GMAT Discount And Profit Questions With Solutions

Ritu's marks = 56% of 875 = $$\frac{56}{100}$$ = 490

Smita's marks = 92% of 875 = $$\frac{92}{100}$$ = 805

Rina's marks = 634

Total marks = 1929

Average = $$\frac{490 + 805 + 634}{3}$$ = $$\frac{1929}{3}$$ = 643

The time taken by her to type 5000 words essay at a speed of 60words/min = 5000/60=83.33
Now when the speef is increased by 15%, the new speed = 69 words/min|
Time taken by her to type 5000 words essay with this speed = 5000/69 = 72.46
Time difference ~ 10 minutes

Let the total number of children be N.

Number of sweets received by each child = N/5

So, N * N/5 = 605 => $$N^2 = 605 * 5$$ = 3025

So, N = 55

Number of sweets received by each child = 55/5 = 11

Let the number of children be C.
So, the number of sweets that each child gets is 20% of C.
Hence, the total number of sweets there are is 20%*C*C which equals 405.
So, $$20 \%C^2 = 405$$ or $$C^2 = 2025$$

Hence, C=45 and the number of sweets each child gets is 9

Population of the state in 1998 = 42 lakhs.
Population of the state in 2001 = 42*120% = 50.4 lakhs
Population of the state in 2004 = 50.40*120% = 60.48 lakhs
Therefore, population of the state in 2007 = 60.48*1.2 = 72.576 lakhs

Let the money Gaurav inherited be equal to X

So, his hostel expenses equal 0.4X
Expenses on books and stationery is 0.2X
Remaining amount is X - 0.4X - 0.2X = 0.4X

Transport expenses is 50% of 0.4X = 0.2X
Remaining amount is 0.4X - 0.2X = 0.2X

He saves 50% of it, that is 0.1X and it equals 450.
So, the value of X is 450/0.1 = 4500

Let A had Rs.400x initially.
He gave 25% of Rs.400x = Rs.100x to B.
B spent, 30% of Rs.100x = Rs.30x on dinner.
Remaining amount with B = Rs.100x - Rs.30x = Rs.70x.
Amount B kept on savings : spent on book = 5 : 2.
Amount that B saved = $$\dfrac{5}{7}\times70x = Rs.50x$$
Amount spent on book = Rs.20x.
Given, 20x = 460 => x = 23.
Therefore, Amount present with  A initially = Rs.400x = Rs.400$$\times$$23 = Rs.9200.

(682% of 782)$$ \div$$856
= $$\frac{682}{100}\times \frac{782}{856}$$
= 6.23

cost of per sq ft. =Rs 6300

Total cost of 1800 sq ft plot = Rs 6300 × 1800 = 11340000

Booking amount = $$\frac{45}{100}\times11340000$$ = Rs 5103000

Let the salary of A be "a" and the salary of B is "b"

$$\frac{a}{b}$$ = $$\frac{8}{7}$$

Now , After 20% increment, salary of A becomes 1.2a

and after increment of 10% , salary of B becomes 1.1b

Given that $$\frac{1.2a}{1.1b}$$ = $$\frac{96}{77}$$, this implies that $$\frac{a}{b}$$ = $$\frac{8}{7}$$

As both the equations are giving same ratio and in no way the exact values of A and B can be found out .

Hence Option D is the correct answer

Let Ganesh, Jassi and Sukhvinder be represented as G,J and S respectively.
G's monthly Income = $$2\times$$ J's monthly Income
$$\frac{2}{3}$$ rd of J's monthly income = S's monthly income.
S's annual Income = 2,34,000
S's monthly income = 19,500

J's monthly income $$ = \frac{19500\times 3}{2} = 29,250$$

G's monthly income $$ = 2 \times 29250 = 58,500$$.
Hence Option C is the correct answer.

Total marks scored by sushil=103+111+98+110+88=510
Let total max. Marks be x.

Now, it is given that, 85% of x=total marks scored by sushil
I.e, 0.85*x=510
Therefore, x=510*100/85
x=600

This is for 5 subjects altogether.
Therefore, max. Marks for one subject = 600/5 =120 marks.
Option B is the right answer.

