Question 14 Marks
A two digit number is four times the sum and $3$ times the product of its digits, find the number.
AnswerLet this two digit number be 'Xi. Which means $X=10 x$ (as it comes in tens digit).
Then as per the question,
$X Y=4(X+Y)$
$\Rightarrow 4(X+Y)=10 X+Y .\ldots(i)$
and $X Y=3(X Y)=10 X+Y$$\ldots(ii)$
From $(i),$ we get, $6 X=3 Y$
$\Rightarrow X =\frac{ Y }{2}$
Putting this in (ii), we get : $\frac{10 Y }{2}+ Y =\frac{3 Y }{2} \times Y$
$5 Y + Y =\frac{3}{2} Y ^2$
$\Rightarrow$ Solving this we get: $3 Y^2=12 Y$
$\Rightarrow Y=4$
Hence $X=\frac{Y}{2}=2$
Hence, number is $24.$
View full question & answer→Question 24 Marks
The difference of the square of two natural numbers is $45.$ The square of the smaller number is $4$ times the larger number. Determine the numbers.
AnswerLet these two numbers be X and Y, Y being the bigger number. Then as per the question,
Y^2 - X^2 =45 ..... (i)
X^2 = 4Y ..... (ii)
From (i), we get X2=Y2- 45, Putting this in (ii), we get
$Y^2 - 45 - 4Y = O$
$\Rightarrow Y^2 - 4Y- 45 = 0$
$\Rightarrow Y^2 - 9Y + 5Y - 45 = 0$
$\Rightarrow Y (Y - 9) + 5( Y - 9) = 0$
$\Rightarrow (Y-9) (Y+S) = 0$
$\Rightarrow Y$ can't be a negative number and hence Y=9
$\Rightarrow$ Putting $Y = 9$ in (ii), we get $X^2 = 4 x 9=36$
$\Rightarrow X = 6$
Hence numbers are $6$ and $9.$
View full question & answer→Question 34 Marks
The sum of the square of $2$ positive integers is $208$. If the square of larger number is $18$ times the smaller number, find the numbers.
AnswerLet these two numbers be $X$ and $Y, Y$ being the bigger number. Then as per the question,
$X^2+Y^2= 208 ..... (i)$
$Y^2 = 18X ..... (ii)$
From (i), we get $Y^2= 208 - X^2.$ Putting this in $(ii),$ we get
$208 - X^2 = 18X$
$\Rightarrow x^2 + 18X - 208 = 0$
$\Rightarrow X^2 + 26X - 8X - 208 = 0$
$\Rightarrow X(X + 26) - 8(X + 26) = 0$
$\Rightarrow (X-8) (X+ 26) = 0$
$\Rightarrow X$ can't be a negative number , hence $X=8$
$\Rightarrow $ Putting $X=8$ in $(ii),$ we get $Y^2 = 18 x 8=144$
$\Rightarrow Y= l 2$
$\Rightarrow X=8$ and $Y = 12.$
View full question & answer→Question 44 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.
AnswerLet these two numbers be $X$ and $Y, Y$ being the bigger number. Then as per the question,
$X + Y = 25 .....(i)$
$2Y^2 = 3X^2 + 29 ..... (ii)$
From (i), we get $Y= 25 - X$. Putting this in (ii), we get
$2{25-X)^2 = 3X^2 + 29$
$\Rightarrow 1250 + 2X^2 -100X= 3X^2 + 29$
$\Rightarrow X^2 + l OO X - 1221 = 0$
$\Rightarrow X^2 - 11 X +111X - 1221 = 0$
$\Rightarrow X (X - 11) +111( X - 11) = 0$
$\Rightarrow (X-11) (X+111) = 0$
$\Rightarrow X$ can't be a negative number and hence $X=11$
$\Rightarrow X=11$, hence $Y = 14 .$
View full question & answer→Question 54 Marks
The angry Arjun carried some arrows for fighting with Bheeshm. With half the arrows, he cut down the arrows thrown by Bheeshm on him and with six other arrows, he killed the chariot driver of Bheeshm. With one arrow each, he knocked down respectively the chariot, the flag and the bow of Bheeshm. Finally, with one more than $4$ times the square root of arrows, he laid below Bheeshm unconscious on the arrow bed. Find the total number of arrows Arjun had.
