harder factorisation

Resolve into factors:

15 y² + 22y – 48

Answer:

Multiply and divide the expression by 15.

So,  ¹/₁₅ {(15y)² + 22.15y – 48.15}

Let 15y = x

Therefore, 1/15 {x² + 22x - 720}

=¹/₁₅{x² + 40x – 18x - 720}

=¹/₁₅(x+40)(x-18)

=¹/₁₅(15y +40)(15y- 18)  substituting x = 15y

= (3y +8)(5y -6)

Resolve into factors:

28 + 31p -5p²

Answer:

28 + 31p -5p²

=28 + 35p – 4p - 5p²

=7(4 + 5p) –p(4 + 5p)

= (7 – p)(4 + 5p)

 

Resolve into factors:

15 + 16p – 15p²

Answer:

15 + 16p – 15p²

Multiply and divide the numerator and denominator by 15

= ¹/₁₅{15.15 + 15.16p – (15p)²}

Let, 15p = y

= -¹/₁₅{y2 – 16y +15} 

=-¹/₁₅{y2 – 15y - y +15} 

=-¹/₁₅{y(y – 15) -1( y -15)} 

= -¹/₁₅(y -1)(y – 15)

= -¹/₁₅(15p -1)(15p -15)

= - (15p -1)(p – 1)

= (15p -1)(1- p)

mixtures, STD V

A, B and C have $ 507 between them. B and C together have $ 345 and C and A together have $ 305. How much does C have?

Answer:

A+B+C =507--------(1)

B+C = 345----------(2)

A+C = 305----------(3)

Adding (2) and (3), we get A+2B+C = 345 + 305------(4)

Substituting (1) in (4), we get

507 + B = 650

Or, B = 143

Therefore, C = 345 – 143 = 202

A Sum of money is divided into A, B and C. C gets twice as much as A, A and B together get $70, and B and C together get $90. How much does each person get?

Answer:

2A = C -------(1)

A + B = 70------(2)

B +C = 90-------(3)

(2) – (3)

A – C = -20,-----(4)

Substituting (1) in (4), we get

A – 2A = -20

Or, A = 20, therefore B = 50 and C = 40

Tougher problems on HCF and LCM

In the HCF of two numbers which were prime to each other, the quotients are 2,2,7,2 on finding HCF by method of division. Find the two numbers.

 Answer:

Since the numbers are prime, the HCF is 1.

Now following the method of finding HCF by the method of division, we get

Final Dividend = 1x2 = 2

In the pen ultimate division, the divisor is 2, the quotient is 7 and the remainder is 1.

So, the dividend is = 2x7 + 1 = 15

In the prior division, the divisor is 15, the quotient is 2 and the remainder is 2.

So, the dividend is = 15x 2 + 2 = 32

Therefore, the smaller of the two numbers is 32.

The next number is the divisor = 32 x 2 + 15 = 79

So, the two numbers are 32 and 79.

                                                               

Check:

                               32) 79 (2

                                     64

                                       15) 32 (2

                                              30

                                                2)15 (7

                                                   14

  1) 2 (2

2

x

tougher problems in HCF and LCM

                                                                                                                                                                                               

The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find the number and the remainder.

                                                                                                                                                                                               

Answer:

Let, Y be the common remainder. So, we have 11284 – y , 7655 – y as the perfectly divisible numbers.

If 2 numbers are exactly divisible by a certain number, then there difference is also exactly divisible.

So, (11284 – y) – (7655 – y), is exactly divisible.

Or, 3629 is divisible by the required number. Let us find the factors of 3629.

We have, 3629 = 191 x 19

Therefore the 3 digit number is 191.

Now let us divide 11284 by 191 to find the remainder

11284/191 = 59 ¹⁵/₁₉₁.

Therefore, the remainder is 15.

simple equations STD VI

Present ages of Anu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.

Answer:

Let the present ages of Anu and Raj be 4x years and 5x years respectively.

After eight years. Anu’s age = (4x + 8) years;

After eight years, Raj’s age = (5x + 8) years.

