QUESTION BANK FOR LINEAR EQUATIONS
1) A garage consist of 2 wheeler and 4 wheeler vehicles. Total number of vehicle in the garage is 10. Find the numbers of 2-wheeler and 4-wheeler vehicles in the garage, if the total numbers of wheels is 28. Solution Let the number of 2-wheeler cars be =$x$ Let the number of 4-wheeler cars be =$y$ Given, total number of vehicle= 10 i.e $x+y=10$ ........(1) Total number of wheels are 28. Since 2-wheeler have 2 wheels and 4-wheeler have 4 wheels, therefore, $2x+4y=28$ ................(2) Solving eqn (1) and (2) From eqn (1) $x+y=10$ => $y=(10-x)$ Putting value of $y$ in eqn (2) $2x+4y=28$ => $2x+4(10-x)=28$ =>$2x+40-4x=28$ =>$2x=12$ =>$x=6$ and $y=(10-x)$=>y=(10-6)=4 Ans: Number of 2 wheelers vehicle = 6 and 4-wheeler vehicle = 42) A crowd consist of humans and donkeys. Total number of head count in the crowd is 50 and total number of legs is 120. Find the number of humans and donkeys. Solution: Let the number of humans be $x$ and number of donkeys be $y$ Given, Total number of head count = 50 i.e $x+y=50$ .........(1) and Total numebr of leg count = 120. (Since humans have 2 legs and donkeys have 4 legs) i,e$(2x+4y=120)$ .......(2) Multiplying eqn (1) by 2 and subracting it with eqn (2) $2x+2y=100$ .......(3) $2x+4y=120$ $2x+2y=100$ ___________ $2y=20$ $y=10$ and $x=40$ Ans: $x=40$ and $y=10$3) 10 years ago, father age was twice the son's age. 10 years hence, son's age will be two-third his father's age. Find the present age of father and son. Solution: Let the father present age be $x$ years and son's present age be $y$ years Given:10 years ago, father age was twice the son's age i.e $x-10=2(y-10)$ i.e $x= 2y-10$ ............(1) and 10 years hence, son's age will be two-third his father's age i.e $2/3(x+10)=(y+10)$ i.e $2x+20=3y+30$ .........(2) Substituting $x$ in eqn (2) i.e $2x+20=3y+30$ => $2(2y-10)+20=3y+30$ =>$4y-20+20=3y+30$ i.e $y=30$ and $x=2y-10$ = $50$ Ans: Father present age is 50 years and son present age is 30 years4) Jacob has two brother's. Jacob present age is twice the sum of his brother's age. 20 years hence, Jacob's age will be equal to the sum of his brother's age. Find the Jacob's and his brother's age, if the difference between the brother's age is 2 years. Solution: Let Jacob's age be $x$ years Let his brothers's age be $y$ and $z$ years Given: $y=z+2$ Given, Jacob present age is twice the sum of his brother's age i.e $x=2(y+z)$ i.e $x=2(2z+2)$ $x=4z+4$ ..........(1) 20 years hence, Jacob's age will be equal to the sum of his brother's age i.e $x+20=(y+20+z+20)$ $x+20=(y+z+40)$ $x+20=(z+2+z+40)$ $x+20=(2z+42)$ $x=(2z+22)$ ...........(2) From eqn (1) and (2) $4z+4=2z+22$ $2z=18$ $z=9$ Therefore $y=z+2$ => $11$ and $x=y+z$ => $20$ Ans:So, Jacob's age is 20 years,and his brother's ages are 11 and 9 years5) The larger of two complementary angles exceeds the smaller by 18 degrees. Find the two angles. Solution: When two angles add up to form right angle, then these angles are called complementary angles. Let the smaller angle be $x$ degree. and the larger angle be $y$ degree. Therefore $x+y=90$ ...........(1) $x-y=18$ ..........(2) Solving eqn (1) and (2), $2x=108$ => $x=54$ and $y=90-x$ => $y=90-54$ => $y=36$ Ans:Therefore the two angles are $36$ $degree$ and $54$ $degree$6) A piggy bank consist of 50 paisa and ₹ 1 coins. Total numbers of coins in the piggy bank is 50 and total value of money in the piggy bank is ₹ 40. Find the numbers of 50 paisa and ₹ 1 coins in the piggy bank. Solution Let total number of 50 paisa coins be $x$ and total number of ₹ 1 coins be $y$ Given: Total numbers of coins in the piggy bank is 50 i.e $x+y=50$ ..........