spent on travelling. (1/6) of the remaining was spent on education. (2R–1)% of the remaining was spent on house rent and after that the remaining amount was saved by him. If the annual savings of Rohit is Rs. 402480. 96000 of [100-(R-4)]% of [100-(R+7)]% of [1-(1/6)] of [100-(2R–1)]% = 402480/12 96000 x [100-R+4]% x [100-R-7]% x (5/6) x [100-2R+1]% = 33540 0.096 x [100-R+4] x [100-R-7] x (5/6) x [100-2R+1] = 33540 0.016 x [104-R] x [93-R] x 5 x [101-2R] = 33540 0.08 x [104-R] x [93-R] x [101-2R] = 33540 [104-R] x [93-R] x [101-2R] = 419250 (−R+18) (2R 2 − 459R + 30979) = 0 So R = 18 Monthly expenditure on travelling = 96000 of [100-(R-4)]% of (R+7)% Put the value of ‘R’ in the above equation. = 96000 of [100-(18-4)]% of (18+7)% = 96000 of [100-14]% of 25% = 96000 of 86% of 25% = 96000 x (86/100) x (25/100) = Rs. 20640
