10-09-2021 Answers

 1. Between 250 and 750,how many integers are divisible by 11?

Answer: 46

Solution: The integers are like 253, 264, ...., 748

Here d=11 and a1 = 253, An=748.

So, 748 = 253 + 11(n-1), Solving n=46

2. Three terms in a geometric progression have their sum as 49 and the product as 2744. Find the smallest among these terms.

Answer: 7

Solution: Let the terms be x/d, x, xd.

Product = x^3 = 2744. So x=14.

Sum = 14/d + 14 + 14d = 49,

Solving d=1/2. The terms are 28, 14, 7.

3. Find the sum of 20 + 40 + 60 + 80 + 100 + ........ + 2000

Answer: 101000

Solution: 20 + 40 + 60 + 80 + 100 + ........ + 2000 can be written as

20 (1+2+3+ .....+100) = 20 * 100/2 (1+100) = 101000  

4. If the sum of 3rd and 15th terms of an arithmetic progression is equal to the sum of 6th, 11th and 13th terms of the same progression, which term of the series should be equal to zero?

Answer: 12th term

Solution: Let the initial term be a and the difference between the terms be d.

3rd term = a+2d and 15th term = a+14d, 6th term = a+5d and so on.

Hence a+2d + a+14d = a+5d + a+10d + a + 12d,

16d = a+27d, a+11d = 0. Hence the 12th term will be zero.

5. Two numbers A and B are such that their GM is 20% lower than their AM. Find the ratio between the numbers

Answer: 4:1

Solution: 4/5 * (A+B)/2 = SQRT(AB)

squaring both sides,

4 * A^2 + 4 * B^2 + 8AB = 25 * AB,

4 * A^2 + 4 * B^2 -17AB =0,

(A-4B)(4A-B) = 0,

Hence either A=4B or B=4A.

So the ratio of A:B is 1:4 or 4:1.

Among the given options 4:1 is the answer.

6. Fourth term of an arithmetic progression is 10. What is the sum of the first 7 terms in the same progression?

Answer: 70

Solution: Fourth term = a1 + (4-1)d = a1 + 3d = 10

Sum of seven terms = a1 + a1 + d + a1 + 2d + .... + a1 + 6d = 7a1 + 21d = 7(a1+3d)

= 7 * 10 = 70

7. Find the value of the expression 1-2+3-4+5-6+ ..... -96+97-98+99

Answer: 50

Solution: There are two series 1+3+5+....+99 and -2-4-6.....-98

Sum of 1+3+5+....+99 = 50/2 * (99+1) = 2500

Sum of -2-4-6.....-98 = - 49/2 * (98+2) = -2450

so net result = 2500 - (2450) = 50

8. A series is given as 

1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4, .........................

What is the number at 2320th position (term) in the sequence?

Answer: 4

Solution: In the series there are 10 terms in the series 1,2,2,3,3,3,4,4,4,4.

Next will be 20 terms, followed by 40 terms, 80 terms and so on.

10+20+40+80+160+320+640 = 1270th term will end with 4.

Out of next 640*2 = 1280 terms, first 128 will be 1s, next 256 terms will be 2s, next 384 terms will be 3s and the next 512 terms will be 4s.

2320 - 1270 = 1050 which is 128+256+384+282. Hence the number will be 4.

9. A man bought 1 cow on the first day. He bought 11 cows on day two, 21 cows on day 3 and 31 cows on day 4. The number of cows bought was increased by ten every day. How many cows in total were bought by him on Day 50?

Answer: 12300

Solution: The formula to find sum of terms in Arithmetic progression is = n/2 (2a1 + (n-1)d) . Here a1 = 1 (as one cow bought on day one, n=50 and difference between terms = 10=d)

= 50/2 * (2 + 49 * 10)= 25 * 492 = 49200/4 = 12300

10. The ninth term and thirteenth term of an arithmetic progression are 15 and 7 respectively. Find the sum upto 30 terms.

Answer: 60

Solution: A9 = A1 + 8d = 15

A13 = A1 + 12d = 7,

So d = -2.

A1 = 31 and A30 = -27

Sum = 30/2 * (31-27) = 60

11. How many terms of the series 1 + 5 + 9 + .... must be taken in order that the sum may be 190?

Answer: 10

Solution: a=1, d=4

Sum = n/2 * (a1 + an) = 190

n/2 * (1 + 1 + (n-1)*4) = 190, (an = a1 + (n-1)d)

Solving n = 10

12. A student wrote all numbers divisible by 7 from 1 to 350. What is the sum of all such numbers?

Answer: 8925

Solution: The numbers form the progression 7,14,21,28 ......, 350

350 = 7 + (n-1)7

n=50. Hence 50 such numbers are there.

