GMAT Ratio and Proportion Questions With Solutions

Parineeta’s present age = 33 - 9 = 24 yrs.
Manisha’s present age = 24 - 9 = 15yrs
Difference between Manisha and Deepali's prresent ages = Parineeta's present age
Deepali’s present age = Parineeta's present age + Manisha's present age = 24 + 15 = 39 yrs.
Ratio of the present age of Manisha and Deepali = 15 : 39 = 5 : 13
x = 13

Total marks = 24 x 56 = 1344
Sum of actual marks = 1344 - (44 + 45 + 61) + (48 + 59 + 67) = 1368
Actual Average = $$\frac{1368}{24}$$ = 57

Let the amount be x
If the amount is divided into 9 parts , each part = $$\frac{x}{9}$$
B's supposed amount = 3 x $$\frac{x}{9}$$ = $$\frac{x}{3}$$
But B's actual amount = 2 x $$\frac{x}{14}$$ = $$\frac{x}{7}$$
$$\frac{x}{3}$$ - $$\frac{x}{7}$$ = 40
x = 210

Required ratio = (40-34)/(34-30)=6/4=3/2

let’s say present age of son is x, hence father’s age will be 3x
4 years later, (x+4)*2 +13 = 3x+4 ; x=17 and 3x=51

The total amount distributed among officers = 45*25000= 11,25,000

If the total amount is x, then this amount is (5/8)x=11,25,000

Hence, x=18,00,000

Amount taken by M = $$\frac{3}{4}*44352$$ = Rs 33264

Remaining amount = 44352 - 33264 = Rs 11088

Amount taken by N = $$\frac{1}{6}*11088$$ = Rs 1848

Remaining amount = 11088 - 1848 = Rs 9240

O gets $$\frac{3}{7}$$th share of Rs 9240 => $$\frac{3}{7}$$*9240 = Rs 3960

Since the ages of A, B, C, D are consecutive

Let the ages of A, B, C, D be n, n+1,n+2,n+3

$$\frac{n+n+1+n+2+n+3}{4} = 49.5$$

4n+6 = 49.5*4 = 198

4n = 192

n = 48

Ages of A, B, C, D = 48, 49, 50 ,51

Product of ages of B and D = 49*51 = (50-1)(50+1) =$$50^2-1$$ = 2500-1 = 2449.

Let the amount recieved by each clerk be x

$$\frac{45*25000}{80*x} = \frac{5}{3}$$

x = 8437.5

Total profit = (8437.5*80)+(25000*45) = 18 lakh

Remaining amount of amount in percent = 100 - 10 - 12 -32 = 46
The amount remaining with him = 46% of (54550*12)
= 3,01,116
The amount of invested in full deposit = 0.5*30116 = 1,50,558
The correct answer is option A.

Let the numerator and denominator be x and y.
(x/y) = (16/21)
Now, if the first number is increased by 30% and the second number is decreased by 20% then the ratio is $$\frac{1.3*16}{0.8*21}$$
Which equals $$\frac{208}{168}$$
Which equals $$\frac{26}{21}$$

Option B is the correct answer.

Let Ravi's age be x and Simar's be y.
x=y+4
After 4 years,
(x+4)/(y+4) = 9/8
8x-9y = 4
After solving we get Ravi's age be 32 years.
15 years before his age is 32-17=17 years.

Number of litres purchased in 45 days = 0.550*45 = 24.75 litres
Amount he has to pay at the rate of Rs 44/litres = 24.75*44 = Rs. 1089

Option D is correct.

Let the average of the six batsmen be x.

