If 5 workers can collect 60 kg wheat in 3 days, how many kilograms of wheat will 8 workers collect in 5 days ?
(A) 80 kg
(B) 100 kg
(C) 120 kg
(D) 160 kg
Solution:
Let the required quantity be x kg. More workers, More quantity (Direct proportion) More days, More quantity (Direct proportion) \[\left. \begin{gathered} {\text{Workers 5}}:8 \hfill \\ \,\,\,\,\,\,\,\,{\text{Days 3}}:5 \hfill \\ \end{gathered} \right\}::60:x\] $$\eqalign{ & \therefore {\text{ }}5 \times 3 \times x = 8 \times 5 \times 60 \cr & \Leftrightarrow x = \frac{{\left( {8 \times 5 \times 60} \right)}}{{\left( {5 \times 3} \right)}} \cr & \Leftrightarrow x = 160 \cr} $$
12.
If 5 spiders can catch five files in five minutes, how many files can hundred spiders catch in 100 minutes ?
(A) 100
(B) 500
(C) 1000
(D) 2000
Solution:
Let the required number of chairs be x. Then, More spiders, More flies (Direct proportion) More time, More flies (Direct proportion) \[\left. \begin{gathered} \,\,\,{\text{Spiders 5}}:100 \hfill \\ {\text{Minutes 5}}:100 \hfill \\ \end{gathered} \right\}::5:x\] $$\eqalign{ & \therefore {\text{ }}5 \times 5 \times x = 100 \times 100 \times 5 \cr & \Leftrightarrow x = \frac{{\left( {100 \times 100 \times 5} \right)}}{{\left( {5 \times 5} \right)}} \cr & \Leftrightarrow x = 2000 \cr} $$
13.
In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?
(A) 20
(B) 30
(C) 40
(D) 50
Solution:
There is a meal for 200 children 150 children have taken the meal Remaining meal is to be catered to 50 children Now,200 children = 120 men ∴ 50 children $$\eqalign{ & = \left( {\frac{{120}}{{200}} \times 50} \right){\text{men}} \cr & = {\text{30 men}} \cr} $$
14.
If a quarter kg of potato costs 60 paise, how many paise will 200 gm cost?
A garrison had provision for a certain number of days. After 10 days, of the men desert and it is found that the provisions will now last just as long as before. How long was that ?
(A) 15 days
(B) 25 days
(C) 35 days
(D) 50 days
Solution:
Initially, Let there be x men having food for y days After 10 days, x men had food for days (y - 10) Also, $$\left( {x - \frac{x}{5}} \right)$$ men had food for y days $$\eqalign{ & \therefore \,x\left( {y - 10} \right) = \frac{{4x}}{5} \times y \cr & \Leftrightarrow 5xy - 50x = 4xy \cr & \Leftrightarrow xy - 50x = 0 \cr & \Leftrightarrow x\left( {y - 50} \right) = 0 \cr & \Leftrightarrow y - 50 = 0 \cr & \Leftrightarrow y = 50 \cr} $$
16.
A wall of 100 meters can be built by 7 men or 10 women in 10 days. How many days will 14 men and 20 women take to build a wall of 600 metres ?
(A) 15
(B) 20
(C) 25
(D) 30
Solution:
Let the required number of days be x 7 men = 10 women (14 men and 20 women) = (20 + 20) women = 40 women More length, More days (Direct proportion) More women, Less days (Indirect proportion) \[\left. \begin{gathered} {\text{Length 100}}:600 \hfill \\ \,\,{\text{Women 40}}:10 \hfill \\ \end{gathered} \right\}::10:x\] $$\eqalign{ & \therefore {\text{ }}100 \times 40 \times x = 600 \times 10 \times 10 \cr & \Leftrightarrow x = \frac{{\left( {600 \times 10 \times 10} \right)}}{{\left( {100 \times 40} \right)}} \cr & \Leftrightarrow x = 15 \cr} $$
17.
An industrial loom weaves 0.128 metres of cloth every second. Approximately, how many seconds will it take for the loom to weave 25 metres of cloth ?
(A) 178
(B) 195
(C) 24
(D) 488
Solution:
Let the required time be x seconds Then, More metres, More time (direct proportion) $$\eqalign{ & \therefore 0.128:25::1:x \cr & \Leftrightarrow x = \frac{{25}}{{0.128}} \cr & \Leftrightarrow x = \frac{{25 \times 1000}}{{128}} \cr & \Leftrightarrow x = 195.31 \cr} $$` ∴ Required time =195 seconds (approx)
18.
4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats would be woven by 8 mat-weavers in 8 days?
(A) 4
(B) 8
(C) 12
(D) 16
Solution:
Let the required number of bottles be x. More weavers, More mats (Direct Proportion) More days, More mats (Direct Proportion) \[\left. \begin{gathered} {\text{Wavers}}\,\,\,\,4:8 \hfill \\ \,\,\,\,\,\,{\text{Days}}\,\,\,\,4:8 \hfill \\ \end{gathered} \right\}::4:x\] $$\eqalign{ & \therefore 4 \times 4 \times x = 8 \times 8 \times 4 \cr & \Rightarrow x = \frac{{ {8 \times 8 \times 4} }}{{4 \times 4}} \cr & \Rightarrow x = 16 \cr} $$
19.
5 persons can prepare an admission list in 8 days working 7 hours a day. If 2 persons join them so as to complete the work in 4 days, they need to work per day for ?
(A) 8 hours
(B) 9 hours
(C) 10 hours
(D) 12 hours
Solution:
Let the number of working hours per day be x More persons, Less working hours ( Indirect proportion) Less days, More working hours (Indirect proportion) \[\left. \begin{gathered} {\text{Persons 7}}:5 \hfill \\ {\text{Quantity 4}}:8 \hfill \\ \end{gathered} \right\}::7:x\] $$\eqalign{ & \therefore {\text{ }}7 \times 4 \times x = 5 \times 8 \times 7 \cr & \Leftrightarrow x = \frac{{\left( {5 \times 8 \times 7} \right)}}{{\left( {7 \times 4} \right)}} \cr & \Leftrightarrow x = 10 \cr} $$
20.
A team of workers was employed by a contractor who undertook to finish 360 pieces of an article in a certain number of days. Making four more pieces per day than was planned, they could complete the job a day ahead of schedule. How many days did they take to complete the job ?
(A) 8 days
(B) 9 days
(C) 10 days
(D) 12 days
Solution:
Let the team take x days to finish 360 pieces Then, number of pieces made each day = $$\frac{{360}}{x}$$ More number of pieces per day, Less days (Indirect proportion) $$\eqalign{ & \therefore \,\left( {\frac{{360}}{x} + 4} \right):\frac{{360}}{x}::x:\left( {x - 1} \right) \cr & \Leftrightarrow \left( {\frac{{360}}{x} + 4} \right) \left( {x - 1} \right) = \frac{{360}}{x} \times x \cr & \Leftrightarrow 360 - \frac{{360}}{x} + 4x - 4 = 360 \cr & \Leftrightarrow 4x - \frac{{360}}{x} - 4 = 0 \cr & \Leftrightarrow x - \frac{{90}}{x} - 1 = 0 \cr & \Leftrightarrow {x^2} - x - 90 = 0 \cr & \Leftrightarrow \left( {x - 10} \right)\left( {x + 9} \right) = 0 \cr & \Leftrightarrow x = 10 \cr} $$