AnswerLet the total number of arrows be a.
As per the question :
$0.5 a +6+1+1+1+(1+4 x -\sqrt{a})= a$
$4 x -\sqrt{a}=0.5 a -10$
Squaring both the sides, we get
$16 a=0.25 a^2+100-10 a$
Solving and multiplying both sides by $4,$ we get
$a^2-104 a+400=0 $
$a^2-100 a-4 a+400=0 $
$a(a-100)-4(a-100)=0$
$(a-100)(a-4)=0$
Hence, $a =100$ or $4 .$
It can't be $4$ as Arjun has thrown more than $4$ arrows.
Hence no. of arrows $=100$
View full question & answer→Question 64 Marks
A scholarship account of $Rs.75,000$ was distributed equally among a certain number of students. Had there been $10$ students more, each would have got $Rs.250$ less. Find the original number of persons.
AnswerLet the original number of students be $S$.
Original amount that the student got $=\frac{75000}{ S }$
In new scenario, $5$ increased to $5+10$, Amount reduced by $250 .$
$\Rightarrow \frac{75000}{S}-\frac{75000}{S+10}=250$
Dividing by $250$ throughout the equation, we get:
$\frac{300}{S}-\frac{300}{S+10}=1 $
$\Rightarrow 300 S+3000-300 S=S^2+105 $
$\Rightarrow S^2+105-3000=0 $
$\Rightarrow S^2+605-505-3000=0 $
$\Rightarrow S(S+60)-50(S+60)=0 $
$\Rightarrow(S+60)(5-50)=0 $
$\Rightarrow S=-60,50$
Number of students cannot be negative.
Hence number of students $=50$
View full question & answer→Question 74 Marks
A piece of cloth cost $Rs.5000.$ If the cost price of the cloth was $Rs.5$ less per meter, SOm more of the cloth would have been purchased. Find the cost price per meter of cloth and length of the cloth purchased.
AnswerLet the cloth meters purchase initially was a meters.
Then, Cost per meter $=\frac{5000}{ a }$
New cost per meter $=\frac{5000}{ a -5}$, which enabled to buy $50$ meter more.
$\Rightarrow$ In new condition, (Cost per meter) $x$ (total length) $=$ total $\operatorname{cost}\left(\frac{5000}{a}-5\right)(a+50)=5000$
$\Rightarrow(5000-5 a)(a+50)=5000 a $
$\Rightarrow 5000 a+250000-5 a^2-250 a=5000 a $
$\Rightarrow 5 a^2+250 a-250000=0 $
$\Rightarrow a^2+50 a-500000=0 $
$\Rightarrow a^2+250 a-200 a-500000=0 $
$\Rightarrow a(a+250)-200(a+250)=0 $
$\Rightarrow(a+250)(a-200)=0 $
$\Rightarrow a=200 \text { Meters } $
$\Rightarrow \text { Cost per meter }=5000 / 200=\text { Rs } 25 / \text { meter }$
Hence answer is $Rs.25/$ meter and $200$ meters.
View full question & answer→Question 84 Marks
An aeroplane takes $1$ hour less for a journey of $1200\ km,$ if its speed is increased by $100\ km/ hr$ from its usual speed. Find the usual speed.
AnswerLet the Usual Speed of the aircraft be S, and time taken be t. Distance $=1200 km$
Time $=$ Distance $/$ Speed.
Time taken is reduced by $1 hr$ when speed increases by $100 km / hr$
$\frac{1200}{S}=\frac{1200}{S+100}+1 $
$\Rightarrow \frac{1200+S+100}{S+100}=\frac{1200}{S} $
$\Rightarrow 1300 S+S^2=1200 S+120000 $
$\Rightarrow S^2+100 S-120000=0 $
$\Rightarrow S^2+400 S-300 S-120000=0 $
$\Rightarrow S(S+400)-300(S+400)=0 $
$\Rightarrow(S+400)(S-300)=0 $
$\Rightarrow S=-400,300$
Speed cannot be negative.