So, runs scored by the captain = x+30

Total score of the team = 6x + x + 30 = 7x + 30 = 310 => x = 40

So, the score of the captain = 70

Given , the ratio of investments of A,B,C is 4:6:5 . So , let them be 4k , 6k, 5k.
Given that A started the business. After 4 months from the start of the business, B and C joined .
i.e. A worked for 12 months while B , C worked for 8 months.
A’s share in annual profit was Rs. 250 more than C’s share.
So , A = C +250
12(4k) = 8(5K) + 250
48K = 40K + 250
k = 31.25
Total amount = 12(4K) + 8(6k) +8(5K)
= 136K
= 136*31.25
= 4250

Let the investment of A, B , and C be y , 3y , 5y

after 4 months A invested same amount and hence for next 8 months in a year his invest is proportional to 2y . where as B and C withdrew half of their amount

So their effective amount after 4 months are 1.5y and 2.5y respectvely

So after a year their profits will be shared in ratio = A:B:C = (4y+16y) : (12y + 12y) : (20y + 20y)

A:B:C = 20:24:40 = 5:6:10

Let the monthly salary of Om be 7x
Monthly salary of Pihu = 9x
Savings of Om = $$7x \times \frac{20}{100} = \frac{7x}{5}$$
Savings of Pihu = $$9x \times \frac{40}{100} = \frac{18x}{5}$$
Given that,
$$\frac{1}{2}\times \frac{7x}{5} + \frac{7}{9}\times \frac{18x}{5} = 17500$$
$$x = 5000$$
Pihu's monthly salary $$= 9x = $$Rs. 45000

let the monthly salary of neil and dipti be 5y and 6y

Now as neil save 40 % of monthly salary and dipti save 25% .so saving of neil = 0.4 x 5y = 2y

Savings of dipti = 0.25 x 6y = 1.5 y

It is given that 2y - 1.5y = 1750

0.5y = 1750

y = 3500

So salary of neil = 5y = 17500

Total amount = Rs 44352

M's share= $$\frac{3}{8}\times44352$$ = Rs 16632

Remaining amount = 44352 - 16632 = Rs 27720

N's Share = $$\frac{1}{6}\times27720$$ = Rs 4620

Remaining amount = 27720 - 4620 = Rs 23100

Ratio in which the remaining amount is divided between O: P = 3:2

P's share = $$\frac{2}{5}\times23100$$ = Rs 9240

As the ratio A:B is given as 3:2 , lets assume that A is 3y and B is 2y

It is given that 6$$(2y)^2$$ = 540 + $$(3y)^2$$

24$$(y)^2$$ - 9$$(y)^2$$ = 540

15$$(y)^2$$ = 540

$$(y)^2$$ = 36

y = 6

hence B = 2 y = 2 x 6 = 12

M's share = 3/8
N's share = 1/6 of remaining = 1/6 of 5/8 = 5/48

Remaining share = 1 - $$\frac{18+5}{48}$$

O's share = $$\frac{3}{7}$$ * $$\frac{25}{48}$$ * 44352

Hence Rs 9900

Let 'x' be the original number of students.

$$\frac{3200}{x-80}-\frac{3200}{x}=2$$

Substituting options, we see that only x=400 satisfies the equation. Hence, option D is the right answer.

Let the no. of sweets be 'x'.
According to question
$$\frac{x}{495}-\frac{x}{495+396}=4$$.
$$\frac{x}{495}-\frac{x}{891}=4$$.
$$x=4455$$.
Hence, Option C is correct.

Let total salary of the man be Rs. $$100x$$

Total amount given to his son and daughter = $$\frac{20}{100} * 100x = 20x$$

Amount given to daughter = $$\frac{2}{5} * 20x = 8x$$

Amount invested in LIC = $$8x * 2 = 16x = ?$$

Remaining amount = $$100x - 20x - 16x = 64x$$

Amount given to wife = $$\frac{1}{4} * 64x = 16x$$

=> Remaining amount = $$64x - 16x = 48x$$

Acc to ques : $$48x = 16800$$

=> $$x = 350$$

$$\therefore$$ Amount invested in LIC = 16 * 350 = Rs. 5600

Let the number of boys initially = $$31x$$

=> Number of girls = $$23x$$

Acc to ques,

=> $$\frac{31x}{23x + 75} = \frac{124}{107}$$

=> $$3317x = 2852x + 9300$$

=> $$3317x - 2852x = 465x = 9300$$

=> $$x = \frac{9300}{465} = 20$$

=> Number of boys = $$31 \times 20 = 620$$

Number of girls = $$23 \times 20 + 75 = 535$$

$$\therefore$$ Number of girls who should join the college to make number of boys and girls equal

