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A garrison of 3300 men had provisions for 32 days, when given at the rate of 850 gms per head. At the end of 7 days, reinforcement arrives and it was found that the provisions will last 17 days more, when given at the rate of 825gms per head. What is the strength of the reinforcement?


A1700

B1600

C1400

D1250

Answer: Option A

Explanation:

The problem becomes:
3300 men taking 850gms per head have provisions for (32-7) or 25 days. How many men taking 825gms each have provisions for 17 days?
Less ration per head, more men (indirect).
Less days, more men (indirect)
Ration 825:850::3300:x
Days 17:25::3300:x
Therefore, 825*17*x= 850*25*3300 or x= 5000.
So, strength of reinforcement = 5000-3300 = 1700.

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A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?


A21

B91

C81

D55

Answer: Option C

Explanation:

Remaining work = 1-4/7 =3/7.
Remaining period = (46-33) days =13 days.
Less work, less men (direct)
Less days, more men (indirect).
More hours per day, less men (indirect)
Therefore, work 4/7:3/7 ::117/x
Days 13:33 :: 117/x
Hrs/day 9:8:: 117/x
Therefore, 4/7*13*9*x= 3/7*33*8*117 or x= 198.
Therefore, additional men to be employed =(198-117) = 81.

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4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?


A35

B40

C45

D50

Answer: Option B

Explanation:

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y.
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10.
Solving the two equations, we get: x = 11/400, y = 1/400.
So, 1 woman's 1 day's work = 1/400.
10 women's 1 day's work = 1/400 x 10 = 1/40.
Hence, 10 women will complete the work in 40 days.

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A garrison of 3300 men has provisions for 32 days, when given at a rate of 850 grams per head. At the end of 7 days a reinforcement arrives and it was found that now the provisions will last 8 days less, when given at the rate of 825 grams per head.How, many more men can it feed?


A1000

B1650

C1700

D1690

Answer: Option C

Explanation:

Initial stock = 3300*32*850
stock consumed in 7 days = 3300*7*850
stock remains = 3300*25*850
a/q
3300*25*850 = (3300+reinforcement) *825*(32-7-8)
gives reinforcement = 1700

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On a map the distance between two mountains is 3 1 2 inches. The actual distance between the mountains is 136 km. Ram is camped at a location that on the map is 3 4 inch from the base of the mountain. How many km is he from the base of the mountain?


A13.50 km

B14.82 km

C12.82 km

DNone of these

Answer: Option B

Explanation:

since 312 inch=136 km
so 1 inch=136/312 km
so 34 inch=136*34/312=14.82 km.

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Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 for 30 cows . How many cows can eat away the same in 96 days?


A18

B20

C21

D19

Answer: Option B

Explanation:

Let g = grass initially, r = rate at which grass grow/day and c= cow eat grass/day.
=> g + 24r = 70 * 24c = 1680c -----------(i)
and,
=> g + 60r = 60 * 30c = 1800c
=> g = 1800c - 60r -----------(ii)
By solving these equations, we have c = (3/10)r -------------(iii)
=> g + 96r = 96nc.
=> 96nc = 1800c - 60r + 96r = 1800c + 36r=1800c+120c=1920c
=>n=20.

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in a grass field, If there are 40 cows, they could eat for 40 days. If there are 30 cows, they could eat for 60 days. Than if 20 cows, How much day they could eat?


A98 Days.

B80 Days.

C67 Days.

D87 Days.

Answer: Option B

Explanation:

40 cows take 40 Days.
and 30 cows take 60 Days.
Its obvious that 10 cows less 20 days More.
So, 20 Cows will take 80 Days.

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A person can buy 50 cds with $135 and 15 dvds with $100 Nikita has a budget of $1680 for cds & budget of $150 for dvds How many cds & dvds in total can she buy?


A624

B688

C644

D642

Answer: Option C

Explanation:

cds = (50*1680)/135=622 apprx
dvds = (150*15)/100=22 apprx
Total=644 apprx

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A stock of rice is enough for 100 men for 60 days. How long will the same stocks of rice last for 500 men:


A9 days

B5 days

C6 days

D10 days

E12 days

Answer: Option E

Explanation:

Men Days
----------------------
100 (u) 60 (d)
500 x

100/500 = x/60
=> x = (100/500) * 60 = 12 days.

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A play school has chocolates which can supply 50 students for 30 days. For the first ten days only 20 students were present. How many more students can be accommodated into the earlier group such that the entire chocolates get consumed in 30 days. Assume each student takes the same number of chocolates.


A45

B60

C55

D70

Answer: Option A

Explanation:

Let each students gets 1 chocolate.
Total chocolates = 50 * 30 = 1500
If first 10 days only 20 students were present, then total chocolates consumed = 10 * 20 = 200
Now we are left with 1500 - 200 = 1300 chololates. These were to be consumed in 20 days.
So each day 1300/20 = 65 chocolates were to be distributed.
So we can add 65 - 20 = 45 students.

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