SSC CGL RATIO AND PROPORTION MCQ ENGLISH

Question – 1 : If 50% of a certain number is equal to one-third of the second number. Then find the ratio between 2 numbers.

Answer – (A) : 02:03

Answer – (B) : 01:01

Answer – (C) : 03:02

Answer – (D) : 02:03

View Answer

Correct Answer : (D)

Explanation : As per the given data,50% of a certain number is equal to one-third of the second number.Let the 1st number be X and 2nd number be Y .X2=Y3⇒3X2=YTo find ratio in between 2 numbers,X:Y=2:3

Question – 2 : In a class, 58 of the total students are boys. Among boys 1725 are taller than 160 cm. If 40 boys are shorter than 160 cm, find the total number of students in the class.

Answer – (A) : 200

Answer – (B) : 160

Answer – (C) : 200

Answer – (D) : 240

View Answer

Correct Answer : (C)

Explanation : Let the total number of students in class be Y .Number of boys =58×YNumber of boys shorter than 160 cm =(1−1725)×58×Y=825×58×Y825×58×Y=40∴Y=40×85×258∴Y=200

Question – 3 : If the ratio of A:B is 3:6 and ratio of B:C is 7:28 , then find the ratio of A:B:C .

Answer – (A) : 01:02:08

Answer – (B) : 01:02:08

Answer – (C) : 02:01:08

Answer – (D) : 06:07:28

View Answer

Correct Answer : (B)

Explanation : Given:A:B=3:6=1:2⋯(1)B:C=7:28=1:4⋯(2)To equate the value of B in both equations we will multiply equation (2) by 2 so we get,⇒B:C=2:8⇒A:B:C=1×2:2×2:2×8⇒A:B:C=2:4:16=1:2:8∴A:B:C=1:2:8

Question – 4 : Two numbers are in the ratio of 5:7 . If 9 is subtracted from each of them, their ratio becomes 7:11 . The difference in the numbers is:

Answer – (A) : 12

Answer – (B) : 12

Answer – (C) : 18

Answer – (D) : 15

View Answer

Correct Answer : (B)

Explanation : Given:Two numbers In ratio =5:7Let the number be 5x and 7xEach of them is subtracted =9Then their ratio =7:11Now,⇒(5x−9)(7x−9)=711 (by cross multiplication)Then,11(5x−9)=7(7x−9)⇒55x−99=49x−63⇒55x−49x=99−63⇒6x=36⇒x=6Difference =7x−5x=2xPut the value of x=2×6=12

Question – 5 : Two numbers A and B are in the ratio 2:3 and that of B and C is 3:5 . If the sum of all the three numbers is 900 , then the first number is what percentage of the third number?

Answer – (A) : 40%

Answer – (B) : 50%

Answer – (C) : 40%

Answer – (D) : 78%

View Answer

Correct Answer : (C)

Explanation : Given:A:B=2:3B:C=3:5A+B+C=900Now,Ratio of number A and B is 2:3 .Ratio of number B and C is 3:5 .Ratio of A.B and C is 2:3:5 .Let A , B and C is 2x,3x and 5x .According to the question,2x+3x+5x=900⇒x=90∴A,B and C are 180,270,450 .Required Percentage =(180450)×100=40%∴A is 40% that of C .

Question – 6 : Two equal vessels are filled with a mixture of water and glycerine in the ratio 2:3 and 5:4 respectively if the mixtures are poured into a third vessel, the ratio of water and glycerine in the third vessel will be?

Answer – (A) : 43:47:00

Answer – (B) : 20:21

Answer – (C) : 15:17

Answer – (D) : 43:47:00

View Answer

Correct Answer : (D)

Explanation : Given:Two equal vessels are filled with a mixture of water and glycerine in the ratio 2:3 and 5:4 respectively.Let the initial quantity of both mixtures be:LCM of [(2+3),(5+4)]=45 litresSo, the amount of water in the first vessel =(25)×45=18 litresThe amount of glycerine in the first vessel =(35)×45=27 litresSo, the amount of water in the second vessel =(59)×45=25 litresThe amount of glycerine in the second vessel =(49)×45=20 litresWhen the two mixtures are mixed together,The resulting quantity of water =18+25=43 litresThe resulting quantity of glycerine =27+20=47 litres∴ The resulting ratio of water and glycerine is 43:47 .

