Square word problems worksheet with solutions | 10th grade math (2023)


Question 1.
A train travels a distance of 63 km at a certain average speed and then travels a distance of 72 km at an average speed of 6 km/h above its original speed. If it takes 3 hours to go the entire distance, what is the original average speed?
solution

The initial speed of the train is x km/h
Time required for 63 km at speed x km/h,
$time = \frac {distance}{speed} =\frac { 63}{x}$hours
After 63 km the speed of the train will be (x + 6) km/h
Time needed to travel 72 km at a speed of (x + 6) km/h
$time = \frac {distance}{speed} =\frac { 72}{x +6}$ hours
now the question
$\frac {63}{x}+ \frac {72}{x +6}= 3 $
$\frac {21}{x}+ \frac {24}{x +6}= 1$
$ \frac {(21x + 126 + 24x) }{x(x+6}= 1$
$45x + 126 = x^2+ 6x$
$x^2- 39x - 126 = 0$
(x + 3)(x - 42) = 0
∴x = 42 y -3 , but x ≠ -3 because velocity cannot be negative
Therefore, the original speed of the train = 42 km/h

Question 2.
Find two consecutive negative even integers whose product is 24?
solution

Let n and n-2 be the two consecutive negative even integers
after
$n(n-2)=24$
$n^2 -2n-24=0$
o
$(n-6)(n+4) =0$
n = 6 or -4
Since we only want negative integers, n=-4
another number =n-2= -4-2 = -6
So the numbers are (-6,-4)


Question 3
A rectangle has a length that is 2 times less than 3 times its width. If the area of ​​this rectangle is 16, find the dimensions and the perimeter.
solution

Let x = the length of the rectangle and y = its width.
according to the question
$x = 3y- 2$
Let the area of ​​this rectangle be 16
now
$area = xy = 16$
o
$ (3a-2)a=16$
$3 years^2- 2 years- 16 = $0

Factoring
$ (3a-8)(a+2)=0$
y = 8/3 e -2
Since we cannot have a negative width, y= 8/3
Now x=3y-2 = 6
Umfang = 2x + 2y = 12 + 16/3 = 64/3

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question 4
The difference of two numbers is 2 and their product is 224. Find the numbers
solution

let x be the smallest number
x+2 can be the largest number
$x(x+2)=224$
$x^2+2x-224=0$
factor the squares
$(x+16)(x-14)=0$
x= -16 o 14
for positive numbers
x = 14
Second number = x+2= 16
so the numbers are 14 and 16
for negative numbers
x=-16
Second number = x+2= -14
So the numbers are -16 and -14


question 5
X and Y working together can complete a given project in 6 days. If X works alone on the project, it will take Y 5 days less than Y to complete it. Can you find the days it takes X and Y to complete the project yourself?
solution

Let p and q be the days it takes X and Y to complete the project alone.
Now p = q-5
In the same way
$\frac{1}{p} + \frac{1}{q} = \frac{1}{6}$
o
$\frac{1}{q-5} + \frac{1}{q} = \frac{1}{6}$
$\frac {q + q-5}{q(q-5)} = \frac{1}{6}$
$6(2q-5) = q^2 -5q$
$q^2 -17q +30=0$
Factoring this quadratic
q=15 o 2
It can't be 2, because then p becomes negative.
So q=15 days and p=10 days


question 6
A speedboat traveling at 18 km/h in calm water takes 1 hour longer to travel 24 km upstream than it does to return to the same location downstream. Find the velocity of the flow.
solution

The speed of the chain is x km/h.
Upstream distance = Downstream distance = 24 km
Upstream boat speed = 18 - x
Boat speed downstream = 18+x
gas time = 24/18-x
Time spent downstream = 24/18+x
now as question
$ \frac{24}{18-x} = 1 + \frac{24}{18+x}$
$\frac{24}{18-x} = \frac {42+x}{18+x}$
$24(18+x) = (42+x)(18-x)$
$x^2 ​​+ 48x-182=0$
x=-54 o 6
If a negative value is rejected, the flow velocity is 6 km/h

question 7
The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm long, find the other two sides.
solution

Let x be the base, the height is x-7
Now by the Pythagorean theorem
$x^2 ​​+(x-7)^2 = 13^2$
$x^2 ​​-7x -69=0$
x=12 o -5
Since x cannot be negative
x=12

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question 8
Find the value of k for which the roots of the quadratic equation are equal.
solution