(1) and,total value of money in the piggy bank is ₹ 40 i.e $0.5x+1y=40$ .........(2) Solving equation (1) and (2) we get $0.5x=10$ => $x=20$ and $y=30$ Ans: Total number of 50 paisa coin is 20 and 1 rupee coin is 307) The sum of the digits of two-digit number is 8. Also , the difference between the number and the number obtained by reversing the order of digits is 54. Find the number. Solution: Let the ten's and one's place digits of two digit number be $x$ and $y$ Therefore the number will be of the form $10x+y$ The number obtained by reversing the digit will be $10y+x$ Given: $x+y=8$ ..............(1) and, the difference between the number and the number obtained by reversing the order of digits is 54 i.e $(10x+y)-(10y+x)=54$ i.e $(9x-9y)=54$ i.e $x-y=6$ ...................(2) Solving eqn (1) and (2) $2x=14$ $x=7$ and $y=1$ Ans: So the number is $71$ or $17$8) The difference of digit of two-digit number is 2. Also , the sum of the number and the number obtained by reversing the order of digits is 110. Find the number. Solution: Let the ten's and one's place digits of two digit number be $x$ and $y$ Therefore the number will be of the form $10x+y$ The number obtained by reversing the digit will be $10y+x$ Given: $x-y=2$ ............(1) $(10x+y)+(10y+x)=110$ $11x+11y=110$ $x+y=10$ .................(2) Solving eqn (1) and (2) we have, $2x=12$ => $x=6$ and $y=4$ Ans: Therefore the number is $64$9) A fraction become $3/5$ if 2 is added to the denominator and it becomes 1 when 4 is subracted from the denominator. Find the fraction. Solution: Let the numerator be $x$ and denominator be $y$ of the fraction $x/y$ Given: fraction become $3/5$ if 2 is added to the denominator i,e $x/(y+2)=3/5$ $5x=3y+6$ $5x-3y=6$ ................(1) and fraction become 1 when 4 is subracted from denominator, i.e $x/(y-4)=1$ $x=y-4$ ...................(2) By method of substitution, $5(y-4)-3y=6$ $5y-20-3y=6$ $2y=26$ $y=13$ and $x=y-4$ => $9$ Ans:So, the fraction is ${9/13}$10) If the income of two person A and B are in the ratio 7:5 and expenditure in ratio 3:2 and each one of them saves ₹ 3,000 find the income of person A and B. Solution: SAVING=INCOME-EXPENDITURE Given , ratio of income of A and B= $7:5$ Therefore income of A = $7x$ and income of B= $5x$ and ratio of expenditure of A and B= $3:2$ Therefore expenditure of A = $3y$ and expenditure of B= $2y$ Saving of A= ₹ 3000 i.e $7x-3y=3000$ i.e $7x-3y-3000 =0$ ............(1) Saving of B= ₹ 3000 i.e $5x-2y=3000$ i.e $5x-2y-3000=0$ .............(2) Solve it by any given method, we are using cross multiplication method for pratice purpose only. ${x/(b_1c_2 − b_2c_1) = y/(c_1a_2 − c_2a_1) = 1/(a_1b_2 − a_2b_1)}$ ${x/((-3) × (-3000) − (-2) × (-3000)) = y/((-3000)×(5) − (-3000)×(7)) = 1/((7)×(-2) − (-3)×(5))}$ ${x/(9000 − (6000)) = y/(-15000 +21000) = 1/(-14 +15)}$ ${x/(3000) = y/(6000) = 1/1}$ Therefore $x=3000$ and $y=6000$ Ans: Income of A= 7x = ₹ 21,000. Ans: Income of B= 5x = ₹ 15,000.11) Ram can do a job in 5 days and Shyam can do the same job in 10 days. If, they do the same job together, find the number of days in which they can complete the job. Solution Ram do a job in 5 days, So in 1 days Ram does $(1/5)^{th}$ of the job, Shyam does a job in 10 days. So in 1 days Shyam does $(1/10)^{th}$ of the job They do the job together, so in 1 days both completes $(1/5 +1/10)^{th}$ of the job. => $(2+1)/10$ => $(3/10)^{th}$ of the job Ans:Hence, they complete the job in $10/3$ days i.e 3${1/3}$ days12) A boat goes 10 km upstream in 10 hours and 10 km downstream in 6 hours. Determine the speed of the stream and speed of the boat in still water.If the boat has to go 10 km upstream again in 6 hours, what shall be the increase in speed of the boat in still water. Solution Let the speed of the boat in still water be $x$ km/h and speed of stream be $y$ km/h Therefore speed of boat upstream = $(x-y)$ km/h Therefore speed of boat downstream = $(x+y)$ km/h Since $Speed$= ${Distance}/{Time}$ $(x-y)$ =$10/10$ => $x-y$ = $1$ ............