Passing percentage=663/1020*100 =65%

Shreya percentage=612/1020*100=60%

Therefore she failed by 5%

Let the total amount be Rs. 'T'

Savings percent = [100 - (23 + 33 + 19 + 16 )]% = 9 %

Given , 9% of T = 504

=> T= 504 * (100/9) = 5600

Expenditure on food and insurance policy together = 56% of 5600

= Rs. $$\frac{56}{100}$$*5600

=Rs. 3136

Let the Total marks is 'T' .

The unreserved guy got 300 but failed by 24 marks i.e. his pass mark is 300 + 24 = 324 .

Given that , the pass mark for unreserved guy is 54% of the total . i.e. $$\frac{54}{100}$$T = 324.
We get , T = 600 .
Also given that the pass mark for reserved guy is 40% of the total .i.e. {40/100} * 600 = 240.

Given that , the salary in august is 6300. And the salary increases by 4% every month. i.e. the new salary will be (1+$$\frac{4}{100}$$) times.
In September , it will be 6300* (1+0.04 ) = 6552

In October , it will be 6552 * (1+0.04) = 6814 rs.

let the number be a

0.37xa= 114.7

a= 310

Now 1.30xa= 1.3x310= 403

Let the number be 'n'.
According to question,
$$\frac{5}{6}$$n-35% of n = 1392.
$$\frac{5}{6}n-\frac{35\times{n}}{100}=1392$$.
$$\frac{29\times{n}}{60}=1392$$.
$$n=2880$$.
55 % of n = $$\frac{55\times2880}{100}$$.
=1584.
Hence, Option B is correct.

Let the amount with Abhay , Vinay, Vinay's mother and grocer be $$A, V , V_m and G $$ respectively
From the data given in question
$$V = \frac{3}{10} A$$
$$ V_m = \frac{2}{3} V = \frac{2}{3} \times \frac{3}{10} A $$
$$C = \frac{5}{8} V_m$$
Vinay's mother has 600 left after giving money to the grocer.
$$\frac{3}{8} V_m =600$$
$$\frac{3}{8} \times \frac{2}{3} \times \frac{3}{10} A =600$$
$$A = 8000$$
Option B is the correct answer.

Shiva gave 20% of her monthly salary to her mother.Remaining is 80%

50% of 80% was invested in insurance and PPF in the ratio of 5:3
PPF = $$\frac{3}{8} \times $$ 40% of salary
Insurance scheme = $$\frac{5}{8} \times $$ 40% of salary
Remaining 40% of 65% was put in Bank account.
20% of his monthly salary and $$\frac{3}{8} \times$$ 40% of salary= 35% of his salary = 12600
Let his monthly salary be Rs.X
35% of X =12600
X = Rs. 36,000
Option A is the nearest answer.

Dhruva gave 35% of her monthly salary to her mother.Remaining is 65%
18% of 65% was given for rent and 42% of 65% was kept for monthly expenses.
Remaining 40% of 65% was put in Bank account.
35% of his monthly salary and 40% of remaining 65% of his monthly salary = 43290
Let his monthly salary be Rs.X
35% of X + 40% of 65% of X =43290
0.35 X + 0.26 X =43290
0.61 X = 43290
X = 43290 / 0.61 $$\approx$$ 71000
Option E is the nearest answer.

Revenue = consumption $$\times$$ tax amount

Let consumption = 10 and tax = 10

=> Revenue = $$10 \times 10 = 100$$

Now, after tax is reduced by 10 %, new tax = $$10 - \frac{10}{100} \times 10$$

= $$10 - 1 = 9$$

Total revenue remains unchanged

=> New consumption = $$\frac{100}{9}$$

$$\therefore$$ % increase in consumption = $$\frac{\frac{100}{9} - 10}{10} \times 100$$

