puzzle

1    A shopkeeper is selling his goods at a profit of 25%. One day a customer bought a toy worth Rs 400. The customer gave the shopkeeper a Rs 500 note. Since the shopkeeper did not have the change, he went to the next shopkeeper for help. His neighbouring shopkeeper gave him 5 notes of Rs 100 after taking the Rs 500 note from the him. The shopkeeper then returned Rs 100 to the customer. In the evening the neighbouring shopkeeper came back to the shop and told the shopkeeper that the Rs 500 note that he had given to him is fake. The shopkeeper promptly returned him a fresh Rs 500 note. How much money did the shopkeeper loose?

A) 500                   B) 920                    C) 600                    D) 1000

JMO, indices, number

In the sum 23¹⁵ + 19¹⁷ + 27³⁵, find the digit in the unit place.

JMO STD V and STD VI

numbers -- JMO std. V and VI

 Find the total number of digits in the number
123456789101112..........20102011

JMO STD V and STD VI

series summation - JMO

Pages of a book are numbered from one onwards. A boy has torn one page of the book. If sum of page numbers of remaining pages of the book is 1620.
Then (1) find page number of torn page and
(2) How many pages the book had ?

JMO ---- STD V and VI

distance and time --- JMO V and VI

If I travel at 30 km/hr I reach  my destination at 10 O clock.
If I travel a 20 km/hr I reach my destination at 4 O' Clock.
How far is the destination from my house

JMO STD V VI

JMO VII and VIII -- series problems

If 

3! = 3 x 2 x 1;

4! = 4 x 3 x 2 x 1,

5! = 5 x 4 x 3 x 2 x 1

Then, find the value of 

1.1 ! + 2.2! + 3.3! + 4.4! + …… + 99.99!               

 JMO VII and VIII

JMO -- DIvisibility

Which are the two numbers less than 260 that exactly divides

 2³² – 1?

JMO – VII and VIII

MAT test paper ---- STD V and STD VI

MAT test paper ---- STD V and STD VI

series problems for STD V and STD VI

Questions on the application of the formulae:

For S = 1 + 2 + 3 + .....n terms = i.e. sum of natural numbers from 1 to n.
S = n (n+1)/2

1. S = ( 1 + 3 + 5 + ... 25 terms) + ( 2 + 4 + 6 ..... 25 terms)
2. S = (3 + 6 + 9 ...... 25 terms)
3. S = ( 2 + 4 + 6 + ...... 25 terms)
4. S = (1 - 1/5) + (1 - 2/5) + (1 - 3/5) + (1 - 4/5) + ..... 25 terms
5. S = (1 - 1/n) + (1 - 2/n) + (1 - 3/n) +........ n terms
6. S = (2a - 3b) + (4a - 6b) + (6a - 9b) +.......... 25 terms, given a = 5 and b = 2

MAT and IPM mega final test paper

Maths Aptitude Test ----- last 15 questions for std. VIII

IPM mega Finals practice Paper for Std. VIII

1.       If 4 kg of coffee is cost as much as 3 kg of tea, and 5 kg of tea cost as much as 10 kg of sugar. What is the price of each when 1 kg of coffee, 1 kg of tea and 1 kg of sugar together cost Rs 112.50?

2.       The price of cooking gas has increased by 20%. By what percent should the consumption come down so as not to increase his expenditure?

3.       A dishonest vendor professes to sell his goods at cost price but he uses a weight of 900 grams for a kg weight. Find his gain Percent?

4.       A reduction in 20% in the price of onions enables a purchaser to obtain 8 kgs more for Rs 80/-. What is the reduced price per Kg?

5.       1 kg of tea and 4 kg of sugar cost Rs 88.20, but if sugar prices rose by 50% and that of tea by 10% they would cost Rs 114.30. Find the price per kg of tea? Answer Rs 45

6.       A sells his goods 5% cheaper than B and 5% dearer than C. A man buys Rs 1000 worth of goods from A. Would it have been more or less profitable for him to have bought half thee goods from B and half from C?

7.       Find the simple interest on Rs 1000 from March 3^rd to July 27^th of the same year @ 5% per annum?

8.       An army consists of 12000 men and women. The average height of the men is 1.9 metres, that of the women is 1.8 metres and that of the army is 1.875 metres. Find the number of women in the army.

9.       If S = (1+ 3 + 5 + ........... 35) + (2 + 4 + 6..........+ 34), Then the Value of S is...........................

