CHAIN RULE - MCQ Test (1)


16 Questions MCQ Test RAS RPSC Prelims Preparation - Notes, Study Material & Tests | CHAIN RULE - MCQ Test (1)


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This mock test of CHAIN RULE - MCQ Test (1) for UPSC helps you for every UPSC entrance exam. This contains 16 Multiple Choice Questions for UPSC CHAIN RULE - MCQ Test (1) (mcq) to study with solutions a complete question bank. The solved questions answers in this CHAIN RULE - MCQ Test (1) quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this CHAIN RULE - MCQ Test (1) exercise for a better result in the exam. You can find other CHAIN RULE - MCQ Test (1) extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
QUESTION: 1

If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the samerate?

Solution:

Option C

Explanation :cost of x metres of wire = Rs. dcost of 1 metre of wire = Rs.(d/x)cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)

QUESTION: 2

In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, howmany men will be catered to with remaining meal?

Solution:

Option B

Explanation :Meal for 200 children = Meal for 120 menMeal for 1 child = Meal for 120/200 menMeal for 150 children = Meal for (120×150)/200 men=Meal for 90 menTotal mean available = Meal for 120 menRenaming meal = Meal for 120 men ‐ Meal for 90 men = Meal for 30 men

QUESTION: 3

36 men can complete a piece of work in 18 days. In how many days will 27 men complete thesame work?

Solution:

Option D

Explanation :Let the required number of days be xMore men, less days (indirect proportion)Hence we can write asMen36:27}::x:18 ⇒36×18=27×x ⇒12×18=9×x⇒12×2=x⇒x=24

QUESTION: 4

A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made21 revolutions, what will be the number of revolutions made by the larger wheel?

Solution:

Option D

Explanation :Let the number of revolutions made by the larger wheel be xMore cogs, less revolutions (Indirect proportion)Hence we can write asCogs 6:14}: x: 21 ⇒6×21=14×x ⇒6×3=2×x ⇒3×3=x ⇒x=9

QUESTION: 5

3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4pumps work in order to empty the tank in 1 day?

Solution:

Option B

Explanation :Let the required hours needed be xMore pumps, less hours (Indirect proportion)More Days, less hours (Indirect proportion)Hence we can write asPumps 3:4::x:8Days 2:1⇒3×2×8=4×1×x⇒3×2×2=x⇒x=12

QUESTION: 6

39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons,working 6 hours a day, complete the work?

Solution:

Option D

Explanation :Let the required number of days be xMore persons, less days (indirect proportion)More hours, less days (indirect proportion)Hence we can write asPersons 39:30::x:12Hours 5:6⇒39×5×12=30×6×x ⇒39×5×2=30×x ⇒39=3×x ⇒x=13

QUESTION: 7

A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how manyseconds will it take for the loom to weave 25 meters of cloth?

Solution:

Option D

Explanation :Let the required number of seconds be xMore cloth, More time, (direct proportion)Hence we can write asCloth 0.128:25} :: 1:x⇒0.128x=25 ⇒x=25/0.128 ⇒25000/128=3125/16≈195

QUESTION: 8

21 goats eat as much as 15 cows. How many goats each as much as 35 cows?

Solution:

Option A

Explanation :15 cows ≡ 21 goats1 cow ≡21/15 goats35 cows ≡ (21×35)/15 goats≡(21×7)/3 goats≡7×7 goats ≡ 49 goats

QUESTION: 9

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat onebag of husk?

Solution:

Option B

Explanation :Assume that in x days, one cow will eat one bag of husk.More cows, less days (Indirect proportion)More bags, more days (direct proportion)Hence we can write asCows 40:1 ::x:40Bags 1:40⇒40×1×40=1×40×x ⇒x=40

QUESTION: 10

. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost?

Solution:

Option D

Explanation :Let 200 gm potato costs x paiseCost of ¼ Kg potato = 60 Paise=> Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = 1000/4 gm = 250 gm)More quantity, More Paise (direct proportion)Hence we can write asQuantity 200:250} :: x:60⇒200×60=250×x ⇒4×60=5×x ⇒4×12=x ⇒x=48

QUESTION: 11

A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After30 days, 2/5 of the work is completed. How many additional persons should be deployed so that thework will be completed in the scheduled time, each person’s now working 9 hours a day.

Solution:

Option D

Explanation :Persons worked = 104Number of hours each person worked per day = 8Number of days they worked = 30Work completed = 2/5Remaining days = 56 ‐ 30 = 26Remaining Work to be completed = 1 ‐ 2/5 = 3/5Let the total number of persons who do the remaining work = xNumber of hours each person needs to be work per day = 9More days, less persons(indirect proportion) More hours, less persons(indirect proportion)More work, more persons(direct proportion)Hence we can write asDays 30:26Hours 8:9 ::x:104Work 35:25⇒30×8×3/5×104=26×9×2/5×x⇒x=(30×8×3/5×104)/(26×9×2/5)=(30×8×3×104)/(26×9×2)=(30×8×104)/(26×3×2)=(30×8×4)/(3×2)=5×8×4=160Number of additional persons required = 160 ‐ 104 = 56

QUESTION: 12

x men working x hours per day can do x units of a work in x days. How much work can becompleted by y men working y hours per day in y days?

Solution:

Option B

Explanation :Let amount of work completed by y men working y hours per in y days = w unitsMore men, more work(direct proportion)More hours, more work(direct proportion)More days, more work(direct proportion)Hence we can write asMen x:yHours x:y ::x:wDays x:y⇒x3w=y3x ⇒w=y3x/x3=y3/x2

QUESTION: 13

A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building,which casts a shadow of length 28.75 m under similar conditions?

Solution:

Option A

Explanation :Let the required height of the building be x meterMore shadow length, More height (direct proportion)Hence we can write asShadow length 40.25:28.75}:: 17.5:x⇒40.25×x=28.75×17.5 ⇒x=(28.75×17.5)/40.25=(2875×175)/40250= (2875×7)/1610=2875/230=575/46=12.5

QUESTION: 14

If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of suchapples?

Solution:

Option A

Explanation :Let the required price be xMore apples, More price (direct proportion)Hence we can write asApples 357:(49×12)} :: 1517.25:x⇒357x = (49×12)×1517.25 ⇒x = (49×12×1517.25)/357=(7×12×1517.25)/51= (7×4×1517.25)/17=7×4×89.25≈2500

QUESTION: 15

9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal isrequired for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as4 engines of latter type?

Solution:

Option D

Explanation :Let required amount of coal be x metric tonnesMore engines, more amount of coal (direct proportion)If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate ofconsumption.If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rateof consumptionMore rate of consumption, more amount of coal (direct proportion)More hours, more amount of coal(direct proportion)Hence we can write asEngines 9:8rate of consumption 13:14 ::24:xhours 8:13⇒9×1/3×8×x=8×1/4×13×24 ⇒3×8×x=8×6×13 ⇒3xX=6×13 ⇒x=2×13=26

QUESTION: 16

In a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people cameand the food last only for 20 more days. How many people came?

Solution:

Option A

Explanation :Given that food was sufficient for 2000 people for 54 daysHence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 ‐ 15 =39)Let x number of people came after 15 days.Then, total number of people after 15 days = (2000 + x)Then, the remaining food was sufficient for (2000 + x) people for 20 daysMore men, Less days (Indirect Proportion) ⇒Men 2000:(2000+x)} :: 20:39⇒2000×39=(2000+x)20 ⇒100×39=(2000+x) ⇒3900=2000+x ⇒ x=3900−2000=1900

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