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Class 7 R S Aggarwal Maths Solutions

Contents

9.        Unitary Methods (Direct and Inverse Variation) 9A, 9B

10.      Percentage 10A, 10B

11.      Profit & Loss 11A

12.      Simple Interest 12A

 

 

UNITARY METHOD  EXERCISE – 9A

Q1.         Cost of 15 oranges = Rs.110

                Cost of 1 orange =                                    [less orange, less cost]

               cost of 39 oranges =            [more orange, more cost]           

Q2.         For Rs.260, sugar bought = 8 kg

                For Rs.1,     sugar bought=

                For Rs.4420, sugar bought =

Q3.         For Rs. 6290, silk bought = 37m.

                For Rs.1, silk bought =   

                For Rs.4625, silk bought =26m.  [hints:37 x 17=629, 17 x 26= 442]

Q4.         For Rs.1110, worker work = 6 days.

                For Rs.1,   worker work =

                For Rs.4625, “        “     =     [5 x 222=1110, 5 x 925 = 4625]

Q5.         In 42ltrs , car covered a distance = 357km

                In 1ltr,      car   “            “     “        =  

                In 12 ltrs,  car   “                “        =  

Q6.         Travel by rail  900km, fare =   Rs.2520,

                Travel by rail  1km ,   fare =      

                Travel by rail  360km ,   fare =   x 360 = Rs.1008.

Q7.         Train covers 51km, in 45min

                Train covers 1km, in min

                Train covers 221km, in  x 221= 195min. = 3hrs 15min.

Q8.         Length of 85.5 kg of rod =22.5m

                Length of 1kg of rod =

                Length of 22.8kg of rod =

Q9.         13.5 kg = 13500g,

                No. of sheets in 162g = 6 sheets

                “            “         “    1g   = .

                “           “         “  13.5   =

Q10.       For 1152 bars of soap, cartoon required = 8

                For 1 bar of soap, cartoon req. =  

                For 3888 bar of soap, cartoon req. =    x 3888 =

Q11.       In 44cm thick pile, no. of cardboards = 16

                 In 1cm thick pile, no. of cardboards =

                In 71.5cm thick pile, no. of cardboards =  x 71.5= 26.

Q12.       If the length of the shadow is 8.2m, height of the building = 7m.

                If the length of the shadow 1m, height of the building =

                If the length of the shadow 20.5m, height of the building =  20.5 = 17.5m.

Q13.       In one day 16.25m wall build by   15 men.

                In one day 1m wall build by    men.

                In one day 26 m wall will build by    x 26 = 24 men.

Q14.       No. of patients accommodated in 1350 ltrs = 60.

                No. of patients accommodated in 1 ltrs =

                No. of patients accommodated in 1710 ltrs = x 1710=76.

Q15.       Weight which produces an extension of 2.8 cm = 150gm.

                Weight which produces an extension of 1 cm =

                Weight which produces an extension of 19.6 cm = x 19.6=1050= 1kg 50g.

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EXERCISE – 9B   UNITARY METHOD    INVERSE VARIATION

Q1.         48 men can dig  a trench in 14 days.

                1 man   can  “          “         “  14 X 48 days    [less men, more days]

28 man   can  “          “         “  = 4 days.

Q2.         A field  can be reaped  in 30days by 16 men.

                “              “              “      1 day      by   30 x 16 men.

                “              “              “      24 days      by    men.

Q3.         A field can graze in 13 days by  45 cows.

                “              “              “    1  day “    45 x 13 cows.

                “              “              “   9 days “ 

Q4.         16 horse consume corn in 25 days.

                1 horse       “                “      16 x 25 days.

                40 horse       “             “      days.

Q5.         If girl reads 18 page per day, she finish it in 25 days.

                “              “       1 page  “     “              “              “  18 x 25 days.

                “              “       15 pages  “     “         “              “  days.

Q6.         Reeta  type 40 words per min and takes time  = 24 mins.

                Reeta  type 1 word  per min and takes time  = 24 x 40 mins.

                Reeta  type 48 word  per min and takes time  = mins.

Q7.         At an average speed of 45km/h , a bus covers a distance in 3 hrs 20 min.

                At an average speed of 1km/h, the bus covers the distance in 3 hrs 20min x 45 min.

                an average speed of 35 km/h, the bus will cover  the distance in

=  min. =  257.14 min  = 4.28 hrs.   Ans.                  

