Online exam in Ratio and Proportions For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT
Ratio and Proportions-Test 6
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Subject :- Quantitative Aptitude
Chapter :- Ratio and Proportion- Test 6
Questions :- 25
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The ratio of present ages of meena and fiona is 16:13,4 yr ago ratio of their ages was 14:11.what will be fiona’s age 4yr from now?
Correct
here,meena:fiona=16:13
4yr ago ratio=14:11
so,182-176=4*(16-13)
6=4)3
1=12/6=2
hence,fiona’s age 4yr from now=13*2+4=26+4=30 yr
Incorrect
here,meena:fiona=16:13
4yr ago ratio=14:11
so,182-176=4*(16-13)
6=4)3
1=12/6=2
hence,fiona’s age 4yr from now=13*2+4=26+4=30 yr
Question 2 of 25
2. Question
1 points
Radha’s present age is 3yr less than twice her age 12yr ago.also the ratio of raj’s present age and radha’s present age is 4:9.what will be raj age after 5yr?
Correct
here,radha’s present age=x yr
12yr ago radha age=x-12 yr
now,x=2(x-12)-3
x=2x-24-3
x=27
then,raj:radha=4:9
raj present age=4/9*27
raj age after 5 yr=12+5=17yr
Incorrect
here,radha’s present age=x yr
12yr ago radha age=x-12 yr
now,x=2(x-12)-3
x=2x-24-3
x=27
then,raj:radha=4:9
raj present age=4/9*27
raj age after 5 yr=12+5=17yr
Question 3 of 25
3. Question
1 points
The ratio between the ages of father and son at present is 5:3.4 yr hence the ratio between the ages of the son and his mother will be 1:2.what is the ratio between the present ages of the father and the mother?
Correct
here,father:son=5:2
let x be common ratio
then,present age of father=5x and son=2x
son age after 4 yr=2x+4 yr
his mother’s age=2(2x+4)=(4x+8)yr
hence,required ratio cannot be determined
Incorrect
here,father:son=5:2
let x be common ratio
then,present age of father=5x and son=2x
son age after 4 yr=2x+4 yr
his mother’s age=2(2x+4)=(4x+8)yr
hence,required ratio cannot be determined
Question 4 of 25
4. Question
1 points
Ratio of rani’s and komal’s age is 3:5.ratio of komal’s and pooja’s age is 2:3.if rani is 2/5th of pooja’s age ,what is rani’s age?
Correct
here,rani:komal=3:5
komal:pooja=2:3
rani:komal:pooja=6:10:15
hence,from given data ,we cannot determine the rani’s age.
Incorrect
here,rani:komal=3:5
komal:pooja=2:3
rani:komal:pooja=6:10:15
hence,from given data ,we cannot determine the rani’s age.
Question 5 of 25
5. Question
1 points
The ages of sulekha and aruna are in the ratio of 9:8.after 5 yr the ratio of their ages will be 10:9.what is the difference in yr between their ages.
Correct
here,sulakeha:aruna=9:8
after 5 yr their ages ratio=10:9
now,1=5*1=5
required difference=5(9-5)=5)1=5yr
Incorrect
here,sulakeha:aruna=9:8
after 5 yr their ages ratio=10:9
now,1=5*1=5
required difference=5(9-5)=5)1=5yr
Question 6 of 25
6. Question
1 points
The present ages of A,B and C are in the ratio of 8:14:22.the present age of B,C and D are in the ratio of 21:33:44.which of the following represents the ratio of the present ages of A,B,C and D?
Correct
here,A:B:C=8:14:22
B:C:D=21:33:44
required ratio =A:B:C:D=12:21:33:44
Incorrect
here,A:B:C=8:14:22
B:C:D=21:33:44
required ratio =A:B:C:D=12:21:33:44
Question 7 of 25
7. Question
1 points
The age of sonal and nitya are in the ratio of 9:5.after 8 yr the ratio of their age will be 13:9.what is the difference in the yr between their ages?
Correct
here,sonal:nitya=9:5
after 8 yr their age=13:9
now,81-65=4*8
16=32 = 1=2
required difference = 2*4=8yr
Incorrect
here,sonal:nitya=9:5
after 8 yr their age=13:9
now,81-65=4*8
16=32 = 1=2
required difference = 2*4=8yr
Question 8 of 25
8. Question
1 points
Present age of seema and naresh are in the ratio of 5:7.5yr hence the ratio of their ages becomes 3:4.what is the present age of naresh?
