Exercise 13 (B)

Solve 1
Let x n y be the number
X be greater than Y.
ATP:
X=12+y.......1
X+y=10.......2
Substitute 1 in 2
(12+y)+y=10
12+2y=10
2y=-2
Y=-1
X=12+(-1)
=11

Solve 2
Let the smaller number be x
let the larger number be y
then according to question
y= 5x............(1)
y-18=3x
5x-18= 3x (from(1) )
5x-3x= 18
2x = 18
x= 9
y= 5x
y= 9*5=45

Solve 3
[1]
36, 35, 37
x-1 + x + x+1 = 108
3x = 108

x = 36

[2]

Given,

sum of three consecutive odd numbers is 93.

Let,1st no.=x

2nd no.=x+2

3rd no.=x+4

According to problem,

x+x+2+x+4=93

==3x+6=93

==3x=93-6

==3x=87

==x=87/3

==x=28

therefore,

1st no.is 29

2nd no.is 31

3rd no.is 33

[3]

Required three consecutive even numbers are 80,82 and 84
Step-by-step explanation:
Let x ,(x+2) and (x+4) are three consecutive even numbers.
According to the problem given,
Sum of these numbers = 246
=> x+x+2+x+4 = 246
=> 3x + 6 = 246
=> 3x = 246 - 6
=> 3x = 240
/* Divide each term by 3, we get
=> x = 80
Therefore,
Required three consecutive even numbers are
x = 80,
x + 2 = 80 + 2 = 82 ,
x + 4 = 80 + 4 = 84

[4]
Let the first consecutive multiples of 7 be 7x.
Then,
second consecutive multiples of 7 = 7(x+1)
third consecutive multiples of 7 = 7(x+2)
According to question,
7x + 7(x+1) + 7(x+2) = 777
=> 7x + 7x + 7 + 7x + 14 = 777
=> 21x + 21 = 777
=> 21x = 777 - 21
=> 21x = 756
=> x = 756/21
=> x = 36
Hence, the first consecutive multiples of 7
= 7 × 36 = 252
Second consecutive multiples of 7 = 7 × 37 = 259
Third consecutive multiples of 7 = 7 × 38 = 266.

Solve 4
Let the first part be x
Then second part = 2x-32
Third part = x+18
Their sum = 534
x+2x-32+x+18 = 534
4x-14 = 534
4x = 534+14
4x = 548
x = 548/4
x = 137

First part = 137
Second part = 2(137)-32
= 274-32 = 242
Third part = 137+18 = 155

Solve 5
Numbers are n, n+2 and n+4 
3n-7 = 2(n+4) 
3n-7 = 2n+8 
3n-2n = 8+7 n = 15 
Answer: 15, 17 and 19 

Solve 6
Let one number be x
other no =x+7
(x+7)² -77 =x²
x²+49+14x-77=x²
x²-28+14x=x²
x²-x²+14x = 28
14x= 28
x= 2
other no = x+7 =2+7 =9
the numbers are 2 and 9

Solve 7

full-width

Solve 8
Let the numerator be x
Then, the denominator will be x+3
ATQ
x+5/x+3+5=4/5
x+5/x+8=4/5
5 (x+5)=4 (x+8)
5x+25=4x+32
5x-4x=32-25
x=7
The fraction is 7/10

Solve 9
Required two integers are ,
Required two integers are ,first integer = 25,
Second integer = 75

Step-by-step explanation:
Given ratio of two integers = 1:3
Let first integer = x,
Second integer = 3x
According to the problem given,
"The difference between two positive integers is 50 "
3x-x = 50
=> 2x = 50
=> x = 50/2
=> x = 25
Therefore,
Required two integers are ,
first integer =x = 25,
Second integer = 3x = 3×25 = 75

Solve 10

The two digit number is $25$

Explanation:

Let tens digit and units digit of the number be

$x\text{and}y;x+y=7;\left(1\right)$

The two digit number is $10x+y$, when reversed,

the two digit number becomes $10y+x$, by given condition,

$10y+x=10x+y+27\text{or}9y-9x=27$ or

$9(y-x)=27\text{or}(y-x)=3\text{or}y=x+3$ Putting

$y=x+3$ in equation (1) we get, $x+x+3=7;$ or

$2x=4\therefore x=2;y=2+3=5$

Hence the two digit number is $25$ [Ans]

Solve 11
Let the age of Vijay be x years.
Then the age of Sushma is 15 years older than Vijay = (x+15) years.
Given that in 3 years she will be 8 times as old as Vijay was 3 years ago.
(x + 15) + 3 = 8(x - 3)
x + 15 + 3 = 8x - 24
x + 18 = 8x - 24
7x = 42
x = 42/7
x = 6.
If Vijay's present age is 6 years, Then the age of Sushma = (x+15) = 21 years.
The age of Vijay = 6 years.
The age of Sushma = 21 years.

Solve 12
Let sanjay  bro's present age= x
Sanjaya's age= x/2
after 6 years, sanjaya bro's age will be = x+6, Sanjay will be = x/2+6
as per question
x/2+6=3/5(x+6)
=>x/2+6=3x/5+18/5
=>6-3.6=3x/5-x/2
=>2.4=x/10
=>x=24
Sanjay's bro is 24 yrs old and sanjay is 12 yrs old.

