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500/23 = 21 17/23 => 21 Numbers

The sum of the digits in 508, 328, 706 and 148 is 13, but not in 608.

x = 14 + 10√2
x/2 = 7 + 5√2
2/x = (7 – 5√2) / (7 + 5√2)(7 – 5√2) = 5√2 – 7
x/2 + 2/x = 10√2 = 2√50

M = 4.5
S = x
DS = 4.5 + x
US = 4.5 + x
4.5 + x = (4.5 – x)2
4.5 + x = 9 -2x
3x = 4.5
x = 1.5

3/8 * 32000 = 12000
5/8 * 32000 = 20000

A:B:C = 2/3:1/6:1/6 = 4:1:1
x * 2/100 * 2/3 = 200
B capital = 15000*1/6 = 2500

Speed Ratio = 1:3/4 = 4:3
Time Ratio = 3:4
1 ——– 1/3
3 ——— ? è 1 hour

H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.

Net part filled in 1 hour = 1/5 + 1/6 – 1/12 = 17/60
The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs.

Except Japanian, all others are appropriate usage of citizenship.

I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.

y^{p(q – r)} . y^{q(r – p)} . y^{r(p – q)}
= y^{pq – pr + qr – qp + rp – rq}
= y^p0 = 1.

Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/39 = 50/3
3x + 900 = 1950 => x = 350 m.

106 * 106 – 94 * 94 = (106)² – (94)²
= (106 + 94)(106 – 94) = (200 * 12) = 2400

Number of men in the colony = 4/7 (70) = 40
Number of women in the colony = 3/7 (70) = 30
Number of educated women in the colony = 1/5 (30) = 6
Number of uneducated women in the colony = 4/5 (30) = 24
Number of educated persons in the colony = 8/35 (70) = 16
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 40 – 10 = 30
Number of educated men and uneducated men are in the ratio 10:30 => 1:3

Let 4207 – x = 3007
Then, x = 4207 – 3007 = 1200

3M = 6W —- 20 days
12M + 8W —–?
24W + 8 W = 32W —?
6W —- 20 32 —–?
6 * 20 = 32 * x => x = 15/4 days

Let the interest for one year be x.
As amount = Principal + Interest, we have
P + 4x = 20720 — (1) ; P + 6x = 24080 — (2)
Solving the equations (1) and (2), we can get
P = Rs. 14000 and x = Rs. 1680
Interest for one year on Rs. 14000 is Rs. 1680
So, R = (100 * 1680)/(14000 * 1) = 12% p.a.

1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500
Shyam = 4/13 * 3250 = 1000
Mohan = 3/13 * 3250 = 750

Required probability = 1 – 1/126 = 125/126

Clearly, (2272 – 875) = 1397, is exactly divisible by N.
Now. 1397 = 11 * 127
The required 3-digit number is 127, the sum of whose digit is 10.

1 to 11 = 11 * 50 = 550
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6^th = 294 + 312 – 550 = 56

20% of 40 + ? % of 80 = 120
=> 800/100 + ?/100 * 80 = 120
=> ? = 140

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
= (⁶C₁ * ⁹C₁)/¹⁵C₂ + ⁶C₂/¹⁵C₂
= (6 * 9 * 2)/(15 * 14) + (6 * 5)/(15 * 14) = 36/70 + 1/7
= 46/70 = 23/35

x + y = 80
x – 4y = 5
x = 65 y = 15

Let the prices of X, Y and Z be 8k, 5k and 3k respectively.
After increase
Price of X = 8k * 125/100 = 10k
Price of Y = 5k * 120/100 = 6k
Price of Z = 3k * (133 1/3)/100 = 4k
Required ratio = 10k : 6k : 4k = 5 : 3 : 2.

Speed = 60 * 5/18 = 50/3 m/sec
Length of the train = speed * time = 50/3 * 9 = 150 m

Let their speed be x km/hr and y km/he respectively.
Then, 120/(x + y) = 1 => x + y = 120 — (i)
Now, when they move in same direction:
(Distance traveled by P in 6 hrs) – (Distance traveled by Q in 6 hrs) = 120 km
6x – 6y = 120 => x – y = 20 — (ii)
Sloving (i) and (ii), we get x = 70, y = 50
P’s speed = 70 km/hr.

S = 4
M = x
DS = x + 4
US = x – 4
6/(x + 4) + 6/(x – 4) = 2
x = 8

27% of 6000 + 45% of 4200 = ?
=> ? = 27/100 * 6000 + 45/100 * 4200
=> ? = 1620 + 1890 = 3510 = 3500

(x – 5)/15 + x/20 = 1
x = 11 3/7 days

Let the total worth of the stock be Rs. x.
The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50
The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50
Total SP = 11x/50 + 38x/50 = 49x/50
Overall loss = x – 49x/50 = x/50
x/50 = 400 => x = 20000

The part of the tank filled by three pipes in one hour = 1/8
=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.
The tank can be filled in 12 hours.

Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
i.e., (120/100 * 7x) and (110/100 * 8x)
i.e., 42x/5 and 44x/5
Required ratio = 42x/5 : 44x/5 = 21:22

6 2/3 * 9 3/5 * 2 1/3 + 1 1/3 = 20/3 * 48/5 * 7/3 + 4/3
= 448/3 + 4/3 = 452/3 = 150 2/3

Except Real, all others are rhyming words.

Time taken by B run 1000 meters = (1000 * 10)/50 = 200 sec.
Time taken by A = 200 – 10 = 190 sec.

42/360 * 22/7 * 12 * 12 = 52.8 m²

US = 25
DS = 35
M = (35 + 25)/2 = 30

Speed of train relative to man = 63 – 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 500 * 3/50 = 30 sec.

2/10 + 4/15 + x/20 = 1
x = 32/3 = 10 2/3

Suppose he has 100 items. Let C.P. of each item be Re. 1.
Total cost = Rs. 100. Number of items left after theft = 80.
S.P. of each item = Rs. 1.10
Total sale = 1.10 * 80 = Rs. 88
Hence, loss % = 12/100 * 100 = 12%

1120 = P [1 + (5*12/5)/100]
P= 1000

one man can consume the same food in 30*40 = 1200 days.
10 more men join, the total number of men = 40
The number of days the food will last = 1200/40 = 30 days.

Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75.

4πr₁² : 4πr₂² = 1 : 4
r₁ : r₂ = 1 : 2
4/3 πr₁³ : 4/3 πr₂³
r₁³ : r₂³ = 1 : 8

Let the number be x. Then,
2/5 of 1/3 of 3/7 of
x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2
40% of 525/2 = (40/100 * 525/2) = 105