ICSE Board: Suggested Books     ICSE Board:  Foundation Mathematics
Class 8: Reference Books               Class 8: NTSE Preparation
--------------------------------------------------------------

 

Q.1 The sum of two numbers is 60 and their difference is 14. Find the numbers.

Answer:

Let the two numbers be x \ and \ y

  x+y=60  ... ... ... ... ... ... (i)

  x-y=14 ... ... ... ... ... ... (ii)

Add  (i) and (ii)

2x=74 \ or\  x=37

Therefore y=37-14=23

Hence the two numbers are 23 \ and\  37

\\

Q.2 Twice a number is equal to thrice the other number. If the sum of the numbers is 85, find the numbers.

Answer:

Let the two numbers are x \ and \ y

2x=3y  ... ... ... ... ... ... (i)

x+y=85 ... ... ... ... ... ... (ii) 

Solving for x \ and\  y 

 2x-3y=0 

\underline {(-)2x+2y=170 }

-5y=-170 

or y=34 

Therefore x=\frac{3}{2}\times 34=51

Hence the two numbers are 51 \ and\  34 .

\\

Q.3 Find two numbers such that twice the first added to thrice the second gives 70 and twice the second added to thrice the first gives 75.

Answer:m57x

Let the numbers be x \ and \ y

2x+3y=70 ... ... ... ... ... ... (i)

3x+2y=75 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiply i) by 2 and ii) by 3 and subtract ii) from i)

 6x+9y=210

\underline{(-)  6x+4y=150}

5y=60

\Rightarrow y=12

Substituting in i)

\Rightarrow x=\frac{70-3y}{2}=\frac{70-36}{2}=17

Hence the two numbers are 17 \ and\  12

\\

Q.4 Find two numbers which differ by 9 and are such that four times the larger added to three times the smaller gives 92.

Answer:

Let the numbers be x \ and \ y

 x-y=9  ... ... ... ... ... ... (i)

4x+3y=92 ... ... ... ... ... ... (ii)

Solving by x \ and\  y

Multiply i) by 4 and subtract ii) from i)

  4x-4y=36

\underline{(-)4x+3y=92}

-7y=-56

\Rightarrow y=8

   Calculating for x

\Rightarrow x=y+9=8+9=17

Hence the two numbers are 8 \ and \  17

\\

Q.5 The sum of two numbers is 30 and the difference of their squares is 180. Find the numbers.

Answer:

Let the two numbers be x \ and \ y

x+y=30  ... ... ... ... ... ... (i)

x^2-y^2=180 ... ... ... ... ... ... (ii)

Simplifying ii)

(x-y)(x+y)=180

\Rightarrow x-y=180/30=6

x-y=6 ... ... ... ... ... ... (iii)

Solving for x \ and\  y

Add i) and iii)

 x+y=30

\underline{(+)   x-y=6} 

2x =36

\Rightarrow x=18

Substituting in i) y=30-18=12

Hence the numbers are 18 \ and \  12

\\

Q.6 The sum of the digits of a two-digit number is 8. On adding 18 to the number, its digits are reversed. Find the number.

Answer:

Let the two digit numbers be xy

x+y=8  ... ... ... ... ... ... (i)

10x+y+18=10y+x ... ... ... ... ... ... (ii)

Simplifying  ii)

9x-9y=-18

\Rightarrow x-y=-2 ... ... ... ... ... ... (iii)

Solving for x \ and\  y

Add i) and iii)

x+y=8

\underline{(+)    x-y=-2}

2x=6

\Rightarrow x=3

Substituting in i)  \Rightarrow y=8-3=5

Hence the numbers is 35

\\

Q.7 Two digit number is three times the sum of its digits. lf 45 is added to the number, its digits are reversed. Find the original number.

Answer:

Let the two digit numbers be xy

3(x+y)=10x+y  ... ... ... ... ... ... (i)

10x+y+45=10y+x ... ... ... ... ... ... (ii)

Simplifying  i) and ii)

7x-2y=0 ... ... ... ... ... ... (iii)

9x-9y=-45

\Rightarrow x-y=-5 ... ... ... ... ... ... (iv)

 Solving  for x \ and\  y

Multiplying iv) by 7 and Subtract iv) from ii)

 7x-2y=0

\underline{(-)7x-7y=-35}

5y=35

\Rightarrow y=7

Hence x=y-5=7-5=2

Therefore the numbers is 27

\\

Q.8 A two-digit number is seven times the sum of its digits. lf 27 is subtracted from the number, its digits get interchanged. Find the number.