 

PERCENTAGE        EX 10A                                                  

Q1. Convert the fraction into percentage.

i)        =                                                                                 

Ii)        

iii)        

v)           =

vii)          = 66    

Q2.         Convert into fraction.

i)                    32%  =      

ii)                  6=                                 

iii)                26%   =                   

iv)                6.25 =

v)                  0.06%  =   =                      

vi)                22.75%  =  =       

Q3.         Express the following in ratio

i)                    43%                 ii)  75% =

Q4.         Convert ratio into percentage

i)                    37:100    =           ii)   16: 25 =  =  iii) 5:4=

Q5.         Convert into decimal form

i)                    45% =                ii)   0.23%    =

Q6.         Convert decimal into %

i)                    0.6 =  =   OR   (0.6 X 100)% = 60%.

ii)                  0.005 =    OR ()% = 0.5% .

Q7.         Find:

i)                    32% 0f 425 =          

ii)                  16 

iii)                136% of 70   =                                                                     

Q8.         Find:

I)                    25% of Rs. 76  =  Rs. = Rs.               19

II)                  7.5% of 600 m =  Rs.     

III)                3 of 90km  =  Rs.      

IV)               8.5% of 5 kg =   = 0.425kg = (0.425 x 1000)g = 425g.

V)                 20% of 12 litres =

Q9.         Find the numbers whose 13% is 65.

                Let the number be y. Then, 13% of y = 65  =>    => y =

Q10.       Find the number whose 6 

                Let the number be y. Then, of 2 =

Q11.       What amount is 10% more than Rs. 90?

                90 + 90 x                                                                                                       

Q12.       What amount is 20% less than Rs 6o?

                60 – 60 x 20%       = 60 – 60 x            

Q13.       If 3% of x is 9, find the value of x.

                3% of x = 9          = x x = 9         => x =                      

Q14.       If 12.5% of x is 6, find the value of x.

                X x => X =

Q15.       What per cent of 84 is 14?

                84 x X% = 14   => 84 x    =  X =                              

Q16.       What percent is

i)                    Rs. 15 of Rs 120?

120 x X% = 15     => 120 x

ii)                  36 min of 2hrs?

2 x 60 x       => X =

iii)                8 hrs of 2 days?

48 x       => x =

iv)                160 m of 4 km?

4km = 4 x 1000 = 4000m

4000 x     =>  x =  = 4% .

v)                  175 ml of 1 ltr?

1 ltr = 1000ml then,    1000 x     => x =  = 17.5%

vi)                25 paise of Rs 4?

Rs.4 = 400 paise. Then, 400 x

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Exercise – 10B 

Q1.         We have,  total marks= 750,  marks obtained = 495.

Then, the percentage marks =

Q2.         15625 x 12%    = 15625 x  = Rs. 1875.  new salary = 15625 + 1875 = 17500.

Q3.         Exercise duty reduced 950 – 760 = 190. Then, reduction % =

Q4.         Let the cost of TV be Rs.y. Then, y x

Q5.         Let the total student be y. Then, boys y x 70%  and girls =y x 30% = 504.

π  Y = . boys = 1680 x      

 

Q6.         Since, 12kg copper in 100kg ore.

                  in 1kg   “               “  

                 In 69 kg  “         “            

Q7.         Let the maximum marks be y.

                Passmarks y x 36% =  . ATQ,  passmarks = 123 + 39= 162. Then,

                Y =

Q8.         Let Fruit seller had y no. of apples. Now, he sell y x 40% = . And remains 420 apples.

                Then,  y =   +  420 =>  y -

Q9.         Let the total no of examinees = y. Failure = 392 and pass = y x 72% =               .

                Y -

Q10.       Let the gross value be Rs. y. Then, y – y x 5% = 15200.  => y - .

                100y – 5y = 15200 x 100 => y =

Q11.       75 + 10 + charcoal = 100.  Charcoal   = 15%.                  (since,% means how much in 100)

                Then, Amount of charcoal in 8 kg will be 15% of 8kg.    8 x

Q12.       In 1kg of chalk = carbon = 1000g x

                Oxygen = 1000g x

Q13.       Let the school was open = y days. Then,ATQ, y x 75% = 219. => y =

Q14.       Let the total value be y . then, y x 3% = 42660. => Y =  .

Q15.       ATQ, total votes polled= 60000 x 80% = 48000.