Hence, $S=300\ km / kr$
View full question & answer→Question 94 Marks
A plane left $40$ minutes late due to bad weather and in order to reach its destination, $1600$ kms away in time, it had to incease its speed by $400$ km/ hr from its usual speed. Find the usual speed of the plane.
AnswerLet the Usual Speed of the aircraft be $S$, and time taken be $t$.
Distance $=1600 km$
Time $=$ Distance $/$ Speed.
Time remains the same in both the cases. When he is late by $40 mts$, his speed is increased by $400 km / hr$
$\frac{1600}{S}=\frac{1600}{S+400}+\frac{40}{60}$
$\Rightarrow \frac{96000+40 S +16000}{60 S +24000}=\frac{1600}{ S }$
$\Rightarrow 96000 S +40 S 2+16000 S =96000 S +38400000$
$\Rightarrow 40 S^2+16000 S-38400000=0$
$\Rightarrow S^2+400 S-960000=0$
$\Rightarrow S^2+1200 S-800 S-960000=0$
$\Rightarrow S(S+1200\}-800(S+1200)=0$
$\Rightarrow(S+1200)(S-800)=0$
$\Rightarrow S=1200,-800$ (Speed cannot be negative)
$S =800 km / hr$
View full question & answer→Question 104 Marks
In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of the flight increased by 30 minutes. Find the scheduled duration of the flight.
AnswerLet the normal Speed of the aircraft be $S$, and time taken be $t$.
Distance $=600 km$
Time $=\frac{\text { Distance }}{\text { Speed }}$
Hence, $t =\frac{600}{S} \ldots . .$. (i)
And $t+0.5=\frac{600}{S-200} \ldots .$. (ii)
Puttiing (i) in (ii), we get
$\frac{600}{S-200}-\frac{1}{2}=\frac{600}{S}$
$\Rightarrow \frac{1200-S+200}{2 S-400}=\frac{600}{ S }$
$\Rightarrow 1200 S-S^2+200 S=1200 S-2400000$
$\Rightarrow S^2-200 S-240000=0$
$\Rightarrow S^2-600 S+400 S-240000=0$
$\Rightarrow S(S-600)+400(S-600)=0$
$\Rightarrow(S-600)(S+400)=0$
Hence, $S=600 km / hr$
$\Rightarrow T =\frac{600}{600}=1$ Hour
View full question & answer→Question 114 Marks
The speed of the boat in still water is 11 km/ hr. It can go 21 km upstream and 12 km downstream in 3 hours. Find the speed of the stream.
AnswerLet the speed of the stream be $Skm / hr$. So in upstream, boat speed will be 11-S (Against the water flow) and downstream will be $S + I 1$ (Towards the water flow and hence speed is added).
Distance travelled $=21 Km$ Upstream and $12 km$ downstream
Total Time $=3$ hours
Time $=$ Distance $/$ Speed.
$\frac{21}{11- S }+\frac{12}{11+ S }=3$
$\Rightarrow 21(11+ S )+12(11- S )=3(11+ S )(11- S )$
$\Rightarrow 231+21 S +132-12 S =-3 S ^2+363$
$\Rightarrow 3 S ^2+9 s =0$
$\Rightarrow 3 S ( S +3)=0$
$\Rightarrow S = O ,-3$
$S=0$
Since, speed cannot be negative, thus, the speed of the stream is $0 km / hr$.
View full question & answer→Question 124 Marks
The time taken by a person to cover $150$ kms was $2.5$ hrs more than the time taken in the return j ourney. If he returned at a speed of $10\ km/ h$ more than the speed when going, find his speed per hour in each direction.
AnswerLet the speed of the person be $S$, Hence return speed of the person $= S +10$.