= $$\frac{10}{9} \times 10 = \frac{100}{9}$$

= $$11\frac{1}{9} \%$$

7% of Ms. Sujata monthly's salary = Rs. 2,170

=> Monthly salary = $$\frac{100}{7} \times 2170$$

= Rs. $$31,000$$

Percentage monthly investment by Sujata = 7 + 18 + 6 = 31 %

$$\therefore$$ Sujata's annual investment

= $$12 \times \frac{31}{100} \times 31,000$$

= Rs. $$1,15,320$$

Let Mary's monthly salary = Rs. $$100x$$

Money spent on EMI = $$\frac{15}{100} \times 100x = 15x$$

Amount left = $$100x - 15x = 85x$$

% amount left after paying internet bill and rent = $$100 - (10 + 20) = 70 \%$$

=> Amount left = $$\frac{70}{100} \times 85x = 24,990$$

=> $$\frac{7 \times 17x}{2} = 24,990$$

=> $$x = \frac{24,990 \times 2}{7 \times 17}$$

=> $$x = 210 \times 2 = 420$$

$$\therefore$$ Mary's monthly salary = $$100 \times 42$$ = Rs. $$42,000$$

Let total population of village = 100

=> Number of literate people = $$\frac{64}{100} \times 100 = 64$$

=> No. of literate male = $$\frac{3}{3 + 1} \times 64$$

= $$3 \times 16 = 48$$

=> Number of illiterate people = 100 - 64 = 36

=> No. of illiterate males = $$\frac{5}{5 + 4} \times 36$$

= $$5 \times 4 = 20$$

$$\therefore$$ Required ratio = $$\frac{48}{20}$$

= 12 : 5

Let the original strength = $$100x$$

Men remaining after war = $$100x - \frac{10}{100} \times 100x = 90x$$

Men remaining after diseases = $$90x - \frac{10}{100} \times 90x$$

= $$90x - 9x = 81x$$

Men remaining excluding the disabled = $$81x - \frac{10}{100} \times 81x$$

= $$81x - 8.1x = 72.9x$$

Acc to ques,

=> $$72.9x = 729000$$

=> $$x = \frac{729000}{72.9} = 10000$$

$$\therefore$$ Original strength of men = $$100 \times 10,000 = 1,000,000$$

Let the number of registered voters in the village = $$100x$$

=> No. of voters who cast their votes = $$\frac{63}{100} \times 100x = 63x$$

Let no. of votes received by winning candidate = $$5y$$

=> No. of votes received by losing candidate = $$4y$$

Acc to ques,

=> $$5y = 1120$$

=> $$y = \frac{1120}{5} = 224$$

=> Total votes = $$5y + 4y = 9y$$

= $$9 \times 224 = 2016$$

Now, total casted votes = $$63x = 2016$$

=> $$x = \frac{2016}{63} = 32$$

$$\therefore$$ Number of registered voters = 32 * 100 = 3200

Let monthly salary = Rs. $$100x$$

Amount spent on buying electronic goods = $$\frac{20}{100} \times 100x = 20x$$

Amount spent on repair work = $$\frac{10}{100} \times 100x = 10x$$

Amount left = $$100x - 20x - 10x = 70x$$

Amount invested in fixed deposits = $$\frac{30}{100} \times 70x = 21x$$

=> Amount left = $$70x - 21x = 49x$$

Acc to ques,

=> $$49x = 18963$$

=> $$x = \frac{18963}{49} = 387$$

$$\therefore$$ Monthly salary = 100 * 387 = Rs. 38,700

Let population of village Q one year ago = $$x$$

=> Population of village P one year ago = $$2x$$

=> $$3528 = x (1 + \frac{20}{100})^2$$

=> $$3528 = x (\frac{6}{5})^2$$

=> $$x = \frac{3528 \times 25}{36} = 2450$$

$$\therefore$$ Present population of village P = $$2 \times 2450 = 4900$$