(1) $(x+y)$ =$10/6$ => $0.6x+0.6y$ = $1$ ............(2) Solving the equation by substitution method, From eqn (1), $x=y+1$ Substituting $x$ in eqn (2) $0.6(y+1)+0.6y$ = $1$ $0.6y + 0.6 + 0.6 y = 1$ $1.2y=0.4$ $y=0.4/1.2$ =>$y=1/3$ km/h $x=y+1$ => $x=1/3+1$ $x=4/3$ km/h Let $x_1$ be the increase in speed of the boat in still water. Since it has to goes upstream in 6 hours, hence $x_1 -y$ = $10/6$ $x_1$ =$10/6 + 1/3$ $x_1$ =$(10+2)/{6}$ $x_1$ =$2$ km/h So, the boat has to travel at a speed of $2$ km/h in still water so as to reach in 6 hours.13). In a right angle ∆, the base of the ∆ is equal to the height of ∆. If the area of the ∆ is 50 $m^2$, find the sides of the ∆. Solution: Let $x$ be the base of the ∆. So, height of ∆=base of ∆= $x$ Since, area of ∆ = $1/2 × base × height$ 50 = $1/2 × x × x$ $x^2=100$ $x=10$ $m$14). In a ∆, the base of the ∆ is 4 more than the altitude . If the area of the ∆ is 16 $m^2$, find the altitude of the ∆. Solution Let the altitude be $x$ m. So, base of ∆ is $x+4$ m. Since, area of ∆ = $1/2 × base × height$ Therefore , 16= $1/2× (x+4)× x$ 32= $x^2+4x$ $x^2+4x-32=0$ $x^2+8x-4x-32=0$ $x(x+8)-4(x+8)=0$ $(x+8)(x-4)=0$ $x=4$ or $-8$ Ans:Altitude of the ∆ is 4 $m$15).The sum of parallel sides of a parallelogram is 3 times that of the height . Find the height if the area is 150 $m^2$. Solution Let the height be $x$ $m$ Let $a$ and $b$ be the shorter and longer parallel sides of the parallelogram. Given, $a+b$ = $3x$ Area of parallelogram =$1/2 × $Sum of parallel sides × height $150$ = $1/2$ ×$(a+b)$ × $h$ $150$ = $1/2$ ×$(3x)$ × $x$ $300$ = $3x^2$ $100$ = $x^2$ $10$ = $x$ Ans:Height of parallelgram = 10 m16) A plane left 30 minutes late than its schedule time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed. Solution Let the usual speed of the plane be $x$ km/h. Time taken to cover 1500 km with usual speed = $1500/x$ km/h Time taken to cover 1500 km with increase speed of $(x+100)$ km/h = $1500/{x+100}$ km/h Therefore $1500/x$ = $1500/{x+100}$ + $1/2$ (since 30 minutes is equal to 1/2 hour) $1500/x-1500/{x+100} = 1/2$ $1/x-1/{x+100} = 1/3000$ ${x+100-x}/{x(x+100)} = 1/3000$ $3000×100 = x^2+100x$ $x^2+100x-300000 = 0$ $x^2+600x-500x-300000=0$ $x(x+600)-500(x+600)=0$ $(x+600)(x-500)=0$ $x=500$ Ans: Usual speed of the plane is 500 km/h17) A motor boat whose speed is 18 km/h in still water, takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Solution Let the speed of still water be $x$ km/h Speed of boat in upstream = $18-x$ km/h Time taken = $24/{18-x}$ hr Speed of boat downstream = $x+18$ km/h Time taken = $24/{18+x}$ hr Difference between the time taken for upstream and downstream = 1 hr $24/{18-x} - 24/{18+x} = 1 $ $1/{18-x} - 1/{18+x} = 1/24 $ ${18+x-18+x}/{(18-x)(18+x)}=1/24$ ${2x}/{324-x^2} = 1/24$ $48x=324-x^2$ $x^2+48x-324=0$ $x^2-6x+54x-324=0$ $x(x-6)+54(x-6)$ $(x-6)(x+54)$ Therefore $x$ = 6 or -54 Ans:Since the speed of still water cannot be negative, so speed of still water is 6 km/h.18) A train travels at a certain speed for a distance of 63 km