Therefore, the ratio of their ages after eight years = (4x + 8)/(5x + 8)

But this is given as 5: 6

Therefore, (4x + 8)/(5x + 8) = 5/6

Cross-multiplication gives 6 (4x + 8) = 5 (5x + 8)

Or, 24x + 48 = 25x + 40

Or, 24x + 48 – 40 = 25x

Or, 24x + 8 = 25x

Or, 8 = 25x – 24x

Or, 8 = x

Therefore, Anu’s present age = 4x = 4 × 8 = 32 years

Raj’s present age = 5x = 5 × 8 = 40 years

The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Answer:

Let the present ages of Hari and Harry be 5x years and 7x years respectively.

After eight years. Hari’s age = (5x + 4) years;

After eight years, Harry’s age = (7x + 4) years.

Therefore, the ratio of their ages after eight years =

(5x + 4)/(7x + 4)

This is given to be 3: 4

Therefore, (5x + 4)/(7x + 4) = 3/4

Cross-multiplication gives  4(5x + 4)=  3(7x + 4)

Or, 20x + 16 = 21x + 12

Or, 16 – 12 = 21x – 20x

Or, 4 = x

Or, x = 4

Therefore, Hari’s present age = 5x = 20 years

Harry’s present age = 7x = 28 years

The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

Answer:

Let the numerator be y

So, the denominator will be 8 + y

Given, ( y + 17)/ (8 + y -1) = 3/2

Or, 2(y + 17) = 3( 7 + y)

Or, 2y + 34 = 21 + 3y

Or, 34 – 21 = 3y – 2y

Or, y = 13

So, the rational number is = 13/21

simple equations STD VI

Half of a herd of deer are grazing in the field and three fourths of the remaining is playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Answer:

Let the number of deer be y

No. of deer grazing = y/2

Remaining no. of deer = 1-y/2 = y/2

No. of deer playing = ¾ x y/2 = 3y/8

Given that, y – (y/2 + 3y/8) = 9

Or, y – y(4+3)/8 = 9

Or, y(1 – 7/8) = 9

Or, y x 1/8 = 9

Or, y = 72.

A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Answer:

Let the age of the granddaughter be y

So, the age of the grandfather is 10y.

Also given that 10y – y = 54

Or, 9y = 54

Or, y = 6 years

So, the age of the granddaughter is 6, and the age of the grandfather is 60.

Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Answer:

Let son’s age by y

So, Aman’s age is 3y

Also given that, 5 x (y-10) = 3y – 10

Or, 5y – 50 = 3y – 10

Or, 5y – 3y = 50 – 10

Or, 2y = 40

Or, y = 40/2 = 20

So, the son’s age is 20 years and Aman’s age is 60 years.

Simple Equations STD VI

Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Answer:

Let Shobo’s present age by y.

So, Shobo’s mother’s present age is 6y.

Also given, (y + 5) = 1/3 x 6y

Or, y + 5 = 2y

Or 2y – y = 5

Or, y = 5

So, Shobo’s age is 5.

And Shobo’s mother’s age is 6y = 30. 

There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per meter it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Answer:

Let the length of the plot be L

Let the breadth of the plot be B

So, given L:B = 11: 4

Or, 4L = 11B

Or, B = 4L /11

Now perimeter of the rectangular plot is

= 2 ( L + B)

= 2 ( L + 4L/11)  because B = 4L/11

Now given that 100 x 2(L + 4L/11) = 75000

Or, 200(L + 4L /11) = 75000

Or, L (1 + 4/11) = 75000/200

Or, L(15/11) = 375

Or, L = 375 x 11/15 = 25 x 11 = 275

So, the length of the plot is 275 meters

And the breadth of the plot is 4x275/11 = 4 x 25 = 100 meters.

Check:

2(275+100) x 100 = 750x100 = 75000

Simple equations STD VI

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Answer:

Let the number be y

Therefore the other number is 5y

Also, 2(21 + y) = 5y + 21

Or, 42 + 2y = 5y + 21

Or, 3y = 42 – 21

Or, 3y = 21

Or, y = 7

So, the two numbers are 7 and 35

Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Answer:

Let the number in the units place be y

So, the other number is 9-y.