Let monthly salary = Rs. $$100x$$

Amount spent on buying electronic goods = $$\frac{18}{100} \times 100x = 18x$$

Amount spent on repair work = $$\frac{32}{100} \times 100x = 32x$$

Amount left = $$100x - 18x - 32x = 50x$$

Amount invested in fixed deposits = $$\frac{42}{100} \times 50x = 21x$$

=> Amount left = $$50x - 21x = 29x$$

Acc to ques,

=> $$29x = 12325$$

=> $$x = \frac{12325}{29} = 425$$

=> Monthly salary = 100 * 425 = Rs. 42,500

$$\therefore$$ Annual salary = 42500 * 12 = Rs. 5,10,000

Let Sarvagya's salary = Rs. 100

=> Sarthak's salary = $$\frac{40}{100} * 100$$ = Rs. 40

=> Harish's salary = $$\frac{60}{100} * 40$$ = Rs. 24

=> Required % = $$\frac{100 - 24}{24} * 100$$

= 317%

Present population of boys = $$\frac{80}{100} \times 610$$ = 488

=> Number of girls = $$\frac{175}{100} \times 488$$

= $$7 \times 122$$ = 854

Let the original marks of the student = $$100x$$

Marks after re-evaluation = $$\frac{60}{100} \times 100x = 60x$$

=> $$60x = 96$$

=> $$x = \frac{96}{60} = 1.6$$

Thus, original marks = 100 * 1.6 = 160

$$\therefore$$ Marks lost after re-evaluation = 160 - 96 = 64

Expression : $$\frac{32}{100} \times 750 = 240$$

(A) : $$\frac{23}{100} \times 600 = 138$$

(B) : $$\frac{46}{100} \times 207 = 95$$

(C) : $$\frac{98}{100} \times 250 = 245$$

(D) : $$\frac{75}{100} \times 320 = 240$$

=> 32% of 750 < 98% of 250

Total C.P. of 60 kg of sugar

= Rs. $$(36 \times 45 + 24 \times 40)$$

= Rs. $$(1620 + 960)$$ = Rs. $$2580$$

Total S.P. = $$2580 \times \frac{120}{100}$$

= Rs. $$3096$$

$$\therefore$$ S.P. per kg = $$\frac{3096}{60}$$

= Rs. $$51.60$$

Let Sonai's monthly salary be Rs. $$100x$$

Sonali's % monthly expenditure = (15 + 55)% = 70 %

=> Savings % = 100 - 70 = 30 %

Acc to ques,

=> $$\frac{30}{100} * 100x = 12750$$

=> $$x = \frac{12750}{30} = 425$$

=> Sonali's monthly income = $$100x$$

= 100 * 425 = Rs. 42,500

Let Sophia's monthly salary be Rs. $$100x$$

Sophia's % monthly expenditure = (25 + 15 + 35)% = 75%

=> Savings % = 100 - 75 = 25%

Acc to ques,

=> $$\frac{25}{100} * 100x = 9050$$

=> $$x = \frac{9050}{25} = 362$$

=> Sophia's annual income = $$12 * 100x$$ = Rs. $$1200x$$

= 1200 * 362 = Rs. 4,34,400

Let CP of DVD player = Rs. $$100x$$

Price charged by owner = $$\frac{122}{100} * 100x$$ = Rs. $$122x$$

=> $$122x = 10980$$

=> $$x = \frac{10980}{122} = 90$$

$$\therefore$$ CP of DVD player = 100 * 90 = Rs. 9,000

Let the max aggregate marks = $$100x$$

The student gets 4% less than 40% = 36%

=> $$\frac{36}{100} \times 100x = 261$$

=> $$x = \frac{261}{36} = 7.25$$

$$\therefore$$ Total marks = 100 * 7.25 = 725

Let the number be $$x$$

Difference of % = (42 - 28)% = 14%

Acc to ques,

=> $$\frac{14}{100} \times x = 210$$

=> $$x = \frac{210 \times 100}{14} = 1500$$

$$\therefore$$ Required answer

= $$\frac{59}{100} \times 1500 = 885$$

There are 40 students and 5 teachers.