Their sum as per the formula = (7+350)/2 * 50 = 357 * 25 = 357*100/4 = 8925

13. Find largest of the three terms in Arithmetic progression such that their sum is 33 and product is 1155.

Answer: 15

Solution: Let the terms be x-d, x, x+d,

Sum = 3x = 33, x=11.

Product = (11-d)11(11+d) = 1155,

(121-d^2) = 105, d=4.

The terms are 7,11,15

14. There is one lily in the pond on 1st june. There are two in the pond on 2nd june. There are four on 3rd june and so on. The pond is full with lilies by the end of the june. On which date the pond is half full?

Answer: 29th June

Solution: As it is doubling every day and June has 30 days, the previous day (29th) the pond will be half full.

15. The 8th term of a geometric progression is 27 times the 5th term. If the third term is 54, the first term is

Answer: 6

Solution: If 5th term is x/d, 8th term = x*d^2 = 27 * x/d.

Thus d^3 = 27 and hence d=3.

First term = Third Term/(d^2) = 54/(3^2) = 6

16. Find the sum of 112 + 126 + 140 + 154 +.... + 980 + 994.

Answer: 35392

Solution: 112 + 126 + 140 + 154 +.... + 980 + 994 can be written as

14 (8+9+10+11+ .....+71) = 14 * 64/2 (8+71) = 35392

17. A book series was published at 7 seven year intervals. When the seventh book was published, the total sum of publication years was 13,524. When was the first book published?

Answer: 1911

Solution: Let the year in which first book was published is a. So second book in the year x+7, third on x+14 and so on (Note that this is arithmetic progression).

The formula is Sum of n terms = n/2 * (2a + (n-1)d), where a is the initial term, d is the difference between terms, n is the number of terms in sequence.

Hence n=7, d=7.

13524 = 7/2 * (2a + 6*7),

2a= 13524*2/7 - 42 , solving a=1911.

18. Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms in the same progression?

Answer: 56

Solution: Fourth term = a1 + (4-1)d = a1 + 3d = 8

Sum of seven terms = a1 + a1 + d + a1 + 2d + .... + a1 + 6d = 7a1 + 21d = 7(a1+3d)

= 7 * 8 = 56

19. How many terms of the geometric progression 2,6,18,... are needed to give a sum of 6560?

Answer: 8

Solution: Here d=3, a=2.

Sum = a * (d^n -1)/(d-1) = 6560,

2 * (3^n -1)/2 = 6560,

3^n = 6561,

Solving n=8. Hence 8 terms are required.

20. Find the smallest number in a GP whose sum is 38 and product 1728

Answer: 8

Solution: Let the numbers in geometric progression be x/a, x and ax.

Now x/a + x + ax = 38

x(1+a+a^2)=38a - Eq1

Product = x^3=1728

x=12.

Using Eq1

12(1+a+a^2)=38a

a^2 -13a/6 +1 =0

Solving a=2/3 or 3/2.

 we get the smallest number when we consider a=3/2. Hence in the series 8,12,18 the smalles number is 8.

21. Find the 20th term of the series 5 + 9 + 13 + 17 + 21 ......

Answer: 81

Solution: a = 5 and d=4

20th term = 5 + 19*4 = 81

22. Find the sum upto ten terms of an arithmetic progression whose fifth term is 5 and seventh term is 3

Answer: 45

Solution: A5 = A1 + 4d = 5

A7 = A1 + 6d = 3,

So d = -1.

A1 = 9 and A10 = 0

Sum = 10/2 * (9) = 45

23. How many 3 digit numbers that leave a remainder of '2' when divided by 3 are there?

Answer: 300

Solution: The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.

The next number will be 104, 107, ....

The largest 3 digit number is 998.

Using the formula for arithmentic progression,

998 = 101 + (n - 1)* 3, n = 300

24. What is the sum of all even integers between 99 and 301?

Answer: 20200

Solution: Sum = 100+102+ ... +300

Using A.P formula, a1=100 and an =300 and d=2.

300 = 100 + (n-1)*2, n=101

Hence using the formula to find the sum in an arithmetic progression,

Required sum = 101/2 * (100+300) = 20200

25. The 9th term of a geometric progression is 64 times the 6th term. If the fourth term is 96, the second term is

Answer: 6

Solution: If 6th term is x/d, 9th term = x*d^2 = 64 * x/d.

Thus d^3 = 64 and hence d=4.

Second term = Fourth Term/(d^2) = 96/(4^2) = 6