Model 7 Working with individual wages Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 7 EXERCISES
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New 299+ Time and Work Aptitude Test On Individual Wages »
The following question based on time & work topic of quantitative aptitude
(a) $343/25$
(b) $49/125$
(c) $25/343$
(d) $125/49$
The correct answers to the above question in:
Answer: (d)
Using Rule 1,
${M_1D_1T_1}/W_1 = {M_2D_2T_2}/W_2$
${7 × 7 × 7}/7 = {5 × 5 × 5}/W_2$
49 × $W_2$ = 125
$W_2 = 125/49$
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Read more working with individual wages Based Quantitative Aptitude Questions and Answers
Question : 1
Seventy-five men are employed to lay down a railway line in 3 months. Due to certain emergency conditions, the work was to be finished in 18 days. How many more men should be employed to complete the work in the desired time ?
a) 375
b) 350
c) 300
d) 325
Answer »Answer: (c)
Using Rule 1,
$M_1D_1 = M_2D_2$
75 × 90 = $M_2$ × 18
$M_2 = {75 × 90}/18 = 375$
Number of additional men
= 375 - 75 = 300
Question : 2
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 ?
a) 15 days
b) 14 days
c) 10 days
d) 13 days
Answer »Answer: (d)
Less persons, more days (Indirect)
More working hours/day, less days (Indirect)
Let required no. of days be x.
Persons | Working hours /day | Days |
39 | 5 | 12 |
↑ | ↑ | ↓ |
30 | 6 | x |
∴ ${30 : 39}/{6 : 5}]$ : : 12 : x
30 × 6 × x = 39 × 5 × 12
$x = {39 × 12 × 5}/{30 × 6}$ = 13 days
Using Rule 1If $M_1$ men can finish $W_1$ work in $D_1$ days and $M_2$ men can finish $W_2$ work in $D_2$ days then, Relation is${M_1D_1}/{W_1} = {M_2D_2}/{W_2}$ andIf $M_1$ men finish $W_1$ work in $D_1$ days, working $T_1$ time each day and $M_2$ men finish $W_2$ work in $D_2$ days, working $T_2$ time each day, then${M_1D_1T_1}/{W_1} = {M_2D_2T_2}/{W_2}$
Here, $M_1 = 39, D_1 = 12, T_1$ = 5
$M_2 = 30, D_2 = ?, T_2$ = 6
$M_1D_1T_1 = M_2D_2T_2$
39 × 12 × 5 = 30 × $D_2$ × 6
$D_2 = {39 × 12 × 5}/{30 × 6}$ = 13 days
Question : 3
Some persons can do a piece of work in 12 days. Two times the number of such persons will do half of the work in
a) 3 days
b) 5 days
c) 9 days
d) 6 days
Answer »Answer: (a)
Using Rule 1,
${M_1D_1}/W_1 = {M_2D_2}/W_2$
${M × 12}/W = {2M × D_2}/{W/2}$
${M × 12}/W = {4MD_2}/W$
$D_2$ = 3 days
Question : 4
If x men can do a piece of work in x days, then the number of days in which y men can do the same work is
a) $x^2y$ days
b) $x^2/y$ days
c) xy days
d) $y^2/x$ days
Answer »Answer: (b)
Using Rule 1,
$M_1D_1 = M_2D_2$
$x . x = y . D_2$
$D_2 = x^2/y$ days
Question : 5
Some carpenters promised to do a job in 9 days but 5 of them were absent and remaining men did the job in 12 days. The original number of carpenters was
a) 18
b) 16
c) 24
d) 20
Answer »Answer: (d)
Using Rule 1,
Let the original number of carpenters be x.
$M_1D_1 = M_2D_2$
x × 9 = (x - 5) × 12
9x = 12x - 60
3x = 60 ⇒ x = 20
Question : 6
‘x’ number of men can finish a piece of work in 30 days. If there were 6 men more, the work could be finished in 10 days less. The original number of men is
a) 15
b) 12
c) 6
d) 10
Answer »Answer: (b)
Men | Days |
x | 30 |
↑ | ↓ |
x + 6 | 20 |
x + 6 : x : : 30 : 20
${x + 6}/x = 30/20 = 3/2$
2x + 12 = 3x
3x - 2x = 12 ⇒ x = 12
Using Rule 1,
Here, $M_1 = x, D_1 = 30, M_2 = x + 6, D_2$ = 20
$M_1D_1 = M_2D_2$
x × 30 = (x + 6) × 20
$3x = 2x + 12$ ⇒ x = 12
time & work Shortcuts and Techniques with Examples
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Model 1 Basics on Time & Work
Defination & Shortcuts … -
Model 2 Formula method ‘M1D1W1 = M2D2W2’
Defination & Shortcuts … -
Model 3 Man leaves & joins
Defination & Shortcuts … -
Model 4 Working with Man, Woman, Child
Defination & Shortcuts … -
Model 5 Split & Fraction of work
Defination & Shortcuts … -
Model 6 Efficiency of the worker
Defination & Shortcuts … -
Model 7 Working with individual wages
Defination & Shortcuts …
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