Correct
here,seema:naresh=5:7
5yr hence,ratio of their ages=3:4
now,21-20=5*(4-3) = 1=5
hence,naresh’s present age=5*5=25yr
Incorrect
here,seema:naresh=5:7
5yr hence,ratio of their ages=3:4
now,21-20=5*(4-3) = 1=5
hence,naresh’s present age=5*5=25yr
Question 9 of 25
9. Question
1 points
If 17 labourers can dig a ditch 20 m long in 18 days,working 8 h a day,how many more labourers should be angaged to dig a similar ditch 39 m long in 6 days ,each labourers working 9 h a day?
Correct
here,let the total no of men to be engaged = x
more length,more labourers (direct proportion)
less days,more labourers(indirect proportion)
more hours per day,less labourers(indirect proportin)
length 26:39
days 6:18 }::17:x
hours/day 9:8
(26*6*9*x)=(39*18*8*17)
x=39*18*8*17/26*6*9=68
no of more labourers=68-17=51
Incorrect
here,let the total no of men to be engaged = x
more length,more labourers (direct proportion)
less days,more labourers(indirect proportion)
more hours per day,less labourers(indirect proportin)
length 26:39
days 6:18 }::17:x
hours/day 9:8
(26*6*9*x)=(39*18*8*17)
x=39*18*8*17/26*6*9=68
no of more labourers=68-17=51
Question 10 of 25
10. Question
1 points
400 persons,working 9h per day complete 1/4th of the work in 10 days.the number of additional persons working 8h per day ,required to complete the remaining work in 20 days is
Correct
let no of persons completing the work in 20 days be x
work done=1/4
remaining work=1–1/4=3/4
less hours per day,more men required(indirect proportion)
more work,more men required(direct proportion)
more work,less men required(indirect proportion)
hours per day 8:9
work 1/4:3/4 } :: 400:x
days 20:10
8*1/4*20*x=9*3/4*10*400
40x=2700
x=675
additional men=675-400=275
Incorrect
let no of persons completing the work in 20 days be x
work done=1/4
remaining work=1–1/4=3/4
less hours per day,more men required(indirect proportion)
more work,more men required(direct proportion)
more work,less men required(indirect proportion)
hours per day 8:9
work 1/4:3/4 } :: 400:x
days 20:10
8*1/4*20*x=9*3/4*10*400
40x=2700
x=675
additional men=675-400=275
Question 11 of 25
11. Question
1 points
12 men working 8h per day complete a piece of work in 10 days.to complete the same work in 8 days ,working 15h a day ,the number of men required is
Correct
here,no of men be x
less days,more men (indirect proportion)
more working hours per day,less men(indirect proportion)
days 8:10
} :: 12:x
working hours 15:10
8*15*x=10*10*12
x=10
Incorrect
here,no of men be x
less days,more men (indirect proportion)
more working hours per day,less men(indirect proportion)
days 8:10
} :: 12:x
working hours 15:10
8*15*x=10*10*12
x=10
Question 12 of 25
12. Question
1 points
3 pumps,working a day,can empty a tank in 2 days.how many hours a day mist 4 pumps work to empty the tank in 1 day?
Correct
here,no of working hours per day be x
more pumps,less working hours per day (indirect proportion)
less days,more working hours per day(indirect proportion)
pumps 4:3
}::8:x
days 1:2
4*1*x=3*2*8
x=12
Incorrect
here,no of working hours per day be x
more pumps,less working hours per day (indirect proportion)
less days,more working hours per day(indirect proportion)
pumps 4:3
}::8:x
days 1:2
4*1*x=3*2*8
x=12
Question 13 of 25
13. Question
1 points
The sum of 3 consecutive odd numbers is 285.what is the ratio of smallest and largest numbers?
Correct
here,let 3 consecutive odd number be x-2 , x and x+2
then,x-2+x+x+2=285
3x=285
x=95
required ratio=93:97
Incorrect
here,let 3 consecutive odd number be x-2 , x and x+2
then,x-2+x+x+2=285
3x=285
x=95
required ratio=93:97
Question 14 of 25
14. Question
1 points
` 5783 is divided among Sherry, Berry, and Cherry in such a way that if t` 28, t` 37 and t` 18 be deducted from their respective shares, they have money in the ratio 4 : 6 : 9. Find Sherry’s share.
Correct
The problem clearly states that when we reduce 28, 37 and 18 rupees respectively from
Sherry’s, Berry’s and Cherry’s shares, the resultant ratio is: 4 : 6 : 9.
Thus, if we assume the reduced values as
4x, 6x and 9x, we will have Æ
Sherry’s share Æ 4x + 28, Berry’s share Æ 6x + 37 and Cherry’s share Æ 9x + 18 and thus we have
(4x + 28) + (6x + 37) + (9x + 18) = 5783
Æ 19x = 5783 – 83 = 5700
Hence, x = 300.