                Now, A’s vote = 48000 x 60% = 28800. And B’s vote = 48000 – 28800 = 19200.

Q16.       Let the price of  shirt be Rs.100. Then, price after deduction 100 – 12 = Rs.88.

                When,  shirt price is Rs. 88          then original price is Rs. 100.

                “              “              “      Rs .1              “              “              “       

                “              “              “       Rs. 1188      “              “              “             

Q17.       Let the original price be Rs. 100 then, by 8% increase price will be Rs. 108.

                Now, since, increased price Rs.108   then     original price   is Rs. 100.

                                             “              “      Rs. 1              “              “              “      Rs. 

                                             “              “      Rs. 1566       “              “              “      Rs.

Q18.       Let the income                 = Rs. 100.

                (-)100 x 80%= 80               =          80

                                                                -----------------

                                                                                20

                Now, (-)20 x 10% = 2       =     -         2

                                                                ----------------

                                                                        Rs.18

                Since, Rs. 18 remains     then income = Rs.100.

                                Rs.1     “                 “         “          = Rs.

                                Rs. 46260 “          “           “         =  = Rs. 257000.

Q19.         Let the number be 100.    Increased  = 100 + 100 x 20% = 120.

Decreased = 120 – 120 x 20% = 120 – 24 = 96. Net decreased = 100 – 96 = 4%.

Q20.       Let the salary be Rs. 100.  Increased salary = 100 + 100 x 20% = Rs.120.

                Since, new salary is Rs 120 to restore it should be reduced by Rs. 20.

                                  “          “      1       “           “              “              “                   

                             “              “              Rs.100   “              “              “             

Q21.       200000 x . Again, Rs. 200000 x  

Remaining amount = 540000 – 400000 = 140000. Now, 140000 x 0.5% = Rs.700.

Total commission = 4000 +2000 + 700 = Rs. 6700.

Q22.       Let the income of Akhil be Rs. 100 then Nikhil income is Rs.80.

                When, Nikhil income is Rs.80 Akhil income is Rs.20 more.

                                “              “              Rs.1        “              “       Rs.          

                                “              “              Rs.100  ”               “              .x 100= 25%.

Q24.       Let 1 yr ago value of machine be y. Then, ATQ, y – y x 10% = 387000.

               

Q25.       ATQ, 450000 x 20% = 90000. Value of car after 1st yr 450000 – 90000 = 360000.

                Again, 360000 x 20% = 72000. Value of car after 2nd yr 360000 – 72000 = Rs.288000.

Q26.       We have, present population 60000. Then, increased in 1st  yr = 60000 x 10% = 6000.

                  60000 + 6000 = 66000. Increase in 2nd  yr = 66000 x 10 % = 6600.

                  population after 2 yrs = 66000 + 6600 = 72600.

Q27.       Let original price = Rs. 100/kg.    New price = Rs. 125/kg.

                  In Rs. 125 we are getting  1kg

                      In Rs 1             “              “     

                   In Rs. 100      “           ‘      .

                 reduce in consumption = 1 -  

               

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EXERCISE – 11A   PROFIT & LOSS

1)    Gain = SP – CP,     2)     Loss = CP – SP,    3)   Gain% =

4)  Loss% =     5)    SP =      6)  SP =

7)  CP =  x SP        8) CP =   x SP 

Q1.         Find the SP when:

i)                    CP =Rs. 950,  gain  = 6% . SP =   = .

ii)                  SP =      =      =

iii)                SP =        =  

iv)                SP =

Q2.         Find the gain or loss per cent when:

i)                    SP>CP it is gain.                    Gain = SP – CP = 2592 – 2400 = 192.

Gain% =

ii)                  CP >SP. It is loss.   Loss = CP –SP= 1650 – 1452 =Rs. 198.

Loss% =     =

Q3.         Find the CP when

i)                    SP=924. Gain=10%    CP =  xSP  = .

ii)                  SP=1755, Gain= CP =  x1755     = 

iii)                SP= 8510, loss = 8%.   CP=  

iv)                CP=

Q4.         Almirah CP= 13600 + 400 = Rs.14000.  SP = 16800. Since, SP> CP. It is gain.

                Gain = 16800 – 14000 = Rs.2800.  Gain % =  =

Q5.         CP 0f house = 765000 + 115000(spent) = Rs.880000. Now, SP = CP + Cp x 5%

                SP =880000 + 880000 x 5%   = 880000 + 44000 = Rs.924000.