Solve 13
Area of rectangle = l*b
                            =8(x-5)
                            =8x-40
given area is 3x cm sq
so 8x-40=3x
8x-3x=40
5x=40
so x=8
if perimeter is 40
perimeter =2(l+b)
                =2(8+(x-5))
                =2(8+x-5)
                =2(3+x)
               =6+2x
6+2x=40
2x=34
x=17

Solve 14
Length = 25 m
breadth = b (say)
area = 225
25 x b = 225
b = 9
now
perimeter = 2( 25 + 9) = 2x34 = 68 m
so
4(x-2) = 68
x - 2 = 68/4
x - 2 = 17
x = 19

Solve 16

In right triangle ABC, angle A= 90°
AB:AC=3:4
Therefore, AB=3x and AC=4x
By pythagoras theorem

BC=√AC×AC+AB×AB
BC=5x
Now, In rectangle BCDE , Length =BC=5x
Since, BC= 4/5×5x=4x
But,
Perimeter = 180cm
2×(5x+4x)=180
x=10
since, the shortest side of right ∆ ABC=3x
= 3×10
=30cm.

Solve 17

Average Speed of Car A=x km/hr

Average speed of Car B=x+5 km/hr

Average Speed of Two Cars A+B=425/5=85km/hr

x+(x+5)=85

2x+5=85

x=40 km/hr

A=40 km/hr & B=45 km/hr

Solve 18
Let the speed of first car be x km/hr
speed of 2nd car =x*8/9 =8x/9 km/hr

after 7 hrs distance travelled
by 1st car=7*x=7x km
by 2nd car=7*8x/9=56x/9
now
A/q
7x+56x/9 =595
63x+56x=595*9
=119x=595*9
x=595/119 *9
=x=45
speed of first car =45 km/hr
speed of second car=45*8/9 =40 km/hr

Solve 19
Let the speed of the boat in still water be x.
Given that the speed of the stream is 2 km/h.
Upstream:
Speed = ( x - 2) km/h
Distance = Speed x Time
Distance = 6 (x - 2) km
Downstream:
Speed = ( x + 2) km/h
Distance = Speed x Time
Distance = 5(x + 2) km
Solve x:
Since both the distance are the same:
6(x- 2) = 5( x + 2)
6x - 12 = 5x + 10
x = 22 km/h
Answer: The speed of the boat in still water is 22 km/h

Solve 20
Step-by-step explanation:
Let s = speed of the boat in still water
then
(s-4) = speed upstream
and
(s+4) = speed downstream
Distance both ways is equal. Make a distance equation; Distance = Time * speed
Up distance = down distance
25(s-4)= 20(s+4)
25s - 100 = 20s + 80
25s - 20s = +100 + 80
5s = 180
s = 180/5
s = 36 mph
Find the distance:
Dist = time * speed
downstream: d = 20(36+4) = 800 mi
and
Upstream: d = 25(36-4) = 800 mi also

Solve 21
She bought:10 rupee cards: x
5 rupee cards: 2/3 x
15 rupee cards:1/5 x
So the equation is 10x + 5*(2/3 x) + 15(1/5 x) = 245
So
x = 15
So
She bought:10 rupee cards: 15
5 rupee cards: 10
15 rupee cards:3

Solve 22
enrolment this year = 552
let the last year's enrolment be x
 so this year's enrolment is 115x/100
 ATQ,
23x/20=552
⇒x=552X20/23=480
so last year's enrolment was 480 students.

Solve 23
let the no of orange bought by the vendor be X
according to question:
no of orange bought by the vendor = no of banana bought by the vendor = X
now;
cost of orange at which vendor bought =5rs
then cost of X no of orange will be = 5X rs
cost of banana at which the vendor bought = 2 rs
then cost of X no of banana will be = 2X rs
also, according to the question:
vendor sells the orange at a profit of 20% i.e, he sells it at the price of =
5X ×20/100 +5X= 6X rs
also vendor sells the banana at a profit of 15% i.e,he sells it at the price of=
2X×15/100 +2X= 23X/10 rs
according to the question his total profit is 390 rs i.e,
(final price of orange- initial price of orange) +(final price of banana - initial price of banana) =390 rs
(6X-5X) + (23X/10-2X) = 390
X +3X/10 =390
13X/10 = 390
X =390 ×10/13
X=300

Solve 27
Answer - The the total money spent by them is Rs 117
Given that:
Nine persons went to a hotel for taking their meals
Eight of them spent rupees 12 each over there meals
Total amount spent by 8 of them is:
= 12 x 8 = 96
So total amount spent by 8 of them is Rs 96
Let "x" be the amount spent by all of them
The ninth spend rupees 8 more than the average expenditure of all the nine
amount spent by ninth person = 8 + x/9
According to given question,

amount spent by all of them = total amount spent by 8 of them + amount spent by ninth person
x = 96+8+x/9
x = 864+72+x/9
9x - x = 936
x = 936/8
Thus the the total money spent by them is Rs 117