[3 marks sum] - Mathematics STD 10 Questions - Vidyadip

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[3 marks sum]

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17 questions · timed · auto-graded

Question 13 Marks

A two digit number is such that the product of the digits is 14. When 45 is added to the number, then the digit are reversed. Find the number.

Answer

Let this two digit number be XY. Then as per the question,
XY = 14 .......(i)
XY + 45 = YX .........(ii)
From (i) , possible combination are : 27 , 72.
From (ii) , it's clear that the number XY is less than YX as we are adding 45 to XY which makes it equal to YX.
Hence , the number is 27 and we add it by 45 , we get a number of 72 which is a number where the digit are interchanged.
hence answer is 27.

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Question 23 Marks

Out of a group of Swans, $3$ and half times the square root of the total number are playing on the shore of a pond. The two remaining ones are swinging in water. Find the total number of swans.

Answer

Let the total number of swans be a.
Then, $3.5 x-\sqrt{a}+2=a$
$\Rightarrow 3.5 x-\sqrt{a}=a-2$
Squaring both sides, we get:
$12.25 a=a^2+4-4 a$
$\Rightarrow a^2-16.25 a+4=0 $
$\Rightarrow(a-16)(a-0.25)=0$
Hence, total number of swans $=16$

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Question 33 Marks

One fourth of herd of camel was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining $15$ camels were seen on the bank of a river. Find the total number of camel.

Answer

Let the total number of camel be a.
Then, $1 / 4 a+2-\sqrt{a}+15=a$
$\Rightarrow 2 \sqrt{ a }= a -15-\frac{1}{4} a $
$\Rightarrow 2 \sqrt{ a } \times 4=3 a -60$
Squaring both sides, we get
$\Rightarrow 9 a^2-424 a+3600=0 $
$\Rightarrow 9 a^2-324 a-100 a+3600=0$
$\Rightarrow 9 a(a-36)-100(a-36)=0$
$\Rightarrow(a-36)(9 a-100)=0 $
$\Rightarrow a=36, a=\frac{100}{9}$
As 'a' has to be a natural number, no. of camels $=36$

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Question 43 Marks

A shopkeeper buys a number of books for $Rs.840.$ If he had bought $5$ more books for the same amount, each book would have cost $Rs.4$ less. How many books did he buy?

Answer

Let the num,ber of books be b.
Amount of each bok $=\frac{840}{ b }$
If the number of books increased to $B +5,$ it would have costed $\frac{840}{b}-4$
As the amount is same, hence
$\left(\frac{840}{b}-4\right)(b+5)=840 $
$\Rightarrow(840-4 b)(b+5)=840 b$
$\Rightarrow 840 b+4200-4 b^2-20 b=840 b$
$\Rightarrow(b+35)(b-30)=0$
Hence, no.of. books $=30$

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Question 53 Marks

The speed of a boat in still water is $15km/ hr.$ It can go $30km$ upstream and return downstream to the original point in $4$ hours $30$ minutes. Find the speed of the stream.

Answer

Let the speed of the stream be $Skm / hr$. So in upstream, boat speed will be $15- S$ (Against the water flow) and downstream will be $S +15$ (Towards the water flow and hence speed is added).
Distance travelled is same $=30 KM$. Total Time $=4.5$ hours
Time $=$ Distance $/$ Speed.
$\frac{30}{15- S }+\frac{30}{15+ S }=4.5 $
$\Rightarrow 30 \times(15+ S )+30 \times(15- S )=4.5 \times(15+ S ) \times(15- S ) $
$\Rightarrow 200=225- S ^2 $
$\Rightarrow S ^2=25$
Hences $=5 Km / hr$.

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Question 63 Marks

A train travels a distance of $300$ kms at a constant speed. If the speed of the train is increased by $10$ km/ hour, the j ourney would have taken $1$ hour less. Find the original speed of the train.

Answer

Let the speed of the train be $S$.
D $=300 Km$, Time $=$ Distance $/$ Speed. Time difference $=!$ Hours
Hence, in these two conditions,
$\frac{300}{S}-\frac{300}{S+10}=1 $
$\Rightarrow 300(S+10)-3005=S(S+10) $
$\Rightarrow S^2+105-3000=0 $
$\Rightarrow S^2+605-505-3000=0 $
$\Rightarrow S(S-60)-50(S-60)=0 $
$\Rightarrow(S+60)(S-50)=0$
As the speed can't be negative, $S =50 km / hr$.

