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Q:

The Age of Ashok , Deepak and Ramu are in the proportion of 3:5:6. If the difference between the ages of Deepak and Ramu is 5 years . what is Ashok's age in years?

Answer

Suppose the ages of Ashok, Deepak and Ramu are 3x, 5x and 6x years respectively.

$6x-5x=5\: \: \Rightarrow x=5$

Hence , Ashok's age = $3x$ = $3\times 5$ = 15 years

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Subject: Problems on Ages

Q:

Ten years ago, A was half of B in age. If the ratio of their present ages is 3:4, then what will be the total of their present ages?

Answer

Let A's age 10 years ago be x years

Then B's age 10 years ago be 2x yeats

$\inline \Rightarrow \frac{x+10}{2x+10}=\frac{3}{4}$

$\inline \Rightarrow 4(x+10)=3(2x+10)$

$\inline \Rightarrow x=5$

$\inline \therefore$ Total of their present ages = $x+10+2x+10$

= $3x+20$

= 35 years

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Subject: Problems on Ages

Q:

One year ago the ratio between Mohan and Ashok's age was 4:3 after one yerar the rato of their ages will be 5:4 . what is the sunm of their present ages in years.

Answer

Let their ages one year ago be 4x and 3x years

$\Rightarrow \frac{4x+2}{3x+2}=\frac{5}{4}\Rightarrow 4(4x+2)=5(3x+2)$

$\Rightarrow x=2$

$\therefore$ Sum of their present ages =(4x +1+3x+1) = 16 years.

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Subject: Problems on Ages

Q:

The Number of times the digit 8 will be written when listing the integers from 1 to 1000 is :

A) 100	B) 200
C) 300	D) 400

Answer & Explanation Answer: C) 300

Explanation:

Since 8 does not occur in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where $0 \leq x, y, z \leq 9 .$

Let us first count the numbers in which 8 occurs exactly once.

Since 8 can occur atone place in $3 C_{1}$ ways. There are $3 x 9 x 9$ such numbers.

Next, 8 can occur in exactly two places in $3 C_{2} \times 9$ such numbers. Lastly, 8 can occur in all three digits in one number only.

Hence, the number of times 8 occur is

$1 \times (3 \times 9^{2}) + 2 \times (3 \times 9) + 3 \times 1 = 300$

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Filed Under: Permutations and Combinations

Q:

If the letters of the word VERMA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word VERMA is :

A) 108	B) 117
C) 810	D) 180

Answer & Explanation Answer: A) 108

Explanation:

The number of words beign with A is 4!

The number of words beign with E is 4!

The number of words beign with M is 4!

The number of words beign with R is 4!

Number of words beign with VA is 3!

Words beign with VE are VEAMR

VEARM

VEMAR

VEMRA

VERAM

VERMA

Therefore, The Rank of the word VERMA = 4 x 4! + 3! + 6 = 96 + 6 + 6 =108

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Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
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Q:

The number of signals that can be generated by using 5 differently coloured flags, when any number of them may be hoisted at a time is:

A) 235	B) 253
C) 325	D) None of these

Answer & Explanation Answer: C) 325

Explanation:

Required number of signals = $5 P_{1} + 5 P_{2} + 5 P_{3} + 5 P_{4} + 5 P_{5}$

= 5 + 20 + 60 + 120 + 120 = 325

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Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT
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Q:

The least value of the expression $2 l o g_{10} (x) - l o g_{x} (\frac{1}{100})$ for x>1 is:

A) 2	B) 3
C) 4	D) 5

Answer & Explanation Answer: C) 4

Explanation:

Hence the least value of is 4

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Filed Under: Logarithms
Exam Prep: AIEEE , Bank Exams , CAT
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Q:

The number of solutions of the equation $l o g_{\frac{x}{2}} (x^{2}) + 40 l o g_{4 x} (\sqrt{x}) - 14 l o g_{16 x} (x^{3}) = 0$ is:

A) 0	B) 1
C) 2	D) 3

Answer & Explanation Answer: D) 3

Explanation:

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Filed Under: Logarithms
Exam Prep: AIEEE , Bank Exams , CAT
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