Question – 7 : A daughter’s and her mother’s age are in the ratio 4:7 respectively. The product of their ages is given to be 1792 . What will be the ratio of mother to daughter’s age after 7 years?

Answer – (A) : 21:13

Answer – (B) : 21:13

Answer – (C) : 22:13

Answer – (D) : 24:13:00

View Answer

Correct Answer : (B)

Explanation : Given:The ratio of daughter and mother’s age =4:7Product of their ages =1792Now,Let the present age of the daughter and mother be 4x and 7x respectively.4x×7x=1792⇒28×2=1792⇒x2=179228⇒x2=64⇒x=8Required Ratio:⇒(7x+7):(4x+7)⇒(56+7):(32+7)⇒63:39⇒21:13

Question – 8 : A man had Rs. 3000 with him. He kept Rs. 250 with him and distributed the rest. The ratio of A and B was 5:6 . The ratio of B and C was 3:4 and the ratio of C and D was 4:3 . Find the amount which C got.

Answer – (A) : Rs. 880

Answer – (B) : Rs. 875

Answer – (C) : Rs. 880

Answer – (D) : Rs. 900

View Answer

Correct Answer : (C)

Explanation : Given:Total amount = Rs. 3000Amount which man kept with himself = Rs. 250Ratio of amount between A and B=5:6Ratio of amount between B and C=3:4Ratio of amount between C and D=4:3Amount which is for distribution = Rs. (3000–250)= Rs. 2750According to the question,A:B=(5:6)×1=5:6B:C=(3:4)×2=6:8C:D=(4:3)×2=8:6So, A:B:C:D=5:6:8:6Now, share of C =825 × Rs. 2750= Rs. 880∴ The amount what C got is Rs. 880

Question – 9 : In what ratio must rice at Rs. 9.40 per kg be mixed with rice at Rs. 11.80 per kg, so that the mixture is worth Rs. 10 per kg?

Answer – (A) : 03:01

Answer – (B) : 06:07

Answer – (C) : 08:05

Answer – (D) : 03:01

View Answer

Correct Answer : (D)

Explanation : First of all, we use given quantities and place them in the formula, to know the required quantities.Cost of 1 kg rice of first type = Cost price of dearer =d= Rs. 11.80Cost of 1 kg rice of second type = Cost price of cheaper =c= Rs. 9.40Desired cost of 1 kg of the mixture = Mean price =m= Rs. 10Required rate = Quantity of Cheaper Quantity of Dearer=d−mm−c=11.80−1010−9.40=1.800.60=31∴ The ratio of the mixture should be 3:1 .

Question – 10 : A bag contains 1 Rs., 2 Rs. and 5 Rs. coins in the ratio 5:3:2 amounting to Rs. 525 . Find the number of coins of each type respectively.

Answer – (A) : 125,75,50

Answer – (B) : 125,75,50

Answer – (C) : 50,75,125

Answer – (D) : 150,25,75

View Answer

Correct Answer : (B)

Explanation : Let the ratio be x .Number of 1 Rs. coin =5xNumber of 2 Rs. Coin =3xNumber of 5 Rs. Coin =2xSo, total amount =5x×1+3x×2+2x×5=21xBut the given amount =525So, 21x=525Thus, x=25So the number of 1 Rs. coin =25×5=125So the number of 2 Rs. Coin =25×3=75So the number of 5 Rs. Coin =25×2=50

Question – 11 : In 81 liters of seawater, there were traces of oil present in the ratio of 1:8 as oil to water. How much water in liters must be added to seawater to create oil to water solution in the ratio of 1:10 ?

