If the work done by p men in (p + 2) days is to the work done by (p + 4) men in (p – 1) days is in the ratio 1 : 1, then the value of p is:
Work done will be directly proportional to number of men and days.
So according to the question:
[(p)(p + 2)] / [(p + 4)(p - 1)] = 1/1
p2 + 2p = p2 + 4p - p - 4
p = 4
If p and q are positive integers then √2 always lies between:
► Suppose you take p = 3 and q = 2. It can be clearly seen that the square root of 2 does not lie between 2 and 3. Hence, option (b) is incorrect.
► Further with these values for p and q option (a) also can be ruled out since it means that the value should lie between 5 and 6 which it obviously does not. Also, Option (d) gives 6 / 5 and 1 / 6.
►This means that the value should lie between 0.1666 and 1.2 (which it obviously does not). Hence, option (c) is correct.
Monthly incomes of E and R are in the ratio 1 : 3 and their expenses are in the ratio 19 : 40. E saves Rs. 18,860 less than that R and in total they save Rs. 36,020. Income of E and R respectively are:
“Total amount saved by E and R is Rs. 36,020”
Es + Rs = 36020 ——— (1)
where EsEs and RsRsrepresent the savings amount of E and R.
“E saves 18,860 less than R”
Es+ 18860 = Rs ———-(2)
solving equation (1) and (2), we get EsEs as Rs. 8580, and RsRs as Rs. 27440
E and R have saved Rs. 8580 and Rs. 27440 respectively.
“Their expenses are in the ratio of 19:40”
From the previous two statements, we have the information on expenses and savings that sum up to the monthly income.
So the incomes of E and R are 19x + 8580 and 40x + 27440 respectively, where ‘x’ defines the number of times the expenses are to be multiplied.
“The monthly incomes of E and R are in 1:3”
19x+858040x+2744019x+858040x+27440 = 1313
Solving the above equation, we get x = 100
As E and R earn a monthly income of 19x + 8580, and 40x + 27440, their incomes are Rs. 10480 and Rs. 31440 respectively.
The incomes of Sheldon, Leonard, and Howard are in the ratio of 4 : 5 : 6 respectively and their spending are in the ratio of 6 : 7 : 8 respectively. If Sheldon saves one fourth his income, then the savings of Sheldon, Leonard, and Howard are in the ratio:
Let the incomes be 4x, 5x, 6x and the spending be 6y, 7y, 8y and savings are (4x–6y), (5x–7y) & (6x–8y)
Sheldon saves 1/4th of his income.
⇒ 4x – 6y = 4x / 4
⇒ 4x – 6y = x
⇒ 3x = 6y
⇒ x / y = 2
∴ y = x / 2
Ratio of Sheldon’s Leonard’s & Howard’s savings:
= 4x – 6y : 5x – 7y : 6x – 8y
= x : 5x – 7y : 6x – 8y
= x : 5x – 7x / 2 : 6x – 8x / 2
= x : 3x / 2 : 2x
= 2 : 3 : 4
The expenses of an all boys’ institute are partly constant and partly vary as the number of boys. The expenses were 10,000 for 150 boys and 8400 for 120 boys. What will the expenses be when there are 330 boys?
► Expenses for 120 boys = 8400
► Expenses for 150 boys = 10000.
Thus, variable expenses are 1600 for 30 boys.
► If we add 180 more boys to make it 330 boys,
we will get an additional expense of 1600 * 6 = 9600.
► Total expenses are Rs 19600.
A vessel contains p litres of wine and another vessel contains q litres of water. r litres are taken out of each vessel and transferred to the other. If r X (p + q) = pq. If A and B are the respective values of the amount of wine contained in the respective containers after this operation, then what can be said about the relationship between A and B.
► The constraint given to us for the values of p, q, and r is X (p + q) = pq
So, if we take p = 6, q = 3 and r = 2, we have 18 = 18 and a feasible set of values for p, q and r respectively.
► With this set of values, we can complete the operation as defined and see what happens.
Wine left in the vessel A = (6 – 2) = 4
Wine in the vessel B = 2
► With these values, none of the first 3 options matches. Thus, option (d) is correct.
The volume of a pyramid varies jointly as its height and the area of its base; and when the area of the base is 60 square dm and the height 14 dm, the volume is 280 cubic dm. What is the area of the base of a pyramid whose volume is 390 cubic dm and whose height is 26 dm?
V = k AH
⇒ 280 = k X 60 X 14
⇒ 280 = 840k
Thus, k = 1 / 3 and the equation becomes:
V = AH / 3 and 390 = 26A / 3
⇒ A = 45
Total expenses of running the hostel at IIM Ahmedabad are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs.70 when there are 25 borders and Rs.60 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?
► When there are 25 boarders, the total expenses are Rs. 1750. When there are 50 borders, the total expenses are Rs. 3000.
► The change in expense due to the coming in of 25 borders is Rs. 1250.
► Hence, expense per boarder is equal to Rs. 50. This also means that when there are 25 borders, the variable cost would be 25 * 50 = Rs.1250.
Hence, Rs.500 must be the fixed expenses.
► So for 100 boarders, the total cost would be: Rs. 500 (fixed) + Rs. 5000 = Rs. 5500
► Average expense per boarder will be 5500/100 = Rs 55
An alloy of gold and silver is taken in the ratio of 1 : 2, and another alloy of the same metals is taken in the ratio of 2 : 3. How many parts of the two alloys must be taken to obtain a new alloy consisting of gold and silver that are in the ratio 3 : 5?
Let x and y be mass of two alloys mixed.
In first alloy:
Gold = x × 1 / (1 + 2) = x/3
Silver = x × 2 / (1 + 2) = 2x/3
In second alloy:
Gold = y × 2 / (2 + 3) = 2y/5
Silver = y × 3 / (2 + 3) = 3y/5
In resulting alloy:
Gold / Silver = 3 / 5
(x/3+2y/5) / (2x/3+3y/5) = 3 / 5
(x/3+2y/5) × 5 = (2x/3+3y/5) × 3
5x/3 + 2y = 2x + 9y/5
5x/3 - 2x = 9y/5 - 2y
-x/3 = -y/5
x / y = 3 / 5
Therefore, two alloys should be taken in ratio of 3 : 5.
The sum of three numbers x, y, z is 5000. If we reduce the first number by 50, the second number by 100, and the third number by 150, then the new ratio of x & y = 4 : 5 & the new ratio of y & z =3 : 4. What is the value of x + y ?
► If new values of x, y, z are x′, y′ and z′, and respectively then x′ : y′ = 4 : 5, y′ : z′ = 3 : 4
⇒ x′ : y′ : z′ = 12 : 15 : 20
⇒ x + y + z = 5000
⇒ x′ + 50 + y′ + 100 + z′ + 150 = 5000 x′ + y′ + z′ = 4700
⇒ 12k + 15k + 20k = 4700 k = 100
► x = 1200 + 50 = 1250
► y = 1500 + 100 = 1600 z = 2000 + 150 = 2150
► x + y = 1250 + 1600 = 2850