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Question 73 Marks

The sum of a number and its reciprocal is $2 \frac{9}{40}$. Find the number.

Answer

Let the number be A. Then as per the question,
$A+\frac{1}{A}=2 \frac{9}{40} $
$\Rightarrow A^2+1-\frac{89}{40} A=0 $
$\Rightarrow 40 A^2+40-89 A=0 $
$\Rightarrow 40 A^2-89 A+40=0 $
$\Rightarrow 40 A^2-64 A-25 A+40=0 $
$\Rightarrow 8 A(5 A-8)-5(5 A-8)=0 $
$\Rightarrow(8 A-5)(5 A-8)=0 $
$\Rightarrow A=\frac{5}{8}, \frac{8}{5}$

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Question 83 Marks

The area of the isosceles triangle is $60 cm^2$, and the length of each one of its equal side is $13cm$. Find its base.

Answer

Area of $a$ isosceles triangle $=\frac{ b \times \sqrt{4 a ^2- b ^2}}{4}$
Here, area $=60, a =13$. Putting this in equation and squaring both sides, we get
$(60 \times 4)^2=b^2\left(4 \times 13 \times 13-b^2\right) $
$\Rightarrow 57600=b^2 \times\left(676-b^2\right)=676 b^2-b^4 $
$\Rightarrow b^4-676 b^2+57600=0 $
$\Rightarrow b^4-100 b^2-576 b^2+57600=0 $
$\Rightarrow b^2\left(b^2-100\right)-576\left(b^2-100\right)=0 $
$\Rightarrow\left(b^2-100\right)\left(b^2-576\right)=0 $
$\Rightarrow b^2=576, b^2=100 $
$\Rightarrow b=24, b=10$
$\Rightarrow$ Hence base can be either $24$ or $10 cm$.

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Question 93 Marks

The area of right-angled triangle is $600cm^2$. If the base of the triangle exceeds the altitude by $10cm$, find the dimensions of the triangle.

Answer

Area of a triangle $=\frac{\text { Height } \times \text { Base }}{2}$
Here, Height and base are $x$ and $(x+10)$ and the area is 600
Hence, $x \times(x+10) \times \frac{1}{2}=600$
$
\Rightarrow x^2+10 x=1200
$
$
\Rightarrow x^2+10 x-1200=0
$
$
\Rightarrow x^2+40 x-30 x-1200=0
$
$
\Rightarrow x(x+40\}-30(x+40)=0
$
$\Rightarrow(x+40)(x-30)=0$, hence $x=30$.
$
\text { Base }=30+10=40 cms .
$
Hence $h^2=40^2+30^2=2500$

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Question 103 Marks

The length of the sides forming a right angle in a triangle are $5x cm$ and $(3x-1) cm$. If the area of the triangle is $60cm^2$, find the hypotenuse.

Answer

Area of a triangle $=\frac{\text { Height } \times \text { Base }}{2}$
Here, Height and base are $5 x$ and $(3 x-1)$ and the area is 60
Hence, $5 x \times(3 x-1) \times \frac{1}{2}=60$
$
\Rightarrow 15 x^2-5 x=120
$
$
\Rightarrow 3 x^2-x-24=0
$
$
\Rightarrow 3 x^2-9 x+8 x-24=0
$
$
\Rightarrow 3 x(x-3)+8(x-3)=0
$
$\Rightarrow(3 x+8)(x-3)=0$, hence $x=3$.
Sides are $5 \times 3$ and $3 \times 3-1=15$ and 8 ems.
$\Rightarrow$ Hence, $h^2=15^2+8^2=289$
Hypotenuse $=17 cms$.

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Question 113 Marks

The product of a girl's age five years ago and her age $3$ years later is $105$. Find her present age.

Answer

Let the present age of the Girl be G year. Then, as per the question description,
$(G-S)(G+3)=105$
$\Rightarrow G^2-2 G-120=0$
$\Rightarrow G^2-12 G+10 G-120=0$
$\Rightarrow G(G-12)+10(G-12)=0$
$\Rightarrow G=12, G=-10 \text { (Agecannotbe negative) }$
$\Rightarrow G=12 \text { Years }$

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Question 123 Marks

The sum of $2$ numbers is $18.$ If the sum of their reciprocals is $\frac{1}{4}$, find the numbers.