Given, 10(9 – y) + y + 27 = 10y + (9 – y)

Or, 90 – 10y + y + 27 = 10y + 9 – y

Or, 117 – 9y = 9y + 9

Or, 18 y = 117 – 9

Or, y = 108/18 = 6

So, the other number is 9- y = 3

So, the 2 digit number is 36.

Simple equations STD VI

Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.

Answer:

Let us take Shriya’s present age to be x years.

Then Arjun’s present age would be 2x years.

Shriya’s age five years ago was (x – 5) years.

Arjun’s age five years ago was (2x – 5) years.

It is given that Arjun’s age five years ago was three times Shriya’s age.

Thus, 2x – 5 = 3(x – 5)

Or, 2x – 5 = 3x – 15

Or, 15 – 5 = 3x – 2x

Or, 10 = x

So, Shriya’s present age = x = 10 years.

Therefore, Arjun’s present age = 2x = 2 × 10 = 20 years.

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?

Answer:

Let us take the two digit number such that the digit in the units place is b. The digit in the tens place differs from b by 3. Let us take it as b + 3. So the two-digit number is,          

10 (b + 3) + b = 10b + 30 + b = 11b + 30.

With interchange of digits, the resulting two-digit number will be,

10b + (b + 3) = 11b + 3

If we add these two two-digit numbers, their sum is

(11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33

It is given that the sum is 143. Therefore, 22b + 33 = 143

Or, 22b = 143 – 33

Or, 22b = 110

Or, b = 110/22

Or, b =5

The units digit is 5 and therefore the tens digit is 5 + 3, which is 8. The number is 85.

Check: On interchange of digits the number we get is

58. The sum of 85 and 58 is 143 as given.

Simple equations STD VI

The organizers of an essay competition decide that a winner in the competition gets a prize of $ 100 and a participant who does not win gets a prize of $ 25. The total prize money distributed is $ 3,000. Find the number of winners, if the total number of participants is 63.

Answer:

Let the number of winners be y

So, number of students who did not win = 63 – y

Given that, 100y + (63 – y )25 = 3000

Or, 100y + 1575 – 25y = 3000

Or, 75y = 3000 – 1575

Or, y = 1425/75

Or, y = 19.

I have a total of $ 300  in notes of denomination $ 1, $ 2 and $ 5. The number of $ 2  notes is 3 times the number of $ 5 notes. The total number of notes is 160. How many notes  of each denomination are with me?

Answer:

Let the number of $5 denomination note be y

So, number of $2 denomination note is 3y

So, number of $1 denomination note = 160 – y – 3y = 160 -4y

Given, 5y + 2x3y +1x (160 – 4y) = 300

Or, 5y +6y + 160 – 4y = 300

Or, 7y = 140

Or, y = 20

The number of $5 denomination note is 20

So, number of $2 denomination note is 3y = 60

So, number of $1 denomination note = 160 - 4y = 80

Simple equations STD VI

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?

Answer:

Let total number of notes be y

So, fraction of notes of Rs 100 denomination = 2/ (2+3+5) = 2/ 10 = 0.2

So, fraction of notes of Rs 50 denomination = 3/ (2+3+5) = 3/ 10 = 0.3

So, fraction of notes of Rs 100 denomination = 5/ (2+3+5) = 5/ 10 = 0.5

Therefore, 0.2y x 100 + 0.3y x 50 + 0.5y x 10 = 400000

Or 20y +15y + 5y = 400000

Or, 40y = 400000

Or, y = 400000/40 = 10000

So, 100 denomination notes = 0.2 x 10000 = 2000

So, 50 denomination notes = 0.3 x 10000 = 3000

So, 10 denomination notes = 0.5 x 10000 = 5000

Check:

2000 x 100 = 200000

3000 x 50 = 150000

5000 x 10 = 50000

SO, total = 200000 + 150000 + 50000 = 400000

simple equations STD V

• Three Consecutive integers add up to 51. What are these integers?