Answer:

Let the two digit number be xy

10x+y=7(x+y)

\Rightarrow 3x-6y=0  ... ... ... ... ... ... (i)

10x+y-27=10y+x

\Rightarrow 9x-9y=27

\Rightarrow x-y=3 ... ... ... ... ... ... (ii)

Solving for x  \ and\  y

Multiplying ii) by 3 and Subtract ii) from i)

3x-6y=0

(-)   3x-3y=9

-3y=-9 

\Rightarrow y=3

Hence x=y+3=6

Therefore the numbers is 63

\\

Q.9 Find a fraction which reduces to 2/3 when 3 is added to both its numerator and denominator; and reduces to 3/5 when 1 is added to both its numerator and denominator.

Answer:

Let the fraction be \frac{x}{y} 

\frac{x+3}{y+3}=\frac{2}{3}

\Rightarrow 3x+9=2y+6

\Rightarrow 3x-2y=-3  ... ... ... ... ... ... (i)

(x+1)/(y+1)=3/5

\Rightarrow 5x+5=3y+3

\Rightarrow 5x-3y=2 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiplying i) by 5 and ii) by 3 and subtract ii) from i)

15x-10y=-15

\underline {(-)15x-9y=-6}

-y=-9

\Rightarrow y=9

Hence x= \frac{2y-3}{3}= \frac{18-3}{3}=5

Therefore the fraction is  \frac{5}{9} 

\\

Q.10 On adding 1 to the numerator of a fraction, it becomes ½. Also, on adding 1 to the denominator of the original fraction, it becomes 1/3. Find the original fraction.

Answer:m54x

Let the fraction be \frac{x}{y} 

\frac{x+1}{y}=\frac{1}{2}   

\Rightarrow 2x-y=-2  ... ... ... ... ... ... (i) 

\frac{x}{y+1}=\frac{1}{3}   

\Rightarrow 3x-y=1 ... ... ... ... ... ... (ii) 

Solving for x \ and \ y

Multiplying i) by 3 and ii) by 2 and subtract ii) from i)

   6x-3y=-6 

\underline{ (-)   8x-2y=2} 

-y=-8 

\Rightarrow y=8 

Hence x=\frac{1+8}{3}=3 

Therefore the fraction is  \frac{3}{8} 

\\

Q.11 In a given fraction, if the numerator is multiplied by 2 and the denominator is reduced by 5, we get 6/5. But, if the numerator of the given fraction is increased by 8 and the denominator is doubled, we get 2/5. Find the fraction.

Answer:

Let the fraction be \frac{x}{y} 

 \frac{2x}{y-5}=\frac{6}{5}    

 \Rightarrow 10x-6y=-30  ... ... ... ... ... ... (i) 

 \frac{x+8}{2y}=\frac{2}{5}    

 \Rightarrow 5x-4y=-40 ... ... ... ... ... ... (ii) 

Solving for x \ and \ y

Multiplying ii) by 2 and subtract ii) from i

 10x-6y=-3 

 \underline{(-)10x-8y=-80} 

 2y=50   

 \Rightarrow y=25 

Hence  x=(4y-40)/5=(100-40)/5=12 

Therefore the fraction in   \frac{12}{25}   

\\

Q.12 5 years ago, a lady was thrice as old as her daughter. 10 years hence, the lady would be twice as old as her daughter. What are their present ages?

Answer:

Let the present age of lady  =x

Let the Present age of Daughter  = y

 x-5=3(y-5)  

 \Rightarrow x-3y=-10  ... ... ... ... ... ... (i)

 x+10=2(y+10)  

 \Rightarrow x-2y=10 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Subtract ii) from i)

 x-3y=-10

 \underline{(-)   x-2y=10}

 -y=-20  

 \Rightarrow y=20

Hence  x=2y+10=50

Therefore :

Lady’s Age  = 50 \ years

Daughter’s Age  = 20 \ years

\\

Q.13 The sum of the ages of A and B is 39 years. In 15 years’ time, the age of A will be twice the age of B. Find their present ages.

Answer:

Let A’s Age =x

Let B’s Age =y

x+y=39  ... ... ... ... ... ... (i)

x+15=2(y+15)  

\Rightarrow x-2y=15 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Subtract ii) from i)

  x+y=39

\underline{(-)  x-2y=15}

3y=24

or y=8

\Rightarrow x= 39-y=39-8=31

Therefore

A’s Age =31 \ years

B’s Age =8\  years

\\

Q.14 A is 15 years elder than B. 5 years ago A was four times as old as B. Find their present ages.

Answer:

Let the age of B =x,

\Rightarrow Age \ of \ A=x+15

+15-5=4(x-5)

\Rightarrow x+10=4x-20

3x=30 \ or\  x=10

Hence A’s Age = 10+15=25

Therefore

B’s Age =10 \ years

A’s Age =25  \ years

\\

Q.15 Six years ago, the ages of Geeta and Seema were in the ratio 3 : 4. Nine years hence, their ages will be in the ratio 6 : 7. Find their present ages.

