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16x + 85 = 17(x + 3)
x = 34 + 3 = 37

Actual price = 95% of 90% of 85% of Rs. 12000
= 95/100 * 90/100 * 85/100 * 12000 = Rs. 8721.

108 * 107 * 96 ≡ 1109400

1/20 + 1/30 = 1/12
1 + 1/3 = 4/3
1 — 12
4/3 — ?
4/3 * 12 = 16 hrs

Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.

(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6

I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b

9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500
50/(3 * 100) of ? = 1500 => ? = 1500 * 6
? = 9000

Given sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)
= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)
= 7 + (765 + 120 – 68)/1020 = 7 817/1020

Number of words which can be formed = 5! – 1 = 120 – 1 = 119.

(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓
=> (6 ÷ 2 ÷ 9) * ? = 100 / 3
=> 3 / 9 * ? = 100 / 3
=> ? = 100

Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.

Let the number be x. Then,
50% of x – 35% of x = 12
50/100 x – 35/100 x = 12
x = (12 * 100)/15 = 80.

Let the two-digit number be 10a + b
a + b = 12 — (1)
If a>b, a – b = 6
If b>a, b – a = 6
If a – b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 – a = 3
Number would be 93.
if b – a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 – b = 3.
Number would be 39.
There fore, Number would be 39 or 93.

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.

(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days.

r₁: r₂ = 1: 3
Πr₁²: Πr₂²
r₁²: r₂² = 1: 9

Let the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 – 20% of 130 = 130 – 26 = Rs. 104.
In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.
But present price of the article is Rs. 2288
for 114.4 —> 100 ; 2288 —> ?
Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000.

x – y = 5
4x – 6y = 6
x = 12 y = 7

Let the number be x. Then,
x + 1/x = 34/8
8x² – 34x + 8 = 0
4x² – 17x + 4 = 0
(4x – 1)(x – 4) = 0
x = 4
required number = 4 * √4 = 4 * 2 = 8.

Let the ten’s digit be x and unit’s digit by y
Then, x + y = 15 and x – y = 3 or y – x = 3
Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6
Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69.
Hence, the number cannot be determined.

100 * 100 = 10000
125 * 80 = 10000
No change

B = 1/16 – 1/24 = 1/48 => 48 days

Let the earnings of P and Q be Rs. 9x and Rs. 10x respectively.
New ration = [9x + 1/4(9x)]/[10x – 1/4(10x)]
=> [9x(1 + 1/4)]/[10x(1 – 1/4)] = 9/10 * (5/4)/(3/4) => 3/2.

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.

The discriminant of the quadratic equation is (-12)² – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.

x * 1/3 * 3/4 =21 => x = 84

WC = 1/6:1/8:1/12 => 4:3:2
3/9 * 1800 = 600

Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec.

CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725.

Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.

We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.
Hence, the required number of straight lines
= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1
= 55 – 15 – 10 + 2 = 32

Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x² – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.

(743.30)² = 552500

9M + 12B —– 12 days
12M + 12B ——- 10 days
10M + 10B ——-?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B
18B + 12B = 30B —- 12 days
20B + 10B = 30B —–? => 12 days

Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125.

Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25%

Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meter

90%
104%
——–
14% —- 140
100% —- ? => Rs.1000

x + y = 80
x – 4y = 5
x = 65 y = 15

Let the rate of interest be R% p.a.
4400{[1 + R/100]² – 1} = 11193.60
[1 + R/100]² = (44000 + 11193.60)/44000
[1 + R/100]² = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]² = 784/625 = (28/25)²
1 + R/100 = 28/25
R/100 = 3/25
herefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560

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

Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.
Then, 4x + 9y = 13400 — (i)
And, 10% of 4x + 20% of 9y = 1880
2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)
Solving (i) and (ii), we get : x = 2000 and y = 600.
Cost price of each horse = Rs. 2000.

Let the distance traveled be x km.
Then, x/10 – x/15 = 2
3x – 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.

Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000.

Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.

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