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Parin Sir > Quantitative Aptitude (Maths) > Chain rule or unitary method > Examples 5 to 9 of chain rule > Examples 5 to 9 of chain rule

Examples 5 to 9 of chain rule

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(5) If  the wages of  12 men for 10 days be Rs. 2500, then find the wages of  15 men for 18 days.

Soln. M1  =  12 men,   D1  =  10 days,  W1  =   Rs.  2500,

M2  =  15 men,   D2  =  18 days,  W2  =   ( ? )   Rs.

\frac{M_1\mathrm{\ } \mathrm{D}_1}{W_1}  =  \frac{M_2\mathrm{\ } \mathrm{D}_2}{W_2}

∴ \frac{12\times10}{2500} = \frac{15\times18}{W_2}

∴ W2  =  \frac{15\times18\times2500}{12\times10}

∴  W2  =  5625   Rs.

(6) If  25 men working 18 hours a day, can reap a field in 32 days, in how many days will 36 men reap the field, working 16 hours a day  ? 

Soln. M1 =  25 men,  T1 = 18 hrs.,  D1 = 32 days,  M2 = 36 men,  T2 = 16 hrs.,  D2 =  ?  days

M1 D1 T1  =   M2 D2 T2

∴  25 × 32 × 18  =  36 × D2 × 16

∴  D2   =  \frac{\mathrm{25\ \ }\times\mathrm{\ 18\ }\times\mathrm{\ \ 32}}{\mathrm{36}\times\mathrm{16}}

∴  D2   =   25   days

(7) If   20 engines consume 28 metric tones of  coal, when each is working  21  hours a day, how much coal will be required for 9 engines, each running 15 hours a day, it being given that 5 engines of  former type consume as much as 3 engines of  later  type  ?

Soln. M1  =  20  engines,  W1  =  28  metric tones coal,  T1  =  21 hours,

M2  =  9  engines,  W2  =  ( ? )  metric tones coal,   T2  =  15 hours,

5 × Former  =  3 × Later       ∴   E1  =  3  and  E2  =  5

Here,    \frac{M_1\mathrm{\ } \mathrm{T}_1\mathrm{\ } \mathrm{E}_1}{W_1}=    \frac{M_2\mathrm{\ } \mathrm{T}_2\mathrm{\ } \mathrm{E}_2}{W_2}

∴ \frac{20\times21\times3}{28}  =  \frac{9\times\mathrm{15}\times5}{W_2}

∴ W2  =  \frac{9\times15\times5\times28}{20\times21\times3}

∴ W2   =      15   metric tones  coal.

(8) A contract is to be completed in 102 days and 324 men were set to work, each working 8 hours a day. After 84 days,  of  the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working   9  hours a day   ?

Soln. M1  =  324 men,  D1  =  84  days,  T1  =  8 hrs.,    W1  =  \frac{7}{9}  of   total  work

M2  =  ( ? )  men,  D2   =  (102 – 84)  =  18 days,  T2  =  9  hrs.,   W2  =  \left(1-\frac{7}{9}\right) = \frac{2}{9}

\frac{M_1\mathrm{\ } \mathrm{D}_1\mathrm{\ } \mathrm{T}_1}{W_1} = \frac{M_2\mathrm{\ } \mathrm{D}_2\mathrm{\ } \mathrm{T}_2}{W_2}

∴ \frac{324\times84\times8}{\frac{7}{9}}   =  \frac{M_2\times18\times9}{\frac{2}{9}}

∴ \frac{324\times84\times8\times2}{18\times9\times7} =   M2

∴ M2   =   384.

∴ Additional men   =   ( 384 – 324 )   =   60.

(9) A garrison of  4500 men had provisions for 50 days, when given at the rate of  950 gms. per head. At the end of  14 days, a reinforcement arrives and it was found that the provisions will last 25 days more, when given at the rate of  900 gms.  per head. What is the strength of  the reinforcement?

Soln. M1  =  4500  men,   D1  =  (50 – 14)  =  36 days,   R1   =   950  gms.

M2  =  ( ? )  men,   D2   =  25  days,   R2   =  900  gms.

M_1 = M_2

∴  4500  × 36  ×  950   =   M2  × 25  ×  900

∴  \frac{4500\times36\times950}{25\times900}   =   M2

∴   M2   =   6840   men.

∴ the  strength  of  the  reinforcement  =   ( 6840 – 4500 )   =   2340  men.

Best of Luck…

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