Ratio and Proportions-Test 6

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

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

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

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.

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

here,A:B:C=8:14:22

B:C:D=21:33:44

required ratio =A:B:C:D=12:21:33:44

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

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

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

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

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

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

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

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.

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.

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.

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

Assume the values as x = 3 and y = 4.
Then we have

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

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.

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.

(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.

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.

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.