An intercity cab service has a lumpsum fixed charges of Rs. 800 upto certain kilometer. Beyond that the charges are per kilometer. When two passengers separately travelled total 450 km, they were charged Rs. 2600 and Rs. 2000 respectively. If a single passenger had travelled all 450 km, he would have been charged Rs. 5000. How many kilometers are allowed for first Rs. 800 ?
When two passengers together travells 450 km, let 1st passenger travells 'x' km and 2nd travells (450-x) km.
Let Rs. 800 are charges upto 'a' km and charge per km beyond 'a' km is Rs. 'r'
Charges of 1st passenger = 800 + (x-a)*r = 2600 ⇒ xr - ar = 1800 ---(i)
Charges of 2nd passenger = 800 + (450-x-a)*r = 2000 ⇒ 450r - xr - ar = 1200 ---(ii)
Charges when single passenger travells 450 km = 800 + (450-a)*r = 5000 ⇒ 450r - ar = 4200 ---(iii)
Adding eqn (i) & (ii), we get 450r - 2ar = 3000 ⇒ 225r - ar = 1500 ---(iv)
Subtracting eqn (iv) from (iii), we get r=12
And substituting value of r=12, in eqn (iv), we get 225(12) - a(12) = 1500
2700 - 12a = 1500
a = 100
Correct Option 7)
A bus covers a certain distance at an average speed of 80 kmph without any stoppages. While returning the same journey the bus covers the distance at an average speed of 60 kmph with stoppages. What is the average stoppage time per hour taken by the bus?
Let’s assume the total journey distance in either direction is 'D' km
Therefore, time taken for the onward journey = D/80 hrs
And time taken for the return journey = D/60 hrs
The difference in journey time indicates the stoppage time = (D/60)-(D/80) = D/240 hrs
In the journey of D/60 hrs, stoppage time is D/240 hrs
Therefore, the average stoppage time per hour = (D/240)/ (D/60) hrs = 1/4 hrs = 15 minutes.
Correct Option 8)
Alloy A and B contains Tin and Copper in the ratio 5:3 and 2:5 respectively. In waht ratio A and B should be mixed to get a new alloy containing Tin and Copper in the ratio 4:3 ?
Let the Tin and Copper in alloy A is 5x/8 and 3x/8 respectively.
And Tin and Copper in alloy B is 2y/7 and 5y/7 respectively.
Now, required ratio = (5x/8 + 2y/7) / (3x/8 + 5y/7) = 4/3
(35x+16y)*3 = (21x+40y)*4
Solving, we get x/y=16/3
Alternate Method (by alligation rule):
Tin in alloys A and B=5/8 and 2/7
Tin in final alloy = 4/7
5/8 ------------------------- 2/7
--------------- 4/7 ------------
(4/7 - 2/7) ------------ (5/8 - 4/7)
= 2/7 ------------------ 3/56
Ratio = (2/7) / (3/56) = 16/3
In the word 'INDEPENDENCE' we have 5 vowels IEEEE and 7 consonants NDPNDNC
We treat vowels IEEEE as one group
Thus, we have letter group NDPNDNC (IEEEE).
This has 8 (7+1) letters of which N occurs 3 times, D occurs 2 times and rest are different.
Number of ways arranging these letters = 8!/(3!x2!) = 3360
Now, 5 vowels in which E occurs 4 times and the rest are different, can be arranged in 5!/4! = 5 ways
.'. Required number of ways = ( 3360 x 5) = 16800
Correct Option 8)
In an arithmetic series, if the sum of third and seventh terms is 8 and the sum of the sixth and the tenth terms is 26, then what is the first term of the sequence?
On the eve of Raksha Bandhan, Veena want to tie a Rakhi to her three brothers.If she has total seven silver and three gold Rakhis, then in how many different ways she can select Rakhis such that at least one gold Rakhi is there?
Veena may select Rakhis as (1 gold and 2 silver) or (2 gold and 1 silver) or (3 gold)
Required number of ways = (3C1 x 7C2) + (3C2 x 7C1) + (3C3) = 63 + 21 + 1 = 85
Corerect Option 6)
Average score of a group of athletes was 42 in certain event. When a new athlete join a group and scores equal to 250% of the average score of a group, then average of a group increases by 30%. How many athletes were there in a group before joining of a new athlete ?
Before joining new athlete, let there were 'x' athletes, so their total score = 42x
Score of a new athlete = (250/100)*42 = 105
New average = (42x + 105) / (x+1) = (130/100)*42
42x + 105 = (x+1)*54.6
12.6x = 50.4
x = 4
.'. There were 4 athletes before joining of a new athlete.
Correct Option 1)
A two-digit number is chosen such that it is 18 more when digits are reversed. What is the probability that sum of digits of this number is more than 10 ?
If x and y are digits of a number, then two-digit number=10x+y
Difference between original number & number formed by digits revered= (10y+x) - (10x+y) =18
⇒ y - x = 2
Values of y,x satisfying above condition for a two-digit numbers are (3,1) (4,2) (5,3) (6,4) (7,5) (8,6) (9,7)
Thus there are total 7 numbers 13, 24, 35, 46, 57, 68, 79
Numbers with sum of digits more than 10 are 57, 68, 79 , Total 3
Hence, required probability = 3/7
Correct Option 4)
Certain work is proposed to be completed in 54 days with 30 men. But after 36 days it is observed that only 40% of the work is completed. How many more men are required to complete the work in stipulated time ?
Total days=54, Work completed with 30 men in 36 days=40%=2/5
Remaining work=1 - 2/5=3/5, Remaining days=54-36=18
Let 'x' extra men are employed to finish the work, then total new men = (30+x)
According to Men, days & work rule, 30 Men*36 Days / (2/5) Work = (30+x) Men*18 Days / (3/5) Work
30*36 / (2/5) = (30+x)*18 / (3/5)
⇒ x=60
.'. 60 extra men are required to finish the work in time of 54 days.
Correct Option 4)
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