15-01-2022 Answers

1. In how many ways 648 can be written as a product of two factors?

Answer: 10

Solution: 648 = 2^3 * 3^4

Hence total number of factors = N(f) = (3+1)*(4+1) = 20

As 20 is even, required ways to express as product of two factors = N(f)/2 = 20/2 = 10

2. The value of log2 (1/64) is

Answer: -6

Solution: log2 (1/64) = log2 (2^-6) = -6

3. If a, b are two perfect square digits and ab is a two digit perfect square number, such that (a*b)+(a+b)=ab then the value of [ba-(b+a)] is

Answer: 81

Solution: Perfect square digits are 1,4,9

ab should be a perfect square number. Hence only 49 is possible (which square of 7).

Hence a=4 and b=9

So ba-(b+a) = 94 - (13) = 81

4. The difference in the simple interest earned on a certain sum at 6% p.a. at the end of two years and the end of four years is Rs. 600. What is the sum?

Answer: Rs. 5000

Solution: Let the sum be x.

x * 6/100 * (4-2) = 600,

x = 5000

5. The present ages of three persons are in proportions 5 : 6 : 7. Five years ago, the sum of their ages was 39. Find the sum of their present ages (in years).

Answer: 54

Solution: (x - 5 ) + ( y - 5 ) + ( z - 5 ) = 39

x + y + z - 15 = 39

x + y + z = 54

6. The pages of a book are numbered starting from 1. A total of 3001 digits are used in numbering the pages. How many pages are in the book?

Answer: 1027

Solution: 1 digit numbers are from 1 to 9 and hence 9 digits.

2 digit numbers are from 10 to 99 = 90 and hence 180 digits.

3 digit numbers are from 100 to 999 = 900 and hence 2700 digits. 

 Total digits till page 999 = 9+180+2700 = 2889.

Number of four digit numbers used further = (3001-2889)/4 = 28 pages.

So total number of pages = 999 + 28 = 1027 pages.

7. A man forgot his 6 digit bank account number but he remembered that it was of the form M515M0 and was divisible by 36. What is the value of M?

Answer: 8

Solution: The digital sum should be a multiple of 9 (as 9 is a factor of 36).

Hence 11 + 2M should give a digital sum 9.

Also as it is divisible by 36, it should be divisible by 4 too.

Hence the last two digits can be 20,40,60,80.

Among these only 8 satisfies the condition as 11+16 = 27 which inturn gives 9 as digital sum.

8. The arithmetic mean of two numbers is 10 and their geometric mean is 6. The difference between the numbers is

Answer: 16

Solution: Numbers be x,y

(x+y)/2 = 10, x+y=20, x=20-y

xy = 36, (20-y)*y = 36,

y^2 -20y +36 = 0, Solving y=18 or y=2

Difference = 18-2 = 16

9. If log3 = 0.47712125, then find the number of digits in 3^27

Answer: 13

Solution: Taking log to base 10,

log 3^27 = 27log3 = 27 * 0.47712125 = 27 * 0.477 appprox = 12.879

Hence ceiling 12.879 number of digits is 13.

10. The sum of five prime numbers (which need not be consecutive) is 1146. Find the cube of the smallest prime number (among these five prime numbers).

Answer: None of these

Solution: .Other than 2, all prime numbers are odd. But the sum of five prime numbers is even which means 2 is one among them. (Sum of five odd numbers will always be odd. This implies one of them should be even).

As 2 is the smallest prime number, cube is 8.

11. If log 150 = 2.17609, find the value of log 3.375.

Answer: 0.5283

Solution: log150 = log (15*10) = log15 + log10

= log15+1 = 2.17609.

Hence log15 = 1.17609.

As log15 = log(3*5), log3+log5 = 1.17609.

Now considering log3.375,

log 3.375 = log (5*5*3*3*15* 1/1000)

= 2log5 + 2log3 + log15 - 3log10

= 2*(log5+log3) + log15 - 3

= 2*1.17609 + 1.17609 - 3

= 0.52827 which is approx 0.5283

12. The difference between the compound interest and the simple interest on a certain sum at 12.5% for two years is Rs.120. What will be the difference between the compound and simple interest at the end of 3 years?

Answer: Rs.375

Solution: The S.I and C.I are same at the end of 1 year.

In the second year the 12.5% on the additional interest accrued is Rs. 120.

 Let the amount invested be x. C.I accrued at the end of first year = x * 12.5/100

x * 12.5/100 * 12.5/100 = 120, x = Rs.7680.

S.I for 3 years = 7680 * 3 * 12.5/100 = Rs.2880

C.I for 3 years = 7680 (1 + 12.5/100)^3 - 7680 = 7680 * (9/8)^3 - 7680 = Rs.3255

Difference = 3255 - 2880 = Rs.375.

13. The average marks of girls in a class is 70 . The average marks of 5 girls among them is 75. The average marks of remaining girls is 67.5, then what is the number of girls in the class ?

Answer: 15

Solution: Let the number of girls be x. 70x = 5*75 + 67.5 (x-5). Solving x = 15.

14. A board has 16 squares arranged in 4*4 format. A given square can be filled with numbers from 1 to 36 (with 1 to 36 repeating any number of times to fill any given square). If for any given square, the average of numbers around the square must be equal to the number present in the square, how many such arrangements are possible?

Answer: 36

Solution: Let x be the smallest number present in a given arrangement.