D $=150 Km$, Time $=$ Distance $/$ Speed. Time difference $=2.5$ Hours
Hence, in these two conditions,
$\frac{150}{ S }-\frac{150}{ S +10}=2.5 $
$\Rightarrow 150 \times( S +10)-150 \times S = S \times( S +10) \times 2.5 $
$\Rightarrow 2.5 s ^2+25 s -1500=0 $
$\Rightarrow 25 s ^2+250 s -15000=0 $
$\Rightarrow S ^2+10 S -600=0 $
$\Rightarrow S ^2+30 S -20 S -600=0 $
$\Rightarrow S ( S +30)-20( S +30)=0 $
$\Rightarrow( S +30)( S -20)=0$
As the speed can't be negative, $S=20 km / hr$ and while return, its $30 km / hr$
View full question & answer→Question 134 Marks
An ordinary train takes $3$ hours less for a j ourney of $360\ kms$ when its speed is increased by $1$ Okm/ hr.Fnd the usual speed of the train.
AnswerLet the usual speed of the train be $S$, Hence speed of the train when speed is increased $=S+I O$.
$D=360 Km$, Time $=$ Distance $/$ Speed. Time difference $=3$ Hours
Hence, in these two conditions,
$\frac{360}{ S }-\frac{360}{ S +10}=3 $
$\Rightarrow 360 \times( S +10)-360 \times S = S \times( S +10) \times 3 $
$\Rightarrow 3 S ^2+30 S -3600=0 $
$\Rightarrow S ^2+10 S -1200=0 $
$\Rightarrow S ^2+40 S -30 S -1200=0 $
$\Rightarrow S ( S +40)-30( S +40)=0 $
$\Rightarrow( S +40)( S -30)=0$
As the speed can't be negative, $S =30\ km / hr$
View full question & answer→Question 144 Marks
A fast train takes $3$ hours less than a slow train for a journey of $600\ kms.$ If the speed of the slow train is $1$ Okm/ hr less than the fast train, find the speed of the fast train.
AnswerLet the speed of the slow train be $S_1$ Hence speed of the fast train $= S\ 10 .$
$D=600 Km$, Time $=$ Distance $/$ Speed. Time difference $=3$ Hours
Hence, in these two conditions,
$\frac{600}{s}-\frac{600}{s+10}=3$
$\Rightarrow 600 \times(s+10)-600 \times S=S \times(S+10) \times 3 $
$\Rightarrow 3 s^2+305-6000=0 $
$\Rightarrow S^2+10 S-2000=0 $
$\Rightarrow S^2+50 S-40 s-2000=0 $
$\Rightarrow S(S+50)-40(S+50)=0 $
$\Rightarrow(S+50)(S-40)=0$
As the speed can't be negative, $S =40 km / hr$
Hence, speed of the fast train $=40+10=50 km / hr$
View full question & answer→Question 154 Marks
A takes 10days less than B to Finish a piece of work , If both A and B work together , they can finish the work in 12 days. Find the time taken by B to finish the work.
AnswerLet $B$ take ' $c$ ' days to finish the work. Hence, A will take 'c - 1 O' days.
$\Rightarrow$ Work done by $B$ in One day $=\frac{1}{ c }$, whereas work done by $A$ in one day $=\left(\frac{1}{ c -10}\right)$
$\Rightarrow$ Work done by $A$ and $B$ combined in one day $=\frac{1}{ c }+\frac{1}{ c -10}=\frac{ c -10+ c }{ c ^2-10 c }$
$\Rightarrow c^2-10 c=24 c-120$
$\Rightarrow c^2-34 c+120=0$
$\Rightarrow c^2-30 c-4 c+120=0$
$\Rightarrow C( C -30)-4( c -30)=0$
$\Rightarrow( c -30)( c -4)=0$.
It can't be 4 as they together complete in 12 days and hence has to be more than that.
Hence no. of days taken by B to finish the work $=30$
View full question & answer→Question 164 Marks
A takes $6$ days less than the time taken by B to finish a piece of work. If both A and B work together they can finish in $4$ days. Find the time taken by B to finish the work.
