class 10 simultaneous equations and quadratic

SE 12:47 PM

Tuesday, November 24, 2020

1a) There years ago the sum of the ages of father and his son was 48
years and three years hence from now fathers age will be three times
that of his son. Find the present ages of the father and his son
solution :
let age of fathers be x and son be y
by first condition ,

(x-3) + (y-3)=48

x=54-y

by second condition

putting value of y in x=54-y

1b) 2 years ago , father's age was nine times the sin's age but 3
years later it will be 5 times only find the present ages of
father and son
solution

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solution

solution :
let age of fathers be x and son be y
by first condition ,

x-2 = 9(y-2 )

by the second conditions

2a) five years ago fathers age 4 times his son's ages . Now the
sum of their ages 45 years . Find their present age.

let age of fathers be x and son be y
by first condition ,

x-5 = 4(y-5 )
by second condition
Substituting value of x

Putting value of y in equation I
x=33

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2b) A year hence a father will be 5 times as old as his son . Two
years ago the father was 3 times as old as his son will be in 4
years hence from now . Find their present ages.

3a) the present age of the father is three times the age of his
son . If the age of the son after 10 years is equal to the age of
the father before 20 years, fin the present ages of father and
son.

let age of fathers be x and son be y
by first condition ,
by second condition

putting value of x from I

putting value of y in equation 1
x=3×15=45

present age of father is

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3b) the ages of two girls are in the ratio of 5:7 . Eight years ago
there ages were in the ration of 7:13 .find their present ages.
Solution ,
Ratio of ages of two girls 5:7
let their ages be and 7x
by question

x=3
3

their present ages are 15 and 21 years respectively

4a) the present age of the father is double of the age of his son.
If the age of the son after 10 years is equal to the age of the
father before 15 years , find their present ages.
solution
x=2y
x-15=y+10

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4b) A father ages was four times and two times of his son at
2032 and 2050 respectively , find the birth year of son.
Solution
let father age be x and son be y at 2032
by first condition

After 18 years ( 2050-2032)

Birth year of son is
2032-9=2023

5a ) A number consisting of two digits is three times the sum
of their digits . If 45 is added to the number , the digits are
interchanged . Find the numbers.
let a two digits numbers be

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7x= 2y …..i

If 45 is added to the number , the digits are interchanged

Substituting the value of y in equations 1

5 x=10
x=2
substituting value of x in equations ii
2+5=y
y=7
the required number =

=
5b) the sum of the digits in a two digits numbers is 11. the
numbers formed by interchanging the digits will be 45 more
than the original numbers. Find the original numbers.
x+y=11

Book page 173 QN 7a

Let a fraction be
Numerator I multiply by 4 & denominator is reduce by 2 result is 2

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Again if the numerator is increased by 15&2 is subtracted from
double of denominator

from i

Require fraction is
Book page 173 QN 7 b
Let a fraction be
Numerator is multiply by 4 & denominator is reduce by 2 result is
4

Again if the numerator is increased by 15&2 is subtracted from
double of denominator

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double of denominator

Y=7 &x=5
Require fraction is

BOOK page 173 QN 6a
A number of two digits is six times the sum of its digits .If 9 is subtracted from
the number the digits are reversed .what is the number

Let number of two digits is
Number of two digits is 6 time the sum of its digits

If 9 is subtracted from the number the digits reverse

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From (i)

y=4
From II
x=5
Required number is

=

=54
BOOK page 173 QN 6b
Let number of two digits is
Sum of its digits is 9

………i
If 27 is added to the number the digits reverse

From (i)

Y=6
From I
X=3
Required number is

=

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=36

PAGE 173 QN8 A
Let speed of boat be x and speed of steams be y
We have ,

A boat goes 12 Km upward in 6 hours

In upward going the speed of boat is

decease by the stream

If the boat goes downward stream the speed increase and it become
Boat goes downward 36 km in 6 hours

From( I)

From( I) x=4

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PAGE 173 QN8 b

Let speed of boat be x and speed of steams be y
We have ,

A boat goes 16 Km upward in 8 hours

In upward going the speed of boat is decease

by the stream

If the boat goes downward stream the speed increase and it become
Boat goes downward 32 km in 8 hours

From( I)