and then travels to a distance of 72 km at an average speed of 6 km/h more than its original speed.If it takes 3 hours to complete total journey, what is the original average speed. Solution: Let the original average speed for 63 km be $x$ km/h Time taken $t_1$ = $63/x$ The avg speed for next 72 km be $(x+6)$ km/h Time taken $t_2$ = $72/{x+6}$ SInce total time taken for the journey is 3 hour. $63/x + 72/{x+6}=3$ $63x+378+72x=3x(x+6)$ $135x+378=3x^2+18x$ $3x^2-117x-378=0$ $x^2-39x-126=0$ $x^2+3x-42x-126=0$ $x(x+3)-42(x+3)=0$ $(x+3)(x-42)=0$ $x= -3$ or $42 $ Since the speed cannot be negative, hence $x$=42 km/h Ans: Original avg speed = 42 km/h19) A father's age is three times the sum of ages of his two sons. After five years, his age will be two times the sum of their ages . Find the present age of the father. Solution Let the father's present age be $x$ Let the sum of two son's age be $y$ Given, $x=3y$ .............(1) After 5 years, father's age will be $x+5$ After 5 years sum of tow son's age will be $y+10$ Also given, $x+5 = 2(y+10)$ ...........(2) Substituting $y=x/3$ in eqn (2), we get $x+5=2(x/3+10)$ $3x+15=2(x+30)$ $3x+15=2x+60$ $x=45$ Ans: Present age of father is 45 years20) A fraction become $1/3$ when 2 is subracted from the numerator and it becomes $1/2$ when 1 is subracted from the denominator. Find the fraction. Solution: Let the fraction be $x/y$ Given , ${x-2}/y=1/3$ i.e $3x-6=y$ ...........(1) also, $x/{y-1}=1/2$ i.e $2x=y-1$..........(2) Substituting the value of $y$ of eqn (1) in eqn (2), we get $2x=3x-6-1$ $2x=3x-7$ $x=7$ and $y=3×7 - 6$ i.e $y=21-6$ = 15 Ans:Hence the fraction is $7/15$21) Two water taps together can fill a tank in $1 7/8$hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tap separately. Find the time in which each tap can fill the tank separately. Solution: Let the time taken by the longer diameter be $x$ hour and time taken by the smaller diameter be $y$ hour When both the taps are open, time taken = $15/8$ hours Since discharge is inversely proportional to time, therefore $1/x+1/y=8/15$ ...........(1) and $x-y$=2 ..................(2) Substituting $x=y+2$ in eqn (1), we get $1/{y+2}+1/y=8/15$ ${y+y+2}/{y^2+2y}=8/15$ $15(2y+2)=8(y^2+2y)$ $30y+30=8y^2+16y$ $8y^2-14y-30=0$ i.e $4y^2-7y-15=0$ $4y^2-12y+5y-15=0$ $4y(y-3)+5(y-3)=0$ $(y-3)(4y+5)=0$ Since value cannot be negative , so $y=3$ hours and $x=5$ hours.22) A boat goes 30 km upstream and 44 km downstream in 10 hours.In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of boat in still water. Solution: Let the speed of stream be $x$ km/h and speed of boat in still water be $y$ km/h Speed of boat in upstream = $(x-y)$ km/h Speed of boat in downstream = $(x+y)$ km/h A boat goes 30 km upstream and 44 km downstream in 10 hours i.e $30/{x-y} + 44/{x+y} = 10$ .............(1) and In 13 hours, it can go 40 km upstream and 55 km downstream. i.e $40/{x-y}+55/{x+y}=13$ .................(2) Let $1/(x-y)=u$ and $1/(x+y)=v$ Therefore $30u+44v=10$ ............(3) $40u+55v=13$ ...........(4) Multiply eqn (3) with 4 and eqn (4) with 3, we get $120u+176v=40$ ...........(5) $120u+165v=39$ ...........