Hence, Sherry’s share is t` 1228.
Incorrect
The problem clearly states that when we reduce 28, 37 and 18 rupees respectively from
Sherry’s, Berry’s and Cherry’s shares, the resultant ratio is: 4 : 6 : 9.
Thus, if we assume the reduced values as
4x, 6x and 9x, we will have Æ
Sherry’s share Æ 4x + 28, Berry’s share Æ 6x + 37 and Cherry’s share Æ 9x + 18 and thus we have
(4x + 28) + (6x + 37) + (9x + 18) = 5783
Æ 19x = 5783 – 83 = 5700
Hence, x = 300.
Hence, Sherry’s share is t` 1228.
Question 15 of 25
15. Question
1 points
Two numbers are in the ratio P : Q. When 1 is added to both the numerator and the
denominator, the ratio gets changed to R/S. Again, when 1 is added to both the numerator and the denominator, it becomes 1/2. Find the sum of P and Q.
Correct
Option A: It has P + Q = 3. The possible values of P/Q are 1/2 or 2/1.
Using 1/2, we see that on adding 2 to both the numerator and the denominator we get
3/4 (Not the required value.)
Similarly, we see that 2/1 will also not give the answer. We should also realise that the numerator has to
be lower than the denominator to have the final value of 1/2.
Next we try Option B, where we have
1/3 as the only possible ratio.
Then we get the final value as 3/5 (Not equal to 1/2) Hence, we reject option B.
Next we try Option C, where we have
1/4 or 2/3
Checking for 1/4 we get 3/6 = 1/2. Hence, the option is correct.
Incorrect
Option A: It has P + Q = 3. The possible values of P/Q are 1/2 or 2/1.
Using 1/2, we see that on adding 2 to both the numerator and the denominator we get
3/4 (Not the required value.)
Similarly, we see that 2/1 will also not give the answer. We should also realise that the numerator has to
be lower than the denominator to have the final value of 1/2.
Next we try Option B, where we have
1/3 as the only possible ratio.
Then we get the final value as 3/5 (Not equal to 1/2) Hence, we reject option B.
Next we try Option C, where we have
1/4 or 2/3
Checking for 1/4 we get 3/6 = 1/2. Hence, the option is correct.
Question 16 of 25
16. Question
1 points
If 10 persons can clean 10 floors by 10 mops in 10 days, in how many days can 8 persons
clean 8 floors by 8 mops?
Correct
Do not get confused by the distractions given in the problem. 10 men and 10 days means 100 man-days are required to clean 10 floors.
That is, 1 floor requires 10 man-days to get cleaned. Hence, 8 floors will require 80 man-days to clean.
Therefore, 10 days are required to clean 8 floors.
Incorrect
Do not get confused by the distractions given in the problem. 10 men and 10 days means 100 man-days are required to clean 10 floors.
That is, 1 floor requires 10 man-days to get cleaned. Hence, 8 floors will require 80 man-days to clean.
Therefore, 10 days are required to clean 8 floors.
Question 17 of 25
17. Question
1 points
Three quantities A, B, C are such that AB = KC, where K is a constant. When A is kept
constant, B varies directly as C; when B is kept constant, A varies directly C and when C is kept constant, A varies inversely as B.
Initially, A was at 5 and A : B : C was 1 : 3 : 5. Find the value of A when B equals 9 at constant C.
Correct
Initial values are 5, 15 and 25.
Thus we have 5 × 15 = K × 25.
Hence, K = 3.
Thus, the equation is AB = 3C.
For the problem, keep C constant at 25. Then, A × 9 = 3 × 25.
i.e. A = 75/9 = 8.33
Incorrect
Initial values are 5, 15 and 25.
Thus we have 5 × 15 = K × 25.
Hence, K = 3.
Thus, the equation is AB = 3C.
For the problem, keep C constant at 25. Then, A × 9 = 3 × 25.
i.e. A = 75/9 = 8.33
Question 18 of 25
18. Question
1 points
If x/y = 3/4, then find the value of the expression, (5x – 3y)/(7x + 2y).
Correct
Assume the values as x = 3 and y = 4.
Then we have
$\frac{15\u201312}{21+8}=3/29$
Incorrect
Assume the values as x = 3 and y = 4.
Then we have
$\frac{15\u201312}{21+8}=3/29$
Question 19 of 25
19. Question
1 points
3650 is divided among 4 engineers, 3 MBAs and 5 CAs such that 3 CAs get as much as 2
MBAs and 3 Engineers as much as 2 CAs. Find the share of an MBA.