Answer:

Let the age of Geeta =x

Let the age of Seema =y

\frac{x-6}{y-6}=\frac{3}{4}

     \Rightarrow 4x-24=3y-18

      \Rightarrow 4x-3y=6  ... ... ... ... ... ... (i)

\frac{x+9}{y+9}=\frac{6}{7}

      \Rightarrow 7x+63=6y+54

7x-6y=-9 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiplying i) by 7 and ii) by 4 and Subtract ii) from i)

28x-21y=42

\underline{(-)28x-24y=-36}

3y=78  

\Rightarrow y=26

Hence x=\frac{6+3\times 26}{4}=21

Therefore

Geeta’s Age =21 \ years

Seema’s Age =26 \ years

\\

Q.16. 4 knives and 6 forks cost Rs. 200, while 6 knives and,7 forks together cost Rs. 264. Find the cost of a knife and that of a fork.

Answer:

Let the cost of knives = x

Let the Cost of fork = y

4x+6y=200  ... ... ... ... ... ... (i)

6x+7y=264 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiplying i) by 3 and ii) 2

12x+18y=600

\underline{(-)12x+14y=528}

4y=72

\Rightarrow y=18

Hence x=\frac{200-6\times 18}{4}=23

Hence Cost of :

Knive =23 \ Rs.

Fork =18 \ Rs.

\\

Q.17 The cost of 13 cups and 16 spoons is Rs. 296, while the cost of 16 cups and 13 spoons is Rs. 284. Find the cost of2 cups and 5 spoons.

Answer:

Let the cost of Cup = x

Let the Cost of Spoon =y

13x+16y=296  ... ... ... ... ... ... (i)

16x+13y=284 ... ... ... ... ... ... (ii)

Solve for x \ and \ y

Multiply i) by 16 and ii) by 13 and subtract ii) from i)

208x+256y=4736

\underline{(-)     208x+169y=3692}

87y=1044

\Rightarrow y=12

Hence x=\frac{296-16\times 12}{13}=8

Hence Cost of:

Cup =Rs.\ 8

Spoon =Rs.\  12

\\

Q.18 Rahul covered a distance of 128 km in 5 hours, partly on bicycle at 16 kmph and partly on moped at 32 kmph. How much distance did he cover on moped?

Answer:m52x

Total Distance =128 \ km

Total Time =5 \ Hrs.

Let the distance covered by cycle =x \ km

Let the Distance covered by moped = (120-x) \ Km.

\Rightarrow    \frac{x}{16}+\frac{120x}{32}=5

Simplify \Rightarrow 2x+128-x=160

\Rightarrow x=160-128=32

Hence Distance Covered by:

Cycle =32\ Km.

Moped = 96\ Km.

\\

Q.19 A boat can go 75 km downstream in 5 hours and, 44 km upstream in 4 hours. Find i) the speed of the boat in still water (ii) the rate of the current.

Answer:

Let the Speed of boat in still water = x

Speed of Stream = y

75/(x+y)=5 

\Rightarrow x+y=17 ... ... ... ... ... ... (i)

4x/(x-y)=4

\Rightarrow x-y=11 ... ... ... ... ... ... (ii)

Solve for x \ and \ y ,  add i) and ii)

2x=28 or x=14 \ Km /hr

Speed of Stream =3 \ Km/hr

\\

Q.20 The monthly incomes of A and B are in the ratio 4 : 3 and their monthly savings are in the ratio 9 : 5. If each spends Rs. 3500 per month, find the monthly income of each.

Answer:

Let A’s Income =x \ Rs.

Let B’s Income  = y \ Rs.

\frac{x}{y}=\frac{4}{3}    

\Rightarrow 3x=4y  ... ... ... ... ... ... (i)

\frac{x-3500}{y-3500}=\frac{9}{5}

\Rightarrow 5x-17500=9y-31500

\Rightarrow 5x-9y=-14000 ... ... ... ... ... ... (ii)

Substituting  x=\frac{4}{3} y in ii) 

\Rightarrow 5(\frac{4}{3} y)-9y=-14000

-7y=-14000\times 3  or y=6000 \ Rs.

Hence x=\frac{4}{3}\times 6000=8000 \ Rs.

Therefore :

A’s Income = Rs.\ 8000

B’s Income = Rs.\ 6000

\\

Q.21 6 nuts and 5 bolts weigh 278 grams, while 8 nuts and 3 bolts weigh 268 grams. Find the weight of each nut and that of each bolt. How much do 3 nuts and 3 bolts weigh together?

Answer:

Let the Weight of Nut = x \ gm.

Let the Height of Bolt = y \ gm.