(6) Solving eqn (5) and (6) we get , $11v=1$ => $v=1/11$ and $30u+44(1/11)=10$ i.e $30u=10-4$ i.e $u=6/30$ i.e $u=1/5$ Since $1/(x-y)=u$ and $1/(x+y)=v$ Therefore $1/(x-y)=1/5$ and $1/{x+y}=1/11$ $x-y=5$ and $x+y=11$ Solving above two equation, we get $2x=16$=>$x=8$ km/h and $y=x-5$ => $y=3$ km/h23) For what value of $k$, will the line represented by the following pair of equations be parallel. $3x-y-5=0$ $6x-2y-k=0$ Solution Given pair of lines are parallel if , ${a_1/a_2=b_1/b_2≠c_1/c_2}$ ${3/6=(-1)/(-2)≠(-5)/(-k)}$ ${3/6=1/2≠5/k}$ $k≠10$ Ans:For all real values except $k≠10$ , the line represent parallel lines.24) For what value of $k$ , will the lines represented by the equation are parallel. $3x+2ky=2$ $2x+5y+1=0$ Solution Given pair of lines are parallel if, ${a_1/a_2=b_1/b_2≠c_1/c_2}$ $a_1/a_2=b_1/b_2$ $3/2={2k}/5$ $k=15/4$25)Find the vaue of $k$ for which the pair of equations $kx-y=2$ and $6x-2y=3$ will have infinitely many solution. Solution Given pair of lines have infinitely many solutions if the lines coincides i.e, ${a_1/a_2=b_1/b_2=c_1/c_2}$ $k/6=(-1)/(-2)=2/3$ i.e $k=3$ or $k=4$26) Ram can complete a job in 10 days and Shyam can complete the same job in 15 days. Ram start the job and work for 5 days and left. Shyam came in and complete the remaining job. How much days does Shyam takes to complete the remaining job. Solution: Ram can complete a job in 10 days, so, in 1 days Ram can do = $(1/10)^{th}$ of the work. In 5 days Ram does = $(5/10)^{th}$ i.e $(1/2)^{th}$ of the work. Therefore amount of job remaining = $(1-1/2)$ = ${1/2}^{th}$ Shyam can complete the same job in 15 days. So in 1 days Shyam does $(1/15)^{th}$ of the work. Let $x$ be the no. of days taken to finish the ${1/2}^{th}$ of the job. So, in $x$ days Shyam does $(x/15)^{th}$ of the job. Hence, $x/15$ = $1/2$ $x$ = $15/2$ days i.e $7{1/2}$ days Ans: Shyam will take $7{1/2}$ days to complete the remaining job27). 2 man and 7 boys can do a piece of work in 4 days.Also 4 men and 4 boys can do the job in 3 days.How long will it take for 1 man and 1 boy to do it. Solution: Let 1 man can do the job in $x$ days and 1 boy can do the job in $y$ days. Work done by 1 man in 1 day = $1/x$ Work done by 1 boy in 1 day = $1/y$ Therefore work done by 2 man and 7 boys in 1 day = $2/x+7/y$ Since they complete the job in 4 days. So in 1 day, $(1/4)^{th}$ of the job is completed. i.e $2/x+7/y$ = $1/4$ ..............(1) Also 4 men and 4 boys can do the job in 3 days. i.e $4/x+4/y=1/3$ ...............(2)27) Solve the following pair of equations. $2/√x + 3/√y = 2 $ $4/√x - 9/√y = -1$ Solution Let $1/√x = p$ and $1/√y = q$ $2p+3q = 2$ ..............(1) $4p-9q = - 1$ ............(2) Solving (1) and (2) Multiply eqn (1) by 3 , we get $6p+9q=6$ $4p-9q = -1$ Therefore $10p=5$ $p=1/2$ and $q$ = $1/9(4p+1)$ = $1/9(4(1/2)+1)$ = $1/3$ Since $1/√x = p$ Squaring both the sides, $1/x = p^2$ $1/x = (1/2)^2$ i.e $x$=4 Also $1/√y = q$ Squaring both the sides $1/y = (q)^2$ $1/y = (1/3)^2$ i.e $y$ =9 Ans: $x$ = 4 and $y$= 928) Anil completes a job in 10 days and Gautam completes the same job in 15 days.Both Anil and Gautam work for 3 days and Gautam leaves the job. Find the days required by Anil the complete the remaining job ?. Option: (i) 6.66 days (ii) 5 days (iii) 4.33 days (iv) 3.33 days29) A man walking in a shopping mall take 10 sec to climb up the escalator. If he come down in the same esclator at his usual speed, he takes 30 sec to come down. Find the speed of the escalator if the length of ecalator is 20 meter. Solution Let the speed of the man be $u$ $m$/$sec$ and the speed of the escalator be $v$ $m$/$sec$ Therefore relative speed of the man going up the escalator = $(u+v)$ m/sec and relative speed of the man going down the escalator = $(u-v)$ m/sec Since Speed=Distance/time Therefore $(u+v)$ = $20/10$ ............(i) and $(u-v)$ = $20/30$ ...........(ii) Solving (i) and (ii) , we get $2u = 2 + 2/3$ => $u = 4/3$ m/sec and $v = 2 - 4/3$ => $v = 2/3$ m/sec Answer: Speed of escalator is 0.67 m/sec