Correct
4E + 3M + 5C = 3650
Also, 3C = 2M, that is, M = 1.5 C
and 3E = 2C that is, E = 0.66C
Thus, 4 × 0.66C + 3 × 1.5 C + 5C = 3650
C = 3650/12.166
That is, C = 300
Hence, M = 1.5 C = 450
Incorrect
4E + 3M + 5C = 3650
Also, 3C = 2M, that is, M = 1.5 C
and 3E = 2C that is, E = 0.66C
Thus, 4 × 0.66C + 3 × 1.5 C + 5C = 3650
C = 3650/12.166
That is, C = 300
Hence, M = 1.5 C = 450
Question 20 of 25
20. Question
1 points
The ratio of water and milk in a 30 litre mixture is 7 : 3. Find the quantity of water to be added to the mixture in order to make this ratio 6 : 1.
Correct
Solve while reading Æ As you read the first sentence, you should have 21 litres of water and 9
litres of milk in your mind.
In order to get the final result, we keep the milk constant at 9 litres.
Then, we have 9 litres, which corresponds to 1
Hence, ‘?’ corresponds to 6.
Solving by using unitary method we have
54 litres of water to 9 litres of milk.
Hence, we need to add 33 litres of water to the original mixture.
Incorrect
Solve while reading Æ As you read the first sentence, you should have 21 litres of water and 9
litres of milk in your mind.
In order to get the final result, we keep the milk constant at 9 litres.
Then, we have 9 litres, which corresponds to 1
Hence, ‘?’ corresponds to 6.
Solving by using unitary method we have
54 litres of water to 9 litres of milk.
Hence, we need to add 33 litres of water to the original mixture.
Question 21 of 25
21. Question
1 points
Divide t` 1870 into three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are equal.
Correct
Solve this question using options. 1/2 of the first part should equal 1/3 rd of the second part and of the third part. This means that the first part should be divisible by 2, the second one by 3, and the third one by 6. Looking at the options, none of the first 3 options has its third number divisible by 6. Thus, option (d) is correct.
Incorrect
Solve this question using options. 1/2 of the first part should equal 1/3 rd of the second part and of the third part. This means that the first part should be divisible by 2, the second one by 3, and the third one by 6. Looking at the options, none of the first 3 options has its third number divisible by 6. Thus, option (d) is correct.
Question 22 of 25
22. Question
1 points
Divide t` 500 among A, B, C and D so that A and B together get thrice as much as C and D together, B gets four times of what C gets and C gets 1.5 times as much as D. Now the value of what B gets is
Correct
(A + B) = 3 (C + D) Æ A + B
= 375 and C + D = 125.
Also, since C gets 1.5 times D we have
C = 75 and D = 50, and B = 4 C = 300.
Incorrect
(A + B) = 3 (C + D) Æ A + B
= 375 and C + D = 125.
Also, since C gets 1.5 times D we have
C = 75 and D = 50, and B = 4 C = 300.
Solve using options. Since x > y > 0 it is clear that a ratio of x:y as 3:2 fits the equation.
Incorrect
Solve using options. Since x > y > 0 it is clear that a ratio of x:y as 3:2 fits the equation.
Question 24 of 25
24. Question
1 points
If 4 examiners can examine a certain number of answer books in 8 days by working 5 hours a day, for how many hours a day would 2 examiners have to work in order to examine twice the number of answer books in 20 days.
Correct
4 × 8 × 5 = 160 man-hours are required for ‘x’ no. of answer sheets. So, for ‘2x’ answer sheets
we would require 320 man-hours = 2 × 20 × n Æ n = 8 Hours a day.
Incorrect
4 × 8 × 5 = 160 man-hours are required for ‘x’ no. of answer sheets. So, for ‘2x’ answer sheets
we would require 320 man-hours = 2 × 20 × n Æ n = 8 Hours a day.
Question 25 of 25
25. Question
1 points
In a mixture of 40 litres, the ratio of milk and water is 4 : 1. How much water must be added to this mixture so that the ratio of milk and water becomes 2 : 3.
Correct
In 40 litres, milk = 32 and water = 8. We want to create 2 : 3 milk to water mixture, for this we would need: 32 milk and 48 water. (Since milk is not increasing). Thus, we need to add 40 litres of water.
Incorrect
In 40 litres, milk = 32 and water = 8. We want to create 2 : 3 milk to water mixture, for this we would need: 32 milk and 48 water. (Since milk is not increasing). Thus, we need to add 40 litres of water.