Answer – (A) : 18

Answer – (B) : 18

Answer – (C) : 20.5

Answer – (D) : 24

View Answer

Correct Answer : (B)

Explanation : Given:81 liters of seawater contain oil to water in the ratio of ∶ 1∶8 .Let’s assume the common multiplication constant is k .∴ Oil contain in 81 liters of sea water is k liters.And, water contain in 81 liters of sea water is 8k .∴ We can say in 81 liters of sea water k out of 9k parts is oil contain and 8k part is water.∴ Total water contain in sea water is 72 liters.Let x be water added to seawater.9(72+x)=110∴x=18 liters∴18 liters of water must be added to seawater to maintain the new ratio.

Question – 12 : An amount of Rs. 7200 is divided. The ratio of amount received by A:B=3:1,B:C=1:2 and C:D=2:3 . Find the amount received by (B+C) .

Answer – (A) : Rs. 2400

Answer – (B) : Rs. 2400

Answer – (C) : Rs. 3400

Answer – (D) : Rs. 1600

View Answer

Correct Answer : (B)

Explanation : Given:Total amount = Rs. 7200Amount received by A:B=3:1Amount received by B:C=1:2Amount received by C:D=2:3Now,A:B:C:D3:1:1:11:1:2:22:2:2:36:2:4:6⇒3:1:2:3Total of ratios =3+1+2+3=9⇒ Amount received by (B+C)=7200×39⇒ Amount received by (B+C)= Rs. 2400

Question – 13 : The salaries of Rakesh and Raghav are in ratio 2:3 . If the salary of each is increased by Rs. 5000 , the new ratio becomes 40:56 . What is Raghav’s increased salary?

Answer – (A) : Rs. 35000

Answer – (B) : Rs. 35000

Answer – (C) : Rs. 36870

Answer – (D) : Rs. 36800

View Answer

Correct Answer : (B)

Explanation : Let the original salaries of Rakesh and Raghav be 2x and 3x respectively.According to the question,2x+50003x+5000=4056⇒56(2x+5000)=40(3x+5000)⇒112x+280000=120x+200000⇒280000−200000=120x−112x⇒80000=8x⇒x=800008=10000∴ Increased salary of Raghav =(3x+5000)=3×10000+5000= Rs. 35000

Question – 14 : Given the ratio of a:b=5:7,b:c=21:6 then find the value of (3a+c):(2b−c) .

Answer – (A) : 17:12

Answer – (B) : 12:17

Answer – (C) : 17:12

Answer – (D) : 05:03

View Answer

Correct Answer : (C)

Explanation : Given:a:b=5:7b:c=21:6After multiplying 3 in both the denominator and numerator of the first ratio we get,a:b=15:21And b:c=21:6So, a:b:c=15:21:6So we have to find the value of (3a+c):(2b−c)=(3×15+6):(2×21−6)∴(3a+c):(2b−c)=51:36=17:12

Question – 15 : The third proportional to P and Q is 24 , while the fourth proportional to P,Q and R is 48 . Find the ratio of Q and R .

Answer – (A) : 01:02

Answer – (B) : 01:02

Answer – (C) : 02:03

Answer – (D) : 01:03

View Answer

Correct Answer : (B)

Explanation : The third proportional to P and Q is 24 .⇒PQ=Q24⇒Q2P=24 equation (1)The fourth proportional to P,Q and R is 48 .⇒PQ=R48⇒QRP=48 equation(2)Dividing the two equations, we get QR=2448=1:2

Question – 16 : Two numbers are in the ratio of 4:5 . If the first number is increased by 50% and the second number is decreased by 27 , the ratio becomes 3:4 . Find the difference between numbers.

