Question: A candidate scoring `25%` marks in an examination fails by `30` marks while another candidate who scores `50%` marks gets `20` marks more than those required to pass. The pass percentage is :
A
`25%`
B
`35%`
C
`40%`
D
`50%`
Note: Let total marks = x
`25% of x + 30`
`= (50% of x) - 20`
`⇒ 25/(100) x + 30`
`= 50/(100) x - 20 ⇒ (x/2 - x/4)`
`= 50`
`⇒ x/4 = 50 ⇒ x = 200.`
:. pass marks
`= (25% of 200) + 30`
`= (25/100 xx 200 + 30)`
`= 80`
:. pass percentage
`= (80/200 xx 100)%`
`= 40%`
Question: Two numbers are less than a third number by` 30%`and `37%` respectively. How much percent is the second number less than the first ?
A
`10%`
B
`7%`
C
`4%`
D
`3%`
Note: Let, third number be x. Then,
First number`= 70% of x = (7x)/10,`
Second number`= 63% of x = (63x)/100`
Required Percentage
`= ((7x)/100 xx 10/(7x) xx 100)%`
`= 10%`
Question: A bag contains `600` coins of `25 p` denomination and `1200` coins of `50p` denomination. If `12%` of `25%` coins and `24%` of 50p coins are removed, the percentage of money removed from the bag is nearly :
A
`30%`
B
`21.6%`
C
`17.8%`
D
`15.6%`
Note: Total money
`= (600 xx 25/100 + 1200 xx 50/100)`
`= Rs. 750.`
`25` paise coins removed
`= (12/100 xx 600)`
`= 72`
`50` paise coins removed
`= (24/100 xx 1200)`
`= 288`
Money removed
`= (72 xx 25/100 + 288 xx 50/100)`
`= Rs. 162`
Required percentage
`= (162/750 xx 100)%`
`= 21.6%`
Question: `5%` of income of A is equal to `15%` of income of B and `10%` of income of B is equal to `20%` of income c. If income of c is Rs.` 2000, then total income of A, B & C is :
A
`Rs. 6000`
B
`Rs. 18000.`
C
`Rs. 20000`
D
`Rs. 14000`
Note: `5/(100) A`
`= 15/(100) B & 10/(100) B`
`= 20/(100) C`
`:. A = 3B and B = 2C`
`= 2 xx 2000`
`= 4000`
`:.A = 3 xx 4000`
`= 12000`
`:. A + B + C`
`= (12000 + 4000 + 2000)`
`= 18000.`
Question: In an examination, A got `10%` marks less than B, B got `25%` marks more than C and C got `20%` less than D. If A got `360` marks out of `500,` the percentage of marks obtained by D was :
A
`70`
B
`75`
C
`80`
D
`85`
Note: `A = 90/(100) B, B`
`= 125/(100) C and C`
`= 80/(100) D`
`:. B = 10/9 A,`
`C = 4/5 B and D`
`= 80/(100) C.`
`B = 10/9 xx 360`
`= 400, C = 4/5 xx 400`
`= 320 & D`
`= 5/4 xx 320`
`= 400`
Percentage of D
`= (400/500 xx 100)%`
`= 80%`
Question: `95` men had provisions for `200` days. After `5` days, `30` men died due to an epidemic. The remaining food will last for how many days ?
A
`136 (16)/(19)` days
B
`180` days
C
`285` days
D
None of these
Note: The remaining food is sufficient for `95` men for `195` days.
But, now remaining men `= (95 - 30) = 65`.
Less men, More days (Indirect)
`:. 65 : 95 :: 195 : x` or
`65x = 95 xx 195` or
`x = (95 xx 195)/(65)`
`= 285`.
Hence, the remaining food will last for `285` days.
Question: A garrison of `500` men had provisions for `27` days. After `3` days, a reinforcement of `300` men arrived. The remaining food will now last for how many days ?
A
`15`
B
`16`
C
`17 1/2`
D
`18`
Note: The remaining food is sufficient for `500` men for `24` days.
But, now the number of men `= (500 + 300) = 800`.
More men, less days (Indirect)
`800 : 500 :: 24 : x` or
`800x = 500 xx 24` or
`x = (500 xx 24)/800`
`x = 15`.
`:.` The food will now last for `15` days.
Question: If the numerator of a fraction be increased by `15%` and its denominator be diminished by `8%` the value of the fraction is `15/16`. The original fraction is :
A
`3/5`
B
`3/4`
C
`3/7`
D
`2/3`
Note: Let the given fraction be `x/y`
Then,`(115% of x)/(92% of y)`
`= 15/16 => x/y`
`= (15/16 xx 92/115)`
`= 3/4`
Question: A garrison had provisions for a certain number of days. After `10` days `1/5`th of the men desert and it is found that the provisions will now last just as long as before. How long was that ?
A
`15` days
B
`25` days
C
`35` days
D
`50` days
Note: Initially, let there be `x` men having food for `y` days.
After `10` days, `x` men had food for `(y - 10)` days.
i.e. `(x - x/5)` or
`(4x)/5` men had food for `y` days.
`:. x (y - 10) = ((4x)/5)y` or
`xy - 50x = 0` or
`x (y - 50) = 0`.
So, `y = 50 [ :. x != 0]`