= $$620 - 535 = 85$$

Total profit = Rs. 13,050

Pihu's share = Rs. 6,750

=> Rani's share = 13,050 - 6,750 = Rs. 6,300

Let Pihu withdrew after $$x$$ months

=> Ratio of investments of Pihu and Rani

= $$(9,000 \times x) : (6,300 \times 24)$$

= $$5x : 84$$

Acc to ques,

=> $$\frac{5x}{84} = \frac{6750}{6300}$$

=> $$\frac{5x}{84} = \frac{15}{14}$$

=> $$x = \frac{15}{14} \times \frac{84}{5}$$

=> $$x = 3 \times 6 = 18$$ months

Let Riddhi's allowance = Rs. $$8x$$

=> Siddhi's allowance = Rs. $$(3800 - 8x)$$

Allowance paid as tuition fees by

Riddhi = $$\frac{2}{8} \times 8x = 2x$$

Siddhi = $$\frac{2}{8} \times (3800 - 8x) = 950 - 2x$$

Acc to ques,

=> $$(950 - 2x) - (2x) = 80$$

=> $$950 - 4x = 80$$

=> $$4x = 950 - 80 = 870$$

=> $$x = \frac{870}{4}$$

$$\therefore$$ Riddhi's allowance = $$8 \times \frac{870}{4}$$

= $$2 \times 870$$ = Rs. $$1,740$$

Let A ,B and C respectively = $$x , 2x , 3x$$

=> Sum of A , B and C = $$x + 2x + 3x = 3 \times 600$$

=> $$6x = 1800$$

=> $$x = \frac{1800}{6} = 300$$

=> $$A = 300 , B = 600 , C = 900$$

Value of A after increment = $$\frac{110}{100} \times 300 = 330$$

Value of B after increment = $$\frac{80}{100} \times 600 = 480$$

New value of average = $$\frac{105}{100} \times 600 = 630$$

Acc to ques,

=> $$\frac{330 + 480 + C'}{3} = 630$$

=> $$810 + C' = 3 \times 630 = 1890$$

=> $$C' = 1890 - 810 = 1080$$

$$\therefore$$ C is increased by = 1080 - 900 = 180

Let B invested the money for = $$x$$ months

=> Ratio of investments of A : B = $$(5 \times 7) : (7 \times x)$$

= $$5 : x$$

Acc to ques

=> $$\frac{5}{x} = \frac{1}{2}$$

=> $$x = 5 \times 2 = 10$$ months

Let the number of boys = $$x$$ and girls = $$y$$

=> Total weight of boys = $$30x$$

Total weight of girls = $$20y$$

Acc to ques,

=> $$30x + 20y = 23.25 (x + y)$$

=> $$30x + 20y = 23.25x + 23.25y$$

=> $$6.75x = 3.25y$$

=> $$\frac{x}{y} = \frac{325}{675} = \frac{13}{27}$$

Hence, possible strength of boys and girls respectively can be 13 and 27

Let the number of boys = $$x$$

=> Number of girls = $$(x + 16)$$

=> Total students = $$x + x + 16 = 40$$

=> $$2x = 40 - 16 = 24$$

=> $$x = \frac{24}{2} = 12$$

=> Number of girls = 12 + 16 = 28

=> Total weight of 28 girls = 28 * 45 = 1260

Total weight of all 40 students = 40 * 47.7 = 1908

=> Total weight of the boys = 1908 - 1260 = 648

$$\therefore$$ Average weight of boys = $$\frac{648}{12} = 54$$ kg

Total initial weight of the class = 44 * 38 = 1672 kg

Total weight of the class after 12 new students joined = $$(38+12) \times (44+4.68)$$

= 50 * 48.68 = 2434 kg

=> Total weight of the new 12 students = 2434 - 1672 = 762 kg

$$\therefore$$ Average weight of new 12 students = $$\frac{762}{12}$$

= 63.5 kg

Let number of students in each class = $$x$$

Total number of students in the 4 class = $$4x$$

Acc. to ques,

=> $$(4x \times 6) + 94 = 1414$$

=> $$24x = 1414 - 94 = 1320$$

=> $$x = \frac{1320}{24} = 55$$

Let the five consecutive odd numbers in Set A be $$(x) , (x+2) , (x+4) , (x+6) , (x+8)$$