Answer – (A) : 9

Answer – (B) : 7

Answer – (C) : 8

Answer – (D) : 9

View Answer

Correct Answer : (D)

Explanation : Given:Ratio of numbers =4:5Increment in 1st number =50%Decrement in 2nd number =27We know that,If X is distributed between A and B in the ratio of a:b , then A=a(a+b)×X and B=b(a+b)×XTo convert any ratio to an exact value, we should have to multiply any constant value.Let, numbers be 4x and 5x respectively.Now, 1st number =4x+50% of 4x=4x+2x=6x2nd number =5x−276x(5x−27)=34⇒6x×4=(5x−27)×3⇒24x=15x−81⇒9x=−81⇒x=−9Numbers are −9×4=−36 and −9×5=−45The difference of numbers =−36−(−45)=9

Question – 17 : The ratio of male to female employees in a company is 7:8 . If 20% male and 30% female employees were reduced in July. Find the new ratio of male to female employees.

Answer – (A) : 01:01

Answer – (B) : 01:02

Answer – (C) : 02:01

Answer – (D) : 03:01

View Answer

Correct Answer : (A)

Explanation : Given:The ratio of male to female employees in a company is ∶ 7∶8 .Let’s assume the common multiplication constant is k .∴ Number of male employees =7kWhen 20% male employees were reduced, new number will be 0.8×7k=5.6kAnd, Number of female employees =8kWhen 30% female employees were reduced, new number will be 0.7×8k∴ New number of female employees =5.6k∴ New ratios of male to female employees,=5.6k:5.6k=1:1

Question – 18 : The Honda company reduces the number of its worker in the ratio 7:4 and increases their wages in the ratio 3:5 then find the ratio of the company’s expenditure on total wages previously and now.

Answer – (A) : 21:20

Answer – (B) : 20:21

Answer – (C) : 04:05

Answer – (D) : 05:04

View Answer

Correct Answer : (A)

Explanation : Given:Ratio of number of worker reduces =7:4We know that,Total Wages = number of worker × wages per personNow,Total wages before reduction of workers =7x×3y=21xyTotal wages after reduction of workers =4x×5x=20xyRatio of wages (previous : now) =21xy:20xy=21:20

Question – 19 : The ratio between the savings of A and B is 3:4 . The difference between the savings of B and A is Rs. 2000 . Find the average savings of A and B .

Answer – (A) : Rs. 7000

Answer – (B) : Rs. 6500

Answer – (C) : Rs. 8000

Answer – (D) : Rs. 8600

View Answer

Correct Answer : (A)

Explanation : Given:The ratio between the savings of A and B is 3:4 . The difference between the savings of B and A is Rs. 2000 .Now,Let savings of A and B be Rs. 3n and Rs. 4n respectively.4n−3n=2000⇒n=2000The savings of A=2000×3= Rs. 6000The savings of B=2000×4= Rs. 8000Total savings of A and B=6000+8000= Rs. 14000Average saving of A and B =140002= Rs. 7000

Question – 20 : Sunil got marks in 3 subjects in the ratio of 2:3:4 . If he scores 15%,10% and 15% more in the next exam in respective subjects, find the new ratio of marks.

Answer – (A) : 23:33:46

Answer – (B) : 24:33:46

Answer – (C) : 23:34:46

Answer – (D) : 23:33:45

View Answer

Correct Answer : (A)

Explanation : Given:Sunil scored marks in the ratio of 2:3:4 .Let’s assume the common multiplication constant is X .∴ Sunil got 2X,3X and 4X marks in 3 subjects.When he scored 15%,10% and 15% more marks in next exam in respective subjects.New marks in 1st subject =2X+2X×(15100)=2.3XNew marks in 2nd subject =3X+3X×(10100)=3.3XNew marks in 3rd subject =4X+4X×(15100)=4.6X∴ Ratio of new marks =2.3X:3.3X:4.6X=23:33:46

Question – 21 : In an apple garden, the number of apples is 1700,10% of them are rotten. The remaining apples are put into the basket in the ratio of 4:5:6 , find the number of apples in the smallest basket.