This implies if a number present in the surrounding squares is greater than x, then some of the numbers present in the surrounding squares must be smaller than x which contradicts the above statement.

Hence for any given number x, the only possibility is that all the squares are filled with a value which is not greater than (or smaller than) x, that is a value which is equal to x.

As there are 36 numbers from 1 to 36, the answer is 36 ways.

15. What is the remainder when 12^408  is divided by 5?

Answer: 1

Solution: Also 12 and 5 are coprime.

12^408 mod 5 = 12^(4*102) mod 5 = 1 (As Euler's Totient for 5 is 4)

16. A cricket team that played 60 games had won 30% of its game played. After a phenomenal winning streak this team raised its average to 60%. How many games must the team have won in a row to attain this average?

Answer: 45

Solution: Total game played= 60

%won =30%

Total won= 60*30/100 i.e. 18

Now team plays x games and win all of those to increase the average to 60%.

So, (60+x)*60/100=18+x

(60+x) * (3/5) =18+x

180 + 3x = 90 + 5x

2x = 90, x = 45; So the final answer is 45

17. Rs. 800 amounts to Rs. 920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to how much?

Answer: Rs. 992

Solution: Interest per year = (920-800)/3 = Rs.40

Let rate of interest be R.

800 * R/100 * 1 = 40, R = 5%.

If R is increased by 3%, it will be 8%.

Hence S.I for 3 years = 800 * 8/100 * 3 = 192.

Hence the amount will be 192+800 = 992

18. A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 2.7  less than the previous one. What was the difference of the two digits a and b?

Answer: 3

Solution: Original number = 10a+b ; Interchanged number = 10b+a

Reduction in the average of 10 numbers = 2.7

Total = 27

10a+b - 10b -a = 27 ; 9a - 9b = 27; a-b=3

19. The difference between a three digit number and it's reverse is always divisible by

Answer: 33

Solution: Let the number be xyz

100x+10y+z - (100z + 10y -x) = 99(x-z)

Hence the difference is always divisible by 99 or it's factors.

Hence among the given options answer is 33.

20. Vishu is younger than Sagul by 6 years. If their ages are in the respective ratio of 5:7, how old is Vishu?

Answer: 15

Solution: Let the age of Sagul and Vishu be S and V respectively

S = V + 6

V/S = 5/7

so, V / (V+6) = 5 / 7

V = 15

21. The average mark of a student in 10 tests is 80. If the highest and lowest scores are not considered, the average is 81. If the highest score is 92, find the lowest score.

Answer: 60

Solution: Highest + Lowest = 80*10 - 81*8 = 152

Lowest = 152-Highest = 152-92 = 60

22. The average age of 40 students of a class is 15 years. When 10 new students are admitted, the average is increased by 0.2 years. Find the average age of the new students.

Answer: 16 years

Solution: Total age of 40 students = 40 * 15 = 600 years

New total age of 50 students = 50 * 15.2 = 760 years

Therefore, total age of 10 new students = 760 - 600 = 160 years

Average age of 10 new students = 160/10 = 16 years

23. Choose the valid option with reference to the below equation:

Answer: a+b = 1

Solution: Answer is a+b=1

log(a/b) + log(b/a) = log(a+b)

log(a/b*b/a) = log(a+b)

log(1) = log(a+b). So, a+b = 1.

24. A team of 8 persons participate in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The team score was 

Answer: 665

Solution: Total points scored by team if the best marksman had scored 92 = (8*84) = 672.

But he scored only 85 which is 7 points less than 672. So actual team score is 672-7=665.

25. There were 35 students in a hostel. When the number of students increased by 7, the expense increased by Rs.42 per day and the average expenditure per day decreased by Re.1. Find the original expenditure.

Answer: Rs.420

Solution: Let original expense per day per student be x.

35x + 42 = 42(x-1),

x=12. Thus original expense = 12*35 = Rs.420

26. If 774958A96B is to be divisible by 8 and 9, then A+B is

Answer: 8

Solution: To be divisible by 8, last three digits should be divisible by 8.

For 96B to be divisible by 8, B must be 0 or 8

To be divisible by 9, digital sum of digits must be divisible by 9.

This 55+A+B digital sum is 9.

So when B=0, A must be 8

when B=8, A must be 0

Hence in both cases A+B=8

27. A hundred digit number is formed by writing numbers starting from 1 in the following manner.

123456789101112.......

What will be the remainder when the number is divided by 8?

Answer: 1

Solution: As there are 100 digits, the number will be like 123456789101112.....515253545 (55 cannot be written completely but first digit alone will be used as it is the 100th digit)

So 545 mod 8 the remainder is 1

28. The average weight of all the members in a family is 65kgs. The average weight of men in the family is 70kgs and that of women in the family is 55kgs. Find the ratio of number of men to that of women.

Answer: 2:1

Solution: Let the num of men be m and women be w.

65(m+w) = 70m+55w,

5m = 10w, Hence ratio = 2:1

29. In a company, each employee gives a gift to every other employee. If the total number of gifts is 992, then the number of employees in the company is

Answer: 32

Solution: Let num of employees be x.

x*(x-1) = 992, Solving x=32

30. In what time will Rs. 4500 amount to Rs. 4770 at 3% per annum?

Answer: 2 years

Solution: S.I earned = 4770-4500 = 270.

Let the time required be N years.

4500 * 3/100 * N = 270, N = 2.