AnswerLet $B$ take 'c' days to finish the work. Hence, A will take $'c- 6 '$ days.
$\Rightarrow$ Work done by $B$ in One day $=\frac{1}{ c }$, whereas work done by $A$ in one day $=\frac{1}{ c -6}$
$\Rightarrow$ Work done by $A$ and $B$ combined in one day $=\frac{1}{ c }+\frac{1}{ c +6}=\frac{ c -6+ c }{c}$
Together they can complete the work in $4$ days, hence:
$\frac{c^2-6 c}{2 c-6}=4 $
$\Rightarrow c^2-6 c=8 c-24 $
$\Rightarrow c^2-14 c+24=0 $
$\Rightarrow c^2-12 c-2 c+24=0 $
$\Rightarrow c(c-12)-2\{c-12)=0 $
$\Rightarrow(c-12)\{c-2)=0 .$
It can't be $2$ as they together complete in $4$ days and hence has to be more than that.
Hence no. of days taken by B to finish the work $=12$
View full question & answer→Question 174 Marks
The length of a hall is $5 \ m$ more than the breadth. If the area of the floor of the hall is $84m2$, find the length and breadth of the hall.
AnswerLet the breadth of the hall be a. Then length = a+S Area = Length x Breadth
$\Rightarrow 84= (a+5) x a$
$\Rightarrow a^2 + 5a -84 = 0$
$\Rightarrow a^2+12a - 7a - 84 = 0$
$\Rightarrow a (a + 12) - 7(a + 12) = 0$
$\Rightarrow (a+12) (a - 7) = 0 ;$
Hence $a = 7$
Hence, sides of the hall are $7$ and $12 m.$
View full question & answer→Question 184 Marks
There is a square field whose side is $44\ m.$ A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and graving the path at $Rs\ 2. 75$ and $Rs. 1.5$ per square metre, respectively, is $Rs. 4,904.$ Find the width of the gravel path.
AnswerLet the side of flower bed be $a$ and that of gravel path be $b$.
Then $a+2 b=44$ (as $44$ is overall size of field and it contains side of flower bed and double side of gravel path) .... (i)
Area of Flower bed $=a^2$
Area of Gravel path $=$ Area of Square - Area of flower bed $=44 \times 44-a^2$
$\Rightarrow$ Area of Gravel path $=1936- a ^2$
Cost of laying flower bed + Gravel path $=$ Area $x$ cost of laying per sq.m
$\Rightarrow 4904=\left(a^2 \times 2.75\right)+\left(1936-a^2\right) \times 1.5 $
$\Rightarrow 4904=2.75 a^2-1.5 a^2+2904 $
$\Rightarrow 1.25 a^2=2000 $
$\Rightarrow a^2=1600, \text { Hence } a=40 .$
Hence gravel path width $=\frac{44-40}{2} m =2 m$
View full question & answer→Question 194 Marks
The perimeter of a rectangular field is $82\ m$ and its area is $400\ m^2,$ find the dimension of the rectangular field.
AnswerArea of rectangular garden with sides $a$ and $b=400=a b$
$\Rightarrow a =\frac{400}{ b } \ldots \text {..(i) }$
Perimeter of rectangular garden with sides $a$ and $b=82=2 a+2 b$
$\Rightarrow a + b =41 \ldots \text { (ii) }$
Putting $(i)$ in $(ii),$ we get
$\Rightarrow \frac{400}{b}+b=41 $
$\Rightarrow b^2-41 b+400=0 $
$\Rightarrow b^2-25 b-16 b+400=0 $
$\Rightarrow b(b-25)-16(b-25)=0 $
$\Rightarrow(b-25)(b-16)=0 $
$\Rightarrow b=25 \text { or } 16 .$
Putting this in $(i),$ we get, $a=16$ or $25$
Hence two sides of the rectangle are $16$ and $25.$
View full question & answer→Question 204 Marks
A farmer wishes to grow a $100\ m^2$ rectangular vegetable garden. Since he was with him only $30\ m$ barbed wire, he fences $3$ sides of the rectangular garden letting the compound of his house to act as the $4^{th}$ side. Find the dimensions of his garden .