 6x+5y=278  ... ... ... ... ... ... (i)

 8x+3y=268 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiply i) by 8 and ii) by 6 and Subtract ii) from i)

48x+40y=2224

\underline{(-)48x+18y=1608}

22y=616)   

\Rightarrow y=28

Hence x=\frac{278-5\times 28}{6}=23

Weight of:

Nut = 23 \ gm.

Bolt = 28\ gm.

Therefore 3 Nuts and 3 Bolts will weight = 3\times 23+3\times 28=153 gm.

\\

Q.22 There are some girls in two classrooms, A and B. If 12 girls are sent from room A to room B, the number of girls in both the rooms will become equal. If 11 girls are sent from room B to room A, then the number of girls in room A would be double the number of girls in room B. How many girls are there in each class room?

Answer:

Let No. of girls in classroom A = x and in B=y

x-12=y+12 

\Rightarrow x-y=24  ... ... ... ... ... ... (i)

2(y-11)=x+11 

\Rightarrow x-2y=-33 ... ... ... ... ... ... (ii)

Solving x \ and \ y ,

Subtract ii) from i)

x-y=24

\underline{(-)  x-2y=-33}

y=57

Hence x=24+57=81

No. of Girls in Class Room:

A=81

B=57

\\

Q.23 4 men and 4 boys can do a piece of work in 3 days, while 2 men and 5 boys can finish it in 4 days. How long would it take 1 man alone to do it?

Answer:

Suppose 1 man finished the work in x days and 1 Boy finished the work in y days

There 1 man’s 1 day’s =  \frac{1}{x}

And, 1 Boy’s 1 Day’s work =  \frac{1}{y}

Now 4 men and 4 Boys finished the work in 3 days

\Rightarrow \frac{ 4}{x}+\frac{5}{y}=\frac{1}{3}  ... ... ... ... ... ... (i)

Similarly 2 men and 5 boys finished in 4 days

\Rightarrow \frac{ 2}{x}+\frac{5}{y}=\frac{1}{4} ... ... ... ... ... ... (ii)

Solving for x  \ and\  y

Multiply ii) by 2 and Subtract ii) from 2

\Rightarrow \frac{ 4}{x}+\frac{4}{y}=\frac{1}{3}

\Rightarrow \underline{(-) \frac{ 4}{x}+\frac{10}{y}=\frac{1}{2}}

\frac{-6}{y}=\frac{-1}{6}

or y=36

Similarly x=\frac{4}{\frac{1}{3}-\frac{4}{36}}=\frac{4 \times 36}{8} = 18

So one men can finished the work in 18 \ days.

\\

Q.24 A takes 3 hours longer than B to walk 30 km. But, if A doubles his pace, he is ahead of B by 1 hour 30 minutes. Find the speeds of A and B.

Answer:

Let the Speed of A = x \ km/Hr.

Let the Speed of B = y \ km/Hr.

Time taken By A =  \frac{30}{x}

Time Taken By B =\frac{30}{y} 

\Rightarrow  \frac{30}{x}=\frac{30}{y}+3  ... ... ... ... ... ... (i)

If A Double him Race Time taken by A =\frac{30}{2x} 

Therefore  \frac{30}{2x}+\frac{3}{2}=\frac{30}{y} 

\Rightarrow \underline{(-) \frac{30}{x}=\frac{30}{y}+3}

\frac{-30}{2x}+\frac{3}{2}=-3

\Rightarrow x=\frac{10}{3}=3\frac{1}{3} Km/Hr.

Hence  \frac{30}{y}=\frac{30}{10} \times 3-3=6  

\Rightarrow y=\frac{30}{6}=5 \ km/Hr.

\\

Q.25 If the length of a rectangle is reduced by 1 m and breadth increased by 2 m, its area increases by 32 m2. If however, the length is increased by 2 m and breadth reduced by 3 m, then the area is reduced by 49 m2. Find the length and breadth of the rectangle.

Answer:m55x

Let the Length of the rectangle =x

Let the Breadth of the rectangle = y

\Rightarrow (x-1)(y+2)=xy+32

\Rightarrow xy-y+2x-2=xy+32

\Rightarrow 2x-y=34  ... ... ... ... ... ... (i)

\Rightarrow (x+2)(y-3)=xy-49

\Rightarrow 3x-2y-3x-6=xy-49

\Rightarrow 3x-2y=43 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

From i) y=2x-34

Substituting in ii) \Rightarrow 3x-2(2x-34)=43

\Rightarrow 3x-4x+68=43

\Rightarrow  x=25\ m

Hence y=2(25)-34=16\ m

Therefore:

Length = 25\ m

Breadth = 16\ m.

 

ICSE Board: Suggested Books     ICSE Board:  Foundation Mathematics
Class 8: Reference Books               Class 8: NTSE Preparation
--------------------------------------------------------------

 

Advertisements