Shubhendu was asked to find the average of first 'n' natural numbers. By mistake, he added a number twice but he didn’t noticed it. As a result, he obtained a wrong average of 31.3. Find the number he added twice?
Sum of first 'n' natural numbers = n(n+1)/2
.'. Their average = [n(n+1)/2]/n = (n+1)/2
Given, average is 31.3 including one number twice.
As we know average lies some way at middle of sum of first 'n' natural numbers.
So check for the average of first 60 numbers = (60+1)/2 = 30.5
Given, average 31.3 is 0.8 more than the average of first 60 number including one number twice.
So, number repeated twice = 0.8x60 = 48
Correct Option 6)
Due to continuous rise in the prices of petrol, Mr. Rohan decided to convert his existing scooter into EV by fitting a compatible kit which costs Rs. 58,500. Present petrol price is Rs. 110 per liter and mileage of existing scooter is 40 km/liter and if converted to EV, it cost Rs. 15 per charge and scooter can run 100 km per charge. If it is presumed that existing petrol, charging costs and mileage remains same over the period, then how much minimum kilometer must Mr. Rohan drive, so that his EV conversion cost gets recovered ?
Cost/km of a petrol vehicle = 110/40 = 11/4 Rs.
Cost/km for an EV = 15/100 = 3/20 Rs.
Cost saved per km by using EV = (11/4) - (3/20) = 13/5 Rs.
⇒ For saving of Rs.13/5, travelling distance is 1 km, so to save Rs. 58500, distance to be travelled = 58500/(13/5) = 22500 km.
Correct Option 5)
Series is formed by sequential multiplication by 2, 3, 4, 5, ... and subtraction of 3, 8, 13, 18, ....
5 x 2 - 3 = 7
7 x 3 - 8 = 13
13 x 4 - 13 = 39
39 x 5 - 18 = 177
177 x 6 - 23 = 1039
Correct Option 4)
When a two-digit number is multiplied by sum of its digits, the product is 715 but when the reverse of a two-digit number is multiplied by sum of its digits the product obtained is 99 less, The sum of the digits of a number is
Let the two-digit number be (10x+y), then (10x+y)*(x+y)=715 ⇒ 10x² + y² +11xy = 715 ---(i)
And (10y+x)*(x+y)=(715-99) ⇒ x² + 10y² + 11xy = 616 --- (ii)
Subtracting eqn (ii) from (i), we get x² - y² = 11
⇒ (x+y) (x-y) = 11
As 11 is prime, only product 11*1 = 11 is possible
.'. (6+5) (6-5) = 11, so x=6, y=5
Hence, the number is 65 and sum of the digits=11
Alternate Method:
If x and y are the digits of a number, then (10x+y)*(x+y)=715 and (10y+x)*(x+y)=616
As we know (x+y) is common factor in both the equations.
Only 11 is the common factor.
Therefore sum (x+y)=11
Correct Option 7)
Company Writewell produces quality ball pens. If on average 4% of the produced pens are always defective so are rejected before packing. Company promises to deliver 8400 pens to its wholesaler at Rs. 8 each. It estimates the overall profit on all the manufactured pens to be 25%. What is the manufacturing cost of each pen?
As 4% pens are defective,so company is able to deliver only 96% of manufactured pens, hence to deliver 8400 pens company has to to produce 8400*(100/96) = 8750 pens
Total income by selling 8400 pens @ Rs. 8 pen pen = 8400 x 8 = 67200
As income 67200 is 125% of the manufacturing cost of total 8750 pens .'. cost of 8750 pens = 67200 x 100/125 = 53760
Therefore, cost of each pen = 53760/8750 = 6.144
Correct Option 5)
The Income of Arjun is 2/3 of Rajan's income and the expenditure of Arjun is 3/4 of Rajan's expenditure. If 1/3 of the income of Rajan is equal to the expenditure of Arjun, then the ratio of the savings of Arjun to that of Rajan is
Let Rajan's income=x, then Arjun's income=(2/3)*x
Let expenditure of Rajan=y, then Arjun's expenditure=(3/4)*y
Given, 1/3 of the income of Rajan is equal to the expenditure of Arjun ⇒ (1/3)*x = (3/4)*y
.'. x/y = 9/4 or y = (4/9)*x
Ratio of savings of Arjun & Rajan = [(2/3)*x - (3/4)*y] : [x - y]
= [(2/3)*x - (3/4)*(4/9)*x] : [x - (4/9)*x]
= 3 : 5
Correct Option 5)
When an article is sold for Rs. 2080, a shopkeeper suffers as much loss as he would have gained by selling it at a profit of 20%. If he sells the same article for Rs. 2470, his profit or loss percentage is
OPtion
1) 5% Gain
2) 8.84% Loss
3) 2.5% Gain
4) 1.2% Loss
5) 1.15% Loss
6) 5% Loss
7) 2.7% Gain
8) 2.5% Loss
9) 4% Gain
10) None of these
Solution
Let the cost be Rs x, then (x - 2080) = (120/100)*x - x
x - 2080 = 0.2x
0.8x = 2080
x = 2600
Loss when SP is 2470 = 2600 - 2470 =Rs. 130
% Loss = (130/2600)*100 = 5%
Correct Option 6)
A boy started from one corner of a rectangular park and walks along the adjacent sides to reach opposite corner. If he had walked through a shortest distance diagonally, he could have saved the distance equal to five-sixth of the shortest side a park, then the ratio of a longest side to the shortest side is:
Let the length=x and breadth=y of a rectangular park.
Distance covered, when boy walked along the adjacent sides = (x+y)
Distance along diagonal=√(x² + y²)
According to given condition: (x+y) - √(x² + y²) = (5/6)*y
(6x+y)/6 = √(x² + y²)
Squaring on both sides, (36x² + y² + 12xy)/36 = x² + y²
12xy = 35y²
x/y = 35/12
Correct Option 3)
On the first day of Ganesh Chaturthi, ladoos and modakas are offered to Lord Ganesha as prasad in the ratio 5:3 respectively which is then mixed and distributed in such a way that each visiting devotee get one count of prasad. If devotees increases daily with a constant number and on last day there are 112 devotees. If prasad last for exactly ten days of festival. Which of the following can be the number of devotees visited on the first day ?
Given, devotees increases constantly on daily basis, which forms an A.P. If number of devotees on 1st day=a, increase per day=d and number of days n=10
Then, number of devotees on last day=a+(n-1)d ⇒ 112=a+9d
Now calculate values of 'a' for different values of 'd'
When (d=1,a=103) (d=2,a=94) (d=3,a=85) (d=4,a=76) (d=5,a=67) (d=6,a=58) (d=7,a=49) (d=8,a=40) (d=9,a=31) (d=10,a=22) (d=11,a=13) (d=12,a=4)
Let total Ladoos=5x and Modakas=3x, then total count of prasad=8x, which last for 10 days.
We know, sum of an A.P. is given by Sn = (n/2)[2a + (n-1)d]
As the total count of prasad is 8x, therefore the sum of an AP should be multiple of 8.
When d=1,a=103, Sum=1075
d=2, a=94 Sum=1030
d=3, a=85 Sum=985
d=4, a=76 Sum=940
d=5, a=67 Sum=895
d=6, a=58 Sum=850
d=7, a=49 Sum=805
d=8, a=40 Sum=760
d=9, a=31 Sum=715
d=10, a=22 Sum=670
d=11, a=13 Sum=625
d=12, a=4 Sum=580
From above, the only sum which is multiple of 8 is 760, hence count of devotees on first day can be 40.
Correct Option 8)