AnswerArea of rectangular garden with sides $a$ and $b=100$.
$a x b=100$
$\Rightarrow a =\frac{100}{ b }\ldots(i)$
also $2 a+b=30$ (two sides of $a$ and one side of b) $... (ii)$
Putting $(i)$ in $(ii),$ we get
$2 \times \frac{100}{b}+b=30 $
$\Rightarrow b^2-30 b+200=0 $
$\Rightarrow b^2-20 b-10 b+200=0 $
$\Rightarrow b(b-20)-10(b-20)=0 $
$\Rightarrow(b-10)(b-20)=0$
$\Rightarrow$ Hence $b =10$ or $20.$ Hence $a =10$ or $5$
Hence sides are either $5$ and $20\ m$ or $10$ and $10\ m$.
View full question & answer→Question 214 Marks
The perimeter of the right angled triangle is 60cm. Its hypotenuse is 25cm. Find the area of the triangle.
AnswerPerimeter $= a + b + h$ where $a$ and bare sides and his hypotenuse. Here $h =25$.
Given that $a+b+25=60$, hence $a+b=35$
$\Rightarrow a =35- b$$\ldots(i)$
Also, $a^2+b^2=h^2=625 \ldots$ (ii)
Putting (i) in (ii), we get,
$(35-b)^2+b^2=625$
$1225+b^2-70 b+b^2=625$
$2 b^2-70 b+600=0$,
Dividing by 2 , we get: $b^2-35 b+300=0$
$\Rightarrow b ^2-20 b -15 b +300=0$
$\Rightarrow b ( b -20)-15( b -20)=0$
$\Rightarrow(b-20)(b-15)=0$
$\Rightarrow b =20$ or 15
Hence if $b=20, a=35-20=15$ and area $=\frac{20 \times 15}{2}=150 cm ^2$
View full question & answer→Question 224 Marks
The hypotenuse of a grassy land in the shape of a right triangle is $1 \ m$ more than twice the shortest side. If the third side is $7\ m$ more than the shortest side, find the sides of the grassy land.
AnswerLet hypotenuse=h, and other sides by x and y (x bigger than y). As per the question,
$h= 2y + 1, x = y + 7 .... (i)$
For right angled triangle, $x^2 + y^2 = h^2 ... (ii)$
Putitng (i) in (ii), we get:
$(y+ 7)^2 + y^2 = (2y+1)^2$
$\Rightarrow y^2 + 49 + 14y + y^2 = 4y^2 +1 + 4y$
$\Rightarrow 2y^2 -10 y - 48 = 0$
$\Rightarrow y^2 - 5y - 24 = 0$
$\Rightarrow y^2 - 8y + 3y - 24 = 0$
$\Rightarrow y (y - 8) + 3 (y - 8) = 0$
$\Rightarrow (y+3) (y-8) = 0$
$\Rightarrow y = 8$
$\Rightarrow x=y+ 7= 15, h = 2 x 8+ 1 = 17$
Hence the sides are $8, 15, 17 cm.$
View full question & answer→Question 234 Marks
The present age of the mother is square of her daughter's present age. $4$ years hence, she will be $4$ times as old as her daughter. Find their present ages.
AnswerLet the present age of the Mother be M years and her daughter age be $D$ years.
Then, as per the question description,
$M = D^2 ...... (i)$
$(M + 4) = 4(D + 4) ...... (ii)$
Putting (ii) in (i) above, we get
$D^2 + 4 = 4D +16$
$\Rightarrow D^2 - 4D -12 = 0$
$\Rightarrow D^2 - 6D + 2D - 12 = 0$
$\Rightarrow D(D-6 )+ 2(D-6)=0$
$\Rightarrow (D+ 2)(D-6)=0$
$\Rightarrow D= -2, D = 6$ (Age cannot be negative)
$\Rightarrow D =6$ years and hence, M =36 years
View full question & answer→Question 244 Marks
The sum of the ages of a father and his son is $45$ years. Five years ago, the product of their ages was $124$. Determine their present ages.