$kx(x-2 \sqrt {5}) + 10 =0$

$kx(x-2 \sqrt {5}) + 10 =0$
o
$kx^2 -2 \sqrt {5} kx +10 =0$
For equal roots, the discriminant must be zero.
$b^2 -4ac=0$
of games $b= -2 \sqrt {5} k$, a=k ,c=10
$( -2 \sqrt {5} k)^2 -4 \times k \times 10=0$
$20.000^2 -40.000=0$
o k=0 o 2
k cannot be zero so the correct value of k = 2

question 9
The numerator of a fraction is 3 less than its denominator. When 2 is added to the numerator and denominator, the sum of the new fraction and the original fraction is 29/20. find the original fraction
solution

Let x be the denominator and x-3 the numerator.
The fraction is = $\frac {x-3}{x}$
Now when 2 is added
Neuer Bruch = $\frac {x-3+2}{x+2}=\frac {x-1}{x+2} $
according to the question
$\frac {x-3}{x} +\frac {x-1}{x+2}= \frac {29}{20}$
$ \frac {(x-3)(x+2) + x(x-1)}{x(x+2)} =\frac {29}{20}$
$\frac{2x^2 -2x-6}{x(x+2)} =\frac {29}{20}$
$20(2x^2 -2x-6) = 29x(x+2)$
$40x^2 -40x -120 =29x^2 + 58x$
$11x^2 -98x -120=0$
factor the squares
x=10 o -12/11
So the fraction is $\frac {x-3}{x} = \frac {7}{10}$


question 10
Two tubes are used to fill a pool when the larger diameter tube can only fill half the pool for 4 hours and the smaller diameter tube for 9 hours. Find out how long it would take to fill each tube in the pool separately if the smaller diameter pool takes 10 hours longer than the larger diameter pool to fill the pool.
solution

Let the hours x and y be the time it takes for the largest and smallest pipes to fill the pool.
Now $y - x = $10 or $y = x + $10 is given
If a smaller pipe takes x hours to fill the pool, it will fill 1/x part in 1 hour
Similarly, if a larger pipe takes y hours to fill the pool, it will fill 1/y part in 1 hour.
depending on the question
$ 4 \times \frac {1}{x} + 9 \times \frac {1}{y} = \frac {1}{2}$
o
$ \frac {4}{x} + \frac {9}{y} = \frac {1}{2}$
Now substitute y = x + 10
$ \frac {4}{x} + \frac {9}{x+10} = \frac {1}{2}$
$ \frac {4x+ 40 + 9x}{x(x+10)} = \frac {1}{2}$
$26x + 80 =x^2+10x$
$x^2 ​​- 16x - 80 = 0$
O (x -20)(x+4) = 0
ignoring the negative root,
x = 20 by default
E y = 30h

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question 11
Five times a positive integer is less than twice its square times 3. Find the integer.
solution

Let x be the number, then
$5x=2x^2 -3$
$2x^2 -5x-3=0$
$2x^2 -6x + x-3=0$
$2x(x-3) + 1(x-3)=0$
o
(2x+1)(x-3)=0
or x=3 since x is a positive integer


question 12
Find the positive values ​​of P such that the equations $x^2+2px+64=0$ and $x^2-8x+2p=0$ have real roots
solution

For an equation to have real roots, its discriminant must be greater than or equal to 0.
$D=b^2 -4ac\geq 0$

For the equation (1) $x^2+2px+64=0$
$(2p)^2- 4 \times 64 \geq 0$
$p²\geq 64$
$p\geq 8$ or $p\leq -8$

For the equation (2)$x^2-8x+2p=0$
$64 -8p \geq0$
$8 \geq p$

The answer is $( -\infty , -8) \cup {8}$

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question 13
If Zeba were 5 years younger than he actually is, his age squared would be 11, more than 5 times his actual age. How old is she now?
solution

the age of the zebra is x years
According to the question $(x-5)^2=11+5x$
$x^2+25-10x=11+5x$
$x^2-15x+14=0$$x^2-14x-x+14=0$
$(x-1)(x-14)
So Zeba's age is 14 years old, because if she were 1 year old, she couldn't be 5 years younger.


question 14
The sum of the squares of two consecutive multiples of 7 is 637, can you find the multiples?
solution

Let x and x+7 be the two consecutive multiples of 7
now as question
$x^2 ​​+ (x+7)^2 =637$
$x^2 ​​+ x^2 + 49 + 14x =637$
$x^2 ​​+ 7x -294=0$
x=-21 y 14
So the answer is -21, -14 or 14.17


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