Sum of these numbers = $$(x) + (x+2) + (x+4) + (x+6) + (x+8) = 185$$

=> $$5x + 20 = 185$$

=> $$5x = 185 - 20 = 165$$

=> $$x = \frac{165}{5} = 33$$

=> Highest number in Set A = $$(x + 8) = 33 + 8 = 41$$

=> Smallest number in Set B = $$41 + 13 = 54$$

Now, 4 consecutive even numbers in Set B starting from 54 = 54,56,58,60

Sum of numbers in Set B = 54 + 56 + 58 + 60 = 228

$$\therefore$$ Required average = $$\frac{228}{4} = 57$$

Let Arati's age 5 years ago = $$2x$$

and Kiran's age 5 years ago = $$3x$$

=> Kiran's present age = $$(3x + 5) = 20$$ [Given]

=> $$x = \frac{15}{3} = 5$$

=> Arati's present age = $$2x + 5$$

= 2*5 + 5 = 15 years

Now, after 5 years Sonal’s age will be twice the present age of Arati

=> Sonal's age after 5 years = 2*15 = 30 years

and Sonal's present age = 30 - 5 = 25 years

$$\therefore$$ Rakshak's present age = 2*25 = 50 years

A : B = 7 : 8

B : C = 4 : 9 = 8 : 18

=> A : B : C = 7 : 8 : 18

Let total profit be $$P$$

=> B's share = $$\frac{8}{33} \times P = 7104$$

=> $$P = \frac{7104 \times 33}{8}$$

Now, C's share = $$\frac{18}{33} * P$$

= $$\frac{18}{33} \times \frac{7104 * 33}{8}$$

= $$18 * 888 = $$Rs. $$15,984$$

Let the number of 1 rupee, 50 paise and 25 paise coins = $$10x , 8x , 5x$$

=> Total amount = $$(10x \times 1) + (8x \times \frac{1}{2}) + (5x \times \frac{1}{4}) = 976$$

=> $$14x + \frac{5x}{4} = 976$$

=> $$\frac{61x}{4} = 976$$

=> $$x = \frac{976 \times 4}{61} = 64$$

$$\therefore$$ No. of 25 paise coins = 5 * 64 = 320

Sum of first 22 numbers = 22 * 61 = 1342

Sum of last 19 numbers = 19 * 44 = 836

Numbers remaining = 68 - 22 - 19 = 27

=> Sum of remaining 19 numbers = 3474 - 1342 - 836 = 1296

$$\therefore$$ Required average = $$\frac{1296}{27}$$

= 48

Total weight of original 40 students = $$40 \times 55 = 2200$$ kg

Weight of 6 students who left = $$6 \times 52 = 312$$ kg

Weight of 6 new students who joined = $$6 \times 42 = 252$$ kg

=> New weight of new group of 40 students = $$(2200 - 312 + 252)$$ kg

= $$2140$$ kg

$$\therefore$$ Required Average = $$\frac{2140}{40}$$

= $$53.5$$ kg

Case I : Milkman sells 120 litres of milk for Rs. 3,360/-

=> S.P. per litre = Rs. $$\frac{3360}{120}$$

= Rs. 28

Case II : Milkman sells 240 litres of milk for Rs. 6,120/-

=> S.P. per litre = Rs. $$\frac{6120}{240}$$

= Rs. 25.5

$$\therefore$$ Required discount = Rs. (28 - 25.5) = Rs. 2.5

Let the 5 consecutive even numbers be = $$(x) , (x+2) , (x+4) , (x+6) , (x+8)$$

Sum of these 5 consecutive numbers = 52 * 5 = 260

=> $$(x) + (x+2) + (x+4) + (x+6) + (x+8) = 260$$

=> $$5x = 260 - 20 = 240$$

=> $$x = 48$$

=> B = $$x + 2$$ = 48 + 2 = 50

E = $$x + 8$$ = 48 + 8 = 56

$$\therefore$$ B $$\times$$ E = 50 * 56 = 2800

Let the present age of Aarzoo = $$11x$$ years

Present age of Arnav = $$13x$$ years

Acc to ques,

=> $$\frac{11x + 7}{13x + 7} = \frac{20}{23}$$

=> $$260x + 140 = 253x + 161$$

=> $$7x = 21$$

=> $$x = 3$$

$$\therefore$$ Difference between their ages

= $$13x - 11x = 2x$$

= 2 * 3 = 6 years

Let B remained in the business for $$x$$ months.