          Let the integers be y-1, y, y +1.

         Given that y-1 + y + y +1 = 51

         Or, 3y = 51

         Or, y = 17

         So, the three numbers are 16, 17 and 18.

• The sum of 3 consecutive multiples of 8 is 888. Find the multiples?

          Let the numbers be y-8, y and y +8

         Given that y-8 + y + y + 8 = 888

         Or, 3y = 888

         Or, y = 296

         So, the three numbers are 288, 296, 304

• The ages of Rahul and Haroon are in the ratio of 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

         Let the age of Rahul four years hence be y.

         So, age of Haroon is 56 – y.

         Present age of Rahul = y – 4

         Present age of Haroon = 56 – y – 4 = 52 – y

        Given y-4 : 52 – y = 5 : 7

         The product of mean = product of extreme. So,

         7(y – 4 ) =5(52 – y)

        Or, 7y – 28 = 260 – 5y

        Or, 12y = 260 + 28 = 288

        Or, y = 288/ 12 = 24

        So, Haroon’s age 4 years hence is = 56 – y = 32

        So, the present age of Rahul and Haroon are = 20 and 28.

• The number of boys and girls in a class are in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

          Answer:

          Let the number of girls in the class be y

          So, the number of boys = y + 8

          Given that, (y+8):y = 7:5

          Product of mean = product of extreme, so,

          5(y+8) = 7y

         Or, 5y + 40 = 7y

         Or, 2y = 40

         Or, y = 20

         So, the total strength of the class is = y + y+8 = 20 + 28 = 48. 

simple equations STD V

• If you subtract ½ from a number and multiply the result by ½, you get 1/8. What is the Number?  

         Answer:

          Let the number be y

          Then, (y- ½ ) x ½ = 1/8

          Or, y – ½ = 2/8 = ¼

          Or, y = ½ + ¼ = ¾ 

• The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 2/15cm. What is the length of either of the remaining equal sides?

          Answer

          Let the side be y.

          Since it is an isosceles triangle, the 2 sides are equal.

          So, the equation would be y + y + 4/3 = 4 ²/₁₅ = 62/15

          Or, 2y = 62/15 – 4/3

          Or, 2y = (62 – 20)/15

          Or, 2y = 42/15

          Or, y = 42 / (15 x 2) = 21/15 = 1 ⁶/₁₅

• The sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

         Answer:

         Let one of the number be y.

         Then the other number is y + 15.

         Given that y + y + 15 = 95

         Or, 2y + 15 = 95

         Or, 2y = 95 -15

         Or, 2y = 80

         Or, y = 40

         Therefore the other number is y + 15 = 40 + 15 = 55.

         The two numbers are 40 and 55.

• Two numbers are in the ration of 5:3. If they differ by 18, what are the numbers?

         Answer

         Let one of the number be y.

         The other number is y-18

         We are given that y: y-18 = 5:3

         Product of mean = product of extreme.

         So, 3y = 5(y-18)

         Or, 3y = 5y -90

         Or, 2y = 90

         Or, y = 45

         Therefore, y -18 = 27

         So, the 2 numbers are 27 and 45.

indicies STD VII

indicies (STD VII)

IPM Mega finals STD IV answers

In the year 2012, Mrs. Gokhale celebrated her 20th marraige anniversary on the 163rd day of the year. On which of the date did she marry?





Answer:
month             days            cummulative 

Jan                  31                      31
Feb                 29                      60
march             31                      91
april               30                     121
may                31                    152

Now 163 - 152 = 11 days
So, her anniverary date is 11th of June
So, she got married on 11th June 1992   (20 years back)

IPM mega final STD IV answers

A 12 litre tank is filled with water at the rate of 1/4 litre in 5 secs. How much time will it take to fill 5/6 of the tank?




Answer

Water to be filled = 5/6 x 12 = 10 litres = 10000 ml 

now 250 ml is filled in 5 sec.

therefore, time taken by  water to fill the tank = (10000 /250) x 5 = 200 secs = 3 mins and 20 sec

rationalization