10.   Q as a percentage of P is equal to P as a percentage of (P + Q). Find Q as a percentage of P. (given square root of 5 = 2.236)

11.   Solve:       ^{(719×719+347×347−719×347)} / _{(719×719×719+347×347×347)} =.....................................

12.   Find the sum:  (1−¹/_n) + (1−²/_n) + (1−³/_n) +.................up to n terms=?

13.   What is the remainder when 17²⁰⁰ is divided by 18?  Answer 1

14.   If 7^{(x - y)} = 343 and 7^{(x + y)} = 16807, what is the value of x?

15.   What is the unit digit in (6324)¹⁷⁹⁷ × (615)³¹⁶ × (341)⁴⁷⁶?

Chain rule

Chain Rule Questions from Pitambar Lal   ( with solution)

how to solve chain rule problems

A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the provisions will now last just as long as before. How long was that?

a)      50 days                b) 30 days          c) 40 days           d) 60 days

Answer:

Let “x” be the number of days the provisions would last for the garrison of “y” men

We can re phrase the question now.

“ there is provision for y men for ( x – 10) days, and there is provision for ( y – y/5 = 4y/5) men for x number of days”

So, x = (x – 10) * {y/(4y/5)}

Or, x = (5/4)( x – 10)

Or, 4x = 5x – 50

Or, x = 50 days. answer is a.

If 40 men can make 30 boxes in 20 days, How many more men are needed to make 60 boxes in 25 days?

a)      28           b) 24                      c) 22                       d) 26

Answer:

40 x (60/30) x (20/25) =  64 men

So, extra men = 64 – 40 = 24. Answer is b.

Note: for making more number of boxes you will need more number of men. So fraction must be more than 1............... (60/ 30)

In more number of days you will need less number of men.

So, the fraction should be less than 1............................(20/25)

In a camp, food was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came?

a)      1900       b) 1800                 c) 1940                  d) 2000

Answer:

We will re-phrase the question

There is food for 2000 people for (54 – 15 = 39) days. Food lasted for 20 days. So, how many people where there.

= 2000 x (39/20) = 3900

So, extra people who joined was = 3900 – 2000 = 1900

answer is a

How to solve problems using chain rule

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?

a.       1              b.  40                     c. 20                       d. 26

Answer:

What do we need to find? ... the number of days.

So, first record the day data given =  40 x ..................       (1)

Next go to the other variables. First, consider the number of cows.

Since the number of cows is decreasing, the husk will last more number of days. Whenever there is more, your multiplication factor should be greater than 1. So, (40/1)  ------ (2)

Second variable is bags of husk

Since the number of bags of husk is decreasing, the husk will last less number of days. Whenever there is less, your multiplication factor should be less than 1. So, (1/40) ------ (3)

Therefore, total number of days = 40 x (40/1) x (1/40) = 40 days   ...............(1) x (2) x (3)

Answer is B.

In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?

a)      50                           b) 40                      c) 30                       d) 10

Answer:

150 children = 120 x (150/200) men = 90 men

So, balance number of men = 120 – 90 = 30

 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day?

a)      10                           b) 12                      c) 8                         d) 15

Answer:

= 8 x (3/4) x(2/1) = 12 hours

Note the factor   ----  ¾

More number of pumps will take less number of hours to do the job. less means fraction should be less than 1. So, ¾ and not 4/3.

Now, note the factor   2/1

In job is to be completed in less number of days, more hour of work is required. More means the fraction should be more than 1. So, 2/1 and not ½.

Remember, that relationship has to be established strictly with the variable you have to find.

21 goats eat as much as 15 cows. How many goats eat as much as 35 cows?

a)      49                           b)32                       c)36                        d)41

Answer:

 21 x (35/15) = 49

More number of cows means more the number of goats. So, the multiplication factor should be more than 1. So, fraction will be (35/15) and not (15/35)

A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. Find out the number of days for which the remaining food will last.

a)      45                           b) 42                      c)40                        d)38

Answer:

We can rewrite the above question as, “the fort has provision for 150 men for (45 – 10 = 35) days. How many days will the provision last for (150 – 25 = 125) men”

 35 x (150/125) = 42 days

Answer is b

Note: less number of men means provision will last more number of days. More means fraction should be greater than 1. So, fraction is (150/125) and not (125/150).

Maths Aptitude test (MAT) for class V and VI

Maths Aptitude test (MAT) for class V and VI ----- Last 15 questions of 50 questions

MAT practice paper for STD V / Std VI

MAT practice paper Std. V/ Std. VI

IPM mega final exam

Hope you fared well in the Mega finals of IPM.
Would you mind sending me the question papers on my email
shubhro.enerjet@gmail.com
I will post them so that everyone can benefit from it.

MAT /IPM Mega Final Practice paper STD VIII

MAT / IPM Mega Final Practice paper STD VIII

MAT/ IPM mega final practice paper std VII

MAT/ IPM mega final practice paper  std VII

MAT/ IPM Mega Final practice paper --- STD VI

MAT / IPM Mega Final STD VI - practice paper

MAT / IPM Mega Finals Practice paper - STD V / STD VI

MAT / IPM Mega Finals Practice paper - STD V / STD VI