Amount invested by A = Rs. 3600

Amount invested by B = Rs. 4000

Ratio of share in profit received by A and B

=> $$\frac{3600 \times 12}{4000 \times x} = \frac{6}{5}$$

=> $$\frac{9 \times 2}{10 \times x} = \frac{1}{5}$$

=> $$x = \frac{18 \times 5}{10} = 9$$

$$\therefore$$ B joined the business after = $$12 - 9 = 3$$ months from the start of the business.

Let the unit's digit of the number = $$y$$ and ten's digit = $$x$$

=> Number = $$10x+y$$

Sum of digits = $$x+y=15$$ ----------(i)

According to ques, => $$(10x+y)-(10y+x)=9$$

=> $$9x-9y=9$$

=> $$x-y=1$$ ----------(ii)

Adding equations (i) and (ii), => $$2x=15+1=16$$

=> $$x=\frac{16}{2}=8$$

Substituting it in equation (i), => $$y=15-8=7$$

$$\therefore$$ Original number = 87

=> Ans - (C)

Let number of males = $$x$$

Average age of males = 20 years

=> Sum of age of males = $$20 \times x = 20x$$ years

Sum of age of females = 240 years

Acc. to ques, => $$\frac{(20 x) + (240)}{15 + x} = 18$$

=> $$20x + 240 = 270 + 18x$$

=> $$20x - 18x = 2x = 270 - 240 = 30$$

=> $$x = \frac{30}{2} = 15$$

Let the smaller number be $$x$$ and the other number be $$y$$

Acc. to ques, => $$x + 2y = 2x + 16$$

=> $$2y - x = 16$$ -------------(i)

Also, difference between the numbers = $$y - x = 6$$ ------------(ii)

Multiplying eqn(ii) by 2 and then subtracting it from (i), we get :

=> $$(2y - 2y) + (-x + 2x) = 16 - 12$$

=> $$x = 4$$

Let the three positive numbers be $$x,y,z$$     (where $$x < y < z$$)

Average of the three numbers = $$\frac{x + y + z}{3}$$

Acc. to ques, => $$\frac{1}{3} \times (\frac{x + y + z}{3}) = z - 8$$

=> $$x + y + z = 9z - 72$$

=> $$x + y = 8z - 72$$

Dividing both sides by 2, we get :

=> $$\frac{x + y}{2} = 4z - 36$$

Also, average of the lowest and the second lowest number is 8, => $$\frac{x + y}{2} = 8$$

=> $$4z - 36 = 8$$

=> $$4z = 8 + 36 = 44$$

=> $$z = \frac{44}{4} = 11$$

Let monthly salary of Pia = $$Rs. 500x$$

=> Monthly salary of Som = $$Rs. 400x$$

Amount given by Pia to her mother = $$\frac{3}{5} \times 500x = 300x$$

Amount for her sister tuition fees = $$\frac{15}{100} \times 500x = 75x$$

Amount for loan = $$\frac{18}{100} \times 500x = 90x$$

=> Amount left = $$500x - (300x + 75x + 90x) = 2100$$

=> $$500x - 465x = 35x = 2100$$

=> $$x = \frac{2100}{35} = 60$$

$$\therefore$$ Som's salary = $$400 \times 60 = Rs. 24,000$$

Let the amount received by Pinku, Rinku and Tinku respectively = $$7x , 8x , 6x$$

Total amount = $$7x + 8x + 6x = 4200$$

=> $$21x = 4200$$

=> $$x = \frac{4200}{21} = 200$$

=> Amount with Pinku = 7*200 = 1400

Rinku = 8*200 = 1600

Tinku = 6*200 = 1200

If Rs. 200 is added to their shares, respective ratio

= $$(1400 + 200) : (1600 + 200) : (1200 + 200)$$

= $$1600:1800:1400 = 8:9:7$$