Answer – (A) : 408

Answer – (B) : 408

Answer – (C) : 420

Answer – (D) : 450

View Answer

Correct Answer : (B)

Explanation : Total number of apples =1700Remaining apples =1700×90100=1530Number of apples in smaller basket =1530×415=408

Question – 22 : A metal rod was cut into three parts A,B and C , such that the length of A and B are in the ratio 4:5 , while that of B and C are in the ratio 3:2 . If the difference between the lengths of A and C is 8 cm, find the length of the metal rod.

Answer – (A) : 148 cm

Answer – (B) : 148 cm

Answer – (C) : 152 cm

Answer – (D) : 164 cm

View Answer

Correct Answer : (B)

Explanation : Given:A:B=4:5B:C=3:2Difference between the lengths of A and C is 8 cm.Let the lengths of A and B be ‘ 4x ‘ cm and ‘ 5x ‘ cm respectively.Ratio of lengths of B and C=3:25xLength of C=32Length of C=23×5x=10×3 cmNow,Difference between lengths of A and C=8 cm⇒4x−10×3=8⇒12x−10x=24⇒x=242=12 cm∴ Length of metal rod =4x+5x+10×3=4(12)+5(12)+103×12=48+60+40=148 cm

Question – 23 : Cost price of shampoo A and B are in ratio 4:3 and the cost price of shampoo C is Rs. 70 , then find out ratio of A,B and C if total cost price of all three shampoos be Rs. 560 .

Answer – (A) : 04:03:01

Answer – (B) : 04:03:01

Answer – (C) : 01:03:02

Answer – (D) : 01:03:07

View Answer

Correct Answer : (B)

Explanation : Given:A:B:C are in ratio 4:3:C .Therefore cost of shampoo C=C(7+C)×560∴C(7+C)=70560=18So, C=1∴A:B:C=4:3:1

Question – 24 : The income of Adarsh, Satpal and Rahim in the ratio of 12 : 9 : 7 and their spendings are in the ratio 15 : 9 : 8. If Adarsh saves 25% of his income. What is the ratio of the savings of Adarsh, Satpal and Rahim?

Answer – (A) : 15:18:11

Answer – (B) : 05:08:07

Answer – (C) : 23:18:11

Answer – (D) : 25:16:13

View Answer

Correct Answer : (A)

Explanation : Given.Income = Expenditure + SavingAdarsh: 12x = 15y + 3x (3x = 25% of 12x)Satpal: 9x = 9y + (9x – 9y)Rahim: 7x = 8y + (7x – 8y)Therefore, 12x – 3x = 15y⇒xy=53y=3×5Therefore, savings = (income – expenditure)Adarsh = 12x – 9x= 3xSatpal = 9x – 9y =9x−275x=185xRahim = 7x – 8y=7x−245x=115xi.e., the ratio of savings of Adarsh : Satpal : Rahim=3x:185x:115x=15:18:11

Question – 25 : In an auditorium, there are a total of 440 people. Amongst them teachers, students and guests are in the ratio of ∶∶ 1∶5∶4 . Find the number of guests present in the auditorium.

Answer – (A) : 176

Answer – (B) : 176

Answer – (C) : 220

Answer – (D) : 240

View Answer

Correct Answer : (B)

Explanation : Given:Total number of people in auditorium =440And teachers, students, and guests are in the ratio of ∶∶ 1∶5∶4Let the common multiplication constant be X .∴ We can say the auditorium is occupied by X teachers, 5X students and 4X guests.Now,Total number of people = Total numbers of teachers + Total numbers of students + Total numbers of guests440=X+5X+4X∴X=44∴ Total numbers of guests =4X=4×44=176

Question – 26 : The ratio of boys and girls in a class is 3:4 . In new session, the number of boys increased by 10 and the number of girls increased by N. Now, their ratio became 5:6 . Find the value of N.