From( I) x=3

BOOK PAGE 174 QN 9a
Let speed of train be x , time taken be y & traveled distance be d
We have ,

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We have ,
If train had been 6 km/ hr faster the it takes 4 hour less

If train had been 6 Km/hr slower than it takes 6 hour mere

Putting value of x in equation (1)

From ii
=30

Total distance travelled =

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BOOK PAGE 174 QN 9 b
Let speed of train be x , tine time be y & traveled distance be d
We have ,
If train had been 5 km/ hr faster the it takes 3 hour less

If train had been 4 Km/ hr slower than it takes 3 hour more

Putting value of x in equation (1)

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From ii =40
Total distance travelled =

PAGE 174 QN 10 a
Let speed of two cars be x & y respectively
If the move on the same direction
A must travelled 120 km more than B
Distance travelled by A=distance travelled by B+120 km

1 hours +12 minutes =
=1.2 hours

If the move on the opposite direction the travelled 120 km jointly

Using equation I

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From equation ( I)

PAGE 174 QN 10 b
Let speed of two cars be x & y respectively
If the move on the same direction
A must travelled 70 km more than B
Distance travelled by A=distance travelled by B+70km

If the move on the opposite direction the travelled 120 km jointly
Using equation I

From equation ( I)

Qn 11 a page 174
Let cost of a pen be x &pencil be y
59 pens and 47 pencil cost Rs 1893
47 pens and 59 pencil cost Rs 1605

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11b)

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QE 3:44 PM

Tuesday, November 24, 2020

1a) The present ages of elder and younger brothers are 13 years and 7 years
respectively . In how many years will the product of their ages be 280? Find it.

Solution,
Let x years after the product of their ages be 280

x=7 or -27
Age cannon be in negative so Required age is 7

1b) the present ages of a fathers and his son are 35 and 12 years respectively . Find how
many years ago the product of their age was 210.
Solution,
Let x years ago the product of their ages be 210

Before 42 years they are not born so required years is 5

2a) the difference of present ages of two brothers is 4 years . 5 years ago , if the product
of their ages was 96 , find their presents.
solution
let the present age of two brothers be x &y respectively
by question

or
by second condition
(

algebra Page 1

(
(

2b) The difference of the present two sister is 5 years .after 4 years if the product of
their ages becomes 104, find their present ages.
solution
let the present age of two sister be x &y respectively
by question
or
by second condition
(
(
(

3a) the difference of the ages of two brothers is 4 and the product of their ages is 221.
Determine the ages of two brothers.
solution
let the present age of two brothers be x &y respectively
by question
or
by second condition

(

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(

PAGE 180 Q4 A
Let age of two sister be x &y
Differenced of their ages is 5

………………..i
Product of their ages is 84

…………………..
Y=7or-12
Age cannot be negative so y =7
From equation I

4b
Let age of two sister be x &y
Differenced of their ages is 7

………………..i
Product of their ages is 44

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………………………………………………
…………………..
Y=4 or-11
Age cannot be negative so y =4
From equation I

Page 180
Qn 5a

5b)

QN 6 PAGE 180
let ages of two sister be x& y respectively
Product of Present age of two sister is 150

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Product of Present age of two sister is 150
Five years ago , elder sister is twice of younger sister

Putting value in equation I

Y=10 x=15
From equation I
Age cannot be negative,

6b) xy=160 & x-4=2(y-4)

Qn 7a page 180
Let present ages of father and son be x &y respectively
6 years ago , father is 6 time of his son

Product of present ages is 396

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Age cannot negative y=11
From equation I x=36

Qn8a

Sum of their square is 261
Putting value of x from equation 1

8b )

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9a)
Let a two digit number be
Product digits (x&y) is 18
Sum of digits is 9

9b)

10 a)

10b)

Q N 11 a PAGE 180
Let the hypotenuse be x
Sides containing right angle are less than hypotenuse by 5&10
Using Pythagoras theorem

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If
Which is impossible
If x=25 then p=20 b= 15
Sides of the right angle are 25, 20 15

Q N 11 b PAGE 180
Let the hypotenuse be x
Sides containing right angle are less than hypotenuse by 2&4
Using Pythagoras theorem

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If
Which is impossible?
If x=10 then p=8 , b= 6
Sides of the right angle are ……..

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