A conveyer belt moves grain at the rate of 3 tons in 5 minutes and a second conveyer belt moves grain at the rate of 2 tons in 9 minutes. How many minutes will it take to move 37 tons of grain using both conveyer belts?

Solution

The first belt lifts ³⁄₅ of a ton per minute and the second one lifts ⁴⁄₉ together will lift -³⁄₅ + ²⁄₉ = \(\frac{37}{45}\)

∴ \(\frac{37}{45}\) ton per minute can be removed using 0 conveyers at a time.

∴ Therefore using both belts, to move 37 tons, it will take \(\frac{37}{\frac{37}{45}}\)= 45 minutes.

n = 4p, where p is a prime number greater than 2. How many different positive even divisors does n have, including n?

Solution

P is a prime number greater than 2.

∴ p is an odd number.

e.g. p can be 3 ∴ 4p = 12

∴ The possible even divisors of pare 2, 4, 2p, 4p.

A = {2, 3, 4, 5}

B = {4,5,6,7,8}

Two integers will be randomly selected from sets above, one integer from set A and one integer from set B. What is the probability that the sum of two integers will equal to 9?

Solution

The number of possible selections from A is 4 and number of possible selections from B is 5. Therefore, the number of different pairs one from A and one from B = 4 × 5 = 20.

∴ Sample space is 20 Of these 20 pairs of numbers, 4 possible pairs sum to 9, namely 2 & 7, 3 & 6, 4 & 5, 5 & 4.

∴ Required probability = \(\frac{4}{20}\) = 0.20

75% of the class answered the first question on a certain test correctly, 55% answered second question on test correctly and 20% answered neither of questions correctly, what percent answered both correctly?

Solution

This is vein diagram with 2 circles and the formula is Total population = Individual Total + Individual Total - Common.

20% student could not solve any of the questions therefore only 80% students could solve either No. 1 or 2 or both.

∴ 80 = 75 + 55 - Common.

∴ Common number is 50.

In a certain calculus class, ratio of number of mathematics majors to number of students who are not mathematics majors is 2:5. If two more mathematics majors were to enter the class, the ratio would be 1 to 2. How many students are in the class?

Solution

Let the number of number of students who are mathematics majors and number of students who are not be 2x and 5x.

\(\frac{2x+2}{5x}=\frac{1}{2}\) ∴ 4x + 4 = 5x ∴ x = 4

∴ Number of students = 7x = 7 × 4 = 28

A lake has a rare breed of lotuses that triple in number every minute. If the lake is full of flowers after 30 minutes when did it have flowers only (¹⁄₃) of the lake?

Solution

After 30 minutes → lake was full of flowers.

Since number of flowers increase 3 times every minute, lake was (¹⁄₃)^rd full 1 minute before.

Therefore lake was (¹⁄₃)^rd full after 29 minutes.

In the next minute number of flowers increased 3 times and the lake was full.

Two cars start together in same direction from same place. The first goes with uniform speed of 10kmph. Second goes at a speed of 6 kmph. in first hour and increases speed by 1 km each succeeding hour. After how many hours will second car overtake the first if they travel non-stop.

Solution

Let both cars meet after n hours.

Distance traveled by the first car = 10n

Distance traveled by the second car follows an

A.P. = ⁿ⁄₂[2a + (n - 1)d]

10n = ⁿ⁄₂[2 × 6 + (n - 1)1]

10n = ⁿ⁄₂[12 + n - 1]

20n = n(11 + n) ∴n = 9

Vessel 1 contains benzene and naphthalene in the ratio 3:2 and vessels 2 contains same liquids in the ratio 1:2. In what proportion should liquids in two vessels be mixed to give a solution with benzene and naphthalene in the ratio 3:4?

Solution

Fraction of benzene in vessel 1 = 3/5

Fraction of benzene in vessel 2 = 1/3

Fraction of benzene in the mixture = 3/7

Taking LCM of 5, 3, 7 i.e. 105

We have,\(\frac{3}{5}=\frac{63}{105},\frac{1}{3}=\frac{35}{105},\frac{3}{7}=\frac{45}{105}\)

∴ The ratio is 10 : 18

There are n friends receiving an average of 5 chocolates each. The friends are standing in a line. Each friend in the row receives one chocolate more than the earlier one. If 5th person in the line receives 5 chocolates, then fmd the total number of friends in the line.

Solution

Method:

The number of chocolates received are in A.P. with common addition of 1. If the fifth person receives 5 chocolates, the first person in the row will receive 1 chocolate.

Total chocolates are 5n

Using formula for the sum of A.P.

Sn = ⁿ⁄₂[2a + (n - 1)d] 5n = ⁿ⁄₂[2 × I + (n -1)1]

5n = ⁿ⁄₂(1 + n) 10n = n(1 + n) 10 = 1+ n

∴ n = 9

Method 2:

Since the first person receives 1 chocolate, it is a total of natural number, we can use formula.

\(\frac{n(n+1)}{2}\)

∴ 5n = \(\frac{n(n+1)}{2}\) ∴ 10n = n(n + 1) ∴ n = 9

In a survey of 100 offices, 80 have computers, 40 have typewriters while 20 offices have neither computers nor typewriters. What percent of offices having any of two have only computers?

Solution

This problem is based on Ven diagram with two circles. Formula is T = I + I - C (Total population = Individual Total + Individual Total - Common in both) 20 offices do not have either computers or typewriters.

∴ Total officers = 80

80 = 80 + 40 - Common

∴ Common = 40, who have computers and typewriters both.

The answer is 50%.

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