Answer – (A) : Cannot be determined

Answer – (B) : 60

Answer – (C) : 77

Answer – (D) : Cannot be determined

View Answer

Correct Answer : (D)

Explanation : Let the number of boys be 3 T and number of girls be 4 T.In new session, the number of boys increased by 10 and the number of girls increased by N. Now, their ratio became 5:6 .⇒ (3T+10)(4T+N)=56⇒ 18T+60=20T+5NThere are two variables and one equation. The value of N cannot be uniquely determined.

Question – 27 : In 240 litre mixture, 200 litre is milk and the rest is water. If 30 litre mixture is taken out and 5 litre milk and 5 litre water is added again, then find the quantity of milk in the final mixture.

Answer – (A) : 180 litre

Answer – (B) : 200 litre

Answer – (C) : 180 litre

Answer – (D) : 40 litre

View Answer

Correct Answer : (C)

Explanation : Given:Initial Mixture =240 litreMilk in initial mixture =200 litreWater in initial mixture =40 litreMixture taken out =30 litreWater added =5 litreMilk added =5 litreNow,The ratio of milk and water in the initial mixture =200:40=5:1In 30 litre mixture taken out ratio of milk and water is 5:1 .Let milk removed be 5x litre and water removed be x litre.5x+x=30⇒6x=30⇒x=5Milk removed =5x=5×5=25 litreWater removed =x=5 litreMilk in mixture after removal =200−25=175 litre5 litre more milk is added.Quantity of milk in final mixture =175+5=180∴ The quantity of milk in the final mixture is 180 litre.

Question – 28 : The present age of Akbar and Birbal are in the ratio 15:12 and the difference between their ages is 12 years. Find the ratio of their ages three years ago.

Answer – (A) : 19:15

Answer – (B) : 15:20

Answer – (C) : 12:15

Answer – (D) : 19:15

View Answer

Correct Answer : (D)

Explanation : Given:The ratio of the present age of Akbar and Birbal =15:12Difference between ages =12 (always remains same)3 units =12 yearsThen, 15 units =60 years12 units =48 yearsThen, 3 years ago, required ratio =57:45=19:15

Question – 29 : In what proportion, should the two kinds of tea at prices Rs. 30 , and Rs. 40 be mixed respectively to get a mixture worth Rs. 34 per Kg?

Answer – (A) : 03:02

Answer – (B) : 04:02

Answer – (C) : 01:02

Answer – (D) : 03:02

View Answer

Correct Answer : (D)

Explanation : Given:The prices of the two varieties of tea = Rs. 30 , Rs. 40 per kgPrice of the final mixture = Rs. 34 per kgLet the amount of the first variety =x kgLet the amount of the second variety =y kgFrom the given conditions, we get the total cost as:30x+40y=34(x+y)⇒30x+40y=34x+34y⇒34x−30x=40y−34y⇒4x=6y⇒2x=3y⇒x:y=3:2∴ The required ratio to mix these two types of mixtures is 3:2 .

Question – 30 : How many liters of a 75% alcohol solution should Rekha mix with 40 liters of a 50% alcohol solution if she wants to produce a 60% alcohol solution?

Answer – (A) : 26.67 liters

Answer – (B) : 23.33 liters

Answer – (C) : 25 liters

Answer – (D) : 30 liters

View Answer

Correct Answer : (A)

Explanation : 40 liters of a 50% alcohol solution will contain:Water =20 litersAlcohol =20 litersLet x liters of 75% alcohol solution is mixed with the above solution.The 75% alcohol solution will contain:Alcohol =75100×x=3×4 liters∴ Water =x−3×4=x4 litersNew volume of mixture =(40+x) litersNew volume of alcohol =(20+3×4) litersNew volume of water =(20+x4) liters∴ New concentration =20+3×440+x×10060=(80+3x)440+x×100⇒60100×(40+x)=80+3×4⇒0.6×4(40+x)=80+3x⇒96+2.4x=80+3x⇒0.6x=16∴x=160.6=803=26.67 liters

SSC CGL RATIO AND PROPORTION MCQ ENGLISH

SSC CGL RATIO AND PROPORTION MCQ ENGLISH