Sweets received by each student = $$\frac{15}{100} * 40$$ = 6

=> Total sweets received by all the students = 40 * 6 = 240

Sweets received by each teacher = $$\frac{20}{100} * 40$$ = 8

=> Total sweets received by both teachers = 5 * 8 = 40

$$\therefore$$ Total sweets distributed in class = 240 + 40 = 280

Let the max marks in the exam = $$100x$$

Passing marks = 310

Now, acc to ques,

=> $$(\frac{28}{100} * 100x) + 93 = 310$$

=> $$28x = 310 - 93$$

=> $$x = \frac{217}{28} = 7.75$$

$$\therefore$$ Max marks = 100 * 7.75 = 775

Let the original fraction be $$\frac{x}{y}$$

Acc to ques,

=> $$\frac{x + \frac{400}{100} * x}{y + \frac{500}{100} * y} = \frac{10}{21}$$

=> $$\frac{5x}{6y} = \frac{10}{21}$$

=> $$\frac{x}{y} = \frac{10}{21} \times \frac{6}{5}$$

=> $$\frac{x}{y} = \frac{4}{7}$$

Let the max marks in the exam = $$100x$$

Passing marks = 380

Now, acc to ques,

=> $$(\frac{32}{100} * 100x) + 76 = 380$$

=> $$32x = 380 - 76$$

=> $$x = \frac{304}{32} = 9.5$$

$$\therefore$$ Max marks = 100 * 9.5 = 950

Let $$x$$ kg of good quality wheat should be added.

Acc to ques,

=> $$\frac{10}{100} \times x = \frac{5}{100} \times (x + 150)$$

=> $$10x = 5x + 750$$

=> $$x = \frac{750}{5} = 150$$ kg

Initially, number of boys = 40 and number of girls = 50

Average weight of boys = 65 kg and average weight of girls = 60 kg

Now, 40% of the girls and 50% of the boys leave

=> Boys left = $$\frac{100 - 50}{100} \times 40 = 20$$

Girls left = $$\frac{100 - 40}{100} \times 50 = 30$$

Since, average weight of the boys and the girls remains constant throughout

$$\therefore$$ New average weight of the class

= $$\frac{(20 \times 65) + (30 \times 60)}{20 + 30} = \frac{1300 + 1800}{50}$$

= $$\frac{3100}{50} = 62$$ kg

Let the five consecutive odd numbers in increasing order = $$(x-4) , (x-2) , (x) , (x+2) , (x+4)$$

Sum of these numbers = $$(x-4) + (x-2) + (x) + (x+2) + (x+4) = 195$$

=> $$5x = 195$$

=> $$x = \frac{195}{5} = 39$$

Thus, the odd numbers are = 35 , 37 , 39 , 41 , 43

Let another series of even numbers in increasing order = $$(y-4) , (y-2) , (y) , (y+2) , (y+4)$$

Also, $$37 = (y + 2) - 9$$

=> $$y = 37 + 9 - 2 = 44$$

Thus, second lowest number of the even series = 44 - 2 = 42

$$\therefore$$ 40% of 42 = $$\frac{40}{100} \times 42 = 16.8$$

Initially, number of boys = 80 and number of girls = 75

Average weight of boys = 64 kg and average weight of girls = 70 kg

Now, 60% of the girls and 30% of the boys leave

=> Boys left = $$\frac{100 - 30}{100} \times 80 = 56$$

Girls left = $$\frac{100 - 60}{100} \times 75 = 30$$

Since, average weight of the boys and the girls remains constant throughout

$$\therefore$$ New average weight of the class

= $$\frac{(56 \times 64) + (30 \times 70)}{56 + 30} = \frac{3584 + 2100}{86}$$

= $$\frac{5684}{86} = 66.09$$ kg

Let Prerna's salary = $$Rs.100x$$

Amount which Prerna decided to donate = $$\frac{15}{100} \times 100x=Rs.15x$$

Amount which she actually donated = Rs. 1896

According to ques,

=> $$\frac{80}{100} \times 15x=1896$$

=> $$\frac{4}{5} \times 15x=1896$$

=> $$12x=1896$$

=> $$x=\frac{1896}{12}=158$$

$$\therefore$$ Prerna's salary = $$100 \times 158=Rs.$$ $$15,800$$

=> Ans - (C)

Let the number = $$100x$$

Acc. to ques,

=> $$(\frac{42}{100} \times 100x) - (\frac{28}{100} \times 100x) = 210$$

=> $$42x - 28x = 14x = 210$$

=> $$x = \frac{210}{14} = 15$$

=> Number = $$100 \times 15 = 1500$$

$$\therefore$$ 59 % of the number = $$\frac{59}{100} \times 1500$$

= 885