Q6.         Since, CP of 12 lemon     Rs.25.

                      “   CP 0f 1  “               Rs. 

                Since, SP of 5 lemon   = Rs. 12

                      “      SP  0f 1 lemon    =        Now, SP > CP. It is gain.

 Gain = SP - CP . = 2.4 – 2.08= 0.32

                Gain =  =  

Q7.         Let the CP of each pens be Rs.1. then, CP of 12 pens = Rs.12.

                SP of 12 pens = CP of 15 pens = Rs.15.  => gain = SP – CP = 15 – 12 =Rs. 3.

                Gain % =       =

Q8.         Let the CP of each spoons = Rs.1.  CP of 16 spoons = Rs.16.

                SP of 15 spoons = Rs. 15.  Loss = CP – SP   = 16 – 15 = Rs.1.

                %loss =    = .

Q9.         Manoj’s  CP = Rs.12000,       SP =12000 + 12000 x 10% = Rs.13200 = Rahul’s CP.

                Rahul SP = 13200 - 13200 x 5% = 13200 – 660 = Rs.12540 = Rakesh’s CP.

Q10.       Let the CP of sofa- set is Rs. y . Then,   y + y x 8% = 21600.  =>  y +

π  100y + 8y = 21600 x 100 =>  108y  = 21600 x 100  => y =

OR,

CP =  x SP        = CP =  x 21600  =     x 21600 = Rs. 20,000.

Q11.       We have, SP = Rs.11400,  loss =  5% ,    CP = ?

                CP =   xSP  => CP =   x11400   => CP =   x11400= Rs.12000.

Q12.       We have SP = 1325, gain = 6%    New SP = ?,  New gain = 12% .

                   CP =  x SP      =  

                SP =      =

Q13.       We have, SP= Rs.24480,   loss = 4%. Then, CP =  x SP .

                CP =  = 25500.      SP =      =     = Rs.26520.

Q14.       Let the CP be y, then, ATQ, gain = 15%  SP =  .

                Now, SP at gain of 20% = .

                Now, 108 more on 15% = 20% gain.

π     + 108=        =>        = -108  => .

π     =>  -y = - 108 x 20.  => y = 2160.

Q15.       Let CP be y.  Then ATQ, Loss = 8%.      =. SP =

Now, SP at 6% gain =  =  now,   when 3360 more added 8% loss become 6% gain.

π       =>    =>  

π  Rs.24000.

Q16.       ATQ, on 10% gain,   CP =  2376 = Rs2160.

                CP=       Total CP = 2160 +2640 = Rs.4800.

                Total SP = 2376 x 2 = Rs.4752. Here CP > SP . It is loss.  Loss = 4800 – 4752 = Rs.48.

                Loss% = .

Q17.       SP = 7350, Let CP = Rs.y.     Profit = .      CP + profit = SP.    Y +

                    y = 7350 x .   CP = Rs. 6300.

Q18.       CP of Rahim = 14300 = SP of karimCP of karim =

                   CP of Mohit =

Q19.       SP of retailer = 37950,  profit = 25%.

                CP of retailer = 37950 x

                Now, SP of wholesaler  = Rs. 30360, profit = 15%.

                  CP of wholesaler  = 30360 x

                SP of manuf. = Rs. 26400, Profit = 10%.

                  CP of manuf. = 26400 x.

Thus the production cost of a washing machine = Rs. 24000.

                OR

                 Cost of production = 37950 x

Q20.       CP of video = 20000, loss = 5%.      CP of TV = 30000, gain = 8%.

                 SP of  video =

                 SP of TV =    Total CP = Rs.50000.   Total SP = Rs.51400.

                  Total gain = 1400.     % gain =

Q21.

Q22.

EXERCISE – 12A     SIMPLE INTEREST

[i]        SI =          [ii]  R =      [iii]  P =     [iv) T = .

[v]       Amount = (principal + Interest)

Q1.         P = 6400, r = 6%,  t = 2yrs.   SI =           =

                Amount = (principal + Interest)  = 6400 + 768 = Rs.7168.

Q2.         P = 2650, r = 8%, t =   SI =  

                Amount = 2650 + 530 = Rs.3180.

Q3.         SI =

Q4.         SI =         amt.  = 9600 +300 = Rs.9900.

Q5.         SI =

Q6.         P= 6400, SI = 1152,   r = 6%p.a., t = ?.