Answer

Let these numbers be $A$ and $B$.
Then as per the question,
$A+B=18,1 / A+1 / B=1 / 4$
$\Rightarrow A=18-B$, Putting this in 2nd equation, we get:
$\Rightarrow \frac{1}{18-B}+\frac{1}{B}=\frac{1}{4} $
$\Rightarrow(B+18-B) \times 4=(18-B) \times B $
$\Rightarrow B^2-18 B+72=0 $
$\Rightarrow B^2-12 B-6 B+72=0 $
$\Rightarrow B(B-12)-6(B-12)=0 $
$\Rightarrow(B-12)(B-6)=0 $
$\Rightarrow B=12,6 .$
Hence the numbers are $6$ and $12.$

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Question 133 Marks

A two digit number is such that the product of its digit is $8.$ When $18$ is subtracted from the number, the digits interchange its place. Find the numbers.

Answer

Let this two digit number be $x y$. Which means $x=10 x$ (as it comes in tens digit).
Then as per the question,
$x y=8, \ldots \ldots . \text { (i) } $
$10 x+y-18=10 y+x $
$\Rightarrow 9 x-9 y-18=0 $
$\Rightarrow x-y-2=0 \ldots . . . \text { (ii) }$
Putting $x=\frac{8}{y}$ from $(i)$ in $(ii),$ we get ,
$\frac{8}{y}-y-2=0 $
$\Rightarrow y^2+2 y-8=0 $
$\Rightarrow(y-2)(y+4)=0$
$\Rightarrow y=2$, hence from (i), $x=4$.
Hence the number is $42$

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Question 143 Marks

The sum of the square of two numbers is $233$. If one of the numbers is $3$ less than twice the other number. Find the numbers.

Answer

Let the numbers be $x, 2 x-3$. Then,
$x^2+(2 x-3)^2=233$
$\Rightarrow x^2+4 x^2+9-12 x=233$
$\Rightarrow 5 x^2-12 x-224=0$
$\Rightarrow 5 x^2-40 x+28 x-224=0$
$\Rightarrow 5 x(x-8)+28(x-8)=0$
$\Rightarrow(5 x+28)(x-8)=0$
$\Rightarrow \mathrm{x}=8$ (As the number have to be natural number)
$\Rightarrow$ Other number $=2 \times 8-3=13$
Hence, the numbers are 8 and 13

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Question 153 Marks

Three consecutive natural numbers are such that the square of the first increased by the product of other two gives $154$. Find the numbers.

Answer

Let the numbers be $x-1, x, x+l$ . Then as per question,
$(x-1)^2 + (x) (x+l ) = 154$
$\Rightarrow x^2 -2x + 1+ x^2 +x =154$
$\Rightarrow 2 x^2 - x - 153 = 0$
$\Rightarrow 2 x^2 - 18x +17x - 153 = 0$
$\Rightarrow 2x (x-9) + 17 (x-9) = 0$
$\Rightarrow (2x+17) (x-9) = 0$
$\Rightarrow$ As the number have to be natural number, $x=9$
Hence, the numbers are $8,9 ,10$ .

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Question 163 Marks

The sum of the square of $2$ consecutive odd positive integers is $290$.Find them.

Answer

Let the numbers be $X$ and $\mathrm{X}+2$. Then as per question,
$X^2+(X+2)^2=290$
$\Rightarrow 2 x^2+4 X-286=0$
$\Rightarrow X^2+2 X-143=0$
$\Rightarrow X^2+13 X-11 X-143=0$
$\Rightarrow X(X+13)-11(X+13)=0$
$\Rightarrow(X+13)(X-11)=0 . X$ can't be negative number as its natural number.
Hence, $X=11$ and $X+2=13$.

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Question 173 Marks

The sum of the square of the $2$ consecutive natural numbers is $481$. Find the numbers.

Answer

Let these numbers be A and $A+ 1$.
Then as per the question, $A^2 + (A+1)^2 = 481$
$\Rightarrow A^2 + A^2 + 2A + 1 = 481$
$\Rightarrow 2 A^2 + 2A - 480 = 0$
$\Rightarrow A^2 + A - 240 = 0$
$\Rightarrow A^2 + 16 A - 15 A - 240 = 0$
$\Rightarrow A (A + 16) - 15 (A + 16) = 0$
$\Rightarrow (A - 15) (A + 16) = 0$
$\Rightarrow A = 15 , -16$ (-16 is not a natural number)
$\Rightarrow A=15$
$\Rightarrow$ Hence the numbers are $15$ and $16$.

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[3 marks sum] - Mathematics STD 10 Questions - Vidyadip