Every 2 out of 3 engines require alterations in the body, every 3 out of 4 need alterations in materials used and every 4 out of 5 need it in the mechanism. Then how many alterations will be required for a batch of 300 engines?

Solution

Alterations in engines ²⁄₃ x 300 = 200

Alterations in materials ³⁄₄ x 300 = 225

Alterations in mechanism ⁴⁄₅ x 300 = 240

Total alterations 665

\(3^{x-7}=\frac{6^{19}\times 9}{2^{16}\times 8}\),find x

Solution

\(3^{x-7}=\frac{2^{19}\times 3^{19}\times ^{2}}{2^{16}\times 2^{3}}=\frac{2^{19}\times 3^{21}}{2^{19}}\)

∴ 3^{x - 7} = 3²¹

Since bases are equal, comparing indices

∴ x - 7 = 21 ∴.x = 28

Mark and Michael started on a journey in the same direction. They traveled 9 km. and 15 km respectively daily. After traveling for 6 days, Mark doubled his speed and both of them finished the journey in the same time. Find the time, taken in days by them to reach their destination?

Solution

Distance traveled by Mark in 6 days = 6 x 9 = 54 km.

Distance traveled by Michael in 6 days

= 6 x 15 = 90 km

∴ Difference in distances = 90 - 54 = 36 km.

Speed of Mark after 6 days

= 9 x 2 = 18 km per day.

Difference between their speeds now is 18 - 15 = 3 km per day.

To gain 36 km, number of days required

= ¹⁄₂ x 36 = 12 days.

Hence total number of days = 12 + 6 = 18 days.

In a city, 36% of people are illiterate and 64% are poor. Among the rich, 25% are illiterate. What fraction of poor population is illiterate?

Solution

Method 1:

Let there be 100 people in the city.

∴ 36% ⇒ 36 are illiterate.

64% ⇒ 64 are poor.

It has to be assumed that people who are not poor are essentially rich.

∴ 36% => 36 people are rich.

∴ 25% of 36 = 9 people are rich and illiterate.

∴ The remaining 27 (i.e. 36 - 9) people are poor and illiterate.

Method 2:

25% of Rich i.e. 25% of 36 are illiterate = 9

∴ Require fraction = \(\frac{27}{64}\)

p² + q² = 74, pq = 35, find the value of \(\frac{p+q}{p-q}\)

Solution

(p + q)² = p² + q² + 2pq

∴ (p + q)² = 74 + 2 x 35

∴ (p + q)² = 144 ∴ p + q = 12

Similarly,

(p - q)² = p² + q² 2pq

∴ (p - q)² = 74-(2 x 35) = 74 - 70 = 4

∴ (p - q) = 4 :∴ p - q = 2

∴ p + q = 12 = 6

∴ \(\frac{p\,+ q}{p\,- q}=\frac{12}{2}=6\)

A box contains 10 balls of which 7 are yellow and 3 are green. Three balls are drawn in succession (without replacement) from the box. What is the probability of getting at least 2 green?

Solution

At least 2 green balls will give us two combinations -2 green and 1 yellow or all 3 green.

\(\frac{^{7}C_{1}\times ^{3}C_{2}+^{3}C_{3}}{^{10}C_{3}}=\frac{(7 \times 3) + 1}{\frac{10\times 9\times 8}{3\times 2}}\)

=\(\frac{22\times 3\times 2}{10\times 9\times 8}=\frac{11}{60}=0.18\)

In a box there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly, what is the probability that it is neither red nor green?

Solution

The ball required is neither red nor green

∴ It has to be blue.

Probability of getting a blue ball is \(\frac{7}{21}=\frac{1}{3}\).

A machine produces 10 units of an article in a day of which 4 are defective. The quality inspector allows release of products if he finds that none of 3 units chosen by him at random are defective. What is the probability of quality inspector allowing the release?

Solution

Quality Inspector will allow the release of products if all the three units chosen by him are non-defective.

Thus, all the three units must be chosen from 6 non-defective units.

∴ Required probability =\(\frac{6c_{3}}{10c_{3}}=\frac{1}{6}\)

A box contains 20 radios, 5 of them being standard. A worker takes out 3 radios at random. Find probability that at least 1 of 3 radios is standard.

Solution

P (at least 1 of 3 parts to be standard)

= 1 - P (not standard)

∴ P(E)=\(1-\frac{15c_{3}}{20c_{3}}=1-\frac{15\times 14\times 13}{20\times 19\times 18}\)

=\(1-\frac{91}{228}=\frac{137}{228}\)

The probability that a manufacturing firm gets car manufacturing contract is ³⁄₇ and that it will not get a truck manufacturing contract is \(\frac{7}{11}\). If the probability of getting at least one contract is \(\frac{5}{11}\),what is the probability that it will get both the contracts?

Solution

Let A : Getting car manufacturing contract.
B : Getting truck manufacturing contract.

A ∩ B : Getting both the contracts.

Given: P(A)= ³⁄₇,P(\(\bar{B}\)) = \(\frac{7}{11}\),P(A ∪ B) = \(\frac{5}{11}\)

∴ P(B) = \(\frac{4}{11}\)

Required probability = P(A ∩ B)

Now, P(A ∪ B) = P(A)+ P(B)- P(A ∩ B)

∴ P(A ∩ B) = ³⁄₇ + \(\frac{4}{11}\) - \(\frac{5}{11}\) = ³⁄₇ - \(\frac{1}{11}\)

=\(1-\frac{91}{228}=\frac{137}{228}\)

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