                SI = .     T =

Q7.            =

Q8.         We know,   SI = A – P,     = 6450 – 5000 = 1450.    And  =

                =  

Q9.         R =      =

Q10.       R =      =

Q11.       A= P + I = 5421.20 = 3560 + I => I = 916.20.  Now, R =   =  9%.

Q12.       SI =    Amt. = P + SI = 6000 + 2640 = Rs. 8640.

Q13.       Given, P = Rs. 12600, R = 15%, T = 3y, Money returned = Rs. 7070, Find – cost of goat.

                SI =  = 126 x 15 x 3 = Rs. 5670. Now, A = P + I = 12600 + 5670 = Rs. 18270.

                Thus cost of goat = amount – money returned = 18270 – 7070 = Rs. 11200.  

Q14.       Given, T = 3yrs, R = 10% SI – 829.50, P = ?

                P =

Q15.       Given, R =

                A = P + SI  =>  3920 = P +    => 3920 = P +

                P x

Q16.       Case – I

                Given, R = 11%, A = Rs.4491, T = 2yrs 3m,  2 +

                A = P + SI   =>   P +    = 4491  => P +

π  P +    

Case – II

                P = Rs.3600, T = 3 yrs., R = 11%,    Si =

                3600 + 1188 = Rs. 4788.

Q17.       Case – I

                R=8%,  A=Rs.12122,  T = 2 yrs.   =>   P +    = 12122 => P +

π   

Case –II

P = Rs. 10450,  R = 9%, T = .  SI =  = 209 x 3 x 4 = Rs.2508.

               
  A = 2508 + 10450 = Rs.12958.

Q18.       Given, A = Rs.4734, P = 3600, T = R = ?.

R =

Q19.       Case – 1

                Given, P = Rs.640, A = Rs.768,  T = 2yrs 6m = 2+

                R =

                Case –II

                P = Rs.850, R = 8%, T = 3yrs, amount= ?

                SI =     =

                 amt. = P + SI = 850 + 204 = Rs.1054.

Q20.       Given P = Rs.5600, A = Rs. 6720, R = 8%.

                SI = (A-P)=  6720  - 5600 = 1120.    T =   =

                = 2yrs

Q21.       Given,  Let Principal = Rs.P, Then amount =    R = ?.

                SI = A – P = .     Now, R =   =

Q22.       Given, Let principal = Rs.P, Rate = R%,  CASE – I  ,   A = 783,   T = 2 yrs.

                Amt. = (P + SI)      => 783 = P +

                  P = 783 x   ------------(1)

CASE – II,      A = Rs.837, T = 3yrs,    we know,  A = P + SI   =>  837 = P +

837 = P +  => 837 =

               From (1)&(2)

                783 x    => 

π  87R – 62R = 3100 – 2900   =>  25R = 200      R =

  P = 738 x

Q23.       Given, CASE – I,     A = Rs. 4745,   T = 3yrs,  Let principal = Rs. P & Rate = R%.

                A = (P + SI) =>  4745 = P +

x  --------------(1)

                CASE – II,    A = Rs.5475,  T = 5 yrs, A = (P + SI)

                5475 = ------------------(2)

                From (1) & (2)

                4745 x     =>  949(100 + 5R) = 1095(100 + 3R)

                94900 + 4745 R = 109500 + 3285 R  => 4745 R – 3285 R = 109500 – 94900.

π   460 R = 4600   =>  R =

From (1)     =>   P  4745 x

Q24.     Let 1st part = Rs.y, then 2nd  part = Rs. (3000 - y).

 For first part  P = Rs. y, R = 8%, T = 4yrs, SI1 =

For 2nd  part ,    P = Rs. (3000 - y), R = 9%,   SI2 = .

Given,    SI1 = SI2           

                  => 32y = 18 x 3000 – 18y  => y =

                1st part = y = Rs.1080.   &   2nd part = 3000 – y = 3000 – 1080 = Rs. 1920.

Q25.       Let 1st part = y   & 2nd part = 3600 – y.

                For 1st part, p = Rs.y.    r=9%, t = 1yr,    SI =

                For 2nd part,   P = Rs.(3600 - y),  R  = 10%,   T = 1yr

                SI =  Since, SI for 1st part + SI for 2nd part = 333.

π 

π 

 1st Part = y = Rs. 2704.    2nd part = 3600 – y = 3600 – 2700 = Rs.900.

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