AnswerLet the present age of the man be M years and his son's age be 5 years. Then, as per the question description,
$M + 5 = 45 ...... (i)$
${M - 5)(5 - 5) = 124 ...... (ii)$
From (i), we get: $M=45-5 ...... (iii)$
Putting (iii) in (ii), we get: {45 - 5 - 5) (5 - 5) = 124
$\Rightarrow (40 - 5)(5 - 5) =124$
$\Rightarrow S^2 - 45 S + 324 = 0$
$\Rightarrow S^2 - 9S - 36S + 324 = 0$
$\Rightarrow S (5 - 9) - 36(5 - 9) = 0$
$\Rightarrow S = 9 , 36$
$\Rightarrow S = 9$ years and hence , $M = 36$ years
View full question & answer→Question 254 Marks
Three years ago, a man was $5$ times the age of his son. Four years hence, he will be thrice his son's age. Find the present ages of the man and his son.
AnswerLet the present age of the man be $M$ years and his sons age be $S$ years.
Then, as per the question description,
$M-3=5(S-3) \ldots (i)$
$M+4=3(S+4) .\ldots(ii)$
From $(i),$ we get: $M+12=5 S$
$\Rightarrow S =\frac{ M +12}{5}\ldots(iii)$
From $(ii),$ we get: $M-8=3 S$
Putting $(iii)$ above, we get: $M -8=\frac{3(M+12)}{5}$
$\Rightarrow 5 M -40=3 M +36$
$\Rightarrow 2 M =76$
$\Rightarrow M=38$ years and hence, $S=10$ years
View full question & answer→Question 264 Marks
A two digit number is $4$ times the sum of its digit and twice the product of its digit. Find the number.
AnswerLet this two digit number be xy. Which means x=10x (as it comes in tens digit).
Then as per the question,
$4(x+y)= 10x+y, ....... (i)$
and $2(xy) =10x + y$
$\Rightarrow 10x + y - 2xy = O ....... (ii)$
From (i), we get: 6x = 3y
$\Rightarrow y = 2x$
Putting this in (ii) , we get
$10x + 2x - (2x) 2x = 0$
$\Rightarrow 4x^2 = 12x$
$\Rightarrow 4x (x-3)=0$
$\Rightarrow x=3$,
hence $y=6$
Hence the number is $36$
View full question & answer→Question 274 Marks
A two digit number is such that the product of the digit is $12.$ When $36$ is added to the number, the digits interchange their places. Find the numbers.
AnswerLet 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=12, \ldots \ldots . . \text { (i) } $
$10 x+y+36=10 y+x $
$\Rightarrow 9 x-9 y+36=0 $
$\Rightarrow x-y+4=0 \ldots . . . \text { (ii) }$
Putting $x=\frac{12}{y}$ from $(i)$ in $(ii),$ we get,
$\frac{12}{y}-y+4=0 $
$\Rightarrow y^2-4 y-12=0 $
$\Rightarrow y^2-6 y+2 y-12=0 $
$\Rightarrow y(y-6)+2(y-6)=0 $
$\Rightarrow(y-6)(y+2)=0$
$\Rightarrow y =6$, hence from $( i ), x =2$.
Hence the number is $26$
View full question & answer→Question 284 Marks
Find two natural numbers which differ by $3$ and whose squares have the sum of $117.$
AnswerLet the numbers be $x$ and $x-3.$ Then,
$x^2 + (x-3)2 = 117,$
$\Rightarrow x^2 + x^2 + 9 - 6x =117$
$\Rightarrow 2 x^2 -6x - 108 = 0$
$\Rightarrow x^2 - 3x - 54 = 0$
$\Rightarrow x^2 - 9x + 6x - 54 = 0$
$\Rightarrow x (x - 9) + 6(x - 9) = 0$
$\Rightarrow (x-9) (x+6 ) = 0$
$\Rightarrow x = 9$ (As the number have to be natural number)
Hence the numbers are $6$ and $9.$
View full question & answer→Question 294 Marks
Two natural numbers differ by $4$. If the sum of their square is $656$, find the numbers.
