- A CNC machine become 10% 20% & 40% more efficient when it is supported by three other supportive machine P,Q& R respectively. Find the net saving of fuel consumed by the CNC machine if it is supported by all the three supportive machine independently?
584966.876.564.3Option C

Improvement in efficiency = Net fuel saving= Reduction in fuel consumed

Consumption of fuel reduces by 10%,20%, and 40% successively.

Let initial consumption of fuel be 100 Liter

10% Improvement in efficiency = 10% Net fuel saving= 10% Reduction in fuel consumed

Thus fuel used by main CNC machine after adding supportive machine P = 100-10 = 90 Liter

20% Improvement in efficiency = 20% Net fuel saving= 20% Reduction in fuel consumed

Thus fuel used by main CNC machine after adding supportive machine Q = 90- 18=72 Liter

40% Improvement in efficiency = 40% Net fuel saving= 40% Reduction in fuel consumed

Thus fuel used by main CNC machine after adding supportive machine R = 72-28.8=43.2 Liter

Thus total fuel saved by main machine = 100-43.2 = 66.8 Liter

= (66.8/100)*100 = 66.80% - Sara buys a Micro oven for Rs. 11000. After using it for an year she sell it at a loss of 15%. Then again she buys a new Oven at a price 15% higher than the previous one. She uses the new Oven for one year and sell it at a loss of 20% then find her average loss per year during this period?
21002090301030002200Option B

Loss in first year = 15% 0f 11000

=1650

Cost of mobile in second year 115% of 11000

= 11000 + 1650

= 12650

Loss in second year = 20% of 12650

= 2530

Total loss = 1650 +2530

=4180

Average loss= 4180/2 = 2090 - A construction can be finished by X labour in 80 days. If 10 labours leave before the work started then it will take 20 extra days to complete the work. Find the number of days to complete the same work if 100 people are working.
1020304050Option D

If X labour can be completed the work in 80 days

Then 1 labor can complete the work in 80*X days ————–I

After 10 labors leave the job total labors = (X-10) and work completed in 80+20= 100 days

Now,

(X-10) labours can complete the work in 100 days

Then 1 labours can complete the work in 100*(x-10) days ————-II

Equating I and II value

80*X = 100*(X-10)

X = 50 labours

Thus 50 labours can complete the work in 80 days

100 labours can complete the work in (80*50)/100 = 40 days - The average speed of a school van is 60km/h excluding its stoppage time and if stoppage time is included its average become 50km/hr. How many minutes does the school van stop in an hour?
510152025Option B

Let the distance travel by school van is 300 km (L.C.M of 60 & 50)

Time taken when the speed is 60km/h = 300/60

=5 hours

Time taken when the speed is 50km/h = 300/50

=6 hours

In 6 hours bus travel only 5 hours and 1 hours stop on its stoppage

Thus in one hour stoppage time =1/6 hour

=(1/6*60) minutes

= 10 minutes - The average salary of 25 executives is 55000 rupees. If 5 executive were dropped from it the average reduced by 5000. What is the average salary of the executives who were dropped?
6000056000700007500055000Option D

Total salary of 25 = 25* 55000

=1375000

Average salary of remaining 20 = 55000 – 5000

= 50000

Total salary of remaining 20 = 50000 * 20

= 1000000

Sum of the salary or 5 who left = 1375000 – 1000000

=375000

Average salary or 5 who left 375000/5 =75000 - 34484.78 ÷ 10.99 + (4.76/9) of 12149.56 + 55.87 % of 8999.5 = ?^2 – 451
120121122123124Option E

(34485/11) + (5/9)*12150 + (56/100)*9000 = x^2 – 4513135 + 6750 + 5040 + 451 = x^2

15376 = x^2

x = 124

- If the radius of the spherical ball is 4.9 cm, what is the surface area of the ball?
301.84 cm^2334.76 cm^2391.22 cm^2201.74 cm^2271.84 cm^2Option A

Radius of the sphere = 4.9 cmSurface area of the square = 4 * 22/7 * r * r

= 4 * 22/7 * 4.9 * 4.9 = 301.84 cm2

- If a man sold the dress for Rs. 1200 and he earned X% of the profit. If he increase the selling price of the dress by Rs. 300 and he can makes the profit of (X + 30) %, then find the value of X?
1020304050Option B

CP * (X + 100)/100 = 1200CP * (X + 130)/100 = 1500

120000/(X + 100) = 150000/(X + 130)

4X + 520 = 5X + 500

X = 20 %

- A blackboard is in rectangular shape.If the ratio of the length of the board to the side of the another square board is 5 : 8 and the ratio of the breadth of the board to the side of the square board is 1: 2. If the difference between the perimeter of the board and the square board is 56 cm, then what is the area of the rectanglular board?
450470340360320Option E

Ratio of length = 5: 8Ratio of the breadth = 1: 2

Ratio of the length and breadth of the rectangle and side of the square = 5: 4: 8

Perimeter of the rectangle = 2 * (l + b) = 2 * (9x) = 18x

Perimeter of the square = 4 * a = 4 * 8x = 32x

32x – 18x = 56

x = 4 cm

Area of the rectangle = l * b

= (5 * 4) * (4 * 4)

= 320 cm2

- Ratio of the number of red, blue and yellow cards in the bag is 4: 3: 5. If 18 yellow cards taken out and 10 blue cards is added, then the ratio of red, blue and yellow cards become 6: 7: 3. What is the total number of cards in the bag at initially?
7274767880Option A

Ratio of the red, blue and yellow cards = 4: 3: 5(5x – 18)/(3x + 10) = 3/7

9x + 30 = 35x – 126

26x = 156

X = 6

Total number of cards = 12x = 12 * 6 = 72