MAT Mock Test Series 3 | MAT Online Test Series 3 | MAT Sample Papers

(112 * 5⁴) = 112 * (10/2)⁴
= (112 * 10⁴)/24 = 1120000/16 = 70000

Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x – 5) and (x – 9) hours respectively to fill the tank.
1/x + 1/(x – 5) = 1/(x – 9)
(2x – 5)(x – 9) = x(x – 5)
x² – 18x + 45 = 0
(x- 15)(x – 3) = 0 => x = 15

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.

15/100 * 195 + 14/100 * 205 = 15/100 * 200 + 14/100 * 200
= 30 + 28 = 58

SI = 20 + 20
CI = 20 + 21
20 —- 1
100 —- ? => 5%

Two men, three women and one child can be selected in ⁴C₂ * ⁶C₃ * ⁵C₁ ways
= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5
= 600 ways.

Part filled in 2 hours = 2/6 = 1/3.
Remaining part = 1 – 1/3 = 2/3
(A + B)’s 1 hour work = 2/21
C’s 1 hour work = [(A + B + C)’s 1 hour work – (A + B)’s 1 hour work]
= (1/6 – 2/21) = 1/14
C alone can fill the tank in 14 hours.

8/12 = 2/3 * 710 = 473.33
460.00
—————
13.33
1/3 uniform 13.33
1 ————— ? => Rs.40

6 * 22 = 132
7 * 19 = 133
————–
1

π (15² – 13²) = 176
176 * 1/4 = Rs.44

I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b

4:5
3:2
——-
12:15:10
12:10
6:5

Clearly, the retailer gets 1 dozen out of 6 dozens free.
Equivalent discount = 1/6 * 100 = 16 2/3%.

9 = (1*40*R)/100
R = 22 1/2 %

Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l² = 324 => l = 18.

Other number = (11 * 7700)/275 = 308.

(8000*3*2)/100 = 480
9200
——–
9680

Let the original earnings of A and B be Rs. 4x and Rs. 7x.
New earnings of A = 150% 0f Rs. 4x = (150/100 * 4x) = Rs. 6x
New earnings of B = 75% of Rs. 7x = (75/100 * 7x) = Rs. 21x/4
6x:21x/4 = 8:7
This does not give x. So, the given data is inadequate.

Given Exp. = (1000)⁹/10²⁴
= (10³)⁹/10²⁴ = 10²⁷/10²⁴ = 10^{(27 – 24)} = 10³ =1000

Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x = 230.

Let the CP of an article be Rs. 100
Given that successive discounts are 20% and 15%.
SP = 85% of 80% of 100 = (85/100)(80/100)(100)
=> SP = Rs. 68
Clearly, single discount is 32%.

B⁺²D⁺²F⁺⁶L, P⁺²R⁺³U⁺⁵Z, J⁺²L⁺³O⁺⁵T, C⁺²E⁺³H⁺⁵M and I⁺²K⁺³N⁺⁵S.
Except BDFL, all other groups follows similar pattern.

5M + 12B —– 4 days
7M + 6B ——- 5 days
20M + 48B = 35M + 30B
18B = 15M => 5M = 6B
M: B = 6:5

P = (P*5*R)/100
R = 20%

24 – [2.4 – {0.24 * 2 – (0.024 – ?)}] = 22.0584
24 – [2.4 – {0.48 – (0.024 – ?)}] = 22.0584
24 – [2.4 – 0.48 + 0.024 – ?] = 22.0584
24 – 2.4 + 0.48 – 0.024 + ? = 22.0584
? = 22.0584 – 24.48 + 2.424
? = 24.2824 – 24.48 => ? = 0.0024

Let x percent of 64 be 80.
64 * x/100 = 80 => x = (80 * 100)/64 => x = 125
80 is 125% of 64.

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

r = 3     πr² = 9
r = 5     πr² = 25
25 π – 16 π
100 —- ? => 64%

A = 8000(21/20)³
= 9261
= 8000
———
1261

(50 – ?/29)% of 4200 = 3√196

Net part filled in 1 hour = (1/4 – 1/9) = 5/36
The cistern will be filled in 36/5 hrs i.e., 7.2 hrs.

Let us say x boys and x girls joined the group.
(64 + x)/(40 + x) = 4/3
192 + 3x = 160 + 4x => x = 32
Number of members in the group = 64 + x + 40 + x
= 104 + 2x = 168.

I. x² + 6x – 3x – 18 = 0
=>(x + 6)(x – 3) = 0 => x = -6, 3
II. y² + 6y – 5y – 30 = 0
=>(y + 6)(y – 5) = 0 => y = -6, 5
No relationship can be established between x and y.

23 + 29 + 31 + 37 + 41 = 161/5 = 32.2

The greatest number which divides three dividends p, q and r leaving the same remainder in each case is given by H.C.F(any two of (q – p), (r – p) and (r – q)).
For the given problem, p = 83, q = 125 and r = 209.
Hence the greatest number which divides these three leaving the same remainder in each case is H.C.F(any two of (125 – 83, 209 – 83, 209 – 125) = 42.

Let the fraction be x/y.
When the numerator decreased by 25% and the denominator decreased by 20%.
(x * 3/4)/(y * 4/5) = 3/4 => x/y = 3/4 * (4/5)/(3/4) = 4/5

Cost price = Rs.500
profit = 20% of 500 = Rs.100
Selling price = Cost price + Profit
= 500 + 100 = 600

(4 + 5 + 2) – (1 + 6 + 3) = 1, not divisible by 11.
(2 + 6 + 4) – (4 + 5 + 2) = 1, not divisible by 11.
(4 + 6 + 1) – (2 + 5 + 3) = 1, not divisible by 11.
(4 + 6 + 1) – (2 + 5 + 4) = 0.
So, 415624 is divisible by 11.

100/7 = 14 2/7 => 14 Numbers

M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4

P^-3M⁺⁶S, R^-3O⁺⁶U, G^-3D⁺⁶J, K^-2I⁺⁴M and E^-3B⁺⁶H
Except KIM, all other groups follows similar pattern.

Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) – (6 * 3) – 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs. 4530

A = P(1 + TR/100)
=> 3600 = 3000[1 + (4 * R)/100] => R = 5%
Now R = 6%
=> A = 3000[1 + (4 * 6)/100] = Rs. 3720.

Let the integers be x and (x + 5). Then,
x(x + 5) = 500
x² + 5x – 500 = 0
(x + 25)(x – 20) = 0
x = 20
So, the numbers are 20 and 25.

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.

First persons investment = x
Second persons investment = x + 1000
Third persons investments = x + 2000
x + x + 1000 + x + 2000 = 9000
3x = 6000
x = 2000
Ratio = 2000 : 3000 : 4000
2:3:4
4/9 * 54000 = 2400

[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + [(2/5 * 3/5 – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + {(6 – 5 /25) * 625 + 7} ÷ 4 = ?
=[8 + {32} ÷ 4] = ? = 16

3x + 5x + 7x = 45
x =3
3x = 9

Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.