AnswerLet the two numbers be x and y. Then, as per the question,
$x^2 + y^2 = 656 ...... (i)$
and $x - y = 4$
$\Rightarrow x = 4 + y ........ (ii)$
Putitng 2nd equation in first, we get:
$(y+4 )^2 + y^2 = 656,$
$\Rightarrow 2y^2 + 8y - 640 = 0$
$\Rightarrow y^2 + 4y - 320 = 0$
$\Rightarrow y^2 + 20y -16y- 320 = 0$
$\Rightarrow y(y+ 20) - 16 (y+20) = 0$
$\Rightarrow (y-16 ) (y+ 20) = 0$
y can't be negative, hence $y= 16$
$x= y+4 = 20$
Hence numbers are $16$ and $20.$
View full question & answer→Question 304 Marks
A two digit number is such that its product of its digit is $18$. When $63$ is subtracted from the number, the digits interchange their places. Find the number.
AnswerLet this two digit number be $XY$.
Then as per the question,
$X Y=18 \ldots . . \text { (i) }$
$X Y-63=Y X .\ldots(ii)$
Let this two digit number be $'Xi$.
Which means $X =10 x \ ($as it comes in tens digit$)$.
Then as per the question, $x \ x y=1 s ... 10 x+y-63=10 y+x$
$\Rightarrow 9 x-9 y-63=0$
$\Rightarrow x-y-7=0$
$\Rightarrow$ Puting $x=\frac{18}{ y }$ in above, we get
$\Rightarrow 18- y ^2-7 y =0$
$\Rightarrow y ^2+7 y -18=0$
$\Rightarrow y^2+9 y-2 y-18=0$
$\Rightarrow( y +2)( y -9)=0$
As y can't be negative, hence $y=9$
$\Rightarrow$ Hence $x=\frac{18}{9}=2\ ($ from $(i))$
$\Rightarrow$ Hence answer is $92$
Alternate Answer:
From $(i),$ possible combinations are : $29, 36, 63, 92.$
From $(ii),$ it's clear that the number ' $Xi$ is more than $63$ as that is the only case when we subtract this number by $63 ,$ we get a positive value.
Hence, the number is $92$ and when we delete it by $63 ,$ we get a number of $29$ which is a numbers where the digits are interchanged.
Hence answer is $92$.
View full question & answer→Question 314 Marks
The hypotenuse of a right$-$angled triangle is $17\ cm.$ If the smaller side is multiplied by $5$ and the larger side is doubled, the new hypotenuse will be $50 \ cm$. Find the length of each side of the triangle.
AnswerLet hypotenuse $=h,$ and other sides by $x$ and $y\ (x$ bigger than $y)$. As per the question,
$h = 17 , X^2 + y^2 = 17 X 17$
$\Rightarrow x^2 + y^2 = 289 ..... (i)$
In second scenario, sides become $Sy$ and $2x,$ new $h$ becomes $50 \ cm$
$\Rightarrow (5y)^2 + (2x)^2 = 50 x 50$
$\Rightarrow 25y^2 + 4x^2= 2500$
$\Rightarrow (21y^2 + 4 y^2 )+ 4x^2 = 2500 ..... (ii)$
Putitng $(i)$ in $(ii),$ we get:
$21y^2 + 4(289) = 2500$
$\Rightarrow 21y^2= 1344$
$\Rightarrow y^2 = 64$
Hence $y = 8\ cm$.
Putting this is $(i),$ we get
$\Rightarrow x^2= 289 - 64 = 225$
$\Rightarrow x= 25\ cm$
Hence, the sides are $8, 15, 17 \ cm.$
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